Francis borceux (auth ) an algebraic approach to geometry geometric trilogy ii springer international publishing (2014)

Fermat’s Analytic Geometry

The French mathematician Pierre de Fermat (1601–1655) had probably developed his analytic geometry not later than 1629 and his results circulated in manuscript form for some fifty years It was only several years after his death, in 1679, that the manuscript was eventually published.

Fermat considers a half-line with originOand a direction other than that of the line (see Fig.1.1) Given a pointP of a curve, he drawsP P parallel to the chosen direction and locates the pointP via the two distancesx=OP andy=P P Let us stress the fact that Fermat considers only positive values for the “distances”x andy This apparently minor limitation will prove to be a serious handicap to an efficient development of analytic geometry.

Fermat’s first preoccupation is to prove that an equation of the form ax=by represents a straight line through the originO Indeed ifP andQare two points of the curve ax=by, let P , Q be their corresponding projections on the base half-line (see Fig.1.2) CallingR the intersection of the lines OP andQ Q, the

Fig 1.2 similarity of the trianglesOP P andOQ Ryields

On the other hand, sinceP andQare on the curveax=by, we obtain

It follows at once thatQ R=Q Qand thusR=Q This proves thatO,P,Qare on the same line.

Next Fermat considers the case of the equationax+by=c 2 which he proves— by analogous arguments—to be the equation of a segment Indeed, let us recall that for Fermat, the letters a,b, c,x,y represent positive quantities: this is why the values ofx andy are “bounded” and the corresponding curve is only a segment, not the whole line Nevertheless, the use ofc 2 in the equation, instead of justc, is worth a comment Fermat wants to introduce algebraic methods, but he is still very strongly under the influence of Greek geometry The quantitiesa,b,x,yrepresent

“distances”, thus the product quantitiesax,byrepresent “areas”; and an “area”ax+ bycannot be equal to a “length”c: it has to be compared with another “area”c 2 !After considering equations of the first degree, Fermat switches to the second degree and intends to show that the equations of the second degree represent precisely the (possibly degenerate) conics Indeed Fermat knows, by the work of the

Greek geometers Menechmus and Apollonius, that every conic admits an equation of degree 2; to prove the converse, Fermat first considers some special cases. First, Fermat considers the case where the projecting direction is perpendicular to the base half-line He shows that the equationb 2 −x 2 =y 2 represents (a quarter of) a circle with centerOand radiusb Indeed ifP,Qare two points of the curve b 2 −x 2 =y 2 with respective projectionsP ,Q , let us draw the circle with center

Oand radiusOP, cutting the lineQ Qat some pointR(see Fig.1.3) Pythagoras’ theorem tells us that

OQ 2 +Q R 2 =OR 2 =OP 2 while, sinceQandP are on the curveb 2 −x 2 =y 2

This proves at once thatQ=R, thus each pointQof the curve is on the circle with centerOand radiusOP.

Still using perpendicular directions of reference, the equationx 2 =ayrepresents a parabola To see this, Fermat uses the work of the Greek mathematician Menech- mus, presented in Sect 2.5 of [7], Trilogy I Cutting a right circular cone by a plane perpendicular to a generatrix, at a point D situated at a distance AD= a 2 from the vertexAof the cone, yields a parabola with equationx 2 =2ãADãy, that is, x 2 =ay.

Analogously, using the work of Apollonius, Fermat observes thatxy=b 2 is the equation of (pieces of) a hyperbola;x 2 ±xy=ay 2 is the equation of two straight lines;b 2 −x 2 =ay 2 , of an ellipse;b 2 +x 2 =ay 2 of a hyperbola again; and so on. Finally, starting with the general equation of degree two, Fermat uses the algebraic methods of Viète to transform the equation in one of the forms indicated above and so is able to conclude that every equation of degree two represents a conic.

It is striking how Fermat’s approach to analytic geometry is still rather close to the elementary treatments of this question considered nowadays.

Descartes’ Analytic Geometry

The French mathematician René Descartes (1596–1650) developed his approach to analytic geometry during the same period as Fermat As everybody knows, Descartes’ name is now closely attached to analytic geometry, since we all speak of Cartesian coordinates.

We mentioned already that Fermat did not publish his work during his lifetime; Descartes did, but as an appendix to a treatise on optics, in 1637! These facts, and the sometimes “obscure” style of both works, did not help to rapidly promote this new approach to geometry.

In contrast to the systematic approach of Fermat, who first considered lines, and then conics, Descartes introduced the new algebraic approach to geometry via the so-called Pappus’ problem.

Papus’ problem Consider four linesd 1,d 2,d 3,d 4 and four anglesα 1,α 2,α 3,α 4.

Through every pointP of the plane, draw the four linesd i (i=1,2,3,4) forming respectively an angleα i with the lined i For each indexi, callP i the intersection of d i andd i Determine the locus of those pointsP such thatP P 1ãP P 3=P P 2ãP P 4

Descartes writesO for the intersection of the linesd 1,d 2 and putsx=OP 1, y=P 1 P In other terms, he considers the system of Cartesian coordinates with originO, the two axis beingd 1and the line making withd 1an angleα 1 Next he computes the values ofP P 2 ,P P 3 ,P P 4 in terms of x,y,α 1 ,α 2 ,α 3 ,α 4 and the various distances between the intersections of pairs of linesd i : a rather tricky job based on the trigonometry of triangles But then all quantitiesP P i are expressed in terms ofx,yand Descartes observes that the equalityP P 1ãP P 3=P P 2ãP P 4now takes the form y 2 =ay+bxy+cy+dx 2 wherea,b,c,dare constants which have been computed from the given ingredients of the problem Descartes uses this equation to construct points of that locus, with ruler and compass To do this, he gives arbitrary values tox and solves, “with ruler and compass”, the corresponding equation of degree 2 iny.

All this has a good flavor of modern analytic geometry, since in contemporary language, Descartes has written the equation of the locus considered in an adequate system of Cartesian coordinates But it is generally accepted that Descartes was not fully aware of having introduced a new general technique—the now so-called

“Cartesian coordinates”—for investigating all geometric problems algebraically In- deed, as the discussion above shows, Descartes was essentially using some algebraic technique to help solve the problem with ruler and compass In any case, it took another century before mathematicians started to recognize and systematically use this new algebraic approach to geometry in terms of coordinates Leonhard Euler(1707–1783) was particularly influential in this respect The interested reader will find in [9] a much more refined discussion on “Descartes” versus “Cartesian coordinates”.

More on Cartesian Systems of Coordinates

While Greek geometry was essentially confined to the study of lines and conics, the new analytic geometry opened the way to the study of an incredibly wider class of curves This is the main reason behind its success However, before reaching its full strength, analytic geometry had to overcome a few hurdles.

We shall not dwell on the trivial way to extend the idea of Cartesian coordinates from the plane to three dimensional space: simply choose three axis of coordinates, not in the same plane The three dimensional case had been considered by Fermat and Descartes from the very beginning of analytic geometry, and later by La Hire (1640–1718) and many others.

Let us instead focus on the evolution of ideas concerning these systems of coordinates.

An important step was the recognition of negative numbers as possible coordinates: a coordinate is not a distance (thus a positive number), it is a distance equipped with a sign—that is, after all—an arbitrary number It seems that the British mathematician Wallis (1616–1703) was the first to use negative coordinates. But it is essentially the work of the famous British mathematician and physicist Isaac Newton (1642–1734), in particular a work on curves of higher degree published in 1676, which popularized the use of negative coordinates In this work, Newton had classified 72 types of cubics, forgetting half a dozen of them.

The idea of “separating the variables” of an equation is due to the Swiss mathematician Leonhard Euler (1707–1783) Instead of considering one equation with two variablesx,y, he writes these two variables separately in terms of some variable parameter.

1.3 More on Cartesian Systems of Coordinates 7

For example, consider the following system of equations: x=Rcosθ y=Rsinθ

We can view it in two different ways (see Fig.1.5).

Ifθ is a varying parameter whileR >0 is fixed, all the pairs(x, y), for all the possible values ofθ, describe a circle of radiusR The classical equation x 2 +y 2 =R 2 is recaptured by eliminating the parameterθbetween the two parametric equations: simply square both equations and add the results.

However, ifθ is constant and R is the variable parameter, the same system of equations describes all the points(x, y)of a line making an angleθ with the horizontal axis! The elimination of the parameterR between the two equations is now straightforward: xsinθ=ycosθ.

So before presenting a system of parametric equations, it is important to clearly identify the parameter.

Another useful technique was to clarify the rules by which coordinates transform when passing to another system of coordinates As we have seen, Fermat had used methods of this type to study the conics empirically But again it was Euler who developed the general theory of “coordinate changes” In particular, he observed the very special form that these formulổ take in the case of rectangular systems of coordinates:

Definition 1.3.1 By a rectangular system of coordinates in the plane or in space is meant a Cartesian system of coordinates in which any two of the axes are always perpendicular.

Notice that in those days the possible choice of different unit lengths on both axis was not considered, since the coordinate along an axis was defined as the distance from the origin to the projection of the point on this axis, with the adequate sign. Let us follow Euler’s argument, in the case of the plane Euler had developed an analogous theory in dimension 3; we omit it here since the general theory, in arbitrary finite dimensionn, will be studied in subsequent chapters, using the full strength of modern algebra Let us also mention that the results of Euler were discovered independently, a little bit later, by the French mathematician Lagrange (1736– 1813).

Consider two rectangular systems of coordinates in the plane (see Fig.1.6) Let

(a, b)be the coordinates ofO in the system of coordinates with originOand(x, y), (x , y ), respectively, the coordinates of a pointP in the systems of coordinates with originO,O We use the notationA,B,X,Y,X ,Y to indicate the corresponding points on the axes Writeθfor the angle between the directions of thexandx axes. Draw the various parallels to the axis and consider the points so obtained on the figure We observe first that the two right angled trianglesO MY andP N X are isometric, proving that

Therefore x=OX=OA+AX=OA+O K=OA+

An analogous argument withy yields the formulổ x=a+x cosθ−y sinθ y=b+x sinθ+y cosθ.

Non-Cartesian Systems of Coordinates

It is now easy to infer the form of the inverse formulổ: multiply the first equation by cosθand the second by sinθand add the results: one findsx in terms ofxandy.

An analogous argument withy yields the formulổ x =a +xcosθ+ysinθ y =b −xsinθ+ycosθ where a = −acosθ−bsinθ, b =asinθ−bcosθ.

In modern terms, Euler observed that the matrix

−sinθ cosθ contains all the information about both changes of coordinates: indeed, the inverse change of coordinates simply uses the transposed matrixM t x y

As already mentioned, Euler exhibited the same property in dimension 3.

1.4 Non-Cartesian Systems of Coordinates

The Cartesian coordinates introduced by Fermat and Descartes sometimes led to very heavy computations, especially in those early days when efficient algebraic techniques were not available Therefore the idea of inventing other systems of coordinates arose These other systems could never compete with the Cartesian coordinates as a universal method of investigation, but they proved to be efficient in the study of some specific problems closely linked with the essence of these new systems of coordinates.

Is was again Isaac Newton who first had the idea of non-Cartesian systems of coordinates The best known of them is the system of polar coordinates You fix a half-line with originOand you locate a pointP by the pair(r, θ ), whereris the distance fromO toP andθ is the angle between the reference half-line andOP

In polar coordinates, the equation of a circle of radiusRwith center the origin is thus simply r=R while the equation of a line through the origin making with the base half-line an angleΘis θ=Θ.

Newton knew how to switch from Cartesian coordinates to polar coordinates and conversely, but again it was Euler who exhibited the modern trigonometric formulổ x=rcosθ y=rsinθ

In three dimensional space, polar coordinates can be generalized in two ways. The spherical coordinates consider a rectangular system with axes OX,OY,

OZ Given a pointP, one considers its projectionP on the planeOXY The point

P is then located by the triple(r, ϕ, θ )whereris the distance fromOtoP,ϕis the angle betweenOXandOP andθis the angle betweenOZandOP (see Fig.1.8).

With the same notation as above, the cylindrical coordinates of the pointP is the triple(r, ϕ, h)wherehis the distanceP P equipped with a positive or negative sign (see Fig.1.9).

A wide range of other systems of coordinates have been invented, but most of them are no longer used today For example, Newton used in various places the so- called bipolar coordinates, where a point of the plane is located via its distances to two given fixed points.

Computing Distances and Angles

Computing the distance between two points in terms of coordinates can easily be done: simply use Pythagoras’ theorem Let us nevertheless write down the details, in order to exhibit the algebraic structure underlying this problem.

Of course we choose a system of coordinates which will make life as easy as possible: a rectangular system of coordinates whose origin is one of the two points to consider (see Fig.1.10) We thus want to compute the distance between the origin

Oand an arbitrary pointP First, let us write(x 1 , x 2 )—instead of(x, y)—for the coordinates ofP Clearly, this notation is intended to allow an easy generalization to then-dimensional case We writeX 1 =(x 1 ,0),X 2 =(0, x 2 )for the projections ofP on the axis By Pythagoras’ theorem, we have

Notice further that writingθfor the angle between the first axis andP Q, we have cosθ=OX 1

Now consider another pointQwith coordinates(y 1 , y 2 )and corresponding an- gleσ; we want to compute the angleτ betweenOP andOQ(see Fig.1.11) We have cosτ=cos(σ−θ )

This tells us that the following operation onR 2 contains all the information that we need to compute distances and angles in the plane:

R 2 ×R 2 −→R, (−→x ,−→y )→x 1 y 1 +x 2 y 2 where of course we have written

Let us formalize this in arbitrary dimension:

1.5 Computing Distances and Angles 13 is called the canonical scalar product onR n We denote it by

Proposition 1.5.2 For all “vectors”−→x ,−→y ,−→z ∈R n and all “scalars”α, β∈R, the following properties hold:

In modern terms, the first two properties express the bilinearity of the scalar product in the sense of the theory of real vector spaces.

The considerations of the beginning of this section can now be re-written as:

Proposition 1.5.3 Let−→x ,−→y ∈R 2 be non-zero vectors; callP =(x 1 , x 2 ),Q (y 1 , y 2 )the corresponding points of the plane, in some rectangular system of coordinates.

1 The length of the segmentOP, also called the norm of the vector−→x and denoted by−→x, is given by

2 The angleθbetween the two segmentsOP andOQis determined by cosθ= (−→x |−→y )

3 In particular, the segmentsOP,OQare perpendicular when(−→x |−→y )=0.

With some effort, one observes that analogous considerations (with analogous notation) hold in three dimensional space, simply replacing the indices 1,2 by the indices 1,2,3 Indeed:

Proposition 1.5.4 Consider two vectors−→x ,−→y ∈R 3 ; callP =(x 1 , x 2 , x 3 ),Q (y 1 , y 2 , y 3 )the corresponding points of space, in some rectangular system of coordinates.

1 The length of the segmentOP, also called the norm of the vector−→x and denoted by−→x, is given by

2 The angleθbetween the two segmentsOP andOQis determined by cosθ= (−→x |−→y )

3 In particular, the segmentsOP,OQare perpendicular when(−→x |−→y )=0.

Proof We start withP =(x 1 , x 2 , x 3 )and consider its projectionP =(x 1 , x 2 ,0)on the(x 1 , x 2 )-plane and its three projections X 1=(x 1 ,0,0),X 2=(0, x 2 ,0),X 3 (0,0, x 3 )on the three axes (see Fig.1.12) The distance betweenOandP is given by

OP 2 =OP 2 +P P 2 =OX 2 1 +OX 2 2 +OX 2 3 =x 1 2 +x 2 2 +x 3 2

To compute the angleθ betweenOP andOQ, let us assume first thatOP and

OQhave length 1 (see Fig.1.13) WritingMfor the middle point ofPQ, we have then cos θ 2 = QM The coordinates ofMare simply y 1 −x 1

Planes and Lines in Solid Geometry

A straightforward computation then yields cosθ=2 cos 2 θ

More generally, given arbitrary non-zero vectors−→x ,−→y ∈R 3 , then the vectors

− → x , − − → → y y have norm 1 Thus the angle between these last vectors, which is of course the same as the angleθbetween−→x and−→y, is given by cosθ −→x

1.6 Planes and Lines in Solid Geometry

The terminology “plane geometry” is still used today to mean “two-dimensional geometry” The term “solid geometry” has long been used to mean “three dimensional geometry”.

Fermat and Descartes were well aware that their analytic geometry could be developed in the three-dimensional case La Hire (1640–1718) pursued this idea, but it was only with Jacob Hermann (1678–1733) that elementary properties of planes and of some surfaces of degree 2 were studied We shall not follow his original approach, but shall rely at once on the considerations of the previous section. Let us work in a rectangular system of axis Given a vector−→v =(a, b, c)inR 3 , we know that all the vectors−→x =(x 1 , x 2 , x 3 )perpendicular to−→v are those such that(−→v|−→x )=0 (see Proposition1.5.4) In other words, ax 1+bx 2+cx 3=0 is the equation of the plane through the origin ofR 3 , perpendicular to−→v

Now if we apply a change of origin:

⎪⎩ x=x −u y=y −v z=z −w the equation takes the form ax +by +cz =d (∗) whered is a constant given by d=au+bv+cw.

Thus (∗) is the general form of the equation of a plane, orthogonal to the direction

(a, b, c), in three dimensional space, when working in a rectangular system of axes.

In our next chapter, we shall study changes of coordinates in full generality As a consequence, we shall observe that every plane of three dimensional space, in whatever Cartesian system of coordinates—not necessarily rectangular—still has an equation of the form ax+by+cz=d.

But it is only in a rectangular system of axes that(a, b, c)is the direction perpendicular to the plane.

Now what about lines in three dimensional space? A line through the origin is given by all the multiples of a fixed non-zero vector−→v =(a, b, c) So, in an arbitrary system of Cartesian coordinates, it admits the parametric equations

Since−→v =0—let us say,c=0—we can eliminate the parametert between these equations by writingt= z c and obtain the system cx−az=0 cy−bz=0.

To obtain the equations of the line in a system of coordinates with an arbitrary origin, simply apply the change of origin as above We get the system cx −az =d 1 cy −bz =d 2

The Cross Product

whered 1andd 2are again constants This is a system of two equations of two planes and the coordinates of the points of the line are those which satisfy both equations.

In other words, the line has been presented as the intersection of two planes. More generally, a line can always be presented—in infinitely many ways—as the intersection of two non-parallel planes, yielding a system of equations of the form ax+by+cz=d a x+b y+c z=d

We have already noticed that in rectangular axes, not being parallel reduces to

(a, b, c)and(a , b , c )(the orthogonal vectors) not being proportional Again as we shall see in the next chapter, this is a general fact: for this system of equations to determine a line (a “one-dimensional subspace”) in three dimensional space, the matrix a b c a b c must be of rank 2.

Let us come back to the equation of a plane in solid space There is another way to define a plane through the origin ofR 3 : the plane containing two non-proportional fixed vectors−→x and−→y As we have seen in the previous section, to exhibit the equation of this plane it suffices to know a vector perpendicular to it, that is, perpendicular to−→x and−→y This is the so-called cross product of−→x and−→y, which is denoted by−→x × −→y Of course, we still have to define the length and the orientation of this vector−→x × −→y

We keep working in a rectangular system of coordinates Writing−→z = −→x × −→y, Proposition1.5.4tells us that the perpendicularity that we are looking for reduces to the conditions x 1 z 1 +x 2 z 2 +x 3 z 3 =0, y 1 z 1+y 2 z 2+y 3 z 3=0.

This is a system of two equations with three unknownsz 1 ,z 2 ,z 3 Since−→x and−→y are linearly independent, this system admits a one-dimensional space of solutions. Considering the two determinants det

⎠=0 and expanding them with respect to the first rows shows at once that

,det x 1 x 2 y 1 y 2 is a possible solution, which is non-zero since−→x and−→y are linearly independent. All solutions are thus given by multiples of this first solution, which we choose to be the cross product of−→x and−→y.

Definition 1.7.1 Given arbitrary vectors−→x ,−→y ∈R 3 , their cross product is defined to be the vector

It is then a long but straightforward task to verify that:

Proposition 1.7.2 Given−→x ,−→y ,−→z ∈R 3 andα∈R, the following equalities hold, whereθindicates the angle between−→x and−→y:

The last formula is the so-called Jacobi identity The first formula shows in particular that−→x and−→y are linearly independent if and only if their cross product is non-zero.

Proof In view of the formula sin 2 θ+cos 2 θ=1, the first equality reduces to proving that

Forgetting the Origin

This equality, as well as all other equalities in the statement, is easily proved by straightforward computations from the definitions; these computations can often be shortened by using the basic properties of determinants

It should be mentioned that the generalization of the cross product to the n- dimensional case consists of defining the cross product ofn−1 vectors, by a trivial extension of the 3-dimensional formula Given the vectors−→x (1) , ,−→x (n − 1) , their cross product is the vector−→z whosei-th component is given by (with the usual convention wheni=1 ori=n)

In particular, in dimension 2, one obtains the cross product of one vector −→x (x 1 , x 2 ), which is thus the vector(x 2 ,−x 1 ), orthogonal to−→x

In this section, let us focus once more on the case of the plane, but the argument carries over as such to arbitrary dimensions.

Plane geometry is the study of the plane and the use of Cartesian coordinates allows us to put the set of points of the plane in bijective correspondence with the set of all pairs of real numbers So—roughly speaking—plane geometry reduces to the study of the geometry ofR 2 ButR 2 is a real vector space, thus the full strength of linear algebra can be used to study plane geometry This constitutes the basic principle of what is called today linear geometry.

While all of this is true, inR 2 there is a privileged point, namely, the origin

O=(0,0)and there are also two privileged axes, namely, thexandyaxes Analytic geometry tells us precisely that if we choose a privileged point and two privileged axes in the geometrical plane, then there is a canonical way to put the plane in bijective correspondence withR 2 , in such a manner that the privileged elements on both sides correspond to each other This is simply rephrasing the introduction of Cartesian coordinates But there are infinitely many ways to privilege a point and two axes in the geometrical plane, thus there are infinitely many ways to express a bijection between the plane andR 2 We know this very well: this is the famous change of coordinates problem So the plane is not exactly the same thing asR 2 , because it does not come equipped with a canonical origin and two canonical axes.

To handle this question elegantly, mathematicians have followed an intuition coming from mechanics and the theory of forces When a force−→x is applied to a pointP in the plane, it is common to represent this force graphically by an arrow starting atP, pointing in the direction where the force is applied, and whose length is given by the amplitude of the force It is well-known in physics that when two forces

→x and−→y are applied simultaneously in the plane to the same pointP, the resulting effect is the same as applying a single force−→z whose graphical representation is the diagonal of the parallelogram constructed on the graphical representations of−→x and−→y (see Fig.1.14).

Of course a force−→x can also be multiplied by a real numberα: the new force is applied in the same direction as−→x whenαis positive, and in the opposite direction whenαis negative; the amplitude of the forceα−→x is that of−→x multiplied by|α|.

As usual, let us use the term “vector of originA”, or “vector with specified origin”, to describe a pair(A, B)of points of the plane, which we visualize by drawing an arrow fromAtoB Of course in physics, the same force−→x can be applied to different points: analogously, the “same” vector can be drawn from different origins in the plane Let us formalize this:

Definition 1.8.1 In the plane, a vector(A, B)with origin Ais equipollent to a vector(C, D)with originC when the quadrilateral(A, B, D, C)in Fig.1.15is a parallelogram.

Let us observe at once that:

Proposition 1.8.2 Equipollence is an equivalence relation on vectors with a specified origin An equivalence class is simply called a vector The equivalence class of (A, B)is written−→

Proof Using the well-known properties of parallelograms,(A, B)is equipollent to

(C, D)when the segmentsABandCDare parallel and have the same length; these two properties are trivially equivalence relations

Now we present the key reason for introducing affine geometry, as in Chap.2.

Proposition 1.8.3 The vectors of the plane constitute a vector space over the field of real numbers.

Proof Taking our inspiration from the theory of forces, we have defined an addition and a scalar multiplication on the vectors with origin a fixed pointO It is a trivial game on parallelograms to observe that this induces corresponding operations on vectors, independently of the origin chosen to perform the operation Let us give the details for addition; the case of scalar multiplication is even easier and reduces at once to Thales’ theorem.

OC, where (O, A, C, B) is a parallelogram Analogously,−−→

O C , where(O , A , C , B )is a parallelogram But for this to make sense, when−→

O B , we must make sure that−→

O C The assumptions imply that(O, A, C, B), (O , A , C , B ),(O, A, A ,0 ) and

B C , thus(B, C, C , B )is a parallelogram as well Finally

AA , proving that(O, C, C , O )is a parallelogram as expected.

The axioms for a vector space are themselves proved by analogous reasoning on parallelograms Let us give a proof for the most involved of them: the associativity of addition (see Fig.1.17).

We thus consider three vectors−→

OC we must prove that

In other words, we must prove that (O, A, Z, Y )is a parallelogram, knowing by assumption that(O, A, X, B),(O, B, Y, C)and(O, X, Z, C)are parallelograms. The assumptions imply

BY , thus(B, X, Z, Y )is a parallelogram as well Therefore

OA and(O, A, Z, Y )is a parallelogram as expected

Now we present the result which underlies the modern definition of affine space on an arbitrary field, as studied in the next chapter.

Theorem 1.8.4 Consider the geometrical planeEand the vector spaceV of vectors inE, as given by Proposition1.8.3 There exist two operations

Proof The first operation is the one defined in Proposition1.8.2 To define the second operation, consider a pointAand a vector−→v =−→

CD Constructing the parallelogram(A, B, D, C)as in Fig.1.18we thus have−→v =−→

AB The second property announced in the statement does not leave us any choice, we must define

This not only takes care of the second property, but also of the third one which simply reduces to−→

AB= −→v The first property is proved analogously: consider again the parallelogram

BDand thus, by the parallelogram rule for adding vectors,

Our next observation is the one which underlies the notion of parallel subspaces in affine geometry (see Definition2.4.1).

Proposition 1.8.5 Consider again the planeEand the vector spaceV of all vectors.

AB|A, B∈d} ⊆V is a vector subspace; it is called the direction ofd.

2 Two lines are parallel if and only if they have the same direction.

Proof Given three points A, B, C on d, the fourth vertex of the parallelogram (A, B, D, C)is still on d ThusW is stable under the addition of vectors, and of course also under scalar multiplication.

For the second assertion, first consider two parallel lines d, d with respective directions W, W Given A, B ∈d and C∈d , construct the parallelogram

(A, B, D, C) ThenD∈d becaused is parallel tod But then−→

Conversely ifW =W , consider againA, B ∈d andC∈d Then D=C +

CD So(A, B, D, C)is a parallelogram and the two lines are parallel.

The Tangent to a Curve

From the very beginning of analytic geometry, both Fermat and Descartes considered the problem of the tangent to a plane curve Descartes’ approach is essentially algebraic, while Fermat’s approach anticipates the ideas of differential calculus, which were developed a century later by Newton and Leibniz.

To compute the tangent to a given curve at some pointP, Descartes writes down the equation of a circle passing throughP and whose center is a point (c,0) of the first axis He computes the intersections of the curve and the circle: in general, he finds two of them Then he determines the value c 0 of c for which “the two intersection points coincide” (compare with Definition7.4.5); the tangent atP to the circle with center(c 0 ,0) is then the tangent to the curve at the pointP (see Fig.1.19).

Let us follow this argument in the case of the curve

(an ellipse) and let us compute the tangent at the pointP =(1,1) A circle with center(c,0)has the equation

(x−c) 2 +y 2 =k; putting x=1, y=1 in this equation we find the value ofk such that the circle passes through(1,1), yielding

Extractingy 2 from this equation and introducing this value into the equation of the given curve yields

The points of intersection of the curve and the circle thus have a first coordinate which is a root of this equation of the second degree inx The points of intersection are equal when the two roots of this equation are equal, and this is the case when c 2 +4(c+1)=0 that is,

The tangent to the curve at the point(1,1)is thus the tangent to the circle corresponding to the valuec= −2 The segment joining(−2,0)and(1,1)is the radius of that circle and points in the direction−→v =(3,1) The tangent is thus the line perpendicular to this radius at the point(1,1), that is, the line with equation

The limitations of Descartes’ method are rather evident: to findc, he has to solve an equation whose degree depends heavily on the degree of the given curve: thus— except in very special cases—this method can hardly be used for curves of higher degrees.

Fermat’s approach is totally different and has the advantage of being applicable to curves of arbitrary degree, without even any need of using rectangular axes To determine the tangent at some point(x, y)of a curve, Fermat considers a point of the curve “close to(x, y)” and whose coordinates are written(x+ x, y+ y).

When the second point is “infinitely close” to the first one, the ratio x y is the slope of the tangent (see Fig.1.20).

Let us make Fermat’s argument explicit in the case of the quartic (i.e the curve of degree 4) with equation x 3 +2x 2 y 2 =3.

The assertion that the point(x+ x, y+ y)is on the curve means

Subtracting the two equations one obtains

Next Fermat puts α= y x, the quantity which will be the slope of the tangent when “ x and yare infinitely small” Dividing the equation above by x yields an expression in whichαnow appears explicitly, namely

Since x and y have to become “infinitely small”, Fermat now sets them both equal to 0, without changing the occurrences ofα The equality thus reduces to

This is an equation of the first degree in α (and notice that this will always be an equation of the first degree, whatever the degree of the equation of the original curve) From this equation, Fermat infers the value ofα: α= −3x 2 +4xy 2

Think what you want of this argument, it is rather far from the present standards of rigour, but for every curve with equation

F (x, y)=0 (in our example,F (x, y)=x 3 +2x 2 y 2 −3), Fermat’s method yields the result α= −∂F /∂x

The Conics

which is indeed the precise way to compute the tangent to an algebraic curve (see Proposition7.4.6or Proposition 2.4.4 in [8], Trilogy III) One ought to recall at this point that differential calculus was developed by Newton and Leibniz one century after Fermat’s method was described.

Coming back to our quartic, the tangent at the point(1,1)(for example) thus has a slope given byα= − 7 4 and therefore an equation of the form

Puttingx=1,y=1 in this equation gives us the value ofd—namely, 11—so that the tangent at the point(1,1)is the line with equation

As mentioned in Sect.1.1, Fermat proved that the equations of degree 2 in the plane correspond exactly to the conics: the sections of a circular cone by a plane The importance of these curves justifies the devotion of a section to them However, instead of going back to Fermat’s arguments, we shall use the general theory which will be developed in subsequent chapters of this book.

Our Theorem2.24.2tells us that, given an equation of degree 2 in an arbitrary system of Cartesian coordinates of the plane, there exists a rectangular system of coordinates with respect to which the equation transforms into one of the three forms ax 2 +by 2 =0, ax 2 +by 2 =1 ax 2 =y, wherea, b∈R We shall discuss the form of the corresponding “curve” in terms of the values ofa,b.

For simplicity, let us make the following convention Every coefficienta,bex- plicitly written will from now on represent a non-zero positive real number We shall thus write−afor a negative coefficient, and omit the term in case of a zero coefficient Moreover, in the list which follows, we avoid considering the unessential variations obtained by interchanging the roles of the coordinates, multiplying by−1 in the first case or applying the change of coordinatesy = −y in the third case.

Let us first investigate the first type of equation.

• ax 2 +by 2 =0; the “curve” reduces to a single point: the origin.

• ax 2 −by 2 =0; the equation can be re-written as

=0; the “curve” comprises two intersecting lines.

• ax 2 =0; the equation is equivalent tox=0 and the “curve” is just a line.

• 0=0; the “curve” is the whole plane.

Let us next investigate the second type of equation.

• ax 2 +by 2 =1; the curve is called an ellipse (see Fig.1.21): we shall study this case in more detail below.

• ax 2 −by 2 =1; the curve is called a hyperbola (see Fig.1.22): again we shall study this case in more detail below.

• −ax 2 −by 2 =1; the equation does not have any solution and the “curve” is the empty set.

• ax 2 =1; the equation can be re-written asx= ± √ 1 a ; the “curve” comprises two parallel lines.

• −ax 2 =1; the equation does not have any solution and the “curve” is again the empty set.

• 0=1; once more the “curve” is the empty set.

Finally, the third type of equation.

• ax 2 =y; the curve is called a parabola (see Fig.1.23) and will be studied in more detail below.

• 0=y; the “curve” is a line, namely, thex-axis.

The Ellipse

In conclusion, there are only three cases where we obtain an actual curve: the ellipse, the hyperbola and the parabola The other cases reduce to one or two lines, or just one point, the empty set or the whole plane.

Let us focus on the three “actual curves” encountered in the previous section We begin with the ellipse.

Proposition 1.11.1 A curve in the plane admits an equation of the form ax 2 +by 2 =1, a >0, b >0 in a rectangular system of axes if and only if it is the locus of those pointsP =(x, y) whose sumd(P , F )+d(P , F )of distances to two fixed pointsF,F is a constant

Fig 1.24 greater than the distance betweenF andF The curve is called an ellipse and the two pointsF,F are called the foci of this ellipse.

Proof First, ifF =F , then of coursed(P , F )=d(P , F ) Writing 2Rfor the constantd(P , F )+d(P , F ), we getd(P , F )=R=d(P , F )and the curve is a circle with centerF =F and radiusR As we have seen, its equation in a rectangular system of coordinates centered atF=F is simply x 2 +y 2 =R 2 that is, an equation of the announced type.

IfF andF are distinct, let us still write 2Rfor the constantd(P , F )+d(P , F ).

Of course for the problem to make sense, 2Rmust be strictly greater than the distance betweenF andF , which we shall write as 2k Let us work in the rectangular system of coordinates whose first axis is the line throughF andF , while the second axis is the mediatrix of the segmentF F (see Fig.1.24).

The coordinates ofF andF thus have the form

The distances from an arbitrary pointP=(x, y)to the pointsF andF are thus d(P , F ) (x−k) 2 +y 2 , d

The equation of the curve characterized by d(P , F )+d

Taking the squares of both sides yields

The Hyperbola

Isolating the square root on the left and squaring again both sides yields, after cancellation of the term 4(x 2 +k 2 +y 2 ) 2 which appears on both sides,

This is an equation of the expected type, since we have chosen 2R >2k.

Conversely, consider an equation as in the statement We can of course assume a≤b: otherwise, it suffices to interchange the roles of both coordinates in the following argument The conclusion then follows by considering the pointsF=(k,0),

It should further be noticed that when we write the equation of the ellipse in the form x a

(±a,0), (0,±b) are the four intersections of the ellipse with the coordinate axes; these four points are generally called the vertices of the ellipse The lengthsa andb are sometimes called the two radii of the ellipse.

The second curve to consider is the hyperbola which, besides having two foci, also admits two asymptotes (see Definition 2.5.1, [8], Trilogy III).

Proposition 1.12.1 A curve in the plane admits an equation of the form ax 2 −by 2 =1, a >0, b >0 in a rectangular system of coordinates if and only if it is the locus of those points

P=(x, y)whose difference 2R= |d(P , F )−d(P , F )|of distances to two distinct

Fig 1.25 fixed pointsF,F is a constant smaller than the distance 2kbetween F and F

The curve is called a hyperbola and the two pointsF,F are called the foci of this hyperbola This hyperbola admits the two lines y= ± k 2 −R 2

Proof Of course this time, we must chooseF=F : otherwised(P , F )=d(P , F ) and the difference is 0, for every point of the plane.

So we chooseF =F at a distance 2kand we write again|d(P , F )−d(P , F )| 2R This time, for the problem to make sense, the “triangular inequality” requires that 2k >2R Let us work in the rectangular system of coordinates whose first axis is the line throughF andF , while the second axis is the mediatrix of the segment

The coordinates ofF andF thus have the form

The distances from an arbitrary pointP=(x, y)toF andF are thus d(P , F ) (x−k) 2 +y 2 , d

The equation of the curve characterized by d(P , F )−d

1.12 The Hyperbola 33 Taking the squares of both sides yields

Isolating the square root on the right and squaring again both sides yields, after cancellation of the term 4(x 2 +k 2 +y 2 ) 2 which appears on both sides,

This is an equation of the expected type, since we have chosen 2R 0, in a rectangular system of coordinates if and only if it is the locus of those points

P =(x, y)whose distances to a fixed point F and a fixed line f not containing

F are equal The curve is called a parabola; the pointF is called the focus of this parabola and the linef, its directrix.

Proof Let us write 2kfor the distance betweenF andf We choose as first axis the line parallel tof, at a distancekfromf andF The second axis is perpendicular to the first one and passes throughF (see Fig.1.26).

The pointF thus has the coordinatesF=(0, k)and the linef admits the equationy= −k A pointP=(x, y)is at a distance x 2 +(y−k) 2 fromF andy+k fromf The equation of the curve is thus given by x 2 +(y−k) 2 =y+k that is, squaring both sides x 2 +(y−k) 2 =(y+k) 2

This reduces to x 2 −2yk=2yk that is y=x 2

Conversely, given an equation as in the statement, the conclusion follows at once by forcinga= 4k 1 , that is by choosingk= 4a 1

This time, we notice that the origin(0,0)is a point of the parabola with equation y=ax 2

This point is called the vertex of the parabola.

The following property of the parabola is widely used in optics: if you send light rays in all directions from the focus of a parabola, the parabola reflects all these rays in a sheaf of parallel rays.

Proposition 1.13.2 In a rectangular system of coordinates, consider the parabola y=ax 2 , a >0 and its focusF The line joining F to a point P of the parabola makes with the tangent atP the same angle as this tangent with the parallel to the y-axis (see

Proof Let us use the notation of Proposition1.13.1and write the equation as y=x 2

Fig 1.27 whereF=(0, k)is the focus of the parabola As will be proved in Proposition 2.4.4 in [8], Trilogy III, the tangent at a pointP =(x 0 , y 0 )to the parabola p(x, y)=y−x 2

4k=0 is given by the equation

As we know, the coefficients of this equation are the components of the vector perpendicular to the tangent, thus this tangent is in the direction of the vector

Of course they-axis is in the direction of−→e 2 =(0,1) Routine computations then give

The Quadrics

Writingθfor the angle between−→

F P and−→t andτ for the angle between−→t and−→e 2 , Proposition1.5.3yields at once cosθ= x 0 x 2 0 +4k 2

Let us now extend our study to the equations of degree 2 in three dimensional space: we obtain the so-called quadrics.

Our Theorem2.24.2 tells us again that given an equation of degree 2 in an arbitrary system of Cartesian coordinates of solid space, there exists a rectangular system of coordinates with respect to which the equation transforms into one of the three forms ax 2 +by 2 +cz 2 =0, ax 2 +by 2 +cz 2 =1, ax 2 +by 2 =z, wherea, b, c∈R We shall discuss the form of the corresponding “surface” in terms of the values ofa,b,c.

For simplicity, as for conics, let us make the following convention Every coefficienta,b,cwhich will now be explicitly written represents a non-zero positive real number We shall thus write−a for a negative coefficient and omit the term in case of a zero coefficient Moreover, in the list which follows, we avoid considering the trivial variations obtained by interchanging the roles of the coordinates, multiplying by−1 in the first case or applying the change of coordinatesz= −z in the third case.

Let us first investigate the equations of the first type.

• ax 2 +by 2 +cz 2 =0; the “surface” reduces to a single point: the origin;

• ax 2 +by 2 −cz 2 =0; we observe that:

1 the intersection with the planez=0 is the point(0,0,0);

2 the intersection with an arbitrary horizontal planez=d is an ellipse ax 2 +by 2 =cd 2 z=d;

3 the intersection with a vertical planey=kxis equivalently given by a+k 2 x+√ cz a+k 2 x−√ cz

=0 y=kx; this is the intersection of two intersecting planes with a third plane, all three of them containing the origin; this yields two intersecting lines.

The corresponding surface is called a cone (see Fig.1.28).

• ax 2 +by 2 =0; the solutions are the points(0,0, z), that is, the “surface” degenerates to thez- axis;

• ax 2 −by 2 =0; this is equivalent to

(√ ax+√ by)(√ ax−√ by)=0; we obtain two intersecting planes through the origin.

• ax 2 =0; this is equivalent tox=0: we obtain the(y, z)-plane.

• 0=0; this is one possible equation of the whole space.

Next, we consider the second type of equation.

Fig 1.29 The ellipsoid the section by a horizontal plane with equationz=dis given by ax 2 +by 2 =1−cd 2 z=d when 1−cd 2 >0, that is ford < √ 1 c , we obtain an ellipse; and whend > √ 1 c we obtain the empty set Analogous observations hold when cutting the surface by planes with equationsx=dory=d The corresponding surface is thus bounded, with all its sections parallel to the planes of coordinates given by ellipses Such a surface is called an ellipsoid (see Fig.1.29).

This time all sections by a horizontal planez=d yield ellipses: ax 2 +by 2 =1+cd 2 z=d.

Cutting by the vertical planey =0 yields the hyperbolaax 2 −cz 2 =1 in the

(x, z)-plane; analogously, if we cut byx=0, we get a hyperbola in the (y, z)- plane The surface thus has the shape depicted in Fig.1.30and is called a hyper- boloid of one sheet.

This time all sections by a horizontal planez=d yield hyperbolas: ax 2 −by 2 =1+cd 2 z=d.

The same conclusion applies when cutting by a planey=d Cutting by a plane x=dyields the curveby 2 +cz 2 =ad 2 −1 in the(y, z)-plane This is an ellipse whenad 2 −1>0 and the empty set whenad 2 −10 and the empty set when d 0 and in the direction of they-axis whend 0 of terms, the field is said to be of characteristic

0 (in other terms, adding 0 times the scalar 1 is the only way to get 0).

Definition 2.9.2 LetKbe a field of characteristic distinct from 2 Given two points

A,Bof an affine space onK, the middle point of the pair(A, B)is the barycenter of(A,1)and(B,1).

IfC is the barycenter ofA,Bas in Definition2.9.2, one has

AB, which justifies the description of C as “the middle point of

Triangles

The first basic geometric figure is the triangle.

Definition 2.10.1 Let(E, V )be an affine space.

1 By a triangle is meant a triple(A, B, C)of affinely independent points.

2 The three pointsA,B,Care called the vertices of the triangle.

3 The three lines through each pair of vertices are called the sides of the triangle (see Fig.2.1).

Even if we do not have a notion of distance in an affine space (see Sect.4.1for more comments), several classical “metric results” on triangles (for example Theo-

Fig 2.2 rem 5.4.1, Proposition 5.2.2 and Theorem 4.12.1 in [7], Trilogy I) can be generalized in terms of “barycenters” to the context of affine spaces.

Theorem 2.10.2 (Ceva’s theorem) In a given triangle A, B, C, consider three points, other than the vertices:

1 Xon the sideBC, with barycentric coordinates(b, c)with respect to(B, C);

2 Y on the sideCA, with barycentric coordinates(c , a )with respect to(C, A);

3 Zon the sideAB, with barycentric coordinates(a , b )with respect to(A, B).

The three linesAX,BY,CZare convergent if and only if they are not parallel and a bc =a b c(see Fig.2.2).

Proof If the three lines converge at a pointP, let(a, b, c)be the barycentric coordinates ofP with respect to(A, B, C).

Notice first thatb+c=0 Otherwise one would havea=1 and thus

BC would imply that the lineP Ais parallel to the line BC But by assumption these two lines meet atX and therefore, by Corollary2.7.7, should be equal This contradicts the fact thatA,B,Care affinely independent Thusb+c=0 and analogously,c+a=0,a+b=0.

The barycenter X of (B, b) and (C, c) thus lies on the line BC (Proposi- tion2.8.2); but by associativity (Proposition 2.8.3.3),P is also the barycenter of

(A, a)and(X , b+c), proving thatX also lies on the line AP Thus X =X, proving that(b, c)=k(b, c)for somek∈K An analogous argument holds forX andY:

This yields at once a bc =kk k abc=a b c.

Conversely assume the equality in the statement and suppose (for example) that

AX and BY intersect at some pointP Again write (a, b, c) for the barycentric coordinates ofP with respect to(A, B, C) The first part of the proof shows that

WritingZ for the barycenter of(A, a)and(B, b), the first part of the proof also shows thatZ is on the lineCP But the equalitya bc =a b cnow becomes a (kb) k c k a b (kc) that is,a b=ab This immediately implies a , b

=k (a, b) withk =a a =b b , sincea=0 (this would forceP to be onBC, thusY onBCand finallyY =C) and analogously,b=0 This shows thatZ=Z and concludes the proof

Definition 2.10.3 LetKbe a field of characteristic distinct from 2 By a median of a triangle in an affine space is meant a line joining a vertex with the middle point of the other two vertices.

Proposition 2.10.4 LetKbe a field of characteristic distinct from 2 and 3 Given a triangle in an affine space onK, the three medians converge (see Fig.2.3).

Proof The proof of Proposition2.10.2shows that the barycenterP of(A,1),(B,1), (C,1)is such that the following three points are on a line:A,P and the barycenter

Xof(B,1),(C,1), that is, the middle point ofBC An analogous argument holds for the two other medians

Problem2.27.3provides an example of an affine space, over a field of characteristic 3, where the three medians of a triangle are parallel, highlighting the necessity of the assumption thatKhas characteristic distinct from 3 in Proposition2.27.3. Theorem 2.10.5 (Menelaus’ theorem) In a triangleA, B, C, consider three points, other than the vertices:

1 Xon the sideBC, with barycentric coordinates(b, c)with respect to(B, C);

2 Y on the sideCA, with barycentric coordinates(c , a )with respect to(C, A);

3 Zon the sideAB, with barycentric coordinates(a , b )with respect to(A, B). The three pointsX,Y,Zare collinear if and only if (see Fig.2.4) a bc = −a b c.

Proof By Proposition2.8.3, the pointXis also the barycenter of

C,b b c while the pointY is the barycenter of

When the equality in the statement holds,

Consider the four weighted points

The barycenter of the first two isZ while the barycenter of the last two isY The sum of the four masses is b +b b c=b b (b+c)=b b =0, so that the barycenter exists by Proposition 2.8.2 By associativity (Proposi- tion 2.8.3.3), the barycenter of the four weighted points is thus on the line Y Z.

But trivially, the barycenter of the four points is the same as that of

, which is the pointX This proves thatX,Y,Zare collinear.

Conversely when X, Y, Z are collinear, consider the barycenter X of the weighted points

By the first part of the proof, the points X , Y, Z are collinear, proving that

X is on the line Y Z But by Proposition 2.8.2, X is also on the lineBC By

Corollary2.7.5,X =Xand therefore, by Proposition2.9.1, b b c= − a a c , that is, a bc = −a b c.

Parallelograms

To finish exhibiting the relation with the considerations of Sect.1.8, let us define a parallelogram in an arbitrary affine space.

Definition 2.11.1 By a parallelogram in an affine space(E, V )is meant a quadruple(A, B, D, C)of pairwise distinct points, not on the same line, such that the lines

AB,DCare parallel, and the linesAD,BCare parallel as well The four lines already mentioned are called the sides of the parallelogram, while the linesAC,BD are called the diagonals of the parallelogram (see Fig.2.5).

The following well-known property of parallelograms holds in every affine space:

Proposition 2.11.2 In an affine space (E, V ), consider four points A, B, C, D, pairwise distinct and not on the same line The following conditions are equivalent:

Proof To prove the equivalence

BC it suffices of course to prove one of the two implications If−→

Condition 2 implies that the lineABis parallel to the lineDC, while condition 3 forces the parallelism of the two linesADandBC Thus the equivalent conditions 2,

Conversely given condition 1, since the direction of a line has dimension 1, there exist scalarsk, k ∈Ksuch that−→

Ifk=1=k , we are done If not—let us sayk=1—thenk =1, otherwise one would have−→

ADare proportional, proving thatA,B,Dare on the same line SinceDCis parallel toAB,Cis on that line as well, which contradicts the assumption in the statement Proposition2.11.2exhibits in particular the link with the “parallelogram rule” for adding vectors in the ordinary plane (see Sect.1.8).

Corollary 2.11.3 Given a parallelogram as in Definition2.11.1, one has

Corollary 2.11.4 Every parallelogram is contained in an affine plane.

Proof By Proposition2.5.1, the direction of the affine subspace generated by the points A, B, C, D is generated by the three vectors −→

AD suffice These are linearly independent becauseA,B,D are not on the same line Thus the affine subspace generated byA,B,C,Dis an affine plane (see Definition2.7.1)

Now let us be very careful: in Fig.2.5, it is “clear” that the diagonals of the parallelogram intersect This is by no means a general fact in affine spaces Let us make this precise.

Proposition 2.11.5 Consider a fieldK The following conditions are equivalent:

2 given a parallelogram(A, B, D, C)in an affine space(E, V )overK, the two diagonals intersect.

Moreover in that case, the two diagonals intersect at their “middle point”.

Proof If 1+1=0, defineI to be the middle point of(A, C), that is,−→

CI=0 andI is also the middle point of(D, B).

We prove the converse by contraposition If 1+1=0, then given a parallelogram

BD proving that the two diagonals are parallel, thus disjoint by Corollary2.4.3

Corollary 2.11.6 Consider a fieldK The following conditions are equivalent:

2 given a parallelogram(A, B, D, C)in an affine space(E, V )overK, the two diagonals are parallel.

Proof This follows by Proposition2.11.5and Corollary2.7.7.

Affine Transformations

We have introduced the “objects of our study”, namely, the affine spaces As for every mathematical structure, two topics must be considered: the parts of the space which inherit the structure of an affine space (the affine subspaces, see Sect.2.3), and the transformations of the space which respect the affine structure The following sections will be devoted to these transformations.

Since an affine space has two components, a set and a vector space, an affine transformation must have two components as well: a mapping which sends points to points, and a linear mapping which sends vectors to vectors Of course, these ingredients must commute with the operations defining an affine space More precisely:

Definition 2.12.1 An affine transformation(f,−→ f ): (E, V )−→(F, W )between two affine spaces over the same fieldKconsists of:

2 aK-linear mapping−→ f :V −→W, such that, for allA, B∈Eandv∈V:

(See also Convention2.12.5concerning the notation.)

A more general notion of affine transformation between affine spaces over different fields can also be given: it involves an additional component, namely, a homo- morphism between the two fields We shall not consider this generalization in this book.

Of course—the field being fixed—Definition2.12.1is the sensible one: an affine transformation is a transformation which preserves the whole structure of an affine space Note that for every pointA, the two operations−→

A•andA+ •on the affine space are the “inverse of each other”, as Proposition2.1.2indicates Therefore, pre- serving one of these operations automatically preserves the other one; more precisely:

Lemma 2.12.2 In Definition2.12.1, axiom [AT2] is redundant.

Proof Let us prove that in Definition2.12.1, axiom [AT2] is a consequence of [AT1] and, of course, the axioms of affine spaces of(E, V )and(F, W ).

Indeed we have, by axioms [AT1] and [AF3],

By Proposition2.1.2, this implies axiom [AT2]

Lemma 2.12.3 An affine transformation(f,−→ f ):(E, V )−→(F, W )is entirely determined by the linear mapping −→ f and the image X=f (A)of a fixed point

Proof Indeed givenB∈E, we have at once

It is perhaps less expected that Definition2.12.1 can be “cooked down” even more drastically, by omitting every reference to the linear mapping−→ f

Proposition 2.12.4 An affine transformation (E, V )−→(F, W ) between two affine spaces over a fieldKcan equivalently be defined as a single mapping f:E−→F satisfying the axiom, forA, B, C, D∈Eandk∈K

Proof First, let us prove that an affine transformation as in Definition2.12.1satisfies the axiom in the present statement By axiom [AT1] and the linearity of−→ f

Next let us observe that axiom [AT1] implies that a mappingf as in the statement can be part of at most one affine transformation(f,−→ f )as in Definition2.12.1 This is simply because every vectorv∈V can be written as v=−→

AB, by choosing an arbitrary pointA∈Eand puttingB=A+v Then axiom [AT1] implies

AB)=−−−−−−→ f (A)f (B) so that the knowledge off forces the form of−→ f

To conclude, it remains to prove that the formula above, expressing−→ f in terms of f, does indeed define an affine transformation, as soon as the axiom of the statement holds.

First, let us verify that the definition of−→ f in terms of f does not depend on the choice of the pointA Thus givenA ∈EandB =A +v, we must prove that

−−−−−−→ f (A)f (B)=−−−−−−−→ f (A )f (B ) This is an immediate consequence of the axiom in the statement, choosingk=1, since −→

A B This observation proves at the same time the validity of axiom [AT1].

Affine Isomorphisms

It remains to prove that−→ f is linear First

Next, givenv, v ∈V, let us writev=−→

Finally, givenv∈V andk∈K, we again writev=−→

ADso that by the axiom in the statement

Convention 2.12.5 In view of Proposition 2.12.4 and when no confusion can occur, we shall freely write f: E−→F to indicate an affine transformation (f,−→ f ):(E, V )−→(F, W ).

Since an affine transformation respects the whole structure of an affine space, the following result is of course expected:

Proposition 2.12.6 Let(f,−→ f ): (E, V )−→(F, W )be an affine transformation. Then:

1 the image under(f,−→ f )of an affine subspace is an affine subspace;

2 (f,−→ f )maps parallel subspaces onto parallel subspaces.

Proof Let(E , V )be an affine subspace; consider(f (E ),−→ f (V ))⊆(F, W ) Of course, sinceE is non-empty, then so is f (E ), while −→ f (V ) is a vector subspace ofW Moreover givenA, B∈f (E ), let us writeA=f (X),B=f (Y )with

XY ∈V The result follows by Lemma2.12.2

The notion of affine isomorphism is the most natural one:

(f,−→ f ):(E, V )−→(E, V ) is called an affine isomorphism when it admits an inverse affine transformation: that is, an affine transformation

(g,−→g ):(E, V )−→(E, V ) with the properties f◦g=id E , g◦f =id E , −→ f ◦ −→g =id V , −→g ◦−→ f =id V

Proposition 2.13.2 Consider an affine transformation

1 f is injective if and only if−→ f is injective;

2 f is surjective if and only if−→ f is surjective;

3 f is bijective if and only if−→ f is bijective;

4 (f,−→ f )is an affine isomorphism if and only iff is bijective.

Proof Letf be injective If−→ f (v)=0, writev=−→

AB)=−→ f (v)=0 proving that−−−−−−→ f (A)f (B)=0, thusf (A)=f (B)by Lemma2.1.5and thusA=B by injectivity off Thereforev=−→

AB=0, proving that the kernel of−→ f is(0) Thus

Let−→ f be injective Iff (A)=f (B), we have

AB=0 by injectivity of−→ f ThereforeA=B by Lemma2.1.5 Thusf is injective.

AB, by surjectivity off we can findC,Dsuch that v=−→

Let−→ f be surjective and considerA∈E FixingO∈E, by surjectivity of−→ f there exists av∈V such that−→ f (v)=−−−−→ f (O)A Writingv=−→

By Proposition2.1.2, it follows thatA=f (C)andf is surjective.

The case of bijections follows at once Whenf and−→ f are both bijective, they admit inverses g and −→g with −→g linear By Lemma 2.12.2, it remains to check Axiom [AT1] for(g,−→g )(see Definition 2.12.1) But givenA, B∈E and putting

C=g(A),D=g(B), we haveA=f (C)andB=f (D) Since(f,−→ f )is an affine transformation,

AB) where the second implication is obtained by applying−→g

In the finite dimensional case, the result becomes even stronger Let us recall that given a finite dimensional vector spaceV, the determinant of a linear mapping

→f :V −→V is the determinant of the matrixAof−→ f with respect to any basis ofV Indeed given another basis where−→ f has the matrixA , ifMis the change of basis matrix, one has

Proposition 2.13.3 Let(E, V )be a finite dimensional affine space The following conditions are equivalent, for an affine transformation

Proof The equivalence of conditions 1, 2, 3, of conditions 4, 5 and of conditions 6, 7 is attested by Proposition2.13.2.(2⇒4),(2⇒6),(3⇒5)and(3⇒7)are trivial.

(3⇔8)is a well-known fact in linear algebra The equivalence(5⇔7)also follows from a classical result in linear algebra Considering the image and the kernel of−→ f , one has dimKer−→ f +dimIm−→ f =dimE proving that

This simultaneously proves(5⇒3)and(7⇒3)and concludes the proof.

Translations

The first type of affine transformation that we shall consider is the translation.

Definition 2.14.1 Let(E, V )be an affine space For every fixed vectorv∈V, the mapping t v :E−→E, A→A+v is called the translation by the vectorv.

Proposition 2.14.2 Let(E, V )be an affine space For every fixed vectorv∈V, the translationt v by the vectorvis an affine isomorphism, whose vector part−→t v is the identity onV.

Proof GivenA, B∈E, we have (see Fig.2.6)

Bt v (B) proving that(A, t v (A), t v (B), B)is a parallelogram (see Proposition2.11.2). Therefore, we have

As a consequence we get trivially

By Proposition2.12.4, the translation is an affine transformation and

Projections

proving that−→t v is the identity.

Of course the translation by the vectorvis an affine isomorphism whose inverse is the translation by the vector−v

Putting together the main results of Sects.2.4and2.6, we can define the notion of parallel projection:

Definition 2.15.1 Let (E, V ) be an affine space Consider an affine subspace

(F, W )⊆(E, V )andW ⊆V, a vector subspace which is supplementary toW By Theorem2.4.2, given a pointA∈E, there exists a unique affine subspace(F A , W ) such thatA∈F A By Theorem2.6.2,F ∩F A is a singleton The unique point of

F∩F A is called the projection ofAonF, parallel toW

Proposition 2.15.2 With the notation of Definition 2.15.1, the projection on

(F, W ), parallel toW , is an affine transformation.

Proof Let us writep(A)for the projection ofA We apply Proposition2.12.4and use its notation Given−→

CD, we must prove that−−−−−−→ p(A)p(B)=k−−−−−−→ p(C)p(D).

SinceWandW are supplementary, this reduces to proving that

But by definition of the projection,p(A), p(B),p(C),p(D) are in F, thus the vectors−−−−−−→ p(A)p(B)and−−−−−−→ p(C)p(D)are inW So the linear combination above is in

Was well It remains to prove that it is inW

Still by definition of the projection,Aandp(A)are both inF A , thus−−−−→

W ; an analogous argument holds forB,C,D It then suffices to write

Dp(D) and this last vector is inW since each of its four terms are inW

Among the affine transformations, the projections admit an interesting characterization.

Proposition 2.15.3 Let(E, V ) be an affine space For an affine transformation(p,−→p ):(E, V )−→(E, V ), the following conditions are equivalent:

1 pis the projection on an affine subspace(F, W ), parallel to a supplementary directionW ;

Proof Assume first thatpis as in Definition2.15.1 To prove thatp(p(A))=p(A), it suffices to observe that, by definition,p(A)∈F∩F p(A)

AB)=−−−−−−→ p(A)p(B), it follows immediately that−→p ◦ −→p = −→p We define(F, W )to be the image of the whole space

(E, V )under(p,−→p )(see Proposition2.12.6), that is,F =p(E)andW= −→p (V ).

We further defineW as the kernel of−→p, that is

Let us prove that the subspacesW andW are supplementary.

Ifv∈W∩W , thenv∈Wimplies thatv= −→p (v )for somev ∈V Therefore v= −→p v

Next givenv∈V, we certainly have v= −→p (v)+ v− −→p (v) with−→p (v)∈W We must prove thatv− −→p (v)∈W , which is the case since

To complete the proof, using the notation of Definition2.15.1, it remains to show thatp(A)is the (necessarily unique) point lying inF ∩F A Certainlyp(A)∈ p(E)=F On the other hand p(A)=A+−−−−→

SinceA∈F A , it suffices to prove that−−−−→

Symmetries

The idea of a symmetry is the following Consider a subspace(F, W )and a supplementary directionW (see Fig.2.7) Given a pointA, first consider its projection p(A)onF, the projection parallel toW The point symmetric to the pointA, with respect toF, in the directionW , is classically defined as the points(A), on the line joiningAandp(A), “on the other side ofF with respect toA, and at the same distance fromp(A)asA”.

In the next chapter, we shall discuss the fact that notions such as “other side” or

“distance” do not make sense in an arbitrary affine space However, in the case of a symmetry, we do not need such notions: in Fig.2.7, the points(A)is equivalently characterized by the equality−−−−→

Ap(A)=−−−−−−→ p(A)s(A), a property which makes perfect sense in every affine space.

Definition 2.16.1 Consider an affine space (E, V ), together with an affine subspace(F, W )and a directionW supplementary toW The symmetry with respect to(F, W ), parallel toW , is the mapping s:E−→E, A→p(A)+−−−−→

Ap(A) wherep indicates the projection on the subspace (F, W )parallel to the direction

Proposition 2.16.2 A parallel symmetry, as in Definition2.16.1, is an affine transformation.

Proof First of all, givenA, B∈E, let us observe that

Let us write 2=1+1∈Kand apply Propositions2.12.4and2.15.2 If−→

Remark 2.16.3 When the fieldK has characteristic 2, every parallel symmetry is the identity.

Proof From 1+1=0 we get 1= −1 Therefore, with the notation of Defini- tion2.16.1, s(A)=p(A)+−−−−→

The following special case of a symmetry is certainly the most popular:

Definition 2.16.4 Given a pointC in an affine space(E, V ), the symmetry with centerC(see Fig.2.8) is the mapping s:E−→E, A→C+−→

Proposition 2.16.5 The central symmetry defined in Definition2.16.4is a special case of a symmetry as defined Definition2.16.1, namely, the symmetry with respect to({C}, (0))parallel toV.

Proof Trivially,({C}, (0))is an affine subspace andV is supplementary to 0 Since the subspace on which we are projecting is reduced to a single point, necessarily p(A)=Cfor everyA∈E

Proposition 2.16.6 Every parallel symmetryson an affine space(E, V )is involu- tive, that is, satisfiess◦s=id E In particular, every parallel symmetry is an affine isomorphism.

Proof We use the notation of Definition2.16.1 By definition,

Homotheties and Affinities

In particularsadmits an inverse—namely, itself—and therefore is bijective

A picture is homothetic to another one when it is an enlargement (or a reduction) of it (see Fig.2.9) This is very easy to formalize in an affine space.

Definition 2.17.1 By a homothety with centerC∈Eand ratiok∈Kin an affine space(E, V )we mean the mapping h:E−→E, A→C+k−→

Comparing with Definition2.16.4, we notice that a homothety of ratio −1 is precisely a central symmetry In fact, a homothety is a special case of the more general notion of an affinity, which itself generalizes the notion of symmetry.

Definition 2.17.2 In an affine space(E, V ), consider an affine subspace(F, W ), a directionW supplementary toW and a scalark∈K The affinity of ratiokwith respect to(F, W ), parallel toW , is the mapping a:E−→E, A→p(A)+k−−−−→

Ap(A) wherep indicates the projection on the subspace (F, W )parallel to the direction

Proposition 2.17.3 Every affinity as defined in Definition2.17.2is an affine transformation.

Proof The proof of Proposition2.16.2 transposes easily to the present situation. GivenA, B∈E,

A symmetry is thus an affinity of ratio 1 and a homothety of ratiokis a central affinity of ratio−k.

The Intercept Thales Theorem

The intercept Thales theorem is another fundamental geometric result which remains valid in the very general context of affine geometry over an arbitrary field.

Theorem 2.18.1 (Intercept Thales Theorem) Let(E, V )be an affine plane Con- sider four parallel linesd 1,d 2,d 3,d 4 and two arbitrary linesd,d , not parallel to the linesd i ,i=1,2,3,4 Each lined,d intersects each lined i by Corollary2.7.8, yielding pointsA i ,A i as in Fig.2.10 One has

Proof WriteW for the direction of thed i ’s andWfor the direction ofd , which are thus one-dimensional vector subspaces ofV Sinced is not parallel tod i ,W=W ThereforeW+W has dimension strictly greater than 1, thus dimension 2, while

W∩W has dimension strictly smaller than 1, thus dimension 0 SoW andW are supplementary and we can consider the projection on(d , W ) parallel toW By construction, the projection of eachA i is A i The result follows immediately by

Corollary 2.18.2 In an affine plane (E, V ), consider a triangle ABC and two pointsY,Z, distinct from the vertices:Z on the sideAB andY on the sideAC.

The following conditions are equivalent:

1 the lineZY is parallel to the lineBC;

2 there exists a scalar 0=k∈Ksuch that

Under these conditions, one has further−→

Proof (1 =⇒ 2)is just a special case of Theorem2.18.1.

To prove(2 =⇒ 1), draw throughZ the line parallel to BC It is not parallel toAC, sinceBCandACare not parallel By Corollary2.7.8it intersectsAC at a pointY By the first part of the proof,−−→

AY andY =Y. This proves thatZY =ZY is parallel toBC.

Finally, observe that under these conditions

Affine Coordinates

In Chap1, we have insisted on the basic idea of analytic geometry: to transform, by the use of coordinates, a geometrical problem into an algebraic one In the present chapter, until now, we haven’t used any coordinates This is because modern algebra—in our case, the theory of vector spaces—is strong enough to allow many developments without introducing coordinates.

However, being able to introduce coordinates when this is helpful is another powerful tool of affine geometry This is easily done via Proposition2.1.2.

Definition 2.19.1 By an affine basis of an affine space (E, V ) is meant a pair

The coordinates of a pointA∈Eare the coordinates of the vector−→

OAwith respect to the basis(e i ) i ∈ I

OA is a bijection, thus specifying the coordinates of−→

When working with coordinates, we shall generally assume that the affine space is finite dimensional, in order to avoid handling “infinite matrices”.

To avoid heavy notation, when we say that−→x “are” the coordinates of a pointA with respect to an affine base(O;e 1 , , e n ), we mean that

⎟⎠ are the coordinates ofA An analogous convention holds for the coordinates of a vectorv∈V.

To begin with, let us describe the structure of the affine space in terms of coordinates.

Proposition 2.19.2 Let (O;e 1 , , e n ) be an affine basis of the affine space (E, V ) Given two pointsA, B ∈E with coordinates −→x,−→y and a vectorv∈V with coordinates−→z:

Change of Coordinates

OA, which proves the first assertion.

For the second assertion, using axiom [AF2] and the equality above, we have v=−−−−−−→

One should compare Proposition2.19.2with Examples2.2.1,2.2.2 Let us conclude this section with a useful observation:

Lemma 2.19.3 Let(E, V )be a finite dimensional affine space An affine isomorphism

(f,−→ f ):(E, V )−→(E, V ) transforms every affine basis(O, e 1 , , e n )into an affine basis f (O);−→ f (e 1 ), ,−→ f (e n )

Proof The vectors−→ f (e 1 ), ,−→ f (e n )are indeed linearly independent: k 1−→ f (e 1 )+ ã ã ã +k n −→ f (e n )=0 =⇒ −→ f (k 1 e 1+ ã ã ã +k n e n )=0

=⇒ k 1= ã ã ã =k n =0, because−→ f is linear, bijective and thee i sare linearly independent

The first problem to solve when working with coordinates is to establish the formulas expressing a change of coordinates.

Proposition 2.20.1 Consider two bases(O;e 1 , , e n )and(O ;e 1 , , e n )of an affine space(E, V )of dimensionn Given a pointA∈Ewith coordinates−→x in the first basis and−→ x in the second basis, the following formulas hold

→x =M−→x + −→v , −→x =M − 1 −→ x +−→ v whereMis the matrix of the change of coordinates between the corresponding bases ofV,−→v are the coordinates ofOin the second basis and−→ v are the coordinates of

Proof The first equality follows at once from

The second equality is obtained analogously

Let us recall that the matrixMis obtained by putting in columns the components of the vectorse i with respect to the basise 1 , , e n Analogously,M − 1 is obtained by putting in columns the components of the vectorse i with respect to the basis e 1 , , e n

Notice further that multiplying the first equality in Proposition2.20.1byM − 1 , and the second one byM, shows that

The Equations of a Subspace

To be able to use coordinates to solve a problem, we must know how to translate the problem in terms of coordinates Our first concern is to characterize an affine subspace by a system of equations: a system, not just a single equation, as we have already seen in Sect.1.6in the case of a line in solid space.

Proposition 2.21.1 Let(O;e 1 , , e n )be a basis of the affine space(E, V ) The points of an affine subspace(F, W )⊆(E, V )can be characterized as those whose coordinates−→x, with respect to the given basis, satisfy a system of linear equations

A−→x =−→ b, whereAis a matrix withn columns and−→ b ∈K n Conversely, every such system admitting a solution is a system of equations of an affine subspace.

Proof Let us consider a pointO ∈F and a basise 1 , , e s ofW Let us further complete that sequence to a basise 1 , , e n ofV The vectors ofW are thus those whose coordinatesx i with respect to the basise 1 , , e n satisfy

But sinceO ∈F,A∈F if and only if −−→

O A∈W, that is, if and only if the coordinates ofAwith respect to the basis(O ;e 1 , , e n )satisfy the system above.

The Matrix of an Affine Transformation

Let us writex i for the coordinates with respect to the first basis,v i for the components of−→v andM=(m i,j ) Applying Proposition2.20.1, we obtain the equivalent conditions

We have indeed obtained a system of linear equations.

Conversely given a system of linear equations

a r,1 x 1+ ã ã ã +a r,n x n =k n admitting a solution, we writeF for the set of points whose coordinates are solutions of the system andWfor the vectors ofV whose coordinates with respect to the basis e 1 , , e n satisfy

The argument of Example2.2.2carries over as such to show that(F, W )is an affine subspace

2.22 The Matrix of an Affine Transformation

Let us now determine the matrix expression of an affine transformation.

Proposition 2.22.1 Choose an affine basis (O;e 1 , , e n ) of an affine space (E, V ) and an affine basis (O ;e 1 , , e m ) of an affine space (E , V ) The following conditions are equivalent, for a mappingf:E−→E :

2 in terms of coordinates with respect to the two affine bases,f can be described by a formula

Under these conditions,Mis the matrix of the corresponding linear mapping −→ f with respect to the bases ofV andW, while−→ b are the coordinates off (O)inE

Proof Whenf is an affine transformation,

This immediately implies the matrix description of the statement ChoosingA=O, we have−→x =0, thusM−→x =0 and−→ b are the coordinates off (O).

Conversely given such a matrix description off, define−→ f to be the linear mapping admittingAas matrix with respect to the given basis ofV andW Given two pointsA, B∈Ewith respective coordinates−→x ,−→y, Proposition2.19.2tells us that the coordinates of−−−−−−→ f (A)f (B)are

(A−→y +−→ b )−(A−→x +−→ b )=A−→y −A−→x =A(−→y − −→x ) where therefore−→y − −→x are the coordinates of−→

AB This yields axiom [AT1] and concludes the proof, by Lemma2.12.2

Let us now review, in terms of coordinates, the examples of affine transformations that we have studied in the previous sections.

Proposition 2.22.2 Let (O;e 1 , , e n ) be an affine basis of the affine space (E, V ) Fixing v∈V, the matrix expression of the translation t v by the vectorv is

→x → −→x + −→v where−→v indicates the coordinates ofvwith respect to the basise 1 , , e n Proof By Proposition2.14.2,−→t v is the identity

Proposition 2.22.3 Let(E, V )be a finite dimensional affine space Consider supplementary subspacesW, W ⊆V with respective basese 1 , , e m ande m + 1 , , e n , and a pointO∈E Write(F, W )for the affine subspace with directionW con- tainingO(see Theorem2.4.2) Under the affine basis(O;e 1 , , e n ), the projection pon(F, W ), parallel toW , admits the matrix expression

→x →A−→x whereAis the diagonal matrix defined by a i,i =1 for 1≤i≤m, a i,i =0 form+1≤i≤n.

Proof SinceO∈F, we havep(O)=Oand the translation vector−→ b is 0 (Propo- sition2.22.1) More generallyp(A)=Afor each pointA∈F, thus−→p (w)=wfor eachw∈W and the firstmcolumns of the matrix are those of the identity matrix. Finally ifw ∈W , write w =−→

OB; the projection of B is then O, proving that

→p (w)=0; therefore then−mlast columns of the matrix are zero.

The Quadrics

Proposition 2.22.4 Under the conditions of Proposition 2.22.3, the symmetrys with respect to(F, W ), parallel toW , admits the matrix expression

→x →A−→x whereAis the diagonal matrix defined by a i,i =1 for 1≤i≤m, a i,i = −1 form+1≤i≤n.

Proof We have seen, in the proof of Proposition2.16.2, that−→s =2−→p − −→ id, with pthe corresponding projection This immediately yields the result

We now switch to the study of those parts of an affine space which can be described by an equation of degree 2 in terms of the coordinates of their points, in some affine basis For convenience, since we are working with coordinates, we shall always assume that the space is finite dimensional.

Of course, when we write something likeX 2 , whereXis an unknown with values inK, the exponent 2 is the natural number 2∈N, not the element 2=1+1∈K.

Indeed,X 2 is just an abbreviation forXX But if we compute

(X+Y ) 2 =X 2 +2XY+Y 2 , the coefficient 2 ofXY is now the element 2∈K Indeed 2XY is an abbreviation for

XY+Y X=XY+XY=1XY+1XY =(1+1)XY.

Thus when working with equations of degree 2, the element 2∈Kwill rapidly enter the story Hence, even if the exponent 2∈Nis not the element 2∈K, we obtain in characteristic 2 formulas like

Such formulas will easily lead to pathologies! This is why we make the following convention:

Convention In the sections of this chapter devoted to quadrics,Kwill always denote a field of characteristic distinct from 2.

Definition 2.23.1 A subsetQ⊆Eof a finitely dimensional affine space(E, V )is called a quadric when there exists an affine basis with respect to which the points ofQcan be characterized as those whose coordinates satisfy some equation n i,j = 1 a i,j X i X j + n i = 1 b i X i +c=0, a i,j , b i , c∈K. of degree 2.

Let us make clear that in Definition2.23.1, all coefficients are arbitrary scalars, thus possibly zero.

Lemma 2.23.2 Let(E, V )be a finite dimensional affine space A quadricQ⊆E can be characterized by an equation of degree 2 in every affine basis.

Proof Consider an equation of degree 2 as in Definition2.23.1 In a change of coordinates, the coordinatesX i are expressions of degree 1 in terms of the new coor- dinatesX i But substituting expressions of degree 1 into an expression of degree 2 leaves an expression of degree 2

Convention In the sections of this chapter devoted to quadrics, by equation of a quadric we always mean equation of degree 2.

Lemma 2.23.3 Under the conditions of Definition2.23.1, the equation of the quadric as in Lemma2.23.2can always be written

Proof Following Lemma2.23.2, consider the general form of an equation of degree 2 Since we are not in characteristic 2, we can always write a i,j X i X j +a j,i X j X i =a i,j +a j,i

2 X j X i and grouping the terms of the equation in this way, we force the matrixAto become symmetric

In Sect.1.10we have listed the “quadrics” ofR 2 , more commonly called “conics” Among these we have found some “degenerate” examples which can hardly be considered as genuine curves of degree 2: the empty set, a point, a line or the whole plane Analogous considerations hold for the quadrics ofR 3 , listed in Sect 1.14 where the empty set, a point, a line, a plane or the whole space cannot be considered as genuine surfaces of degree 2 Let us formalize this in a definition.

Definition 2.23.4 A quadricQ⊆Ein a finite dimensional affine space(E, V )is non-degenerate when

1 Qis not contained in a proper subspace;

Counterexample 2.23.5 For a quadric, “being non-degenerate” also depends on the choice of the fieldK, not just on the form of the equation and certainly not on the rank of the matrixAin Lemma2.23.3.

The Reduced Equation of a Quadric

X 1 2 +X 2 2 =0 is degenerate since it reduces to a single point: the origin However the corresponding matrixAis the identity matrix On the other hand, over the field of complex numbers, the same equation is that of a non-degenerate quadric, since it factors as

(X 1 +iX 2 )(X 1 −iX 2 )=0 that is, the equation of two intersecting lines

Our definition of a quadric in terms of an equation in a given affine basis suggests at once the question:

Can a quadric have several equations in the same affine basis?

The answer is of course yes: it suffices to multiply an equation by a non-zero constant! So a more sensible question should rather be:

Are two equations of a quadric in the same affine basis necessarily proportional?

The answer is of course no: both equations

X 1 2 +X 2 2 =0, X 2 1 +2X 2 2 =0 describe the same quadric ofR 2 , namely, the origin However, the answer is yes for non-degenerate quadrics We shall prove this result later (see Theorem2.26.9).

2.24 The Reduced Equation of a Quadric

Convention In this section,Kis a field of characteristic distinct from 2 and equation of a quadric always means equation of degree 2.

We are going to show that an appropriate choice of affine basis allows us to considerably simplify the equation of an arbitrary quadric.

Lemma 2.24.1 LetQ⊆Ebe a quadric in a finite dimensional affine space(E, V ). There exists an affine basis(O;e 1 , , e n )in which the quadric has an equation of the form n i = 1 a i X 2 i + n i = 1 b i X i +c=0.

Proof Fix an arbitrary affine basis (O;e 1 , , e n ) in which the quadric has the equation (see Proposition2.23.3)

X +c =0 withAa symmetric matrix The following mapping is thus a symmetric bilinear form onV (see DefinitionG.1.1) ϕ:V ×V −→K, (v, w)→−→

Y whereX ,Y indicate the coordinates ofv,wwith respect to the basise 1 , , e n Applying Corollary G.2.8, we have another basis e 1 , , e n in which the matrix ofϕ is diagonal WritingM for the matrix of change of coordinates, the equation of the quadric with respect to the affine basis(O;e 1 , , e n ) thus becomes (see CorollaryG.1.4)

X +c =0 where the matrixA=M t A Mis now diagonal

Theorem 2.24.2 Let Q⊆E be a quadric in a finite dimensional affine space (E, V ) There exists an affine basis(0;e 1 , , e n )with respect to which the equation of the quadric takes one of the following forms:

Such an equation is called a reduced equation of the quadric.

Proof Let us begin with an affine basis(P;ε 1 , , ε n )with respect to which the equation of the quadric has the form n i = 1 α i Y i 2 + n i = 1 β i Y i +γ=0

(see Lemma2.24.1), whereY i indicates the coordinates of a point.

If all coefficientsα i =0 are zero, the equation reduces to an equation of lower degree, namely n i = 1 β i Y i +γ=0.

If further, all coefficientsβ i are zero, we end up with the equationγ =0: this is an equation of type 2 whenγ =0 (the equation of the whole space) and—up to division byγ—the equation 1=0 of type 1 whenγ=0 (the equation of the empty set) If not all coefficientsβ i are zero, we get the equation of a hyperplane; choosing

2.24 The Reduced Equation of a Quadric 95 a new basis with its origin and then−1 first vectors in this hyperplane, the equation becomesZ n =0 (the last coordinate): this is an equation of type 3 Notice in particular that when allα i are zero, the quadric is degenerate.

If not all coefficientsα i are zero, up to a possible renumbering of the vectors of the basis, there is no loss of generality in assuming that α i =0 fori≤m, α i =0 fori > m, 1≤m≤n.

Notice that the casem=nis the case where allα i ’s are non-zero.

Let us study the effect of a change of origin, keeping the same basisε 1 , , ε n ofV If the new originOhas coordinatesδ i and the new coordinates are writtenZ i , the change of basis formulổ are simply

Y i =Z i +δ i and the equation of the quadric becomes m i = 1 α i Z i 2 + m i = 1

For every index 1≤i≤m, the coefficient ofZ i is an expression of degree 1 inδ i , withα i =0 as coefficient ofδ i Choosing δ i = − β i

This time, each variableZ i appears at most once: in degree 2 or in degree 1. Write k= − m i = 1 β i 2

4α i + n i = m + 1 β i δ i +γ for the constant in this equation.

Ifk=0, this is an equation of type 2 Ifk=0, dividing by−kyields an equation of type 1 In those cases, the theorem is proved.

Let us now consider the remaining case:m < n If β i =0 for alli≥m, then again equation (∗) takes the form n i = 1 α i Z i 2 +k=0 and the same conclusions apply, whatever the values given toδ m + 1 , , δ n Finally, we address the remaining case wherem < n, withβ i =0 for somei≥m.

Again, up to a possible renumbering of the last vectors of the basis, there is no loss of generality in assuming thatβ n =0 Fix then arbitrary values (for example, 0) for δ i , for allm+1≤i≤n−1; it remains to chooseδ n Butkis now an expression of degree 1 inδ n , withβ n =0 as coefficient ofδ n Choose forδ n the “root” ofk=0, regarded as an equation of degree 1 inδ n Equation (∗) becomes m i = 1 α i Z i 2 + n i = m + 1 β i Z i =0.

To conclude the proof, it remains to find another affine basis—with respect to which the coordinates will be writtenX i —and giving rise to a change of coordinates with the properties

We thus want a matrix of change of coordinates which has the form

But a matrixM is the matrix of a change of coordinates if and only if it is in- vertible Themfirst lines of the matrix above are trivially linearly independent; and sinceβ n =0, so too is the system of themfirst lines together with the last one.

To obtain the expected matrix of change of coordinates, it remains to complete this sequence ofm+1 linearly independent lines to a system ofnlinearly independent lines: this is simply completing a basis to a system ofm+1 linearly independent vectors ofK n

The Symmetries of a Quadric

We recall again the following

Proposition 2.25.1 Let(O;e 1 , , e n )be a basis of an affine space(E, V ) Con- sider a non-degenerate quadricQ⊆E which, with respect to this basis, has a reduced equation as in Theorem2.24.2 Consider a vector subspaceW ⊆V which:

1 in the case of an equation of type 1 or 2, is generated by some of the vectors e 1 , , e n ;

2 in the case of an equation of type 3, is generated by some of the vectors e 1 , , e n − 1.

Consider the supplementary vector subspaceW⊆V generated by the other vectors of the basis and the corresponding affine subspace(F, W )containingO (see The- orem2.4.2) Under these conditions, the quadricQis stable under the symmetry with respect to(F, W ), parallel toW

Proof This follows at once from Proposition2.22.4: if a point is on Q, its symmetric partner has coordinates obtained by changing the sign of those coordinates corresponding to vectors ofW ; these coordinatesX i appear under the formX i 2 in the equation

We shall be particularly interested in the following special case:

Definition 2.25.2 LetQ⊆Ebe a non-degenerate quadric in a finite dimensional affine space(E, V ) A pointO∈Eis called a center of symmetry ofQwhenQis stable under the symmetry with centerO(see Definition2.16.4).

Of course a center of symmetry can be a point of the quadric: for example, the vertex of a cone inR 3 A quadric can also have a center of symmetry “outside” of it: for example, the center of a circle inR 2 Furthermore, a quadric can have many centers of symmetry: for example, all the points of the axis of a circular cylinder inR 3

Proposition 2.25.3 Let(0;e 1 , , e n )be an affine basis of an affine space(E, V ). For a non-degenerate quadricQ⊆E, the following conditions are equivalent:

1 Ois a center of symmetry ofQ;

2 Qadmits in(O;e 1 , , e n )an equation without any term of degree 1;

3 all equations ofQin(O;e 1 , , e n )are without any term of degree 1.

Proof (1⇒3) Consider an equation n i,j = 1 a i,j X i X j + n i = 1 b i X i +c=0 of the quadric in the given basis For every pointP=(X i ) i = 1, ,n of the quadric, by assumption, the points(P )=(−X i ) i = 1, ,n is on the quadric as well (see Proposi- tion2.22.4) This yields n i,j= 1 a i,j X i X j − n i= 1 b i X i +c=0.

Subtracting the two equations shows that every point of the quadric satisfies the equation

Since 2=0 by assumption, this would be the equation of an hyperplane, as soon as at least one of theb i ’s is non-zero But this would contradict the fact that the quadric is non-degenerate (Definition2.23.4) Thusb i =0 for each indexiand the equation of the quadric does not have any term of degree 1.

(3⇒2)is obvious.(2⇒1)is obvious as well, since when all non-constant terms are of degree 2, changing the sign of the unknowns does not affect the result

Proposition 2.25.4 Consider a non-degenerate quadric in a finite dimensional affine space If the quadric admits a center of symmetry belonging to the quadric, then all the centers of symmetry belong to the quadric.

Proof Consider an affine basis(O;e 1 , , e n )withOa center of symmetry of the quadric Write n i,j = 1 a i,j X i X j +c=0 for the corresponding equation (see Proposition2.25.3) Since the originOsatisfies this equation, we have in fact c=0 If O is another center of symmetry of the quadric, the change of coordinates to the affine basis(O ;e 1 , , e n )takes the form

Y + −→v with−→v the coordinates ofO with respect to the original basis The new equation thus becomes

2.25 The Symmetries of a Quadric 99 since these are 1×1 matrices So the new equation becomes

SinceO is a center of symmetry, by Proposition2.25.3we have−→v t A=−→

0 But then the constant term disappears as well and the equation becomes

This is trivially satisfied byO , the origin of the new basis Here is now a “geometric” classification of quadrics:

Theorem 2.25.5 Consider a non-degenerate quadric Q⊆E in a finite dimensional affine space(E, V ) With the terminology of Theorem2.24.2:

1 all the reduced equations of the quadric are of the same type;

2 the quadric admits a reduced equation of type 1 if and only if it has a center of symmetry which is not a point of the quadric;

3 the quadric admits a reduced equation of type 2 if and only if it has a center of symmetry which is a point of the quadric;

4 the quadric admits a reduced equation of type 3 if and only if it does not have a center of symmetry.

Proof Consider an affine basis(O;e 1 , , e n )with respect to which the quadric admits a reduced equation.

• If the equation is of type 1, the origin is a center of symmetry (Proposition2.25.3) and it does not belong to the quadric.

• If the equation is of type 2, the origin is a center of symmetry (Proposition2.25.3) and it belongs to the quadric.

• If the equation is of type 3, the origin is not a center of symmetry, again by Proposition2.25.3.

Let us prove further that in the third case, the quadric does not admit any center of symmetry.

Indeed, let n − 1 i = 1 a i X i 2 =X n be the equation of the quadric with respect to the affine basis(O;e 1 , , e n ) Given an arbitrary pointO ∈E, the equation of the quadric with respect to the affine basis (O ;e 1 , , e n )is obtained by a change of coordinates

X i =Y i +v i where−→v are the coordinates ofO with respect to the original basis The new equation becomes n − 1 i = 1 a i Y i 2 +2 n − 1 i = 1 a i v i Y i + n − 1 i = 1 a i v i 2 =Y n +v n

By Proposition2.25.3, the presence of the termY n implies thatO is not a center of symmetry.

In view of Proposition2.25.4, we can now rephrase the implications at the beginning of the proof:

• If the equation is of type 1, the quadric admits a center of symmetry and all the centers of symmetry belong to the quadric.

• If the equation is of type 2, the quadric admits a center of symmetry and none of the centers of symmetry belongs to the quadric.

• If the equation is of type 3, the quadric does not admit any center of symmetry.

In these implications, the statements in terms of the centers of symmetries cover all possible cases (again by Proposition2.25.4) and trivially exclude each other; thus all three implications are equivalences In particular, the type of a reduced equation is entirely determined by the properties of the centers of symmetry

It is probably worth concluding this Section with a comment Being a center of symmetry of a quadric is definitely a geometric property of the quadric Admit- ting an equation of some given type is an algebraic property of the equation of the quadric Theorem2.25.5thus provides the link between those geometric and algebraic properties It remains to further reinforce this theorem by somehow proving the “uniqueness” of the equation of a non-degenerate quadric.

The Equation of a Non-degenerate Quadric

We continue to adopt the

Let us conclude this chapter by proving that with respect to a given basis, a non-degenerate quadric can only have “one” equation (of course, up to a non-zero multiplicative constant) This section relies heavily on the theory of the resultant of two polynomials (see AppendixD).

Lemma 2.26.1 LetQbe a quadric in an affine space of dimensionn The following conditions are equivalent:

1 two equations of this quadric with respect to a given basis are proportional;

2 these two equations, transformed via a change of basis, are proportional.

2.26 The Equation of a Non-degenerate Quadric 101 Proof Trivially if in one basis two equations take the form p(X 1 , , X n )=0, kp(X 1 , , X n )=0, 0=k∈K withpa polynomial, every change of coordinates respects this proportionality, since it occurs “inside”p(X 1 , , X n )

Lemma 2.26.2 LetQbe a non-degenerate quadric in an affine space of dimensionn Given an equation p(X 1 , , X n )=0 of this quadric in some affine basis, one of the two following possibilities holds:

2 p(X 1 , , X n )is the product of two non-proportional factors of degree 1.

Proof Ifpis not irreducible, it factors uniquely as a product of irreducible factors (see TheoremB.4.9), which are thus necessarily of degree 1 If these factors were proportional, the equation of the quadric would take the form k(a 0 +a 1 X 1 + ã ã ã +a n X n ) 2 =0, 0=k∈K.

Thus the quadric would be the hyperplane with equation a 0+a 1 X 1+ ã ã ã +a n X n =0.

This would contradict the non-degeneracy of the quadric

Lemma 2.26.3 LetQbe a non-degenerate quadric in an affine space of dimensionn If for some affine basis, an equation ofQdoes not contain the variableX n , thenX n does not appear in any equation ofQwith respect to this same affine basis.

Proof The assumption implies that given a pointP of the quadric and an arbitrary scalark∈K, the corresponding pointP k

⎟⎠ is still on the quadric.

Consider an arbitrary equation n i,j = 1 a ij X i X j + n i = 1 b i X i +c=0 of the quadric with respect to the given affine basis GivenP∈Q, the pointsP 1and

P − 1are still on the quadric This means that each point of the quadric satisfies the equations n− 1 i,j = 1 a ij X i X j +2 n− 1 i = 1 a in X i +a nn + n− 1 i = 1 b i X i +b n +c=0 n − 1 i,j = 1 a ij X i X j −2 n − 1 i = 1 a in X i +a nn + n − 1 i = 1 b i X i −b n +c=0.

Subtracting these two equations, we conclude that every point of the quadric satisfies the equation

If one coefficienta in is non-zero, this is the equation of a hyperplane in which the quadric is thus contained This contradicts the non-degeneracy of the quadric Thus all coefficientsa in are equal to zero As a consequence, since the quadric is not empty,b n =0 as well Thus the equation of the quadric has the form n − 1 i,j = 1 a ij X i X j +a nn X n 2 + n − 1 i = 1 b i X i +c=0.

Let us next express the fact that given a pointP of the quadric, the two pointsP 1 andP 0are still on the quadric This means that every point of the quadric satisfies the equations n − 1 i,j = 1 a ij X i X j +a nn + n − 1 i = 1 b i X i +c=0 n − 1 i,j = 1 a ij X i X j + n − 1 i = 1 b i X i +c=0.

Subtracting, we conclude that every point of the quadric satisfies the equationa nn 0, or in other words, since the quadric is non-empty,a nn =0

Now comes the key step for proving the uniqueness of the equation of a non- degenerate quadric:

Lemma 2.26.4 LetQbe a non-degenerate quadric in an affine space(E, V )of dimensionn Consider another quadric Q⊆Q E Assume the existence of an affine basis with respect to which the quadricQ admits an equation of the form

2.26 The Equation of a Non-degenerate Quadric 103 p(X 1 , , X n )=0, where one of the coordinates appears with exponent 1 but not with exponent 2 ThenQ=Q and two equations of this quadric with respect to the same basis are necessarily proportional.

Proof By Lemma 2.26.1, it suffices to prove the result for the basis where the quadric Q admits the equation p(X 1 , , X n )=0 as in the statement Let q(X 1 , , X n )=0 be an equation of the quadricQ with respect to this same basis. Notice first thatQ is a non-degenerate quadric Indeed it is not the whole space

Eby assumption and it is not contained in a hyperplane, since this is already the case forQ Thus the equations ofQandQ are actual equations of degree 2.

Ifp(X 1 , , X n )is irreducible, by PropositionD.2.2,pdividesq Since both have degree 2, this implies the proportionality of p and q This forces in particular

Ifp(X 1 , , X n )is not irreducible, by Lemma2.26.2, it is the product of two non-proportional factors of degree 1. p(X 1 , , X n )=r(X 0 , , X n )s(X 0 , , X n ).

Ifr(a 1 , , a n )=0, thenp(a 1 , , a n )=0 and thusq(a 1 , , a n )=0 By Corol- laryD.2.3,r(X 1 , , X n )is then a factor ofq(X 1 , , X n ) An analogous argument holds fors(X 1 , , X n ) Since these two factors are not proportional and the decomposition ofq(X 1 , , X n )into irreducible factors is unique (see TheoremB.4.9), it follows that q(X 1 , , X n )=kr(X 1 , , X n )s(X 1 , , X n )=kp(X 1 , , X n ).

Let us recall that for a non-degenerate quadric, the type of a reduced equation is fixed (see Theorems2.25.5and2.24.2).

Proposition 2.26.5 LetQbe a non-degenerate quadric of type 3 in an affine space of dimensionn Two equations of this quadric with respect to the same basis are necessarily proportional.

Perhaps more surprising is that the quadrics of type 2 also enter into the context of Lemma2.26.4.

Proposition 2.26.6 LetQbe a non-degenerate quadric of type 2 in an affine space of dimensionn Two equations of this quadric with respect to the same basis are necessarily proportional.

Proof Choose as origin a centerP 0of symmetry, which is thus a point of the quadric (see Theorem2.25.5) Since the quadric is non-degenerate, there arenother points

P 1 , , P n such thatP 0 , P 1 , , P n are affinely independent, that is,

P 0 P n ) is an affine basis With respect to this basis, an equation takes the form n i,j = 1 a ij X i X j =0 because the origin is a center of symmetry belonging to the quadric (see Proposi- tion2.25.3) Introducing the coordinates of the vectore i into this equation, we get a ii =0 Thus finally the equation takes the form i = j a ij X i X j =0 where all appearing variablesX i have the exponent 1 The result again follows by

It remains to handle the case of quadrics of type 1 In this case, it is generally impossible to find an affine basis such that the equation contains one of the variables with only the exponent 1 (see Problem2.27.14) But nevertheless, this case can once more be reduced to the situation in Lemma2.26.4.

Lemma 2.26.7 LetQbe a non-degenerate quadric of type 1 in an affine space of dimensionn Suppose that with respect to some affine basis,Qadmits a reduced equation of the form n i = 1 a i X 2 i =1 where the quadratic form φ (X 1 , , X n ) n i = 1 a i X i 2 does not admit any non-zero isotropic vectors (see DefinitionG.2.1) Then two equations ofQwith respect to the same affine basis are necessarily proportional.

Proof By Lemma2.26.1, it suffices to consider the case of a second equation ofQ with respect to the same affine basis By Proposition2.25.3, the origin is a center of symmetry thus this second equation takes the form n i,j = 1 b ij X i X j =1.

2.26 The Equation of a Non-degenerate Quadric 105

Let us then consider the following two quadricsQ 1andQ 2inK n+ 1 n i = 1 a i X i 2 =X n 2 + 1 , n i,j = 1 b ij X i X j =X 2 n + 1

Up to multiplication by an arbitrary scalark, these are thus all the points ofQ 1and

Q 2admitting a non-zero last component.

Let us next observe thatQ 1does not have any other point, that is, a point with coordinates

Indeed having such a point onQ 1would mean φ (k 1 , , k n ) n i = 1 a i k 2 i =0.

This cannot exist because by assumption, the form φ does not admit a non-zero isotropic vector.

The arguments above thus prove already that Q 1⊆Q 2 They also prove that these quadrics are not the whole spaceK n + 1 : given a point not onQ, it suffices to add a last coordinate 1 to get a point which is neither onQ 1nor onQ 2.

Problems

As a convention, in all the problems concerning quadrics, the fieldKhas a characteristic distinct from 2.

2.27.1 In the affine spaceZ2×Z2, find a non-empty subset which is not an affine subspace, but contains the line generated by any two distinct points of it.

2.27.2 Prove that an affine subspace generated by three non-collinear points is an affine plane.

2.27.3 In the affine spaceZ3×Z3, consider the triangle with vertices(0,0),(2,0) and(0,2) Prove that the medians of this triangle are parallel.

2.27.4 Consider a triangle in an affine space over a field of characteristic distinct from 2 and 3 Give sense to (and prove) the statement: The medians intersect each other at a point situated 2 3 of the distance between a vertex and the opposite side.

2.27.5 In an affine plane over a field of characteristic distinct from 2, the middle points of the sides of an arbitrary quadrilateral constitute a parallelogram (Varignon’s theorem).

2.27.6 Consider a tetrahedronABCD(i.e four affinely independent points) in an affine space(E, V ) of dimension 3, over a field of characteristic 0 Consider the barycenterA of(B,1),(C,1),(D,1), and define analogouslyB ,C ,D Prove the existence of a homothety mappingABCDontoA B C D

2.27.7 Let(E, V )be an affine space of dimension 3 over a fieldKof characteristic distinct from 2 Given a tetrahedronA 1 A 2 A 3 A 4 , a bimedian is a line through the middle point of one pairA i A j and the middle point of the other pair Prove that the four bimedians converge.

2.27.8 In an affine plane over a field of characteristic distinct from 2, consider four intersecting lines as in Fig.2.12 Prove that the middle points of the pairs(A, A ), (B, B ),(C, C )are collinear.

2.27.9 Prove that every affine transformation is the composite of a translation and an affine transformation with a fixed point.

2.27.10 Consider two affine spaces(E, V ),(F, W ) Prove that a mappingf:E−→

F is an affine transformation if and only if it preserves barycenters.

2.27.11 Let (E, V ) be a finite dimensional affine space Prove that two equations of a hyperplane with respect to the same affine basis are necessarily proportional.

2.27.12 LetQ⊆Ebe a quadric in a finite dimensional affine space Given an affine subspace(F, W ), prove thatQ∩F is a quadric of(F, W ).

2.27.13 Consider a quadricQ⊆E in a finite dimensional affine space When Q admits a center of symmetry, the set of all centers of symmetry is an affine subspace.

2.27.14 In an affine space (E, V ) of dimension n, consider a non-degenerate quadricQof type 1 admitting with respect to some affine basis an equation of the form n i = 1 a i X 2 i =1.

Suppose that the quadratic form φ (X 1 , , X n ) n i = 1 a i X i 2 does not admit an isotropic vector (see DefinitionG.2.1) Prove that in every equation ofQwith respect to any affine basis, n i,j = 1 a ij X i X j + n i = 1 b i X i +c=0 all the coefficientsa ii are necessarily non-zero.

2.27.15 Let(E, V )be an affine space of dimensionnover a fieldKwith at least five elements Consider a quadricQwhich is not contained in an affine subspace of dimensionn−2 Prove that two equations ofQin a given affine basis are necessarily proportional (see [5]).

Exercises

2.28.1 Determine if the following triples (E, V ,−−−→

(−|−)) are affine spaces If so, describe the+operation between points and vectors.

4 E=C,V =C(as a complex vector space),−→ zz =z −z.

5 E= {♣,♦,♥,♠},V =(Z2 ) 2 ; the mapping−−−→(−−)is defined by the following table:

2.28.2 Let(E, V )be an affine space,P , Q, R, S∈Eandu, v∈V Prove that

2.28.3 LetP andQbe fixed points in an affine space(E, V ) Consider the mapping

2.28.4 Let(E, V )be aK-affine space andO∈E Show that the following conditions are equivalent, fornpointsA 1 , , A n ofE:

1 the pointsA 1 , , A n are affinely dependent;

3 there exist elementsλ 1 , , λ n ∈K, not all zero, such that n i= 1 λ i −−→

2.28.5 If S, A 1 , , A n are points in a real affine space and α 1 , , α n are real numbers such that n i = 1 α i =1, show that the point

, (2.2) is an affine subspace of the vector spaceR 2 × 2 of matrices, viewed as an affine space over itself How can one determineW from the sole knowledge of F and the fact that(F, W )is an affine subspace?

2.28.7 Consider the vector spaceF(R,R)of all mappingsR−→R, viewed as an affine space over itself Knowing that(F, W )is an affine subspace such that

• the identity functionf (x)=xbelongs toF, prove that

2.28.8 In the vector space(Z3 ) 2 viewed as an affine space over itself, consider the pointsA=(0,0)andB=(1,2).

2 On this picture, emphasize the affine subspace generated byAandB(the “affine line” throughAandB);

3 Determine the “middle point” of the pairAB.

2.28.9 Consider the vector spaceR 3 viewed as an affine space over itself; we work in the canonical basisR c

1 Consider the pointQ=(20,14,12)and the three vectors e 1 =(4,0,1), e 2 =(1,−1,3), e 3 =(2,4,2).

Find the coordinates in R c of a point P such that, with respect to the basis

2 A planeπ admits inR the equationx +5y −3z =5 What is “its” equation with respect to the canonical basisR?

3 If possible, answer the two questions above in the case whereRis substituted by

2.28.10 Let(E, V )be an affine space of dimension n We consider a hyperplane (H, W )with equation a 1 x 1 + ã ã ã +a n x n =b in some affine basis(P;e 1 , , e n ) What is, with respect to the basis(e 1 , , e n ), the equation of the directionW of the hyperplane?

2.28 Exercises 113 2.28.11 We view the vector spaceR 4 as an affine space over itself Consider

With respect to the canonical basis, give the parametric and Cartesian equations of

1 the hyperplane containingP and parallel tou,v,w;

2 the plane containingP and parallel touandv;

3 the line containingP and parallel tou;

2.28.12 In the vector spaceR 4 viewed as affine space over itself, give an example of two planes whose intersection is a singleton.

2.28.13 We work in the canonical basis of(Z5 ) 2

1 Give the parametric equations and the Cartesian equation of the line passing through(1,2)and(3,1).

2 Determine—when it exists—the intersection point of

(a) the lines with equationsx+y=3 andx−y=1;

(b) the lines with equations 4x+y=3 andx−y=1.

2.28.14 LetKbe a field of characteristic zero We work inK 3 , viewed as an affine space over itself We consider four pointsA,B,C,D not in the same plane; we consider further the two affine bases

DC) and the pointP with coordinates

⎠ with respect to the basisR Write

⎠ for the respective coordinates of a pointXwith respect to these two affine bases.

2 Determine the Cartesian equations, with respect to the basisR, of the line con- tainingP and intersecting the two linesADandBC, if such a line exists!

2.28.15 Consider the vector spaceE=R 2 × 2 of 2×2-matrices, viewed as an affine space over itself Consider further two affine subspaces(F, W )and(G, V )such that

, G= {M∈E|m 11 =0, m 21=1} whereTrindicates the trace andm ij the elements of the matrixM.

2 If it makes sense, give for a basis of your choice the matrix representation of the projection on(F, W ), parallel toV.

3 IsF∩Gan affine subspace ofE?

2.28.16 Consider the spaceR (2) [X]of real polynomials of degree at most 2, viewed as an affine space over itself Consider

1 Prove thatR =(0 ;e 1 , e 2 , e 3 )is an affine basis.

2 Consider the plane with equation 5x +2y +5z =1 with respect to the basisR and compute its equation with respect to the canonical basisR=(0,1, X, X 2 ).

2.28.17 InR 3 viewed as affine space over itself, consider the two lines d 1 ≡

Determine the parametric equations and the Cartesian equation of the planeπ con- tainingd 1and parallel tod 2.

2.28.18 LetK be a field of characteristic zero InK 3 , viewed as an affine space over itself, consider a tetrahedronABCD Write respectivelyM 1,M 2,M 3for the middle points of the pairs(A, B),(A, C),(A, D)andN 1 ,N 2 ,N 3 for the middle points of the pairs(C, D),(D, B),(B, C) In the affine basis(A;−→

AD), calculate the coordinates of the points M k ,N k and the equations of the linesd k containingM k andN k

2.28.19 Consider four affinely independent pointsA,B,C,Din a real affine space of dimension 3 Consider the affine basis(A;−→

AD) Give a vectorial equation of the planeBCDand infer a Cartesian equation of this plane.

2.28.20 Consider the vector spaceR 2 × 2 of(2×2)-real matrices, viewed as an affine space over itself Consider the affine subspace(F, W )whereF is the set of those matrices whose trace is equal to 1 Give a Cartesian equation ofF with respect to the canonical basis ofR 2 × 2

2.28.21 LetR=(O;u, v)be an affine basis of an affine space(E, V ) of dimension 2 Show that(O+u;u−v, u+v)is still an affine basis of(E, V )if and only if the characteristic ofKis distinct from 2.

E (x, y, z)∈C 3 x+y=z+2 ofC 3 viewed as affine space overC 2 via the operation

(x 1 , y 1 , z 1 )(x 2 , y 2 , z 2 )=(x 2−x 1 , z 2−z 1 ) with(x 1 , y 1 , z 1 )∈E,(x 2 , y 2 , z 2 )∈E Give an affine basis of this affine space.

2.28.23 Consider the vector spaceR (2) [X]of real polynomials of degree at most

2, viewed as affine space over itself Calculatef, given that the pair(f,−→ f )is an affine transformation ofR (2) [X]such that:

How would you have calculated−→ f from the sole knowledge off? When a mapping f is given, is it always the case that−→ f exists?

2.28.24 Consider an affine transformation(f,−→ f )on an affine space(E, V )and let

A, B∈Ebe two distinct points Write

Prove that whenA, B∈Fix (f ), then the whole line containingAandBis contained inFix (f ).

2.28.25 Consider an affine space(E, V )of dimension 3 and letA, B, C, D∈Ebe four affinely independent points Show that there exists a unique affine transformationf such that f (A)=B, f (B)=C, f (C)=D, f (D)=A.

2.28.26 Find all the affine transformationsf on an affine space(E, V )such that

∀u∈V t u ◦f =f◦t u wheret u indicates translation by the vectoru.

2.28.27 LetKbe a field of characteristic zero ViewK 3 as an affine space over itself and calculate, in its canonical basis, the matrix representation of the affine transformation(f,−→ f )mapping respectively the pointsO, A, B, ContoO , A , B , C , where

2.28.28 Give—if possible—the matrix representation of the symmetry with respect to the planeπ with equation x+y +z=1, parallel to the direction d given by x=y=z:

1 when interpreting this statement in(Z2 ) 3 with its canonical basis;

2 when interpreting this statement in(Z3 ) 3 with its canonical basis.

2.28.29 ViewR 3 as affine space over itself and consider the affine transformation

(f,−→ f )given by the matrix formula f

Prove that−→ f is a projection Give the characteristic ingredients of this projection.

2.28.30 Consider four affinely independent pointsA,B,C,Din an affine space of dimension 3 over a field of characteristic distinct from 2 One determines an affine transformation of this affine space by f (A)=B, f (B)=C, f (C)=D, f (D)=A.

1 Give the matrix representation off with respect to the affine basis(A;−→

2 Calculate all the fixed points off andf ◦f.

3 Show thatf is bijective and its inverse isf − 1 =f◦f ◦f.

2.28.31 InR 3 viewed as affine space over itself, consider the vectors u=(2,1,1), v=(1,0,2), w=(0,1,−1) and the pointP =(2,0,1) Letπbe the plane containingP and parallel touandv.

Letd be the line containingP and parallel tow Give, in the canonical basis, the

Fig 2.13 matrix representation of the symmetry with respect todand parallel to the direction ofπ.

2.28.32 ViewR 3 as an affine space over itself Consider the planesπ 1andπ 2with respective equations

2x+y−2z=0, 3x−z=0 and the lined=π 1 ∩π 2 Consider further the planeπ containing the pointP and parallel to the vectorsu,v:

With respect to the canonical basis, determine the matrix representation of the projection on the lined parallel to the planeπ.

2.28.33 InR 3 , consider a parallelepipedABCDEF GH, as in Fig.2.13 With respect to the affine basis(A;−→

AE), give the matrix representation of the affine transformationtsuch that t (A)=G, t (B)=F, t (D)=C, t (E)=H.

2.28.34 Show that when a quadric and a line have three common points, the line is entirely contained in the quadric.

2.28.35 LetQ be a quadric in an affine space(E, V ) Let C∈Q be a center of symmetry ofQ Prove that for each pointA∈Q, the line joiningCandAis entirely contained in the quadricQ.

2.28.36 Is the empty subset always a quadric?

2.28.37 Is the whole affine space always a quadric?

2.28.38 InR 3 , regarded as an affine space over itself, consider the quadric with equation αyz+x 2 +x+βy−z+γ=0 with respect to the canonical basis Determine the values of the parametersα,β,γ so that the quadric contains the line with equations x−1=y−3

More on Real Affine Spaces

In this short chapter we focus our attention on the additional properties of an affine space when the ground field is that of the real numbers The first additional property that we have in the reals, compared with the case of an arbitrary field, is the existence of an order In particular, we can distinguish between positive and negative scalars.This allows us to define an orientation of the affine space We shall also make an important use of the existence of square roots of positive numbers, to investigate additional properties of quadrics.

About Left, Right and Between

Up to now, for example in Definition2.10.1when introducing triangles and in Def- inition2.11.1, when introducing parallelograms, we have defined the sides of these geometrical figures as being “full lines”, not segments Of course when we consider the perimeter of a triangle, we refer generally to the three segments joining two vertices, not to the full lines! Furthermore, when we mention the surface of the triangle, we mean the interior of that figure: a point is in the interior of the triangle when it lies on the segment joining two points of the perimeter (see Fig.3.1).

This idea of the segment joining two points is a very basic notion in “ordinary” geometry Does it make sense in an affine space?

Rather clearly, in real geometry, we can define:

Definition 3.1.1 Let(E, V )be a real affine space Given two pointsAandBofE, the segment[A, B]is defined as

Such a definition refers explicitly to the existence of an order (also called a total order) on the real line:

F Borceux, An Algebraic Approach to Geometry, DOI 10.1007/978-3-319-01733-4_3, © Springer International Publishing Switzerland 2014

∀k∈R k≥0 or k≤0, thus yielding a notion of a positive or negative number.

Choosing an affine basis of an affine real line, we get a bijection between the line and the base fieldR: we can thus “run along the line” following the “increas- ing coordinates”, or following the “decreasing coordinates” We get two possible orientations of the line.

The idea of “two possible orientations” has its counterpart in the real plane, where it is possible to turn clockwise or counter-clockwise, and also in three dimensional real space as well, where we can consider screws with a right-hand thread and screws with a left-hand thread So in dimensions 1, 2 or 3, we are familiar with two opposite ways of “moving”, what one calls two possible orientations of the space The orientation should be carefully considered, especially when we want to compose various movements: according to their orientations, the composition can be considered to “add” or “subtract” the movements.

So if we want to use segments in an affine space, or opposite directions, and so on, we should reduce our attention to affine spaces on a totally ordered field

K: a fieldK provided with an order≤which is “sufficiently compatible with the field structure” and where “every two elements are comparable” The fieldRof real numbers is such a field, thus so are all the subfieldsK⊆R.

It is well-known that the field of complex numbers does not admit such an order: would you say that 1−iis positive or negative? Analogously, the fieldZ p of integers modulo a prime numberpdoes not admit such an ordering: inZ5, where 3= −2, would you say that this element is positive or negative? Of course these arguments are not proofs that an adequate order cannot possibly exist, just hints to suggest that it is so.

Thus working with a subfieldK⊆Rwill provide a notion of orientation, related to the order onK But the corresponding geometry will still remain rather far from

“the ordinary geometry of the ordinary plane” Take the exampleK=Q, the field of rational numbers In “ordinary geometry” you certainly want to consider the “circle”

Orientation of a Real Affine Space

Fig 3.2 with equation x 2 +y 2 =1 and the line with equation (see Fig.3.2)

All the coefficients are rational numbers (and even integers) But in the “rational plane”Q 2 the line does not intersect the circle! Indeed the two “intersection points” if we can call them that should be

—they do not have rational coordinates!

Finally, we seem to end up with the single caseK=R! This is in fact too severe a conclusion, but we shall not insist on the possible generalizations.

Let us conclude this section with a classical notion, valid as soon as the notion of a segment makes sense:

Definition 3.1.2 Let(E, V )be a real affine space A subsetX⊆Eis convex (see Fig.3.3) when

A, B∈X =⇒ [A, B] ⊆X i.e.Xcontains the whole segment[A, B]as soon as both extremitiesAandB lie inX.

3.2 Orientation of a Real Affine Space

Let us first introduce the basic ingredient for defining an orientation:

Definition 3.2.1 Let(E, V )be a finite dimensional real affine space Two affine bases

O ;e 1 , , e n have the same orientation when in the corresponding change of basis formula

→x =M−→x + −→v the determinant of the matrixMis positive (See Proposition2.20.1.)

The key observation is then:

Proposition 3.2.2 Let(E, V )be a finite dimensional real affine space The property “Having the same orientation” is an equivalence relation on the set of affine bases For non-zero dimensional spaces, there are exactly two equivalence classes.

Proof Under the conditions of Definition3.2.1, the inverse change of basis formula has the form

(see Proposition2.20.1) Since detM − 1 = 1 detM we get detM − 1 >0 and the “having the same orientation” relation is symmetric. Next if

O , e 1 , , e n also have the same orientation, with change of basis formula given by

3.2 Orientation of a Real Affine Space 123

Fig 3.5 then the change of basis formula from the first basis to the third one has the form

But since detN MNãdetM this quantity remains positive and the “having the same orientation” relation is also transitive, thus it is an equivalence relation.

Replacinge n by−e n in the first basis yields a change of basis matrix with determinant−1: thus these two bases are not in the same equivalence class, proving that there are at least two equivalence classes Given an arbitrary affine basis, the determinants of the change of basis formulổ to the two bases above (withe n and with−e n ) are opposite real numbers, proving that the arbitrary basis lies in one of the two equivalence classes

We can thus make the following definition:

Definition 3.2.3 An orientation of a finite dimensional real affine space consists in choosing one equivalence class of an affine basis, with respect to the equivalence relation “having the same orientation” The bases in the chosen equivalence class are said to have direct orientation and the bases in the other class, to have inverse orientation.

Keeping in mind the last paragraph of the proof of Proposition3.2.2, let us ex- amine the situation in the 1, 2 and 3-dimensional spaces, with the “usual choice” for the direct orientation.

When one draws the real line horizontally, one generally chooses the left to right orientation as the direct one (Fig.3.4).

In the real plane, turning counter-clockwise, passing frome 1 toe 2 without cross- ing the first axis, is generally taken as the direct orientation (Fig.3.5).

In three dimensional space, the direct orientation is generally chosen according to the famous rule of the cork-screw If you align a cork-screw alonge 3 and “turn it frome 1 toe 2 ”, it moves positively in the direction ofe 3 (see Fig.3.6) A formal translation of this cork-screw rule, in a special case of interest, is given by Exam- ple3.2.4.

Example 3.2.4 Given two linearly independent vectorsx,yinR 3 , the three vectors x, y, x×y constitute a basis having the same orientation as the canonical basis.

Proof It suffices to observe that the matrix whose columns are the coordinates ofx, y,x×y has a strictly positive determinant (see Definition3.2.1) Developing this determinant with respect to the third column, we obtain (see Definition1.7.1) det x 2 y 2 x 3 y 3

This quantity is of course positive; it is strictly positive becausexandyare linearly independent, thus at least one of the 2×2 matrices is regular

Of course in the very special case of the vector spaceR n viewed as an affine space, one generally considers the “canonical basis” e 1=(1,0, ,0), e 2=(0,1,0, ,0), , e n =(0, ,0,1) as having direct orientation.

The choice of which orientation is to be the direct one is just a matter of personal taste: neither of the two choices is mathematically more “canonical” than the other. Adopting either orientation will result in an equivalent theory: the only important fact is to keep in mind that both orientations exist simultaneously.

There exists an alternative (and equivalent) topological approach to the notion of orientation (see Problem3.7.4) The setB of all bases of ann-dimensional real vector spaceV can be regarded as a subset ofV n and therefore can be provided with the structure of a metric space Two bases ofV have the same orientation precisely when they are connected by a continuous path in the metric spaceB.

Direct and Inverse Affine Isomorphisms

3.3 Direct and Inverse Affine Isomorphisms

First of all, with the considerations of Sect.2.13in mind:

Definition 3.3.1 Let(E, V )be a finite dimensional real affine space and

• The isomorphism(f,−→ f )is direct when det−→ f >0.

• The isomorphism(f,−→ f )is inverse when det−→ f 0, we obtain the same subsetsF − andF + and whena n,n 0 is a direct isomorphism Can a homothety of ratiok 0 for all indicesi As in Example4.5.1, we conclude thatϕ is positive and definite

Example 4.5.3 Consider a closed interval[a, b]of the real line and the vector space

C([a, b],R)of continuous functionsf: [a, b] −→R The formula

(f|g) # b a f (x)g(x) dx defines a scalar product onC([a, b],R).

Proof Only the last axiom requires a comment Iff (x 0 )=0 for somex 0∈ [a, b] then by continuity|f (x)|> 1 2 |f (x 0 )| on a neighborhood ofx 0 in[a, b] Writing ε >0 for the length of an interval on which this is the case we conclude, since f 2 (x)is positive for allx∈ [a, b], that

One should observe that in this example the distance between two functions is given by d(f, g) # b a f (x)−g(x) 2 dx which is not the same as “the area separating the two graphs”

It should be mentioned here that “the area separating the two graphs” nevertheless yields a good notion of distance in the sense of metric spaces: but this notion of distance is not inherited from a scalar product.

Let us also mention that if ω: [a, b] −→R, ∀x∈ [a, b]ω(x) >0 is a strictly positive continuous function, the argument above can at once be adapted to prove that

(f|g) # b a ω(x)f (x)g(x)dx is another scalar product onC([a, b],R) At the end of Sect.4.9, we shall remark on the benefits of introducing such a weight

Example 4.5.4 Consider the vector spaceR (n) [X]of polynomials with real coefficients and degree at mostn Consider furthern+1 distinct real numbers a 0 < a 1 0, the formula

(p|q) n i = 1 ω i p(a i )q(a i ) still defines a scalar product onR (n) [X]

Example 4.5.5 Every affine subspace of a Euclidean space is itself a Euclidean space with respect to the induced scalar product.

Proof Trivially, given a scalar product on a real vector spaceV, its restriction to any vector subspaceW⊆V remains a scalar product.

Going back to Example2.2.2and considering the canonical scalar product on

R n (see Example4.5.1), we conclude that the affine space of solutions of a system

A−→x =−→ b of linear equations can be provided with the structure of a Euclidean space.

Analogously, considering Examples2.2.4and4.5.3, we conclude that the affine space of solutions, on an interval [a, b], of a differential equation ay +by + cy=d, can be provided with the structure of a Euclidean space

In a Euclidean space, using an affine basis closely related to the Euclidean structure allows drastic simplifications As usual when we work with bases and coordinates, we reduce our attention to the finite dimensional case, even if various results hold (often with the same proofs) in arbitrary dimensions.

Definition 4.6.1 By an orthonormal basis of a finite dimensional Euclidean space

(E, V )is meant an affine basis(O;e 1 , , e n )such that:

Let us first list some advantages of working with orthonormal basis.

Proposition 4.6.2 Let (O;e 1 , , e n ) be an orthonormal basis of a Euclidean space(E, V ) The coordinates of a vectorx∈V with respect to this basis are

Proof If x=x 1 e 1+ ã ã ã +x n e n computing the scalar product withe i yields preciselyx i

Proposition 4.6.3 Let (O;e 1 , , e n ) be an orthonormal basis of a Euclidean space(E, V ) Given two vectorsx, y∈V, their scalar product is

Proof By Proposition4.2.2, the matrix of the scalar product is the identity matrix because the basis is orthonormal

Proposition 4.6.4 Let (O;e 1 , , e n ) and (O ;e 1 , , e n ) be two orthonormal bases of a Euclidean space(E, V ) In the change of basis formula (see Proposi- tion2.20.1)

→x =M−→x +−→ b the matrixMis orthogonal, that is,M − 1 =M t

Proof The matrixMis obtained by putting in columns the coordinates of the vectors e i with respect to the basis(e 1 , , e n ) By Proposition4.6.2,

Considering the inverse change of basis formula

→x =M − 1 −→ x +−→ b we obtain in the same way

The conclusion follows at once, by symmetry of the scalar product

This last result is certainly the most striking one, since computing an inverse matrix is generally rather hard work, especially when the dimension is high However, all of these beautiful properties will only be made available to us if we can prove the existence of orthonormal basis For that we observe first:

Proposition 4.6.5 Let(E, V )be a Euclidean space Given non-zero pairwise orthogonal vectors e 1 , , e n , i=j =⇒ e i ⊥e j these vectors are necessarily linearly independent.

Computing the scalar product withe i yields x i (e i |e i )=0 by perpendicularity of the vectors But (e i |e i )=0 because e i =0; therefore x i =0

Theorem 4.6.6 (Gram-Schmidt process) Let(O;e 1 , , e n )be an arbitrary basis of a Euclidean space (E, V ) There exists an orthonormal basis (O;v 1 , , v n ) with the additional property that for every indexk, the two subspaces e 1 , , e k and v 1 , , v k generated by the firstkvectors of each basis are equal.

Proof We prove the result by induction onn Whenn=1, it suffices to put v 1 = e 1 e 1

Assuming the result up to the dimension n−1, let us apply it to the vector subspacee 1 , , e n − 1 and its basise 1 , , e n − 1 We obtain an orthonormal basisv 1 , , v n− 1of this subspace, which satisfies the condition of the statement up to the indexn−1 Consider then v n =e n −(e n |v 1 )v 1 − ã ã ã −(e n |v n − 1 )v n − 1

We get at once, for 1≤i≤n−1 v n |v i

Putting v n = v n v n thus yields a sequencev 1 , , v n pairwise orthogonal vectors of length 1 By Propo- sition4.6.5, this is a basis ofV.

Polar Coordinates

In this section, we want to stress the fact that in Euclidean spaces, all the classical techniques mentioned in Chap.1(polar coordinates in the plane, polar or cylindrical coordinates in the three-dimensional space, and so on) now make perfect sense We shall not dwell on these straightforward aspects Just as an example, we focus on the case of polar coordinates in a Euclidean space.

Given a triangle (A, B, C) in a Euclidean space, we have defined the angle (BAC)(see Definition4.2.6) Trivially, by symmetry of the scalar product

Of course we might be tempted to say instead that

In order to be able to do this, we need to provide each angle with a sign This is possible only in the special case of a Euclidean plane.

Definition 4.7.1 When an orientation of a Euclidean plane has been fixed, the relative angle(v, w)between two linearly independent vectorsv,wis the angle (v, w) of Definition4.2.6 provided with the sign+when the basis (v, w) has direct orientation and with the sign−when this basis has inverse orientation When two non-zero vectorsv andw are linearly dependent, their relative angle is their ordinary angle as in Definition4.2.6, that is, 0 orπ(see Proposition4.3.3).

Of course choosing the opposite orientation of the plane interchanges the signs of all relative angles Moreover, since an angle is defined via its cosine, the following convention certainly does not hurt:

Convention 4.7.2 Under the conditions of Definition4.7.1, we shall freely identify a relative angleθwith any angleθ+2kπ, for every integerk∈Z.

The reader is invited to explain why such definitions do not make sense in a three dimensional Euclidean space.

The existence of polar coordinates in every Euclidean plane is then attested by the following result:

Proposition 4.7.3 Let(E, V )be a Euclidean plane provided with an orthonormal basis(O;e 1 , e 2 )considered as having direct orientation The coordinates of a point

OP ) whereθis the relative angle as in Definition4.7.1.

Proof Given real numbersa,bsuch thata 2 +b 2 =1, thena 2 ≤1 thus−1≤a≤ +1 Thereforea=cosτ for a uniqueτ∈ [0, π], while sinτ≥0 But then b 2 =1−a 2 =1−cos 2 τ=sin 2 τ and thusb= ±sinτ Ifb≥0, we have at once a=cosτ, b=sinτ.

Ifb≤0 we have a=cosτ=cos(−τ ), b= −sinτ=sin(−τ ).

In both cases, we end up with

(a, b)=(cosσ,sinσ ) for a unique relative angleσ.

Now given 0=P ∈E, we thus have a unique relative angleσsuch that

OP cosσ sinσ since the vector −→ OP

−→ OP has norm 1 It remains to show thatσ is also the relative angle θbetweene 1and−→

The matrix having as columns the coordinates ofe 1 and−→

; its determinant is simply sinσ Thus by Definition4.7.1,θ is positive or negative according to the sign of sinσ, that is, it has the same sign asσ Sinceσ andθhave the same cosine and the same sign, they are equal

Analogous arguments can be developed for the other systems of coordinates considered in Chap.1 We are not really interested here in these considerations The only reason for introducing the straightforward observations above is to emphasize the fact that we have now gathered all the necessary ingredients to make the link with “ordinary” geometrical notions: we need to have the affine structure, a notion of orientation and a notion of “measure” of angles and distances.

Orthogonal Projections

To avoid any ambiguity, let us make the following definition:

Definition 4.8.1 Let (E, V ) be a Euclidean space and (F 1 , W 1 ), (F 2 , W 2 ) two affine subspaces These subspaces are called orthogonal when every vector ofW 1is orthogonal to every vector ofW 2.

Observe that this definition is more restrictive than the notion of perpendicularity “in real life” For example, you will probably say that a wall of your room is

“perpendicular” to the floor However this is not the situation described in Defini- tion4.8.1! Consider the line of intersection between the wall and the floor: a “vector” in this intersection is both on the wall and on the floor, but is not orthogonal to itself! This is a general fact:

Lemma 4.8.2 Let(E, V )be a Euclidean space and(F 1 , W 1 ),(F 2 , W 2 )two orthogonal affine subspaces ThenW 1∩W 2= {0}and thusF 1∩F 2 is either the empty set or is reduced to a singleton.

Proof Indeedw∈W 1 ∩W 2 is such that(w|w)=0, thusw=0 The result follows by Propositions2.3.4and2.1.2

Let us warn the reader: the following theorem is generally not valid for Euclidean spaces of infinite dimension.

Theorem 4.8.3 Let(E, V )be a finite dimensional Euclidean space andWa vector subspace ofV The set

W ⊥ = {v∈V|∀w∈W v⊥w} is a vector subspace ofV, orthogonal toW The subspacesW andW ⊥ are supplementary ThereforeW ⊥ is called the orthogonal supplementary ofW.

Proof The setW ⊥ is a vector subspace by bilinearity of the scalar product; it is trivially orthogonal toW By Lemma4.8.2we know already thatW∩W ⊥ = {0}; it remains to prove thatW+W ⊥ =V.

To prove this, consider a basise 1 , , e k ofWand extend it to a basise 1 , , e n ofV Apply the Gram–Schmidt construction (Theorem4.6.6) to get an orthonormal basisv 1 , , v n such that in particular,v 1 , , v k is still a basis ofW The vector subspacev k + 1 , , v n is contained inW ⊥ and is supplementary toW Thus

Observe further (even if not needed for the proof) that since v k + 1 , , v n andW ⊥ are two supplementary subspaces of W, they have the same dimension n−dimW Sincev k + 1 , , v n is contained inW ⊥ , these two subspaces are necessarily equal

Corollary 4.8.4 Let(E, V )be a finite dimensional Euclidean space with orthonormal basis(O;e 1 , , e n ) Fix a pointA=(a 1 , , a n )∈Eand consider the vector lineW generated by a non-zero vectorw=(w 1 , , w n )inV The affine hyperplane(F, W ⊥ )containingAand of directionW ⊥ (see Theorem4.3.5) admits the equation n i = 1 w i (x i −a i )=0.

Proof SinceA∈F, the pointP =(x 1 , , x n )∈Elies inF when−→

AP is perpendicular tow The result follows by Proposition4.6.3

Definition 4.8.5 Let (E, V ) be a finite dimensional Euclidean space and

(F, W ) an affine subspace The projection on W, parallel to W ⊥ (see Defini- tion2.15.1) is called the orthogonal projection on(F, W ).

Here is a key property of orthogonal projections:

Proposition 4.8.6 Let(E, V )be a finite dimensional Euclidean space and(F, W ) an affine subspace Given a pointA∈Eand its orthogonal projectionP ∈F, one has, for every other pointQ∈F (see Fig.4.6) d(A, P ) < d(A, Q), P =Q∈F.

QP∈W, thus the triangleAP Qis right angled By Pythagoras’ Theorem (see4.3.5) d(Q, P ) 2 +d(A, P ) 2 =d(A, Q) 2

SinceQ=P,d(P , Q)=0 and it follows thatd(A, P ) < d(A, Q) Proposition4.8.6can thus be rephrased in the following way:

The orthogonal projection of a point A on a given subspace is the best approximation of A by a point of the subspace

Our next section will take full advantage of this observation.

It remains to establish an efficient formula for computing orthogonal projections.

Proposition 4.8.7 Let(E, V )be a finite dimensional Euclidean space and(F, W ) an affine subspace Given an orthonormal basis (O;e 1 , , e k ) of the subspace (F, W )and a point A∈E, the orthogonal projectionP ofAon(F, W )is given by

Proof Extend(e 1 , , e k )to an orthonormal basis(e 1 , , e n )ofV, as in the proof of Theorem4.8.3 As observed at the end of that proof, (e k + 1 , , e n ) is an orthonormal basis ofW ⊥ By Proposition4.6.2,

OA|e i ) is thus the unique decomposition

ButP =O+w(see the proof of Theorem2.6.2): this yields the formula of the statement.

Some Approximation Problems

Consider a subspace(F, W )of a finite dimensional Euclidean space(E, V ) Given a pointA∈E, what is the best approximation ofAby a point ofF? This is the point

B∈Fsuch that the distanced(A, B)is as small as possible! Such a point exists and is unique: by Proposition4.8.6, it is the orthogonal projection ofAon(F, W ).

Example 4.9.1 (Overdetermined systems) How can one find the “best approximation” of a solution for a systemA−→x =−→ b ofmequations withnunknowns when mis much bigger thann?

Proof This situation occurs when we want to determine the values of some physical quantitiesX 1 , , X n which are impossible to isolate experimentally, but we are able to measure experimentally the result of some linear combination of these quantities: a 1 X 1+ ã ã ã +a n X n

Repeating the experiment with different values of the coefficientsa i , we obtain a systemA−→x =−→ b of equations We want to “statistically” correct the experimental imprecisions by performing a large numbermof experiments (i.e of equations), a numbermwhich is much bigger than the numbernof quantities to measure Due to imprecisions in the measurements, there is no hope that the systemA−→x =−→ b will still have an “algebraic solution”, but of course the problem has a “physical solution”: the actual values of the quantitiesX 1 , , X n

Without any experimental error,−→ b would be of the formA−→x , that is, would be a linear combination of the columns of the matrixA Consider the canonical scalar product onR m (see Example4.5.1) and the vector subspaceW⊆R m generated by the columns ofA It remains to replace−→ b by its “best approximation by a vector

→c ∈W”, that is, by its orthogonal projection onW The systemA−→x = −→c now has a solution

Example 4.9.2 (Approximation by the law of least squares) How can one find the polynomialp(X)of degreenwhose valuesp(a i )are “as close as possible” from prescribed valuesb i , when the numbermof indicesiis much bigger that the degree nofp(X)?

Proof Assume that some physical law is expressed by a formula of degree 2, for example: the resistance of the air is proportional to the square of the speed We want to determine the proportionality coefficient, under some specific conditions of pressure or shape.

More generally, the theory tells us that some physical quantityY can be expressed by a polynomial of degreenin terms of the physical quantityX

We want to determine experimentally the coefficients of the polynomialp(X) For this we perform a large numbermof experiments, for different valuesX=a i ∈R, measuring the corresponding valuesY =b i We are looking for the polynomial p(X)of degree n such that each p(a i )is as close as possible to b i Figure4.7 presents an example withn=2 andm

First of all, observe that there is a polynomialq(X)of degreem−1 such that q(a i )=b i for each indexi: this is simply q(X) m i = 1 b i $

Consider the Euclidean space R (m− 1) [X] of all polynomials of degree at most m−1, provided with the scalar product of Example4.5.4 α(X)|β(X) m i = 1 α(a i )β(a i ).

The orthogonal projectionp(X)of q(X)on the subspaceR (n) [X]of polynomials of degree at mostnis the polynomial of degreensuch that the quantity m i = 1 p(a i )−q(a i ) 2 n i = 1 p(a i )−b i

2 is the smallest possible (Proposition4.8.6).

This polynomialp(X)is thus the solution to our problem according to the law of least squares: the sum of the squares of the “errors” has been made as small as possible

Example 4.9.3 (Fourier approximation) How can one find a “best approximation” g(X)of a periodic functionf (X)by a linear combination of sine and cosine functions?

Proof This time we need to realize a periodic electrical signal y =f (x) with a prescribed shape For example the signal in Fig.4.8, which is the typical signal for the horizontal scanning of a screen.

The basic electrical signals that one can produce are continuous (a constant function) or alternating (a sine or cosine function, with an arbitrary frequency) electrical signals We need to determine how to add such signals in order to get a result as close as possible to the prescribed periodic function.

Of course—up to a possible change of variable—there is no loss of generality in assuming that the period off (X)is equal to 2π Notice that each function sinkx or coskx, fork∈N, itself admits 2π as a period, even if this is not the smallest possible period But a linear combination of functions with period 2π remains a function with period 2π This proves that it suffices to compute the approximation on the interval[−π, π]: the approximation will automatically remain valid on the whole real line A priori, f (X)is not necessarily continuous, as in the example above For simplicity, let us nevertheless assume that f (X)is continuous on its period] −π, π[, with continuous extension to[−π, π](again, as in the example above).

We can now consider the Euclidean spaceC([−π, π],R)of Example4.5.3 To switch back to a finite dimension, consider the Euclidean subspaceV generated by f (X)and the functions

1,sinX, cosx, sin 2X, cos 2X, ,sinnX, cosnX, n∈N.

WriteW for the subspace ofV generated by these last functions The orthogonal projectiong(X)of f (X)on W thus yields the best approximation off (X)by a linear combination of sinkXand coskXfunctions, fork≤n.

It is interesting to observe that in this specific case, the orthogonal projection can be computed very easily Indeed let us recall that

This can be rephrased by saying that the functions

√π cos 2x, constitute an orthonormal sequence of functions in C([−π, π],R) Therefore, by

Proposition4.8.7, the functiong(X)above is simply g(x)= 1 2π

This is a so-called Fourier approximation off (x).

For the “horizontal scanning” function above f (x)=x, −π≤x≤π

Figure4.9gives the Fourier approximation obtained when choosingn.

Of course a Fourier approximation is always a continuous function, being a linear combination of continuous functions Therefore the possible discontinuityf (x) atπ can imply a lower quality of the Fourier approximation around this point This is a typical case where one might want to use a weight functionω(x)as in Exam- ple4.5.3: a strictly positive function whose values around−π and+π are slightly greater than at the middle of the interval Doing this will improve the quality of the approximation at the extremities of the interval, but to the detriment of the quality of the approximation elsewhere!

Isometries

As the name indicates, an isometry is “something” which leaves the “measures” unchanged.

(f,−→ f ):(E, V )−→(F, W ) between Euclidean spaces is an affine transformation such that−→ f respects the scalar product:

Proposition 4.10.2 An isometry between two Euclidean spaces respects distances and angles and in particular, is injective.

Proof Distances and angles are defined in terms of the scalar product (see Defini- tions4.2.4and4.2.6) MoreoverA=B precisely whend(A, B)=0 (see Proposi- tion4.3.2)

Proposition 4.10.3 Let(E, V )and(F, W )be Euclidean spaces of respective finite dimensionsnandm Consider an affine transformation

→x →A−→x +−→ b with respect to orthonormal bases of(E, V )and(F, W ) The following conditions are equivalent:

2 the columns ofAconstitute an orthonormal sequence of vectors inR m

O ;e 1 , , e m be the two orthonormal bases The columns ofAare the coordinates of the vectors −→ f (e i )in the second base By Proposition 4.10.1these vectors constitute an orthonormal sequence inW and since the second base is orthonormal, their coordinates are orthonormal vectors inR m (see Proposition4.6.3).

Conversely, the assumption onAcan be rephrased asA t A=Id m , whereId n is then×n-identity matrix Given two vectors ofV with coordinates−→x ,−→y , we get once more by Proposition4.6.3

Taking full advantage of Corollary4.3.4, let us now give an interesting characterization of isometries.

Theorem 4.10.4 Let(E, V )and(F, W )be finite dimensional Euclidean spaces. There is a bijection between:

Proof By Proposition4.10.2, it remains to show that a mappingf:E−→F pre- serving distances is the first component of a unique isometry(f,−→ f ) The uniqueness is immediate since in an affine transformation(f−→ f ), the linear mapping−→ f is entirely determined byf (see Proposition2.12.4).

First observe that given pointsA, B, C∈Eand a scalarr∈R:

When 0≤r≤1, Corollary4.3.4reduces the condition−→

Such a property is thus preserved byf Whenr /∈ [0,1], a permutation of the roles ofA,B,Creduces the problem to the first case For example ifr >1, then

Thus we know already thatf transforms an affine line into an affine line But by Pythagoras’ Theorem4.3.5,f also transforms a right triangle into a right triangle. Thusf respects the perpendicularity of two affine lines.

Fix now an orthonormal basis(O;e 1 , , e n )of(E, V ) Write furthere i =−−→

OA i The vectors−−−−−−−→ f (O)f (A i )then constitute an orthonormal sequence inW and we can complete it to an orthonormal basis f (O);−−−−−−−→ f (O)f (A 1 ), ,−−−−−−−→ f (O)f (A n ), e n + 1 , , e m of(F, W ).

Classification of Isometries

Consider a pointP ∈Eand itsi-th coordinatex i with respect to the orthonormal basis(O;e 1 , , e n ) The pointX i such that

OA i is thus the orthogonal projection ofP on the lineOA i , that is, the unique point of the line throughOA i such that the triangleOX i P is right angled But thenf (X i ) is the orthogonal projection off (P )on the linef (O)f (A i )and thei-th coordinate off (P )with respect to the orthonormal basis of(F, W )is the scalarx i such that

Since we already know thatf preserves the proportionality of vectors with the same origin, we conclude thatx i =x i

We can summarize our results by saying that, with respect to the two bases indicated,f admits the following matrix description:

→x →M−→x whereMis anm×n-matrix whosenfirst lines are those of then×n-identity matrix and whosem−nlast lines are zero lines Of course we define −→ f to be the linear mapping−→ f :V −→W admitting the matrixM with respect to the two orthonormal bases ofV andWas above The columns ofMare trivially orthonormal so that by Proposition4.10.3,(f,−→ f )will be an isometry as soon as it is an affine transformation.

This last fact is obvious Working in terms of coordinates in the orthonormal bases indicated,f and−→ f act simply by addingm−nzero coordinates Therefore axioms [AT1] and [AT2] are trivially satisfied

In this section, we focus our attention on the isometries from a Euclidean space to itself Proposition4.10.3can at once be rephrased as:

Proposition 4.11.1 Let(E, V )be a finite dimensional vector space and

(f,−→ f ):(E, V )−→(E, V ) be an affine transformation, with matrix expression

→x →A−→x +−→ b with respect to an orthonormal basis(O;e 1 , , e n ) The following conditions are equivalent:

In that case, one has detA= ±1 and the isometry is an affine isomorphism.

In order to determine all possible isometries on(E, V ), let us first review two well-known examples.

Example 4.11.2 Every translation on a Euclidean space is an isometry.

Proof This follows by Proposition2.14.2: of course the identity mapping preserves the scalar product

Example 4.11.3 Let(E, V )be a finite dimensional Euclidean space Every orthogonal symmetry is an isometry.

Proof Consider an affine subspace (F, W )⊆(E, V ) and the orthogonal supple- mentW ⊥ ofW (see Theorem4.8.3) Choose a pointO∈F, an orthonormal basis e 1 , , e k of W and an orthonormal basise k + 1 , , e n ofW ⊥ We thus obtain an orthonormal basis

(O;e 1 , , e k , e k+ 1 , , e n ) of(E, V ) With respect to this basis the orthogonal symmetry with respect to(F, W )

The matrix description of(s,−→s )is thus

→x →A−→x whereAis a diagonal matrix with the firstkentries on the diagonal equal to+1 and the following entries equal to−1 The result follows by Proposition4.11.1

Notice that the identity on(E, V )is both the translation by the vector 0 and the orthogonal symmetry with respect to(E, V ) This observation helps us to better understand the following statement:

Proposition 4.11.4 Let(E, V )be a finite dimensional Euclidean space The isometries(f,−→ f )on(E, V )are precisely the composites of

Rotations

• a direct isometry admitting a fixed point (see Definition2.13.1).

Of course, each composite of such mappings is an isometry.

Proof The last statement holds by Examples4.11.2and4.11.3, since a composite of isometries is trivially an isometry.

Now let(f,−→ f )be an isometry on(E, V ) Fix a pointP ∈E and consider the translation by the vector−−−−→ f (P )P We have

This proves that g=t −−−−→ f (P )P ◦f is an isometry admittingP as a fixed point.

Ifgis a direct isometry, we obtain f =t Pf (P ) ◦id E ◦g andf is expressed as the composite of a translation, an orthogonal symmetry and a direct isometry with fixed pointP.

Ifgis an inverse symmetry andV has dimensionn, let(F, W )be an affine subspace of dimensionn−1 such thatP ∈F Writesfor the orthogonal symmetry with respect to(F, W ) SinceP ∈F, we haves(P )=P The proof of Example4.11.3 tells us that in an ad-hoc orthonormal basis, the matrix ofsis diagonal with the first n−1 diagonal entries equal to+1 and the last one equal to−1 The determinant is thus equal to−1 andsis an inverse isometry But thenh=s◦gis a direct isometry still admittingP as a fixed point Furthermore, sinces◦s=id E f=t Pf (P ) ◦s◦h expressesf as the composite of a translation, an orthogonal symmetry and a direct isometry with fixed pointP

To describe all isometries of(E, V ), it thus remains to determine the form of the direct isometries with a fixed point.

In this section we investigate the form of the direct isometries admitting a fixed point, in dimensions 0, 1, 2 and 3.

Proposition 4.12.1 On a Euclidean space of dimension 0 or 1, the only direct isometry with a fixed point is the identity.

Proof In dimension 0 there is nothing to prove, since the only mapping from the singleton to itself is the identity.

In dimension 1, the only orthogonal matrix with positive determinant is the identity matrix(1); the result follows by Proposition4.11.1

The notion of a rotation of angleθwith centerOin the usual plane makes perfect sense in a Euclidean plane: “all vectors−→

OP turn around the centerO by the same angleθ, in the same orientation” More precisely:

Definition 4.12.2 Let(E, V )be a Euclidean plane Consider a relative angleθ∈ ]−π,0, π](see Definition4.7.1) and a pointO∈E A mappingf: E−→Eis a rotation of angleθwith centerOwhen:

Let us clarify the situation concerning the two trivial casesθ=0 andθ=π.

Proposition 4.12.3 Let(E, V )be a Euclidean plane andOa point ofE Then:

1 a rotation of angle 0 with centerOis the identity mapping onE;

2 a rotation of angleπ with centerOis the central symmetry with respect toO.

These two rotations are direct isometries.

Proof The first two assertions follow immediately from Proposition4.3.3 Of course the identity is a direct isometry In an orthonormal basis(O;e 1 , e 2 )the central symmetry admits as matrix

0 −1 which is an orthogonal matrix with determinant+1 The result follows by Proposi- tion4.11.1

Observe that a central symmetry is always an isometry (Example4.11.3): but it is a direct isometry in even dimensions and an inverse isometry in odd dimensions, as the proof of4.12.3immediately suggests.

The key result is then:

Theorem 4.12.4 Let(E, V )be a Euclidean plane andf:E−→Ean arbitrary mapping The following conditions are equivalent (see Theorem4.10.4):

1 f is a direct isometry with a fixed point;

Proof Let us work in an orthonormal basis(0;e 1 , e 2 )considered as having direct orientation.

Suppose thatf is a rotation with centerOand relative angleθ IfP=O, working in polar coordinates (see Proposition4.7.3) the rotationf is simply described by

But trivially cos(τ+θ ) sin(τ+θ ) cosτcosθ−sinτsinθ sinτcosθ+cosτsinθ cosτ −sinτ sinτ cosτ cosθ sinθ

Thereforef can be described by the matrix formula f x y cosτ −sinτ sinτ cosτ x y which is also trivially valid for the originO, which is a fixed point By Proposi- tion2.22.1,f is thus an affine transformation Since its matrix is trivially orthogonal with determinant+1, it is a direct isometry (see Proposition4.10.3and Defini- tion3.3.1) By assumption, it admits the fixed pointO.

Conversely, let(f,−→ f )be a direct isometry with fixed pointO The matrix expression of the isometry is thus x 1 x 2

→ a 1 b 1 a 2 b 2 x 1 x 2 where the columns of the matrix are the coordinates of −→ f (e 1 )and −→ f (e 2 ) The matrix is orthogonal (see Proposition4.11.1) with determinant+1 Thus a 1 2 +a 2 2 =1, b 1 2 +b 2 2 =1, a 1 b 1+a 2 b 2=0, a 1 b 2−a 2 b 1=1.

In particular (see the proof of Proposition4.7.3) a 1 =cosθ, a 2 =sinθ for a unique relative angleθ The resolution of the system b 1cosθ+b 2sinθ=0

−b 1sinθ+b 2cosθ=1 yields at once b 1 = −sinθ, b 2 =cosθ.

Thus the matrix expression of (f,−→ f ) with respect to the orthonormal basis

This is precisely the form of a rotation of angleθ, as observed in the first part of the proof

Let us now switch to dimension 3 The intuitive notion of a rotation about an axis can easily be formalized:

Definition 4.12.5 Let(E, V )be an affine space of dimension 3 Consider an affine line(, L)⊆(E, V )and a relative angleθ A mappingf:E−→Eis a rotation of axisand angleθwhen:

2 for every pointO∈ ,f restricts as a rotation of angleθ and centerO in the affine plane orthogonal to(, L)and containingO.

Proposition 4.12.6 Let(E, V )be a Euclidean space of dimension 3 and(, L)⊆

• a rotation of axisand angle 0 is the identity onE;

• a rotation of axisand angleπis the orthogonal symmetry with respect to(, L).

These two rotations are direct isometries.

Proof As for Proposition4.12.3, via Proposition 4.3.3and Example 4.11.3 The matrix of a rotation of angleπ, with respect to an orthonormal basis(0;e 1 , e 2 , e 3 ) now has the form

In dimension 3, the striking point about rotations is perhaps the “non-existence” of a rotation about a point More precisely:

Theorem 4.12.7 Let (E, V )be a Euclidean space of dimension 3 Consider an arbitrary mappingf: E−→E The following conditions are equivalent:

1 f is a direct isometry with a fixed point;

2 f is a rotation about an axis.

Before giving the proof, let us recall that an orthogonal matrixAcan only have +1 and−1 as eigenvalues Indeed

Choosingvof length 1 we get, sincev t Avis a(1×1)-matrix, λ=λv t v=v t Av v t Av t

Proof Assume first thatf is a rotation of axis(, L)and relative angleθ Let us work in an orthonormal basis(O;e 1 , e 2 , e 3 )withO∈ ande 1∈L We consider (e 2 , e 3 )as having direct orientation in the subspace that these two vectors generate. The considerations in the proof of Theorem4.12.4indicate at once thatf can be described by the matrix formula f

By Proposition2.22.1,f is thus an affine transformation Since its matrix is trivially orthogonal with determinant+1, it is a direct isometry (see Proposition4.10.3and Definition3.3.1).

Conversely, consider a direct isometry(f,−→ f ):(E, V )−→(E, V )and a point

O∈E such thatf (O)=O Fix an orthonormal basis(O;e 1 , e 2 , e 3 ) The matrix expression off becomes

→x →A−→x withAan orthogonal matrix with determinant+1 (see Proposition4.11.1). The characteristic polynomial of the matrixAhas the form p(λ)(A−λId)= −λ 3 +αλ 2 +βλ+detA, α, β∈R.

By continuity ofp(λ), the Intermediate Value Theorem forces the existence of a positive rootλ, that is, a positive eigenvalue ofA As we know, this eigenvalue must be+1.

Lete 1 be an eigenvector with eigenvalue 1; we choose it to be of length 1 Writing

L⊆V for the vector subspace generated by e 1, the line (, L) through O (see Theorem2.4.2) is then entirely composed of fixed points off Indeed,

We introduce an orthonormal basis(O;e 1 , e 2 , e 3 )of(E, V ), where(O;e 2 , e 3 )is an orthonormal basis of the affine plane(F, L ⊥ )passing throughOand orthogonal to(, L)(see Proposition4.10.2).

Since (f,−→ f ) is an isometry, −→ f respects the orthogonality Therefore (f,−→ f ) restricts as an isometry on(F, L ⊥ ) Thus the matrix expression of (f,−→ f ) with respect to the orthonormal basis(O;e 1 , e 2 , e 3 )has the form

WriteBfor this 3×3-matrix SinceBis orthogonal with determinant+1, the same conclusion applies to the sub-matrix

B b 22 b 23 b 32 b 33 which is the matrix of the restriction of(f−→ f ) to (F, L ⊥ ) By Theorem4.12.4, the restriction of(f,−→ f )to(F, L ⊥ )is a rotation with centerO, for some relative angleθ Thus, as observed in the proof of Theorem4.12.4,

⎠ and as we have seen in the first part of the proof, this is the matrix of a rotation of angleθabout the axis(, L).

Similarities

When representing geometrical objects, one often applies a scaling factor, just to get the picture at a reasonable size on the page The scaling factor is somehow irrelevant: two pictures at two different scales “are the same, except for the size”.

We shall say that they are similar.

Definition 4.13.1 Let(E, V )be a Euclidean space andk >0 a scalar A similarity of ratiokis an affine transformation(f,−→ f ):(E, V )−→(E, V )such that:

Proposition 4.13.2 A similarity on a finite dimensional Euclidean space is an affine isomorphism.

Proof With the notation of Definition4.13.1, by Proposition4.3.2

Thusf is injective and the result follows by Proposition2.13.3 Example 4.13.3 Every isometry on a Euclidean space is a similarity.

Proof This follows by Proposition4.10.2

Example 4.13.4 Let(E, V )be a finite dimensional Euclidean space A homothety (see Definition2.17.1) of ratiok=0 is a similarity of ratiok.

Proof Consider a homothety with centerOand ratiok With respect to an orthonormal basis(O;e 1 , , e n ), the homothety admits the matrix expression

Thus all scalar products are multiplied byk 2 , which forces at once the conclusion.

A homothety is also called a central similarity The two examples above are highly representative since:

Proposition 4.13.5 A similarity on a finite dimensional Euclidean space is the composite of an isometry and a homothety.

Proof If the similarity(f,−→ f ):(E, V )−→(E, V )has ratiok, fix an arbitrary point

O∈Eand write the similarity as

(f,−→ f )=(h k ,−→ h k )◦(h 1 k ,−→ h 1 k )◦(f,−→ f ) whereh k andh 1 k indicates respectively the homotheties with centerO and ratios k, 1 k By Example4.13.4,h 1 k multiplies all distances by 1 k , thus h 1 k ◦f respects distances and therefore is an isometry, by Theorem4.10.4

Proposition 4.13.6 Let(E, V )be a finite dimensional Euclidean space A mapping f:E−→Ewhich multiplies all distances by a fixed scalark >0 is necessarily a similarity.

Proof The same argument as for Proposition4.13.5 applies: one can write f h k ◦h 1 k ◦f whereh 1 k ◦f respects distances and hence is an isometry by Theo- rem4.10.4 By Examples4.13.4and4.13.3,f is a similarity, being a composite of two similarities

We obtain a nice characterization theorem for similarities, which can be regarded as an extension of Thales’ (see2.18.1).

Theorem 4.13.7 Consider a finite dimensional Euclidean space (E, V ) and an affine isomorphism(f,−→ f ):(E, V )−→(E, V ) The following conditions are equivalent:

4 f multiplies all distances by a fixed scalark >0.

Proof (1⇒2⇒3)are obvious and(4⇒1)is Proposition4.13.6 It thus suffices to prove(3⇒4).

Consider an orthonormal basis(O;e 1 , , e n ) For each pairi=j of indices, the four points

O, X=O+e i , Y =O+e j , Z=O+e i +e j are the four vertices of a parallelogram, since

This parallelogram is a square sincee i is orthogonal toe j and both have length 1 (see Sect.4.4) But being a square reduces to the perpendicularity of the sides and the diagonals (see Proposition4.4.5) Sincef preserves the perpendicularity, it preserves squares and therefore,e i ande j are mapped by−→ f to orthogonal vectors with the same length: let us say, lengthk Of coursek=0 sincef is an isomorphism.

We thus obtain a new orthonormal basis f (0);

The matrix expression off with respect to the original orthonormal basis and the new orthonormal basis is thus simplykId, whereIdis the identity matrix All distances are thus indeed multiplied byk.

Euclidean Quadrics

In the Euclidean case, Theorem2.24.2can be improved in the expected way:

Theorem 4.14.1 LetQ⊆Ebe a quadric in a finite dimensional Euclidean space (E, V ) There exists an orthonormal basis(0;e 1 , , e n )with respect to which the equation of the quadric takes one of the reduced forms:

Proof The proof is an easy adaptation of that of Theorem2.24.2 We focus only on the necessary changes.

Applying TheoremG.4.1instead of CorollaryG.2.8in the proof of the prelim- inary Lemma2.24.1, we begin with an orthonormal basis(P;ε 1 , , ε n )with respect to which the equation of the quadric has the form n i = 1 α i Y i 2 + n i = 1 β i Y i +γ=0.

The arguments in the proof of Theorem 2.24.2 apply as such to prove the existence of another origin O so that with respect to the orthonormal basis

(O;ε 1 , , ε n ), the equation of the quadric now takes one of the three forms: n i = 1 α i Z i 2 =1 n i = 1 α i Z i 2 =0 m i = 1 α i Z i 2 + n i = m + 1 β i Z i =0.

To conclude the proof, it suffices to find another orthonormal basis giving rise to a change of coordinates with the properties

Multiplying the equation bykwill yield the expected result Notice that if the change of coordinates matrixMis orthogonal, the new basis will automatically be orthonormal Indeed the vectors of the new basis, expressed in terms of the old orthonormal basis, will be the columns of the inverse change of coordinate matrixM − 1 , that is,the lines ofMsinceM − 1 =M t

We must therefore prove the existence of an orthogonal matrix of the form

Choosing k=%%(0, ,0,−β m + 1 , ,−β n )%%=0, the firstmlines of this matrix, together with the last line, constitute an orthonormal sequence of vectors inR n It suffices to complete this sequence to an orthonormal basis ofR n to get an orthogonal matrix

Proposition2.25.1yields immediately, in the Euclidean context:

Proposition 4.14.2 Let (O;e 1 , , e n ) be a given orthonormal basis in some Euclidean space (E, V ) Consider a quadricQ⊆E which, with respect to this basis, has a reduced equation as in Theorem4.14.1 Consider a vector subspace

1 in the case of an equation of type 1 or 2, is generated by some of the vectors e 1 , , e n ;

2 in the case of an equation of type 3, is generated by some of the vectors e 1 , , e n − 1

Write(F, W ⊥ )for the affine subspace with directionW ⊥ passing through the origin O The quadric Qis stable under the orthogonal symmetry with respect to(F, W ⊥ ).

Problems

4.15.1 In a Euclidean space, prove that the sum of the angles of an arbitrary triangle equalsπ.

4.15.2 Let(0;e 1 , , e n )be a finite dimensional Euclidean space and(F, W )a hyperplane with equation n i = 1 a i X i =b Find a formula giving the distance between a pointP and the subspaceF.

4.15.3 In a Euclidean plane, prove that a direct isometry is a rotation or a translation.

4.15.4 In a Euclidean plane(E, V )consider a rotation(r,−→r )by angle 0< θ < π.

Prove that for every vector v∈V, there exists a unique point P ∈E such that

4.15.5 Consider a triangleABC in a Euclidean plane and the medianAM of the sideBC(see Fig.4.10) Prove that d(A, B) 2 +d(A, C) 2 =1

4.15.6 Consider a quadrilateralABCDin a Euclidean plane, together with the middle pointsM,Nof the two diagonals (see Fig.4.11) Prove that d(A, B) 2 +d(B, C) 2 +d(C, D) 2 +d(D, A) 2 =d(A, C) 2 +d(B, D) 2 +4d(M, N ) 2

4.15.7 Let(O;e 1 , , e n )be an arbitrary basis of a Euclidean space Prove that the basis is orthonormal if and only if the coordinates−→x of a point P are such that x i =(−→

4.15.8 Consider two pointsA=Bin a finite dimensional Euclidean space(E, V ).

Prove that the locus of pointsP such thatd(P , A)=d(P , B)is a hyperplane This hyperplane is called the mediatrix hyperplane of the segment[AB].

4.15.9 In a Euclidean plane(E, V ), prove that two distinct affine lines are parallel if and only if they are perpendicular to the same third affine line.

4.15.10 In a finite dimensional Euclidean space, prove that every translation is the composite of two orthogonal symmetries with respect to parallel hyperplanes.

4.15.11 Let (E, V )be a finite dimensional Euclidean space Every similarity on

(E, V )of ratiok=1 has exactly one fixed point.

Exercises

1 If in Fig.4.11,ABCDis a parallelogram, prove that

2 Infer the median theorem from this equality (see Fig.4.10): In a triangleABC, ifMis the middle point of the sideBC, then d(A, B) 2 +d(A, C) 2 =2d(A, M) 2 +1

3 Prove vectorially that a triangleABC “inscribed in a circle” (i.e the three vertices of the triangle are points of a given circle) is a right triangle if and only if two of its vertices are on a diameter of this circle.

4.16.2 Consider the Euclidean spaceE 2 (R), that is,R 2 with its usual scalar product.

Give the matrix of the scalar product with respect to these two bases.

, C 2 4 with respect to the basisR Calculate the scalar product of−→

ACand the angle between these two vectors.

4.16.3 ConsiderR 2 as an affine space over itself.

1 Determine a scalar product such that:

(b) the vectors(1,1)and(0, 1 2 )are orthogonal.

2 In the so-obtained Euclidean space, compute the angle between the two lines with equationsx=1 andy.

3 In the same Euclidean structure, determine the circle of radius 1 centered at the origin.

4.16.4 Consider the Euclidean space R (2) [X] of real polynomials of degree at most 2, where the scalar product is defined by a 2 X 2 +a 1 X+a 0|b 2 X 2 +b 1 X+b 0

Consider the canonical basisR c =(0;1, X, X 2 )ofR (2) [X]and the other basisR (P;e 1 , e 2 , e 3 )given by

P=2X 2 +2X+2, e 1 =X+1, e 2 =X 2 +X, e 3 =X 2 +1. Consider next the pointA, admitting the coordinates

1 2 3 with respect to the canonical basisR c , and the pointB, admitting the coordinates

2 3 with respect to the basisR Find the vectorial and Cartesian equations, with respect to the canonical basisR c , of the planeπ containingAand perpendicular to−→

4.16.5 Let(E, V )be a real affine space of dimension 2 FixO∈Eand two linearly independent vectorse 1 , e 2∈V.

1 Give the formulas of change of coordinates between the following two bases:

2 Is it possible to provideV with a scalar product with respect to which both bases are orthonormal?

4.16.6 InE 3 (R)consider a tetrahedronABCD Prove that, when the edgesAB andCDare orthogonal, as well as the edgesACandBD, then the edgesADand

4.16.7 In a real affine space (E, V ) of dimension 2, consider a parallelogram

ABCD and denote byO the intersection point of its diagonals Consider further the two bases

Show that for any scalar product onV, the basisRis orthonormal if and only if the basisS is orthonormal.

4.16.8 Consider the vector spaceR (2) [X]of real polynomials of degree at most 2, viewed as an affine space over itself Provide this space with the scalar product

1 Calculate the angle(P QR)when

F X 2 +aX+ba, b∈R is an affine subspace ofR (2) [X]and compute the orthogonal projection of the zero polynomial on this subspace.

4.16.9 Let(f,−→ f )be an affine transformation ofE 3 (R) Suppose that−→ f preserves the scalar product Suppose further that−→ f admits 1 as eigenvalue and that the corresponding subspace of eigenvectors of eigenvalue 1 is a vectorial planeπ 0 Suppose finally thatf admits a fixed pointP Explain whyf is necessarily the orthogonal symmetry with respect to the planeπ of directionπ 0 passing throughP.

4.16.10 ConsiderE 3 (R); we work in the canonical basis Prove that the affine transformation defined by f

⎠ is an isometry Is this a symmetry? If so, determine the plane, the line or the center of symmetry Is this a rotation? If so, determine the axis and the angle of rotation.

4.16.11 Consider the affine transformationf onE 3 (R)described, with respect to the canonical basis, by the formula f

Prove that this is an isometry Determine the type of isometry and its geometric elements.

4.16.12 Consider the affine transformationf onE 3 (R)described, in the canonical basis, by the formula f

Prove that this is an isometry Determine the type of this isometry according to the values of the parametera Determine the geometric elements of this isometry for the valuesa=1 anda=2.

4.16.13 Let (E, (−|−)) be a Euclidean space andO∈E Consider an isometry (f,−→ f )of this space to itself and putv=−−−−→

Of (O) Prove thatf admits a fixed point if and only ifvis perpendicular to every vector ofFix(−→ f ).

4.16.14 In the Euclidean spaceE 3 (R)with its canonical scalar product, consider the basisR =(O;e 1 , e 2 , e 3 )with

O=(0,0,0), e 1=(1,0,0), e 2=(1,1,0), e 3=(1,1,1). Prove that the affine transformation defined in the basisR by the formula f

⎠ is an isometry Determine its nature and its geometric elements.

4.16.15 InE 2 (R), find the matrix representation, in the canonical basis, of an isom- etryf mapping:

• the point(0,1)to the point( √ 1

• the point(− √ 1 3 ,0)to the point(0,1).

Give the nature and the geometric elements of this isometry.

4.16.16 Consider the vector spaceR (2) [X]of real polynomials of degree at most 2 viewed as an affine space over itself Provide it with the scalar product a 2 X 2 +a 1 X+a 0 b 2 X 2 +b 1 X+b 0

Determine the matricesA∈R 3 × 3 andB∈R 3 × 1 such that the affine transformation defined by the formula f

⎠+B in the canonical basis is the inverse isometry mapping

4.16.17 Let(E, V , (−|−))be a finite dimensional Euclidean space Consider two isometries(f,−→ f )and(g,−→g )ofEto itself Suppose that bothf andgadmit fixed points and commute with each other, that is,f◦g=g◦f Prove, via the following steps, thatf andghave at least one common fixed point.

1 Prove first that(Fix(f ),Fix(−→ f ))is an affine subspace.

2 Prove that this subspace is invariant under(g,−→g ), that is(g,−→g )restricts to an affine isomorphism

3 LetΩ be a fixed point ofg andwthe orthogonal projection of Ω onFix(f ).

Prove thatwis a common fixed point off andg (Hint: using the definition of an orthogonal projection, show thatg(w)is also the orthogonal projection ofΩ onFix(f ).)

4.16.18 InE 3 (R), consider the orthogonal symmetry with respect to the plane with equationz+2=0 followed by the rotation by a half turn about the line with equationsx=1, y=z Is this composite an isometry? If so, what type of isometry is it?

4.16.19 Consider the quadricQofR 3 whose equation with respect to the canonical basis is

Find an orthonormal basis with respect to which this quadric admits a reduced equation Infer the nature of this quadric.

4.16.20 InE 3 (R)consider the quadricQwith equation

2x 2 +4x−y 2 −2yz−z 2 −3z+5=0 with respect to the canonical basis LetR=(0;e 1 , e 2 , e 3 )be an orthonormal basis with respect to which the equation ofQ is reduced and let π be the plane with equation−x+y+z=0 in the canonical basis What is the equation of the direction ofπwith respect to the basis(e 1 , e 2 , e 3 )?

4.16.21 InE 3 (R)and its canonical basis, consider the pointA=(1,0,1)and the planeπwith equationy=x−1 LetQbe the locus of those points whose distance to the pointAis equal to√

2 times the distance to the planeπ Prove thatQis a quadric.

This short chapter describes how to extend, to affine spaces over the field of complex numbers, some Euclidean notions such as “length” and “perpendicularity” We do not dwell on the developments which admit a straightforward generalization from the Euclidean to the Hermitian case, but we underline some major differences between the real and the complex cases.

Hermitian Products

Observe firstly that the operation onC n

(−→x |−→y ) n k= 1 x k y k can by no means be a good generalization of the notion of scalar product Indeed, already in the casen=2, one has

=1 2 +i 2 =1−1=0 with of course,(1, i)=(0,0) The sensible definition is the following one, where x=a−biindicates the conjugate of the complex numberx=a+bi(a, b∈R):

Proposition 5.1.1 Given 0=n∈Nthe Hermitian product of two vectors−→x ,−→y ∈

This Hermitian product has the following properties, for scalarsλ k ∈Cand vectors

The spaceC n provided with this Hermitian product is writtenH n (C).

Proof This is an immediate calculation The main difference between the present situation and the “non-example” of the beginning of this section is that we now have

|x k | 2 and this quantity is zero when all|x k |are zero, that is, when allx k are zero

We can now formalize this situation.

Definition 5.1.2 A Hermitian space consists of a complex affine space(E, V )together with a Hermitian product onV, that is, a mapping

V ×V −→C, (x, y)→(x|y) satisfying the axioms, forx,y,y k ∈V andλ k ∈C x n k = 1 λ k y k n k = 1 λ k (x|y k )

Proposition 5.1.3 Let(E, V )be a Hermitian space Given vectorsx k andyinV and scalarsλ k ∈C: n k = 1 λ k x k y n k = 1 λ k (x k |y).

Proof This follows simply because n k = 1 λ k x k y y n k = 1 λ k x k y n k = 1 λ k (y|x k ) n k = 1 λ k (y|x k ) n k = 1 λ k (x k |y).

The Hermitian product is therefore said to be sesquilinear (etymologically: 1 1 2 times linear), since it is linear in the second variable and “linear up to conjugation” in the first variable

The Euclidean notions of norm and distance admit easy generalizations in the Hermitian case:

Definition 5.1.4 Let(E, V )be a Hermitian space Then:

1 the norm of a vectorv∈V is v (v|v)∈R+;

2 the distance between two pointsA,B∈Eis d(A, B)= −→

A distance in a Euclidean space is thus always a positive real number, not an arbitrary complex number! On the other hand the definition of an angle as in Def- inition4.2.6no longer makes sense as such, even if the Schwarz inequality can be generalized to Hermitian spaces (see Proposition5.3.2)! The difficulty is the fact that(x|y)is no longer a real number.

Nevertheless there is no problem in defining a right angle:

Definition 5.1.5 Let(E, V ) be a Euclidean space Two vectorsv,w∈V are orthogonal (or perpendicular) when(v|w)=0.

The orthogonality of vectors is trivially a symmetric relation, since one has

(w|v)=(v|w); thus one quantity vanishes if and only if the other one does But one should be aware of some important differences between orthogonality in Eu- clidean spaces versus Hermitian spaces For example

Theorem 5.1.6 (Pythagoras’ theorem) Consider a right triangleBAC (i.e.−→

AC) in a Hermitian space(E, V ) One then has d(B, A) 2 +d(A, C) 2 =d(B, C) 2

Proof As in Theorem4.3.5, one computes

=d(B, A) 2 +d(A, C) 2 since the other two terms are zero by assumption

Let us stress that the converse implication in Theorem5.1.6does not hold! If the equality in the statement holds, the computation in the proof yields

This means that, writing Rexfor the real part of the complex numberx (see Defi- nitionF.1.2)

AC) thus we can only conclude that the real part of this Hermitian product vanishes, not the product itself.

An analogous observation shows that the Hermitian product cannot be reconstructed from the norms, as in the proof of Theorem4.13.7 Indeed from

Once more only the real part of the Hermitian product can be reconstructed from the norm Thus there is no hope of obtaining an immediate generalization of results such as Theorem4.10.4.

Orthonormal Bases

Since length and orthogonality make sense in a Hermitian space, so does the notion of an orthonormal basis.

Definition 5.2.1 Let(E, V )be a finite dimensional Hermitian space By an orthonormal basis is meant an affine basis(O;e 1 , , e n )such that:

It is then routine, provided we take care concerning the lack of symmetry, to generalize most results concerning orthonormal bases in the Euclidean case (see Sect.4.6).

Proposition 5.2.2 Let (O;e 1 , , e n ) be an orthonormal basis of a Hermitian space(E, V ) The coordinatesx k of a pointP ∈Eare x k =(e k |−→

Proof Simply expand the product

Of course, it is compulsory in Proposition5.2.2to writee k as the left hand factor and−→

OP as the right hand factor (see Proposition5.1.3).

Proposition 5.2.3 Let (O;e 1 , , e n ) be an orthonormal basis in a Hermi- tian space (E, V ) Given two points with coordinates P =(x k ) 1 ≤ k ≤ n and Q (y k ) 1 ≤ k ≤ n with respect to this basis

Proof Simply expand the product n k = 1 x k e k k l = 1 y l e l

Proposition 5.2.4 Let(E, V )be a finite dimensional Hermitian space The change of coordinates matrix M between two orthonormal bases is unitary, that is,

O ;e 1 , , e n for the two orthonormal bases and consider the change of coordinates formulas

The matrixMis obtained by putting in columns the coordinates of the vectorse k in the second basis: thusm lk =(e l |e k )by Proposition5.2.2 Analogously n kl e k |e l e l |e k

Proposition 5.2.5 Let(E, V )be a Hermitian space and(e k ) k ∈ K a family of non- zero vectors When these vectors are pairwise orthogonal, they are linearly independent.

Multiplying on the left bye k j yieldsλ k j (e k j |e k j )=0, thusλ k j =0 sincee k j =0. Now comes the main result:

Theorem 5.2.6 (Gram-Schmidt process) Let(O;e 1 , , e n )be an arbitrary basis of a Hermitian space(E, V ) There exists an orthonormal basis(O;v 1 , , v n ) with the additional property that for every indexk, the two subspaces e 1 , , e k = v 1 , , v k generated by the firstkvectors of each basis are the same.

Proof We prove the result by induction onn Whenn=1, it suffices to put v 1= e 1 e 1

Assuming the result up to dimension n−1, let us apply it to the vector subspace e 1 , , e n − 1 and its basis e 1 , , e n − 1 We obtain an orthonormal basis v 1 , , v n − 1 of this subspace, which satisfies the condition of the statement up to the indexn−1 Consider then v n =e n −(v 1 |e n )v 1 − ã ã ã −(v n − 1 |e n )v n − 1

We get at once, for 1≤i≤n−1 v i |v n

Putting v n = v n v n thus yields a sequence v 1 , , v n of pairwise orthogonal vectors of length 1 By

Proposition5.2.5, this is a basis ofV

The Metric Structure of Hermitian Spaces

Proposition 5.2.7 Let(E, V )be a finite dimensional Hermitian space and(F, W ) an affine subspace Then

W ⊥ v∈V∀w∈W (w|v)=0 is a vector subspace ofV, supplementary toW.

Proof The setW ⊥ is a vector subspace by linearity of the Hermitian product in the second variable.

A vectorw∈W∩W ⊥ is such that(w|w)=0, thusw=0 It remains to prove thatW+W ⊥ =V.

Consider a basise 1 , , e k ofW and extend it to a basise 1 , , e n ofV Ap- ply the Gram–Schmidt construction (Theorem5.2.6) to get an orthonormal basis v 1 , , v n such that in particular,v 1 , , v k is still a basis ofW The vector subspacev k + 1 , , v n is contained inW ⊥ and is supplementary toW Thus

Although it is not necessary for the proof, observe that moreover the vector sub- spacesW ⊥ ande k + 1 , , e n have the same dimensionn−k, so are equal

Proposition5.2.7thus allows us to define an orthogonal projection or an orthogonal symmetry in a finite dimensional Hermitian space.

5.3 The Metric Structure of Hermitian Spaces

This section shows that the notion of distance given by a Hermitian product satisfies all the requirements for a metric space (see Proposition4.3.2) A weak version of the Schwarz inequality is sufficient for this.

Lemma 5.3.1 Let(E, V )be a finite dimensional Hermitian space Given two vectorsx,y∈V

−x ã y ≤Re(x|y)≤ x ã y where Re(x|y)indicates the real part of(x|y).

Proof The proof of Proposition4.2.5also applies to the present result Given a real parameterk∈R, we have x+ky 2 = x 2 +

This is a real polynomial of degree 2 with constant sign, thus

In fact, the usual form of the Schwarz inequality is valid in every Hermitian space Let us prove this in finite dimensions.

Proposition 5.3.2 (Schwarz inequality) Let(E, V )be a finite dimensional Hermi- tian space Given two vectorsx,y∈V

(x|y)≤ x ã y where|(x|y)|indicates the modulus of(x|y)(see DefinitionF.2.1).

Proof By Theorem5.2.6, let us choose an orthonormal basis (O;e 1 , , e n ) By

Proposition5.3.3, and taking the square of both sides, the formula we need to prove reduces to n k = 1 x k y k n l = 1 x l y l

Observe that the “diagonal” terms (i.e withk=l) are the same on both sides There- fore it suffices to prove that for every pairk=lof indices x k y k x l y l +x l y l x k y k ≤x k x k y l y l +x l x l y k y k

This inequality can be re-written as

Proposition 5.3.3 Given a Hermitian space(E, V ), the corresponding distance providesEwith the structure of a metric space (see4.3.2).

Proof Only the Minkowski inequality requires a comment Given three pointsA,

Complex Quadrics

≤d(A, B) 2 +2d(A, B) d(B, C)+d(B, C) 2 d(A, B)+d(B, C) 2 where the inequality holds by Lemma5.3.1

Let us conclude this chapter with some observations, but essentially a warning, concerning quadrics in a Hermitian space First of all, a purely affine result:

Proposition 5.4.1 LetQ⊆E be a quadric in a finite dimensional complex affine space There exists an affine basis(O;e 1 , , e n )with respect to which the equation of the quadric takes one of the following reduced forms: k j = 1

X i 2 =X n withk < nin the last case.

Proof Under the conditions of Theorem2.24.2, up to a possible renumbering of the coordinates, there is no loss of generality in assuming that in the left hand side of the equation, the non-zero coefficients come first Whena j =0, simply apply the change of basis e j

Otherwise, keepe j =e j With respect to the basis(0;e 1 , , e n )the quadric now has a reduced equation of the form indicated in the statement

The main point that we want to stress, in the Hermitian case, is the non-existence, in general, of an orthonormal basis with respect to which the quadric admits a reduced equation.

The reader may wonder why the proof in the real case does not have its counterpart here The answer is that, in the real case, the result is based on TheoremG.4.1, which is itself a consequence of TheoremG.3.2 A closer look at these proofs shows (see5.5.1and5.5.2) that they generalize immediately to the complex case, provided that the assumption on the matrixAisA=A t , notA=A t The matrixA t is called the adjoint matrix ofA Of course in the real case, being self-adjoint is the same as being symmetric.

Having made this observation, the reader might then want to focus his/her attention on those quadrics admitting an equation of the form x t Ax+b t x+c=0 withAself-adjoint ButA=A t forces in particulara kk =a kk for each indexk: thus the diagonal elements are real numbers Moreover given indicesk=j a kj X k X j +a j k X j X k =(a kj +a kj )X k X j

This proves thatx t Ax=x t Bx, whereBis obtained fromAby replacing each entry by its real part Thus, in the diagonalization problem, reducing one’s attention to the self-adjoint case would essentially take us back to the real case.

As far as complex symmetric matrices are concerned, the appropriate generalization of TheoremsG.4.1andG.3.2is PropositionG.5.1: it refers to what we might call eigenvectors up to conjugation Of course when the restricted conditions of this PropositionG.5.1are satisfied, it is routine to adapt the proof of Theorem4.14.1to the Hermitian case (see5.5.5): the equation is a priori diagonalized and it remains simply to take care of the terms of degree at most 1 However, our purpose here is to make clear that the conditions of PropositionG.5.1are generally not satisfied.

Counterexample 5.4.2 There exist quadrics, in finite dimensional Hermitian spaces, which do not admit a reduced equation with respect to any orthonormal basis.

Proof To produce a counterexample, it suffices to exhibit a complex symmetric matrix which does not satisfy condition 2 in PropositionG.5.1 We shall prove that given the matrix

, there does not exist a non-zero vectorvand scalarλ∈Csuch thatAv=λv.

We work by reductio ad absurdum Suppose that

5.4 Complex Quadrics 191 withx,y∈Cnot simultaneously zero This yields the equations x+ix+y=λx (5.1) x−iy+y=λy (5.2)

The first equation shows thatx=0 impliesy=0, while the second one proves that y=0 impliesx=0 Both possibilities are excluded, thus we have bothx=0 and y=0.

Multiply equation (5.1) by y and equation (5.2) by x; subtracting the results yields

Let us now write more explicitly x=a+bi, y=c+di, a, b, c, d∈R.

The real and imaginary parts ofxyare thus ac+bd=0, bc−ad=a 2 +b 2 −c 2 −d 2

The first equality implies that

Introducing these quantities into the second equality yields a 2 +b 2 k 2 +2k−1

Writing further λ=α+βi, α, β∈R, the two equations (5.1) and (5.2) now yield, identifying on both sides the real and imaginary parts, a−(2∓√

2)and comparing with equation (5.6) yields

Since multiplyingx andy by the same non-zero real constant does not change the situation, there is no loss of generality in assuming thata=1=b.

Equations (5.3) to (5.6) now take the form

Solving the system (5.7)–(5.8) of linear equations inα,βyields α= −(2−√

Introducing these values into equation (9) yields 1=5−4√

2, which is the expected contradiction

Let us conclude with a comment Given a complex symmetric(n×n)-matrixA, the equation

Av=λv is a system of n equations with n+1 unknowns, namely, the n components of the vectorv and the scalarλ This is not a linear system of equations, as the right hand side of the equations indicates Nevertheless, most often, a system which has more unknowns than equations admits solutions It can even be proved that for a complex symmetric matrix A chosen at random, the probability that there is an orthonormal basis of solutions of the above system is 1! This does not contradict our counterexample5.4.2, because the set of symmetric complex matrices is infinite.

Problems

5.5.1 Prove that the eigenvalues of a self-adjoint complex matrixA=A t are real.(Hint: mimic the proof of TheoremG.3.2.)

Exercises

5.5.2 Prove that for every(n×n)-self-adjoint complex matrixA=A t ,C n admits an orthonormal basis constituted of eigenvectors ofA (Hint: mimic the proofs of

5.5.3 Adapt the proof of Theorem2.26.9to show that in a finite dimensional affine space overC, two equations of a non-degenerate quadric with respect to the same basis are necessarily proportional.

5.5.4 In a finite dimensional Hermitian space, consider a quadric with equation x t Ax+b t x+c=0 with respect to an orthonormal basis WhenAhas only real entries, prove the existence of an orthonormal basis with respect to which the quadric has a reduced equation (Hint: mimic the proof of Theorem4.14.1.)

5.5.5 In a finite dimensional Hermitian space, consider a quadric with equation x t Ax+b t x+c=0 in an orthonormal basis IfAsatisfies the conditions of PropositionG.5.1, prove the existence of an orthonormal basis with respect to which the quadric has a reduced equation (Hint: mimic the proof of Theorem4.14.1.)

5.6.1 InH 3 (C), consider the three points

Find a Cartesian equation of the planeπcontainingA,BandC What is the distance from the pointP =(1+2i,1+i,−1+i)to the planeπ?

5.6.2 InH 3 (C), compute the distance from the pointP =(1+i,1,−i)to the line containing the two pointsA=(1−i,1+i,0)andB=(1, i, i).

5.6.3 Find a basis ofH 3 (C)in which the following quadric admits a reduced equation: x 2 +2ixy+(1+i)y 2 +3ix+(1−i)y+2z=4.

5.6.4 In the canonical basis ofH 3 (C), describe the affine transformation mapping a point to the point orthogonally symmetric to it with respect to the planeπ with equation x+iy+z=2+i.

5.6.5 Characterize, in terms of their matrices, the so-called unitary transformations ofH 2 (C), that is, the affine transformations respecting the Hermitian product What can one say about their determinant?

5.6.6 Consider the unitary transformations ofH 2 (C)with determinant−1.

1 Do they all admit a line of fixed points?

2 Are those transformations which do admit a line of fixed points always symmetries?

Before studying this chapter, the reader is invited to review Sect 6.1 in [7], Trilogy I, where the origin of projective ideas is discussed, thus providing a good intuitive base for the algebraic presentation which follows This intuition is nevertheless very briefly recalled in Sect.6.1.

This chapter is thus devoted to the study of projective spaces over a fieldK Of course, we shall study projective subspaces, projective transformations, projective quadrics and projective (i.e homogeneous) coordinates, eventually clarifying the link with the corresponding affine notions.

We also show that the projective plane over a fieldK of characteristic distinct from 2 satisfies all the axioms of synthetic projective geometry, as studied in Chap 6 of [7], Trilogy I.

The anharmonic ratio of four points on a line can be regarded as the projective substitute of the notion of “proportionality” in affine spaces We clarify this point and show how powerful this notion is.

Above all, projective spaces satisfy duality principles which do not have their counterparts in affine geometry For example in a projective plane, these principles allow us to interchange the roles of points and lines in a statement Such a general duality principle exists a priori in each projective space, but there is also a specific duality principle attached to each non-degenerate projective quadric The duality principle inferred from a quadric leads in particular to an elegant treatment of the tangent spaces to the quadric.

Convention 6.0.1 In this chapter, the fieldKis always commutative.

Projective Spaces over a Field

Imagine (see Fig.6.1) that you are standing on a horizontal plane and (with just one eye) you observe the various points of this plane When you look at a pointP of the plane, you can represent your line of sight by the line passing throughP and your

Fig 6.1 eye In doing so, you describe a bijection between all the points of the plane and all the non-horizontal lines passing through your eye But what about the horizontal lines through your eye?

Imagine that a lined is drawn in the horizontal plane on which you stand Fol- lowing that line, looking further and further away, as you focus on a very distant point on the lined, your line of sight becomes almost horizontal In fact, your line of sight tends to the horizontal as you look further and further away alongd: your line of sight approaches the horizontal lined ∞ through your eye, which is parallel tod.

Now imagine that two parallel linesd andd are drawn in the horizontal plane, like the two rails of a train track When you look very far along these lines, you have the impression that they meet somewhere “on the horizon” In fact your line of sight, when following these two linesd andd , tends as we have seen to two horizontal linesd ∞ andd ∞ passing through your eye and respectively parallel tod andd But sinced is parallel tod , we must haved ∞ =d ∞ Is this not an elegant way to explain why you have the impression that parallel rails meet “on the horizon”?

In conclusion, we have described a bijection between:

• all the lines through your eye; and

• all the points of the horizontal plane, augmented by those “points at infinity” where you have the impression that parallel lines meet.

Historically, the projective plane was defined as the ordinary plane to which one has added “points at infinity” where parallel lines meet (see Sect 6.1 in [7], Tril- ogy I) An alternative approach is to present the projective plane as the set of all lines through a point (which in the above illustration corresponds to your eye) of

6.1 Projective Spaces over a Field 197 three dimensional space Indeed, why not take your eye as origin of three dimensional space?

Definition 6.1.1 The projective spaceP n (K)of dimensionnover a fieldKis the set of all vector lines of the spaceK n + 1 Each such vector line is called a point of the projective space.

Of course “vector line” means sub-vector space of dimension 1 In general, to make the language less heavy, we shall simply refer toP n (K) without repeating each time “the projective spaceP n (K)of dimensionnover a fieldK”.

Definition6.1.1is definitely the correct definition of a projective space Perhaps, from the intuitive discussion above, you might be tempted to define the projective plane instead as the set of all the half lines starting from your eye and passing through a point of the horizontal plane, plus all the horizontal half lines starting from your eye Given two parallel linesd andd in the horizontal plane, they would then meet at two distinct points at infinity (two opposite horizontal half lines through your eye), depending on whether your line of sight followsdandd one way or the other This is not the definition of the projective plane and moreover, such a definition does not lead to a good geometric theory Indeed, through two distinct points (your “opposite” points at infinity) you would be able to draw infinitely many distinct lines, namely, all the “parallel lines” of the horizontal plane in this direction.

So you would lose at once the very basic geometrical axiom attesting that there is exactly one line passing through two distinct points.

On the other hand let us observe that:

Proposition 6.1.2 The projective spaceP n (K)of dimensionnover a fieldK can equivalently be defined as the set of all vector lines of an arbitraryK-vector space

Proof Indeed, every choice of a basis inV provides an isomorphism betweenV and

Having read this section carefully, the reader cannot be blamed for thinking of a projective point as being a vector line, but this is unlikely to develop into a useful intuition After all, when one imagines a point of the real planeR 2 , one certainly doesn’t view it intuitively as a pair of equivalence classes of Cauchy sequences, or as a pair of Dedekind cuts Instead, our intuitive view of such a point is precisely that of a point on a sheet of paper, exactly the intuitive idea which motivated the technical definitions in the first place Of course when you have to prove the very first properties of the real line—such as its completeness, or the existence of an addition, and so on—you have to switch back to your formal definition of what a real number is However one tends to forget such formal definitions, simply maintaining the intuitive picture of “actual points on a sheet of paper”, ensuring of course that your arguments only use those results that you have already formally proved What you draw on the sheet of paper is not part of the proof: it is just a way to support the intuition.

The same is true with projective spaces Your basic intuition of a projective plane should remain that of a sheet of paper, which extends infinitely, far away from the concrete edges of the sheet of paper, just as one thinks of the plane R 2 , but in the projective case, the sheet of paper should extend even further away in order to include the “points at infinity”, those points where the lines that you draw parallel on the sheet of paper eventually meet! Of course, although pictures on a sheet of paper remain the best way to support our intuition when working in a projective plane, one should not forget that the only acceptable arguments are those which rely formally on the definitions and on anterior results One can rapidly learn to appreciate this intuition of the “projective sheet of paper”: its power and its limitations Even when the fieldKis a finite field, or the field of complex numbers, or any fieldKsuch that

P2 (K)and evenK 2 do not resemble a sheet of paper, our intuition, as in the affine case, is still guided by the properties of this “projective sheet”.

Projective Subspaces

The notion of a projective subspace is at once suggested by that of a general projective space:

Definition 6.2.1 LetP n (K)be the projective space of dimensionnover a fieldK.

A projective subspaceS⊆P n (K)of dimensionmis the set of all vector lines of a vector subspaceV ⊆K n + 1 of dimensionm+1.

As usual one makes the following definition:

• By the projective plane overK is meant the projective spaceP2 (K) of dimension 2.

• By a projective line in the projective spaceP n (K)is meant a projective subspace

• By a projective hyperplane in the projective spaceP n (K)is meant a projective subspace of dimensionn−1.

Trivially we have the following:

Proposition 6.2.3 Every projective subspaceS⊆P n (K)of dimensionmis a projective space of dimensionmoverK.

Proof This follows by Proposition6.1.2

Proposition 6.2.4 InP n (K), the intersection of two projective subspaces is again a projective subspace, provided it is non-empty.

Proof If the projective subspacesS, T ⊆P n (K)are the sets of vector lines of the vector subspacesV , W⊆K n + 1 , thenS∩T is the set of vector lines ofV ∩W

Proposition 6.2.5 In the projective space P n (K), consider a non-empty family(P i ) i∈I of points There exists a smallest projective subspace containing these points; it is called the projective space generated by the points(P i ) i ∈ I

Proof EachP i is a vector line inK n + 1 , thus& i ∈ I P i is a non-zero subset ofK n + 1 Consider the vector spaceV generated by this subset: the required projective sub- spaceSis the set of vector lines ofV

Thus the following definition makes sense:

Definition 6.2.6 Consider two projective subspacesS, T ⊆P n (K) The projective subspaceS+T is the smallest projective subspace containingSandT.

As far as dimensions of subspaces are concerned, we have:

Proposition 6.2.7 LetSandT be two projective subspaces of the projective space

1 The two subspacesSandT intersect if and only if dimS+dimT =dim(S+T )+dim(S∩T ).

2 The two subspacesSandT are disjoint if and only if dimS+dimT =dim(S+T )−1.

Proof LetS andT be the sets of projective lines of the vector subspacesV , W ⊆

K n + 1 of respective dimensionsk+1 andm+1 We know from linear algebra that dimV +dimW=dim(V+W )+dim(V ∩W ).

WhenV ∩W =(0), this can be rephrased as (see the proofs of Propositions6.2.4 and6.2.5)

+ dim(S∩T )+1 from which dimS+dimT =dim(S+T )+dim(S∩T ).

On the other hand ifS∩T is empty, thenV∩W=(0)and we obtain

+0 from which dimS+dimT =dim(S+T )−1.

Since these two equalities exclude each other, the two implications that we have just proved are in fact equivalences

Corollary 6.2.8 Consider two projective subspacesS, T ⊆P n (K) The following conditions are equivalent:

Proof Being a singleton is equivalent to having projective dimension 0 The result follows by Proposition6.2.7

Proposition 6.2.9 Through two distinct points of a projective space over a field passes exactly one projective line.

Proof A point is a projective subspace of dimension 0 Two distinct pointsP,Q have an empty intersection, thus by Proposition6.2.7the smallest subspaceP +Q containing them has dimension 1 It is thus a projective line If another projective linedcontainsP andQ, then it containsP+Qand by equality of the dimensions, d=P +Q

Let us conclude this section with a key result of projective geometry, which is a first occurrence of the duality principle which will be studied in Sect.6.3.

Theorem 6.2.10 In the projective planeP2 (K)over a fieldK:

1 through two distinct points passes a unique line;

2 two distinct lines intersect in a unique point.

Proof Assertion 1 holds by Proposition6.2.9 Next if two projective lines d and d are distinct, their sumd+d contains strictly each of them, thus has dimension strictly greater than 1 Since we are in a projective plane, this dimension can only be 2 The result follows by Corollary6.2.8.

The Duality Principle

We are now going to establish a very appreciated principle: in projective geometry, when you prove a theorem, you get a second one for free! Unfortunately, this principle does not apply to all aspects of everyday life!

Let us recall (see AppendixH) that given a vector spaceV over a fieldK, its dual vector space is

6.3 The Duality Principle 201 Given vector subspacesX⊆V ∗ andY ⊆V, one defines

These constructions satisfy the following properties (see TheoremH.2.5):

Proposition 6.3.1 LetV be a vector space of finite dimensionnover a fieldKand letV ∗ be its dual Given vector subspacesX⊆V ∗ andY ⊆V, the two mappings

1 define inverse bijections between the two posets of vector subspaces ofV ∗ andV;

2 reverse the ordering of every inclusion;

3 transform a subspace of dimensionkinto a subspace of dimensionn−k.

These considerations yield the famous “duality principle” for finite dimensional vector spaces:

Theorem 6.3.2 (Linear duality principle) Consider a statement ϕ about vector spaces of dimensionmover a fieldK: a statement which expresses some inclusions, equalities or inequalities between some vector subspaces of prescribed dimensions.

Write ϕ ∗ for the dual statement obtained from ϕ by reversing all the inclusions and replacing each dimensions of a vector subspace by the dimensionm−s If ϕ is valid in everyK-vector space of dimension n, thenϕ ∗ is also valid in every K-vector space of dimensionn.

Proof IfV has dimension n, by Lemma H.1.3, its dual V ∗ has dimension n as well By Proposition6.3.1, provingϕ ∗ inV is equivalent to provingϕ inV ∗ By assumption,ϕis indeed valid inV ∗ since this is aK-vector space of dimensionn.

Theorem 6.3.3 (Projective duality principle) Consider a statementϕ about projective spaces of dimensionnover a fieldK: a statement which expresses some inclusions, equalities or inequalities between some projective subspaces of prescribed dimensions Writeϕ ∗ for the dual statement obtained fromϕby reversing all the inclusions and replacing each dimensionkof a projective subspace by the dimension n−k−1 Ifϕis valid in every projective space of dimensionnoverK, thenϕ ∗ is also valid in every projective space of dimensionnoverK.

Proof The projective dimensionsnandkcorrespond to the vector dimensionsm n+1 ands=k+1 Thus the vector dimension m−s=(n+1)−(k+1)=n−k=(n−k−1)+1 corresponds to the projective dimensionn−k−1 The result follows by Theo- rem6.3.2

Let us give an example By Proposition6.2.9

Through two distinct points of P 3 (K) passes exactly one projective line

This can be rephrased as

Consider a projective space of dimension 3 Given two projective subspaces P and Q of dimension 0, such that P = Q, there exists a projective subspace d of dimension 1 such that

P ⊆ d and Q ⊆ d Moreover if d is a projective subspace of dimension 1 such that P ⊆ d and Q ⊆ d , then d = d

By the duality principle6.3.3, we obtain

Consider a projective space of dimension 3 Given two projective subspaces P and Q of dimension 3 − 0 − 1 = 2, such that P = Q, there exists a projective subspace d of dimension 3 − 1 − 1 = 1 such that P ⊇ d and Q ⊇ d Moreover if d is a projective subspace of dimension 3 − 1 − 1 = 1 such that P ⊇ d and Q ⊇ d , then d = d

In a projective space of dimension 3, two distinct projective planes contain exactly one common projective line

The duality principle takes a particularly elegant form in the case of projective planes For that purpose, let us introduce a useful point of terminology:

Definition 6.3.4 In a projective plane over a fieldK, one says that a pointP and a linedare incident whenP ∈d.

Theorem 6.3.5 (Plane duality principle) Consider a statementϕabout projective planes over a fieldK: a statement which expresses some incidences, equalities or inequalities between points and lines Writeϕ ∗ for the dual statement obtained from ϕby interchanging the words point and line Ifϕis valid in the projective plane over

K, thenϕ ∗ is also valid in the projective plane overK.

Proof Applying Theorem6.3.3, the notion of “incidence” is self-dual, while dimension 0 becomes dimension 2−0−1=1 and dimension 1 becomes dimension 2−1−1=0

Observe in particular that in Theorem6.2.10, the two assertions are the duals of each other.

Homogeneous Coordinates

Fix an arbitrary basise 1 , e 2 , e 3 of R 3 A vector line ofR 3 —that is, a projective point of the projective plane P2 (R)—is entirely determined by one of its points

(a, b, c)=(0,0,0) Of course, all the other points of the vector line have coordinates a , b , c

The property that two non-zero triples are proportional is trivially an equivalence relation and thus all the members of an equivalence class determine the same vector line, that is, the same projective point.

More generally, letKbe a field On

⎥⎦ to indicate the equivalence class of(a 0 , , a n ).

Definition 6.4.1 Consider an arbitrary basise 0 , , e n ofK n + 1 , for some fieldK.

(a 0 , , a n )=(0, ,0) with respect to this basis, the equivalence class

⎥⎦ is called the system of homogeneous coordinates with respect to the basise 0 , , e n of the corresponding projective point ofP n (K).

Let us stress the following point: homogeneous coordinates are not defined in- trinsically via the choice of n+1 points E 0 , , E n of P n (K) Indeed consider another basis ofK n having the particular form e 0 =k 0 e 0 , , e n =k n e n , 0=k i ∈K.

The various vectorse i ande i thus determine exactly the same projective pointsE i ofP n (K) But if a projective pointP has homogeneous coordinates

⎥⎦ with respect to the basise 0 , , e n it has homogeneous coordinates

⎥⎦ with respect to the basise 0 , e n

These two systems of homogeneous coordinates are distinct, except of course in the very special case where k 0 =k 1 = ã ã ã =k n

We shall investigate further (and overcome) this difficulty in Sect.6.4.

For the time being, let us simply observe the following:

Proposition 6.4.2 Fix a system of homogeneous coordinates inP n (K) Every projective subspace can be characterized as the set of those points whose homogeneous coordinates satisfy some homogeneous system of linear equations Such a system is unique up to a non-zero multiplicative constant Conversely, the non-zero solutions of such a system (provided such solutions exist) are always the homogeneous coordinates of the points of a projective subspace.

Proof It is well-known that for a given basis ofK n + 1 , every vector subspace can be characterized by a homogeneous system of linear equations in terms of the coordinates of its points, a system which is unique up to a non-zero multiplicative constant.

The result follows by Definition6.2.1

Corollary 6.4.3 Given a system of homogeneous coordinates in a projective plane, every projective line admits an equation of the form k 0 X 0+k 1 X 1+k 2 X 2=0 and such an equation is unique up to a non-zero multiple Conversely, the solutions of every such non-zero equation are the homogeneous coordinates of the points of a unique projective line.

Proposition 6.4.4 A projective line over a fieldK is in bijective correspondence with the setK∪ { }, whereis an arbitrary element not inK.

Proof Given a system of homogeneous coordinates on a projective line, all the pairs

, k∈K correspond bijectively with the points of the line.

Projective Basis

In Sect.6.4, we underlined the fact that a system of homogeneous coordinates in

P n (K), constructed from a basise 0 , , e n ofK n + 1 , is not determined by the corresponding pointsE 0 , , E n ofP n (K).

This is a major problem! Recall that our “program”—as suggested in Sect.6.1— was to be able to regard the projective points and the projective lines as the basic geometric entities as soon as possible, forgetting how they have been defined algebraically This is analogous to the situation in analysis, where one tends to rapidly regard the real numbers as the basic entities, forgetting that they have been introduced via Cauchy sequences or Dedekind cuts in the rational numbers.

A basis inK n + 1 is a sequence ofn+1 linearly independent vectors What does this means in “intrinsic” projective terms? This is fairly easy to express:

Definition 6.5.1 PointsP 1 , , P k in a projective space P n (K)are projectively independent when none of them is contained in the projective subspace generated by the other points (see Proposition6.2.5).

Example 6.5.2 Two points ofP n (K) are projectively independent if and only if they are distinct.

Example 6.5.3 Three points ofP2 (K)are projectively independent if and only if they are pairwise distinct and none of them is contained in the line passing through the other two points.

The link with linear independence is immediate:

Lemma 6.5.4 Non-zero vectorse 0 , , e k ofK n + 1 are linearly independent if and only if the corresponding projective points ofP n (K)are projectively independent.

Proof A family of vectors is linearly independent when none of them is a linear combination of the other vectors, that is, when none of them belongs to the vector subspace generated by the other vectors

Let us now exhibit the result which allows us to handle homogeneous coordinates in an “intrinsic” projective way.

Theorem 6.5.5 InP n (K), consider a familyP 0 , , P n + 1 of points such thatn+1 of these points are always projectively independent Then there exists a unique system of homogeneous coordinates such that

Proof First choose non-zero vectorse 0 , , e n + 1 on the corresponding vector lines

P 0 , , P n + 1 By assumption and Lemma6.5.4, the vectorse 0 , , e n constitute a basis ofK n + 1 With respect to this basis write e n + 1 =k 0 e 0 + ã ã ã +k n e n , k i ∈K.

By assumption and Lemma6.5.4again,e n + 1 is not a linear combination ofnvectors of the basis; in other words,k i =0 for alli=1, , n+1 This provides another basis ofK n + 1 e 0 =k 0 e 0 , , e n =k n e n , with eache i still on the vector lineP i and this time e n + 1=e 0 + ã ã ã +e n

The corresponding homogeneous system of coordinates thus satisfies the conditions of the statement.

Conversely if a system of homogeneous coordinates satisfies the conditions of the statement, it is obtained from a basise 0 , , e n ofK n + 1 , where eache i is on the vector lineP i whilee 0 + ã ã ã +e n is on the vector lineP n + 1 In particular we have e 0 =a 0 e 0 , , e n =a n e n , a i =0 while e 0 + ã ã ã +e n =a n + 1 e n + 1

Introducing the values ofe 0 , , e n , e n + 1into this equality we obtain a 0 e 0 + ã ã ã +a n e n =a n + 1 e 0 + ã ã ã +a n + 1 e n

Writingafor this common value, we thus have e 0 =ae 0 , , e n =ae n

The Anharmonic Ratio

proving that the two corresponding systems of homogeneous coordinates are the same

Example 6.5.6 InP1 (K), given three distinct pointsA,B,C, there exists a unique system of homogeneous coordinates such that

Proof This follows by Example6.5.2and Theorem6.5.5

Example 6.5.7 InP2 (K), given four pointsA,B,C,Dsuch that three of them are never on a same line, there exists a unique system of homogeneous coordinates such that

Proof This follows by Example6.5.3and Theorem6.5.5

We make the following definition:

Definition 6.5.8 A projective basis of P n (K)is a family (P 0 , , P n ;P n + 1 ) of points such thatn+1 of these points are always projectively independent.

Proposition 6.5.9 Every projective spaceP n (K)over a fieldKadmits a projective basis.

Proof Choose a basis(e 0 , , e n )ofK n + 1 together with the point e n + 1=e 0+ ã ã ã +e n

The corresponding vector lines constitute a projective basis ofP n (K)

This section introduces one of the most powerful tools in projective geometry:

Definition 6.6.1 Consider four distinct pointsA,B,C,Don a projective line in

P n (K) By Theorem6.5.5, consider the unique system of homogeneous coordinates on this projective line yielding

The anharmonic ratio of these four points is the scalar

Observe that this definition makes sense, because b=0 Indeedb=0 would forceA=D, which is not the case.

Lemma 6.6.2 The anharmonic ratio of four distinct points on a projective line is neither equal to 0 nor 1.

Proof With the notation of Definition6.6.1,B=Dforces the anharmonic ratio to be non-zero, whileC=Dimplies that it is not equal to 1

It should be noted that whenK=Z2= {0,1}, the field of integers modulo 2, each projective line contains only three points (see Proposition6.4.4), thus the notion of anharmonic ratio does not apply On the other hand whenK=Z3= {0,1,2}is the field of integers modulo 3, again by Proposition6.4.4, each projective line contains exactly four points But since by Lemma6.6.2an anharmonic ratio never takes the values 0 or 1, the anharmonic ratio of the four points inP1 (Z3 ), in whatever order, is necessarily always equal to 2= −1∈Z3.

Let us now investigate the action on their anharmonic ratio of a permutation of four pointsA,B,C,Don a line Of course, it suffices to compute what happens for the elementary permutations (permutations of two consecutive elements) in order to be able to infer the result for any one of the 24 possible permutations.

Lemma 6.6.3 Consider four distinct points A,B,C, D on a projective line of

Proof Let us writeρ=(A, B;C, D) By Definition6.6.1there is thus a basise A , e B ofK 2 , withe A on the vector line A,e B on the vector line B,e A +e B on the vector lineCandρe A +e B on the vector lineD.

Starting with the basise B ,e A , we still havee B +e A inC ande B +ρe A inD.

Starting with the basis−e A inA,e A +e B inC, we now have

Projective Transformations

Finally starting withρe A inAande B inB, we haveρe A +e B inDwhile

Let us now investigate what a projective transformation can be The definition of projective spaces (see Definition6.1.1) suggests at once that a projective transformation f: P n (K)−→P m (K), →f () should be the mapping induced at the level of the vector lines ⊆K n + 1 by aK- linear mapping f:K n + 1 −→K m + 1

Of course a linear mapping preserves the proportionality of vectors, thus it maps a vector line onto a vector line, or onto(0)⊆K m + 1 ! So the general definition should be:

Definition 6.7.1 Given a linear mapping f:K n + 1 −→K m + 1 the function f: P n (K)−→P m (K), P →f (P ), P ⊆K n + 1 \Ker f, defined only on those vector linesP not contained in the kernel of the linear mappingf, is called a projective function The projective function is called a projective transformation when it is defined on each projective pointP ∈P n (K).

A projective transformation is thus the case whereKerf =(0), that is, the linear mappingf is injective Of course, two proportional linear functionf andkf, 0 k∈K, define the same projective function.

Example 6.7.2 In the projective planeP2 (K), consider a lined and a pointP /∈d.

For every pointA=P in the projective plane, consider the intersectionp(A)ofd and the unique line throughP andA The function p:P2 (K)−→P2 (K), A→p(A), A=P is a projective function called the central projection ondwith centerP.

Proof Observe first that the statement makes sense, by Theorem6.2.10 The line throughP andAis indeed distinct fromd, sinceP /∈d.

The projective lined corresponds to a vector planeΔ⊆K 3 and the projective pointP corresponds to a vector lineπ⊆K 3 SinceP /∈d, we haveπΔ, thus dim(π∩Δ) dimΔ=2.

This proves that dim(π∩Δ)=0, dim(π+Δ)=3 so thatπandΔare supplementary vector subspaces (see Sect.2.6).

InK n + 1 , consider the projectionponΔparallel toπ(see Sect.2.15) By Propo- sition2.15.2, this is a linear mapping sincep(0)=0 To get the imagep(α)of a vector lineα, through each point ofαwe must draw the affine subspace parallel to πand compute its intersection withΔ In other words,p(α)is the intersection with Δof the vector plane containingπ andα Viewingαas a projective pointA,p(A) is thus the intersection withd of the projective line throughP andA This is the expected central projection

Proposition 6.7.3 Given two systems of homogeneous coordinates inP n (K)and

P m (K), every projective transformation f:P n (K)−→P m (K) can be described by a matrix formula

⎥⎦ whereAis an(n+1)×(n+1)matrix, defined uniquely up to a non-zero multiplicative constant.

Proof Choose a basise 0 , e n ofK n+ 1 ande 0 , , e m ofK m+ 1 inducing the given systems of homogeneous coordinates (see Definition6.4.1) Simply choose forA the matrix, with respect to these bases of a linear mappingf: K n + 1 −→K m + 1 inducing the projective transformationf This proves the existence Of course multiplying the matrixAby a non-zero constant does not change its action at the level of homogeneous coordinates.

Assume now that another matrixBdescribes the same projective transformation f in the same systems of homogeneous coordinates By assumption,

⎥⎦ describe the same projective point of P m (K) This can be rephrased as saying that the first column of A is proportional to the first column of B; notice that these columns are non-zero becausef is a projective transformation (see Defini- tion 6.7.1) Let us write 0=k 0∈K for the proportionality factor Of course an analogous argument holds for the other columns.

It remains to show that the same proportionality factor applies to all the columns. Again sincef is a projective transformation, the homogeneous coordinates

⎥⎦ describe the same projective point ofP m (K) This can be rephrased by saying that the sum of the all the columns ofAis proportional to the sum of all the columns of

B; writek=0 for this proportionality factor We thus obtain, for every indexi a i0+ ã ã ã +a in =k(b i 0+ ã ã ã +b in )=k(k 1 a i0+ ã ã ã +k n a i n ).

In other words, the homogeneous system of equations

Once more sincef is a projective transformation, it is induced by an injective linear mapping and the matrix Athus has rank n+1 It follows that the dimension of the space of solutions of the homogeneous system above is zero: the number of unknowns minus the rank of the system This proves that 1−kk i =0 for each index i, that is,k i = 1 k for each indexi

Lemma 6.7.4 Letf:P n (K)−→P m (K)be a projective function InP n (K), consider two distinct pointsA,Bsuch thatf (A),f (B)are defined and remain distinct.

Thenf is defined everywhere on the lined throughAand B, is injective on that lined, andf (d)is a projective line inP m (K).

Proof The two distinct vector linesA, B⊆K n + 1 generate a vector planeΠmapped by the linear mappingf: K n + 1 −→K m + 1 on a vector subspacef (Π )of dimension at least 2, just because the vector linesf (A)andf (B)remain distinct Con- sidering the linear mapping f: Π−→f (Π ) we know that dimΠ=dimIm(f )+dimKer(f ).

Since we know already that dimΠ=2, dimIm (f )≥2 it follows that dimIm(f )=2, dimKer(f )=0.

Thus the linear mappingf is injective onΠ, from which the conclusion follows immediately

Here is now a key property of projective functions:

Proposition 6.7.5 Letf:P n (K)−→P m (K)be a projective function InP n (K), consider four distinct pointsA,B,C,Don a lined on whichf is defined and is injective Then f (A), f (B);f (C), f (D)

=(A, B;C, D) that is,f preserves the anharmonic ratios.

Proof By assumption and Lemma 6.7.4, f (A), f (B), f (C), f (D) are distinct points on the projective linef (d)inP m (K) Choose projective bases ofdandf (d) such that (see Theorem6.5.5)

Of course with respect to these projective bases,f admits the identity matrix There- fore - a b

Fig 6.2 and thus by Definition6.6.1

The following corollary will prove to be extremely useful:

Corollary 6.7.6 In the projective planeP2 (K), consider two distinct linesd and d and a pointP which is on neither of these Then the projection ofd ond, with centerP, respects the anharmonic ratios.

Proof The projection ond with centerP is defined everywhere, except onP (see Example6.7.2) Thus it yields by restriction a projective transformation p:d −→d to which it remains to apply Proposition6.7.5

As a consequence, let us observe the key role played by central projections.

Proposition 6.7.7 In the projective planeP2 (K), letf: d−→d be a projective transformation between two distinct lines Thenf is a composite of central projections.

Proof Every field has at least two elements 0=1, thus every projective line has at least three points by Proposition6.4.4 WriteP for the intersection ofdandd

Iff (P )=P, choose two other pointsA=Bondand writeSfor the intersection of the two linesd A.f (A) andd B.f (B) (see Fig.6.2) For every other pointC ond, writeC for its projection on d with center S The quadruple(P , A, B, C)on d is then transformed by that central projection into(p=f (P ), f (A), f (B), C ) By

Proposition6.7.5and Corollary6.7.6, we have the equality of anharmonic ratios f (P ), f (A);f (B), f (C)

Fig 6.3 proving thatC =f (C) Thusf is (the restriction of) the central projection with centerS.

Iff (P )=P, consider the two points f (P )∈d andf − 1 (P )∈d, which are thus both distinct fromP Fix a pointA∈ddistinct fromP andf − 1 (P )and con- siderf (A)∈d which is thus distinct fromP andf (P ) Write furtherA for the intersection ofd A.f (A) andd f (P ).f − 1 (P ) For every other pointBond, writeB for the intersection ofd B.f (A) andd f (P ).f − 1 (P ) andB for the intersection ofd and d AB (see Fig.6.3) The projection ofd ond f (P ).f − 1 (P ) with centerf (A) maps the quadruple(P , f − 1 (P ), A, B)onto(f (P ), f − 1 (P ), A , B ) A second projection of d f (P ).f − 1 (P ) on d with center A further transforms this quadruple into

(f (P ), P , f (A), B ) By Proposition6.7.5and Corollary6.7.6, we have the equality of anharmonic ratios f (P ), P;f (A), f (B) f (P ), f f − 1 (P )

P , f − 1 (P );A, B f (P ), P;f (A), B proving thatB =f (B) Thusf is indeed obtained as a composite of the two central projections indicated

The proof of Proposition6.7.7shows that (in general) the decomposition of the projective transformationf as a composite of central projections is by no means unique: in the second part of the proof, the decomposition depends on the choice of the pointA.

Desargues’ Theorem

We have now set out all the useful techniques which enable us to elegantly prove results in the projective plane over a fieldK Our first example will be Desargues’ theorem, already considered (in a different context) in Sect 6.4 of [7], Trilogy I.

Definition 6.8.1 By a triangle in a projective planeP2 (K)is meant a set{A, B, C} of three distinct points, not on the same line The three lines joining each pair of points are called the sides of the triangle.

Observe at once that the notion of triangle is “self-dual”:

Lemma 6.8.2 In a projective planeP2 (K), three lines not having a common point of intersection are always the sides of a unique triangle.

Proof By Theorem6.2.10, the three linesd,d d intersect pairwise in three points

Observe thatA=Bwould mean that this point is on the three linesd,d ,d , which is not the case by assumption Thus the three intersection points are distinct.

On the other hand ifA,B,Cwere on the same lineδ, the two linesd andδwould pass throughAandB, thus would be equal, again by Theorem6.2.10 Eventually one would haved=d =d =δwhich again contradicts the assumptions.

Of course,d,d ,d are the sides of the necessarily unique triangle{A, B, C}.

For simplicity, let us writed AB for the projective line passing through two distinct pointsAandB(see Theorem6.2.10).

Theorem 6.8.3 (Desargues’ Theorem) LetK be a field In the projective plane

P2 (K), consider six distinct points A, B, C, A , B , C constituting two triangles

{A, B, C}and{A , B , C } Suppose that the linesd AA ,d BB andd CC are distinct and intersect at some pointP Under these conditions, the three points

L=d BC ∩d B C , M=d AC ∩d A C , N=d AB ∩d A B are on the same line (see Fig.6.4).

Proof By assumption, the points A, B, C are not on the same line (see Defini- tion6.8.1) On the other handP,A,Bcannot possibly be on the same line, otherwise this line would also containA (which is on the lined P A ) andB (which is on the lined P B ); this would implyd AB =d A B , which is excluded by assumption An analogous argument shows that, respectively,P,A,C andP,B,Care not on the same line Thus three of the four pointsP,A,B,Care never on the same line, proving that(A, B, C;P )is a projective basis of the projective plane (see Example6.5.3

Fig 6.4 and Definition6.5.8) With respect to this projective basis

⎦ for the homogeneous coordinates of a point with respect to this projective basis.

The pointA is on the lined P A , thus its homogeneous coordinates are a linear combination of the homogeneous coordinates ofP andA

SinceA =A, we haveα=0 and sinceA =P,a=0 Since homogeneous coordinates are defined up to a non-zero multiple, there is no loss of generality in choosing α=1 The same argument applies toB andC , proving that the homogeneous coordinates of theses points are

The homogeneous coordinates of the pointsAandB are such thatX 2=0 By Corollary6.4.3,X 2=0 is a homogeneous equation of a projective line and is thus the homogeneous equation of the lined AB Analogously, the homogeneous equation ofd AC isX 1=0 and that ofd BC isX 0=0.

A point ofd A B has homogeneous coordinates α

This point lies ond AB when its last coordinate is zero, that is, whenβ= −α The pointN=d AB ∩d A B thus has the homogeneous coordinates

An analogous argument holds forLandM, proving that

The three projective pointsL,M,N are on the same projective line when the corresponding vectors ofK 3 are linearly dependent, that is, when the determinant det

⎠=0 is equal to zero This is easily seen

Applying the duality principle (see Theorem6.3.5), with Lemma6.8.2in mind, we obtain further:

Theorem 6.8.4 In a projective plane, consider six distinct linesa, b, c, a , b , c such that{a, b, c}and{a , b , c }constitute the sides of two triangles Suppose that the pointsa∩a ,b∩b ,c∩c are distinct and belong to a linep Under these conditions, the three lines l=d b ∩ c,b ∩ c , m=d a ∩ c,a ∩ c , n=d a ∩ b,a ∩ b intersect at the same point (see Fig.6.5).

Figures6.4and6.5have intentionally been drawn identically to stress the fact that putting together Desargues’ Theorem and its dual statement, we obtain an equivalence:

Pappus’ Theorem

Theorem 6.8.5 In the projective plane, consider six distinct points constituting two triangles{A, B, C}and{A , B , C } The following conditions are equivalent:

1 the three lines d AA , d BB , d CC are distinct and intersect at the same pointP;

L=d BC ∩d B C , M=d AC ∩d A C , N=d AB ∩d A B are distinct and lie on the same linep.

Pappus’ Theorem is another fundamental result of projective geometry, already considered in a different context, in Sect 6.4 of [7], Trilogy I.

Theorem 6.9.1 (Pappus) In the projective planeP2 (K) over a fieldK, consider three distinct pointsA,B,C on a lined and three other distinct pointsA ,B ,C on another lined Under these conditions, the three points

L=d BC ∩d B C , M=d AC ∩d A C , N=d AB ∩d A B are well-defined and lie on the same line (see Fig.6.6).

Proof Notice first that the two linesd BC andd B C are distinct Indeedd BC =d B C would imply that the four pointsB, C , B , Care on the same line Sinced=d BC andd =d B C , this would forced=d , which is not the case This shows thatLis correctly defined, by Theorem6.2.10 An analogous argument holds forY andZ.

Still by Theorem6.2.10, letP be the intersection point of the two linesd andd

IfP=A, thend AB =d =d AC , thus

Then of courseLandM=N lie on the same line Thus the only relevant case is that where the seven pointsP,A,B,C,A ,B ,C are distinct This is what we assume from now on.

In the proof, we shall also consider four additional points, namely

Again these points are correctly defined: for exampled AC =d A B would imply that

A,C ,A ,Bare on the same line, thusd=d And so on.

Now observe that(A, C, B ;M)is a projective basis (see Definition6.5.8) Cer- tainly the three points A, C, B are not on the same line because B is not on d AC =B AnalogouslyMis not ond AC , otherwise one would haveA ∈d AC =d, and so on With respect to this projective basis, we thus obtain the homogeneous coordinates

⎦ for the homogeneous coordinates of a point with respect to this projective basis To complete the proof, we now have to compute the coordinates ofLandN.

The pointS is the intersection of the lined CB , with equation X 0=0, and the lined AM , with equation X 1=X 2 Since homogeneous coordinates are never the zero triple and are defined up to a non-zero multiplicative constant,

Analogously the pointRis the intersection of the lined AB , with equationX 1 =0, and the lined CM , with equationX 0 =X 2 Thus

6.9 Pappus’ Theorem 221 Let us now consider the lined CB The three points

⎦ are distinct, thus constitute a projective basis ofd CB The pointLis on this line and is distinct fromC, otherwiseC would be ond Thus the homogeneous coordinates ofLhave a non-zero last component, which we can therefore choose to be equal to 1; finally

Considering only the one-dimensional projective spaced CB and its projective basis

By Definition6.6.1,ρis the anharmonic ratio ρ C, B ;S, L

Anharmonic ratios are preserved by central projections, as attested by Corol- lary6.7.6 Three consecutive projections with respective centersC ,B,A , followed by successive applications of Lemma6.6.3, then yield

In other words, we have proved that ρ= τ τ−1.

The collinearity of the three pointsL,M,N thus reduces to the nullity of the determinant det

⎠ which is straightforward to check

Applying the duality principle (see Theorem6.3.5) to Pappus’ theorem, we obtain:

Theorem 6.9.2 In the projective planeP2 (K), consider three distinct linesa,b, cintersecting at a pointDand three other distinct linesa ,b ,c intersecting at another pointD Under these conditions, the three lines l=d b ∩ c ,b ∩ c , m=d a ∩ c ,a ∩ c , n=d a ∩ b ,a ∩ b are well-defined and intersect at the same point (see Fig.6.7).

Fano’s Theorem

A “good” theory of quadrilaterals in a projective plane requires us to restrict our attention to fields of characteristic distinct from 2: fields where 1+1=0.

Definition 6.10.1 By a quadrilateral in a projective plane P2 (K) is meant a quadruple(A, B, C, D) of distinct points such that none of them belongs to the line passing through two other points (see Fig.6.8).

• The four pointsA,B,C,Dare called the vertices of the quadrilateral.

• The linesd AB ,d BC ,d CD andd DA are called the sides of the quadrilateral.

• The pairs(d AB , d CD )and(d AD , d BC )are called the pairs of opposite sides.

• The linesd AC andd BD are called the diagonals of the quadrilateral.

• The pointsL=d AB ∩d CD ,M=d AC ∩d BD ,N =d AD ∩d BC are called the diagonal points of the quadrilateral.

Definition6.10.1 of a quadrilateral is thus exactly the same as that of a projective basis (see Definition6.5.8) In this definition, working with a quadruple

(A, B, C, D)and not just with a set{A, B, C, D}is what allows us to make the dis- tinction between sides and diagonals Certainly, a more “refined” definition would have been to define a quadrilateral as an equivalence class of such quadruples, where one identifies(A, B, C, D),(D, C, B, A)and all the cyclic permutations of these. This is unessential for our purposes.

Proposition 6.10.2 In a projective planeP2 (K)over a fieldK, the following conditions are equivalent:

2 there exists a quadrilateral whose diagonal points are collinear;

3 in every quadrilateral, the diagonal points are collinear.

Proof As we have just observed, the vertices of an arbitrary quadrilateral constitute a projective basis The corresponding homogeneous coordinates are thus (see Theorem6.5.5)

⎦ for the homogeneous coordinates of a point.

Let us next compute the coordinates of the three diagonal points (see Fig.6.8):

L=d AB ∩d CD , M=d AC ∩d BD , N=d AD ∩d BC

The lined AB admits the equationX 2=0 and the lined CD ,X 0=X 1 The pointL thus has the coordinates

The three pointsL,M,N are collinear if and only if the determinant det

⎠= −2 is equal to zero, that is, if 2=0 inK

Contraposing Proposition6.10.2yields the classical form of Fano’s theorem:

Theorem 6.10.3 (Fano’s theorem) In a projective planeP2 (K)over a field of characteristic distinct from 2, the diagonal points of a quadrilateral are never on the same line.

Harmonic Quadruples

As noted in Sect.6.6, an anharmonic ratio never takes the values 0 or 1 The case where it takes the value−1 will prove to be quite important (see also Sects 6.3 and 4.9 in [7], Trilogy I) Of course for this to make sense, it is compulsory that

−1=1, since an a harmonic ratio never takes the value 1 In other words, we must have 2=0.

Definition 6.11.1 LetK be a field of characteristic distinct from 2 A quadruple

(A, B, C, D)of distinct points on a projective line inP n (K)is said to constitute a harmonic quadruple when the anharmonic ratio(A, B;C, D)is equal to−1. The importance of harmonic quadruples comes from the following result.

Proposition 6.11.2 In the projective planeP2 (K)over a fieldKof characteristic distinct from 2, consider a quadrilateral(A, B, C, D) With the notation of Fig.6.9, writeSandT for the intersections of the two diagonalsd AC andd BD with the line d LN joining the intersectionsLandN of the pairs of opposite sides The quadruple (L, N;S, T )is harmonic.

Proof By Fano’s Theorem6.10.3, the pointsL,M,Nare not on the same line Thus the linesd LN ,d AC andd BD are distinct and the four pointsL,N,S,T are correctly defined and distinct.

Anharmonic ratios are preserved by central projections, as attested by Corol- lary6.7.6 Four consecutive projections with respective centers A, L,N,A, followed by an application of Lemma6.6.3, then yield

In other words, we have proved that

Thus(L, N;S, T )∈Kis a root of the equationX 2 =1.

X 2 −1=(X−1)(X+1)=0 admits exactly the two roots+1= −1 in the fieldK(see CorollaryA.6.7) Since we know (see Sect.6.6) that(L, N;S, T )is never equal to+1, it is necessarily equal to−1.

The Axioms of Projective Geometry

Just to exhibit the link with axiomatic projective geometry, this short section recalls some results established in Chap 6 of [7], Trilogy I.

Definition 6.12.1 An abstract projective plane is a pair(π,L)where:

• πis a set whose elements are called “points”;

• Lis a family of subsets ofπwhole elements are called “lines”.

These data satisfy the following axioms.

P1 Each line contains at least three points.

P2 There exist three points not on the same line.

P3 Two distinct points belong to a unique line.

P4 Two distinct lines have at least one common point.

For an abstract projective plane, we have considered further the two axioms (see Sect 6.6 in [7], Trilogy I):

P5 Desargues’ axiom (the statement of Theorem6.8.3).

P6 Pappus’ axiom (the statement of Theorem6.9.1).

Theorem 6.6.5 in [7], Trilogy I attests that:

Theorem 6.12.2 In an abstract projective plane, Pappus’ axiom implies Desar- gues’ axiom.

The link with the present chapter is then given by Theorem 6.8.3 in [7], Trilogy I: Theorem 6.12.3 An abstract projective plane satisfies Pappus’ axiom if and only if it is the projective planeP2 (K)over a fieldK.

Projective Quadrics

Proposition 6.12.4 The abstract projective planes satisfying Pappus’ axiom P6 and Fano’s axiom P7

P7 The diagonal points of a quadrilateral are never collinear. are precisely the projective planesP2 (K)over a fieldK of characteristic distinct from 2.

We have also seen in Sect 6.7 of [7], Trilogy I, that the theory of projective spacesP n (K)over a skew fieldK can be developed as well Of course it is based on the less familiar theory of vector spaces over a skew field Theorem 6.2.8 in [7], Trilogy I, tells us that:

Theorem 6.12.5 An abstract projective plane satisfies Desargues’ axiom if and only if it is the projective planeP2 (K)over a skew fieldK.

As in the affine case, let us make the following convention in the study of quadrics:

Convention Convention In all sections concerning projective quadrics,Kis a field of characteristic distinct from 2.

Of course, a projective quadric inP n (K)should be a subsetQ⊆P n (K)which, in a system of homogeneous coordinates, can be described by an equation of degree 2: n i,j = 0 a ij X i X j + n i = 0 b i X i +c=0.

But for this to make sense, the quadric inK n + 1 given by the equation above must be entirely constituted of projective points, that is, of vector lines Thus when an

(n+1)-tuple of coordinates satisfies the equation, every multiple of it must satisfy the equation as well (see Sect.6.4).

We are essentially interested in the study of non-degenerate quadrics: a degenerate quadric is better studied as a non-degenerate quadric in a convenient subspace (see Definition2.23.4) But a non-degenerate quadric inK n + 1 , entirely constituted of vector lines, admits the origin as center of symmetry belonging to the quadric By Proposition2.25.3the equation does not contain any term of degree 1 and of course c=0, since the origin belongs to the quadric.

So, given a basise 0 , , e n of K n + 1 , we are interested in the equations of the form n i = 0 a ij X i X j =0.

In the corresponding system of homogeneous coordinates inP n (K), this will therefore be the equation of a projective quadricQ⊆P n (K) As observed in Sect.2.23, when the characteristic ofK is distinct from 2, there is no loss of generality in assuming that the matrixA=(a ij ) i,j is symmetric Then the mapping

K n + 1 ×K n + 1 −→K, (x, y)→x t Ay is a symmetric bilinear form with corresponding quadratic form (see Defini- tionG.1.1) φ:K n + 1 −→K, v→φ (v)=ϕ(v, v).

A vectorv∈K n + 1 is on the quadric precisely whenϕ(v, v)=φ (v)=0, that is, whenvis a so-called isotropic vector of the bilinear formϕ(see DefinitionG.2.1). This discussion shows that in the projective case, quadrics can be defined at once in terms of quadratic forms, without any reference to a particular system of homogeneous coordinates This immediately makes the theory of projective quadrics much more “conceptual” than the theory of affine quadrics.

Convention 6.13.1 Given a symmetric bilinear formϕ onK n + 1 and the corresponding quadratic formφ (v)=ϕ(v, v), we shall freely write ϕ(P , Q)=0, φ (P )=0, P , Q∈P n (K) instead of

Of course this convention cannot possibly hurt since by bilinearity of ϕ, ϕ(v, w)=0 immediately implies ϕ(kv, k w)=0 for any two scalars k, k ∈K.

With this convention in mind we make the following definition.

Definition 6.13.2 By a quadricQin the projective spaceP n (K)overK is meant a set

⊆P n (K) whereφis some fixed non-zero quadratic form onK n + 1

Proposition 6.13.3 With respect to every projective basis, a quadric of P n (K)is described by a homogeneous equation of degree 2 n i,j = 0 a i j X i X j =0 whereA=(a ij ) i,j is a symmetric matrix.

Proof With the notation of Definition6.13.2, simply choose forAthe matrix ofφ with respect to a basis(e 0 , , e n )ofK n + 1 which generates the system of homogeneous coordinates of the projective basis (see Sect.6.5) Let us at once emphasize the fact that:

Proposition 6.13.4 For every quadric of P n (K), there exists a projective basis with respect to which the quadric admits a diagonal equation n i = 0 a i X 2 i =0.

Proof By CorollaryG.2.8, there exists a basise 0 , , e n of K n + 1 with respect to which the matrix of the bilinear form defining the quadric is diagonal Putting e n + 1 =e 0 + ã ã ã +e n and considering the vector lines generated bye 0 , , e n + 1yields the expected projective basis

The following definition is certainly the expected one:

Definition 6.13.5 Two quadricsQ,Q ofP n (K)are projectively equivalent when there exists a projective isomorphismf:P n (K)−→P n (K)such thatf (Q)=Q

Of course by a projective isomorphism is meant a projective mapping induced by a linear isomorphism onK n + 1 (see Definition 6.7.1) In the spirit of Defini- tions2.23.4andG.1.5, let us define further:

Definition 6.13.6 Under the conditions of Definition6.13.2:

• the quadricQis non-degenerate ifQis not the whole spaceP n (K) and is not contained in a projective subspace of dimensionn−1;

• the quadricQis regular when it can be described by a regular quadratic formφ.

The “opposite” notions will be called degenerate and singular quadric, respectively.

Proposition 6.13.7 Two quadratic forms onK n + 1 which determine the same non- degenerate quadric ofP n (K)are necessarily proportional In particular, they are either both regular or both singular.

Proof This follows at once from Theorem 2.26.9 applied to the corresponding quadric ofK n + 1

In an arbitrary projective spaceP n (K):

Counterexample 6.13.8 For a projective quadric, neither of the two properties

“non-degenerate” and “regular” implies the other one.

Proof OnR 3 , the quadratic form φ (X 1 , X 2 , X 3 )=X 1 2 +X 2 2 +X 2 3 is certainly regular, since its matrix is the identity matrix But the corresponding quadric inP2 (R)is the empty set, sinceφvanishes only on(0,0,0).

OnR 3 , the quadratic form φ (X 1 , X 2 , X 3 )=X 2 1 −X 2 2 is not regular, since its matrix with respect to the canonical basis is

The corresponding projective quadric is not the whole space, sinceφ (1,0,0)=0. But it contains the three projectively independent points with homogeneous coordinates

Thus the quadric is non-degenerate

Nevertheless, in the cases of most interest in this book, “regularity” implies “non- degeneracy”:

Proposition 6.13.9 WhenKis an algebraically closed field, every regular quadric ofP n (K)is non-degenerate.

Proof By Proposition 6.13.4, there is no loss of generality in assuming that the quadric is given by a diagonal equation n i = 0 a i X i 2 =0; by regularity,a i =0 for each indexi.

SinceKis algebraically closed, the equation a 0 +a 1 X 2 1 =0

Duality with Respect to a Quadric

− a a 0 1 This means that the pointsP 1andP 1 with homogeneous coordinates

⎦ are distinct points on the quadric The same argument holds for any other index i=0, yielding n points P 1 , , P n which are trivially projectively independent Again trivially,P i cannot be a linear combination ofP 1 , , P n Thus the n+1 pointsP 1 , P 1 , , P n are projectively independent and the quadric is non- degenerate

6.14 Duality with Respect to a Quadric

Convention Convention In this section,Kis a field of characteristic distinct from

2 and equation of a quadric always means homogeneous equation of degree 2.

Theorem6.3.5establishes the duality principle for projective planes Very roughly speaking, this principle tells us that we can interchange the words “point” and

“line” in a statement However, given a point in the projective plane, the principle does not state that there is a precise line corresponding to that point by duality Given a projective point, that is a vector line in a three dimensional vector spaceV, the duality principle instead associates with a vector plane ⊥ in the dual vector space V ∗ To switch back to a vector plane in V, that is a projective line in the projective plane, we have to fix an isomorphism betweenV andV ∗ As we have seen, there is no such canonical isomorphism, but there is one for each choice of a basis inV Analogous comments hold in higher dimensions.

The consideration of a regular non-degenerate quadric in a projective space will force the validity of a much more precise duality principle This time, given a specified subspace, the principle will identify the precise subspace which corresponds to it by duality Moreover, the fact of being a point of the corresponding quadric will also admit an interesting dual notion, namely, being a tangent space to the quadric (see Sect.6.16).

With the notation of Convention6.13.1we define first:

Definition 6.14.1 LetQ⊆P n (K)be a regular non-degenerate quadric Given a projective subspaceX⊆P n (K), the conjugate ofX with respect to the quadricQ is the subspace

X ⊥ Q∈P n (K)∀P ∈X ϕ(P , Q)=0 whereϕis the symmetric bilinear form corresponding to the quadratic formφdefin- ing the quadric (see Definition6.13.2).

Notice that Definition 6.14.1 makes perfect sense By non-degeneracy, two quadratic forms defining Q are necessarily proportional (see Proposition6.13.7) thus by LemmaG.1.2, determine proportional symmetric bilinear forms Of course, proportional bilinear forms define the same notion of conjugation.

Proposition 6.14.2 Under the conditions of Definition6.14.1, ifX is a projective subspace of dimensionk,X ⊥ is a projective subspace of dimensionn−k−1 and

Proof The projective subspaceX corresponds to a vector subspace of dimension k+1 (see Definition6.2.1) By PropositionG.2.4,X ⊥ is a projective subspace corresponding to a vector subspace of dimension(n+1)−(k+1)=n−k Thus it is a projective subspace of dimensionn−k−1 The result follows by CorollaryG.2.5.

Theorem 6.14.3 (Duality principle with respect to a quadric) LetQ⊆P n (K)be a regular non-degenerate quadric In the projective spaceP n (K), consider a state- mentθ which expresses some inclusions, equalities or inequalities between some specified projective subspaces of prescribed dimensions Writeθ ∗ for the dual statement obtained fromθby replacing each projective subspaceXby its conjugateX ⊥ , by reversing all the inclusions and by replacing each dimensionk of a projective subspace by the dimensionn−k−1 Ifθis valid, thenθ ∗ is valid as well.

Proof In view of Proposition6.14.2, it remains to observe that given two projective subspacesX⊆Y, thenY ⊥ ⊆X ⊥ This is trivial.

Poles and Polar Hyperplanes

Proposition 6.15.1 LetQ⊆P n (K)be a regular non-degenerate quadric The corresponding operation of conjugationX→X ⊥ (see Definition6.14.1) defines a bijection between the points and the hyperplanes ofP n (K).

Proof This follows at once from Proposition6.14.2 A point has dimension 0, thus its conjugate has dimensionn−1; analogously a hyperplane has dimensionn−1,

6.15 Poles and Polar Hyperplanes 233 thus its conjugate has dimension 0 Since moreoverX ⊥⊥ =X for every projective subspace, we have indeed described inverse bijections

It will be convenient to use the following classical terminology.

Definition 6.15.2 LetQ⊆P n (K)be a regular non-degenerate quadric.

• Given a pointP ∈P n (K), the hyperplaneP ⊥ is called the polar hyperplane ofP.

• Given a hyperplaneH⊆P n (K), the pointH ⊥ is called the pole ofH.

InP2 (K), a polar hyperplane is more commonly called a polar line.

Lemma 6.15.3 InP n (K), if a line and a quadric have three common points, the line is entirely contained in the quadric.

Proof LetR,S,T be three distinct points of intersection between a quadricQand a lined Consider a projective basis of the form

(P 0=R, P 1=S, P 2 , , P n ;P n+ 1 ) and the corresponding equation of the quadric (see Proposition6.13.3) n i,j = 0 a ij X i X j =0.

The three pointsR,S,T have coordinates, withu, v=0,

⎥⎦ whileX, with of course(x 0 , x 1 )=(0,0), is the general form of a point ofd Since

R andS satisfy the equation, we havea 00 =0 anda 11 =0 SinceT satisfies the equation, we obtain further a 01 uv+a 10 uv=0.

This forcesa 01 +a 10 =0 sinceu, v=0 By symmetry of the matrixA, this reduces toa 01 =0=a 10 It follows at once that the coordinates ofXsatisfy the equation of the quadric

Let us now exhibit the following very important link between the anharmonic ratio and conjugation with respect to a quadric.

Theorem 6.15.4 LetQ⊆P n (K)be a regular non-degenerate quadric Consider a pointP /∈Qand a lined throughP having exactly two intersection pointsR,S with the quadric The lined intersects the polar hyperplane ofP in a unique point

Qand the quadruple(R, S;P , Q)is harmonic (see Definition6.11.1).

Proof On the lined, we have a system of homogeneous coordinates such that

(see Example6.5.6) Write (e 0 , e 1 )for the two vectors of K n + 1 which generate this system Let us extend(e 0 , e 1 )to a basis(e 0 , , e n )ofK n + 1 and consider the corresponding system of homogeneous coordinates inP n (K) We obtain

The equation of the quadric has the form n i,j = 0 a ij X i X j =0.

SinceR∈Q, we havea 00 =0 and sinceS∈Q, we geta 11 =0 as well Saying that

Q∈P ⊥ meansϕ(P , Q)=0, withϕthe symmetric bilinear form which determines the quadric This means

On the other hand sinceP /∈Q, we have a 10 +a 01 =0.

By symmetry of the matrixA, we conclude thata 01=a 10=0 and thusx 0+x 1=0,that is,x 1= −x 0.

Tangent Space to a Quadric

On the lineD, we thus have a projective basis(R, S;P )yielding the homogeneous coordinates

We recall once more our

The notion of polar hyperplane (see Definition6.15.2) yields in particular the notion of tangent space to a quadric For this we observe first that:

Lemma 6.16.1 LetQ⊆P n (K)be a regular non-degenerate quadric Given a point

Aof the quadricQ, the polar hyperplaneA ⊥ contains that pointAand, if it contains a second pointB of the quadric, the line throughAandB is entirely contained in the quadric Moreover,A ⊥ is the unique hyperplane with that property.

Proof Write φ for the quadratic form on K n + 1 defining the quadric and ϕ for the corresponding symmetric bilinear form By definition of a quadric (see Defini- tion6.13.2), we haveφ (A)=ϕ(A, A)=0 for a point of the quadric, which means

A∈A ⊥ (see Definition6.14.1) If moreoverB∈A ⊥ is such thatB∈Q, we have ϕ(A, A)=0, ϕ(A, B)=0, ϕ(B, B)=0.

Fork, k ∈K, and using obvious notation, the bilinearity ofϕthen forcesϕ(A, kA+ k B)=0 Thus the whole line throughAandB lies inA ⊥

Now let H be another hyperplane with the same property In particular, A∈

H∩Q Fix another point B∈H∩Q Identifying a point and its coordinates in some fixed projective basis, consider the pointX=A+B on the line joiningA andB With the notation of Convention 6.13.1, this point lies on the quadric Q when

Since by assumptionφ (A)=0 andφ (B)=0, this reduces toϕ(A, B)=0 and proves thatB∈A ⊥ ThusH⊆A ⊥ , from which we obtain the equality since both have projective dimensionn−1

The consideration of the affine quadrics ofR 3 (see Sects.1.14and1.15) suggests at once that the property ofA ⊥ emphasized in Lemma6.16.1 is characteristic of what a tangent space should be Therefore we define:

Definition 6.16.2 LetQ⊆P n (K)be a regular non-degenerate quadric Given a pointAof the quadricQ, the polar hyperplaneA ⊥ is called the tangent hyperplane to the quadric.

Observe that the duality principle6.14.3can then been made more explicit by stating it as:

Theorem 6.16.3 (Duality principle with respect to a quadric) LetQ⊆P n (K)be a regular non-degenerate quadric In the projective spaceP n (K), consider a state- mentθ which expresses some inclusions, equalities or inequalities between some specified projective subspaces of prescribed dimensions Writeθ ∗ for the dual statement obtained fromθby replacing each projective subspaceXby its conjugateX ⊥ , by interchanging the locutions point of the quadric and tangent hyperplane to the quadric, by reversing all the inclusions, and by replacing each dimensionk of a projective subspace by the dimensionn−k−1 Ifθ is valid, thenθ ∗ is valid as well.

Projective Conics

We continue to follow our

Let us now pay attention to the special case of quadrics in the projective plane, where the following terminology is more commonly used:

Definition 6.17.1 By a projective conic is meant a projective quadric in a projective planeP2 (K).

Proposition 6.17.2 For a non-degenerate projective conicQ, the following conditions are equivalent:

2 the conic does not contain any line;

3 three points of the conic are never on the same line;

4 in an equationp(X 0 , X 1 , X 2 )=0 of the conic, the polynomialp(X 0 , X 1 , X 2 )is irreducible, as well as the conditions:

2 the conic is the union of two distinct lines.

Proof Notice at once that a line has at least three points, because a field has at least two elements 0 and 1 (see Proposition 6.4.4) Together with Lemma 6.15.3, this proves the equivalence of conditions 2 and 3 in the first set of equivalences.

If a lined is contained in the conicQ, consider a projective basis of the form

(P 0 , P 1 , P 2;P 3 )withP 0andP 1ond The three points

⎦ are ond, thus on the conic Introducing their coordinates into the equation n i,j = 1 a ij X i X j =0 we get a 00 =0, a 11 =0, a 10 +a 01 =2a 01=2a 10=0.

The matrix of coefficients then has the form

⎠ and is thus singular This proves(1⇒2)in the first set of equivalences and(2⇒1) in the second set of equivalences.

Next let us assume that the conic is singular and let us choose a projective basis

(P 0 , P 1 , P 2;P 3 )in which its equation is diagonal (see Proposition6.13.4) Since the corresponding diagonal matrix is singular, one of its diagonal entries is zero Up to possibly renumbering the variables, the equation of the conic thus has the form a 1 X 1 2 +a 2 X 2 2 =0.

If one had furthera 1 =0, then the equation of the conic would reduce toX 2 2 =0, which is equivalent toX 2 =0, the equation of a line This is impossible, because the conic is non-degenerate Thus both coefficients are non-zero; there is no loss of generality in assuming thata 1=1.

Trivially, the point P 0 satisfies the equation of the conic By non-degeneracy, there are also other pointsQon the conic:

Sincea 1 =1 anda 2 =0, necessarily the two coordinatesv,wof such a pointQare both non-zero Thus there is no loss of generality in assuming thatw=1 One then has, sinceQ∈Q, v 2 +a 2 =0 that is v=√

Furthermore, since that particular square root turns out to exist

−a 2 X 2 and the conic is the union of the two distinct lines with equations

This proves(1⇒2)in the second set of equivalences.

This implication proves in particular that when the conic is singular, it contains a line By contraposition, if the conic does not contain any line, it is not singular This proves(2⇒1)in the first set of equivalences.

Finally consider an equationp(X 0 , X 1 , X 2 )=0 of the quadric in an arbitrary basis (see Proposition6.13.7) If the polynomialpis reducible, it is the product of two polynomials of degree 1 and the quadric is the union of the corresponding lines. Conversely if the conic is the union of two lines, its equation is given by the product of the corresponding equations of degree 1

Next, we investigate the case where the lines that we consider are tangent to the conic.

Proposition 6.17.3 LetQ⊆P2 (K)be a regular non-degenerate conic.

1 A line is tangent to the conic if and only if it intersects the conic at a unique point.

2 A line non-tangent to the conic intersects the conic at zero or two points.

3 Through a point not on the conic pass zero or two tangents to the conic.

Proof Statement 1 follows from Lemma6.16.1and Proposition6.17.2 Therefore if a line is not tangent to the conic and has a first point of intersection with it, it must necessarily have a second point of intersection By Proposition6.17.2, there cannot be a third point of intersection This proves Statement 2 and also Statement 3, which is the dual statement (see Theorem6.16.3)

Proposition 6.17.4 LetQbe a non-degenerate regular conic inP2 (K) Given two distinct pointsA, B∈Q:

1 the two tangentsaandbat these points are distinct;

2 the intersection pointP ofaandbis the pole of the linepthroughAandB;

3 every other line throughP intersecting the conic at a pointR intersects it at a second pointS;

4 writing Q for the intersection of the lines d P R and d AB , the quadruple

Proof The dual statement (see Theorem6.16.3) of the points A and B are distinct is precisely the tangents a and b are distinct.

The dual statement of the point P is incident to the lines a = A ⊥ and b = B ⊥ is precisely the line p = P ⊥ is incident to the points a ⊥ = A ⊥⊥ = A and b ⊥ = B ⊥⊥ = N.

By Proposition6.17.3.3, the line through P andR is not tangent to the conic, becausea andbare already tangents, from which we infer the existence of a second pointS of intersection, by Proposition 6.17.3.2 The result follows by Theo- rem6.15.4

Let us observe now the existence of “enough” non-degenerate regular conics.

Proposition 6.17.5 In a projective plane, five points, no three of which are on the same line, are contained in a unique conic That conic is regular and non- degenerate.

Proof Given the five pointsA,B,C,D,E, choose a projective basis such that

(see Example6.5.7) Notice thatu=0 would imply thatEis on the line throughB andC, which is not the case Thusu=0 and analogously,v=0,w=0 Moreover u=vwould imply thatEis on the line throughCandD, which is not the case So u=vand analogously,u=wandv=w.

We have to prove the existence—up to a non-zero multiplicative constant—of a unique non-trivial equation

2 i,j = 0 a ij X i X j =0 satisfied by the coordinates of the five points Introducing the coordinates ofA,B,

Cin this equality forces a 00 =0, a 11 =0, a 22 =0 so that by symmetry of the matrix(a ij ) ij we must in fact look for an equation of the form a 01 X 0 X 1+a 02 X 0 X 2+a 12 X 1 X 2=0.

Requiring further that the coordinates ofDandEsatisfy this equation means a 01+a 02+a 12=0 a 01 uv+a 02 uw+a 12 vw=0.

Viewing this system as a system of equations in the unknownsa 01,a 02anda 12, we must thus prove that the vector space of solutions has dimension 1 Since there are three unknowns, this reduces to proving that the matrix of coefficients has rank 2, that is

Sinceu, v, w=0, there is no loss of generality in puttingw=1, yielding the requirement

This is the case sinceu=v.

There is thus a unique conic containing the pointsA,B,C,D,E Since these points are not on the same line, the conic is non-degenerate By Proposition6.17.2, the conic cannot be singular, otherwise it would be the union of two lines and thus at least three of the given points would be on one of these two lines The number “5” in Proposition6.17.5cannot be improved:

Counterexample 6.17.6 Six points inP2 (C), no three of them on the same line, which are not contained in any conic.

Proof We consider the projective conicQwith equation

No three of the following points are on the same line:

The first five of these points lie onQ By Proposition6.17.5,Qis thus the unique conic containing these points It does not contain the last point Observe next that a non-degenerate regular conic always has “enough” points.

Proposition 6.17.7 In a projective planeP2 (K), a non-degenerate regular conic

Qis in bijective correspondence with every projective line.

Proof Let us recall that by Proposition6.4.4, all projective lines are in bijective correspondence withK∪ { }.

Since the conicQis non-degenerate, it contains at least three pointsA,B,Cnot on the same line We shall exhibit inverse bijections f:Q−→d, g:d−→Q between the conic Q and the projective line d passing through B and C (see Fig.6.11).

First, we consider the tangentA ⊥ atA, which is of course distinct fromd, since

A /∈d Therefore there exists a unique intersection point ofd andA ⊥ (see Theo- rem6.2.10) which we callZ.

We definef (A)=Z IfQ=Ais another point of the conic, the lined AQ through

QandAis again distinct from d since it containsA: we define f (Q) to be the unique intersection point ofd andd AQ

Conversely we defineg(Z)=A IfP ∈d=Zis another point ofd, the lined AP throughP andAis not the tangent atA, thus by Proposition6.17.3it cuts the conic

Qat a second point which we define to beg(P ).

Trivially,f andgare the inverses of each other

The bijection in Proposition6.17.7relies heavily on both assumptions of non- degeneracy and regularity Indeed Counterexample6.13.8provides an example of an empty (thus degenerate) regular conic On the other hand the union of two distinct lines is a non-degenerate singular conic (see Proposition6.17.2) and when working over a finite field, this is definitely not in bijection with a single projective line.

The Anharmonic Ratio Along a Conic

We continue to adopt our

Central projections preserve the anharmonic ratio (see Corollary6.7.6); it thus makes perfect sense to define, without any restriction on the characteristic of the fieldK:

Definition 6.18.1 In the projective planeP2 (K)over an arbitrary fieldK, consider four distinct linesa,b,c,d passing through a fixed pointP The anharmonic ratio

(a, b;c, d)of these four lines is defined as being the anharmonic ratio(A, B;C, D) of the four corresponding intersection points obtained when cutting these four lines by a line not containingP (see Fig.6.12).

Indeed in Fig.6.12, cutting by another line yields at once

6.18 The Anharmonic Ratio Along a Conic 243

Fig 6.13 by Corollary6.7.6, showing that Definition6.18.1makes perfect sense.

The following result exhibits an important link between conics and anharmonic ratios.

Theorem 6.18.2 In P2 (K), consider a non-degenerate regular conic having at least five points Consider four distinct pointsA,B,C,Dof this conic and a fifth distinct pointP, still on the conic The anharmonic ratio of the four lines

(d P A , d P B ;d P C , d P D ) is independent of the choice of the fifth pointP (see Fig.6.13).

Proof By Proposition6.17.2, three of the five points considered on the quadric cannot be on the same line In particular,(A, B, C;D)is a projective basis (see Exam- ple6.5.7) and with respect to this basis, we get the homogeneous coordinates

Notice thatu=0 sinceP is not on the line joiningBandC; analogously,v=0 and w=0 Moreoveru=vbecauseP is not on the line joiningC andD; analogously u=wandv=w There is of course no loss of generality in assuming thatu=1. Let us now compute the coordinates of the two pointsB andC of intersection of the linesd P B andd P C with the lined AD The lined AD admits the equationX 1=X 2 whileB has coordinates of the form

ThusB ∈d AD meansv+k=w, that is,k=w−v Therefore

(A, B;C, D) A, B ;C , D where the second quadruple is constituted of the four points

⎦ on the line with equationX 1=X 2 On this line, let us now consider the projective basis(A, B ;C ) The corresponding system of homogeneous coordinates is based on two vectors ofK 3 such that e 0 =α

This yields at onceα+β=1 andβw=v, that isβ= w v andα=1− w v Thus e 0 ⎛

We next have to determineγ,δsuch that γ

6.18 The Anharmonic Ratio Along a Conic 245

Our challenge is to prove that this quantity does not depend onP, that is, it does not depend onv andw For this let us consider the equation of the conic in the projective basis(A, B, C;D):

Since the coordinates ofA,B,Csatisfy this equation, we have a 00=a 11=a 22=0.

After division by 2, the equation reduces to a 01 X 0 X 1+a 02 X 0 X 2+a 12 X 1 X 2=0.

SinceDsatisfies this equation, we have further a 01 +a 02 +a 12 =0.

Since not all coefficients are equal to zero, let us assume thata 01=0; there is then no loss of generality in choosinga 01=1 Let us further puta 02=k∈K so that a 12= −1−k The matrix(a ij ) ij thus has the form

Notice that necessarilyk=0 and−1−k=0, otherwise the matrix would be singular Expressing the fact thatP is on the quadric yields

It remains to observe that this implies at once k

1+k and this quantity indeed depends only on the matrix of the quadratic form defining the conic in the projective basis(A, B, C;D)

Theorem6.18.2thus allows the following definition:

Definition 6.18.3 In P2 (K), consider a non-degenerate regular conic having at least five points Consider four distinct pointsA,B,C,D of this conic and a fifth distinct pointP, still on the conic The anharmonic ratio of the four pointsA,B,C,

Dis by definition the anharmonic ratio of the four corresponding lines

(see Fig.6.13) This definition does not depend on the choice ofP.

An alternative definition, which does not require the existence of a fifth point, consists of writing the equation of the conic with respect to the projective basis

(A, B, C;D) As observed in the proof of Theorem6.18.2, it takes the form

As the proof of Theorem6.18.2shows, the anharmonic ratio of the four pointsA,

B,C,Dof the quadric can then equivalently be defined as

However, this cannot take us very far Indeed notice that by Proposition6.17.7, it is only when working over the fieldZ3= {0,1,2}of integers modulo 3 that the conic does not have a fifth point Since in the formula above,k=0 andk= −1=2, then necessarilyk=1 So that in all cases

Thus computing the anharmonic ratio of four points on a non-degenerate regular conic inP2 (Z3 )does not provide any additional information.

The Pascal and Brianchon Theorems

6.19 The Pascal and Brianchon Theorems 247

This section is devoted to proving two famous theorems about projective conics, which—in the real case—were discovered during the 17th century They concern the notion of a hexagon, which as such makes sense without any restriction on the characteristic of the fieldK.

Definition 6.19.1 By a hexagon in a projective planeP2 (K)over an arbitrary field

K is meant a sextuple(A, B, C, D, E, F )of distinct points, such that no three of them are on the same line.

A, B, C, D, E, F are called the vertices of the hexagon.

• The three pairs of vertices

(A, D), (B, E), (C, F ) are called the pairs of opposite vertices.

• The six lines d AB , d BC , d CD , d DE , d EF , d F A are called the sides of the hexagon.

• The three pairs of sides

(d AB , d DE ), (d BC , d EF ), (d CD , d F A ) are called the pairs of opposite sides.

• The three lines d AD , d BE , d CF joining opposite vertices are called the diagonals of the hexagon.

Figure6.14presents two pictures of a hexagon and its diagonals, where of course the sides and the diagonals should be full lines.

Theorem 6.19.2 (Pascal) In a projective planeP2 (K), consider a hexagon whose six vertices are on a non-degenerate regular conic The pairs of opposite sides of the hexagon meet respectively in three points which are on the same line.

(See Fig.6.15, which is intentionally slightly distorted in order to make it easier to follow the argument in the proof.)

Proof We keep the notation of Definition6.19.1and write

X=d AB ∩d DE , Y=d BC ∩d EF , Z=d CD ∩d F A

6.19 The Pascal and Brianchon Theorems 249 for the points of intersection of the pairs of opposite sides We must prove thatX,

Y,Zare on the same line By definition of a hexagon, we observe at once that these points are correctly defined and distinct The same straightforward observation also holds for all other points considered in this proof.

To prove the expected result, we consider the intersections

If Pascal’s theorem holds, thenZ =Z=Z Conversely ifZ =Z , then this point is on both linesd CD andd F A , thus it is the pointZ Since thenZ=Z , we conclude thatZis indeed on the lined XY as expected It thus suffices to prove thatZ =Z For this we consider the point

The three pointsX,Y,Rare distinct, thus constitute a projective basis of the line d XY (see Example6.5.6) By Definition 6.6.1, the points Z andZ admit with respect to this basis the homogeneous coordinates

This shows that the equalityZ =Z , sufficient to conclude the proof, reduces to proving

To prove this equality, we shall use two additional points, namely

A succession of central projections and corresponding applications of Corol- lary6.7.6, Definition6.18.3and Theorem6.18.2, then yields

=(X, Y;R, Z ) where the last two equalities follow from Lemma6.6.3

Theorem 6.19.3 (Brianchon) In a projective plane P2 (K), consider a hexagon whose six sides are tangent to a non-degenerate regular conic The three diagonals of the hexagon intersect at the same point.

Proof Brianchon’s theorem is the dual statement of Pascal’s theorem (see Theo- rem6.16.3).

Affine Versus Projective

In a sense, this section “closes the circle” that we opened at the very beginning of this chapter We introduced the projective plane via the following example:

Imagine that you are standing on a horizontal plane looking at the various points of this plane When you look at a point P of the plane, you can represent your line of sight by the line passing through P and your eye In doing so, you describe a bijection between all the points of the plane and all the non-horizontal lines passing through your eye

Now that the theory of the projective plane (and more generally, of projective spaces) has been developed, let us formalize this intuition:

The real affine plane can be identified with the set of all “non-horizontal” vector lines of R 3 , that is, the set of all points of the real projective plane not lying on a specific projective line

Theorem 6.20.1 Consider a projective linein a projective planeP2 (K)over an arbitrary fieldK Define π =P2 (K)\ P∈P2 (K)P /∈

1 The setπ is in bijective correspondence with the affine space of dimension 2 overK.

2 This bijection is such that the affine lines ofπ are precisely the traces onπ of the projective linesd=ofP2 (K).

3 An affine line inπ thus has the formd\ {Q}, withd=a projective line and

4 Two distinct affine lines inπ are parallel precisely when the intersection point of the corresponding projective lines lies on.

5 The points ofare in bijective correspondence with all the possible directions of all the affine lines inπ

The projective lineand the pointsQ∈are respectively called the line at infinity and the points at infinity of the affine planeπ

Proof InK 3 , the projective line is the set of all vector lines of a vector plane, which we continue to write as Fix a basis (e 1 , e 2 ) of this vector plane and extend it to a basis(e 0 , e 1 , e 2 )ofK 3 In the corresponding system of homogeneous coordinates, the projective pointsQ∈ are those whose first coordinate is zero. Thus the pointsP∈π are those whose first coordinate is non-zero; of course, there is no loss of generality in choosing this coordinate to be equal to 1.

Consider then the affine plane π ⊆K 3 with equation X 0 =1 (see Proposi- tion2.21.1) Notice that the vector plane with equationX 0=0 is precisely the direction of this affine planeπ (see Example 2.2.2) In particular, the projective points of are the directions of the affine lines ofπ The affine subspaceπcontains the pointe 0, thus admits the affine basis(e 0;e 1 , e 2 )(see Definition2.19.1) Working in terms of coordinates, we trivially obtain a bijection ϕ:π−→π , a b

⎦ between the affine planeπand the setπ of the statement.

Every affine lineδ⊆π can be presented asδ=Δ∩π, whereΔ⊆K 3 is the vector plane containingδand the originO∈K 3 Thus—via the bijectionϕ—every affine lineδ⊆π is the trace onπof the corresponding projective lineΔ.

This correspondenceδ→Δis trivially injective: ifΔ is associated withδ and Δ=Δ , then δ=Δ∩π=Δ ∩π=δ

Moreover ifΣ= is an arbitrary projective line, the intersectionΣ∩ is a projective point (see Theorem6.2.10), that is, a vector subspace of dimension 1 Choose

0=v∈Σ∩ SinceΣ has dimension 2, there also exists a vectorw∈Σ\ , thus a vector whose first coordinate is non-zero SinceΣ is a vector subspace, there is no loss of generality in choosingwwith first coordinate equal to 1, that is,w∈π.

Then the affine lineσ passing throughwand of directionvis both inΣand inπ, proving thatσ=Σ∩π.

We have thus already observed that the correspondenceδ→Δinduces a bijection between the affine lines ofπand the projective lines ofP2 (K), distinct from

The projective lineΔis described by an equation a 0 X 0+a 1 X 1+a 2 X 2=0 while the affine lineδis described by the system of equations a 0+a 1 X 1+a 2 X 2=0

Thus a pointQ∈Δdoes not correspond to a point ofδ if and only if its coordinate of index 0 is equal to zero, that is, if it is a projective point of Again by

Theorem6.2.10there is just one such point, namely, the intersection ofΔand

It remains to consider the case of distinct parallel affine linesδandσ inπ, that is, lines which are disjoint (see Corollary2.7.7) Since the corresponding projective lines intersect in a unique point, again by Theorem6.2.10, this means that this intersection point is not inπ , thus is in

The observant reader will have noticed that the arguments in Theorem6.20.1 carry over at once to the case of a projective spaceP n (K)of arbitrary finite dimension, provided one replaces the word “line” by the word “hyperplane”.

Of course in Theorem6.20.1, the choice of the line is arbitrary Choosing another line drastically changes the situation: for example, the projective lines having parallel affine traces are different This flexibility in the choice of the “line at infinity” will prove to be an efficient tool of investigation As an example, let us apply this process to Desargues’ Theorem.

Choose first as line “at infinity” a line containing the pointP, but none of the other points involved in Fig.6.4 The statement6.8.3translates as:

Proposition 6.20.2 (Desargues’ Theorem) In an affine plane over a field K, consider six pairwise distinct pointsA, B, C, A , B , C constituting two triangles

{A, B, C}and{A , B , C } Suppose that the linesd AA ,d BB andd CC are parallel.

Under these conditions, if the three points

L=d BC ∩d B C , M=d AC ∩d A C , N=d AB ∩d A B of intersection exist, they are on the same line (see Fig.6.17).

Choose now as line “at infinity” a line passing throughLand no other point involved in Fig.6.4 The statement6.8.3translates as:

Proposition 6.20.3 (Desargues’ Theorem) In an affine plane over a field K, consider six pairwise distinct pointsA, B, C, A , B , C constituting two triangles

{A, B, C}and{A , B , C } Suppose that the linesd AA ,d BB andd CC intersect at some pointP and that the linesd BC andd B C are parallel When the intersection points

M=d AC ∩d A C , N=d AB ∩d A B exist, the lined MN is parallel tod BC andd B C (see Fig.6.18).

And so on for all the possible affine variations of the statement of Desargues’ theorem.

Let us observe further that an analogous link exists between projective conics and their “affine traces” We shall come back to interesting examples of such situations in the next section (see Example6.21.3).

Proposition 6.20.4 Under the conditions of Theorem6.20.1, the trace onπ of a projective conic ofP2 (K)is an affine conic.

Proof Let us work in the system of homogeneous and affine coordinates described in the proof of Theorem6.20.1 A projective conic is given by an equation of the form

The trace of that conic onπ , expressed with respect to the affine basis(e 0;e 1 , e 2 ) ofπ, is obtained by simply puttingX 0 =1 in the equation above This indeed yields an equation of degree at most 2 inX 1,X 2(see Definition2.23.1)

One can also study the “affine trace” of a projective transformation, paying close attention to the possible domains of definition The following case is probably the most important one (see Example6.7.2).

Proposition 6.20.5 Under the conditions of Theorem6.20.1, consider a pointQ∈ and a projective linednot containingQ The central projection ond with centre

Q(see Example6.7.2) induces onπ a projection on the affine trace ofd, parallel to the direction inπ determined by the projective pointQ∈

Proof Given a pointP ∈π , one thus has P =Qso that the central projection p(P )∈d is defined One cannot havep(P )∈ , otherwise the line throughQand p(P )—which containsP—would be itself; andP /∈ Thusp(P )∈π and the trace of the central projection on the affine planeπ is correctly defined The result follows by Theorem6.20.1

There is another important link to be made between the projective and the affine settings: the consideration of the anharmonic ratio.

Proposition 6.20.6 Under the conditions of Theorem6.20.1, consider four distinct pointsA,B,C,Don some affine lined ofπ Suppose that for some affine basis

(P; −→e )of that affine line, these points admit the coordinates

The anharmonic ratio of the corresponding projective points is equal to

On the other hand there exist vectors−→e A ,−→e B such that

Considering successively the first two and the last two equalities, we get α−→e A −β−→e B a−b = −→e =(β−1)−→e B − −→e A b−c

The equality of the first and the third terms gives α a−b+ 1 b−c

Since−→e A and−→e B are linearly independent, both coefficients are equal to zero, which provides the following values ofαandβ α=b−a b−c, β=a−b a−c.

P D=α−→e A −a−→e +d−→e and replacingαand−→e by their values computed above, we obtain

Going back to [7], Trilogy I, Sect 6.3, we observe that the projective definition of the anharmonic ratio, in the special case of four “affine points”, coincides with the corresponding “historical” definition.

Real Quadrics

The rest of this chapter focuses on some particular aspects of projective spaces over the fieldRof real numbers, with special attention paid to the case of the real projective plane Chapter7will be entirely devoted to the other important case of interest: the projective plane over the fieldCof complex numbers.

Counterexample6.13.8gives an example of a projective regular quadric over the reals which is degenerate, and even empty! Of course the field of real numbers is not algebraically closed, but nevertheless we obtain an “almost as good” result as that of Proposition6.13.9 In fact, Counterexample6.13.8is the only possible one. Proposition 6.21.1 A non-empty regular quadric in the projective spaceP n (R)is non-degenerate.

Proof Again by Proposition6.13.4, there is no loss of generality in assuming that the quadric is given by a diagonal equation n i = 0 a i X i 2 =0; by regularity,a i =0 for each indexi Since we are working over the reals and the projective quadric is non-empty, not all coefficientsa i have the same sign There is no loss of generality in assuming that a 0 , , a m >0, a m + 1 , , a n m, k=i, j, the pointsP i,j with coordinates

, X k =0 are all on the quadric, as well as the pointsP i,j with coordinates

It remains to observe that the points

The following result concerning projective real conics underlines once more the

“unifying” nature of projective methods.

Theorem 6.21.2 All non-empty regular projective real conics are projectively equivalent.

Proof By Proposition6.13.4, every conicQadmits an equation of the form a 0 X 0 2 +a 1 X 2 1 +a 2 X 2 2 =0 with respect to some convenient projective basis The regularity of the matrix of coefficients means thata 0 ,a 1 anda 2 are all non-zero The non-emptiness of the conic implies that these three coefficients do not have the same sign Up to possibly multiplying by−1 and changing the ordering of the variables, we can suppose that a 0 anda 1 are strictly positive while a 2 is strictly negative A further change of coordinates

−a 2 X 2 further reduces the equation to the form

(compare with Proposition3.6.1) Write(e 0 , e 1 , e 2 )for a basis ofR 3 inducing the system of homogenous coordinates with respect to which the equation of the conic

Given a second non-empty regular conic Q , there is analogously a basis

(e 0 , e 1 , e 2 )ofR 3 with respect to whichQ admits the equation

Trivially, the linear isomorphism ϕ:R 3 −→R 3 defined by ϕ(e 0 )=e 0 , ϕ(e 1 )=e 1 , ϕ(e 2 )=e 2 exhibits the equivalence of the two conicsQandQ

In the spirit of Sect.6.20and in particular Proposition6.20.4, let us now look at the “affine traces” of the projective non-empty regular real conics.

Example 6.21.3 In the real affine plane, the ellipse, the hyperbola and the parabola are precisely the affine traces of the non-empty regular projective conics.

Proof As observed in the proof of Proposition6.21.2, a non-empty regular real conic can be described with respect to a convenient projective basis by an equation of the form

The corresponding quadric inR 3 is thus a cone (see Sect.1.14) The announced result was thus already known to the Greek geometers (see Sects 2.5 and 4.6 in [7], Trilogy I):

Cutting a cone by a plane not passing through its vertex yields an ellipse, a hyperbola or a parabola

Let us nevertheless give an algebraic proof of this fact.

Given the non-empty regular projective conic with equation

X 0 2 +X 2 1 −X 2 2 =0. and a plane inR 3 with equation a 0 X 0+a 1 X 1+a 2 X 2=1,

Fig 6.21 we must prove that the intersection of the cone and the plane is an ellipse, a hyperbola or a parabola (see Figs.6.20,6.21and6.22) Observe that all points

⎦ satisfy the equation of the projective conic and thus there is no loss of generality in considering only those pairs(u, v)such thatu 2 +v 2 =1 If a 0 u+a 1 v+a 2 =0

Fig 6.22 the point ofR 3 with coordinates

⎠ is on the intersection of the plane and the cone Notice that having both u 2 +v 2 =1, a 0 u+a 1 v+a 2 =0 means that inR 2 , the pair(u, v)is both on the circle and the line admitting these respective equations Thus at most two pairs (u, v) must be excluded Since the projective quadric is non-degenerate by Proposition6.21.1, it immediately follows that its affine trace on the plane considered is also non-degenerate.

Going back to Sect.1.10, we conclude already that the affine trace of the projective conic is an ellipse, a hyperbola, a parabola or the union of two lines But this last case must be excluded, because otherwise by PropositionC.2.2, the projective conic would itself be the union of two projective lines and this would contradict Proposition6.17.2.

Observe that takingX 2=0 as the “line at infinity” (see Theorem6.20.1) yields the affine conic with equation

X 0 2 +X 1 2 =1; this is an ellipse Analogously choosingX 0=0 as the “line at infinity” yields the affine conic with equation

X 2 2 −X 1 2 =1; this is a hyperbola On the other hand the change of variables

The Topology of Projective Real Spaces

Choosing this timeY 2 =0 as the “line at infinity” we end up with the affine conic with equation

Y 0 2 =Y 1 which is a parabola We recapture the fact that all three cases occur effectively, for each non-empty regular projective conic.

Conversely, every ellipse admits an equation of the form

X 0 2 +X 2 1 =1 and is thus the affine trace of the projective conic with equation

X 0 2 +X 2 1 −X 2 2 =0 when choosingX 2 =0 as the line at infinity The other two cases are analogous

Thus by Theorem6.21.2, an ellipse, a hyperbola and a parabola, which by Theo- rem3.6.4are three affinely non-equivalent real conics, are nevertheless affine traces of projectively equivalent conics.

Counterexample 6.21.4 Theorem6.21.2does not generalize to quadrics in higher dimensions.

Proof The two quadrics ofR 4

X 0 2 +X 1 2 −X 2 2 −X 2 3 =0 are not affinely equivalent (see Theorem3.6.4), thus the corresponding projective quadrics inP3 (R)are not projectively equivalent Trivially, they are non-empty and regular

6.22 The Topology of Projective Real Spaces

This section appears at the end of the chapter to underline the fact that it is not part of the algebraic treatment of projective spaces The point here is essentially to make some informal comments (for the interested reader) on the topology of projective real spaces All the topological notions used in this section can be found in Appendix A in [8], Trilogy III.

By Definition6.1.1, the real projective planeP2 (R)is the set of all vector lines ofR 3 There is an alternative equivalent way to describe it, by exhibiting a natural

Fig 6.23 bijection between this set of vector lines and another useful setQconstructed from a sphere Consider the sphere of radius 1 centered at the origin ofR 3 and perform the quotient of the sphere by the equivalence relation identifying two diametrically opposite points One obtains a quotientQwhose elements are trivially in bijective correspondence with the vector lines ofR 3 , that is, with the points ofP2 (R) The intersections of the sphere with the vector planes ofR 3 are just the great circles; thus viewingQas the real projective plane, the projective lines inQ∼=P2 (R)are simply the traces, in the quotient, of the great circles of the sphere (see Definition6.2.1). The sphereS 2 has a natural topology as a subspace ofR 3 SinceQis defined as a quotient of the sphere, q:S 2 −→Q∼=P2 (R) we can put on it the corresponding quotient topology: that is, a subsetU⊆Qis open if and only ifq − 1 (U )is open This trivially turnsQinto a topological space andq into a continuous mapping Furthermore, since a quotient map is surjective, we have q(S 2 )=Q, thus the real projective plane is in particular compact as a continuous image of the compact spaceS 2 It is trivial from the definition ofQto infer that the real projective plane is in fact compact Hausdorff.

Let us now have a look at an even more elementary example: the projective real lineP1 (R) This is the set of vector lines ofR 2 (see Fig.6.23) Of course, as above, we can equivalently defineP1 (R)by considering first all the points of the circle of radius 1 inR 2 and next perform the quotient by the equivalence relation which identifies two diametrically opposite points Again via the quotient topology, the real projective line becomes a compact Hausdorff space, a quotient of the circle. Let us next “concretely” construct this quotient (see Fig.6.24) For this, view the circle of radius 1 as being a piece of wire First twist this wire to give it the form of an “infinity sign”, the two pointsDandD being the place where the curve “cuts itself”; adjust the form of the two loops of the curve a little to give them the form of two circles Then fold the curve on itself along the vertical line passing through

D=D Observe thatAnow coincides withA and analogously forB,B and all the other points We have obtained precisely the quotient of the original circle by the equivalence relation identifying two diametrically opposite points This quotient

Problems

Fig 6.25 is another circle This shows (or at least suggests) that the real projective line is topologically homeomorphic to a circle.

However, do not imagine that this fact can be carried over to higher dimensions: this is by no means the case! Consider Fig.6.25, where for clarity only half of the sphere has been drawn As we have seen, the projective planeP2 (R)is the quotient of the sphere obtained when identifying diametrically opposite points We have just seen how to realize this operation “concretely” on a circle Try to apply the process to the equator of the sphere You first glue together the two pointsXandX , then you have to “twist” the equator to make it an “infinity sign” cutting itself atX=X Finally you have to fold the equator on itself But at this point you are stuck: in

R 3 , there is no way to “concretely” realize this operation because of the presence of the rest of the sphere “attached to the equator”, which presents an impassable obstruction So the real projective plane is certainly not homeomorphic to a sphere. One last comment on real projective spaces In Sect.3.1, we have taken full advantage of the fact that the ordering on the set of real numbers allows us to define segments in real affine spaces In other words, given three distinct pointsA,B,C on an affine real line, one of them is between the other two We haven’t mentioned anything like this in the projective case There is a good reason for this: such a property does not hold Indeed the real projective line is topologically homeomorphic to a circle: and on a circle, two distinct points do not define one “segment”, but two!

6.23.1 Given triples of distinct points(A 1 , A 2 , A 3 ),(B 1 , B 2 , B 3 )inP1 (K), prove the existence of a unique projective transformation mapping each pointA i to the corresponding pointB i

6.23.2 Letdandd be distinct lines inP2 (K) Prove that a projective transformation f:d−→d is a central projection if and only if the intersection point ofd andd is fixed byf.

6.23.3 Letdandd be distinct lines inP2 (K)andf: d−→d , a central projection. Prove the existence of a lined passing through the intersection ofdandd and such that, for every two pointsA=Bond, the linesd AB andd A B intersect ond (This lined is called the axis of the projective transformation.)

6.23.4 Describe a projective transformation of the projective plane admitting as affine trace a given parallel symmetry of the affine plane.

6.23.5 Prove that every projective transformation ofP2 (R)admits a fixed point. 6.23.6 State all the affine versions of Desargues’ theorem.

6.23.7 State all the affine versions of Pappus’ theorem.

6.23.8 State all the affine versions of Pascal’s theorem.

6.23.9 Consider a fieldK of characteristic distinct from 2, admitting at least five elements Given four distinct pointsA,B,C,Don a regular non-degenerate conic ofP2 (K), prove that (A, B;C, D)= −1 if and only if the line d AB contains the pole of the lined BC

6.23.10 Show that a sphere inR 3 is topologically homeomorphic to a projective complex line.

Exercises

6.24.1 Determine the number of points of the projective spaceP n (K), whenKis a field withmelements.

6.24.2 LetK be a field of characteristic distinct from 2 in which−1 admits three distinct cubic roots Prove the existence inP2 (K)of nine points, four of them never on the same line, such that each line containing two of these points contains a third one.

6.24.3 InP n (K), if two distinct projective subspaces of dimensionn−2 contain the same projective subspace of dimensionn−3, both of them are contained in the same, necessarily unique, projective hyperplane.

6.24.4 What is the dual statement of: InP3 (K), if two pointsA,Blie in a planeπ,all points of the line joiningAandBare inπas well.

6.24.5 In a projective spaceP n (K)(n≥3), three points not on the same line are contained in a unique plane What is the dual statement?

6.24.6 InP2 (K), consider the three lines with equations a 0 X 0 +a 1 X 1 +a 2 X 2 =0 b 0 X 0+b 1 X 1+b 2 X 2=0 c 0 X 0 +c 1 X 1 +c 2 X 2 =0.

Determine the condition on the coefficients which forces these three lines to intersect at the same unique point.

6.24.7 LetA,B,C,D,Ebe distinct points on a projective line Prove that

6.24.8 In the Euclidean space R 2 , consider two intersecting lines and their bi- sectors Prove that the corresponding four points inP1 (R)constitute a harmonic quadruple.

6.24.9 Given three distinct lines inP2 (K), which intersect at the same point, give a “geometric” construction of the fourth line which yields a harmonic quadruple of lines.

6.24.10 InP3 (K), consider two linesd andd which do not intersect Determine all the fixed points of a projective transformationf: P3 (K)−→P3 (K)which fixes all the points ofdandd

6.24.11 InP2 (C), consider two distinct linesdandd and two pointsA∈d,A ∈d distinct from the intersection ofd andd Consider further a pointP neither ond nord Prove the existence of a regular non-degenerate conic, passing throughP, tangent todatAand tangent tod atA

6.24.12 InP n (R), consider two regular non-degenerate quadricsQandQ With every pointP ∈P n (R), let us associate its polar hyperplaneH P with respect toQ and the polef (P )ofH P with respect toQ Prove thatf is a projective transformation.

6.24.13 Prove that with respect to a convenient projective basis, a regular conic of

P2 (C)can be described by an equation of the formX 2 2 =X 0 X 1.

6.24.14 LetQbe a regular non-empty conic ofP2 (R) Fix two triples of distinct points(A 1 , A 2 , A 3 ), (B 1 , B 2 , B 3 ) on Q Prove the existence of a unique projective transformation mappingQonto itself and each pointA i to the corresponding pointB i

We conclude this book with an elementary study of algebraic plane curves: a first step towards the fascinating world of algebraic geometry This chapter assumes the reader is familiar with the theory of polynomials in several variables (see the Ap- pendices to this book).

An algebraic plane curve is one which can be described by a polynomial equation, such as a conic, which can be described by a second degree equation As we have seen, we can study conics in affine space over a field or in the projective plane over that field Moreover, depending on the field that we consider, we can end up with rather different conclusions.

The fact is that deep and elegant theorems of algebraic geometry generally work in the projective plane over an algebraically closed field We first give some hints to grasp (or at least, guess) the reasons for this In this first introduction to the topic, we therefore choose to restrict our attention to the projective plane over the field of complex numbers.

We proceed then through the study of various intuitive geometric or algebraic notions: the degree of a curve, the tangent to a curve, the multiple points of a curve, the inflexion points of a curve, various possible singularities of a curve We give examples of each notion.

We next focus on the Bezout theorem, which gives information on the number of intersection points of two algebraic curves There are various versions of this theorem, each depending on whether or not one takes into consideration the multiplicities of the intersection points on each curve or the tangency properties between the various “branches” of the two curves Unfortunately, the notion of branch of a curve requires the use of sophisticated algebraic tools which are beyond the scope of the present book Therefore we limit our study to a Bezout theorem which involves only the intersection points and their multiplicities.

Our Bezout theorem is nevertheless sufficient to study the existence of curves passing through some given points, or to give bounds for the number of multiplicities of a curve.

After revisiting some results on conics, we switch to the study of cubics, the striking properties of their inflexion points and the topological group associated with an irreducible cubic.

We conclude with the study of rational curves: projective curves which admit a parametric representation in terms of polynomials.

Notice that since we are working inP2 (C), we shall often find it more convenient to writeX,Y,Z for the three homogeneous coordinates, instead ofX 0,X 1,X 2as in Chap.6.

Looking for the Right Context

Let us introduce this chapter with the following very simple problem:

Consider two distinct circles in the plane How many intersection points do they have?

Of course since it concerns circles, the most natural context in which to pose the question is the Euclidean plane It does not take long to realize (see Fig.7.1) that the answer is:

You will probably accept that in the case of the middle picture in Fig.7.1, where there is exactly one intersection point, this point should be counted twice! Indeed, this second case is obtained as the limit of a process where two distinct points of intersection eventually coincide (see Fig.7.2).

This is true, but the middle picture in Fig.7.1is also obtained as a limit of a process where there was no intersection point at all! (See Fig.7.3.) Is this not a little embarrassing?

Let us overcome the problem by investigating things algebraically Up to a change of coordinates, there is no loss of generality in assuming that the first circle is centered at the origin, while the second one has its center on theX-axis The two circles then have equations of the form

The intersection points are thus obtained for

This is a perfectly decent real number as soon as a=0, that is, as soon as the two circles are not concentric Of course, the study of the intersection of concentric circles is rather trivial!

Thus whena=0, the corresponding values ofY are given by

• when the quantity under the square root is strictly positive, we obtain two distinct real values forY; this corresponds to the right hand picture in Fig.7.1;

• when the quantity under the square root turns out to be precisely zero, i.e when

R=a±r, we findY = ±0, that is a double root of the original equation inX andY; this corresponds to the middle picture in Fig.7.1;

• when the quantity under the square root is strictly negative, we find two distinct complex values forY: this corresponds to the right hand pictures in Figs.7.2 and7.3.

Thus, provided that the two circles are not concentric and if we are working over the field of complex numbers, there are always two intersection points, which turn out to coincide in the “tangent case”.

We have thus already reached a rather elegant conclusion:

In C 2 , two non-concentric circles always meet at exactly two points, distinct or equal

Of course the precise meaning of “two equal points” of intersection should be formalized clearly in terms of a “double root” of an equation Furthermore, notice that the argument above remains valid ifa,R,rare chosen to be complex numbers.

However, we can improve on this, also incorporating the case of concentric circles In that case we have furthera=0 and trivially, the two equations

X 2 +Y 2 =R 2 , X 2 +Y 2 =r 2 do not have a common solution, not even in the complex numbers, as soon as the two circles are distinct, that is, as soon asR=r In that case, the number of intersection points is definitely zero.

However, applying the techniques of Sects.C.2and6.20, now view the two circles as living in the complex projective plane, that is, as the affine traces of the two projective conics

Trivially, the intersection of these two projective conics is given by the two points with homogeneous coordinates

So we eventually conclude that

In P 2 ( C ), two distinct circles always meet at exactly two points, distinct or equal

Thus in the complex projective plane, we end up with an elegant result where the two possible intersection points always exist.

These first simple considerations underline clearly the unifying role of the complex projective plane This will thus be the context in which we shall work in this chapter As we shall see later (see Sect.7.7), the observations above are not especially related to the properties of circles: they can be extended to the case of two arbitrary algebraic curves.

Indeed (although we shall not enter into these considerations in this book), many results proved in this chapter remain valid in the projective plane over an arbitrary algebraically closed field.

The Equation of an Algebraic Curve

The natural way of defining an algebraic curve in the complex projective plane is clearly:

Definition 7.2.1 By an algebraic curve is meant a subsetC⊆P2 (C)of the complex projective plane which, in some system of homogeneous coordinates, can be described as the set of those points whose coordinates satisfy an equation

P (X, Y, Z)=0 whereP (X, Y, Z)is a non-zero homogeneous polynomial.

Of course, Definition7.2.1does not depend on the choice of the system of homogeneous coordinates:

Lemma 7.2.2 In every system of homogeneous coordinates, an algebraic curve can be described as the set of those points whose coordinates satisfy an equation

P (X, Y, Z)=0 whereP (X, Y, Z)is a non-zero homogeneous polynomial.

Proof A change of homogeneous coordinates is expressed by linear formulas These formulas transform a non-zero homogeneous polynomial into another non-zero homogeneous polynomial

However, if we define an algebraic curve as in Definition7.2.1, we should pay attention to which properties depend only on the subsetC⊆P2 (C), and which properties depend explicitly on the polynomialP (X, Y, Z)used to describeC Does this make a difference?

In other words, the following question arises:

Is the equation of an algebraic curve, in a given system of homogeneous coordinates, necessarily unique?

Of course the answer is “no”, as we discovered already in the study of quadrics (see Sect.2.26)! The two equations

P (X, Y, Z)=0, k P (X, Y, Z)=0, 0=k∈C trivially determine the same curve.

Thus a more sensible question would rather be:

Are two equations of an algebraic curve in the same system of homogeneous coordinates necessarily proportional?

Once more the answer is trivially “no”! The two equations

P (X, Y, Z)=0, P (X, Y, Z) n =0, 0=n∈N still determine the same subsetC⊆P2 (C).

This last case is worth more attention By TheoremB.4.9, we know that the poly- nomialP (X, Y, Z)factors uniquely (up to multiplicative constants) as a product of irreducible polynomials and by PropositionC.1.3, all these factors remain homogeneous Clearly, if we replace one of these irreducible factors by a power of it, we do not change the roots ofP (X, Y, Z) So clearly, if we want to reach sensible results, it would be wise to work only with equationsP (X, Y, Z)=0 whereP is a non-zero homogeneous polynomial without any multiple irreducible factors.

We therefore adopt the following definition:

Definition 7.2.3 By a simple equation of an algebraic curve inP2 (C)we mean an equation

P (X, Y, Z)=0 whereP (X, Y, Z)a non-zero homogeneous polynomial without any multiple factors.

Notice that determining whether an equation of a curve is simple is a rather easy task: it suffices to apply CorollaryD.1.5and compute a resultant.

This time, due to the fact that the field of complex numbers is algebraically closed, we get the following important result:

Theorem 7.2.4 Two simple equations of an algebraic curve in a given system of homogeneous coordinates are necessarily proportional.

Proof Consider two simple equations

P (X, Y, Z)=0, Q(X, Y, Z)=0 of the algebraic curve C in a given system of homogeneous coordinates Given an irreducible factor R(X, Y, Z), all its roots are points of C, thus are roots of Q(X, Y, Z) By Proposition D.2.4, R(X, Y, Z) divides Q(X, Y, Z) Since all the irreducible factors of P (X, Y, Z) are simple, this immediately implies that

P (X, Y, Z)itself dividesQ(X, Y, Z) Analogously, the polynomialQ(X, Y, Z)di- videsP (X, Y, Z), proving that both polynomials are proportional

Of course a special case of interest is:

Definition 7.2.5 By an irreducible algebraic curve we mean an algebraic curve whose simple equation is an irreducible polynomial.

Lemma 7.2.6 Every algebraic curve is in a unique way the union of finitely many irreducible curves.

Proof By TheoremB.4.9, factor the simple equation of the curve into its irreducible factors

Lemma7.2.6 reduces—for many purposes—the study of an arbitrary curve to that of irreducible curves The Eisenstein criterionB.4.11is certainly a useful tool to check the irreducibility of a curve On the other hand, finding the irreducible factors of a given polynomialP (X, Y, Z)is generally quite a difficult task.

We shall also use the following terminology:

Definition 7.2.7 An algebraic curveCis called a component of an algebraic curve

Proposition 7.2.8 Consider two algebraic curvesCandDwith respective simple equationsP (X, Y, Z)=0 andQ(X, Y, Z)=0 The following conditions are equivalent:

Proof (2⇒1)is obvious Conversely, by PropositionD.2.4, each irreducible factor ofP (X, Y, Z)is an irreducible factor ofQ(X, Y, Z)

Let us conclude this section by mentioning that, at some places, we will find it convenient to use possibly non-simple equations For example in some arguments, given a curve with simple equationP (X, Y, Z)=0, we shall want to consider the curve (when it turns out to be one) with equation ∂P ∂X (X, Y, Z)=0 Of course this partial derivative can very well admit a multiple irreducible factor.

Counterexample 7.2.9 The equation X 3 +Y 2 Z=0 is simple, but its partial derivative with respect toXis a non-simple equation of a line.

By the Eisenstein criterionB.4.11, the polynomialP, viewed as a polynomial in

Xwith coefficients inC[Y, Z], is at once irreducible, thus a fortiori simple; in the Eisenstein criterion, simply choosed=Z Of course the partial derivative is a non- simple equation of the lineX=0.

The Degree of a Curve

We make the following natural definition.

Definition 7.3.1 The degree of an algebraic curve is the degree of its simple equation.

The purpose of this section is to exhibit an alternative “geometric” approach to this notion of degree.

Lemma 7.3.2 If a lined is not a component of a curveCof degreen, it intersects the curve in at mostnpoints.

274 7 Algebraic Curves Proof Consider two distinct points with coordinates

⎦ on the lined IfP (X, Y, Z)=0 is the simple equation of the curveC, the intersection points are obtained when

This equation is not satisfied by all pairs(s, t )since the line is not a component of the curve Thus the equation is not identically zero and therefore its left hand side is a homogeneous polynomial of degreen ins andt By PropositionC.1.4, this polynomial can be written in a unique way (up to multiplicative constants) as a product ofnpolynomials of the formas−bt Each intersection point corresponds to the unique root (up to a multiplicative constant) of one of the factorsas−bt

Of course one intends to define further:

Definition 7.3.3 Consider a lined not contained in an algebraic curveCof simple equationP (X, Y, Z)=0 The multiplicity of an intersection point of the curve and the line is the algebraic multiplicity of the corresponding root of the equation

P (sa 1+t b 1 , sa 2+t b 2 , sa 3+t b 3 )=0 as in the proof of Lemma7.3.2.

For this to make sense let us observe that:

Lemma 7.3.4 Definition7.3.3does not depend on the choice of the pointsA,Bin the proof of Lemma7.3.2nor on the choice of the system of homogeneous coordinates.

Proof Working instead with two other points

⎦ on the lined, one has

⎦ for some constantsα,β,γ,δ It follows at once that s=αs +γ t , t=βs +δt and introducing this linear change of variables into the equation yielding the intersection points (see the proof of Lemma7.3.2) does not change the multiplicity of the various factors.

An analogous argument holds when changing the system of coordinates, since again such a change gives rise to linear formulas

Corollary 7.3.5 If a lined is not a component of a curveCof degreen, they admit exactlynintersection points counted with their multiplicities.

Let us observe further that

Lemma 7.3.6 Given an algebraic curveCof degreen, through every pointAnot onC, there pass linesd admitting exactlyndistinct intersection points withC.

Proof We choose a homogeneous system of coordinates such that

Following the considerations of Sect.6.20, we chooseZ=0 as the “line at infinity” and work in the affine plane(X, Y ) We investigate the affine intersections of the curveC and a lined through the point at infinityAand some affine pointB By

Lemma7.3.2, we know already that there are at mostnof them.

Let us writeP (X, Y, Z)=0 for the simple equation of C with respect to the chosen system of homogeneous coordinates Since Ais at infinity, an affine line in the direction ofAadmits the equationY=kfor some constantk∈C Moreover sinceAis not onC, by Corollary7.3.5thenintersection points withC, counted with their multiplicities, are all affine points These affine intersections are given by the roots of the equationP (X, k,1)=0 which is thus of degreen We must prove that for some adequate choice ofk, thenroots are distinct We show this by a reductio ad absurdum.

Applying CorollaryD.1.5several times, we obtain the following If for eachk the polynomialP (X, k,1)has a multiple root, then the resultantR(k)ofP (X, k,1) and ∂P ∂X (X, k,1)is zero, that is, the resultantR(Y )ofP (X, Y,1)and ∂P ∂X (X, Y,1) as polynomials with coefficients inC[Y]is zero This is equivalent toP (X, Y,1) having a multiple factor This is a contradiction, because by assumption,P (X, Y, Z) does not have any multiple factors (see Sect.C.2)

Theorem 7.3.7 The degree of an algebraic curve is equal to the greatest number of distinct intersection points between this curve and a line not contained in the curve.

Proof This follows by Lemmas7.3.2and7.3.6

Tangents and Multiple Points

Section1.9has pointed out that, already in the 17th century, the tangent to a curve could be computed using the idea of a double point of intersection between the line and the curve However, let us be careful: having a double point of intersection with a curve does not mean a line is tangent to the curve.

Counterexample 7.4.1 The curve with equation

X 3 +Y 2 Z−X 2 Z=0 has a double point of intersection with the line of equationX=0.

Proof Consider the polynomial in the statement and its partial derivative with respect toY

Viewing these as polynomials with coefficients inC[X, Z]; their resultant is

By CorollaryD.1.5,P (X, Y, Z)is a simple equation of the curve.

⎦ in the proof of Lemma7.3.2, we end up with the equation t 2 s=0 which indeed admitst=0 as a double root The line throughAandB is that with equationX=0 andt=0 corresponds to the pointA.

An easy adaptation of this proof shows that in fact, every line throughAadmits

Aas (at least) a double point of intersection with the curve

Figure7.4represents the real affine part of the curve and the line involved in Counterexample7.4.1, when choosingZ=0 as the line at infinity Of course, no- body would describe this line as “tangent to the curve” Intuitively, there is a double point of intersection because the line cuts the curve at a point where two “branches” of the curve meet: and the line has a “single” intersection point with each of these two “branches” Thus being tangent is more subtle than just having a double point of intersection.

Investigating this example further—in a purely intuitive way, for the time being— we would rather say that the curve admits two tangents at the point considered: one tangent for each branch, as pictured in Fig.7.5 However, if we consider that a tangent to a “branch” has (at least) a double point of intersection with that branch, since this tangent also cuts the other branch, its point of intersection with the curve should thus be (at least) triple Indeed, consider the following.

Example 7.4.2 The two lines with equations

X=Y, X= −Y admit a triple point of intersection with the curve with equation

278 7 Algebraic Curves Proof Consider the case of the line with equationX=Y Choosing on this line

⎦ in the proof of Lemma7.3.2, we obtain the equation t 3 +t 2 (s+t )−t 2 (s+t )=0 that is simply t 3 =0 which indeed admitst=0, corresponding to the pointA, as triple root An analogous argument holds for the line with equationX= −Y.

An easy adaptation of this proof shows that in fact, only the two lines indicated admit with the curve a triple intersection at the pointA The considerations above are the key for defining the tangent(s) to a curve.

Lemma 7.4.3 Consider an algebraic curve C admitting the simple equation

P (X, Y, Z)=0 in some system of homogeneous coordinates Suppose that this curve contains the pointAbelow and consider further the polynomialp(X, Y ) p(X, Y )=P (X, Y,1), A ⎡

The following conditions are equivalent:

1 all the partial derivatives ofP (X, Y, Z), up to the orderk, are zero at the pointA;

2 all the partial derivatives ofp(X, Y ), up to the orderk, are zero at(a, b).

Proof (1⇒2)is obvious, since the partial derivatives ofp(X, Y )are obtained by puttingZ=1 in the partial derivatives ofP (X, Y, Z).

The converse implication follows at once from Euler’s formula (see Theo- remC.1.5) Indeed if

∂Y(a, b)=0 then Euler’s formula implies at once

It remains to repeat the argument, replacingP by one of its partial derivatives

Proposition 7.4.4 Consider an algebraic curveC admitting the simple equation

P (X, Y, Z)=0 in some system of homogeneous coordinates Suppose that:

• all the partial derivatives ofP (X, Y, Z), up to the orderk−1, at a given point

• at least one partial derivative ofP (X, Y, Z), of orderk, at the pointA, is non- zero.

Under these conditions, there are exactlyklines, when counted with their multiplicities, having with the curve an intersection of multiplicityk+1 at A(see Defini- tion7.3.3).

Proof At least one of the coordinates of the pointAis non-zero; let us assume that it is the third one, which we can then choose to be equal to 1 ViewingZ=0 as the line at infinity, an arbitrary line throughAis entirely determined by its pointBat infinity:

Going back to the proof of Lemma7.3.2, we have to consider the polynomial

P (sa+t u, sb+t v, s) and study the multiplicity of its root (s, t )=(1,0), corresponding to the point A The considerations of Sect C.2 indicate that we can equivalently puts=1 and study the multiplicity oft=0 as root of the polynomial p(a+t u, b+t v)=P (a+t u, b+t v,1) Moreover, by Lemma7.4.3, the nullity of all the partial derivatives ofP (X, Y, Z)atAis equivalent to the nullity of all the partial derivatives ofp(X, Y )at(a, b).

Let us consider the Taylor expansion of the polynomialp(a+t u, b+t v)around t=0 (see TheoremB.5.3): p(a+t u, b+t v)=p(a, b)

We know thatp(a, b)=0 sinceAis on the curve Thus in any case the right hand side contains at least one factort.

• The right hand side contains at least two factorstif and only if u∂p

If the partial derivatives of p(X, Y )of order 1 are not all zero at (a, b), this is a homogeneous equation of degree 1 inuandv Up to a constant multiple,

280 7 Algebraic Curves this equation admits a unique solution(u, v), that is, there exists a unique line admitting with the curve an intersection of multiplicity at least 2 atA.

• If now the partial derivatives ofp(X, Y )of order 1 are all zero at(a, b), the right hand side contains at least two factorst It contains at least three factorstif and only if u 2 ∂ 2 p

If the partial derivatives of order 2 are not all zero at(a, b), this is a homogeneous equation of degree 2 inuandv By PropositionC.1.4, this equation admits exactly two roots counted with their multiplicities That is, there are exactly two (equal or distinct) lines admitting with the curve an intersection of multiplicity at least 3 atA.

We thus adopt the following definition:

Definition 7.4.5 Consider a pointAof an algebraic curveCadmitting the simple equation P (X, Y, Z)=0 in some system of homogeneous coordinates Suppose that:

• all the partial derivatives ofP (X, Y, Z), up to the orderk−1, at the pointA, are zero;

• at least one partial derivative of orderkofP (X, Y, Z)is non-zero at the pointA.

• The pointAis called a point of multiplicitykof the curve.

• Thek(equal or distinct) lines throughAhaving with the curve an intersection of multiplicityk+1 are called the tangents to the curve atA.

Of course we shall as usual use the terminology:

• simple point for point of multiplicity 1;

• double point for point of multiplicity 2;

• triple point for point of multiplicity 3; and so on.

Our next concern is to exhibit the equation of the tangent at a simple point.

Proposition 7.4.6 Consider an algebraic curveC admitting the simple equation

P (X, Y, Z)=0 in some system of homogeneous coordinates Given a simple point

⎦ on this curve, the equation of the tangent toCatAadmits the equation

Proof By Euler’s formula (see TheoremC.1.5), we have a∂P

∂Z(a, b, c)=nP (a, b, c)=0 thusAis indeed a point on the line in the statement.

Now at least one of the coordinates ofAis non-zero Thus there is no loss of generality in assuming thatc=1, so that we are back in the situation of the proof of Proposition7.4.4, the notation of which we shall use freely The two quantities

(u, v)corresponding to the tangent thus satisfy the equation u∂P

Multiplying this last equation byt and adding it to the other equation above (with c=1), we obtain

However, again as the proof of Proposition7.4.4indicates, the points with coordinates

⎦ are precisely the affine points of the tangent Thus the homogeneous equation of the tangent is indeed that indicated in the statement (see Sect.C.2)

Corollary 7.4.7 All the points of a line are simple and the tangent to the line at each point is the line itself.

Proof Given a line with equation aX+bY+cZ=0 at least one of the coefficients—that is, at least one first partial derivative—is non- zero So all points are simple The tangent at any point is thus, by Proposition7.4.6

Xa+Y b+Zc=0 that is, the line itself

Proposition 7.4.8 Consider a simple pointAof a curveC IfChas a linear component containingA, this is also the tangent atAto the curveC.

Proof The equation ofChas the form

(aX+bY+cZ)P (X, Y, Z)=0 where the first factor corresponds to the linear component throughA Write

The first partial derivative atAis auP (u, v, w)+(au+bv+cw)∂P

∂X(u, v, w)=auP (u, v, w) becauseau+bv+cw=0 and analogously for the other two partial derivatives. Since the point Ais simple, at least one of these partial derivatives is non-zero, proving thatP (u, v, w)=0 Thus by Proposition7.4.6, the equation of the unique tangent atAreduces to a P (u, v, w) X+b P (u, v, w) Y+c P (u, v, w) z=0.

Simplifying byP (u, v, w)we recapture aX+bY+cZ=0 as the equation of the tangent

Of course, when a line is a component of a curve, its points on this curve are no longer necessarily simple.

Example 7.4.9 Consider the curveCcomprisingnlines passing through the same pointA ThenAis a point of multiplicitynonC.

Proof The equation ofChas the form

Each term of each partial derivative up to the leveln−1 still contains at least one of the factors appearing in the equation ofC, thus vanishes atA SoAis at least of multiplicityn.

Since each factor contains at least one non-zero coefficient, the corresponding partial derivative at the levelnis the product of all these coefficients, thus is non- zero ThusAhas multiplicity exactlyn

In the spirit of the comments at the end of Sect.7.2, we shall sometimes need to consider the following more technical notion, which is not intrinsic, but depends heavily on the choice of a particular—possibly non-simple—equation describing the curve:

Definition 7.4.10 Consider a pointAof an algebraic curveC admitting the arbitrary equationP (X, Y, Z)=0 in some system of homogeneous coordinates Sup- pose that:

• all the partial derivatives ofP (X, Y, Z), up to the orderk−1, at the pointA, are zero;

• at least one partial derivative of P (X, Y, Z), at the order k, at the point A, is non-zero.

Then the pointA is called a point of multiplicity k with respect to the equation

Examples of Singularities

This section presents some examples of curves admitting singularities, that is, multiple points Each time we provide the graphic representation of the real affine trace of the curve, that is, the curve obtained by puttingZ=1 in the equation and considering only points with real coordinates This very poor “sketch” of the actual situation inP2 (C)can be useful in some cases, and possibly misleading in other cases.

As a convention, when by “the origin”, we shall always mean the point with coordinates(X, Y )=(0,0)in the real affine plane We also simply state the results and leave to the reader the routine task of computing the partial derivatives or the equations of the tangents.

We shall use the following terminology:

Definition 7.5.1 A pointAof multiplicitykon an algebraic curveCis said to be ordinary when thektangents atAto the curveCare all distinct.

Of course, every simple point is ordinary.

Example 7.5.2 The Descartes folium with equation

X 3 +Y 3 =3XY Z admits the origin as an ordinary double point (see Fig.7.6).

Proof The two tangents areX=0 andY=0

Example 7.5.3 The clover leaf with equation

+3X 2 Y Z admits the origin as an ordinary triple point (see Fig.7.7).

Proof The three tangents areY=0 andY= ± √ 2 3 X

Example 7.5.4 The quadrifolium with equation

=4X 2 Y 2 Z 2 admits the origin as a quadruple non-ordinary point (see Fig.7.8).

Proof The two axes are both a double tangent

Example 7.5.5 The bifolium with equation

Fig 7.9 The bifolium admits the origin as a non-ordinary double point (see Fig.7.9).

Proof The double tangent isY=0

Example 7.5.6 The semi-cubic parabola with equation

Y 2 Z=X 3 admits the origin as a non-ordinary double point (see Fig.7.10).

Example 7.5.7 The curve with equation

X 4 +X 2 Y 2 +Y 2 Z 2 =2X 2 Y Z+XY 2 Z admits the origin as a non-ordinary double point (see Fig.7.11).

Fig 7.10 The semi-cubic parabola

Examples7.5.6and7.5.7already give the unpleasant impression that when considering the real affine trace of the curve, we “miss” some relevant part of the curve around the multiple point Examples 7.5.8and 7.5.9provide even more striking evidence for this.

X 3 +X 2 Z+Y 2 Z=0 admits the origin as an ordinary double point (see Fig.7.12).

Proof The origin is an “isolated” point in the real plane; the two tangents at this point are the lines with equationsY= ±iX

X 6 =Y 5 Z+X 2 Y 3 Z admits the origin as a non-ordinary quintuple point (see Fig.7.13).

Proof There is a triple tangentY =0 and two simple tangentsY= ±iX.

Inflexion Points

In Sect.7.4, we have seen that the tangent at a simple point is the unique line having with the curve at that point a contact of multiplicity at least 2 Can this multiplicity be higher than 2? To demonstrate that the answer is clearly ‘yes’, let us consider the projective extensions of the curvesY =X n which, in the real plane, are well-known to admit theX-axis as tangent.

Example 7.6.1 The origin is a simple point of the projective extension of the curve

Y =X n , with n≥3 The line with equationY =0 is tangent to the curve at the origin, the intersection point having multiplicityn≥3.

Proof The curve involved admits the equation Y Z n − 1 =X n The polynomial

P (X, Y, Z)=X n −Y Z n − 1 and its partial derivative ∂P ∂X =nX n − 1 , viewed as polynomials with coefficients inC[Y, Z], admit the non-zero resultantR(X, Y ) ±n n (−Y Z n − 1 ) n − 1 By CorollaryD.1.5, the given equation of the curve is simple.

288 7 Algebraic Curves The partial derivative ∂P (X,Y,Z) ∂Y = −Z n− 1 does not vanish at the point

⎦ corresponding to the affine origin of the axes ThusAis a simple point of the curve (see Definition7.4.4).

The line with equationY =0 is that through the two points

The equation to consider to determine the multiplicity of the intersection point then reduces tot n =0 (see the proof of Lemma7.3.2) It admitst=0 as a root of multiplicityn

This situation where the contact of the tangent with the curve has a “higher multiplicity” is worth some attention.

Definition 7.6.2 By an inflexion point of an algebraic curve of degree at least 3 is meant a simple point, not belonging to a linear component of the curve, and where the tangent intersects the curve with a multiplicity of at least 3 (see Definition7.3.3).

The multiplicity of the intersection makes sense only when the line is not a component of the curve (see Definition7.3.3) Since the tangent to a linear component is the line itself (see Proposition7.4.8), it is thus compulsory, in Definition7.6.2, to require that the point is not on a linear component of the curve Moreover by Lemma7.3.2, to obtain a possible triple intersection, it must also be the case that the curve is of degree at least 3.

Warning 7.6.3 The notion of inflexion point as in Definition7.6.2does not recapture, on the real affine trace of an algebraic curve, the notion of inflexion point of a real function, as in analysis.

Figure7.14represents the real traces of the graphs of the two functionsY =X 3 andY =X 4 (see Example7.6.1) and their tangents at the origin The origin is thus an inflexion point in the sense of Definition7.6.2 However, the second case clearly underlines the fact that the origin is not an inflexion point in the sense of that word as it is used in analysis:

A function f : R −→ R of class C 2 has an inflexion point at X = a when f (X) changes its sign at X = a.

In such a situation, it is well-known that the tangent to the graph atX=a“crosses” the graph of the function at that point; this is the precise geometric meaning of an

Fig 7.14 inflexion point in real analysis Of course iff (X)changes sign atX=a, by continuity,f (a)=0 Now in the complex case, “changing sign” no longer makes any sense In fact, in the special case of the graph of a function, the notion of inflexion point in the sense of Definition7.6.2recaptures only the fact thatf (a)=0.

Proposition 7.6.4 Consider a polynomialp(X)∈C[X]of degreen≥3 The following conditions are equivalent:

2 the corresponding point of the projective extension of the curveY=p(X)is an inflexion point in the sense of Definition7.6.2.

Proof The algebraic curve is given by p(X)−Y Z n− 1 =0, A ⎡

⎦ whileAis the point described in the statement The tangent atAis given by (see Proposition7.4.6)

It is the line through

Going back to the proof of Lemma7.3.2, we have to investigate the multiplicity of

(s, t )=(1,0)as a root of p(sa+t )− sp(a)+tp (a) s n − 1 that is, the multiplicity oft=0 as a root of p(a+t )−p(a)−tp (a).

290 7 Algebraic Curves Writing down the Taylor expansion ofp(X)ata (see TheoremB.5.3) p(a+t )=p(a)+tp (a)+t 2

Let us now introduce an efficient technique for studying the inflexion points.

It will be convenient, for this specific result, to switch back to the notation

Theorem 7.6.5 Consider an algebraic curveCof degree at least 3, without a linear component, admitting the simple equationP (X 0 , X 1 , X 2 )=0 The inflexion points ofCare its intersection points with the so-called Hessian curve with equation det

Proof LetAbe a simple point of the curve By Euler’s formula (see TheoremC.1.5), at least one second partial derivative is non-zero atP, otherwise all the first partial derivatives would be zero as well and the point would be multiple Thus

0 ≤ i,j ≤ 2 is a homogeneous polynomial of degree at least 1.

The tangent to the curve atAis given by (see Proposition7.4.6)

Given another pointB of this tangent, the multiplicity of the intersection of the tangent and the curve atAis the multiplicity oft=0 as a root of

(see Definition7.3.3) By Taylor’s formula (see TheoremA.8.6)

(a 0 , a 1 , a 2 )=0 becauseBis on the tangent atA The multiplicity oft=0 as a root of this equation will thus be at least 3 if and only if

In the discussion above, the pointB=Ais arbitrary on the tangent atA What we have just seen is thatAis an inflexion point if and only if all the pointsB=A of the tangent lie on the conic with equation

By Lemma6.15.3, this can further be rephrased as the fact thatAis an inflexion point if and only if the tangent atAis a component of this conic.

IfA is an inflexion point, the conic above is thus the union of two (equal or distinct) lines By Proposition6.17.2the conic is then singular and thus det

=0; thereforeAalso lies on the Hessian curve of the statement.

Conversely, ifAlies on the Hessian curve of the statement, again by Proposi- tion7.10.2the conic indicated above is the union of two lines By CorollaryC.1.6

SinceA lies on the original algebraic curve,P (a 0 , a 1 , a 2 )=0 The equality just mentioned indicates then that A belongs to the conic Again by Euler’s formula (see TheoremC.1.5) applied to the first partial derivatives ofP, the second partial derivatives ofP cannot all vanish atA, otherwise the same would hold for the first partial derivatives: this is not the case becauseAis supposed to be a simple point. Since the conic degenerates into two lines, the tangent to the conic atAis the linear component of the conic containing the pointA(see Proposition7.4.8) By Proposition7.4.6, this is the line with equation

Once more by Euler’s formula (see TheoremC.1.5) applied to the first partial derivatives, the equation of that component of the conic can be rewritten as

By Proposition7.4.6, this is precisely the equation of the tangent atAto the original algebraic curve This tangent is thus as expected a component of the conic.

The Bezout Theorem

The Bezout theorem studied in this section provides information on the number of intersection points of two algebraic curves We shall give two versions of this theorem and remark on a possible third version, which is beyond the scope of the present book.

Proposition 7.7.1 Two algebraic curves always have an intersection point.

Proof Consider the corresponding simple equations

P (X, Y, Z)=0, Q(X, Y, Z)=0 of these curves, of respective degreesnandm Compute the resultantR(X, Y )of these two polynomials, viewed as polynomials inZ with coefficients in C[X, Y] (see DefinitionD.1.3).

IfR(X, Y )=0, the two polynomials have a common irreducible factor and thus the two curves have a whole common component.

If R(X, Y )=0, it is a homogeneous polynomial of degree nm (see Theo- rem D.3.3) Such a polynomial admits non-trivial roots (see Proposition C.1.4). Choose such a non-trivial root(a, b)=(0,0) By TheoremD.1.4, sinceR(a, b)=0, the two polynomialsP (a, b, Z)andQ(a, b, Z)have a common factor, thus since

Cis algebraically closed, they have a common rootc(see Theorem F.3.3) Thus

(a, b, c)=(0,0,0)is a common non-trivial root of both polynomials

Corollary 7.7.2 An algebraic curve always contains infinitely many points but is never the whole spaceP2 (C).

Proof By LemmaA.6.8, the only polynomial P (X, Y, Z) which vanishes for all values of the variables is the zero polynomial Thus an algebraic curve is never the whole space.

Choose a point Anot on the curve There exist infinitely many lines through

Aand by Proposition7.7.1, all of them intersect the curve This exhibits infinitely many distinct points on the curve

Corollary 7.7.3 A curve without a multiple point is always irreducible.

Proof Let the simple equation of a curve have the form

By Proposition7.7.1, the two components with equations

∂X and since this partial derivative vanishes atP andQ, it also vanishes atA An analogous argument holds for the other two partial derivatives andAis thus a multiple point

Theorem 7.7.4 (Bezout) Two algebraic curves of respective degreesnandm, without any common components, admit at mostnmintersection points.

Of course, two curves with a common component always have infinitely many intersection points (see Corollary 7.7.2) Thus the restriction in the statement is necessary.

Proof We work by contraposition We thus assume that the two curves have at least nm+1 intersection points and we shall show that this forces them to have a common component.

Let us first make a selection ofnm+1 intersection points between the two curves. Fix a pointAwhich is on none of the two curves, nor on any line joining two of the selected intersection points This is possible since the union of these finitely many curves is still an algebraic curve, thus is not the whole space, by Corollary7.7.2 In addition, fix a system of homogeneous coordinates such that

P (X, Y, Z)=0, Q(X, Y, Z)=0 for the two simple equations of the curves, of respective degreesnandm We write R(X, Y )for the resultant of these two polynomials, viewed as polynomials inZ with coefficients inC[X, Y],

Consider now the various selected intersection points

For each indexi, the two polynomials

P (a i , b i , Z), Q(a i , b i , Z) have a common root c i , thus a common linear component Z −c i Therefore

Observe next that the pairs(a i , b i )and(a j , b j ), for distinct indicesi=j, are never proportional If this were the case, since we are working in homogeneous coordinates, we could as well assume that(a i , b i )=(a j , b j ) But then one would have ⎡

⎦ and the pointAwould be on the line joining B i andB j This is not the case, by choice ofA.

The resultantR(X, Y )is equal to zero or is a homogeneous polynomial of degree nm(see TheoremD.3.3) SinceR(X, Y )admits at leastnm+1 non-proportional roots(a i , b i ), it cannot be of degreenm(see PropositionC.1.4), thus it is equal to zero By TheoremD.1.4, this forces the two curves to have a common component. Notice at once that the “bound”nmin Theorem7.7.4cannot be improved:

Example 7.7.5 There exist algebraic curves of respective degreesnandm, without any common components, admitting exactlynmintersection points.

Proof The intersection of two distinct lines is always a single point (see Theo- rem6.2.10)

On the other hand, there is no reason that the boundnmin Theorem7.7.4can be attained.

Counterexample 7.7.6 There exist algebraic curves of respective degreesn and m, without any common components, admitting strictly fewer thannmintersection points.

Proof Consider the curveY Z 2 =X 3 of Example7.6.1 The lineY =0 has at the

“origin” an intersection with the curve of multiplicity 3 By Corollary7.3.5, this is thus the only intersection point So in this example,nm=3 while there is a unique point of intersection

Thus if we are to improve Theorem7.7.4it cannot be by improving the bound nm(Example7.7.5): we shall instead improve the way of counting the intersection points, taking care of the possible “multiplicities” A first step towards doing this is:

Lemma 7.7.7 Consider two algebraic curvesC,C , admitting arbitrary equations

P (X, Y, Z)=0 andQ(X, Y, Z)=0 of respective degreesnandm, in some system of homogeneous coordinates such that the pointA below is on neither of the two curves.

LetBbe an intersection point ofCandC such that (see Definition7.4.10):

• B is a point of multiplicity r on the curve C with respect to its equation

• B is a point of multiplicity s on the curve C with respect to its equationQ(X, Y, Z)=0.

WriteR(X, Y )for the resultant ofP andQviewed as polynomials inZwith coefficients inC(X, Y ) ThenR(X, Y )=0 orR(X, Y )admits the pair(a, b)as a root of multiplicity at leastrs.

Proof We suppose thatR(X, Y )=0 and prove the last assertion.

Our first concern is to prove that there is no loss of generality in assuming that

First of all, sinceBis on the curves andAis not, necessarily(a, b)=(0,0) Let us say,a=0 There is then no loss of generality in assuming thata=1, which we do at once The change of coordinates

X =X, Y =Y−bX, Z =Z yields the new coordinates

The resultant of the new equations

X , Y +bX thus admits the root(1,0)with the same multiplicity asR(X, Y )admits the root

(1, b) This proves already that there is no loss of generality in assuming thata=1 andb=0, which we do from now on.

To show that we can further assumec=0, consider the resultant of the two polynomials

P (X, Y, Z+λX), Q(X, Y, Z+λX) viewed as polynomials inZwith coefficients inC[X, Y, λ] View this resultant as a polynomial inλwith coefficients inC[X, Y]:

In particular, s 0 (X, Y ) is the resultant R(X, Y ) of P (X, Y, Z) and Q(X, Y, Z); by assumption at the beginning of the proof, this is a non-zero polynomial Thus

S(X, Y, λ)itself is a non-zero polynomial.

Let us now prove that k =0 We do this by reductio ad absurdum By LemmaA.6.8, the only polynomial which vanishes for all values of the variables is the zero polynomial Ifs k (X, Y )=0 withk >0, then the products 0 (X, Y )s k (X, Y ) is a non-zero polynomial, thus there is a point(u, v)∈C 2 such that s 0 (u, v)=0, s k (u, v)=0.

SinceCis algebraically closed (see TheoremF.3.3), the non-constant polynomial

S(u, v, λ)has a rootλ 0 This proves that the two polynomials inZ

P (u, v, Z+λ 0 a), Q(u, v, Z+λ 0 a) admit a common irreducible factor, thus a common rootw again sinceCis algebraically closed But thenw+λ 0 ais a common root of

P (u, v, Z), Q(u, v, Z) that is, these two polynomials have a common linear factor By TheoremD.1.4, their resultantR(u, v)=s 0 (u, v)is equal to zero, which is a contradiction.

So we have indeed proved thatk=0, that is,S(X, Y, λ)=s 0 (X, Y ) Therefore, considering this time the change of coordinates

X =X, Y =Y, Z =Z−cX, the pointBhas the coordinates

S(X, Y,−c)=s 0 (X, Y )=R(X, Y ) of the new equations remains unchanged thus, trivially, still admits(1,0)as root of the same multiplicity.

So indeed, there is no loss of generality in assuming that

Let us then chooseX=0 as the “line at infinity” (see Sect.6.20) and work in affine coordinates If we can prove that 0 is a root of multiplicity at leastrs ofR(1, Y ), then(1,0)will be, as expected, a root of multiplicity at leastrsofR(X, Y ).

SinceAis not on the first curve,P (0,0,1)=0 Furthermore, sinceP (X, Y, Z) is a homogeneous polynomial of degreen,Z n , the only term not containingXorY, must have a non-zero coefficient In other words,p(Y, Z)=P (1, Y, Z) remains a polynomial of degreeninZ Analogously,q(Y, Z)=Q(1, Y, Z)remains a polynomial of degreeminZ Writing these as polynomials inZwith coefficients inC[Y], we obtain p(Y, Z)=α 0 (Y )+α 1 (Y )Z+ ã ã ã +α n (Y )Z n q(Y, Z)=β 0 (Y )+β 1 (Y )Z+ ã ã ã +β m (Y )Z m

298 7 Algebraic Curves whereα n (Y )andβ m (Y )are constant Of course, the resultant of these two polynomials inZis simplyR(1, Y ).

SinceBis a point of multiplicityrfor the first curve, all the partial derivatives of p(Y, Z)up to rankr−1 vanish at(Y, Z)=(0,0) But this forces each term of the polynomialp(Y, Z)to be of degree at leastr Indeed, assume the existence of a term d Y u Z v =0 withu+v < r Differentiatingutimes with respect toY andv times with respect toZyields for this specific term the valued=0 What about the other terms? They have the formα Y u Z v with(u , v )=(u, v) Ifu > u, there remains a factorY after differentiation and therefore that result vanishes when puttingY =0.

Ifu < u, we immediately obtain 0 after differentiatingutimes with respect toY An analogous argument holds with respect toZ Thus eventually, only the termd Y u Z v will yield a non-zero value after the differentiating process indicated However, this contradicts the fact that the partial derivatives of rank u+v < r are all zero So there cannot exist a termd Y u Z v =0 withu+v < r and all the terms have degree at leastr An analogous argument holds forq(Y, Z).

We can thus be more precise when writingp(Y, Z)andq(Y, Z)as polynomials inZwith coefficients inC[Y]: p(Y, Z)=γ 0 (Y )Y r +γ 1 (Y )Y r − 1 Z+ ã ã ã +γ n (Y )Z n q(Y, Z)=δ 0 (Y )Y s +δ 1 (Y )Y s − 1 Z+ ã ã ã +δ m (Y )Z m

The resultantR(1, Y )of these two polynomials is then

Let us now multiply the last row containing the coefficientsγ i byY s , the previous row byY s − 1 , and so onstimes Let us also multiply the last row containing theδ j byY r , the previous row byY r − 1 , and so onr times We have thus multiplied the resultant by s i = 1 i+ r j = 1 j=s(s+1)

After all these multiplications, the factorY r+s appears in all the terms of thes- th column, the factorY r + s − 1 in all the terms of the(s−1)-th column, and so on. Altogether, after the various multiplications, the resultant contains at least r + s i = 1 i=(r+s)(r+s+1)

2 factorsY This shows that the original resultantR(1, Y )contained at least

Considering several pointsB i instead of one in Lemma7.7.7, we then obtain:

Proposition 7.7.8 Consider two algebraic curvesC,C , without a common component, admitting arbitrary equations

P (X, Y, Z)=0, Q(X, Y, Z)=0 of respective degrees n and m Let these curves admit the intersection points

B 1 , , B k Suppose that for each indexi:

• B i is a point of multiplicity r i on the curve C with respect to its equation

• B i is a point of multiplicity s i on the curve C with respect to its equation Q(X, Y, Z)=0.

Proof The proof is an easy adaptation of the proof of the first version of the Bezout theorem (see Theorem7.7.4).

Curves Through Points

This section investigates the existence of curves passing through some given points.

Proposition 7.8.1 The general form of a homogeneous polynomial of degreenhas n(n + 3)

Proof A homogeneous polynomial of degreencan be written as

• αcan take alln+1 values from 0 ton;

• onceαis fixed,βcan take alln−α+1 values from 0 ton−α;

• onceαandβ are fixed,γ is fixed as well sinceγ=n−α−β.

This implies that there are exactly n α = 0

Theorem 7.8.2 Consider finitely many pointsA 1 , , A k (withk≥1) inP2 (C). There exist algebraic curves of degree at mostnpassing through all these pointsA i provided k≤n(n+3)

Proof Let us fix a system of homogeneous coordinates and consider an algebraic curve with equationP (X, Y, Z)=0, whereP (X, Y, Z)is a homogeneous polynomial of degreen With the notation of Proposition7.8.1, the condition that the curve passes through the pointA i is equivalent to the condition that the linear equation α + β + γ = n a α,β,γ X i α Y i β Z i γ =0 is satisfied by the coefficientsa α,β,γ , whereX i ,Y i ,Z i are the coordinates ofA i , and thus are constants The condition that the curve passes through all the points

A i thus means that the n(n 2 + 3) +1 quantitiesa α,β,γ are solutions of a homogeneous system ofklinear equations A non-zero solution exists as soon as k 3×1 which would contradict the Bezout Theorem7.7.9.

One can be even more precise and prove thatCis necessarily reduced to the line throughAandB.

The Number of Multiplicities

Another important consequence of the Bezout Theorem is to exhibit a limitation on the possible number of multiple points of a curve.

Proposition 7.9.1 If a curveCof degreen≥2 admits the pointsA 1 , , A k with the respective multiplicitiesr 1 , , r k , then k i = 1 r i (r i −1)≤n(n−1).

Notice at once that in the case of a simple point we haver i =1 and sor i −1=0.

So the presence of infinitely many simple points does not hurt at all.

Proof WriteP (X, Y, Z)for a simple equation of the curve in a system of homogeneous coordinates such that the point

308 7 Algebraic Curves is not on the curve (see Corollary7.7.2) This forcesP (X, Y, Z)to contain a non- zero term of the form aZ n , thus to be of degree n≥2 as a polynomial in Z with coefficients in D=C[X, Y] By CorollaryD.1.5 and TheoremD.1.4, since

P (X, Y, Z)does not have any multiple factors, its resultant with ∂P ∂Z (X, Y, Z) is non-zero and thus the two curves with equations

∂Z(X, Y, Z)=0 do not have any common components The first curve is of degreenand the second one, of degree at mostn−1 (and exactlyn−1, if it does not have any multiple factors).

Trivially, if all the partial derivatives ofP (X, Y, Z)up to orderr−1 vanish at a pointB, all the partial derivatives of ∂P ∂Z (X, Y, Z)up to orderr−2 vanish atB By

Lemma7.7.8, we obtain the announced equality

Remark 7.9.2 In Proposition7.9.1, the upper boundn(n−1)cannot be improved.

Proof In Example7.4.9, we have a point of multiplicitynon a curve of degreen, thus we can chooser i =nin Proposition7.9.1 Notice that Proposition7.9.1also indicates that there cannot be any other multiple point

However, when the curve is irreducible, the upper bound can definitely be improved.

Proposition 7.9.3 If an irreducible curve C of degree n≥2 admits the points

A 1 , , A k with the respective multiplicitiesr 1 , , r k , then k i = 1 r i (r i −1)≤(n−1)(n−2).

Proof Of course there is no loss of generality in considering only those points of multiplicity at least r i =2, because r i =1 forces r i (r i −1)=0 We write

P (X, Y, Z)=0 for the simple equation of the curveC.

Consider the finitely many pairs

(A 1 , r 1−1), , (A k , r k −1), (B 1 ,1), , (B 2(n− 1) ,1) where the pointsB j are arbitrarily chosen onCand all the pointsA i ,B j are distinct.

By Proposition7.8.6, there exists a curveC

• admitting an equationQ(X, Y, Z)=0 of degreen−1;

• passing through all the pointsA i andB j ;

• such that each pointA i is of multiplicity at leastr i −1 with respect to the equation

In particular, the intersection ofCandC contains all the pointsA i andB j

SinceCis irreducible and the degree ofC is strictly smaller that the degree ofC, these two curves do not have any common components A priori, we do not know anything about the pointsB j , except of course that they are at least of multiplicity 1 with respect toP (X, Y, Z)=0 andQ(X, Y, Z)=0 Applying Proposition7.7.8we thus get: k i = 1 r i (r i −1)+2(n−1)≤n(n−1) or equivalently, k i = 1 r i (r i −1)≤n(n−1)−2(n−1)=(n−1)(n−2).

Remark 7.9.4 In Proposition7.9.3, the upper bound(n−1)(n−2)cannot be improved.

X n +Y n − 1 Z=0, n≥2 is irreducible Indeed, viewing the left hand side as a polynomial inX with coefficients in the polynomial domainD=C[Y, Z], the elementZ∈D is irreducible and trivially satisfies all the properties required of d=Z in the Eisenstein crite- rionB.4.11.

All the partial derivatives up to the leveln−2 vanish at

⎦ while the partial derivative of ordern−1 with respect toY does not ThusAis a point of multiplicityn−1 The left hand side in Proposition7.9.3thus already

310 7 Algebraic Curves yields the quantity r i (r i −1)=(n−1)(n−2) atA, proving that the bound(n−1)(n−2)cannot be improved and proving at the same time that there are no other multiple points.

Conics

Various results in this section have already been obtained, in a more general context, sometimes via different methods, in Sect.6.17 It is nevertheless useful to collect them here, together with some new results.

Convention 7.10.1 In this chapter, by a conic we always mean an algebraic curve of degree 2.

Thus when we use the term conic in this chapter, we shall not mean—for example—the case of a single point or a single line Let us first revisit the notion of an irreducible conic.

Proposition 7.10.2 For a conic in the complex projective planeP2 (C), the following conditions are equivalent:

2 the determinant of the symmetric matrix of the coefficients is non-zero;

(1⇒3) In Proposition7.9.3, choosingn=2 yields (n−1)(n−2)=0, thus one must haver i −1=0 at any given point; in other words, all points are simple.

(3⇒1) If the conic is not irreducible, it is the union of two distinct lines, thus admits a double point by Example7.4.9

Corollary 7.10.3 A conic admits a multiple point if and only if it is the union of two lines.

Proposition 7.10.4 Through five points ofP2 (C)always passes a conic.

Proof Whenn=2 in Theorem7.8.2, we get n(n+3)

2 =5 so that we can choosek=5 If the curve so obtained has degree 2, the result is proved Otherwise, it is a line and its union with an arbitrary other line yields a conic

By Counterexample6.17.6, the bound 5 in Corollary7.10.4cannot be improved.

Proposition 7.10.5 If two conics have five common points, they have a common component.

Proof By the Bezout Theorem7.7.4, two conics without a common component can have at most four intersection points

Let us also recall Proposition6.17.5, which, in view of Proposition7.10.2, translates as:

Proposition 7.10.6 Through five points ofP2 (C), no three of which are on the same line, passes exactly one conic and this conic is irreducible.

Proof By Proposition7.10.4 there exists a conic through these five points If the conic is not irreducible, it is the union of two lines, thus at least three of the five points lie on one of these lines This is a contradiction.

Cubics and the Cramer Paradox

We switch now to the case of cubics:

Definition 7.11.1 By a cubic we mean an algebraic curve of degree 3.

The Bezout Theorem7.7.4tells us in particular that two curves of degree 3 without a common component have at most nine intersection points Furthermore, this bound nine can easily be reached.

Example 7.11.2 Two cubics admitting exactly nine intersection points.

Proof Simply consider inR 2 the two curves with equations

Y =X(X−1)(X+1), X=Y (Y−1)(Y+1) pictured in Fig.7.16 They already have nine intersection points, thus these become intersection points of their projective extensions

Y Z 2 =X(X−Z)(X+Z), XZ 2 =Y (Y−Z)(Y +Z) further viewed as curves inP2 (C)

On the other hand Theorem7.8.2tells us that through nine points always passes a cubic This bound of nine cannot possibly be improved.

Counterexample 7.11.3 Ten points of P2 (C)through which it is impossible to draw a cubic.

Proof We are looking for a cubicCpassing through the pointsA 1 , , A 10as pictured in Fig.7.17 The four pointsA 1 , A 2 , A 3 , A 4are on the same lined, thus by the Bezout Theorem7.7.4, that linedmust be a component ofC For the same reason, the lined throughA 1 , A 5 , A 6 , A 7 must be a component ofC But then the simple equation ofCis necessarily the product of three linear factors andCis the union of dandd and a third lined SinceA 8andA 9are not ondord , they are ond ; thus d is the line throughA 8 andA 9 But the last pointA 10 is not onC=d∪d ∪d

Putting together Example7.11.2and Counterexample7.11.3we get:

Nine points do not suffice to characterize a cubic, but ten points are too many

If we consider the case of a curve of degree 4, which is called a quartic, we have this time in Theorems7.7.4and7.8.2 n 2 , n(n+3)

Moreover adapting in a straightforward way Example 7.11.2 and Counterexam- ple7.11.3, we get:

Sixteen points do not suffice to characterize a quartic, but fifteen points are too many

This paradox was first pointed out by Colin MacLaurin (1698–1746) To dissipate this paradox, Gabriel Cramer (1704–1752) created the theory of determinants. So: how do we resolve this paradox? To understand what is happening, it suffices to go back to the easier case of conics We have this time in Theorems7.7.4and7.8.2 n 2 =4, n(n+3)

Four points do not suffice to characterize a conic, and through five points one can always draw a conic

In this case, the paradox does not appear Furthermore, Proposition7.10.6provides additional precision, which we could rephrase intuitively as:

Five points “in general position” determine a unique conic

Indeed, the precise nature of the family of points is certainly needed to characterize a unique conic: even a million points are not sufficient to characterize a conic, if you put all these points on the same line!

Thus clearly, if we intend to improve Theorem7.8.2in order to obtain the existence of a unique curve of degreenpassing through the given points, it is necessary to give a precise meaning to the expression “a family of points in general position”. Notice that Warning7.8.3tells us at once the answer to this question: the system of equations considered in the proof of Theorem7.8.2must be of rank n(n 2 + 3) With these observations and this terminology in mind, we can rephrase Cramer’s paradox as:

Nine points do not suffice to characterize a cubic, but nine points “in general position” do characterize a unique cubic

This time we are no longer amazed The explanation is most probably that in the first part of this “paradox”, the nine points are not “in general position” Notice that indeed, in Example7.11.2used to illustrate the first line of the paradox, the nine

314 7 Algebraic Curves points are by no means an “arbitrary family of points”, but a family of points obtained as the intersection of two cubics As the following theorem indicates, passing through all these nine points is a redundant condition.

Theorem 7.11.4 If two cubics intersect in exactly nine points, every cubic passing through eight of these points necessarily passes through the ninth point.

Proof Let us writeP (X, Y, Z)=0 andQ(X, Y, Z)=0 for the simple equations of two cubicsCandC intersecting in the nine pointsA 1 , , A 9 Consider further a cubicC with simple equationH (X, Y, Z)=0 passing throughA 1 , , A 8 We shall prove by reductio ad absurdum thatC also passes throughA 9.

Suppose thatC does not contain A 9 Consider all the cubics C admitting an equation of the form α P (X, Y, Z)+β Q(X, Y, Z)+γ H (X, Y, Z)=0.

Among these we haveC andC whose equations are not proportional, since they have only nine intersection points ButC is also a cubic of this form and its equation is not a linear combination of the equations ofCandC , sinceA 9is onCandC , but not onC Thus identifying a homogeneous equation of degree 3 inX,Y,Z with its list of coefficients, we conclude that all the equations above constitute a vector space of dimension at least 3 Imposing that a cubic of this formCpasses through two specified additional points is equivalent to imposing two linear equations on the coefficients ofC, and since we are in a vector space of dimension 3, there is always a non-zero solution, thus there always exists a cubicC passing throughP 1 , , P 8 and two arbitrarily specified pointsBandC.

We observe next that four of the pointsA 1 , , A 9 can never be on the same lined, otherwise by the Bezout Theorem7.7.4, the lined would be a component of bothCandC , which would then have infinitely many intersection points In an analogous way, seven of the pointsA 1 , , A 9 can never be on a conic, because again this conic would be a component of bothCandC

Let us now split the problem into three cases, which we shall treat separately:

1 three of the pointsA 1 , , A 9are on the same line;

2 six of the pointsA 1 , , A 9are on the same conic;

3 the first two possibilities never occur.

Thus in each case we aim to reach a contradiction.

First, assume thatA 1 , A 2 , A 3are on the same lined By Proposition7.10.4, we consider a conicQpassing throughA 4 , A 5 , A 6 , A 7 , A 8 Such a conic is necessarily unique Indeed ifQ is another such conic, once more by the Bezout Theorem7.7.4,

QandQ have five common points, thus a common component Therefore

Q=d 0∪d 1 , Q =d 0∪d 2 for three linesd 0 , d 1 , d 2 Notice that outsided 0 , the only possibility for one of the five pointsA 4 , , A 8 to be on bothQandQ is to be at the intersection ofd 1 and d 2 This means that all the other points—thus at least four—are ond 0 But as we have seen earlier in this proof, it is impossible to have four of the pointsA 1 , , A 9 on the same line Thus indeed, the conicQas above is unique.

Let us now choose a pointBon the lined, distinct fromA 1 , A 2 , A 3, and a point

Cwhich is not on the linedor on the conicQ Choose a cubicCas above which contains these two additional pointsB andC SinceCcontainsBand alsoA 1 , A 2 , A 3 it intersects the lined at four distinct points, thus by the Bezout Theorem 7.7.4, the lined must be a component of the cubicC ThusC is the union of the lined and a conicQ SinceA 4 , , A 8 andC are not on d, they must be onQ But sinceQ containsA 4 , , A 8, by uniqueness ofQ, we must haveQ =Q This is a contradiction, sinceC∈Q.

Second, let us now assume thatA 1 , , A 6 are on a conicQ and let us write d for the line throughA 7 andA 8 Choose now forB a point onQdistinct from

A 1 , , A 6and forC, a point which is not onQor ond We consider again a cubic

Cas above containing these two additional pointsBandC The conicQthen has at least seven intersection points with the cubicC, thus by the Bezout Theorem7.7.4,

Qis a component ofC The second component ofCmust then be a line containing the remaining pointsA 7 ,A 8 ,C, which is impossible sinceC is not on the lined throughA 7andA 8.

It remains to consider the last case: three of the pointsA 1 , , A 9 are never on the same line and six of these points are never on the same conic This time choosed to be the line throughA 1 andA 2 Choose further the pointsB,C ond, distinct fromA 1 ,A 2 The conicCas above, containing the pointsBandC, has four intersection points withd; thusdis a component ofCby the Bezout Theorem7.7.4.

By assumption of this third case, no other pointA 3 , , A 8can be ond The other component ofCmust therefore be a conic containing the six pointsA 3 , , A 8 But again, this contradicts the assumption of this third case

As corollary we get at once:

Corollary 7.11.5 If two cubics intersect in exactly nine points and if exactly six of these points are on a conicQ, the remaining three points are on a lined.

Inflexion Points of a Cubic

We continue our study of cubics by investigating the amazing properties of their inflexion points.

In the Euclidean plane, Alexis Clairaut (1713–1765) proved that a cubic always has one, two or three inflexion points and Jean-Pierre de Gua (1712–1785) showed that when there are three of them, they are on the same line Once more, inP2 (C), the situation is much richer.

Lemma 7.12.1 InP2 (C), a cubic without a multiple point always has an inflexion point.

Proof By Corollary7.7.3, the cubic is irreducible By Proposition7.7.1, the cubic and its Hessian curve meet in at least one point, which is thus an inflexion point by

Lemma 7.12.2 Let C be a cubic ofP2 (C) without any multiple points and admitting the inflexion point A (see Lemma 7.12.1) In some convenient system of homogeneous coordinates:

• the cubicCadmits an equation of the form

• the inflexion pointAadmits the coordinates

⎦ and is the unique point of the cubic on the line with equationZ=0; moreover that line is tangent to the cubic atA.

Proof Let us choose a second pointBon the tangent to the cubic atA Let us first start with an arbitrary system of homogeneous coordinates in which

The tangent atAis thus the line with equationZ=0 SinceAis an inflexion point, this line admits A as a triple point of intersection with the cubicC (see Defini- tion7.6.2) and is not contained in this cubic, because C is irreducible by Corol- lary7.7.3 By Corollary7.3.5, the lineZ=0 thus admitsAas unique intersection point with the cubicC In particular,B∈C Let us write P (X, Y, Z)=0 for the equation of the cubic in the chosen system of coordinates.

Saying thatAis an inflexion point is thus equivalent to saying thatZ=0 has a triple intersection with the cubic atA, that is,t=0 is a triple root of the polynomial

P (X, Y, Z)=αX 3 +βY 3 +γ X 2 Y+δXY 2 + ã ã ã where the terms not appearing all have a factorZ The condition thatt=0 is a triple root ofP (1−t, t,0)=0 thus reduces to

This means that the coefficients of the terms of degree at most 2 are zero, that is successively,α=0 thusγ =0 thusδ=0 On the other handβ=0, becauseB is not a point ofC So finally, choosingβ=1, the equation of the cubic becomes

The three partial derivatives ofP (X, Y, Z)are thus:

∂Z =aX 2 +bXY+cY 2 +2dXZ+2eY Z+3f Z 2

The first two partial derivatives vanish atA, thus the third one does not, because the cubic does not admit any multiple points Thereforea=0.

ChoosingZ=0 as the line at infinity, the affine trace of the cubic is obtained by puttingZ=1 and admits the equation

Y 3 +aX 2 +bXY+cY 2 +dX+eY +f =0.

Viewing this as an equation of degree 2 inX aX 2 +(bY+d)X+

Let us now switch to a more adequate system of homogeneous coordinates, via the change of coordinates

This new system of coordinates thus admits the same “line at infinity”Z=0 as the previous one The homogeneous coordinates of the inflexion pointA remain unchanged since they are now

Let us write further p(Y )=(bY+d) 2 −4a(Y 3 +cY 2 +eY +f )

In the new system of coordinates, we end up with an equation

X 2 =p(Y ) for the affine trace of the curve, wherep(Y )is a polynomial of degree 3.

Let us prove by reductio ad absurdum that the three roots ofp(Y )are distinct.

Ifris a double root ofp(Y ), it is also a root ofp (Y )(see PropositionA.8.3) But then the point with affine coordinates(0, r)is on the cubic, sincep(r)=0, and the two partial derivatives at this point of the polynomial q(X, Y )=X 2 −p(Y ) vanish, sincep (r)=0 WriteQ(X, Y , Z)for the homogeneous polynomial associated withq(X, Y )(see Sect.C.2); the equationQ(X, Y , Z)=0 is thus that of the cubic in the chosen system with homogeneous coordinates We know that the point with homogeneous coordinates

⎦ is on the cubic, thusQ(0, r,1)=0 On the other hand we have trivially

By Euler’s formulaC.1.5, we obtain further

But thenCis a double point of the cubic, which contradicts the assumption onC.

We have thus already obtained for the cubic an equation of the form

X 2 Z=αY 3 +βY 2 Z+γ Y Z 2 +δZ 3 where the polynomial p(Y )=αY 3 +βY 2 +γ Y+δ has three distinct roots Observe that choosingεto be a root of p(Y )(see Theo- remF.3.3), the change of coordinate

1Y=√ 3 α Y+ε transforms the polynomialp(Y )into a polynomial of the form p(11Y )=1Y 3 +u1Y 2 +v1Y=1Y1Y 2 +u1Y+v

Thus the new change of coordinates

320 7 Algebraic Curves transforms the equation of the cubic into

Trivially, this change of coordinates does not change the coordinates ofAor the

Since the three roots ofp(Y )are distinct, so are the three roots ofp(11Y ) Thus the two roots of1Y 2 +u1Y+vare distinct, proving thatu 2 −4v=0 Since these two roots are also distinct from the third root 0 ofp(11Y ), we have furtherv=0

Theorem 7.12.3 A cubic ofP2 (C), without any multiple points, possesses exactly nine inflexion points Every line through two of these inflexion points contains a third one.

Proof Let us work in the system of coordinates given by Lemma7.12.3, withAan arbitrary inflexion point By Corollary7.7.3, the cubic is irreducible, thus certainly, does not contain any linear components The inflexion points ofCare thus the intersection points of the cubic and its Hessian curve with equation (see Theorem7.6.5) det

=0 that is, after division by 8,

The pointAgiven in Lemma7.12.3is an inflexion point and is the only point of the cubic on the “line at infinity” Z=0 Thus the other inflexion points—if any—will be found among the “affine” points, that is, the points admitting the last coordinateZ=1 We therefore have to solve the system of equations

Introducing into the second equation the value ofX 2 given by the first equation, we obtain

Writingq(Y )for the polynomial on the left hand side, we have q (Y )Y 3 +12uY 2 +12vY Y

(Observe at once the similarity with the right hand side in the equation of the cubic.) The resultant ofq(Y )andq (Y )(see DefinitionD.1.3) is then

Since by Lemma7.12.2,v=0 andu 2 −4v=0, this resultant is non-zero Therefore q(Y )andq (Y )do not have any common irreducible factors (see TheoremD.1.4) andq(Y ) does not have any multiple irreducible factors (see CorollaryD.3.1) In other words, q(Y )and q (Y )do not have any common roots andq(Y ) does not have any multiple roots In particular,q(Y )has four distinct roots which we denote byb 1 ,b 2 ,b 3 ,b 4

To find the inflexion points, it remains to re-introduce these valuesb i into the first equation of the system and successively solve the four equations

But these equations can be re-written precisely as

Since none of theb i ’s is a root ofq (Y ), the right hand side is a non-zero constant, so that the eight affine inflexion points are ± q (b i )

Together with the inflexion pointAat infinity, we obtain the nine announced inflexion points.

Considering now the three inflexion points

⎦ we observe at once that

Thus these three points are on the same line In other words, the line passing through

Aand an affine inflexion point always contains a third inflexion point SinceAwas an arbitrarily chosen inflexion point of the cubic, this concludes the proof.

The Group of a Cubic

Our next concern regarding cubics is to prove that a group can be naturally associated with a cubic Such groups are of particular interest in coding theory.

Lemma 7.13.1 LetCbe an irreducible cubic.

1 IfAandBare two distinct simple points of the cubic, the line throughAandB admits a triple intersection with the cubic, corresponding to three simple points of the cubic:A,Band a third point that we write asA B; possibly,A B can be one of the two pointsAorB.

2 IfC is a simple point of the cubic, the tangent atC to the cubic admits a triple intersection with the cubic, all corresponding to simple points of the cubic:C, where the tangent has a (at least) double intersection with the cubic and a second point which we write asC C; possibly,C C=C.

Proof Since the cubic is irreducible, it does not contain a line Thus in both cases, by Corollary7.3.5, the line indicated has with the cubic a triple intersection By the second version of the Bezout Theorem (see7.7.9), the pointsA BandC Cof the statement are necessarily simple

Notice thatA B=Ameans that the line throughAandBis tangent to the cubic atA(see Definition7.4.5), whileC C=Cmeans thatCis an inflexion point (see Definition7.6.2).

Lemma 7.13.2 Let C be an irreducible cubic of P2 (C)and , the operation of Lemma7.13.1 Given four simple pointsB,A,C,DofC

Proof We refer to Fig.7.19 Consider the two cubics, each comprising the three lines indicated

The cubicCcontains eight of the intersection points ofC andC , namely

By Theorem7.12.3,Calso contains the ninth intersection point ofC andC , which is thus the intersection of the two lines

However this additional intersection point with the cubicC, considered successively on the first line, a component ofC , and the second line, a component of C , is respectively given by

So these two quantities are equal

Theorem 7.13.3 LetCbe an irreducible cubic ofP2 (C) WriteC 1 for the set of its simple points and consider onC 1 the operationof Lemma7.13.1 For every fixed pointO∈C 1, the operation

+: C 1×C 1−→C 1 , (A, B)→A+B=O (A B) is an abelian group operation onC 1 with zero element O Moreover, two distinct choices ofOprovide isomorphic groups.

Proof The commutativity of+follows at once from the commutativity of On the other hand the three points

A, O, A O are collinear by definition, thus

A=O (A O)=A+O andOis a neutral element for the operation+.

Let us next observe that

−A=A (O O) is the inverse ofAwith respect to the operation+ We must prove that

(see Fig.7.20) The three points

A+(−A)=O (O O)=O because the line throughOandO Ois tangent to the cubic atO.

It remains to prove the associativity of+ Since

(A+B) C it suffices to prove that

This is the case by Lemma7.13.2.

Let us now prove that choosing another “zero element”O1∈C 1, the corresponding group structure onC 1

A1+B=O (A B)1 is isomorphic to that constructed fromO The isomorphism is defined by ϕ(A)=A (O O).1 Notice at once that

This proves thatϕis its own inverse and thus, is a bijection.

Observe next that since the three points

Thusϕrespects the zero elements.

It remains to show that ϕ(A+B)=ϕ(A)1+ϕ(B) that is

Applying Lemma7.13.2twice, we get the following first two equalities

(O O)1 while the last equality holds because

Let us view the spaceP2 (C)as the set of equivalence classes inC 3 \(0,0,0)for the equivalence relation identifying two proportional vectors (see Definition6.1.1). ProvidingC 3 with its usual topology, this induces a corresponding quotient topology onP2 (C); this quotient topology further induces a topology on each algebraic curve

C⊆P2 (C)(see Appendix A in [8], Trilogy III) It turns out that on a cubic, the group operations of Theorem7.13.3are continuous with respect to this topology:the group of a cubic is then called a topological group However, we shall not enter into these considerations here.

Rational Curves

The goal of this section is simply to state Definition7.14.1: the notion of a rational curve The somewhat unexpectedly unpleasant form of this definition calls for some easy examples to explain the motivation behind the technical conditions involved there These examples will also give us an opportunity to justify the terminology.

In Sect.1.3we already mentioned Euler’s idea of representing curves via parametric equations In the Euclidean plane, the idea is clearly that “when the parameter runs along the real line, the corresponding point of the curve runs along that curve”. Let us investigate further the very simple example of the circle with equation

X 2 +Y 2 =1 in the Euclidean plane, which is generally described by the system of parametric equations

Of course the system above, which uses trigonometric functions, drastically es- capes the purely algebraic context But using the trigonometric formulas cosθ=1−tan 2 θ 2

1+tan 2 θ 2 these parametric equations can be re-written

Further puttingt=tan θ 2 , we obtain the system of parametric equations

This time each coordinate X,Y is expressed as a rational fraction—that is, the quotient of two polynomials—in terms of a parametert We are now safely back in a purely algebraic environment.

Let us now view the Euclidean plane as being embedded in real projective space (see Sect.6.20) In homogeneous coordinates, the points of the circle are characterized by

Z=1 that is, since homogeneous coordinates are defined up to a multiple

This time, in homogeneous coordinates, the three coordinates are even expressed as actual polynomials in terms of a parametert.

More generally consider an arbitrary rational parametric representation

Y =γ (t ) δ(t ) withα(t ), β(t ), γ (t ), δ(t ) polynomials This always gives rise in homogeneous coordinates—multiplying by β(X)δ(X)—to a polynomial parametric representa-

Z=θ (t ) in homogeneous coordinates translates as a rational parametric representation

Y =ψ (t ) θ (t ) of the affine trace So the slogan could be

Of course, this “slogan” should be considered with care Indeed for it to make sense, the “affinely rational” character requires the values of the denominators always to be non-zero, while the “projectively polynomial” character requires the three component never to be simultaneously zero Trivially the “slogan” holds only if we accept to “overlook” finitely many points.

However, the observant reader will have noticed that we have not been very careful on another point in the discussion above Whent=tan θ 2 runs along the real line, θ

2varies in the interval]− π 2 ,+ π 2 [, thusθvaries in the interval]−π,+π[, while we started with a parametric representation whereθ was running along the whole real line Notice nevertheless an interesting improvement: while in the original representation, infinitely many values ofθ give rise to the same point of the circle, in the “rational representation”, to each value oft corresponds a unique point of the circle On the other hand, we “miss” the point(−1,0)of the circle, which should correspond tot= ±∞!

Notice further that when working in the complex case, the (non-constant) denominator of a rational fraction always has roots by TheoremF.3.3, thus the rational representation is not defined for the finitely many values oft which are roots of the denominator In the case of the “circle” viewed as a complex curve, fort= ±i Of course, in the projective case, given a polynomial representation

Y=γ (t ) if the three polynomialsα(t ),β(t ),γ (t )have a common roott 0, the corresponding triple

⎦ does not correspond to any point of the projective plane.

So indeed, when looking for the existence of a possible rational polynomial parametrization of a curve, we should look for such a parametrization which “works everywhere except possibly at finitely many points”.

However, there is another difficulty Consider the curve already considered in Counterexample7.4.1

Writing the affine trace of this curve in the form

1−X, we observe at once that a polynomial parametrization of this curve is given by

Notice again that this parametrization misses the point

“at infinity” on the curve However, a new phenomena occurs here: this time, the two valuest= ±1 of the parameter correspond to the same affine point(0,0)of the curve This was of course “predictable”: the origin is a double point, a point through which the curve passes twice, thus a point obtained for two distinct values oft. Thus, when looking for the existence of a possible rational/polynomial parametrization of a curve, we should also require the uniqueness of the parametert describing a point of the curve, everywhere, except possibly at the multiple points: there are again only finitely many of them by Proposition7.9.1.

All this suggests that an efficient definition of a rational curve should take the following form:

Definition 7.14.1 An algebraic curveC of P2 (C)is rational when there exists a polynomial parametric representation

Z=γ (t ) withα(t ),β(t ),γ (t )complex polynomials such that, in a given system of homogeneous coordinates:

1 for each valuet∈C, except possibly for finitely many of them,

⎦ are the homogeneous coordinates of a point ofC;

2 each point ofC, except possibly finitely many of them, admits homogeneous coordinates of the form

⎦ for a unique valuet∈Cof the parameter.

The considerations of this section have in particular provided the following two examples:

Example 7.14.2 The algebraic curve with equation

Example 7.14.3 The algebraic curve with equation

A Criterion of Rationality

Let us now establish a sufficient condition for a curve to be rational In view of Proposition7.9.3, the condition can be rephrased as: the curve has as many multiplicities as possible.

Theorem 7.15.1 LetC be an irreducible curve of degree n admitting the points

A 1 , , A k with respective multiplicitiesr 1 , , r k A sufficient condition forC to be rational is that k i= 1 r i (r i −1)=(n−1)(n−2).

Proof Of course there is no loss of generality in assuming that r i >1 for each indexi Let us write P (X, Y, Z)=0 for a simple equation of the curve in some system of homogeneous coordinates: we keep it arbitrary for the time being, but we shall choose it more precisely during the proof For the sake of clarity, we split the proof into several steps.

Step 1 Let us consider the finitely many pairs

C ,1 where the pointsA i ,B j ,C are distinct and the pointsB j ,C are arbitrarily chosen on the curveC The assumption in the statement immediately implies k i = 1

In view of Proposition7.8.6, there exists an algebraic curveC

• admitting an equationQ(X, Y, Z)=0 of degreen−1;

• admitting the pointsA i as points of multiplicitiesr i with respect toQ;

• passing through the pointsB j andC

Step 2 A curveC satisfying the conditions just indicated does not contain any other intersection point withCthan the pointsA i ,B j andC Indeed, if there were another intersection pointD, considering the multiplicities of the intersections of the curveCand the curveC , the formula in Proposition7.7.8would become k i = 1 r i (r i −1)+(2n−3)+1+1=(n−1)(n−2)+2(n−1)+1=n(n−1)+1

332 7 Algebraic Curves so thatCandC would have a common component This is impossible sinceChas a simple irreducible equation of degreenandC has an equation of lower degree. Step 3 In the discussion above, replacingC by another point C we find an algebraic curveC

• admitting an equationS(X, Y, Z)=0 of degreen−1;

• admitting the pointsA i as points of multiplicitiesr i with respect toS;

• passing through the pointsB j andC

Notice that by Step 2 of this proof, the two curves C andC are necessarily distinct, otherwiseC would have the additional intersectionC withC So the poly- nomialsQ(X, Y, Z)andS(X, Y, Z)are not proportional.

Step 4 If follows at once that the curveC γ of equation

• has an equation of degreen−1;

• admits the pointsA i as points of multiplicitiesr i with respect toQ+γ S;

• has at most one additional intersection pointD γ withC

• does not have a common component withC.

The last but one assertion is once more a consequence of Step 2 The last assertion again follows from the fact thatP (X, Y, Z)is an irreducible polynomial of degreen, which cannot be a factor of a polynomial of lower degree.

We shall now prove that—except possibly for finitely many values ofγ—the additional intersection pointD γ always exists and its coordinates can be expressed as polynomials in terms ofγ This will yield the expected polynomial parametrization. Step 5 Let us now choose a system of homogeneous coordinates such that:

• the line “at infinity” with equationZ=0 does not contain any of the pointsA i ,

This is clearly possible since there are only finitely many points to be avoided in the second requirement.

SinceU andV are not on C, the polynomial P (X, Y, Z)contains both a term inX n and a term inY n Analogously, the polynomialsQ(X, Y, Z)andS(X, Y, Z) contain a term inX n − 1 and a term inY n − 1

Step 6 Let us now verify that given two valuesγ,γ such that the additional intersection pointsD γ ,D γ of Step 4 exist, ifD γ =D γ but is not the pointC ,thenγ=γ

By Step 2,C andCintersect only atA i ,B j andC ThusD γ =D γ which is onC does not lie onC Therefore writing(u, v)for the affine coordinates of this point,S(u, v,1)=0 But since this point is both onC γ andC γ , we have

Step 7 Let us writeR(Y, Z, γ )for the resultant of

P (X, Y, Z), Q(X, Y, Z)+γ S(X, Y, Z) viewed as polynomials inXwith coefficients inC[Y, Z] This resultant is non-zero because the two curvesC andC γ do not have a common component (see Theo- remD.1.4) ThusR(Y, Z, γ )is a homogeneous polynomial of degreen(n−1)in

Step 8 Consider nowP (X, Y,0)andQ(X, Y,0)as polynomials inX with coefficients inC[Y] Of course, these remain homogeneous polynomials of respective degreesnandn−1 inX, since they contain respective terms inX n andX n − 1 Their resultant is thus simplyR(Y,0,0) If this resultant is zero, the two polynomials have a non-constant common componentt (X, Y )(see TheoremD.1.4) Thus

Sincet (X, Y )is non-constant, it contains at least one of the two variables: let us say, the variableX FixingY 0arbitrarily, sinceCis algebraically closed (see The- oremF.3.3) we infer the existence of anX 0 such thatt (X 0 , Y 0 )=0 But then the point with coordinates

⎦ is both onCandC , which contradicts the choice of the system of coordinates Thus

R(Y,0,0)is non-zero This proves that the homogeneous polynomialR(Y, Z, γ ) contains a term inY n(n − 1)

Step 9 All the multiple pointsA 1 , , A k ofCare in the “affine plane”, that is, can be presented with a last coordinateZ=1 (see Sect.6.20) We shall therefore work “affinely” and consider the polynomials

ClearlyP (X, Y,1)still has a term inX n and a term inY n , whileQ(X, Y,1)and

S(X, Y,1)have a term inX n − 1 andY n − 1 , since this is the case forP (X, Y, Z), Q(X, Y, Z)andS(X, Y, Z) Therefore the resultant of

P (X, Y,1), Q(X, Y,1)+γ S(X, Y,1) as polynomials inXwith coefficients inC[Y]is simplyR(Y,1, γ ) This is thus a polynomial of degreen(n−1)inY, since this is the case forR(Y, Z, γ ) We shall write

Step 10 Let us now exclude the finitely many roots of the polynomiala n(n − 1) (γ ), that is, let us restrict our attention to thoseγ such thatR(Y,1, γ )remains a polynomial of degreen(n−1) It is then the resultant of the two polynomials

By Lemma7.7.7, the pointsA i andB j with coordinates

⎦ are such that(d i ,1)and(d j ,1)are roots ofR(Y, Z, γ ), thus d i , d j are roots of

SinceR(Y,1, γ )hasn(n−1)roots counted with their multiplicities (see Proposi- tionC.1.4), there remains an additional last root or an additional multiplicity of one of the roots already mentioned.

But the sum of all the roots of the polynomialR(Y,1, γ )is simply

Thus the last root is given by Φ(γ )= −a n(n − 1) − 1 (γ ) a n(n − 1) (γ ) − k i = 1 r i (r i −1)d i −

Observe thatΦ(γ )is a rational fraction in terms of the variableγ.

Step 11 Let us now observe that—except possibly for finitely many values of γ—Φ(γ )is theY-coordinate of the only additional intersection pointC γ considered in Step 4.

First, ifR(Y 0 ,1, γ )=0 for someY 0∈C, then the two polynomials

P (Y 0 ,1, γ ), Q(Y 0 ,1, γ )+γ S(Y 0 ,1, γ ) have a common factor by TheoremD.1.4, that is, a common rootX 0 Therefore

⎦ is an intersection point ofCandC γ This is in particular the case forY 0 =Φ(γ ).

If it turns out that this intersection point is someA i , thend i is a root of multiplicityr i (r i −1)+1 ofR(Y,1, γ ), not just of multiplicityr i (r i −1) Then the

A i and theB j already exhaust all the multiplicities of the roots ofR(Y,1, γ ) By CorollaryA.8.4, this is equivalent to the additional requirement thatd i is also a root of the derivative of orders=r i (r i −1)+2 ofR(Y,1, γ ) Let us simply writeR (s) for that derivative The quantityR (s) (d i ,1,0)is not zero, otherwise theA i andB j would exhaust all the possibilities of intersection points ofCandC 0=C , which is not the case: there exists the additional intersection pointC But thenR (s) (d i ,1, γ ) is not the zero polynomial inγ, thus it can take the value zero only for finitely many values ofγ Let us exclude those finitely many values ofγ Furthermore, let us exclude in the same way the finitely many values ofγ obtained by the same process, using successively all the other pointsA i andB j

After having excluded these finitely many values ofγ, we can thus assert that Φ(γ )is theY-coordinate of the only additional intersection pointD γ ofCandC γ Step 12 We can now repeat the same argument, considering instead the resultant of P (X, Y, Z) andQ(X, Y, Z)+γ S(X, Y, Z) viewed as polynomials inY with coefficients inC[X, Z] This leads to another rational fractionΨ (γ )which is theX- component of the pointD γ , possibly after having eliminated another finite number of values ofγ.

Step 13 We are now ready to prove that

Y =Φ(γ ) is a rational parametrization of the affine trace ofC Writing further Ψ (γ )=ψ 1 (γ ) ψ 2 (γ ), Φ(γ )=φ 1 (γ ) φ 2 (γ ), ψ i (γ ), φ j (γ )∈C[γ],

336 7 Algebraic Curves we shall obtain, as already observed in Sect 7.14, the expected polynomial parametrization

By construction, for all but finitely many values of γ, we know that the pair

(Ψ (γ ), Φ(γ ))yields the affine coordinates of a pointD γ ofC(in fact, of an intersection point ofCandC γ ).

SinceCis irreducible, it does not contain any line, thus in particular the “line at infinity” of equationZ=0 is not a component ofC By Corollary7.3.5,Cthus has at mostnpoints “at infinity” on the lineZ=0 We can therefore ignore those points to check the validity of Condition 2 in Definition7.14.1 We shall also ignore the various particular pointsA i ,B j ,C ,C and all the additional intersection points

D γ (if they exist) in the case of the finitely many “non-acceptable” valuesγ. Consider now a remaining affine pointEofC; since it is distinct fromA i ,B j andC , by Step 2 it does not lie onC Let us write(u, v)for the affine coordinates ofE SinceEis not onC , we haveS(u, v,1)=0 so that the equation

Problems

7.16.1 Consider an algebraic curveC with simple equationF (X, Y, Z)=0 The curveCis irreducible if and only if the ideal generated by the polynomialF (X, Y, Z) is prime in the ring of homogeneous polynomials.

7.16.2 Consider an algebraic curveC admitting an equationF (X, Y, Z)=0 and suppose that

Consider the algebraic curvesC i admitting the equations F i (X, Y, Z)=0 If for each indexi,A∈P2 (C)is a point of multiplicityr i ≥1 with respect to the equation

F i (X, Y, Z)=0, thenAis a point of multiplicityr 1 + ã ã ã +r k with respect to the equationF (X, Y, Z)=0 (See Definition7.4.10.)

7.16.3 Consider a curveCof degreenadmittingkdistinct componentsC i Suppose that for each indexi,A i ∈P2 (C)is a point of multiplicityr i onC i Then

7.16.4 An algebraic curve of degreenhas at mostn(n−1)tangents passing through a given point (This is a result of Gaudin–du Séjour.)

7.16.5 InP2 (C), the number of conics which are tangent to five given conics “in general position” is equal to 3264 (This is a result of Chasles.)

7.16.6 There exist curves of arbitrary degreenwithout multiple points.

7.16.7 If inP2 (C)the opposite sides of a hexagon intersect at three collinear points, the six vertices of the hexagon are on a conic.

7.16.8 Every cubic ofR 2 , in an adequate system of coordinates, admits one of the following equations (this is a result of Newton):

XY 2 +eY=aX 3 +bX 2 +cX+d

7.16.9 In the real affine plane, a cubic possesses one, two or three inflexion points. When there are three inflexion points, they are collinear (This is a result of Clairaut– de Gua.)

7.16.10 InP2 (C), an irreducible cubic with an ordinary double point has three inflexion points and these are collinear The equation of the cubic, in an appropriate system of coordinates, has the formY 2 =X 2 (X+1).

7.16.11 InP2 (C), an irreducible cubic with a non-ordinary double point admits only one inflexion point The equation of the cubic, in an appropriate system of coordinates, takes the formY 2 =X 3

7.16.12 Consider the group structure on the simple points of an irreducible cubic ofP2 (C) When the zero element for the group structure is an inflexion point, prove that the inflexion points constitute a sub-group comprising exactly those pointsP such thatP +P +P =P.

7.16.13 InP2 (C), prove that the group structure on the simple points of an irreducible cubic is a topological group structure.

7.16.14 Consider an algebraic curveC with simple equationF (X, Y, Z)=0 As- sume the existence of three homogeneous polynomialsG i (u, v)of the same degree (i=1,2,3) such that:

• for all but finitely many ratios u v ,

• for all but finitely many pointsAofC, there exists a unique ratio u v such that

Then the curveCis irreducible and rational.

7.16.15 The mapping ϕ:P2 (C)−→P2 (C), φ (X, Y, Z)=(Y Z, XZ, XY ) transforms a rational curve into a rational curve.

X 2 +Y 2 transforms a rational curve into a rational curve.

7.16.17 All irreducible conics are rational curves.

Exercises

7.17.1 Consider the algebraic curve with equationX 2 +Y 2 =Z 2 What are the real affine traces obtained when choosing as line at infinity, respectively,Z=0,X=0,

7.17.2 Find the multiple points of the curves

(X+Y+Z) 3 'XY Z and the corresponding tangent(s) at these points.

7.17.3 For which values ofkdoes the curve

7.17.4 For which values ofkdoes the curve kXY Z=XY 2 +XZ 2 +Y X 2 +Y Z 2 +ZX 2 +ZY 2 admit multiple points?

340 7 Algebraic Curves 7.17.5 Find the inflexion points of the cubic

7.17.6 InP2 (C), find the intersection points of the two conics

7.17.7 InP2 (C), find the intersection points of the two cubics

7.17.8 InP2 (C), find the intersection points of the two quartics

7.17.9 Determine the conic ofP2 (C)passing through the points

Why is such a conic necessarily irreducible?

7.17.10 Show by an example that all quartics passing through thirteen given points do not necessarily have a fourteenth common point.

7.17.11 Prove that the cubic with equationX 3 =Y 2 Zhas a unique inflexion point inP2 (C).

7.17.12 Prove that the cubic with equation

X 3 −Y 2 Z+X 2 Z=0 admits inP2 (C)three inflexion points which are on the same line.

In this Appendix, we assume some basic familiarity with the notion of polynomial over a fieldK: in particular how to add, subtract or multiply polynomials and the fact that these operations provide the set K[X] of polynomials over K with the structure of a commutative ring with unit Let us make clear that we always assume the multiplication of a field to be commutative.

Polynomials Versus Polynomial Functions

A polynomial with coefficients in a fieldKis thus a “formal expression” p(X)=a n X n +a n − 1 X n − 1 + ã ã ã +a 1 X+a 0 , n∈N, a i ∈K.

More precisely, this means that a polynomial is an infinite sequence

(a i ) i ∈N =(a 0 , a 1 , a 2 , , a n , a n + 1 , ) of elements ofK, with the property that

The traditional notation p(X)=a n X n +a n − 1 X n − 1 + ã ã ã +a 1 X+a 0 makes certain computations with polynomials easier, but let us be clear that two polynomials are equal when they are so, viewed as the infinite sequences(a i ) i ∈N and(b i ) i ∈N of their coefficients.

The largest integernsuch thata n =0 is called the degree of the polynomial of the polynomialp(X) As a matter of convention, the zero polynomialp(X)=0 has degree−∞ The polynomials reduced to their terma 0 are called constants.

F Borceux, An Algebraic Approach to Geometry, DOI 10.1007/978-3-319-01733-4, © Springer International Publishing Switzerland 2014

Lemma A.1.1 The product of two non-zero polynomials remains a non-zero polynomial.

Proof The term of highest degree of a non-zero polynomialp(X)∈K[X]has a non-zero coefficienta n ∈K; analogously the term of highest degree of a non-zero polynomialq(X)∈K[X]has a non-zero coefficientb m ∈K The term of highest degree inp(X)q(X)has the coefficienta n b m , which is non-zero becausea n andb m are non-zero

In view of LemmaA.1.1, it is sensible to recall the following classical terminology:

Definition A.1.2 A commutative ring with unit in which the product of two non- zero elements remains non-zero is called an integral domain.

Lemma A.1.3 Given polynomialsp(X)=0,q(X),r(X)over a fieldK, p(X)q(X)=p(X)r(X) =⇒ q(X)=r(X).

=0 and sincep(X)=0, this forcesq(X)−r(X)=0 by LemmaA.1.1 With the polynomialp(X)is associated a corresponding polynomial function p:K−→K, k→a n k n +a n − 1 k n − 1 + ã ã ã +a 1 k+a 0 where this time, the expression on the right hand side is no longer a formal one, but is an actual combination of sums and products of elements ofK For arbitrary fields

K, different polynomials can give rise to the same polynomial function: for example the polynomialsXandX 2 over the fieldZ2= {0,1} However, in this book we are mainly interested in polynomials over the fields of real or complex numbers and in those cases, there is no problem in identifying a polynomial with the corresponding polynomial function (see PropositionA.6.9).

Euclidean Division

First of all, let us recall the Euclidean division of polynomials.

Theorem A.2.1 Consider a fieldK, an arbitrary polynomialp(X)and a non-zero polynomiald(X)inK[X] There exists a unique pair(q(X), r(X))of polynomials inK[X]with the properties:

The polynomialsq(X)andr(X)are respectively called the quotient and the remain- der of the division ofp(X)byd(X).

Proof The proof is by induction on the degreenofp(X) Whenn=0,p(X)is a constanta Ifd(X)itself is a constantb, the only possibility is a=ba b +0.

Furthermore, whend(X)is not constant, the only possibility is a=d(X)ã0+a.

Let us next assume the result in degreenand considerp(X)with degreen+1. When the degree ofd(X)is strictly greater thann+1, the only possibility is p(X)=d(X)ã0+p(X).

Let us now suppose that the degree ofd(X)is less than or equal ton+1 Let us write p(X) n + 1 i = 1 a i X i , d(X) m j = 1 b j X j , m≤n+1, a n + 1=0, b m =0.

The polynomial s(X)=p(X)−a n + 1 b m X n+ 1 −m d(X) no longer has a term of degreen+1 By induction, we can thus write s(X)=d(X)t (X)+r(X), degreer(X)

Tiêu đề	An Algebraic Approach To Geometry
Tác giả	Francis Borceux
Trường học	Universitộ Catholique De Louvain
Chuyên ngành	Mathematics
Thể loại	Book
Năm xuất bản	2014
Thành phố	Louvain-la-Neuve

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Số trang	440
Dung lượng	8,28 MB