π−1(π(U ))= &
t∈R×
t U.
Since multiplication byt ∈ R× is a homeomorphism ofRn+1− {0}, the sett U is open for anyt. Therefore, their unionπ−1(π(U ))is also open.
Corollary 7.15.The real projective spaceRPnis second countable.
Proof. Apply Corollary 7.10.
Proposition 7.16.The real projective spaceRPnis Hausdorff.
Proof. LetS=Rn+1− {0}and consider the set
R= {(x, y)∈S×S|y=t xfor somet ∈R×}.
If we writexandyas column vectors, then[x y]is an(n+1)×2 matrix, andRmay be characterized as the set of matrices[x y]inS×Sof rank≤1. By a standard fact from linear algebra, this is equivalent to the vanishing of all 2×2 minors of[x y] (see Problem B.1). As the zero set of finitely many polynomials,Ris a closed subset ofS×S. Since∼is an open equivalence relation onS, andRis closed inS×S, by Theorem 7.8 the quotientS/∼ RPnis Hausdorff.
7.7 The Standard C∞Atlas on a Real Projective Space
Let[a0, . . . , an]be the homogeneous coordinates on the projective spaceRPn. Al- thougha0is not a well-defined function onRPn, the conditiona0=0 is independent of the choice of a representative for[a0, . . . , an]. Hence, the conditiona0=0 makes sense onRPn, and we may define
U0:= {[a0, . . . , an] ∈RPn|a0=0}. Similarly, for eachi=1, . . . , n, let
Ui := {[a0, . . . , an] ∈RPn|ai =0}. Define
φ0:U0−→Rn by
[a0, . . . , an] → a1
a0, . . . ,an a0
. This map has a continuous inverse
(b1, . . . , bn)→ [1, b1, . . . , bn]
and is therefore a homeomorphism. Similarly, there are homeomorphisms for each i=1, . . . , n:
φi:Ui −→Rn [a0, . . . , an] →
a0
ai, . . . ,a$i
ai, . . . ,an ai
,
where the caret sign$overai/ai means that that entry is to be omitted. This proves thatRPnis locally Euclidean with the(Ui, φi)as charts.
On the intersectionU0∩U1,a0=0 anda1=0, and there are two coordinates systems
[a0, a1, a2, . . . , an]
a0 a1
,a2 a1
, . . . ,an a1
. a1
a0
,a2 a0
, . . . ,an a0
φ1
φ0
Let us call the coordinate functions onU0,x1, . . . , xn, and the coordinate functions onU1,y1, . . . , yn. OnU0,
xi = ai
a0, i=1, . . . , n, and onU1,
y1= a0
a1, y2=a2
a1, . . . , yn =an a1. Then onU0∩U1,
y1= 1
x1, y2= x2
x1, y3= x3
x1, . . . , yn=xn x1; that is,
φ1◦φ0−1(x)= 1
x1,x2
x1,x3
x1, . . . ,xn x1
.
This is aC∞function becausex1 = 0 onφ0(U0∩U1). On any otherUi∩Uj an analogous formula holds. Therefore, the collection{(Ui, φi)}i=0,...,nis aC∞atlas for RPn, called thestandard atlas. This concludes the proof thatRPnis aC∞manifold.
7.7 The StandardC∞Atlas on a Real Projective Space 73
Problems
7.1.* Quotient space by a group
Suppose a left action of a topological groupGon a topological spaceSis continuous;
this simply means that the mapG×S −→ S describing the action is continuous.
Define two pointx, yofSto be equivalent if there is ag∈Gsuch thaty =gx. Let G\Sbe the quotient space. Prove that the projection mapπ: S−→G\Sis an open map.
7.2. The GrassmannianG(k, n)
The GrassmannianG(k, n)is the set of allk-planes through the origin inRn. Such ak-plane is a linear subspace of dimensionkofRnand has a basis consisting ofk linearly independent vectorsa1, . . . , akinRn. It is therefore completely specified by annìkmatrixA= [a1ã ã ãak]of rankk, where therankof a matrixA, denoted by rkA, is defined to be the number of linearly independent columns ofA. This matrix is called amatrix representativeof thek-plane. (For properties of the rank, see the problems in Appendix B.)
Two bases a1, . . . , ak andb1, . . . , bk determine the samek-plane if there is a change of basis matrixg= [gij] ∈GL(k,R)such that
bj =
i
aigij, 1≤i, j ≤k.
In matrix notation,B=Ag.
LetF (k, n)be the set of alln×kmatrices of rank k, topologized as a subspace ofRn×k, and∼the equivalence relation:
A∼B iff there is a matrixg∈GL(k,R)such thatB=Ag.
In the notation of Problem B.3,F (k, n)is the setDmaxinRn×k and is therefore an open subset. There is a bijection betweenG(k, n)and the quotient spaceF (k, n)/∼.
We give the GrassmannianG(k, n)the quotient topology onF (k, n)/∼.
(a) Show that∼is an open equivalence relation. (Hint: Mimic the proof of Propo- sition 7.14.)
(b) Prove that the GrassmannianG(k, n)is second countable. (Hint: Mimic the proof of Corollary 7.15.)
(c) LetS=F (k, n). Prove that the graphRinS×Sof the equivalence relation∼is closed. (Hint: Two matricesA= [a1ã ã ãak]andB = [b1ã ã ãbk]inF (k, n)are equivalent iff every column ofBis a linear combination of the columns ofAiff rk[AB] ≤kiff all(k+1)×(k+1)minors of[AB]are zero.)
(d) Prove that the GrassmannianG(k, n)is Hausdorff. (Hint: Mimic the proof of Proposition 7.16.)
Next we want to find aC∞atlas on the GrassmannianG(k, n). For simplicity, we specialize toG(2,4). For any 4×2 matrixA, letAijbe the 2×2 submatrix consisting of itsith row andjth row. Define
Vij = {A∈F (2,4)|Aij is nonsingular}.
Because the complement ofVij inF (2,4)is defined by the vanishing of detAij, we conclude thatVij is an open subset ofF (2,4).
(e) Prove that ifA∈Vij, thenAg∈Vij for any nonsingular matrixg∈GL(2,R).
DefineUij =Vij/∼. Since∼is an open equivalence relation,Uij =Vij/∼is an open subset ofG(2,4).
ForA∈V12,
A∼AA−121=
⎡
⎢⎢
⎣ 1 0 0 1
∗ ∗∗ ∗
⎤
⎥⎥
⎦= I
A34A−121
.
This shows that the matrix representatives of a 2-plane inU12have a canonical form Bin whichB12is the identity matrix.
(e) Show that the mapφ˜12:V12−→R2×2,
φ˜12(A)=A34A−121, induces a homeomorphismφ12:U12−→R2×2.
(f) Define similarly a homeomorphismφij:Uij −→R2×2. Computeφ12 ◦φ23−1, and show that it isC∞.
(g) Show that{Uij |1≤i < j≤4}is an open over ofG(2,4)and thatG(2,4)is a smooth manifold.
Similar consideration shows thatF (k, n)has an open cover{VI}, whereI is an ascending multi-index
1≤i1< . . . ik ≤n.
ForA∈F (k, n), letAIbe thek×ksubmatrix ofAconsisting ofi1th,. . . , ikth rows ofA. Define
VI = {A∈G(k, n)|detAI =0}. Next defineφ˜I:VI −→R(n−k)×kby
φ˜I(A)=(AA−I1)I,
where( )I denotes the (n−k)×k submatrix obtained from the complement I of the multi-indexI. LetUI =VI/∼. Thenφ˜ induces a homeomorphismφ: UI
−→ R(n−k)×k. It is not difficult to show that{(UI, φI)}is a C∞atlas forG(k, n).
Therefore the GrassmannianG(k, n)is aC∞manifold of dimensionk(n−k).
7.3.* The real projective space
Show that the real projective spaceRPnis compact. (Hint: Use Exercise 7.11.)