In the previous section, we needed the area of a spherical triangle; this formula represents the spherical version of the Gauss–Bonnet theorem. The Euclidean version of Gauss–Bonnet is just the familiar statement that the angles of a Euclidean triangle add up toπ.
Proposition 2.14 If is a spherical triangle with angles α,β,γ, its area is (α+β+γ )−π.
Proof Adouble lunewith angle 0 < α < π consists of the two regions on S cut out by two planes passing through two given antipodal points onS, with the angle
2.5 GAUSS–BONNET AND SPHERICAL POLYGONS 35 between the planes beingα. In view of the fact that the area ofS2is 4π, it is clear that the area of the double lune is 4α.
α
A spherical triangle =ABCis the intersection of three single lunes — in fact two suffice. Therefore,and its antipodal triangleare in all three of the double lunes (with areas 4α, 4β, 4γ) but any other point of the sphere is in only one of the double lunes, as may be seen with the aid of the diagram below.
γ C
A
α
β B
Thus
4(α+β+γ )=4π+2×2×A
where 4πis the total area ofS2andA=area =area. Hence the result follows.
Remark 2.15
(i) For a spherical triangle, α+β +γ > π. In the limit as area → 0, we obtain α+β+γ =π, that is the Euclidean case.
(ii) We may in fact relax our definition of a spherical triangle, by omitting the stipulation that the sides are of length less thanπ. This is only a minor change, since only one side
could have length≥ π (otherwise adjacent sides would meet twice, and we would not have a triangle). If however one of the sides has length≥ π, we can subdivide the triangle into two smaller ones, whose sides have length less thanπ. Applying Gauss–Bonnet to the two smaller triangles and adding, the area of the original triangle is still
α+β+γ+π−2π =α+β+γ−π.
We now extend the Gauss–Bonnet to spherical polygons onS2. Suppose we have a simple closed (spherically) polygonal curveC on S2, the segments ofC being spherical line segments. Let us suppose that the north pole does not lie onC, and we consider the imageofCunder stereographic projection (as defined in the next section), a simple closed curve inC. By Remark2.24, the segments ofare arcs of certain circles or segments of certain lines.
Applying Propositions1.17and1.19, or rather Remarks1.18and1.20, to, we deduce that the complement ofinChas two components, one bounded and one unbounded. Thus, the complement ofCinS2also has two path connected components;
each of these corresponds to the bounded component in the image of an appropriately chosen stereographic projection. The data of the polygonal curveCand a choice of a connected component of its complement inS2determines aspherical polygon. The Gauss–Bonnet formula for polygons will play a crucial role in later chapters, in our study of the Euler number and its topological invariance.
A subsetAofS2is calledconvex if, for any pointsP,Q ∈ A, there is a unique spherical line segment of minimum length joiningPtoQ, and this line segment is contained inA. In particular, minimum length spherical line segments inAmeet in at most one point. We observe that any open hemisphere is a convex open subset ofS2. We prove below, in Theorem2.16, a formula for the area of a sphericaln-gon contained in an open hemisphere, namelyα1+ ã ã ã +αn−(n−2)π, where theαi
are the interior angles. This is a combinatorial proof, proceeding by induction on the number of vertices of the polygon. Let us remark first that it is easy to see the validity of the formula forconvex spherical polygons. This follows immediately from the Gauss–Bonnet formula for spherical triangles, since a convex sphericaln-gon may be split inton−2 spherical triangles. For clarity, we draw below our spherical polygons as Euclidean ones (for which our arguments also hold).
2.5 GAUSS–BONNET AND SPHERICAL POLYGONS 37 More generally, the formula for area given is easily checked to beadditive. If1and 2are spherical polygons, meeting along a common side but otherwise disjoint, then the union is also a spherical polygon; if the smaller polygons haven1andn2sides respectively, the large polygon hasn1+n2−2 sides. The expression given for the area ofis then just the sum of the corresponding expressions for1and2. Thus, if the formula for the area of a spherical polygon is true for both1and2, it will be true for the union.
Theorem 2.16 If⊂S2is a spherical n-gon, contained in some open hemisphere, with interior anglesα1,ã ã ã ,αn, its area is
α1+ ã ã ã +αn−(n−2)π.
Proof The property of the hemisphere we use is that of it being convex, in the sense defined above. We prove the formula by induction onn, the casen =3 following from Gauss–Bonnet. We show that there is always an internal diagonal (that is, a spherical line segment joining non-adjacent vertices whose interior is contained in the interior Intof); this diagonal then dividesinto two polygons, both with strictly less thannsides, and the result follows by induction.
We assume without loss of generality thatis in the southern hemisphere. We first claim that there is a vertexPwhich islocally convex; this means that for nearby pointsP,P on the boundary ofeither side ofP, we obtain a spherical triangle PPPwhich is contained in. To see that such a vertex exists, consider a pointP of at maximum distance from the south pole. For any spherical line segment in the southern hemisphere, the maximum distance from the south pole will be at an end-point; hencePmust be a vertex of(with interior angleα < π) andmust be locally convex atP(see diagram below). Note that both these facts may fail ifis not contained in an open hemisphere.
α P
Having found a locally convex vertex, the proof is essentially just a combinatorial argument, and as such it will be valid in other geometries for geodesic n-gons contained in a suitable convex open set. We shall need the following two obvious
properties of spherical lines, both of which generalize appropriately for the more general geometries we study later.
(i) Two distinct spherical lines have at most one point of intersection in any given open hemisphere.
(ii) Two distinct spherical lines have distinct tangents at any given point of intersection (i.e. intersecttransversely).
Let us consider adjacent vertices P1,P2,P3 of , with P2 locally convex, constructed as above. We may assume therefore thatP2is not on the spherical line segmentP1P3. Letldenote the spherical line segment of shortest length joiningP1
toP3. In this way we obtain a spherical triangle =P1P2P3. Since we know the result for spherical triangles contained in a hemisphere, we may assumen >3 and so= . If the interior oflis contained in Int, then it is an internal diagonal and there is nothing more to prove; we suppose therefore from now on that this is not the case.
LetQt denote the point on the spherical line segmentP2P3with d(P2,Qt)=t d(P2,P3),
wheredhere denotes the standard metric onS2and where 0≤t≤1. We letltdenote the minimal length spherical line segmentP1Qt; thusl1 =l. Fort >0 sufficiently small, we have thatltmeets the boundary ofonly at the end-pointsP1andQt. This follows from property (i) above, since fort >0 sufficiently small,lt can only meet the two sides ofthat haveP1as an end-point at the pointP1, and by a continuity argument, the only other side that it can intersect is thenP2P3atQt. Moreover, by the locally convex property of the vertexP2, the interior points oflt nearQt will be in Int, whentis small. Since the interior ofltis connected, it follows that fort>0 sufficiently small, the interior oflt is contained in Int(since otherwise the interior oflt may be expressed as a disjoint union of two non-empty open subsets, namely those points which are in Intand those which are in the complement of).
P1 l P3
ls Qs R
P2
We now consider the supremumsof the valuest for which the interior oflt is contained in Int; thus 0 <s≤ 1. Note thatlsis contained in, but will contain
2.6 MệBIUS GEOMETRY 39 points of the boundary∂ofother thanP1andQs. In fact, using the fact that distinct spherical lines can only meet transversely, we see that iflscontains an interior point of a side of, it must contain the whole side; otherwise there would existε >0 such that the interior ofltintersects∂fors−ε <t ≤ s, contradicting the assumption thatswas a supremum. Thus, in addition to the end-pointsP1andQs, we have that ls ∩∂ consists of vertices of and possibly also some sides of ; in all cases therefore it contains a vertexRofother thanP1andP3.
In summary therefore, the interior of the spherical triangle=P1P2Qswill be contained in Int, andsis the largest number for which this is true. It is then clear that the spherical line segmentP2Rhas its interior contained in Int, and hence in Int. This then is the internal diagonal we require, dividing into two spherical
polygons with strictly less thannsides.