Proof We may rescale the metric, and so by Lemma8.9, we may assume that the curvatureK=1, 0 or−1. We may write the metric in local geodesic polar coodinates
dρ2+G(ρ,θ)dθ2. We saw in Remark7.17thatG→0 and(√
G)ρ →1 asρ→0.
Let us take a fixed value forθ and setf(ρ) =√
G(ρ,θ). From the formula for curvature, we know that the functionf(ρ)satisfies the differential equation
fρρ+Kf =0.
Given thatf →0 andfρ →1 asρ→0, we deduce in the three cases thatf =sinρ, f =ρandf =sinhρ. From this, it is an easy exercise (Exercise8.4) to show that there exists locally an isometry to an open subset of the appropriate standard model
above.
8.4 Gauss–Bonnet for general closed surfaces
LetS be a compact abstract smooth surface, equipped with a Riemannian metric.
We may define (as we did in Chapter3for the sphere and torus) the Euler number of a triangulation onS as e =F −E+V, whereF = # faces,E = # edges and V = # vertices. We shall show that this is a topological invariant of the surface in an almost identical way to that we used for the sphere and torus, by replacing any topological triangulation of S by a polygonal decomposition with the same Euler number, and by proving a generalized global version of the Gauss–Bonnet theorem.
If we start from just a compact abstract smooth surfaceS, then usually the existence of both a Riemannian metric and a triangulation onS will be straightforward. For instance, an embeddedg-holed torus inR3carries an obvious Riemannian metric, and we argued in Chapter3that there exists a triangulation. The existence in general of a Riemannian metric onSin fact follows by an easy argument, which simply patches together local Riemannian metrics on charts ([6], Lemma 2.3.3 or [12], page 309). The existence in general of a triangulation onSis perhaps most easily proved by using such a metric, and the convexity arguments from this section, in order to produce ageodesic triangulation. One proves first that there is a (geodesic) polygonal decomposition of S(see [6], Theorem 2.3.A.1), from which by further decomposition we may obtain a geodesic triangulation.
The proofs given earlier in the case of the sphere and torus for the global Gauss–
Bonnet theorem and the topological invariance of the Euler number (in particular the proof of Theorem2.16and the detailed arguments from the appendix to Chapter3) apply essentially unchanged to the general case of a compact surface S with a
Riemannian metric, once we have shown the existence of suitable convex open neighbourhoods. We shall therefore first need to say something about convexity.
In Chapters2and3, we defined what it meant for a subsetAof the sphere or locally Euclidean torus to beconvex. This definition may be generalized in an obvious way to subsetsBof a general surfaceS.
Definition 8.11 Let S denote an abstract smooth surface, equipped with a Riemannian metric. A subsetBofSis calledconvexif, for anyQ1,Q2∈B, there is a unique length minimizing geodesic inSjoiningQ1toQ2, and this curve is contained inB. A subsetBof S is calledstrongly convexif it is convex, and for any points Q1,Q2∈B, the length minimizing geodesic is the only geodesic inBjoining the two points.
For example, the open balls of radius strictly less thanπ/2 for the sphere and 1/4 for the locally Euclidean torus are easily seen to be strongly convex. We now prove the existence of strongly convex open neighbourhoods for any surfaceS equipped with a Riemannian metric. We comment that the crucial part of this argument appeared already at the end of Chapter7, in Lemma7.19.
Proposition 8.12 Let P be any point on an abstract surface S equipped with a Riemannian metric. For allε >0sufficiently small, we have that the following two properties hold.
(i) The open geodesic ball B(P,ε)is strongly convex, and
(ii) for any Q ∈ B(P,ε), the open geodesic ball B(Q, 2ε) is a strong normal neighbourhood of Q.
Proof We choose a normal neighbourhoodW0 = σ(Bδ)aroundP, as defined in Theorem7.13, so that the metric takes the formdr2+G(r,θ)dθ2with respect to geodesic polar coordinates(r,θ); we assume furthermore thatδhas been chosen so thatGr >0 onBδ\{0}(possible, since by Remark7.17we haveGr/r→2 asr→0).
We consider the closed geodesic ballσ (B¯δ/2), consisting of the points of S whose geodesic distance fromPis at mostδ/2. By Theorem7.13and the argument we used to prove property (i) after the definition of Gaussian curvature in Section8.2, there existsεwith 0<2ε < δ/2 such that, for allQ∈σ(B¯δ/2), the geodesic ball centred onQwith radius 2εis a normal neighbourhood ofQ, contained inW0=σ (Bδ).
We show that the normal neighbourhoodW =σ (Bε)ofPis convex. Suppose we have pointsQ1,Q2∈W; then
ρ(Q1,Q2)≤ρ(Q1,P)+ρ(P,Q2) <2ε.
Considering the geodesic ballU of radius 2εcentred onQ1, this contains a unique geodesic γ : [0, 1] → U of minimum length from Q1 toQ2, namely the radial geodesic (Corollary7.18). Since any curve fromQ1to a point on the boundary ofU¯ has length at least 2ε, this shows thatγ is also the curve inSof absolute minimum length joining the two points. Note thatU is contained inW0.
8.4 GAUSS–BONNET FOR GENERAL CLOSED SURFACES 167 If now the above curveγcontainsP, it is a radial geodesic inW. Assume therefore thatγdoes not containP; our assumptions imply thatGr >0 at all points ofγ. Since ρ(P,Q1) < εandρ(P,Q2) < ε, we deduce from Lemma7.19thatρ(P,γ (t)) < ε for 0≤t ≤1, and hence the curveγ is always contained inW; henceW isconvex.
Since howeverWis contained in the normal neighbourhoodU =B(Q1, 2ε), in which the radial geodesicγis the unique geodesic joiningQ1toQ2, we deduce that it is the unique geodesic joiningQ1toQ2inW; henceW isstronglyconvex.
If now we repeat this argument, starting instead from a strongly convex normal neighbourhoodW0=σ(Bδ)ofP(just shown to exist), we obtain a strongly convex neighbourhoodW =σ(Bε)ofP, with the additional property that, for anyQ∈W, the open geodesic ballB(Q, 2ε)is a strong normal neighbourhood ofQ.
We shall be arguing below with polygons contained in such strongly convex balls, and we shall need the convexity of any geodesic triangle contained in such a ball.
Lemma 8.13 If W = B(P,ε) is as in Proposition8.12, then any three distinct points of W determine a unique geodesic triangle ⊂W , andis itself strongly convex.
Proof If the three points areA,BandC, we choose one of them, sayA. The open geodesic ball B(A, 2ε)is then a strong normal neighbourhood of A, and contains W. The minimum length geodesicjoiningBtoC is in the strongly convex open setW, and hence inB(A, 2ε). We saw in the proof of Lemma8.6that the three points determine a unique geodesic triangle ⊂ B(A, 2ε), which is moreover contained in the sector with angleα < π determined by the geodesic rays containingABand AC. The description given there forin terms of a union of geodesic rays fromA, together with the convexity ofW, ensures that ⊂W.
Suppose now that we have two distinct points P,Qof ; the minimum length geodesicγfromPtoQis contained inW, by convexity. We apply the argument from Lemma8.6again, deducing thatγmust remain in the sector ofB(A, 2ε)determined by the geodesic rays containing AB andAC (since otherwise γ would intersect a diameter in more than one point). If thereforeγdoes not remain in, it must cross the third side of in more than one point, contradicting thestrong convexity of W. Hence is convex, and therefore automatically strongly convex (as it is
contained inW).
The Gauss–Bonnet theorem for geodesic triangles contained in such a strongly convex open ball onS follows immediately from Proposition8.7, since any such triangle is contained in a strong normal neighbourhood of any of its vertices. This now generalizes to geodesic polygons.
Corollary 8.14 Ifis a geodesic n-gon on an abstract smooth surface S equipped with a Riemannian metric, andis contained in a strongly convex open ball W of
the type constructed in Proposition8.12, then
K dA=
i
αi −(n−2)π,
whereα1,. . .,αnare the internal angles of.
Proof This now follows from the case of geodesic triangles, and the induction argument we used for spherical polygons contained in a hemisphere. To start that argument, we needed to find a locally convex vertexP2; again this is found by taking a point of the polygon at maximum distance from the centrePof the ballW =σ (Bε);
this point is a vertex, sinceW was chosen with the property that, for any geodesic segment contained inW, the maximum distance fromPoccurs at an end-point (this follows from Lemma7.19, given thatGr was assumed to be positive onW \ {P}).
If we take pointsZ1andZ2on the boundary of, either side ofP2and sufficiently close to it, the geodesic triangle =Z1P2Z2will either be contained in(the case when P2is a locally convex vertex), or will have its interior disjoint from . We note in passing that Lemma7.19implies that, for such pointsZ1andZ2, the points ofhave distance fromP at mostρ(P,P2). Moreover, in the second case, points sufficiently close to P2but not inwill lie in. If we consider points Qon the geodesic rayPP2just beyondP2, these will be points ofwithρ(P,Q) > ρ(P,P2), contradicting our initial choice ofP2.
We note that is also contained in a strong normal neighbourhood of any of its vertices, namely the geodesic ball of radius 2εwith centre at the vertex. Apart from standard convexity properties, the rest of the proof of Theorem2.16only used the properties of geodesics in the hemisphere that distinct geodesics meet in at most one point, and they have distinct tangents at any point of intersection; otherwise the proof was purely combinatorial. These two properties hold also in the case being considered, the second fact following from the uniqueness clause in Proposition7.10;
as our assumptions ensure that we have the convexity properties needed, including in particular Lemma8.13, the proof of Theorem2.16therefore applies in the general case. We are thus able to express our polygon as the union of two simpler polygons meeting along a common side, for both of which we may assume that the required formula holds, and hence the general formula follows by induction.
Using this last result, we can now prove the global Gauss–Bonnet theorem, in an exactly analogous way to the method used for the sphere and torus in Chapter3.
In summary, we subdivide the triangulation so that each topological triangle is contained in a suitable strongly convex open set, replace the resulting triangulation by a polygonal decomposition of the surface, and then use the Gauss–Bonnet formula we have just proved for geodesic polygons.
Theorem 8.15 (Gauss–Bonnet theorem) Suppose S is a closed (i.e. compact) surface equipped with a Riemannian metric. Assuming the existence of a triangulation on S,
8.4 GAUSS–BONNET FOR GENERAL CLOSED SURFACES 169
we have
S
K dA=2πe,
where e is the Euler number. In particular, the Euler number depends on neither the choice of triangulation nor the choice of Riemannian metric, and so may be written as e(S).
Proof The surface has a cover by open geodesic ballsB(P,ε(P)/2), where for each P∈S, the open ballB(P,ε(P))is a strongly convex open ball of the type constructed in Proposition8.12. Using compactness, we choose a finite subcover, and letεbe the minimum of the finite set of numbersε(P)occurring. If nowis any subset ofSof diameter less thanε/2, then it must be contained in one of the corresponding finitely many strongly convex geodesic ballsB(P,ε(P)).
Given then any topological triangulation of S, we can subdivide it using Construction3.9, without changing the Euler number, so that each topological triangle has diameter less thanε/2, and hence is contained in a strongly convex open ballW of radiusε, of the type constructed by Proposition8.12. We now use Construction 3.15to polygonally approximate the edges of this triangulation by simple polygonal curves, and then arguing as in Proposition3.16, we see that this yields a polygonal decomposition ofS. We shall need the fact that the complement of a simple closed polygonal curve in S has at most two components, but since our geodesics (in appropriate geodesic polar coordinates) correspond to radial lines, the argument of Proposition1.17still applies (Remark1.18), proving this fact.
Therefore, we have replaced the triangulation by a polygonal decomposition with the same Euler number, with each of the polygons being contained in a strongly convex open ball of the type constructed in Proposition8.12. Hence, the Gauss–
Bonnet formula holds for the integral of the curvature over each of these polygons, by Corollary8.14. This then implies the required result, since by the argument from Proposition3.13,
n≥3
n-gons
interior angles−(n−2)π
=2πe.
Example The torusTwith locally Euclidean metric clearly hasKidentically zero, and soe(T)=0. If however we take the metric onT obtained from considering it as an embedded surface, then we saw in Chapter6that the curvatureK takes positive, negative and zero values; nonetheless, Theorem8.15still says that
TK dA=0 (see Exercises8.2and8.3).
The global Gauss–Bonnet theorem implies directly the topological invariance of the Euler number. We suppose that S is a compact smooth surface, equipped with a Riemannian metric (the particular choice of Riemannian metric being irrelevant).
Corollary 8.16 If X is a metric (or topological) space which is homeomorphic to S, then any (topological) triangulation on X has Euler number e(S).
Proof A topological triangulation onX gives rise, via the homeomorphism, to one with the same Euler number onS. This Euler number is however juste(S), by the
global Gauss–Bonnet theorem onS.
Remark 8.17 If one is prepared to restrict attention to compactorientablesurfaces, and to triangulations of the surface whose edges are piecewisesmoothcurves, then there is another proof of the global Gauss–Bonnet theorem in terms of integrating the geodesic curvatureround the edges of each triangle (Section 4.5 of [5], Chapter 12 of [8], or Chapter 11 of [9]). This proof does not yield the topological invariance of the Euler number, which would therefore need to be proved separately, for instance by the theory of homology groups from elementary Algebraic Topology.