The Gauss Bonnet Theorem

Transcription

1 Chapter 27 The Gauss Bonnet Theorem That the sum of the interior angles of a triangle in the plane equals π radians was one of the first mathematical facts established by the Greeks. In 1603 Harriot 1 showed that on a sphere of radius 1 the area of a spherical triangle (that is, a triangle whose sides are parts of great circles) with angles α, β, γ is given by α + β + γ π. For example, an octant of the unit sphere S 2 (1) has area π/2, and comprises the interior of a triangle with angles all equal to π/2. Both the planar and spherical facts are special cases of the Gauss Bonnet 2 Theorem, which will be proved in this chapter. The Gauss Bonnet Theorem consists of a formula for the integral of the Gaussian curvature over all or part of an abstract surface. For example, we know that the curvature and area of a sphere S 2 (a) are 1/a 2 and 4πa 2. Thus 1 K da = a 2 da = 1 a 2 4πa2 = 4π, S 2 (a) S 2 (a) 1 Thomas Harriot ( ). English mathematician, adviser to Sir Walter Raleigh. He made a voyage to Virginia in and reported on the Native American languages and customs. His telescopic observations were the earliest in England; in particular, he discovered sunspots in 1610 and used them to deduce the period of the Sun s rotation. 2 Pierre Ossian Bonnet ( ). French mathematician, who made many important contributions to surface theory, including the Gauss Bonnet Theorem. Bonnet was director of studies at the École Polytechnique in Paris, professor of astronomy in the faculty of sciences at the University of Paris, and member of the board of longitudes. 901

2 902 CHAPTER 27. THE GAUSS BONNET THEOREM showing that the integral of the Gaussian curvature over a sphere is independent of the radius a of the sphere. It is remarkable that for any ellipsoid, or more generally for any convex surface M in R 3, we have KdA = 4π. M This fact derives ultimately from the presence of K in formulas for area in the image of the Gauss mapping defined on page 333. Still more generally, for any compact surface M, the Gauss Bonnet Theorem expresses the integral of the Gaussian curvature in terms of a topological invariant, the Euler characteristic χ(m). This important number is always an integer, and is expressible in terms of the number of vertices, faces and edges of any triangulation of M. We begin by generalizing to abstract surfaces the notions of turning angle (given in Section 1.5) and of total signed curvature of a plane curve (given in Section 6.1). This enables us to consider integrals of geodesic curvature in the abstract setting, though the results of this chapter can also be appreciated by using the extrinsic definitions given in Section In Section 27.2, we prove a local version of the Gauss Bonnet Theorem, using Green s Theorem which is of independent interest. We improve the results in Section A summary of topological facts about surfaces is given in Section 27.5 in order to establish global versions of the Gauss Bonnet Theorem in Section Finally, Section 27.7 is devoted to applications Turning Angles and Liouville s Theorem Since in general there is no notion of horizontal for a general abstract surface, we must choose a reference vector to measure turning. In the next two lemmas, we suppose that M is an oriented surface, α: (a, b) M is a regular curve, and that X is an everywhere nonzero vector field along α. Lemma Fix t 0 with a < t 0 < b. Let θ 0 be a number such that α (t 0 ) α (t 0 ) = cosθ X(t 0 ) 0 X(t 0 ) + sinθ JX(t 0 ) 0 X(t 0 ). Then there exists a unique differentiable function θ[α,x]: (a, b) R such that θ(t 0 ) = θ 0 and (27.1) α α = 1 ( ) X cosθ[α,x]x + sinθ[α,x]jx, at all points on the curve. We call θ[α,x] the turning angle of α with respect to X determined by θ 0 and t 0.

3 27.1. TURNING ANGLES AND LIOUVILLE S THEOREM 903 Proof. Corollary 1.24 on page 18 implies the existence of a unique function θ[α]: (a, b) R such that θ[α](t 0 ) = θ 0 and cosθ[α](t) = α (t) X(t) α (t) X(t), (27.2) sin θ[α](t) = α (t) JX(t) α (t) X(t). Then (27.2) is equivalent to (27.1). We can now generalize Lemma 1.26, page 20, to abstract surfaces. Lemma The turning angle θ[α, X] is related to the geodesic curvature κ g [α] of α by the formula (27.3) Proof. (27.4) θ[α,x] (t) = κ g [α](t) α (t) α X, JX (t) X(t) 2. Using the analog of (1.23), we have ( ) α (t) α (t) α (t), Jα (t) α α (t) (t), Jα (t) = α (t) 2 = κ g [α](t) α (t). On the other hand, (writing α = α(t), X = X(t) and θ = θ[α,x]) we use (27.1) to compute ( ) ( ) α X cosθ + JX sin θ α α = α X ( ) = X ( ) X sinθ + JX cosθ X cosθ + JX sin θ + X 2 θ X ( ( α X)cos θ + (J α X)sin θ ) = X X α α + Jα α θ + + X ( ( α X)cosθ + (J α X)sin θ ) X, so that (27.5) α ( α α Now (27.3) follows from (27.4) and (27.5). ), Jα α = θ α X, JX + X 2. A practical choice for the vector field X is x u, where x is a patch on an abstract surface. This will lead us to an analogue of Euler s Theorem on page 397. Suppose now that the surface M is oriented, and denote its metric by (27.6) ds 2 = Edu 2 + 2Fdudv + Gdv 2.

4 904 CHAPTER 27. THE GAUSS BONNET THEOREM We shall say (referring to page 879) that a patch x is coherent with the orientation of M if the equation (27.7) Jx u = F x u + Ex v EG F 2 holds, rather than one with an opposite sign. Lemma Let α: (a, b) M be a regular curve whose trace is contained in x(u), where x: U M is a patch coherent with the orientation of M. Write α(t) = x(u(t), v(t)) for a < t < b. Then F κ g [α] α = θ[α,x u ] + E (E uu + E v v ) E v u + G u v + 2F u u 2. EG F 2 Proof. (27.8) We have α = u x u + v x v ; thus from (27.7) it follows that α x u, Jx u x u 2 = 1 E u xu x u + v xv x u, F x u + Ex v, EG F 2 which equals u F xu x u,x u + u E xu x u,x v v F xv x u,x u + v E xv x u,x v divided by E EG F 2. Since xu x u,x u = 1 2 E u, xv x u,x v = 1 2 G u and xu x u,x v = Fu xv x u,x u = Fu 1 2 E v, we see that (27.8) equals 1 E EG F 2 ) ( u 2 (FE u 2EF u + EE v ) + v 2 ( FE v + EG u ). Combining this equation with (27.3) completes the proof. Corollary With the hypotheses of Lemma 27.3, suppose in addition that F = 0 in (27.6). Then (27.9) κ g [α] α = θ[α,x u ] EG (G uv E v u ). Next, we find a formula for the geodesic curvature of a curve on a patch x in terms of the geodesic curvatures of the coordinate curves u x(u, v) and v x(u, v). We denote these geodesic curvatures by (κ g ) 1 and (κ g ) 2 respectively.

5 27.1. TURNING ANGLES AND LIOUVILLE S THEOREM 905 Theorem (Liouville 3 ) Let M be an oriented surface, and suppose that α: (a, b) M is a regular curve whose trace is contained in x(u), where x: U M is a patch coherent with the orientation of M for which F = 0. Then (27.10) κ g [α] = (κ g ) 1 cosθ + (κ g ) 2 sin θ + θ α, where θ = θ[α,x u ] is the angle between α and x u. Proof. Applying (27.9) to the curves u x(u, v) and v x(u, v), we obtain (κ g ) 1 E = E v 2 EG For a general curve α, we have and (κ g ) 2 G = G u 2 EG. so that (27.11) u = α E cosθ and v = α G sin θ, E vu 2 EG = α (κ g ) 1 cosθ and Then (27.10) follows from (27.9) and (27.11). G u v 2 EG = α (κ g ) 2 sin θ. Notice the deceptive similarity between (27.10) and Euler s formula (13.17). Total Geodesic Curvature The analog of Definition 6.1 on page 154 for a curve on a surface is Definition Let M be an abstract surface. The total geodesic curvature of a curve α: [a, b] M, denoted by TGC[α], is the number (27.12) α κ g [α]ds = b where κ g [α] is the geodesic curvature of α. a κ g [α](t) α (t) dt, The abstract definition of the geodesic curvature was given on page 880, though if the surface lies in R 3, one may use the more elementary definition on page 541 to make sense of (27.12). 3 Joseph Liouville ( ). French mathematician, who worked in number theory, differential equations, dynamics, differential geometry and complex variables. Liouville played a large role in introducing Gauss s ideas to France.

6 906 CHAPTER 27. THE GAUSS BONNET THEOREM The generalization of Lemma 6.2 on page 154 is Lemma The total geodesic curvature of a curve on a surface remains unchanged under a positive reparametrization, but changes sign under a negative reparametrization. The proofs of Lemmas 27.7 and 6.2 are essentially the same (see Exercise 8). We also have the following generalization of Lemma 6.3, obtained by integrating (27.3): Lemma Let M be an oriented surface, α: (a, b) M a regular curve, and X be an everywhere nonzero vector field along α. Then the total geodesic curvature can be expressed by α X, JX TGC[α] = θ[α,x](b) θ[α,x](a) + b a X 2 dt The Local Gauss Bonnet Theorem First, we need some facts about piecewise-differentiable curves on surfaces. Let M be an abstract surface, and let α: [a, b] M be a piecewise-differentiable curve. Associated with α is a subdivision a = t 0 < t 1 < < t k = b of [a, b] for which α is differentiable and regular on [t j, t j+1 ] for j = 0,...,k 1. The appropriate left and right derivatives of α are required to exist at the end points of each subinterval. The points α(t 0 ),..., α(t k ) are called the vertices of the curve α. The following theorem is an immediate consequence of Corollary Theorem Let M be an abstract surface, and suppose α: [a, b] M is a piecewise-differentiable curve whose trace is contained in x(u), where x: U M is a patch on M for which F = 0. Denote by α(t 0 ),..., α(t k ) the vertices of α. Then where TGC[α] = = k 1 tj+1 j=0 t j k 1 tj+1 ( Gu j=0 t j κ g [α](t) α (t) dt 2 EG dv dt E v 2 EG ) du dt k 1 ( dt + θ j+1 θ + ) j, j=0 θ j = lim t t j θ[α,x u ] ( α(t) ) and θ + j = lim t t j θ[α,x u ] ( α(t) ).

7 27.2. LOCAL GAUSS BONNET THEOREM 907 The numbers θ + j are only well defined up to integer multiples of 2π, although θ + j completely determines the number θ j+1. Typically, ε j = θ + j θ j is called the exterior angle of α at α(t j ) and ι j = π ε j the interior angle at α(t j ), though we require 0 < ι j < 2π for these definitions to be valid. In Section 12.4 we defined the notion of area of a closed subset of a surface; that notion admits the following generalization: Definition Let M be an abstract surface, x: U M an injective regular patch, S x(u) a closed subset, and f : S R a continuous function. Then the integral of f over S is given by f da = (f x) EG F 2 dudv. S x 1 (S) Just as in Section 12.4, we need to show that this definition is geometric. The proof of the following lemma is similar to that of Lemma on page 374 (see Exercise 1). Lemma The definition of the integral of f is independent of the choice of patch. We make precise the meaning of some common words from general topology. A neighborhood of a point p of an abstract surface M is any open set containing p. The boundary of a subset S of M is the set, denoted S, consisting of points p M for which every neighborhood of p contains both a point in S and a point not in S. A region of M is the union of an open connected subset with its boundary. A key tool for proving the Gauss Bonnet theorem is Green s Theorem. (See, for example, [Buck, page 406].) This can be stated as follows. Theorem (Green 4 ) Let R R 2 be a simply connected region, and let P, Q: U R be differentiable functions, where U is an open set containing R. Then ( Q R u P ) (27.13) dudv = P du + Qdv. v R The line integral is performed by parametrizing the boundary in a counterclockwise sense in the plane. To check this is correct, we may merely take (P, Q) = (0, u) and R = {0 u 1, 0 v 1}, so that (27.13) becomes 1 dudv = u dv, R R 4 George Green ( ). English mathematician, owner of a windmill in Nottingham. Although he published only 10 papers, he made profound contributions to mathematical physics. He invented Green s functions and coined the term potential function. Where Green acquired his mathematical skills remains a mystery.

8 908 CHAPTER 27. THE GAUSS BONNET THEOREM and both sides are equal to 1. More generally, the boundary of a region of an abstract surface will be traversed in the sense determined by the operator J defining the orientation. We are now ready to prove the first version of the Gauss Bonnet Theorem. Theorem (Gauss Bonnet, local version) Let M be an oriented abstract surface and let x: U M be a regular patch coherent with the orientation of M. Let R = x([a, b] [c, d]), where [a, b] [c, d] U. Then (27.14) R K da + κ g ds + R 3 ε j = 2π, where ε 0, ε 1, ε 2, ε 3 are the exterior angles at the vertices p 0,p 1,p 2,p 3 of R. Proof. First, suppose that x is a patch with F = 0. We suppose that the vertices are chosen so that j=0 p 0 = x(a, c), p 1 = x(b, c), p 2 = x(b, d), p 3 = x(a, d). The portion of R between p 0 and p 1 is then parametrized by u x(u, c), and we may choose θ + 0 = θ 1 = 0. We explain next that the hypothesis that x be consistent with the orientation ensures that the external angle at each vertex lies between 0 and π. Ε 3 Ε 2 Ε 1 Ε 0 Figure 27.1: A quadrilateral with exterior angles ɛ i

9 27.2. LOCAL GAUSS BONNET THEOREM 909 On the portion of R between p 1 and p 2, we deduce from (27.1) and (27.7) that the turning angle θ = θ[x v,x u ] of the boundary relative to x u satisfies E sin θ = xv, Jx u = EG. We are therefore free to choose 0 < sin θ + 1 < π, so that the internal angle ι 1 at p 1 also lies between 0 and π. Continuing around the quadrilateral, the portion of R between p 2 and p 3 is parametrized by u x(u, d), and we need to choose θ + 2 = θ 3 = π and π < θ+ 3, θ 4 < 2π. The resulting exterior angles are ε 1 = θ + 1 θ 1 = θ + 1, ε 2 = θ + 2 θ 2 = π θ 2, ε 3 = θ + 3 θ 3 = θ + 3 π, ε 0 = θ + 0 θ 4 + 2π = 2π θ 4. It is necessary to add 2π in the last line so that the internal angle ι 0 = π ε 0 lies between 0 and 2π (indeed, 0 and π). Therefore, 3 j=0 ( θ j+1 ) θ+ j = ( θ 2 ) ( θ+ 1 + θ 4 θ + ) 3 = π ε 2 ε 1 + 2π ε 0 ε 3 π = 2π 3 ε j. By Green s Theorem (Theorem 27.12) and formula (17.6), page 534, we have ( Gu R 2 dv EG dt E ) v 2 du dt EG dt b d (( ) ( ) ) Ev = a c 2 Gu + EG v 2 dudv EG u b d = K EG dudv = KdA. a c Substituting the equations above into Theorem 27.9, it follows that the total geodesic curvature of R is given by 3 tj+1 κ g ds = κ g [α](t) α (t) dt R j=0 = 2π t j 3 j=0 ε j K da. R Since all terms in (27.14) are invariantly defined, the result holds for all patches irrespective of the assumption F = 0. R j=0

10 910 CHAPTER 27. THE GAUSS BONNET THEOREM It will be convenient to rewrite (27.14) using interior angles. Corollary Under the hypotheses of Theorem we have (27.15) R K da + κ g ds = 2π + R 3 ι j, j=0 where ι 0, ι 1, ι 2, ι 3 are the interior angles at the vertices of R An Area Bound This short section provides an independent application of Green s theorem to the area of surfaces of constant negative curvature in R 3, such as the pseudosphere and other examples studied in Chapters 15 and 21. It uses terminology from Section 21.2, in particular Definition on page 690. Lemma Let D = [a, b] [c, d], and let x: D R 3 be a Tchebyshef principal patch of constant negative curvature a 2. Then ( area y(d) ) (27.16) = da 2πa 2. Proof. D Let ω denote the angle function of the associated asymptotic patch y(p, q) = x(p + q, p q), described on page 686. Then the principal patch x has metric ds 2 = a 2 (cos 2 θdu 2 + sin 2 θdv 2 ), where θ = ω/2. The infinitesimal area element of y is given by da = EG F 2 dpdq = a 4 a 4 cos 2 ω dpdq = a 2 sin ω dpdq = a 2 ω pq dpdq, the last line with the help of Lemma By Theorem 27.12, we have ( area y(d) ) = a 2 ω pq dpdq D = a2 ( ωp dp + ω q dq ) 2 D (27.17) = a 2( ω(a, c) + ω(b, d) ω(a, d) ω(b, c) ). Since neither cosθ nor sinθ can vanish, we can assume that 0 < θ < π/2, so 0 < ω(p, q) < π for all p, q. Hence area ( y(d) ) 2a 2 π.

11 27.4. MORE COMPLICATED REGIONS 911 More generally, suppose that we can cover the surface M by one large rectangle divided into a grid of smaller rectangular Tchebyshef principal patches of the type above. When we sum the contributions arising in (27.17), these cancel out in pairs, and we are left with only external vertices. Corollary Let M R 3 be a connected regular surface with constant negative curvature a 2. Then area(m) 2πa 2. This result extends Theorem 26.24, since it follows that one cannot find any complete surfaces of constant negative Gaussian curvature in R A Generalization to More Complicated Regions In order to generalize Theorem 27.13, we introduce the following notion. Definition Let M be an abstract surface. A regular region is a compact region whose boundary is the union of a finite number of piecewise-regular simple closed curves that do not intersect each other. Note that a compact regular surface can be considered as a regular region with empty boundary. Each of the piecewise-regular simple closed curves that constitute the boundary Q of a regular region Q has vertices and exterior angles; therefore, we can speak of the vertices p 1,...,p m and the external angles ε 1,...,ε m of the regular region Q. Let us also call the portion of Q between consecutive vertices an edge of Q. Definition A polygonal region is a regular region with a finite number of vertices, each of which has a nonzero external angle. A triangular region is a polygonal region with exactly 3 vertices. It will be necessary to decompose regular regions into smaller regions. Definition A polygonal decomposition of a regular region Q of an abstract surface is a finite collection R of polygonal regions R called faces such that (i) each R R is homeomorphic to a disk; (ii) the union of all regions in R is Q; (iii) if R 1, R 2 R, then the intersection R 1 R 2 is either a common edge or a common vertex of R 1 and R 2. An oriented polygonal decomposition of Q is a polygonal decomposition in which the faces and edges are assigned orientations as follows:

12 912 CHAPTER 27. THE GAUSS BONNET THEOREM (iv) each face of R is assigned the orientation of M; (v) each edge of a face R R, when parametrized as a curve α with respect to R, is such that Jα points toward the interior of R. If all the faces are triangular regions, then R is called a triangulation of Q. Note that if γ is an edge of polygonal regions R 1 and R 2, the orientation of γ with respect to R 1 is the opposite of the orientation of γ with respect to R 2. As a first step to generalizing Theorem 27.13, we derive the formula for the integral of the Gaussian curvature over a triangular region. Corollary Let M be an oriented abstract surface, and suppose that T is a triangular region of M such that T x(u), where x: U M is a regular patch coherent with the orientation of M. Then (27.18) K da + κ g ds = π + ι 1 + ι 2 + ι 3, T where ι 1, ι 2, ι 3 are the interior angles at the vertices of T. T Proof. We construct an oriented polygonal decomposition of T, consisting of quadrilaterals, as follows. Choose a point in the interior of the triangular region and points on each of the edges between the vertices. Join the three new points on the edges to the point in the center. Three quadrilaterals R 1, R 2, R 3 are formed; we choose them to have their orientations coherent with that of T. Since T is contained in the trace of the patch x, the three quadrilaterals can be chosen as images under x of ordinary quadrilaterals in R 2. Therefore, because of additivity of integration over closed subsets of R 2 and (27.15), we have 3 3 (27.19) K da = K da = 6π + I κ g ds, T R k R k k=1 where I is the sum of all interior angles. The boundaries R 1, R 2, R 3 contribute six interior edges and six exterior edges. The integrals κ g ds cancel in pairs over the interior edges and combine in pairs over the exterior edges; hence 3 (27.20) κ g ds = κ g ds. R k k=1 Included in the sum I are the interior angles ι 1, ι 2, ι 3, together with the interior angles at the newly-created vertices. It is easily seen from Figure 27.2 that the interior angles at the newly created vertices add up to 5π, and so (27.21) T I = 5π + ι 1 + ι 2 + ι 3. k=1

13 27.4. MORE COMPLICATED REGIONS 913 Now (27.18) follows from (27.19) (27.21). Ι 3 R 3 Π 2Π R 1 R 2 Π Π Ι Ι 1 2 Figure 27.2: A triangle decomposed into three quadrilaterals Now we can state the original version of the theorem that Gauss proved in [Gauss2, Section 20]. Corollary Suppose that the sides of the region T in Corollary are geodesics. Then (27.22) K da = π + ι 1 + ι 2 + ι 3, T where ι 1, ι 2, ι 3 are the interior angles at the vertices of T. When K = 0, we obtain the familiar theorem that the sum of the angles of a triangle in the plane equals π. More generally: Corollary Suppose that the sides of the region T in Corollary are geodesics, and that K has the constant value λ 0. Then the area of T is given by area(t ) = π + ι 1 + ι 2 + ι 3 (27.23), λ where ι 1, ι 2, ι 3 are the interior angles at the vertices of T.

14 914 CHAPTER 27. THE GAUSS BONNET THEOREM When λ = 1, equation (27.23) reduces to Harriot s formula for the area of a spherical triangle. Thus on a unit sphere, the sum of the interior angles of any geodesic triangle is greater than π, and the excess over π equals the area of the triangle. By contrast, any triangle on a surface of constant Gaussian curvature 1 has area strictly less than π. Using the upper half-plane model (Figure 26.3 on page 883), we may choose three vertices on the real axis (thus strictly speaking outside the upper half-plane) and join them by three semicircles. The resulting asymptotic geodesic triangle has interior angles all zero and area π relative to the Poincaré metric (26.6) The Topology of Surfaces In this section, we give a rudimentary account of the topology of surfaces. For more information see [AhSa, Chapter 1], [Arm, Chapter 7], [Massey, Chapter 1], [Springer, Chapter 5] and [FiGa]. The most important theorem we need is Theorem Let M 1 and M 2 be compact abstract surfaces. The following conditions are equivalent: (i) M 1 and M 2 are homeomorphic; (ii) M 1 and M 2 are diffeomorphic; (iii) M 1 and M 2 are both orientable or both nonorientable and have the same Euler characteristic. Suppose that (i) holds. A map Φ: M 1 M 2 defining a homeomorphism will not necessarily be a diffeomorphism; nonetheless the theorem asserts that a diffeomorphism Ψ: M 1 M 2 will exist. Here is the definition of Euler characteristic: Definition Let R be a polygonal decomposition of a regular region Q of an abstract surface. Put V = the number of vertices of R, E = the number of edges of R, F = the number of faces of R. Then χ(q) = F E + V is called the Euler characteristic or Euler number of Q. It is implicit in Theorem that the Euler characteristic χ(m) of a compact abstract surtface M does not depend on the polygonal decomposition chosen. This fact really follows from the array of propositions below, details of which can be found in the books cited on the previous page.

15 27.5. TOPOLOGY OF SURFACES 915 Proposition The Euler characteristic of a regular region of an abstract surface is invariant under the following processes: (i) subdivision of an edge by adding a new vertex at the interior point of an edge; (ii) subdivision of a polygonal region by connecting two vertices by a new edge; (iii) introducing a new vertex into a polygonal region and connecting it by an edge to one of the vertices of the polygonal region. The proof of proposition is instructive and not difficult (see Exercise 2). Clearly, it can be used to prove that any two polygonal decompositions of a regular region of an abstract surface have the same Euler characteristic, provided each edge of one of the two decompositions intersects each edge of the other in only a finite number of points and a finite number of closed intervals. However, it can happen that two polygonal decompositions intersect each other in bizarre ways. Nevertheless: Proposition The Euler characteristic of a regular region of an abstract surface is independent of the polygonal decomposition used to define it. Proposition A regular region of an abstract surface always admits a triangulation. If we take these results for granted, the classification is best understood by listing examples in the table below. The sphere S 2 and torus T 2 are orientable surfaces in R 3, unlike the projective plane RP 2 and Klein bottle K that we met in Chapter 11. A polygonal decomposition of the sphere is obtained by dividing it into eight triangles of the type mentioned on page 901. This results in 6 vertices, 12 edges and 8 regions, whence χ(s 2 ) = = 2. We can do the same job with far fewer regions, but Proposition tells us that the Euler characteristic will not change. The Euler characteristics of T 2, RP 2, K can be deduced from the diagrams in Section 11.2, by merely taking F = 1 face, corresponding to the square. The number of vertices and edges will depend on how the perimeter of the square has been identified with itself. For example Figure 11.4 on page 337 represents both the torus and Klein bottle with 1 vertex and 2 edges, whence χ(t 2 ) = χ(k) = = 0. The difference of course is that T 2 is orientable, whereas K is not.

16 916 CHAPTER 27. THE GAUSS BONNET THEOREM M orientable χ(m) sphere S 2 yes 2 projective plane RP 2 no 1 torus T 2 yes 0 klein bottle K no 0 sphere with g 1 handles yes 2 2g sphere with p 1 cross caps no 2 p It is easy to dream up more general surfaces. Starting from the sphere, one can attach one or more handles. If the number of handles is denoted by g (g is called the genus), then the resulting surface M g is homeomorphic to a torus with g holes. There is a short cut to computing the Euler characteristic of such a surface: adding a handle or extra hole effectively reduces it by 2, and we obtain the famous formula χ(m g ) = 2 2g. Note that the sphere has genus 0, and the torus genus 1. Figure 27.3: Orientable surfaces with χ = 6 Figure 27.3 shows the surface M 4. The two representations are topologically equivalent, and were indeed created by the same program, varying only the number of sample points. We can check the value of χ(m 4 ) by counting the

17 27.5. TOPOLOGY OF SURFACES 917 numbers of vertices, edges and faces on the right. Each of the four polyhedral tori contributes at least 12 vertices, at least 28 edges, and exactly 16 faces. Inspection of the central part from different angles (there is an animation in Notebook 27 to help, using the quaternion techniques of Chapter 23) reveals that the whole surface admits a triangulation with V = 62, E = 132, F = 64, confirming that χ = = 6. Figure 27.4: A representation of a nonorientable surface with χ = 1 Compact nonorientable surfaces cannot be represented inside R 3 without self-intersections. Nevertheless, it is customary to think of a nonorientable surface as a sphere with a certain number p 1 of cross caps glued on. The projective plane corresponds to p = 1, since we can shrink the sphere so as to obtain the model shown in Figure on page 346. Adding a cross cap actually lowers the Euler characteristic by 1, so χ(rp 2 ) = 2 1 = 1. Figure 27.4 shows a sphere with one cross cap and one handle attached; the result is a surface S with χ(s) = = 1. Since S incorporates at least one cross cap it must be nonorientable, and Theorem tells us that S is homeomorphic to a sphere with 3 cross caps attached. Visualizing this equivalence is however best done with planar models of the sort discussed in [FiGa]. The last two results of this section will enable us to globalize the discussion of the Gauss Bonnet theorem.

18 918 CHAPTER 27. THE GAUSS BONNET THEOREM Proposition Let M be an oriented abstract surface, and suppose that A = {(x α, U α )} is a family of parametrizations covering M, compatible with the orientation of M. Then there is an oriented triangulation T of M such that every triangular region T T is contained in some coordinate neighborhood U T for some (x T, U T ) A. Furthermore, if the boundary of every triangular region in T is positively oriented, then adjacent triangles determine opposite orientations in common edges. Proposition Assume that M, A and T are as in Proposition 27.28, and let ds 2 α = E αdu 2 α + 2F αdu α dv α + G α dvα 2 be the expression of the metric of M in terms of the coordinates given by x α. Let Q be a regular region, and let f : Q R be a continuous function. Then (f x α ) E α G α Fα 2 du α dv α T T x 1 α (T ) is the same for any triangulation T of Q such that every region T contained in some neighborhood U α with (x α, U α ) A. T is 27.6 The Global Gauss Bonnet Theorem The Gauss Bonnet Theorem is concerned with subsets of abstract surfaces whose boundaries are piecewise-regular curves. More precisely: Theorem (Gauss Bonnet, global version) Let Q be a regular region of an oriented abstract surface M, and let C 1,...,C n be piecewise-regular curves which form the boundary of Q. Suppose that each C i is positively oriented. Let ε 1,..., ε p be the external angles of C 1,...,C n. Then (27.24) Q K da + κ g ds + Q p ε j = 2πχ(Q). Proof. Let T be a triangulation of Q such that every triangular region in T is contained in a coordinate neighborhood of a family of orthogonal patches. We give the positive orientation to each triangular region in T T. In this way, adjacent triangular regions give opposite orientations to their common edge. We apply Corollary to every triangular region in T and add up the results. The integrals of the geodesic curvature along the interior edges (that is, those edges that are not contained in one of the boundary curves) cancel one another. We obtain n (27.25) K da + κ g ds = π F + ι ij, R C i i=1 j=1

19 27.6. GLOBAL GAUSS BONNET THEOREM 919 where F is the number of triangular regions in T and ι ij is the sum of all interior angles. Let V e = number of external vertices of T, V i = number of internal vertices of T, E e = number of external edges of T, E i = number of internal edges of T. The angles around each internal vertex add up to 2π; hence the sum of all internal interior angles is 2πV i. A similar calculation computes the sum of the external interior angles, and so p ιij = 2πV i + πv e ε j. Thus (27.25) can be rewritten as n (27.26) K da + κ g ds + R C i We have the relations i=1 j=1 p ε j = π( F + 2V i + V e ). j=1 (27.27) V e = E e, V = V e + V i, E = E e + E i. Furthermore, 3F = 2E i + E e, which can be written as (27.28) F = 2F 2E i E e. From (27.26) (27.28), we compute n p K da + κ g ds + ε j = π(2f 2E i E e + 2V i + V e ) R C i i=1 j=1 = π [ ] 2F 2(E E e ) E e + 2(V V e ) + V e = π(2f 2E + 2V) = 2πχ(Q), proving (27.24). In the special case of a compact surface we obtain: Corollary Let M be a compact orientable abstract surface. Then (27.29) K da = 2πχ(M), M where K is the Gaussian curvature with respect to any metric on M. Proof. A compact surface has no boundary. Hence (27.24) reduces to (27.29).

20 920 CHAPTER 27. THE GAUSS BONNET THEOREM In 1828, Gauss proved Theorem for the special case of polygonal regions whose sides are geodesics, and Bonnet generalized it to the form above in 1848 (see [Gauss2] and [Bonn1]). The use of Green s Theorem in proving Theorem was clarified by Darboux ([Darb2, volume 3, pages ]) Applications of the Gauss Bonnet Theorem A simple but important application of Corollary is the determination of the diffeomorphism type of compact orientable surfaces with positive Gaussian curvature. Theorem A compact orientable surface M with positive Gaussian curvature is diffeomorphic to a sphere. Proof. If M has positive Gaussian curvature, Corollary implies that M has positive Euler characteristic. Since M is orientable, it follows from Theorem that M is diffeomorphic to a sphere. Next, we prove a theorem of Jacobi (see [Jacobi]). Theorem (Jacobi, 1843) Let α: [a, b] R 3 be a closed regular curve with nonzero curvature. Assume that the curve described by the unit normal N (considered as a curve in the unit sphere S 2 (1) R 3 ) is a simple closed curve. Then N([a, b]) divides S 2 (1) into two regions of equal areas. Proof. Without loss of generality, we can assume that α has unit speed. Let s denote the arc length function of α, and let s denote the arc length function of N, considered as a curve on the unit sphere. The second Frenet formula of (7.12), page 197, implies that and so dn ds = κ T + τ B, d 2 N ds 2 = dκ ds T + dτ ds B (κ2 + τ 2 )N. From these two equations we compute the vector triple product and we also deduce that [ N dn ds 1 = dn d s dn d s = d 2 ] N ds 2 = κ dτ ds τ dκ ds, ( ) 2 ds (κ 2 + τ 2 ). d s

21 27.8. EXERCISES 921 We then obtain κ g (N) = dn ds 3[ N dn ds d 2 ] N ds 2 = = d ds = d d s κ dτ ds τ dκ ds κ 2 + τ 2 ds d s ( arctan τ ) ds κ d s ( arctan τ ). κ Let C denote the curve N([a, b]). Since this is closed, d ( κ g (N)d s = arctan τ ) (27.30) d s = 0. d s κ C C Let R be the region enclosed on one side by C ; then χ(r) = 1 because C is a simple curve. Therefore, from (27.24) and (27.30) we obtain 2π = 2πχ(R) = K da = area(r). Since the total area of the sphere is 4π, the theorem follows. R Finally, we state without proof an important generalization of Corollary due to Cohn-Vossen [CV]. Theorem Let M be a complete orientable abstract surface. Then (27.31) K da 2πχ(M), M provided both sides of (27.31) are finite Exercises 1. Prove Lemma and Corollary Prove Proposition Let M R 3 be a compact orientable surface not homeomorphic to a sphere. Show that there are points where the Gaussian curvature is positive, negative and zero. 4. Let H denote mean curvature. Show that H 2n da. S 2 (a) is independent of the radius of the sphere S 2 (a) if and only if n = 1.

22 922 CHAPTER 27. THE GAUSS BONNET THEOREM M 5. Let T(a, b) = { (x, y, z) R 3 z 2 + ( x2 + y 2 a ) 2 = b 2 } be a torus of wheel radius a and tube radius b (see Section 10.4). Use the parametrization of T(a, b) on page 406 to compute K da and H 2 da. T(a,b) T(a,b) Do the calculations both by hand and using techniques from Notebook 27. What does the Gauss Bonnet Theorem tell us about the integral of the Gaussian curvature over T(a, b)? Is the integral over a surface of the square of the mean curvature a topological invariant? 6. Let M be an abstract surface homeomorphic to a cylinder with Gaussian curvature K < 0. Show that M admits at most one simple closed geodesic.