Lecture 0. The Gauss-Bonnet Theorem In this lecture we will prove two important global theorems about the geometr and topolog of two-dimensional manifolds. These are the Gauss-Bonnet theorem and the Poincaré-Hopf theorem. Let us begin with a special case: Suppose is a compact oriented -dimensional manifold, and assume that there eists a vector field V on which is nowhere zero. Now let g be an Riemannian metric on. Then we can produce an orthonormal frame globall on b taking e = V/g(V,V ) /, and taking e to be the unique unit vector orthogonal to e such that e and e are an oriented basis at each point. The dual basis ω, ω is then also globall defined, and we have a globall defined -form ω on such that dω = ω ω and dω = ω ω. Then we have K(g)ω ω = dω. Stokes Theorem then applies to give K(g)dV ol(g) = Ω =0. So we have the remarkable conclusion that the integral of the Gauss curvature over is zero for ever Riemannian metric on. Remark. In fact the eistence of a non-vanishing vector field implies is diffeomorphic to a torus S S.Acorollar of the computation above is that S does not admit a nonvanishing vector field (as we have alread seen), since if it did the integral of Gauss curvature would be zero for an metric, but we know that the standard metric on S has Gauss curvature.. The result we proved above is a special case of the famous Gauss-Bonnet theorem. The general case is as follows: Theorem 0. The Gauss-Bonnet Theorem Let be acompact oriented two-dimensional manifold. Then for an Riemannian metric g on, K(g) dv ol(g) =πχ() where ξ() is the Euler characteristic of.
8 Lecture 0. The Gauss-Bonnet Theorem Here the Euler characteristic is an integer associated to which depends onl on the topological tpe of. If admits a triangulation (i.e. a decomposition into diffeomorphic images of closed triangles, such that the intersection of an two is either empt, a common verte, or a common edge) then the Euler characteristic is equal to V +F E, where V is the number of vertices in the triangulation, F is the number of faces, and E is the number of edges. The second important theorem we will prove gives an alternative method of computing the Euler characteristic. To state the result we first need to introduce some new definitions. Definition 0. Let be a two-dimensional oriented manifold, and V a vector field on. Suppose, and assume that there eists an open set U containing such that there are no zeroes of V in U\{}. Then the inde I(V,) ofv about is equal to the winding number of V around the boundar of a simpl connected neighbourhood of in U. Eample 0.3 Here are some vector fields in the plane with nontrivial inde about 0: 0 V =(, ), I(V,0) = + 0 V =(, ), I(V,0) = 0 V =(, ), I(V,0) = +
Lecture 0. The Gauss-Bonnet Theorem 83 0 V =(, ), I(V,0) = 0 V =( 3 3, 3 3 ), I(V,0) = +3 0 V =( 3 3, 3 3 ), I(V,0) = 3 Now we can state the Poincaré-Hopf theorem: Theorem 0.4 Let be a compact oriented two-dimensional manifold, and let V be an vector field on which has all zeroes isolated that is, if is an zero of V then there is a neithbourhood U of in such that V is nonzero on U\{}. Label the zeroes of V as,..., N. Then I(V, i )=χ(). i= We will prove that the sum of the indices of the vector field V at its zeroes is independent of V, and the result can then be taken to define the Euler characteristic. To see that this definition of the Euler characteristic agrees with the one given above in terms of triangulations, suppose that we are given a triangulation of. Then choose a vector field as sketched below with zeroes at the centre of each face, the middle of each edge, and at each verte. This vector field points outwards from the centre of each face, so has inde there, and points inwards at each verte, so also has inde there. At the middle of each edge the vector field has inde. Therefore the sum of the indices equals F + V E, agreeing with our previous definition of the Euler characteristic.
84 Lecture 0. The Gauss-Bonnet Theorem We will prove the Gauss-Bonnet theorem and the Poincaré-Hopf theorem at the same time, b showing that for an Riemannian metric g on and an vector field V with isolated zeroes, we have K(g) dv ol(g) =π I(V, i ). If we keep V fied and var g,wededuce that the left-hand side is independent of g, and if we keep g fied and var V,wededuce that the right-hand side is independent of V,sowecan define the Euler characteristic and deduce both results. So let V and g be given, and let,..., N be the zeroes of V.About each of these zeroes we can choose a small neighbourhood U i of i (sa, given b the image of a ball under some chart) such that V is nonvanishing on = \ N i= U i. Let γ i be the boundar of U i, parametrised anticlockwise in some oriented chart. Now is a compact oriented manifold with boundar, and V is nonvanishing on. Define e = V/g(V,V ) /, and take e to be the unit vector orthogonal to e which is given b a positive right-angle rotation of e. Denote the dual frame b ω and ω, and let ω be the corresponding connection one-form. Then we have Ω = dω, and Stokes theorem gives K(g) dv ol(g) = Ω = ω. i=
Lecture 0. The Gauss-Bonnet Theorem 85 The boundar of is the union of the circles γi, parametrized clockwise. Therefore K(g) dv ol(g) = ω. i= γ i Now on each of the regions U i we can choose a non-vanishing frame ē (i) ē (i). This gives a corresponding connection one-form ω(i) each i K(g) dv ol(g) = Ω = U i U i Combining these results, we get K(g) dv ol(g) = i= U i d = γ i ω.,, and we have for γ i. To interpret the integral of ω around γ i,weneed to consider the relationship between the frames e and ē (i). Since these are both unit vectors, the are related b a rotation. Therefore there eists a map from γ i to SO() S R,sa z =(α, β) S, such that ē (i) = αe + βe ē (i) = βe + αe. Then we have = αω + βω = βω + αω. Since α + β =,wehave β dα = α dβ = dθ. Appling the eterior derivative we find d = dα ω + αdω + dβ ω + βdω = (ω β dα) βω +(ω + α dβ) αω =(ω + dθ). and similarl d = (ω + dθ). It follows that ω = dθ, and so ω = dθ γ i γ i is π times the winding number of the map z : γ i S. This is equal to the difference betweeen I(ē (i), i) and I(e, i ). But ē (i) is nonvanishing in U i,so I(ē (i), i)=0, and b construction I(e, i )=I(V, i ). Thus
86 Lecture 0. The Gauss-Bonnet Theorem ω = I(V, i ). γ i We have proved K(g) dv ol(g) = π I(V, i ), and the proofs of the Gauss Bonnet and Poincaré-Hopf theorems are complete. i=