Exercise 1 (Formula for connection 1-forms) Using the first structure equation, show that

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1 1 Stokes s Theorem Let D R 2 be a connected compact smooth domain, so that D is a smooth embedded circle. Given a smooth function f : D R, define fdx dy fdxdy, D where the left-hand side is the integral of a 2-form and the right-hand side is the Riemann integral of f. Let α = P dx + Qdy be a 1-form on D where P and Q are smooth functions on D). Then dα = dp dx + dq dy = P Q dy dx + dx dy y x Q = x P ) dx dy. y Parametrize D counterclockwise by a map c : S 1 D. The integral of α on D is given by α = α c u) ) du. D S 1 Then Stokes s Theorem, which says that dα = α, is equivalent to D D D D Q x P ) dxdy = P dx + Qdy. y D 2 Formula for connection 1-forms Exercise 1 Formula for connection 1-forms) Using the first structure equation, show that ω k i e j ) = dω i e j, e k ) + dω j e i, e k ) dω k e j, e i ). 1) Note the similarity between this and the formula for the Christoffel symbols Γ k ij = 1 n 2 l=1 gkl g x i jl + g x j il ) g x l ij. 1

2 Solution. We compute using dω i = n j=1 ωj ω i j that dω i e j, e k ) + dω j e i, e k ) dω k e j, e i ) ) = ω l ωl i e j, e k ) + ω l ω j l l=1 ) e i, e k ) ) ) ω l ωl k e j, e i ) = 1 ) ωj i e k ) ωk i 2 e j) + ω j i e k) ω j k e i) ωj k e i ) + ωi k e j ) = ωi k e j ) by cancelling terms using ω j i = ωi j. 3 Index of a vector field In this section we discuss the index of a vector field on a 2-dimensional manifold. See [1] and the references therein for details. We first consider a vector field V on an open subset U of R 2 with an isolated 0 at a point p U. Let S 1 p, ε) = { q R 2 : q p = ε } = S 1 the isomorphism is given by q q p ε ) and assume that ε > 0 is small enough so that V q) 0 for all q S 1 p, ε). Consider the map defined by V : S 1 p, ε) S 1 V q) = V q) V q). Then the degree i.e., winding number) of V is independent of ε for ε sufficiently small. The index of V at p is defined to be this degree and is denoted by index p V ). Here are some gradient vector fields with zeros at the origin: 2

3 x 2 + y 2) = 2x, 2y) has index +1 at the origin 0, 0) x 2 y 2) = 2x, 2y) has index +1 at 0, 0) y 2 x 2) = 2x, 2y) has index 1 at 0, 0) Let V be a vector field on a 2-dimensional differentiable manifold with isolated zeros and let p be a zero of V. Let x : U M be a parametrization with p x U). The index of V at p is defined to be the index of the push forward dx 1 )V xu) ) at x 1 p). This definition is independent of the choice of parametrization. Some nongradient vector fields with zeros at the origin: y, x) has index +1 at the origin 0, 0) x 2 y 2, 2xy ) has index +2 at 0, 0) Let f : M n R be a Morse function, i.e., for each critical point i.e., zero of the gradient f with respect to a Riemannnian metric g on M) p of f f, the symmetric matrix p)) is nonsingular. Then there are n real x i x j eigenvalues, all which are nonzero. The index of f at p is defined to be the f number of negative eigenvalues of p)), denoted by index x i x j p f). Then index p f) = 1) indexpf). f Note that if det p)) > 0, then 1) indexpf) = 1, whereas if det p)) < x i x j x i x j 0, then 1) indexpf) = 1. 3 f

4 A main special case of the Poincaré Hopf theorem says that for a Morse function f on a closed manifold M with critical points p 1, p 2,..., p k, the Euler characteristic of M may be expressed as χ M) = index pi f) = 1) indexp i f). 4 The Gauss Bonnet formula In this section we apply the method of moving frames, in particular, the Cartan structure equations, to prove the Gauss Bonnet formula for Riemannian surfaces. For a Riemannian surface M 2, g ), since ω1 1 = ω2 2 = 0, we have that ω2 1 = ω2 1 is the only nonzero connection 1-form. The Cartan structure equations simplify to and dω 1 = ω 2 ω 1 2, dω 2 = ω 1 ω 2 1, Ω 1 2 = dω 1 2. Hence, the Gauss curvature,defined as the only sectional curvature, is given by K Rm e 1, e 2 ) e 2, e 1 = 2Ω 1 2 e 1, e 2 ) = 2dω 1 2 e 1, e 2 ) using Ω j i X, Y ) e j 1 2 Rm X, Y ) e i. Note that the scalar curvature is R = 2 Rm e i, e j ) e j, e i = Rm e 1, e 2 ) e 2, e 1 + Rm e 2, e 1 ) e 1, e 2 = 2K. i,j=1 Exercise 2 Let U R 2 be an open set with coordinates r, θ). Suppose that f : U R + depends only on r and write it as f r). Show that the Gauss curvature of a rotationally symmetric) metric on U of the form g = dr dr + f r) 2 dθ dθ is given by K = f r) f r). 4

5 Solution. A moving coframe is given by ω 1 = dr and ω 2 = f r) dθ, which is dual to the moving frame e 1 = r and e 2 = 1 fr) θ. From 1) we calculate that ω2 1 e j ) = dω 2 e j, e 1 ) + dω j e 2, e 1 ) dω 1 e j, e 2 ), so that So ω 1 2 e 1 ) = dω 1 e 2, e 1 ) = 0, ω 1 2 e 2 ) = dω 2 e 2, e 1 ) = f r) f r). We conclude that Exercise 3 Show that g u = ω 1 2 = f r) dθ, dω 1 2 = f r) dr dθ K = 2dω 1 2 e 1, e 2 ) = f r) f r). e i e i u)) ω j i e i) e j u). 2) Solution. We compute using the formula X α) Y ) X α Y )) α X Y ) that g u = = = ei du) e i ) j=1 e i du e i )) du ei e i )) e i e i u)) ω j i e i) e j u). Exercise 4 Show that if M 2, g ) is a Riemannian surface and u : M R is a function, then where g is the Laplacian. j=1 R e u g) = e u R g) g u), 5

6 Solution. Let {e i } 2 be a moving frame for g with dual coframe { ω j } 2 j=1. Then ẽ i = e u/2 e i and ω j = e u/2 ω j comprise a frame and coframe { } } for e u g. Let ω j i and { Ωi j denote the connection 1-forms and curvature 2-forms of e u g. Since n = 2, 2) yields g u = e 1 e 1 u)) + e 2 e 2 u)) ω 2 1 e 1 ) e 2 u) ω 1 2 e 2 ) e 1 u). 3) Next, using 1) we compute that the connection 1-forms for e u g are given by ω 2 1 ẽ 1 ) = 2d ω 1 ẽ 2, ẽ 1 ) = 2e u d e u/2 ω 1) e 2, e 1 ) since du e 2 ) = e 2 u). Similarly, = 2e u/2 dω 1 e 2, e 1 ) + e u/2 du ω 1 e 2, e 1 ) = e u/2 ω 1 2 e 1 ) e u/2 e 2 u) ω 1 2 ẽ 2 ) = 2d ω 2 ẽ 1, ẽ 2 ) = 2e u/2 ω 2 1 e 2 ) 1 2 e u/2 e 1 u) = e u/2 ω 1 2 e 2 ) 1 2 e u/2 e 1 u). Hence ω 2 1 = ω 2 1 ẽ 1 ) ω 1 + ω 2 1 ẽ 2 ) ω 2 = e u/2 ω2 1 e 1 ) + 1 ) 2 e u/2 e 2 u) e u/2 ω 1 + e u/2 ω2 1 e 2 ) 1 ) 2 e u/2 e 1 u) e u/2 ω 2 = ω e2 u) ω 1 e 1 u) ω 2). We obtain Ω 1 2 = d ω 2 1 = dω d e 2 u) ω 1 e 1 u) ω 2) = Ω e2 e 2 u)) ω 2 ω 1 + e 2 u) dω 1) e1 e 1 u)) ω 1 ω 2 e 1 u) dω 2) 2 6

7 Hence, using K g = 2Ω 1 2 e 1, e 2 ), we obtain K e u g = 2 Ω 1 2 ẽ 1, ẽ 2 ) = e u K g e u e 2 e 2 u)) e 2 u) ω 1 2 e 1 ) ) e u e 1 e 1 u)) e 1 u) ω 2 1 e 2 ) ) = e u K g 1 2 e u g u. since dω 1 e 1, e 2 ) = 1 2 ω1 2 e 1), dω 2 e 1, e 2 ) = 1 2 ω2 1 e 2), and by 3). The exercise follows since the scalar curvature is twice the Gauss curvature. We use the method of moving frames to prove one of the most fundamental results in Riemannian geometry, the Gauss Bonnet formula, which says that the integral of the Gauss curvature on a closed Riemannian surface M 2, g ) is equal to 2π times the Euler characteristic of M 2. Theorem 5 Gauss Bonnet) If M 2, g) is a closed oriented Riemannian surface, then KdA = 2π χm). M The proof of this formula will occupy the rest of the subsection. Let {e 1, e 2 } be a local positively-oriented orthonormal basis for T M in an open set U M so that the area form is da = ω 1 ω 2. Note that for any 2-form on a Riemannian surface, β = 2β e 1, e 2 ) ω 1 ω 2. The Gauss Bonnet integrand is locally the exterior derivative of the connection 1-form ω 2 1 : KdA = 2dω 1 2 e 1, e 2 ) ω 1 ω 2 = dω ) We would like to apply Stokes s theorem: for a compact surface N with possibly nonempty) boundary N, for any 1-form α, dα = α. N N 7

8 Since ω1 2 is only defined locally, we cannot apply Stokes s theorem globally on M. However, we can choose an orthonormal frame field {e 1, e 2 } on M minus a finite number of points. Indeed, let V C M) be a smooth vector field with isolated zeros at a finite number of points p 1,..., p k for example we may take V to be the gradient of a Morse function on M). Remark 6 A Morse function on M is a C function f : M R such ) that each critical point p are nondegenerate, i.e., the matrix f p) is x i x j nonsingular. In this case, the critical points are isolated see Corollary 2.3 in [?]), and assuming M is closed, there only finite many critical points. On any manifold there exists a Morse function see Corollaries 6.7 and 6.8 in [?]). On M 2 {p 1,..., p k } let e 1 = V V, and then e 2 is determined uniquely by the condition that {e 1, e 2 } be a positively-oriented orthonormal frame. Lemma 7 For ε < inj M 2, g ), 1 let Bp i, ε) {x M : dx, p i ) < ε} for i = 1,..., k. The Gauss Bonnet integrand may be rewritten as M KdA = lim ε 0 Proof. By 4), Stokes s theorem, and taking into account the orientation on k Bp i, ε) induced as the boundary of M k Bp i, ε), we have KdA = lim M ε 0 M dω1 2 = lim k Bp i,ε) ε 0 k ω1. 2 5) Bp i,ε) ω 2 1. We shall show that the boundary integral on the rhs of 5) tends to 2π times the index of the vector field V see the next section for the definition and properties of index). Lemma 8 lim ω1 2 = 2π index pi V ). 6) ε 0 1 inj M 2, g ) is the injectivity radius of g. 8

9 Assuming this lemma, we can complete the Proof of the Gauss Bonnet formula. Using 5), 6) and the Poincaré Hopf theorem, which says that the index of a vector field is equal to the Euler characteristic of the underlying manifold, we have KdA = 2π index pi V ) = 2π index M V ) = 2π χm). M We conclude this section with the Proof of Lemma 8. Given 1 i k, let ν denote the unit outward normal vector field to Bp i, ε) and let T denote the unit tangent vector field to Bp i, ε) such that the frame {ν, T } is positively-oriented. Parametrize Bp i, ε) by a unit speed path γ i : [0, L i ] Bp i, ε) M such that dγ i /ds = T. We define the angle function by θ : [0, L i ] R T γ i s)) cos θs) e 1 γ i s)) + sin θs) e 2 γ i s)). We require that θ be smooth so that it is well-defined up to an integer multiple of 2π. The unit outward normal vector field ν is given by ν γ i s)) = sin θs) e 1 γ i s)) cos θs) e 2 γ i s)). The change in the angle θ as you go around γ i is θl i ) θ0) = 2π index pi V ) 1). The above discussion concerning the angle θ is related to the boundary integral in 6) by the following. Sublemma. ω 2 1T ) = κ T θ, 7) where κ dt T, ν is the geodesic curvature of Bp i, ε). Proof of the sublemma. We compute dt T = dt cos θ e 1 + sin θ e 2 ) = T cos θ) e 1 + T sin θ) e 2 + cos θ dt e 1 + sin θ dt e 2 = sin θ T θ e 1 + cos θ T θ e 2 + cos θ ω1t 2 ) e 2 + sin θ ω2t 1 ) e 1 = ν T θ + ω1t 2 ) ) 9

10 since dt e i = T e i, which implies κ = dt T, ν = T θ + ω1t 2 ). Let ds denote the arc length element of Bp i, ε). By 7) and the fundamental theorem of calculus, we have ω1 2 = ω1t 2 ) ds = κ T θ) ds Since we conclude that = = lim ε 0 κ ds θl i ) + θ0) κ ds + 2π index pi V ) 1). lim κ ds = 2π, ε 0 ω 2 1 = 2π index pi V ), and both the lemma and the Gauss Bonnet formula are proved. References [1] Milnor, John W. Topology from the differentiable viewpoint. The University Press of Virginia Charlottesville,