HOMEWORK 2 - RIEMANNIAN GEOMETRY ANDRÉ NEVES 1. Problems In what follows (M, g will always denote a Riemannian manifold with a Levi-Civita connection. 1 Let X, Y, Z be vector fields on M so that X(p Z(p for some p in M. Show that ( X Y (p ( Z Y (p. Find an example that shows that ( Y X (p might be different from ( Y Z (p. By linearity it suffices to show that if X is a vector field with X(p 0 then ( X Y (p 0. Choose coordinates centered at p, in which case we have, with respect to those coordinates, X(x u i (x. x i The condition that X(p 0 becomes u i (0 0 for all i 1,..., n because we can assume without loss of generality that p becomes the origin when using these coordinates. From linearity we have that X Y (x u i (x Y. x i Hence, direct substitution shows X Y (0 0. Consider the connection D on R 2 and set Y x (1, 0, Z (0, 0, and X x x (x, 0. At the origin we have X(0 Z(0 (0, 0. Now which are different. D Y X(0, 0 (1, 0 and D Y Z(0, 0 (0, 0 2 Suppose that M F 1 (0, where F : R n+k R k and 0 is a regular value of F. Following the discussion I wrote on the solutions to HW 1, 1
2 ANDRÉ NEVES problem 3, we see that we can identify T p M with a subspace of R n+k. More precisely T p M { v R n+k DF p ( v 0.} Thus for every point x in M we can consider the map P x : R n+k T x M where P x ( v is the projection of the vector v on T p M. Define a connection on M to be ( X Y (x P x (D X Y, where X and Y are vector fields on M and D denotes the Euclidean connection (i.e., standard differentiation. Show that this is the Levi-Civita connection for the induced metric on M. We need to check that satisfies all the axioms of being a Levi-Civita connection. Uniqueness implies that is indeed the Levi-Civita connection. For the sake of simplicity we denote P x (X simply by X T. The axioms regarding linearity and Leibniz rule are easy. I will just check the symmetry of the connection and the compatibility with the induced metric. Let φ : U M be a coordinate chart where U is a subset of R n. Denote by y i, i 1,..., n the coordinate vectors. Need to see that y j y i y i y j. Denote by i : M R n+k the inclusion map. Using this we can regard the coordinate vectors y i (y as being a vector in R n+k for every y U by identifying it with ( (i φ (i φ 1 (y (y, (i φ2 (y,..., (i φn+k (y y i y i y i y i n+k j1 (i φ j (y y i. x j What this mean is that if f is function defined in a neighborhood of p φ(y on M and if we consider another function f which is defined in a neighborhood of p on R n+k such that f f on M then y i (f φ y i (y (i φ (y, y f(p. i We can now finish what we are trying to do, i.e., check that y j y i y j. Set f l to be the function defined in a neighborhood of p φ(y on M such that f l φ (i φ l for l 1,..., n + k. Likewise, define j f l on a neighborhood of p φ(y on M such that ( j f l φ (i φl y j, where j 1,..., n and l 1,..., n + k.
Lemma 1.1. HOMEWORK 2 - RIEMANNIAN GEOMETRY 3 y i ( j f l Proof. By definition, we have that y l ( i f l ( j f l ( jf l φ (( jf l φ 2 (i φ l y i y i y i y i y j 2 (i φ l y j y i y j ( i f l Denote by j f l a function defined on a neighborhood of p in R n+k which coincides with j f l when restricted to M. In view of the lemma above we have y i ( (i φ D (i φ y j y i y j ( n+k r1 D (i φ y i ( n+k r1 ( T n+k r1 ( n+k D (i φ y i ( (i φ r y j x r T ( T (i φ j f r, x r y j f r r1 i x r T ( n+k T (( j f r φ ( j f r y i x r y r1 i x r ( n+k r1 ( j f r y j x r T T y j y i. We need to check compatibility of the connection with the Riemannian metric. Given X, Y, Z vector fields on M we can see them as vector fields in R n+k defined on M. For the sake of rigour denote them by X, Ȳ, and Z. X(g(Y, Z X( Ȳ, Z X( Ȳ, Z D XȲ, Z + Ȳ, D X Z (D XȲ T, Z + Ȳ, (D X Z T g( X Y, Z + g(y, X Z 3 Let X, Y denote the standard Euclidean inner product. On a unit ball B n { x < 1} consider the metric given by g(x, Y 4 (1 x 2 X, Y, 2 where X, Y are vectors at T x B n. This metric is called the hyperbolic metric and it will show up countless many times in the future.
4 ANDRÉ NEVES i Compute the Christoffel symbols for this metric with respect to the Euclidean coordinates. With respect to the coordinates (x 1,..., x n, the matrix that stands for the metric becomes g ij (x f(xδ ij, where f(x 4(1 x 2 2. In this case g ij 1 f δ ij and so the Christoffel symbols are Γ k ij 1 ( gjk + g ik g ij 1 ( f δ jk + f δ ik f δ ij 2f x i x j x k 2f x i x j x k 2(1 x 2 1 (δ jk x i + x j δ ik x k δ ij ii Show that the line γ(t t(1, 0,..., 0 with 0 t < 1 can be parametrized so that it becomes a geodesic for the hyperbolic metric. What is the length of this geodesic? First we parametrize the geodesic by unit length. We find t f(s so that α(s γ(f(s (f(s, 0,..., 0 has α (s 1. Because α (s 2 f (1 f 2 1 we find f so that f (s 1 (f(s2 2 and f(0 0. Integration shows that the solution is f(s es 1 e s +1. The equation above also implies that f ff by differentiating both sides. Therefore α α α (f x1 α (f x1 + (f 2 k1 Γ k 11 xk f x1 + 2(f 2 f(1 f 2 1 x1 ( ff + 2(f 2 f(1 f 2 1 x1 f f( 1 + 2f (1 f 2 1 x1 0. The parameter s ranges from 0 to infinity and so the length γ with respect to the hyperbolic metric is infinite. 4* Consider a curve α(s (r(s, z(s on the plane R 2 where 0 < s < b and α is parametrized by arc length, i.e., α (s 1 for all 0 < s < b. Moreover r(s is always positive and the curve α has no self-intersections. With this curve we can consider a surface in R 3 obtained by rotating this curve around the z-axis which will be parametrized by S : (0, b [0, 2π] R 3 where S(s, θ (r(s cos θ, r(s sin θ, z(s. Make yourself a drawing to make sure you understand what this is.
HOMEWORK 2 - RIEMANNIAN GEOMETRY 5 i Show that, with respect to the coordinates (s, θ, the induced metric becomes g ss 1, g sθ 0, g θθ r 2 (s. Consider the coordinates (s, θ. The coordinate vectors are s (r cos θ, r sin θ, z and θ ( r sin θ, r cos θ, 0. Hence we have that g sθ s, θ r r cos θ sin θ r r cos θ sin θ 0 and g θθ θ, θ r 2. Likewise g ss s, s (r 2 + (z 2 α 2 1. ii If γ is a geodesic on S which, using the coordinates (s, θ, becomes γ(t (s(t, θ(t, then the following equations hold: d 2 θ 2 + 2r r d 2 s 2 rr dθ ds 0 ( dθ 2 0, where r denotes differentiation with respect to the s-variable. Moreover, when you see something like rr what this really means is r(s(tr (s(t. I did not include this because then the formulas would become a mess. Let s compute the Christoffel symbols first. The formula to use is Γ k 1 ij 2 gkl ( i g jl + j g il l g ij. l1 Using the fact that g ss 1, g sθ 0, and g θθ 1 r 2, we have Γ s ss 0 Γ θ ss 0 Γ s sθ 0 Γθ sθ r r which means that Γ s θθ rr Γ θ θθ 0 s s 0, s θ θ s r r θ θ θ r r s. Hence γ γ γ ( ds s + dθ ( ( ds dθ γ s + γ γ ( ds ( ( ds dθ θ γ s +γ ( ds 2 θ + s s + 2 dθ s + γ ( dθ θ + 2 dθ θ + ds γ s+ dθ ( dθ ds s θ + ds r r θ ( dθ γ θ 2 θ θ 2 r r s
6 ANDRÉ NEVES We note that γ ( ds d 2 s and γ ( dθ 2 d 2 θ. This has been some 2 source of confusion so I will explain why. When we write γ ( ds, what we mean is that we consider a function f defined on a neighborhood of γ(t 0 in S so that f(γ(t ds for all t close to t 0. This is because, technically speaking, γ is a vector and so it differentiates functions on S. Thus by definition of vector we obtain that γ d(f γ (f d ( ds Having said that we finally have ( γ γ d 2 ( s dθ 2 2 r r s + d2 s 2. ( d 2 θ 2 + dθ 2r r Thus γ γ being zero implies the desired result. ds θ. iii Conclude that the meridian lines (i.e. curves with θ constant are geodesics and that the parallels (curves with s constant are geodesics when r (s 0. Draw a picture to convince yourself that this makes sense. (NOTE: I am freely identifying a geodesic with its image. More precisely, when I ask you to show that meridians are geodesics what I mean is that you need to find a parametrization so that they become geodesics (same thing for parallels. Parametrize a meridian by γ(t (t, θ 0. Then it is straightforward to see that the equation is satisfied. Regarding the parallels, parametrize them by γ(t (s 0, t. Because r (s 0 r(s 0 0, it is simple to see that γ is a geodesic as well. iv Conclude from ii that r 2 dθ is constant as a function of t and that ( ds 2 ( dθ 2 + r 2 is constant as a function of t. For the first identity we use ( d r 2 dθ 2r dr dθ +r2 d2 θ ds 2rr 2 ( dθ +r2 d2 θ d 2 2 r θ 2 + dθ 2r r ds 0.
HOMEWORK 2 - RIEMANNIAN GEOMETRY 7 Thus f(t r 2 (s(t dθ is a constant function and the result follows. For the second one we argue in the same way. ( (ds d 2 ( dθ 2 ( + r 2 2 d2 s ds dr dθ 2 2 + 2r + 2r 2 d2 θ dθ 2 ( 2 d2 s ds ds dθ 2 ( ( 2 + 2rr 4r 2 r dθ 2 ds r 2ds d 2 ( s dθ 2 2 rr 0 and thus f(t ( ds 2 + r 2 (s(t ( dθ 2 is constant as a function of t. What this shows is that the geodesics oscillate between two parallels (case A except in the case that one of these parallels has r (s 0, in which case the geodesic approximates this parallel as time t goes to infinity (case B, which itself is a geodesic. Make two nice little pictures where you show case A and case B happening. Let α(s (r(s, z(s be a curve which is defined for all s, the function r has exactly three critical points at s 1, s 0, and s 1 with r(0 1, and r(1 r( 1 1/10. Moreover lim r(s +, lim s ± z(s ±. s ± Note that s 0 is necessarily a local maximum of r, while s ±1 are absolute minimums of r. Pick a geodesic γ with initial conditions γ(0 (0, 0 and γ (0 ( s + θ / 2. Note that γ (0 is a unit vector. Hence, we must have for every t (1 r 2 dθ 1/ 2 and ( ds 2 ( dθ 2 + r 2 1 ( ds 2 + 1 2r 2 1 1 Note that (1 implies that 1, which means that r 2 (s(t 1/2 2r 2 for all t. Furthermore there is exactly one 0 < a < 1 and 1 < b < 0 so that r(a 2/2 and r(b 2/2. It follows from what we have just said that the s component of γ must belong to [a, b] and we argue that indeed γ oscillates between the parallels s a and s b. We start by arguing that there is a first time t 0 for which s(t 0 b (which implies from (1 that ds (t 0 0. If not, then this would mean that ds is never zero (because of (1 and thus s(t would be an increasing function of t (it has to be increasing and not decreasing because s (0 1/ 2. Because s(t is bounded we must have that ds d lim t 0 and lim 2 s t 0 (this is an easy fact of one 2 variable calculus. Hence, we can make t tend to infinity in the
8 ANDRÉ NEVES (2 second equation of ii and conclude that lim t r (s(t 0. This is impossible because then s(t should converge to 1 (we have argue previously than s(t b < 1 for all t. The conclusion is then that there is a first time t 0 for which s(t 0 b. It is not hard to recognize that for times t immediately after t 0 the function s(t must be decreasing and so we could repeat the above arguments with some obvious modifications to conclude the existence of some other time t 1 for which s(t 1 a. Arguing inductively one can then see an infinite sequence (t n n N where s(t n b if n is even and s(t n a if n is odd. In other words, γ oscillates between s a and s b. To find a geodesic which satisfies case B we consider γ with initial condition γ(0 (0, 0 and γ (0 cos α s + sin αθ (previously α π/4 where α is chosen so that sin α 1/10. In this case, (1 becomes r 2 dθ sin α and ( ds 2 ( dθ 2 + r 2 1 ( ds 2 + sin2 α r 2 1. This equations imply that sin2 α 1, which means that r 2 (s(t r 2 sin 2 (α 1/100 for all t. Hence the s component of γ must belong to [ 1, 1] because r(±1 1/10. Note that is this case ds can never be zero in finite time because that would imply that r would be 1 or 1 which would mean that γ would intersect tangentially the geodesic given by the parallel r ±1 and thus contradict the uniqueness of geodesics for the same initial conditions. If ds can never be zero in finite time, this means that (like we argued before γ should approach the parallel s 1 at infinity. 5 Show that all geodesics on a sphere S n are just great circles, i.e., intersection of S n with a plane passing through the origin. Pick a point p in S n and a unit vector v T p M R n+1. We are going to show that the geodesic with initial condition p and v is given by γ(t cos tp + sin t v. First, v T p M R n+1 means that v, p 0. Thus γ is indeed a curve in S n because γ(t, γ(t cos 2 t p 2 + sin 2 t v 2 1.
HOMEWORK 2 - RIEMANNIAN GEOMETRY 9 We need to check that γ γ 0. Denoting by D the Euclidean connection we have that γ γ (D γ γ T (γ T ( cos tp sin t v T (γ T 0. We now justify somewhat carefully the second and fourth equality above. The second inequality comes from the fact that if X γ (t 0 is a vector in T γ(t0 R n and f a function defined on a neighborhood of γ(t 0 in R n so that f(γ(t a(t for all t close to t 0, then X(f(γ(t 0 d(f γ (t 0 da (t 0 a (t 0. If we apply this with a(t x i (t, where γ(t (x 1(t,..., x n (t, we see that γ (x i x i and so D γ γ (γ (x 1,..., γ (x n γ. For the fourth inequality we recall that the tangent space T q S n is nothing but all the vectors which are orthogonal to the vector q. Thus the tangential projection of q on T q S n is zero. A moment of thought shows γ(t as above is just the intersection of S n with the plane spanned by p and v. 6* We can consider S 3 as the following subset of C 2 (which as a real vector space is the same thing as R 4. S 3 {(z, w C 2 z z + w w 1}. The advantage of using complex notation is that for every prime p we can consider the diffeomorphisms F : S 3 S 3 where F (z, w (e 2iπ/p z, e 2iπ/p w. i Show that G {I, F, F 2,..., F p 1 } is a group isomorphic to Z p, where F j denotes composition with itself j-times. Hence, we can consider the manifold L p S 3 /Z p, where x is identified with y if F k (x y for some k. When p 2, L 2 is just the projective plane RP 3. The map F is a linear map of C 2 with determinant non-zero and hence it is an isomorphism when restricted to S 3. In order to conclude the desired result we have to see that F p is the identity map and that F j being the identity map implies that j is a multiple of p. It is simple to see that F j (z, w (e 2iπj/p z, e 2iπj/p w and thus F j is the identity map if and only if j is indeed a multiple of j.
10 ANDRÉ NEVES ii Show that F is an isometry for the induced metric on S 3. implies that we can induce this metric on L p as well This It is enough to see that F is an orientation preserving isometry of C 2 R 4 because if F preserves the Euclidean metric then it will preserve the induced metric on S 3 as well. Because F is a linear map, this corresponds to see that if A denotes its matrix representation then A T A 1 and det A is positive (in which case it has to be one. The matrix A is given by A cos(2π/p sin(2π/p 0 0 sin(2π/p cos(2π/p 0 0 0 0 cos(2π/p sin(2π/p 0 0 sin(2π/p cos(2π/p Direct computation shows that A T A 1 and that det A1. iii Show that all the geodesics of L p are closed curves. On L 3 find two closed geodesics with different lengths. Pick a point [x] in L p and a unit vector v T [x] L p. Denote by α the geodesic which has α(0 [x] and α (0 v. We will show that α(0 α(2π and so α is closed. The projection map π : S 3 L p is a local diffeomorphism and so we can identify v T [x] L p with a unit vector v T x S 3 R 4. Consider the curve given by γ(t x cos t + sin t v which, as we saw in exercise 6, is a geodesic in S 3. Because the map π is a local isometry then [γ(t] π γ(t is a geodesic in L p as well. Uniqueness of geodesics for a given initial condition and velocity shows that α(t [γ(t]. Because γ(2π γ(0 we have α(0 α(2π. We now do the second part of the exercise. We first remark that if the geodesic is parametrized by arclength and t 0 is the first time for which α(t 0 α(0, then the length of α is t 0. Set [x] [(1, 0, 0, 0] and v (0, 0, 1, 0. Consider the geodesic α(t [cos tx + sin t v] [(cos t, 0, sin t, 0]. We claim that if α(t 0 α(0 then t 0 is a multiple of 2π. Indeed we must have (1, 0, 0, 0 F j (α(t 0 (cos t 0 e 2iπj/3, sin t 0 e 2iπj/3, in which case sin t 0 e 2iπj/3 (0, 0. Thus t 0 is a multiple of π. If t 0 is not a multiple of 2π then cos t 0 e 2iπj/3 e 2iπj/3 and this can never be (1, 0. Therefore the length of the closed geodesic is 2π.
HOMEWORK 2 - RIEMANNIAN GEOMETRY 11 Let us find now a closed geodesic with length less than 2π. Set [x] [(2 1/2, 0, 2 1/2, 0] and v (0, 2 1/2, 0, 2 1/2. Consider the geodesic α(t [cos tx + sin t v] [2 1/2 (cos t, sin t, cos t, sin t]. We will show that α(4π/3 α(0 and so the length of α is less than 4π/3. This is because F (α(4π/3 2 1/2 F (e 4πi/3, e 4πi/3 2 1/2 (e 4πi/3 e 2πi/3, e 4πi/3 e 2πi/3 2 1/2 (1, 0, 1, 0 α(0, where we are freely identifying α(t with cos tx + sin t v (technically speaking they are different things because one is a curve in L 3 and the other a curve in S 3. 7 Given a smooth function f on a compact manifold with consider its gradient defined in the following way. Given p in M we define f(p to be the only vector for which g ( f(p, γ (0 for every curve γ on M with γ(0 p. d(f γ (0 i Show that f is well defined and compute its expression on charts (φ α, U α. Show that there is some point q such that f(q 0. Given a chart (φ α, U α we consider the function f α f φ α. Set a i (x g ij (x f α (x x j j1 We are going to check that if X a i x i then for every curve γ with γ(0 φ α (x we have that (3 g φα(x(x, γ d(f γ (0 (0. In coordinates, we have γ (0 n j1 b i x i which means that for evert function h on M we have γ (h φ α (0(h b i (x. x i
12 ANDRÉ NEVES Then g φα(x(x, γ (0 a i (xb j g ij (x i,j1 k1 i,j,k1 f α x k (xb j g ki g ij (x f α x k (xb k γ (0(f d(f γ (0. The expression for X is unique because if Y n c i x i other vector satisfying property (3 we have that c i g ij g φα(x ( Y, x j f α x j (x is any and so c i (x j1 g ij (x f α x j (x. Let q be a point where the absolute maximum of f is achieved. Choosing an appropriate chart, we see that f α defined as above will have an absolute maximum at the origin, in which case all of its partial derivatives vanish. From the expression for f(q given above we see that this vector must be zero. ii For any p M, define the Hessian 2 f to be 2 f : T p M T p M R where 2 f(x, Y g (( X f (p, Y. Show that 2 f is symmetric ( 2 f(x, Y 2 f(y, X and bilinear. The bilinearity is easy. We show that is symmetric. By bilinearity it suffices to see that for any coordinate vectors of a given chart (U, φ. For short, we denote them by xi, i 1,..., n. We also use the notation f f φ. Note that by the definition of f we have that g( f, xi (φ(x d ( f(x + t(0,..., 0, 1, 0,..., 0 f x i (x
HOMEWORK 2 - RIEMANNIAN GEOMETRY 13 Thus 2 f( xi, xj g ( xi f, xj xi (g( f, x j g( f, xi x j ( f xi (g( f, x j g( f, xj x i xi g( f, xj x i x j 2 f x i x j g( f, xj x i 2 f x j x i g( f, xj x i 2 f( xj, xi. iii Suppose that p is an absolute minimum of f. nonnegative definite at p Show that 2 f is Pick a vector V γ (0 T p M. We need to see that 2 f(v, V 0. The function a(t f γ(t has an absolute minimum at the origin and so its second derivative a (0 0. From the first question we have that f(p 0. Using this we have 2 f(v, V g( V f, V V (g( f, V g( f(p. V V V (g( f, V iv Define the Laplacian of f to be d (g( f, V γ(0 d2 a(t 2 (0 a (0 0 f(p 2 f(e i, e i where e 1,...e n is an orthonormal basis for T p M. Show that this is well defined, i.e., does not depend on the basis chosen and compute this for the Euclidean metric on R n. Say that (u j n j1 is another orthonormal basis for T pm. Then u i n j1 a ije j, where the fact that both bases are orthonormal implies that the matrix A (a ij i,j1,...,n has A 1 A T. Hence, from bilinearity we have ( 2 f(u i, u i a ik a ij 2 f(e j, e k a ik a ij 2 f(e j, e k i,j,k1 j,k1 j,k1 δ jk 2 f(e j, e k 2 f(e j, e j. j1
14 ANDRÉ NEVES 2 f In R n, we have that f n ( x i, x i D x i f, x i f x i x i and so 2 f j1 x j x i x j, x i 2 f x i x i. Thus f 2 f x i x i. 8* Consider the exponential map exp p : B ε (0 M which we have seen in class that it is a diffeomorphism onto its image and hence it is a chart. Denote by x i, i 1,..., n the coordinate vectors. (See at the end to see what I mean with this. i Show that the matrix that represents ( the metric at origin is the identity matrix, i..e, if g ij g x i, x i, then g ij (0 is 1 if i j and 0 otherwise. For simplicity we denote x i by xi. We proved in class that D(exp p 0 : T P M T p M is the identity map. The argument is super simple and so I will do it again. Recall that if γ v (t γ(v, t denotes the geodesic with γ v (0 p and γ v(0 v T p M, then exp p (tv γ(tv, 1 γ(v, t. Hence D(exp p 0 (v d (exp p (tv d v (t γ t0 t0γ v(0 v. If ē i denotes the vector in R n with zeros everywhere expect i- th position, we have by definition that xi Dφ(ē i, where φ is defined at the end of this exercise. Keeping with the notation at the end of this exercise we have DF 0 (ē i F (ē i e i (F is a linear map!. Thus xi Dφ(ē i D(exp p 0 (DF 0 (ē i D(exp p 0 (e i e i. Because (e i n was chosen to be an orthonormal basis we have that g ij (0 g(e i, e j δ ij. ii Show that g ij x k (0 0 for all i, j, k 1,..., n. Conclude that the Christoffel symbols Γ k ij vanish at the origin for every i, j, k 1,..., n. HINT: First show that x x i (0 for every i. Then understand i why does this imply that g ii x i (0 0 for every i. The others should follow likewise...
HOMEWORK 2 - RIEMANNIAN GEOMETRY 15 Pick any vector V a i xi, where a i are constants (we will call this a constant vector. We are going to show first that V V (0 0. Indeed γ(t φ(tv exp p (tv is a geodesic with γ (0 V. Thus, by definition, V V (0 0. This implies that for any constant vector V and W we have V W (0 0 because V W W V (easy to see because it is a constant vector and 0 V +W (V + W (0 V V (0 + W W (0 + 2 V W (0 2 V W (0. Therefore we have that V (g(x, Y (0 g( V X, Y +g(x, V Y 0 if X, Y, Z are constant vectors. Applying this with V xk, X xi, and Y xj we obtain g ij x k (0 0. To see that the Christoffel symbols are zero use V xi and W xi and 0 V W (0 n k1 Γk ij (0 x k. What the first and second question showed is that the metric in the coordinates induced by the exponential map coincides with the Euclidean metric up to order two, i.e., not only the matrix (g ij i,j1,...,n is the identity at the origin as all its first derivatives are zero at the origin as well. iii Show that if φ α is a coordinate chart of M with U α {x R n x < 1} such that the metric (g ij i,j1,...,n is the identity at the origin and the Christoffel symbols Γ k ij (x are all zero for every x U α, then the metric (g ij i,j1,...,n (x is the identity for every x U α, i.e., the manifold φ α (U α is isometric to the unit ball with the Euclidean metric. Because the Christoffel symbols are all zero we have xi xj 0 for all x U. Hence g ij g( xk xi, xj + g( xi, xk xj 0. x k Therefore, the function g ij is constant and so g ij (x g ij (0 δ ij. Therefore, we have that φ αg is the Euclidean metric because if U n u i xi and V n v i xi then φ αg(u, V u i v j g ij u i v j. i,j1 i,j1 Let s be more precise on the definition of coordinates on the exponential map. A priori what we have is that exp p is defined on a subset B ε (0 {V T p M g p (V, V < ε 2 } of the tangent plane T p M. In order to see exp p as a chart I need it to be define on an open subset of R n. We do this as following. Pick an orthonormal
16 ANDRÉ NEVES basis e 1,..., e n for T p M, where being orthonormal is defined with respect to the dot product g p. Consider the map F : R n T p M, F (x x i e i and set U to be F 1 (B ε (0. Now we consider the chart φ : U M where φ(x exp p (F (x. As soon we have a chart we can talk about the coordinate vectors x i and the metric g ij as we did in the beginning of the course. What I want you to see is that at the origin (which is contained in U, this matrix is the identity matrix. There is no reason to expect the matrix g ij to be the identity at any other point.