1 Covariant Derivatives along Curves

Transcription

1 Math 6397 Riemannian Geometry II The Jacobi Field 1 Covariant Derivatives along Curves Lemma 4.9 (see P. 57 on the textbook) Let be a linear connection on M. For each curve γ : I M, determines a unique operator D t : T (γ) T (γ) where T (γ) is th eset of smooth vector fields along γ, such that it is linear over R, it satisfies the product rule, and if V is extendible, then D t V (t) = γ (t)ṽ, where Ṽ is an extension of V. D t is called the Covariant Derivatives along the Curve γ. The motivation of D t comes from the following: if f(t) is a function defined on γ, then, we can express f(t) = f γ(t), where f is some function in the neighborhood of γ (assuming that f is extendible). Then f (t) = df(γ (t)) = γ (t)f = D t f. Similarily, if X is a vector field along the curve γ, if we write X = X i i then D t X = dxi dt i + X i D t i. Thus we see that D t is a natural differential operator along the curve γ. 2 Jacobi field Introduction: The Jacobi fields arise naturally from the variation of the geodesic γ through neighborhouring geodesics. Let now γ : I M be a geodesic. A variation of γ is a differential map Γ(s, t) : ( ɛ, ɛ) I M with Γ(0, t) = γ(t) for all t I. Γ is said to be a geodesic variation if for every s ( ɛ, ɛ), the curve F (, s) is geodesic. 1

2 Write T (s, t) = t Γ(s, t), S(s, t) = s Γ(s, t). tells us that D t T = 0. The geodesic equation We can take the covariant derivative of this equation to get D s D t T = 0. The variation vector field of the variation Γ is defined as V = S(0, t). Using the fact that D s D t T = 0 and two lemmas (Lemma 6.3 (symmetry lemma): D s t Γ = D t s Γ and Lemma 10.1: D s D t V D t D s V = R(S, T )V for any vector field V along γ), we can prove that Theorem 1 Let γ : [a, b] M be a geodesic on an m-dimensional Riemmanian manifold (M, g). Then the variation vector field V of its geodesic variation satisfies the Jacobi equation D 2 t V + R(V, γ )γ = 0. Definition of a Jacobi field along a geodesic Let J = J(t) be a vector field along γ. Denote by J (t) = D t J and J (t) = D 2 t J. Definition. Let γ : [a, b] M be a geodesic on an m-dimensional Riemmanian manifold (M, g). A vector field J = J(t) along γ is called Jacobi field if it satisfies the following Jacobi equation: where J (t) = D 2 t J. Remarks: J (t) + R(J(t), γ (t))γ (t) = 0, 2

3 1. It is obvious that if J is a Jacobi field along γ, then J := J < J, T > T is also a Jacobi field along γ where T = γ. 2. One of the important result about Jacobi field is the following: d 2 dt 2 < J, γ > 0, so so < J(t), γ (t) > is a linear function of t, i.e. < J(t), γ (t) >= λt + µ for some constants λ and µ. The proof is as follows: Using the compatibility with the metric and the fact that γ (t) = 0, we t compute D 2 t < J, γ > = < J (t), γ > = < R(J, γ )γ, γ > = R(J, γ, γ, γ ) = 0. Next, we want to find a local expression the the Jacobi equation. Let γ be a geodesic on M. We express the Jacobi field equation in terms of the local coordinate. Choose an orthonormal basis {e i p } for T p M, and extend it to a parallel orthonormal frame along all γ. Then e i(t) = e i (t) = 0. t Let J(t) be a Jacobi field along γ. Write Then m J(t) = J i (t)e i (t). i=1 m J (t) = J i (t)e i (t), m J (t) = J i (t)e i (t). i=1 i=1 Write γ (t) = m k=1 γ k e k, then Thus J (t) + R(J(t), γ (t))γ (t) = i {J i (t) + j R i ljk(γ(t))j j γ k γ l }e i (t). 3

4 Hence the Jacobi equation is reduced to J i (t) = j R i ljk(γ(t))j j γ k γ l, 1 i m. Theorem 2. Let γ : [a, b] M be a geodesic on an m-dimensional Riemmanian manifold (M, g). Then for every v, w Tγ(a)M, there exists a unique Jacobi field J(t) along γ satisfying J(a) = v, J (a) = w. There are always two trivial Jacobi fields along any geodesic γ, which can be written down immediately: Because D t γ (t) = 0 and R(γ, γ )γ = 0 by antisymmetry of R, the vector field J 0 (t) = γ (t) satisfies the Jacobi equation with J 0 (0) = γ (0), J 0(0) = 0. Similarly, J 1 (t) = tγ (t) is a Jacobi field satisfying J 1 (0) = 0, J 1(0) = γ (0). It is easy to see that J 0 is the variation field of the variation Γ(s, t) = γ(t + s), and J 1 is the variation field of the variation F (t, u) = γ(te u ). Therefore, these two Jacobi fields just reflect the possible reparametrizations of γ, and don t tell us anything about the behavior of geodesics other than γ itself. To distinguish these two trivial cases from more informative ones, we make the following definitions. Definition If J is orthogonal to γ everywhere, then J is called a normal Jacobi field. Theorem 3. Let γ : I M be a geodesic on an m-dimensional Riemmanian manifold (M, g), and let a I. (a) A Jacobi field J along γ is normal if and only if J(a) γ (a), J (a) γ (a). (b) Any Jacobi field orthogonal to γ at two points is normal. 4

5 Proof. As we showed earlier, f(t) :=< J(t), γ (t) > is a linear function of t. Note that f(a) =< J(a), γ (a) > and f (a) =< J (a), γ (a) >. Thus J(a) and J (a) are orthogonal to γ (a) if and only if f and its first order derivative vanish at a, which happens if and only if f 0. Similarly, if J is orthogonal to γ at two points, then f vanishes at two points, so f 0. This proves the theorem. Geodesic variations, Gauss lemma, normal coordinates, and conjugate points. We first recall the definition and some important statement regading the differential of the exponent map. From chapter 4 in the textbook (John M. Lee) we have that any initial point p M and any initial tangent vector v T p M determine a unique maximal geodesic γ v with γ v (0) = p and γ v(0) = v. It has the important property that γ cv (t) = γ v ct, whenever it is defined. By taking t = 1 and using c = t, we get γ tv (1) = γ v t, whenever it is defined. Let E := {V T M; γ v is defined on an interval containing [0, 1]} and then we define the exponential map exp : E M by exp(v) = γ v (1). Then we have, using the property which we just mentioned, exp(tv) = γ v (t). (this is one of the most important property of exp. Another importnat result is as follows: (see Lemma 5.10 on the textbook P. 76) Under the natural identification T 0 (T p M) = T p M, we have (exp p ),0 : T 0 (T p M) = T p M T p M is the identity map. Recall the proof of the above result: We write (exp p ) to denote (exp p ),0. To compute (exp p ) (v) for v T p M, we just need to choose a curve τ in T p M starting at 0 whose initial tangent vector is v and compute 5

6 the inital tangent vector of the component of the curve exp p τ. An obvious such curve is τ(t) = tv. Thus (exp p ) (v) = d dt t=0 exp p τ(t) = d dt t=0 exp p (tv) = d dt t=0γ v (t) = γ v(0) = v. Hence the result is proved. We now give a new proof of Gauss lemma using Jacobi field. Lemma (Gauss). Let p M and v T p M such that exp p v is defined. Let w T p M = T v (T p M). Then < (exp p ) v (v), (exp p ) v (w) >=< v, w >. New Proof. Let v T p (M) and consider the geodesic γ(t) := exp p (tv), then γ(0) = p and γ (0) = v. Identifying T v (T p M) with T p M as usual, we can compute (exp),v (w) = d ds s=0 exp p (v + sw). To compute this, we define a geodesic variation of γ, as before, by Γ(s, t) = exp p t(v +sw). (we shift w to sw and connects v and sw to get a straightline in T (M) an then use the exponential map to project down to M, see the picture on P. 184 in the textbook). Then, its variation field J W (t) = s Γ(s, t) s=0 is a Jacobi field and J W (1) = exp,v W. We now compute J(t) and J (t). Hence J (0) = D t J t=0 = (D t s Γ(s, t)) s=0,t=0 = D s t Γ(s, t)) s=0,t=0, note that the purpose to switch the order of differentiation is that we want to use the result aboue (so we need to differentiate Γ in t). Due to the fact that (exp p ) (see the result above) is the identity at the origin of Tγ(s)M, we have that, by definition and the result above, t Γ(s, t) t=0 = (exp p ),0 (v + sw) = v + sw. 6

7 Hence Therefore, J (0) = J (1) = D s (v + sw) s=0 = w < v, w >=< γ (0), J (0) >. Using the property for the Jacobi fields we proved earlier, d 2 dt 2 < J(t), γ (t) >= 0, so < J(t), γ (t) > is a linear function, say < J(t), γ (t) >= at + b, so, < J (t), γ (t) >= a which is a constant (using γ is a geodesic) and b =< J(0), γ (0) >= 0. Therefore, < v, w > = < γ (0), J (0) >=< J (1), γ (1) >= a =< J(1), γ (1) > = (expγ(0)),v (w), (exp p ),v v >. This proves Gauss lemma. In the above new proof of Gauss lemma, we used Γ(s, t) = exp p (t(v + sw)) as the variation of the curve γ v (t) = exp p (tv). Its variation field J(t) = s Γ(s, t) s=0 is a Jacobi field along the curve γ v which satisfies J(0) = 0, J(1) = exp,v (w) and J (0) = w. Using the normal coordinates, so γ v (t) = (v 1 t,..., v n t), we have the following simple expression of J: Theorem Let p M and let (x i ) be a normal coordinate on a neighborhood of p, and let γ be a radical geodesic starting at p. For any W = W i i T p M, the Jacobi field J along γ with J(0) = 0 and J (0) = W is given by J(t) = tw j i. The geodesic variation Γ(s, t) = exp p (t(v + sw)) as the variation of the curve γ v (t) = exp p (tv) and its variation field J(t) = s Γ(s, t) s=0 is also used in the study of conjugate points (see the section about conjugate points ). 7

8 Every Jacobi field arises from the variation of some geodesic variation of γ. We now prove our statement as the following theorem. Theorem Let γ : [a, b] M be a geodesic on a complete Riemmanian manifold (M, g). Let J = J(t) be a Jacobi field along γ. Then J must be a variation vector field of some geodesic variation of γ. Proof. WLOG, we assume that a = 0. The key is to construct a geodesic variation Γ of γ such that its variation field V satisfies V (0) = J(0) and V (0) = J (0). In the proof of Gauss lemma, we used the following natural variation of γ: Γ(s, t) = exp p t(γ (0) + sw), then we see that V (0) = w. By letting w = J (0). Then V (0) = J (0). The problem is that, for Γ(s, t) = exp p t(γ (0) + sw), we always have Γ(s, 0) = p (the initial point is always fixed, so V (0) = 0. To get V (0) = J(0), we need to let the initial point moving. So we replace the initial point with a curve σ = σ(s), s ( ɛ, ɛ) such that σ(0) = γ(0), σ (0) = J(0). After that, take a parallel translation of w := J(0) and γ (0) along the curve σ to get two vector fields W (s) and T (s), s ( ɛ, ɛ) which are parallel along σ. Now we define that Γ(s, t) = expσ(s) t(t (s) + sw (s)) then, for every u ( ɛ, ɛ), the curve Γ(, s) is geodesic, and Γ(0, t) = γ(t), so Γ is a geodesic variation of the geodesic γ. Let V (t) = s Γ(s, t) s=0 be the variation vector field. Then V is a Jacobi field. We verify that V (0) = J(0), V (0) = J (0). To see this, V (0) = s Γ(s, 0) s=0. Using Γ(s, 0) = expσ(s)(0) = σ(s), we have V (0) = σ (0) = J(0). To see V (0) = J (0), we compute V (0) = D t V t=0 = (D t s Γ(s, t)) s=0,t=0 = D s t Γ(s, t)) s=0,t=0. Due to the fact that (expσ(s)) (see the result above) is the identity at the origin of Tσ(s)M, we have that t Γ(s, t)) t=0 = (expσ(s)) ((T (s) + sw (s)) = T (s) + sw (s). 8

9 Hence V (0) = D s (T (s) + sw (s)) s=0 = W (0) since D s T = 0, D s W = 0 due to the fact that T and W are parallel along σ. Hence V (0) = J (0). By the uniqueness result, we have J(t) = V (t). Hence J(t) is a variation vector field of the geodesic variation F of γ. This finishes the proof. As another application, we prove Corollary 3.3 Let p M and v T p M be contained in the domain of definition of exp p, and c(t) = exp p tv. Let the piecewise sommth curve γ : [0, 1] T p M be likewise contained in the domain of definition of exp p, and assume that γ(0) = 0, γ(1) = v. Then v = L(exp p tv t [0,1] ) L(exp p γ), and the equality holds if and only if γ is different from the curve tv, t [0, 1] only by a reparameterization. Proof. We shall show that any piecewise sommth curve γ : [0, 1] T p M with γ(0) = 0 satisfies L(exp p γ) γ(1). Write γ(t) = r(t)φ(t), r(t) R and φ(t) T p M, with φ(t) 1 (polar coordinates in T p M). We have γ (t) = r (t)φ(t) + r(t)φ (t) and < φ, φ > 0. Thus, by Corollary 3.2, also < exp,γ(t)φ(t), exp,γ(t)φ (t) >= 0, exp,γ(t)φ(t) = φ(t) = 1, and it follows that Hence L(exp p γ) = 1 0 exp,γ(t)γ (t) r (t). exp,γ(t)γ (t) dt r (t) dt r(1) r(0) = γ(1).

10 This proves the statement. Remark: Corollary by no means implies that the geodesic c(t) = exp +ptv is the shortest connection between its end points. It only is shorter than any other curve that is the exponential image of a curve with the same initial and end points as the ray tv, 0 t 1. 3 Jacobi Fields of the Manifolds with Constant Sectional Curvature Let (M, g) be a Riemannian manifold with constant sectional curvature C. Recall that the sectional curvature at p M with respect to X p, Y p T p M ie defined by r(x p, Y p, Y p, X p ) K(X p, Y p )) = X p 2 Y p 2 < X p, Y p >. 2 We can show that if (M, g) be a Riemannian manifold with constant sectional, then R(Z, W, X, Y ) =< R(X, Y )Z, W >= c(< X, Z >< Y, W > < X, W >< Y, Z >), or equivalently, R(X, Y )Z = c(< X, Z > Y < Y, Z > X) (Here is the outline of the proof: We first prove the following result: Let V be a m-dimensional vector space and let f : V V V V be multilinear map (i.e. f is a (0, 4)-tensor). f is called curvature-type tensor if it satisfies, for every u, v, z, w V, (i) Antisymmetric: f(u, v, w, z) = f(v, u, w, z) = f(u, v, z, w); (ii) First Bianchi identity: f(u, v, w, z) + f(w, v, z, u) + f(z, v, u, w) = 0, (iii) Symmetric: f(u, v, w, z) = f(w, z, u, v). Now let f, g : V V V V be two curvature-type (0, 4)-tensors. Assume that, for every u, v V, f(u, v, u, v) = g(u, v, u, v). Prove that f g. Hint: Let F (u, v, w, z) = f(u, v, w, z) g(u, v, w, z). The goal is to show that F 0. Step 1: From Note that F (u, v, u, v) = 0 implies that F (u, v + z, u, v+z) = 0. From here to show that F (u, v, u, z) = 0 for every u, v, z V. Step 2: From step 1, we get F (u + v, w, u + v, z) = 0. From here, try to 10

11 get (by expanding it) F (u, w, v, z) = F (v, w, z, u), for every u, v, z, w V. Similarily, one can get F (u, w, v, z) = F (v, w, z, u) = F (z, w, u, v). Step 3: Use the First Bianchi identity and the previous identity to get 3F (u, v, w, z) = F (u, v, w, z) + f(w, v, z, u) + f(z, v, u, w) = 0. This finishes the proof. Then we prove the foolwoign statement: We say that M is wondering at p M if the sectional curvature K p (E) at p is constant(i.e. it is independent of E). Hence R(X p, Y p, X p, Y p ) = K p G(X p, Y p, X p, Y p ) for every X p, Y p T p (M). Prove that if M is wondering at p M, then for any X p, Y p, Z p, W p T p (M), R(X p, Y p, Z p, W p ) = K p G(X p, Y p, Z p, W p ). Hint: Use #2 by verifying that both R and G are curvature-type (0, 4)- tensors. ) Let γ be a geodesic with unit-speed in M. Substitute this into the Jacobi equation, we find a normal Jacobi field J satisfies 0 = J C(< J, γ > γ < γ, γ > J) = J + CJ. Choosing a parallel normal vector field E along γ and setting J(t) = u(t)e(t), and assuming that J(0) = 0, then the Jacobi equation becomes that u (t) + cu(t) = 0 has fundamental solutions with u(0) = 0 as follows: if c > 0, then u(t) = 1 c sin ct, if c = 0, then u(t) = t, if c < 0, then u(t) = 1 c sinh ct. So we proved the following statement: Let (M, g) be a Riemannian manifold with constant sectional curvature C. Let γ be a geodesic with unit-speed in M. Then the normal Jacobi field along along γ with J(0) = 0 are precisely the vector fields J(t) = u(t)e(t), where E is any parallel normal vector field along γ, and u(t) is as follows: if c > 0, then u(t) = 1 c sin ct, if c = 0, then u(t) = t, if c < 0, then u(t) = 1 c sinh ct. 11

12 4 Conjugate points Let (M, g) be a complete m-dimensional Riemannian manifold. According to the Hopf-Rinow theorem, for every p M, the exponential map is defined on T p M everywhere. Because, (exp p ) 0 the derivative of exp p at the zero vector 0, is the identity map on T p (M) = T 0 (T p M), so it is non-singular in a neighborhood of 0. But, in general, sometimes, it may be (exp p ) may be singular at some points other than zero. Let us look at the case of unit-sphere in R n+1. Let M = S n = {x R n+1 n+1 α=1 (x α ) 2 = 1}. Then (M, g) is a space with constant curvature (the sectional curvature is 1). We know that for every p S n, all geodesic curves starting from p converge to q = p, and the length of these geodesics are all equal to π. Hence, on S n 1 (π) T p (M), the image under the map exp p is the single point set {q}. In particular, for every smooth curve σ(t) on S n 1 (π), hence exp p (σ(t)) = q, (exp p ) σ(t)(σ (t)) = 0. This means that the map exp p is degenerate on S n 1 (π) everywhere. Motivated from above example, we introduce the following definition. Definition. Let (M, g) be a complete m-dimensional Riemannian manifold. Let p M, v T p M. If the exponential map exp p is degenerate at v, i.e. there exists w T p M = T v (T p M), such that (exp p ) v (w) = 0, then we call q = exp p (v) M is the conjugate point of p along the curve γ(t) = exp p (tv). Hence, by the inverse function theorem, 12

13 Theorem. Suppose p M, v T p M, and q = exp p v. Then exp p is local diffeomorphism in a neighborhood of v if and only if q is not conjugate to p along the geodesic γ(t) = exp p (tv), t [0, 1]. The following theorem uses Jacobi field to describe the conjugate points. Theorem. Let γ : [0, b] M be a geodesic on a complete Riemannian manifold (M, g). p = γ(0), q = γ(b). Then q is a conjugate point of p along the curve γ if and only if there exists a nonzero Jacobi field J = J(t), satisfying J(0) = J(b) = 0. Proof. We assume that b = 1. Identifying T v (T p M) with T p M as usual, we can compute (exp),v (w) = d dt s=0 exp p (v + sw). To compute this, we define a geodesic variation of γ, as before, by Γ(s, t) = exp p t(v + sw). Then, its variation field J W (t) = s Γ(s, t) s=0 is a Jacobi field and J W (1) = exp,v W. Therefore, exp,v fails to be an isomorphism when exp,v W = 0 for some W, which occurs precisely when there is a nonzero Jacobi field J W, satisfying J W (0) = J W (1) = 0. Conversely, assume there exists a nonzero Jacobi field J = J(t), satisfying J(0) = J(b) = 0. We let w = J (0), and consider Γ(s, t) = exp p t(v + sw). Then, its variation field J W (t) = s Γ(s, t) s=0 is a Jacobi field with J w(0) = w. By the uniqueness, we have J = J w. Hence J W (1) = exp,v w = J(1) = 0. So q = γ(1) is a conjugate point of p. This proves the theorem. Corollary. Let p, q M. If q = γ(b) is the conjugate point of p = γ(0) along the geodesic γ(t), t [0, b], then p is the conjugate point of q along the geodesic γ(t) = γ(b t). Theorem. Let γ : [0, b] M be a geodesic on a complete Riemannian manifold (M, g). p = γ(0), q = γ(b). If q is not a conjugate point of p along 13

14 the curve γ, then for every v T p M, w T q M, there exists a unique Jacobi field with J(0) = v, J(b) = w. Proof. Recall that from Corollary 3.2, we know that J(t) = t(expγ(0)) tγ (0)(J (0)) is a Jacobi field along γ with J(0) = 0. So to construct Jacobi field with J(0) = v, J(b) = w, we construct J 1 with J 1 (0) = 0, J 1 (b) = w and J 2 (0) = v, J 2 (b) = 0 respectively. To do so, we need pass the condition J 1 (b) = w to J 1(b) = w. From the assumption, since γ(t) = exp p (tγ (0)) so q = γ(b) = exp p (bγ (0)), the map (exp p ) bγ (0) is non-singular, so it is a linear isomorphism. So there exists w T bγ (0)(T p M), such that For every t [0, b], let w = (exp p ) bγ (0)( w). J 1 (t) = t b (exp p) tγ (0)( w), Then J 1 is Jacobi field along the curve γ with J 1 (0) = 0, J 1 (b) = (exp p ) bγ (0)( w) = w. On the other hand, since p is not a conjugate point of q, there exists a Jacobi field J 2 along the curve γ with J 2 (0) = v, J 2 (b) = 0. Let J(t) = J 1 (t) + J 2 (t). Then J is Jacobi field along the curve γ with J(0) = v, J(b) = w. We now prove the uniqueness. If J J. Then J = J J is a Jacobi field along the curve γ with J(0) = J(b) = 0. This contradicts with the assumption that q is a conjugate point of p. This finishes the proof. 14

15 5 The first and second variation formulas Let γ : [a, b] M be a differentiable curve with γ (t) = 1 (unit-speed). The lenght of γ is l = L(γ) = b a γ (t) dt = γ (0) (b a). A variation of γ is a differential map Γ(s, t) : ( ɛ, ɛ) [a, b] M with Γ(0, t) = γ(t) for all t [a, b]. Write γ s := Γ(s, ), and its arc-length is Let L(s) = L(γ s ) = b a γ s(t) dt. T (s, t) = t Γ(s, t), S(s, t) = s Γ(s, t) and let V (t) = S(0, t) be the variation vector field of the variation of Γ. Then the first variation formula is, assuming that Γ is smooth on [a i 1, a i ], d ds L(γ ai < D t S, T > s) [ai 1,a i ] = dt. a i 1 < T, T > 1/2 In particular, it shows that a geodesic is the critical point of the arc-length functional. The geodesic is normal (namely, the parameter t is its arc-length) if γ (t) = 1, i.e. l = b a. Using the the first variation formula, we can show that every minimizing curve is geodeisc when it it given by a unit-speed parametrization. Assume Γ is smooth on [a i 1, a i ]. Then = = d 2 ds L(γ s) 2 [ai 1,a i ] ( ai < Ds D t S, T > + < D ts, D s T > 1 ) < D t S, T > 2 < D s T, T ) dt a i 1 < T, T > 1/2 < T, T > 1/2 2 < T, T > 3/2 ( ai < Dt D s S + R(S, T )S, T > + < D ts, D t S > < D ts, T > 2 ) < D s T, T ) dt. < T, T > 1/2 < T, T > 1/2 < T, T > 3/2 a i 1 15

16 Now restricting to s = 0, where T = 1: ai d 2 ds s=0l(γ 2 s ) [ai 1,a i ] = (< D s D t S, T > R(S, T, T, S) a i 1 + D t S 2 < D t S, T > 2 )dt s=0. Because D t T = 0 when s = 0, the first term above can be integrated as follows: ai ai < D s D t S, T > dt = a i 1 t < D ss, T > dt =< D s S, T > t=a i t=a i 1. a i 1 Notice that S(s, t) = 0 for all s at the end points t = a 0 and t = a k = b, so D s S = 0 there. Moreover, along the boundaries {t = a i } of the smooth regions, D s S = D s ( s Γ) depends only on the value of Γ when t = a i, and it is smooth up to the line {t = a i } from both sides; therefore D s S is continuous for all (s, t). Thus when we inerst the above identity, the boundary contributions from the first term all cancel, and we get d 2 ds 2 s=0l(γ s ) = = b a b a ( D t S 2 < D t S, T > 2 R(S, T, T, S))dt s=0 ( D t V 2 < D t V, γ > 2 R(V, γ, γ, V ))dt. Let V be the normal component of V, which is orthogonal to γ, i.e. U = U < U, γ > γ. Then So V = V + < V, γ > γ. < V, V > = < V, V > + < V, γ >< V, γ > + < V, γ > 2 < γ, γ > = < V, V > + < V, γ > 2, where we used < V, γ >= t < U, γ > < U, D t γ >= 0. 16

17 Hence, we have that the second variation formula is d 2 b ds s=0l(γ 2 s ) = ( D t V 2 R(V, γ, γ, V ))dt. a It should come as no surprise that the second variation depends only on the normal component of U; intuitively, the tangential components of U contributes only to a reparametrization of γ, and the length is independent of parametrization. For this reason, we general apply the second variation formula only to variations whose variation fields are proper (fix-end points variation) and normal. We define a symmetric bilinear form I, called the index form, on the space of proper normal vector fields along γ by I(V, W ) = b a {< D t V, D t W > + < R(γ, V )γ, W >}dt. You should think of I(V, W ) as sort of Hessian or second order derivative of the length functional. Then, for every proper variation of γ, L (0) = I(V, V ). Because every proper normal vector field along γ is the variation field of some proper variation, it can be rephrased on terms of the index form in the following way Corollary. If F is a proper variation of a geodesic γ whose variation field is a proper normal variation field V, the second variation of L(γ) is I(V, V ), i.e.l (0) = I(V, V ). In particular, if γ is minimizing, then I(V, V ) 0 for any proper normal vector field along γ. The next Proposition gives another expression for I, which makes the role of the Jacobi equation more evident. Proposition 5.1. For any pair of proper normal vector fields V, W along a geodesic segment γ, b I(V, W ) = < Dt 2 V + R(V, γ )γ, W > dt a 17 k < i D t V, W (a i ) >, i=1

18 where {a i } are the points where V is not smooth, and i V is the jump on V at t = a i. Proof. On any interval [a i 1, a i ] where V and W are smooth, d dt < D tv, W >=< D 2 t V, W > + < D t V, D t W >. Thus, by the fundamental theorem of calculus, ai a i 1 < D t V, D t W > dt = ai a i 1 < D 2 t V, W > + < D t V, W > a i a i 1. Summing over i, and noting that W is continuous at t = a i, and W (a) = W (b) = 0, we get the identity. This finishes the proof. 6 Geodesics do not minimize past conjugate points In this section, we use the second variation to prove another extremely important fact about conjugate points: No geodesic is minimizing past its first conjugate point. Theorem. If γ is a geodesic segment from p to q that has an interior conjugate point to p, then there exists a proper normal vector field X along γ such that I(X, X) < 0. In particular, γ is not minimizing. Proof. Suppose γ : [0, b] M is a unit-speed parametrization of γ and γ(a) is the first conjugate to γ(0) for some 0 < a < b. This means that there is a non-trivial normal Jacobi field J along γ [0,a] that vanishes at t = 0 and t = a. Define a vector field V along γ by V (t) = J(t) for t [0, a], V (t) = 0 for t [a, b]. and X(t) = 0 for t [a + ɛ, b]. This is a proper, normal, piecewise smooth vector field along γ. Let W (t) be a smooth proper normal vector field along γ such that W (b) is equal to the jump V at t = a. Such a vector field is easily constructed in a local coordinates and extended to all γ by a bump function. Note that V = J (a) is not zero, because otherwise J would be a Jacobi field satisfying J(a) = J (a) = 0, and thus would be identically zero. 18

19 For small positive ɛ, let X ɛ = V + ɛw. Then I(X ɛ, X ɛ ) = I(V, V ) + 2ɛI(V, W ) + ɛ 2 I(W, W ). Since V satisfies the Jacobi equation on each subinterval [0, a] and [a, b], and V (a) = 0, by the Proposition above, a b I(V, V ) = < V + R(V, γ )γ, V > dt < V + R(V, γ )γ, V > dt 0 a < V, V (a) >= < V, V (a) >. Similarly, Thus I(V, W ) = < V, W (b) >= W (b) 2. I(X ɛ, X ɛ ) = 2ɛ W (b) 2 + ɛ 2 I(W, W ). If we choose ɛ small enough, this is strictly negative. This proves the theorem. In contrast to the above theorem, we prove Theorem. Let γ : [a, b] M be geodesic. It there does not exist a point conjugate to γ(a) along γ, then there exists ɛ > 0 with the property that for any piecewise smooth curve g : [a, b] M with g(a) = γ(a), g(b) = γ(b), d(g(t), γ(t)) < ɛ for all t [a, b], we have that L(g) L(c) with equality if and only if g is a reparametrization of c. Proof. We want to apply Corollary WLOG, we assume that a = 0, b = 1 and v := γ(0). Since there are no points conjugate to γ(a), the exponential map exp p is of maximal rank along any radical curve tv, 0 t 1. Thus, by inverse function theorem, for each t, exp p is diffeomorphism in a suitable neighborhood of tv. We cover {tv, 0 t 1} by finitely many such neighborhoods {Ω i, 1 i k} and let U i := exp p Ω i. Let us assume that tv Ω i for t i 1 t t i. If ɛ > 0 is sufficiently small, we have for any curve g : [0, 1] M satisfying the assumption d(g(t), γ(t)) < ɛ, g([t i 1, t i ]) U i. Let c(t) = (exp p Ωi ) 1 (g(t)) for t i 1 t t i, then c : [0, 1] T p M with exp p c = g, c(0) = 0, c(1) = v. Thus Corollary implies our statement. 19

20 7 Curvature and topology By means of the second variation of geodesics we can prove the following Theorem (Myers and Bonnet). Let (M, g) be a complete Riemannian manifold. If (i) (Myers) r(m) (n 1)k for some constant k > 0, or (ii) (Bonnet) K M k > 0, then the diameter of M satisfying d(m) π/ k, so that M is actually compact. Proof. Suppose the contrary. There there are points p, q M and (Hopf- Rinow theorem) a minimal unit-speed curve γ connecting p and q with length l > π/ k. Fix a normalized minimal geodesic γ : [0, l] M (with unitspeed) and let {e i } be an orthonormal basis of parallel fields along γ such that e 1 = γ = T. Let W i (t) = sin(πt/l)e i (t) be vector fields along γ. Then W i (t) = π cos(πt/l)e l i(t), W i (t) = π2 sin(πt/l)e l 2 i (t). Hence l I(W i, W i ) = < 2 T W i R(T, W i )T, W i > dt = 0 l from which it follows that n l I(W i, W i ) = i=2 0 0 (sin(πt/l)) 2 { π2 l 2 (sin(πt/l)) 2 {(n 1) π2 l 2 + R(T, e i)t, e i >}dt r(t, T )}dt. So if r(m) (n 1)k > 0 and l > π/ k, then the sum therefore at least one summand must be negative. But this is impossible since some I(W i, W i ) < 0 implies that the second variation of arclength is in some direction negative, so that γ cannot be minimal. This proves the theorem. Remark: (a) In case (ii) (i.e. the Bonnet s condition), we can just take W (t) = sin(πt/l)e(t), where E(t) is any parallel normal unit vector field along γ. (b) The sphere S n (r) := {x R n+1 x = r} of radius r has (sectional) curvature 1, hence Ricci curvature n 1 and diameter πr. The theorem above r 2 r 2 20

21 means that if If M has Ricci curvature not less than the one of S n (r), then the diameter of M is at most one of S n (r). Theorem (Cartan-Hadama). Let (M, g) be a complete Riemnannian manifold with nonpositive sectional curvature. Then M has no conjugate points. Proof. Let γ : [0, l] M be a normalized geodesic, and X be a vector field, normal to γ, then I(X, X) = l 0 { T X 2 X 2 K M (X T )}ds 0. Then by the corollary above, M has no conjugate points. This proves the theorem. 8 Some Comparison Theorems We ll prove, in this section, that an upper bound on sectional curvature produces a lower bound on Jacobi fields. In particular, we recall that we three model space S n, R n and H n of curvature 1, 0, -1. Let γ(t) be a geodesic with γ (t) = 1, w Tγ(0)M where M {S n, R n, H n }. Then the Jacobi field J(t) along γ with J(0) = 0, J (0) = w is given by (sin t)w, tw, (sinh t)w respectively, where W is the parallel vector field along γ with W (0) = w. So it is natural to compare with these spaces. Our starting point is the following very classical comparison theorem for ordinary differential equations. Theorem (Sturm Comparison Theorem). Suppose u and v are differentiable real-valued functions on [0, t], twice differentiable on (0, T ), and u > 0 on (0, T ). Suppose further that u and v satisfy u (t)+a(a)u(t) = 0, v (t)+a(t)v(t) 0, u(0) = v(0) = 0, u (0) = v (0) > 0 for some function a : [0, t] R. Then v(t) u(t) on [0, T ]. Proof. Consider the function f(t) = v(t)/u(t) defined on (0, T ). It follows from l Hopital s rule that lim t 0 f(t) = v (0)/u (0) = 1. Since f is differentiable on (0, T ), if we could show that f 0 there it would follow from 21

22 elementary calculus that f 1 and therefore v u on (0, T ), and by continuity also on [0, T ]. Differentiating ( ) d v = v u vu. dt u u 2 Thus to show that f 0 it would suffice to show that v u vu 0. Since v (0)u(0) v(0)u (0) = 0, we need only show this expression has no nonnegative derivative. Differentiating again and substituting the ODE for u, d dt (v u vu ) = v u + v u v u vu = v u + avu 0. This proves the theorem. Theorem (Jacobi Field Comparison Theorem). Suppose that (M, g) is a Riemannian manifold with all sectional curvature bounded above by a constant C. If γ is a unit speed geodesic in M, and J is any normal Jacobi field along γ such that J(0) = 0, then (1) If C = 0, J(t) t J (0) for t 0; (2) If C = 1 R 2, J(t) R(sin t R ) J (0) for 0 t πr; If C = 1 R 2, J(t) R(sinh t R ) J (0) for t 0; Proof. The function J(t) is smooth wherever J(t) 0. Using the Jacobi equation, we compute d 2 dt J = d < J, J > 2 dt < J, J > 1/2 = < J, J > < J, J > 1/2 + < J, J > < J, J > 1/2 < J, J > < J, J > 3/2 = < R(J, γ )γ, J > J + J 2 J < J, J > 2 J 3. Using the Schwarz inequality, < J, J > 2 J 2 J 2, so the sum of the last two terms above is nonnegative. Thus d 2 dt J R(J, γ, γ, J). 2 J 22

23 Since < J, γ >= 0 and γ = 1, R(J, γ, γ, J)/ J 2 is the sectional curvature of the plane spanned by J and γ. Therefore our assumption on the sectional curvature of M guarantees that R(J, γ, γ, J) C J 2, so d 2 J C J dt2 wherever J > 0. We wish to use the Sturm comparison theorem to compare J with the solution u to u + Cu = 0, where u(t) = t if C = 0, u(t) = R sin t if R C = 1/R 2, and u(t) = R sinh t if C = R 1/R2. To do so, we need to arrange d J /dt = 1 at t = 0, because u (0) = 1. Multiplying J by a positive constant, we may assume, WLOG, that J (0) = 1. Also J can be written near t = 0 as J(t) = tw (t) where W is a smooth vector field (the proof is omitted). Therefore d dt J J(t) J(0) t=0 = lim t 0 t t W (t) = lim = W (0) = J (0) = 1. t 0 t Now the Sturm comparison theorem implies that J u, provided J is nonzero (to ensure that it is smooth). The fact that d J /dt = 1 at t = 0 means that J > 0 on some interval (0, ɛ), and J cannot attain its first zero before u does without contradicting the estimate J u. Thus J u as long as u 0, which proves the theorem. 9 The Hessian and Laplace Comparison Theorem The Comparison theorem is one of the most important tools in the study of analysis on manifolds. Basically it is through the connections between the Jacobi field and the curvature of the manifold, as well as its curvature properties, to give more general properties of the manifold itself. On the other hand, from the Jacobi equation point view, it is also the application of the differential equations to the geometry. For example, the proof of the Bonnet theorem is the application of the Sturm-Liouville theory. 23

24 The purpose of the section is to study the curvature and harmonic functions, so we first derive the Hessian and Laplace Comparison Theorem, and then give some of its consequences. Let M be an n-dimensional complete Riemannian manifold with the metric ds 2 = g ij dx i dx j. The Laplace-Bertrami operator is defined as where G = det(g ij ), (g ij ) = (g ij ) 1. = 1 ( ) Gg ij, G x i x j There is a very natural function on M which is the distance function on M with a point fixed, i.e. fix a point O M, define, for x M, ρ(x) = dist(o, x). Obviously, ρ is not only continuous, but it also satisfies the Lipschitz condition. From the measure theory, it is differentiable everywhere. For O M, consider the exponential map exp o. From the Hopf-Rinow theorem, exp o is defined on T o (M), i.e. exp o : T o (M) M. For every X T o (M), let γ(t) = exp(tx), then γ : R M is a smooth geodesic from X, when t is small, γ is the unique minimal geodesic connection γ(t) and O, and d exp o tx : T tx (T o M) T expo (tx)m is a diffeomorphism. When t increases, there ll be two possibilities: (i) γ(t) is no longer the minimal geodesic connecting γ(t) and O, (ii) d exp o tx is no longer a diffeomorphism, in this case, we say γ(t) is a conjugate point to O. Let t 0 = sup{t γ is the unique minimal geodesic connecting γ(t) and O, i.e.on [0, t]}. Then t 0 R {+ }. If t 0 < +, we call γ(t 0 ) is a cut point along γ, relative to O. The set of all cut points along γ, relative to O is called a cut locus, and is denoted by cut(o). From the definition, if x Cut(O), then either (i) x is the first conjugate point to O along γ, or (ii) there are at least two minimal geodesics with the same arc-length connecting x and O. Obviously, for X T o M with X = 1, there is at most one cut point along exp o (tx) (t > 0), hence cut(o) is the image of the a closed set of 24

25 S n 1 under the map exp o, so its n-dimensional measure is 0. Further, let µ(x) = dist(o, γ(t 0 )), where X S n 1 T o M. Define E = {tx 0 t < µ(x), X S n 1 T o M}, then exp o : E exp o (E) is a diffeomorphism, hence it induces a normal coordinate frame. Obviously, it is the largest possible normal coordinate frame with o as the origin. Furthermore, M = exp o (E) cut(o). It is known that the set cut(o) has measure zero in M. From the definition, we can see that exp o (E) is a star domain with origin O, and ρ is smooth on exp o (E). As the boundary of exp o (E), there may have X 1 X 2 cut(o), but exp o (µ(x 1 )X 1 ) = exp o (µ(x 2 )X 2 ). ρ(x) is smooth on E, since the geodesic has arc-length as parameter, and ρ = 1, i.e. g ij ρ i ρ j direction. = 1, where ρ i is the covariant derivative of ρ along the i Also, recall that Ric : T p M T p M R defined by Ric(X, Y ) = < R(e i, X)Y, e i > where {e i } is an orthonormal basis for T p M. When e = e n, then Ric(e, e) = where K is the sectional curvature. n 1 i=1 K(e i, e), Let f C 2 (M), we now define the Hessian of f. For X, Y T x M, extend X, Y to vector fields X, Ỹ, we define H(f)(X, Y ) = ( XỸ ((f)(x) ( XỸ f)(x). Remark: H(f) = 2 f = df which is a (2,0)-tensor field. Also, (f) = tr( 2 f) = tr(h(f)). 25

26 We now calculate H(ρ) (which is also H(r)). We first recall the Gauss lemma(see the textbook of John Lee): is orthogonal to the geodesic sphere. r Hence we can take the normal (polar) coordinates as ds 2 = dr 2 + f(r) 2 dθ i dθ j. Gauss lemma implies that <, >= 0. So when we write, for any X r θ i T x M, X = h + g r i, then we have θ i <, X >= h = X(r), r So grad r = r. Fix p M, for any point x which is in the cut locus, let σ be the minimal geodesic connecting p and x with σ(0) = p, σ(r) = x. Take X T x M, < X, / r > (x) = 0. Since x is not a conjugate point, we can extend X to a Jacobi field X along σ with X(σ(0)) = 0, X(σ(r)) = X. Since every Jacobi field X is the variation field of some variation of σ, by the symmetry lemma, we have [ X, ] = 0 (0 t r). t Hence H(r)(X, X) = X Xr ( X Xr) = X X, r X X, r = X, X = X, X, r r [ where in the last equality, we used X, ] = 0 (0 t r). Hence, at x t M, we have H(r)(X, X) = = r 0 r 0 d X, dt dt t ( Dt X 2 + < X, Dt 2 X > ) dt. 26

27 Because X is a Jacobi field, Dt 2 X + R( X, t ) t = 0. Hence, we get the expression of the Hessian of the distance function r as H(r)(X, X) = ( r 0 D t X 2 R( X, t ) ) t, X > dt = I( X, X). Before we prove the theorem, we first recall the following lemma: Lemma (Fundamental Index Lemma) Let γ : [0, l] M be a unit-speed geodesic. Assume that there is no conjugate point on γ to p = γ(0), Let J = J(t) be a Jacobi field. Then for any piecewise smooth vector field X along γ with X(0) = J(0), X(l) = J(l), we have I(J, J) I(X, X) and the equality holds if and only if X = J. Proof. Since X J vanishes at the ends, we have I(X J, X J) 0 because γ has no conjugate points to p = γ(0). On the other hand, since J is a vector field, I(J, J) =< D t J, J > l 0 and I(X, J) =< D t J, X > l 0 =< D t J, J > l 0= I(J, J) since X(0) = J(0), X(l) = J(l). Therefore I(X J, X J) = I(X, X) 2I(X, J) + I(J, J) = I(X, X) I(J, J). Thus I(X, X) I(J, J). Hessian Comparison Theorem Let M 1, M 2 be two n-dimensional Riemannian manifolds, γ i : [0, a] M i (i = 1, 2) be two unit-speed geodesics. Denote by ρ i the distance function on M i with initial points γ i (0). Assume that γ i (a) be within the cut locus of γ i (0). Assume that for 0 t a, ) ( ) K 1 (X, K 2 Y,, r 1 r 2 27

28 for X T γ1 (t)m 1, Y T γ2 (t)m 2 which are the unit-vectors orthogonal to Then we have H(ρ 1 )(X 1, X 1 ) H(ρ 2 )(X 2, X 2 ) where X i are the unit vectors in T γi (t)m i and < X i, r i > (γ i (a)) = 0. Proof. Take parallel vector fields E (i) 1,, E n (i) along γ i with E (i) n = r i, i = 1, 2. r i. Then H(ρ i )(X i, X i ) = ( r D (i) t 0 X 2 R (i) ( X i, t ) t, X i > ) γi dt, where X i are the Jacobi field along γ i with X i (0) = 0, Xi (a) = X i. Because < X i, r i >= 0, X2 is orthogonal to E n (2) at every point on γ i. Hence X 2 = n 1 j=1 λ j (t)e (2) j, λ j (0) = 0. Since E (1) j (a), 1 j n 1 can be arbitrary chosen (and then do a parallel translation), we can assume that X 1 = X 1 (a) = n 1 j=1 λ j (a)e (1) j (a). Along the geodesics γ i we define a vector field Z by Z(t) := n 1 Then Z(0) = 0, Z(a) = X 1 = X 1 (a), n 1 j=1 λ j (t)e (1) j (t). Z 2 = (λ j (t)) 2 = X 2 2, j=1 28

29 and D (1) n 1 t Z 2 = λ j(t)e (1) j (t) 2 j=1 n 1 = λ j(t)e (2) j (t) 2 = Dt (2) X2 2. j=1 From the fundamental theorem of index, Hence I (1) ( X, X) I (1) (Z, Z). H(ρ 1 )(X 1, X 1 ) = I (1) X, X) I (1) (Z, Z) ( r = D (1) t Z 2 < R (1) (Z, t ) ) t, Z > γ1 dt This proves the theorem. 0 ( r 0 D (2) t X 2 2 < R (2) ( X 2, t ) t, X 2 > = I (2) ( X 2, X 2 ) = H(ρ 2 )(X 2, X 2 ). As a direct consequence of the above theorem, we have ) γ2 dt Corollary(Laplace Comparison Theorem) Let M be an n-dimensional complete Riemannian manifold. Assume that Ric(M) (n 1)k 2 (k 0). Let N be an n-dimensional simply connected manifold with constant curvature ( k 2 ) (it is called the space-form). Denote by ρ M and ρ N the distance function with relative fixed points. If x M, y N with ρ M (x) = ρ N (y). Then when x is a differentiable point of ρ M, we have ρ M (x) ρ N (y). In order to use the above theorem, we compute ρ in the case of constant curvature ( k 2 ). From above, ( ) H(ρ) = ρ 0 D t X2 2 < R( X, t ) t, X > dt ( ) 29

30 where X is the Jacobi field along the minimal geodesic γ with X(0) = 0 and tildex(γ(ρ)) = X. Therefore, the computation of ρ is reduced to the computation of the Jacobi field. In the case of space of constant curvature k 2, let p = γ(0), q = γ(ρ), X T q M which is perpendicular to γ (ρ). We then obtain an parallel vector field X(t), 0 t ρ by parallel translation. Then any Jacobi field J(t) with Y (0) = 0, Y (ρ) = X has the form Y (t) = f(t)x(t) where f(t) satisfies f k 2 f(t) = 0, f(0) = 0, f(ρ) = 1. Hence f(t) = 1 sinh kt, where R = 1. Substituting it into (*) to get ρ = (n R k 1)k coth(kρ). Corollary On (M, g) with dim M = n, assume that Ric(M) (n 1)k 2, then at every point which is differentiable, we have ρ n 1 (1 + kρ). ρ it Proof. By Laplace comparison theorem, ρ (n 1)k coth(kρ) = n 1 kρ coth(kρ). ρ Since kρ coth(kρ) (1 + kρ), we have ρ n 1 (1 + kρ). ρ Note that the above estimate holds provided the points are not in the cut-locus. However, in the study of the global properties of M, since M = exp o (E) cut(o), and since exp o (E) is a star-shaped domain, the topological properties are good near cut(o). Thus we still hope that ρ behaves well near cut(o). The Lipschitz property of ρ leads us to consider the ρ is the sense of distribution. We do have the following theorem: TheoremAssume M is a complete Riemannian manifold with negative Riccicurvature. For every p M, denote by ρ the distance function of p (with a fixed reference point). Then ρ n 1 ρ 30

31 in the sense of distribution. Proof. Notice that M = Ω cut(o) where Ω is a star-shaped domain. The Lipschitz function ρ is differentiable on Ω, so, by the theorem above, ρ n 1 ρ. In order to prove the theorem, we take φ C0 (M), φ 0. mes(e) = 0, ρ φ = ρ φ. M Ω Because Take a family of star-shaped domain Ω ɛ with Ω ɛ Ω, lim Ω ɛ = Ω, and Ω ɛ is obtained by shrinking Ω in the ρ-direction. Note that Stokes theorem holds for Lipschitz function, and notice that φ is with compact support, M ρ φ = φ ρ = lim( 1) ρ φ, M ɛ 0 Ω ɛ where in the last step we used the fact that ρ = 1 holds almost everywhere and φ is bounded. From Green s formula, ρ φ = ρ φ φ ρ Ω ɛ Ω ɛ Ω ɛ n, Since φ 0 and Ω ɛ is obtained by shrinking Ω in the ρ-direction. we have > 0, thus ρ n Therefore, M This finishes the proof. n 1 ρ φ ρ φ φ. Ω ɛ Ω ɛ Ω ɛ ρ n 1 n 1 ρ φ lim φ = φ = ɛ 0 Ω ɛ ρ Ω ρ M n 1 ρ Remark: There is an alternative proof of the above result by directly computing Ric(, ) and using the maximal principle. Here is the details of r r φ. 31

32 computation: By Gauss lemma, we can assume the Riemannian metric to be ds 2 = dr 2 + h ij (r, θ)dθ i dθ j where (θ 2,..., θ n ) is the local coordinate system of S n 1. We write ds 2 = ω ω 2 n where ω 1 = dr. When we restrict ω 2,..., ω n on {r =cosntant}, we have dω i = η ij ω j, i, j 2 where η ij = η ji are the connection forms for the metric h ij (r, θ)dθ i dθ j. Define ω i1 = ω i r = a ij ω j ω ij = η ij, i, j 2. Then it is easy to verify that dω i = ω ij ω j (note, the Uhlenbecks trick told us that by an orthogonal change of the co-frame, we can assume that (a ij ) is symmetric). To calculate the curvature, we use, by the definition, dω ij + ω ik ω kj = 1 2 R ijklω k ω l. Taking j = 1, dω i1 + ω ik ω k1 = 1 2 R i1pqω p ω q. If we only count the terms ω 1 ω q we have Thus Ric On the other hand, dr ω i1 r = R i11lω 1 ω l. ( r, ) = R 11 = R 1i1i = ( n ) a ii r r i=2 r (ω 1 ω n ) = a ii (ω 1 ω n ). n a 2 ij. i,j=2 32

33 Therefore, Let f := log det h ij, then Hence Thus, 2 f r n 1 In the special case when det h ij = a ii det h ij. r f r = a ii. ( ) 2 f ( n ) a ii + r r i=2 2 f r n 1 n i,j=2 a 2 ij = Ric ( ) 2 ( f Ric r r, ). r ( r, ). r ds 2 = dr 2 + h 2 (r, θ)η ij (r, θ)dθ i dθ j where η ij (r, θ)dθ i dθ j is the standard metric of S n 1, then for some function g. Thus a ij = gδ ij a 2 ij = (n 1)g 2 = 1 n 1 ( a ii ) 2. On the other hand, the Laplacian can be written as Under the assumption that = 1 ( g ij g ). g x i x j ds 2 = dr 2 + h ij (r, θ)dθ i dθ j we have r = r log det h ij = f r. 33

34 Hence, we have r r + 1 ( n 1 ( r)2 Ric r, ). r The above formula also can be derived from the Ricci identity (Bochner s formula, see below): Taking f = r on M, 1 2 r 2 = 2 r 2 + r + Ric( r, r). r Notice that, r 2 = 1 so r 2 = 0, and 2 f 2 i fii 2 1 ( ) 2 1 fii = n 1 n 1 ( f)2, which gives r r + 1 ( n 1 ( r)2 Ric r, ). r We can use the above inequality to prove the comparision theorem: The corresponding function on N (the space form of constant curvature k 2 ) satisfies 2 f 0 r n 1 The initial conditions on f and f 0 are Since and using f r n 1, r Ric 2 f r n 1 by the maximum principle, ( ) 2 f0 (n 1)k 2. r f 0 r n 1. r ( r, ) (n 1)K 2 r ( ) 2 ( f Ric r r, ), r f r f 0 r. 34

35 This proves the comparison theorem at the smooth points of f. Another proof: (from Li and Wang: J. Differential Geometry, 69(2005), 43-74): Define, for w T p M, and V T p M, l KM(w, l V ) = < R(w, e i )w, e i >, i=1 where {e 1,..., e l } being an o.n. basis for V. To set up our model, we let MK l+1 to be the (l + 1)-dimensional, simply connected, space form of constant sectional curvature K. Theorem (Li and Wang). Let M be a complete Riemannian manifold of dimension n. Assume that l-sectional curvatures of M satisfy KM l lk (when l = 1, this is equivalent to K M K, when l = n 1, this is equivalent to Ric (n 1)K). Then, within the cut locus of a fixed point O M and for any V T x M perpendicular to r(x), l l D 2 (r)(e i, e i ) i=1 i=1 D 2 ( r)(ē i, ē i ). where {e 1,..., e l } being an o.n. basis for V and {ē 1,..., ē l } being an o.n. basis of T p Mk l+1 which are orthogonal to r. Proof. Take x exp o (E) {O}, let γ be the minimal normal geodesic joining O to x. At x, we choose an o.n. frame {e 1,..., e n }, such that e 1 = r. By parallel translation, we obtain the frame {e i } along γ with e 1 = r. By taking the covariant derivative of the equation r = 1, we obtain n n 0 = ( r 2 ) αα = 2 r iα r iα + 2 r i r iαα i=1 i=1 for each 2 α n. Since γ is geodesic and each e i is parallel along γ, each term on the right-hand side of above equation can be interpreted as covariant derivatives. The commutation formula for covariant derivative then implies n n n r i r iαα = r i r αα + R iαjα r i r j. i=1 i=1 i,j=1 35

36 Substituting it into the above equation and using the fact that r = 1 = r 1, we obtain 0 2rαα (r αα) + 2K M (e 1, e α ). r Suppose V T x M is spanned by {e 2,..., e l+1 }, then summing over α = 2,..., l + 1, Using l+1 0 rαα 2 + α=2 r l+1 ( l+1 α=2 r 2 αα 1 l α=2 r αα ) + K l M(e 1, V ). ( l+1 ) 2 r αα α=2 and setting f(t) = l+1 α=2 r αα (γ(t)), then we have 1 l f 2 (t) + f (t) + lk 0. Note that since a smooth Riemannian metric is locally Euclidean, lim tf(t) = l. t 0 We will now consider the three separate cases when K = 0, K > 0 and K < 0. Case 1: When K = 0, then we have f (t) + 1 l f 2 (t) 0. This implies that f (t) 0 and f is decreasing. Let (0, T ) be the largest interval such that f(t) > 0, then we have ( ) 1 = f f f 1 2 l and f(t) l on (0, T ). Since f(t) 0 for t T, we can conclude that t f(t) l on (0, ρ(θ)) (the whole interval where γ is defined. t Case 2: When K > 0. Then lf (t) f 2 (t) + l 2 K 1. 36

37 This implies that Integrating from 0 to t we have ( tan 1 ( ) d f dt tan 1 l K 1/2. K f l K ) π 2 K1/2 t, implying that f(t) l K cot( Kt). Case 3. When K < 0, let T be the first time such that f 2 (t) + l 2 K = 0. Then, on (0, T ), we have f 2 (t) + l 2 K > 0 and This implies that lf (t) f 2 (t) + l 2 K 1. ( ) d f dt coth 1 l K 1/2 K and f(t) l K coth( K t) on (0, T ). For t T, we claim that f(t) l K. Indeed, it f(t 1 ) > l K for t 1 > T, then there exists t 2 (T, t 1 ) such that f (t 2 ) 0 and f(t 2 ) > l K. In this case which is a contradiction. Thus f (t 2 ) + 1 l f 2 (t 2 ) + lk > 0, f(t) l K for T t < ρ(θ), and we conclude that f(t) l K coth( K t) 37

38 for 0 < t < ρ(θ). The Theorem follows by observing that r 11 = 0 and the the above inequalities becomes equalities on a simply connected space from with constant sectional curvature. Q.E.D. Observe that the standard Laplacian comparison theorem and Hessian comparison theorem follow from theorem 1.2 by setting l = n 1 and l = 1, respectively. In fact, let ds 2 = dr 2 + h αβ (r, θ)dθ α dθ β. If we denote J(r, θ) = det(h αβ ) to be the area element of the geodesic sphere B O (r), then Thus M = 2 r 2 + r (log J) r + B O (r). M r = (log J) r on exp O (E) {O}. 10 The Gradient Estimate We first introduce the Ricci identity. Let {e 1,..., e m } be locally defined o.n. frame fields of the tangent bundle. Let us denote the dual frame fields by {ω 1,..., ω m }. The connection 1-forms ω ij are given by dω i = ω ij ω j. Cartan s second structural equation yield the curvature tensor ω ij + ω ji = 0. Let f be a smooth function on M. Define the derivative of f using the following formula f i ω i = df. Using the same idea, we define f ij ω j = df i + f j ω ji. 38

39 Using 0 = d 2 f = df i ω i + f i dω i = df i ω i + f i ω ij ω j = (df i + f j ω ji ) ω i = f ij ω j ω i. This implies that f ij = f ji = 0 for all i, j. The symmetric 2-tensro given by f ij ω j ω i is called the Hessian. The 3rd order co-variant derivative of f is defined as A carefully computation gives Thus we have In particular, we have Remark: we have f = f ii. f ijk ω k = df ij + f kj ω ki + f ik ω kj. f ijk ω k ω j = 1 2 f sr sikj ω k ω j. f ijk f ikj = f s R sijk. f iik f kii = f s Ric sk. Bochner Formula Assume (M, g) is complete, f C 3 (M). p M, let {x i } be a normal coordinate system, then For every f 2 (p) = 2 ij f ij R ij f i f j + 2 f i ( f) i, where f i = f i and R ij is the Ricci curvature, or we can state it as: 1 2 f 2 = 2 f 2 + f + Ric( f, f) r Proof. Because f 2 = f 2 i, at the point p, ( f 2 ) = j ( f 2 ) jj = j ( i f 2 i ) jj = 2 j ( i f i f ij ) = 2 f 2 ij + 2 i,j f i f ijj. 39