LECTURE 18: IMMEDIATE APPLICATIONS OF JACOBI FIELD TO CURVATURE Last time we studied Jacobi fields along a geodesic γ, which control all possible geodesic variations of γ. Today we give some immediate applications. 1. Geometric interpretations of various curvatures ¶ Taylor’s expansion of metric tensor in Riemannian normal coordinates. Recall that if V, W are Jacobi fields along a geodesic γ with V (0) = 0, ∇γ˙ (0)V = Xp ∈ TpM and W(0) = 0, ∇γ˙ (0)V = Yp ∈ TpM, then the function f(t) = hV (t), W(t)i has the Taylor’s expansion f(t) = hXp, Ypit 2 − 1 3 Rm( ˙γ(0), Xp, γ˙(0), Yp)t 4 + O(t 5 ). For the first application, we calculate the next term in the Taylor’s expansion of any Riemannian metric tensor in any Riemannian normal coordinate system. Recall that with any Riemannian normal coordinate system centered at p, gij (p) = δij and ∂kgij (p) = 0. We now prove that the next term encodes the curvature information: Theorem 1.1. With respect to Riemannian normal coordinates near p, the functions gij ’s admit the following Taylor expansion at x = 0, (1) gij (x) = δij − 1 3 Rikjl(p)x kx l + O(|x 3 |). Proof. Let (U; x 1 , · · · , xm) be a Riemannian normal coordinate system near p. Fix x i ’s and let γ be the geodesic starting at p in the direction Xp = x i∂i , γ(t) = (tx1 , · · · , txm), 0 ≤ t ≤ ε. For each 1 ≤ i ≤ m, consider a geodesic variation fi(t, s) = (tx1 , · · · , t(x i + s), · · · , txm). Its variation field Vi = t∂i is thus a Jacobi field along γ, which satisfies Vi(0) = 0, ∇γ˙ (0)Vi = ∂i . So if we let h(t) = t 2 gij (tx1 , · · · , txm) = hVi(t), Vj (t)i, 1
2 LECTURE 18: IMMEDIATE APPLICATIONS OF JACOBI FIELD TO CURVATURE then gij (tx1 , · · · , txm) = 1 t 2 hVi(t), Vj (t)i = 1 t 2 δij t 2 − 1 3 Rm(Xp, ∂i , Xp, ∂j )t 4 + O(t 5 ) = δij − 1 3 Rm(∂i , Xp, ∂j , Xp)t 2 + O(t 3 ) = δij − 1 3 Rm(∂i , ∂k, ∂j , ∂l)(txk )(txl ) + O(t 3 ). This proves the theorem. Remark. One can continue to calculate ∇ (k) γ˙ (0)Vi ’s and get a full expansion of gij in Riemannian normal coordinates. For example, the next two terms are 1 6 Riklj;rx kx lx r + ( 1 20 Riklj;rs + 2 45 R m kil Rrjsm)x kx lx rx s . Taking derivative of (1), we get ∂rgij = − 1 3 Rirjlx l − 1 3 Rikjrx k + O(|x| 2 ). Taking derivative again and evaluate at p, we get ∂s∂rgij (0) = − 1 3 Rirjs(p) − 1 3 Risjr(p). As a consequence, we get Riemann’s original definition of the curvature tensor: Corollary 1.2. With respect to Riemannian normal coordinates, one has Rijkl(p) = 1 2 (∂i∂lgjk + ∂j∂kgil − ∂i∂kgjl − ∂j∂lgik)(0). Proof. The right hand side equals 1 6 (Rjlik + Rjilk + Rikjl + Rijkl − Rjkil − Rjikl − Riljk − Rijlk)(p), which equals Rijkl(p) by using symmetries of the Riemann curvature tensor. ¶ Geometric meaning of sectional curvature. Now we are ready to give geometric interpretations of curvatures. We start with sectional curvature: Theorem 1.3. Let Πp ⊂ TpM be a 2-dimensional plane. Denote by C 0 r the circle of radius r in Πp centered at p, and Cr = expp (C 0 r ). Let Lr be the length of Cr. Then limr→0 2πr − Lr r 3 = π 3 K(Πp).
LECTURE 18: IMMEDIATE APPLICATIONS OF JACOBI FIELD TO CURVATURE 3 Proof. Take an orthonormal basis {e1, · · · , em} of TpM so that Πp is spanned by e1, e2, and consider the normal coordinate system with respect to {ei}. Then for r small, the circle Cr has equation Cr : x 1 (t) = r cost, x2 (t) = r sin t, xk (t) = 0 (k ≥ 3). It follows that |C˙ r(t)| 2 =gij (Cr(t)) ˙x i (t) ˙x j (t) = 1− 1 3 R1212x 2x 2 x˙ 1x˙ 1+ 1− 1 3 R2121x 1x 1 x˙ 2x˙ 2−2 1 3 R1221x 1x 2x˙ 1x˙ 2+O(r 5 ) =r 2 − r 4 3 K(Πp) + O(r 5 ). So Lr = Length(Cr) = Z 2π 0 |C˙ r|dt = r Z 2π 0 r 1 − r 2 3 K(Πp) + O(r 3 )dt = 2πr − π 3 K(Πp)r 3 + O(r 4 ). This implies the theorem. So the sectional curvature measures the deviation of the length of small geodesic circles centered at p to the standard circles of the same radius in Euclidean plane. ¶ Geometric meaning of Ricci curvature. With the Taylor’s expansion of gij at hand, it is easy to get the Taylor’s expansion of det(gij ): According to Theorem 1.1 we get (gij ) = I + − 1 3 Rikjl(p)x kx l + O(|x 3 |) which implies 1 log(gij ) = − 1 3 Rikjl(p)x kx l + O(|x 3 |) and thus det(gij ) = det(e log(gij ) ) = e tr(log(gij )) = e − 1 3Rckl(p)x kx l+O(|x| 3 ) = 1 − 1 3 Rckl(p)x kx l + O(|x| 3 ). As an immediate consequence, we get the Taylor’s expansion for the volume element: Corollary 1.4. In Riemannian normal coordinates centered at p, one has q det(gij ) = 1 − 1 6 Rckl(p)x kx l + O(|x| 3 ). 1Here we use log(I + A) = A + O(|A| 2 ) and e A = 1 + A + O(|A| 2 ) for matrix A.
4 LECTURE 18: IMMEDIATE APPLICATIONS OF JACOBI FIELD TO CURVATURE In particular, we can prove that the Ricci curvature measures the change of the volume element in the given direction: Corollary 1.5. Let up ∈ SpM be a unit tangent vector at p, and let γ(t) be the geodesic starting at p with γ˙(0) = up. Then q det(gij (γ(t))) = 1 − Ric(up) 6 t 2 + O(t 3 ). Proof. Take an orthonormal basis of TpM with e1 = up. With respect to the associated Riemannian normal coordinates, the geodesic γ(t)= expp (tup) is given by γ : x 1 (t) = t, x2 = · · · = x m = 0. It follows q det(gij (γ(t))) = 1 − 1 6 Rc11(p)t 2 + O(t 3 ) = 1 − 1 6 Ric(up)t 2 + O(t 3 ). ¶ Geometric meaning of scalar curvature. Finally we study the scalar curvature S. We have Proposition 1.6. For r small enough, Vol(B(p, r)) = ωmr m 1 − S(p) 6(m + 2)r 2 + O(|r 3 |) , where ωm is the volume of the unit ball in R m. Proof. By definition Vol(B(p, r))) = Z Br(0) q det(gij )dx1 · · · dxm = Z Br(0) 1 − 1 6 Rckl(p)x kx l dx1 · · · dxm + O(r 3 ) = ωmr m − Rckl(p) 6 Z Br(0) x kx l dx1 · · · dxm + O(r 3 ). An elementary computation shows Z Br(0) x kx l dx1 · · · dxm = ωm m + 2 r m+2δ kl and the conclusion follows. So the scalar curvature measures the deviation of the volume of a geodesic ball to the standard Euclidean ball with the same radius. Corollary 1.7. The surface area of geodesic sphere S(p, r) is Area(S(p, r)) = mωmr m−1 − S(p) 6 ωmr m+1 + O(r m+2).
LECTURE 18: IMMEDIATE APPLICATIONS OF JACOBI FIELD TO CURVATURE 5 2. Cartan’s local isometry theorem ¶ Cartan’s local isometry theorem. In view of Theorem 1.1, one may anticipate that “curvature determines the Riemannian metric”. Although this is not true in the most general sense, there are many theorems supporting this philosophy. In what follows we prove a theorem of Cartan in this direction. More precisely, let (M, g) and (M, f g˜) be two Riemannian manifolds, p ∈ M and ˜p ∈ Mf. Let B(p, r) and B(˜p, r) be normal neighborhoods of p and ˜p respectively. Given a smooth map φ : B(p, r) → B(˜p, r), one may ask: under what condition will φ be an isometry? Of course if φ is an isometry, then • L = dφp : (TpM, gp) → (Tp˜M, f g˜p˜) is a linear isometry, • “the curvature tensor at corresponding points are the same”. Cartan’s theorem claims that the converse is true. We need more explanation for the second condition above. How to compare the curvature tensor at q and ϕ(q)? We need to identify the tangent spaces TqM and Tφ(q)Mf first. How? We already have the map L which identifies TpM with Tp˜Mf. For q 6= p we simply parallel transport vectors in TqM and in Tφ(q)Mf along geodesics to get vectors in TpM and in Tφ(p)Mf respectively, and then apply the map L. Now we start to state Cartan’s theorem. Suppose L : (TpM, gp) → (Tp˜M, f g˜p˜) is a linear isometry. Then one may define a map φ = expp˜ ◦L ◦ (expp ) −1 : B(p, r) → B(˜p, r). For any q ∈ B(p, r), we let γ = γq : [0, 1] → M be the unique geodesic in B(p, r) with γ(0) = p, γ(1) = q, and let ˜γ = φ ◦ γ. Note that ˜γ is the unique geodesic in B(˜p, r) ⊂ Mf with ˜γ(0) = ˜p = φ(p), ˜γ(1) = ˜q = φ(q) and γ˜˙(0) = L( ˙γ(0)). Define Lq = P γ˜ ◦ L ◦ (P γ ) −1 : TqM → Tφ(q)M, f Theorem 2.1 (Cartan’s local isometry theorem). If for any q ∈ B(p, r) and any u, v, w ∈ TqM, one has Lq(R(u, v)w) = Re(Lq(u), Lq(v))Lq(w), then φ is an isometry, and dφq = Lq for all q ∈ B(p, r). Note that for constant curvature spaces, the condition holds trivially. So we get Corollary 2.2. If both (M, g) and (M, f g˜) has constant sectional curvature k, then for any p ∈ M and p˜ ∈ Mf, one can find a neighborhood U 3 p and Ue 3 p˜ so that (U, g) and (U, e g˜) are isometric. which implies Riemann’s theorem for constant sectional curvature spaces, namely, Theorem 2.1 in Lecture 10
6 LECTURE 18: IMMEDIATE APPLICATIONS OF JACOBI FIELD TO CURVATURE ¶ Proof of Cartan’s local isometry theorem. We first prove |dφq(v)| = |v| for any v ∈ TqM. By using polarization this implies that dφq preserves inner products. Since φ is already a diffeomorphism, we conclude that φ is an isometry. The idea to prove |dφq(v)| = |v| is: realize both v and dφq(v) as Jacobi field vector at endpoints, and compare the length of two Jacobi fields at each point. We first construct a Jacobi field V along γ with V (0) = 0, V (1) = v. By Corollary 2.6 in Lecture 17, if V is such a Jacobi field, then V (t) = (d expp )tγ˙ (0)(t∇γ˙ (0)V ), which implies v = (d expp )γ˙ (0)(∇γ˙ (0)V ). As a result, V is the unique Jacobi field with V (0) = 0 and (∇γ˙ (0)V = (d expp ) −1 tγ˙ (0)(v). Next we construct the Jacobi field Ve along ˜γ with Ve(0) = 0 and Ve(1) = dφq(v). In fact, we may simply take Ve to be the Jacobi field along ˜γ with Ve(0) = 0 and ∇e γ˜˙ (0)V = L(∇γ˙ (0)V ). Since expp ( ˙γ(0)) = γ(1) = q, it follows dφq(v) = (d expp˜ )L( ˙γ(0)) ◦ L ◦ d exp−1 p )q(v) = (d expp˜ )L( ˙γ(0)) ◦ L ◦ (d expp ) −1 γ˙ (0)(v) = (d expp˜ )γ˜˙ (0) L(∇γ˙ (0)V ) = Ve(1), where the last equality follows from Corollary 2.6 in Lecture 17. To prove |V (1)| = |Ve(1)| we apply a standard trick: Let e1(t) = ˙γ(t), e2(t), · · · , em(t) be an orthonormal frame that is parallel along γ. Let V (t) = V i (t)ei(t), then |V (1)| 2 = X(V i (1))2 . Moreover, V i (t) is the solution to the Jacobi equation V¨ i (t) − hR( ˙γ, ek) ˙γ, eiiV k = 0, with initial conditions V i (0) = 0 and ∇γ˙ (0)V = V˙ i (0)ei(0). Similarly we let ˜e1(t) = γ˜˙(t), e˜2(t), · · · , e˜m(t) be the parallel orthonormal frame along ˜γ with ˜ei(0) = L(ei(0)). Then Lγ(t)(ej (t)) = ˜ej (t) for all j, and thus hRe(γ, ˜˙ e˜k)γ, ˜˙ e˜ii = hR( ˙γ, ek) ˙γ, eii. As a result, if Ve = Vei e˜j (t), then Vei ’s satisfy exactly the same equations and the same initial conditions as V i ’s, and thus Vei = V i for all 1 ≤ i ≤ m. It follows |Ve(1)| 2 = X(Vei (1))2 = X(V i (1))2 = |V (1)| 2 . Finally, the fact Ve = V i e˜i that we just proved also implies (dφq)(v) = Ve(1) = V i (1)˜ei(1) = V i (1)Lq(ei(1)) = Lq(V (1)) = Lq(v) and thus the proof is completed