《黎曼几何》课程教学资源（英文讲义）Lec17 JACOBI FIELDS

LECTURE17:JACOBIFIELDSAs we have seen, in the second variational formula the curvature term appears.As a result, the formula will play a crucial role in studying the relation betweencurvature and topology of Riemannian manifolds. Usually the first step will be: startwith a geodesic, and take a special variation (e.g. a geodesic variation, sometimeswith one endpointfixed).Thus thevariationfield of ageodesicvariation willbevery importantfortheremaining of this course.1.DEFINITIONOF THEJACOBIFIELDI The Jacobi field.Let be a geodesic in (M,g). Suppose f : [a,b] × (-e,e) → M is a geodesicvariation of , i.e. each curve = f(,s)is a geodesic in M. Then for any s,Va/at ft =Va/ot=0As a consequence,fTaking s = O, we see that the variation field V of any geodesic variation satisfies(1)V,V,V = R(%, V).Definition 1.1. Let X be a smooth vector field X along a geodesic . We call Xa Jacobi field along if the equation (1) holds.Remark. Let be a geodesic. There are two trivial Jacobi fields along :. Obviously X = is a Jacobi field. It is the variation field of f(t, s) = (t+s).·X=tisa Jacobifield sinceV,V(t) = V(+tV) =0and R(,t) = 0. It is the variation field of f(t, s) = (t + st),.ButX=t2is NOT a Jacobi field sinceV,V(t) = V(2t) = 2 0.It is not amazing that t? is no longer a Jacobi field along :Lemma 1.2. Let X be a Jacobi field along , then f(t) =(X, ) is a linear function

LECTURE 17: JACOBI FIELDS As we have seen, in the second variational formula the curvature term appears. As a result, the formula will play a crucial role in studying the relation between curvature and topology of Riemannian manifolds. Usually the first step will be: start with a geodesic, and take a special variation (e.g. a geodesic variation, sometimes with one endpoint fixed). Thus the variation field of a geodesic variation will be very important for the remaining of this course. 1. Definition of the Jacobi field ¶ The Jacobi field. Let γ be a geodesic in (M, g). Suppose f : [a, b] × (−ε, ε) → M is a ✿✿✿✿✿✿✿✿ geodesic ✿✿✿✿✿✿✿✿✿ variation of γ, i.e. each curve γs = f(·, s) is a geodesic in M. Then for any s, ∇e ∂/∂tft = ∇e ∂/∂tγ˙ s = 0. As a consequence, ∇e ∂/∂t∇e ∂/∂tfs = ∇e ∂/∂t∇e ∂/∂sft = ∇e ∂/∂t∇e ∂/∂sft − ∇e ∂/∂s∇e ∂/∂tft = Re( ∂ ∂t, ∂ ∂s)ft . Taking s = 0, we see that the variation field V of any geodesic variation satisfies (1) ∇γ˙ ∇γ˙ V = R( ˙γ, V ) ˙γ. Definition 1.1. Let X be a smooth vector field X along a geodesic γ. We call X a Jacobi field along γ if the equation (1) holds. Remark. Let γ be a geodesic. There are two trivial Jacobi fields along γ: • Obviously X = ˙γ is a Jacobi field. It is the variation field of f(t, s) = γ(t+s). • X = tγ˙ is a Jacobi field since ∇γ˙ ∇γ˙(tγ˙) = ∇γ˙( ˙γ + t∇γ˙ γ˙) = 0 and R( ˙γ, tγ˙) ˙γ = 0. It is the variation field of f(t, s) = γ(t + st). • But X = t 2γ˙ is NOT a Jacobi field since ∇γ˙ ∇γ˙(t 2 γ˙) = ∇γ˙(2tγ˙) = 2 ˙γ ̸= 0. It is not amazing that t 2γ˙ is no longer a Jacobi field along γ: Lemma 1.2. Let X be a Jacobi field along γ, then f(t) = ⟨X, γ˙⟩ is a linear function. 1

2LECTURE17:JACOBIFIELDSProof. According to the Jacobi field equation,0f"(t) =(X,) = (V,V4X,) = (R(%, X)%,) = 0.口It follows that(X,)is alinearfunction along.I Existence and uniqueness of Jacobi field.So the variation field of any geodesic variation is a Jacobi fields. As a result,the second variation formula for a geodesic variation is very simple. We will showthat conversely, any Jacobi field on can be realized as the variation field of somegeodesic variation of .Before we prove it, we need some basic properties of Jacobifields.Let's take a closer look of the equation for Jacobi fields. Since it is a differentialequation,it is enough to study the equation in a coordinate chart.Although onemay work on a general frame, to simply the computation one may use a specialframe that are parallel along [so that the covariant derivatives of the frame are as simple aspossible]l. So we start with an orthonormal basis [ei, .: ,em) of T,M, with ei = (a),where p = (a). Lete,(t) := the parallel transport of e along ,1≤i<m.According to Proposition 2.1 in Lecture 6,(e;(t),e;(t))(t) = (ei,ej)(a) = ijIn other words, we get an orthonormal frame [ei(t), ., em(t) along with ei(t) (t), and this frame is parallel along , i.e.V(t)es(t) = 0, 1≤k ≤m.Let X be a Jacobifield along , then with respect to this orthonormal frame we canwrite X = X"(t)e;(t), and we getandV,V,X=X(t)e;(t).V,X = Xi(t)ei(t)ItfollowsthattheJacobifield equation becomesX'(t)e;(t) - X'(t)Rn'e;(t) = 0.So we arrived at a system of second order homogeneous ODEs,Xi(t)-Xi(t)Rijt =0, 1≤i≤m,Using the basic theory for second order homogeneous ODEs, we getTheorem 1.3. Let : [a, b] → M be a geodesic, then for any X(a), Yy(a) E T(a)M,there erists a unique Jacobi field X alongso thatX(a) = X(@)andV(a)X = Y(a):Moreover, the set of Jacobi fields along is a linear space of dimension 2m (whichis canonically isomorphic to T(a)M T(a)M)

2 LECTURE 17: JACOBI FIELDS Proof. According to the Jacobi field equation, f ′′(t) = d 2 dt2 ⟨X, γ˙⟩ = ⟨∇γ˙ ∇γ˙ X, γ˙⟩ = ⟨R( ˙γ, X) ˙γ, γ˙⟩ = 0. It follows that ⟨X, γ˙⟩ is a linear function along γ. □ ¶ Existence and uniqueness of Jacobi field. So the variation field of any geodesic variation is a Jacobi fields. As a result, the second variation formula for a geodesic variation is very simple. We will show that conversely, any Jacobi field on γ can be realized as the variation field of some geodesic variation of γ. Before we prove it, we need some basic properties of Jacobi fields. Let’s take a closer look of the equation for Jacobi fields. Since it is a differential equation, it is enough to study the equation in a coordinate chart. Although one may work on a general frame, to simply the computation one may use a special frame that are parallel along γ [so that the covariant derivatives of the frame are as simple as possible]. So we start with an orthonormal basis {e1, · · · , em} of TpM, with e1 = ˙γ(a), where p = γ(a). Let ei(t) := the parallel transport of ei along γ, 1 ≤ i ≤ m. According to Proposition 2.1 in Lecture 6, ⟨ei(t), ej (t)⟩γ(t) = ⟨ei , ej ⟩γ(a) = δij . In other words, we get an orthonormal frame {e1(t), · · · , em(t)} along γ with e1(t) = γ˙(t), and this frame is parallel along γ, i.e. ∇γ˙ (t)ek(t) = 0, 1 ≤ k ≤ m. Let X be a Jacobi field along γ, then with respect to this orthonormal frame we can write X = Xi (t)ei(t), and we get ∇γ˙ X = X˙ i (t)ei(t) and ∇γ˙ ∇γ˙ X = X¨i (t)ei(t). It follows that the Jacobi field equation becomes X¨i (t)ei(t) − X i (t)R j 1i1 ej (t) = 0. So we arrived at a system of second order homogeneous ODEs, X¨i (t) − X j (t)R i 1j1 = 0, 1 ≤ i ≤ m, Using the basic theory for second order homogeneous ODEs, we get Theorem 1.3. Let γ : [a, b] → M be a geodesic, then for any Xγ(a) , Yγ(a) ∈ Tγ(a)M, there exists a unique Jacobi field X along γ so that X(a) = Xγ(a) and ∇γ˙ (a)X = Yγ(a) . Moreover, the set of Jacobi fields along γ is a linear space of dimension 2m (which is canonically isomorphic to Tγ(a)M ⊕ Tγ(a)M)

3LECTURE17:JACOBIFIELDSThefollowing consequenceis fundamental:Corollary 1.4. If X(t) is a Jacobi field along , and X is not identically zero, thenthe zeroes of X are discrete.Proof. If X has a sequence of zeroes that converges to to, then X'(t) =m(t) = 0 for a sequence of points t converging to (to). It follows that Xi(to) = 0口and Xi(to) = 0 for all i, i.e. X(to) = 0, V(to)X = 0. By uniqueness, X = 0. I Jacobi fields as variational fields of geodesic variation.Now we prove that each Jacobi field X along a geodesic can be realized asthe variation field of a geodesic variation of (So the space of all the Jacobi fields along describes all possible ways that can vary in "the space of all geodesics" infinitesimally):Theorem 1.5. A vector field X along a geodesic is a Jacobi field if and only ifX is the variation field of some geodesic ariation of .Proof. We have seen that the variation field of any geodesic variation of is aJacobi field. Now we suppose X is a Jacobi field along and construct the desiredgeodesic variation. For simplicity we parameterize as : [0, 1] → M, so (t) =exp(o)(t(O))is definedfor 0 ≤ t≤1.Itfollowsthatfor any (p,Yp)ina smallneighborhood of ((0), (0)), the exponential map exp,(tY,) is defined for 0 ≤t ≤1.Let : (-e,e)-→ M be the geodesic with initial conditions(0) = (0), (0) = X%(0):Let T(s),W(s) be parallel vector fields along withT(0) =(0)andW(0) = V(o)X.Define f : [0, 1] × (-e,e) → M byf(t, s) = expe(s)(t(T(s) + sW(s)).As we mentioned above, for small enough, f is well-defined. Moreover, f(t, o) =(t), so f is a geodesic variation of .Let V be the variation field of f.Since bothV and X are Jacobi fields along , to show V = X, it is enough to showV(0) = X(0) andV(o)V= V(0)X.The first one follows fromddV(0) = fs(0, 0) =f(0, s) =(s) = X(0):dsdss=0For the second one, we start with the fact Va/atfs=Va/asft. Evaluate the left handside at (0,0) we get(Va/af.). = (Va/atf.(t,0) /t-0 = V(0)V

LECTURE 17: JACOBI FIELDS 3 The following consequence is fundamental: Corollary 1.4. If X(t) is a Jacobi field along γ, and X is not identically zero, then the zeroes of X are discrete. Proof. If X has a sequence of zeroes that converges to t0, then X1 (t) = · · · = x m(t) = 0 for a sequence of points tk converging to γ(t0). It follows that Xi (t0) = 0 and X˙ i (t0) = 0 for all i, i.e. X(t0) = 0, ∇γ˙ (t0)X = 0. By uniqueness, X ≡ 0. □ ¶ Jacobi fields as variational fields of geodesic variation. Now we prove that each Jacobi field X along a geodesic γ can be realized as the variation field of a geodesic variation of γ(So the space of all the Jacobi fields along γ describes all possible ways that γ can vary in “the space of all geodesics” infinitesimally): Theorem 1.5. A vector field X along a geodesic γ is a Jacobi field if and only if X is the variation field of some geodesic variation of γ. Proof. We have seen that the variation field of any geodesic variation of γ is a Jacobi field. Now we suppose X is a Jacobi field along γ and construct the desired geodesic variation. For simplicity we parameterize γ as γ : [0, 1] → M, so γ(t) = expγ(0)(tγ˙(0)) is defined for 0 ≤ t ≤ 1. It follows that for any (p, Yp) in a small neighborhood of (γ(0), γ˙(0)), the exponential map expp (tYp) is defined for 0 ≤ t ≤ 1. Let ξ : (−ε, ε) → M be the geodesic with initial conditions ξ(0) = γ(0), ˙ξ(0) = Xγ(0). Let T(s), W(s) be parallel vector fields along ξ with T(0) = ˙γ(0) and W(0) = ∇γ˙ (0)X. Define f : [0, 1] × (−ε, ε) → M by f(t, s) = expξ(s) (t(T(s) + sW(s))). As we mentioned above, for ε small enough, f is well-defined. Moreover, f(t, 0) = γ(t), so f is a geodesic variation of γ. Let V be the variation field of f. Since both V and X are Jacobi fields along γ, to show V = X, it is enough to show V (0) = Xγ(0) and ∇γ˙ (0)V = ∇γ˙ (0)X. The first one follows from V (0) = fs(0, 0) = d ds     s=0 f(0, s) = d ds     s=0 ξ(s) = Xγ(0). For the second one, we start with the fact ∇e ∂/∂tfs = ∇e ∂/∂sft . Evaluate the left hand side at (0, 0) we get  ∇e ∂/∂tfs  0,0 =  ∇e ∂/∂tfs(t, 0)   t=0 = ∇γ˙ (0)V

4LECTURE17:JACOBIFIELDSand evaluate the right hand side at (0, O) and use the facta(t(T(s) + sW(s) = T(s) + sW(s)fi(0, s) = (d exPe(s)o dtweget(Va/asf.)0。= (Va/asf(0, s)/ = Vx(0(T(s) + sW(s) = W(0) = V4(0)X.口So we get V(o) V = V(o)X and thus completes the proof.Notethat given any Jacobifield V alongageodesic ,thereexist many geodesicvariations of whose variation fields are V [analogue: given any vector u at a point p, thereexist many curves whose tangent vector at p is ul. In the proof above we give an explicitformula for one such geodesic variations, namely,(2)f(t, s) = exPe(s)(t(T(s) + sW(s),where is a geodesic with (0) = (0) and E(0) = V(0), and T, W are parallel vectorfields along with T(0) = (0) and W(0) = V(o) V.2.JACOBIFIELDSWITH SPECIAL CONDITIONSI Normal Jacobi fields.The obviously Jacobi fields , t[and their linear combinations] along are bothtangent to and are not so interesting in applications.Very often we need to ruleout them and mainly consider normal Jacobi fields.Definition 2.1.A Jacobi field along is called a normal Jacobi field if it is per-pendicular to along .It turns out that for any Jacobi field, the tangential components must be a linearcombination of and t:Proposition 2.2. For any Jacobi field X along , there erists cl, dl e R so thatX+=X-c-d'is a normal Jacobi field along .Proof.ByLemma1.2,(X,)isa linear function along,i.e.(X, ) = Cit +difor some constant Ci,di e R. Now we letX+=X-ct-d'with cl =,d' = 郎。 Then it is a Jacobi field along since it is a linearcombination of Jacobi fields along , and it is normal since(X+,)=cit+di-ct2-d2=0

4 LECTURE 17: JACOBI FIELDS and evaluate the right hand side at (0, 0) and use the fact ft(0, s) = (d expξ(s) )0 d dt     t=0 (t(T(s) + sW(s))) = T(s) + sW(s) we get  ∇e ∂/∂sft  0,0 =  ∇e ∂/∂sft(0, s)    s=0 = ∇Xγ(0) (T(s) + sW(s)) = W(0) = ∇γ˙ (0)X. So we get ∇γ˙ (0)V = ∇γ˙ (0)X and thus completes the proof. □ Note that given any Jacobi field V along a geodesic γ, there exist many geodesic variations of γ whose variation fields are V [analogue: given any vector v at a point p, there exist many curves whose tangent vector at p is v]. In the proof above we give an explicit formula for one such geodesic variations, namely, (2) f(t, s) = expξ(s) (t(T(s) + sW(s))), where ξ is a geodesic with ξ(0) = γ(0) and ˙ξ(0) = V (0), and T, W are parallel vector fields along ξ with T(0) = ˙γ(0) and W(0) = ∇γ˙ (0)V . 2. Jacobi fields with special conditions ¶ Normal Jacobi fields. The obviously Jacobi fields ˙γ, tγ˙ [and their linear combinations] along γ are both ✿✿✿✿✿✿✿✿ tangent to γ and are not so interesting in applications. Very often we need to rule out them and mainly consider ✿✿✿✿✿✿✿ normal Jacobi fields. Definition 2.1. A Jacobi field along γ is called a normal Jacobi field if it is per￾pendicular to ˙γ along γ. It turns out that for any Jacobi field, the tangential components must be a linear combination of ˙γ and tγ˙ : Proposition 2.2. For any Jacobi field X along γ, there exists c 1 , d1 ∈ R so that X ⊥ = X − c 1 tγ˙ − d 1 γ˙ is a normal Jacobi field along γ. Proof. By Lemma 1.2, ⟨X, γ˙⟩ is a linear function along γ, i.e. ⟨X, γ˙⟩ = c1t + d1 for some constant c1, d1 ∈ R. Now we let X ⊥ = X − c 1 tγ˙ − d 1 γ˙ with c 1 = c1 |γ˙ | 2 , d1 = d1 |γ˙ | 2 . Then it is a Jacobi field along γ since it is a linear combination of Jacobi fields along γ, and it is normal since ⟨X ⊥, γ˙⟩ = c1t + d1 − c 1 t|γ˙ 2 | − d 2 |γ˙ | 2 = 0

5LECTURE17:JACOBIFIELDS口Note that if X+ is a normal Jacobi field along , then(7--) - (x+,.) -x-,/45) - 00and thus V,X+ I %.ItfollowsCorollary 2.3.A Jacobi field X alongis normal if and onlyif(X(a), (a)) = (V(a)X, (a)) = 0.In particular, the set of normal Jacobi fields form a linear space of dimension 2m-2.Proof.With X =X++c't+d', we have<X(a), (a)) = (c'a +d'),(V(a)X, (a)) = (V(a)(c't+d'), (a)) = cl/2口The conclusionfollows.Corollary 2.4.Let X be a Jacobi field so that(X(t1), (t1)) = (X(t2), (t2)) = 0for two distinct numbers ti,t2. Then X is a normal Jacobi field.Proof. This follows from Lemma 1.2, i.e. (X, ) is a linear function along , and thefact that a linear function has no more than one zero unless it is identically zero.I Normal Jacobi fields on spaces with constant sectional curvature.Let (M,g) be a Riemannian manifold with constant sectional curvaturek,i.e.R(X,Y)Z = -k((X,Z)Y - (Y,Z)X)Let be a normal geodesic in M, and X a normal Jacobi field along .ThenR(%, X)=-k(<%, )X -(X,)) =-X.Sotheequationfora normal JacobifieldXalongbecomesV,V,X+kX =0.Again we take an orthonormal frame (e;(t)) along so that. ei(t) =(t),. each e;(t) is parallel along ,as we did in the proof of Theorem 1.3, and writeX=xi(t)e;(t),12

LECTURE 17: JACOBI FIELDS 5 □ Note that if X⊥ is a normal Jacobi field along γ, then ⟨∇γ˙ X ⊥, γ˙⟩ = d dt⟨X ⊥, γ˙⟩ − ⟨X ⊥, ∇γ˙ γ˙⟩ = 0 and thus ∇γ˙ X⊥ ⊥ γ˙ . It follows Corollary 2.3. A Jacobi field X along γ is normal if and only if ⟨X(a), γ˙(a)⟩ = ⟨∇γ˙ (a)X, γ˙(a)⟩ = 0. In particular, the set of normal Jacobi fields form a linear space of dimension 2m−2. Proof. With X = X⊥ + c 1 tγ˙ + d 1γ˙ , we have ⟨X(a), γ˙(a)⟩ = (c 1 a + d 1 )|γ˙ | 2 , ⟨∇γ˙ (a)X, γ˙(a)⟩ = ⟨∇γ˙ (a)(c 1 tγ˙ + d 1 γ˙), γ˙(a)⟩ = c 1 |γ˙ | 2 . The conclusion follows. □ Corollary 2.4. Let X be a Jacobi field so that ⟨X(t1), γ˙(t1)⟩ = ⟨X(t2), γ˙(t2)⟩ = 0 for two distinct numbers t1, t2. Then X is a normal Jacobi field. Proof. This follows from Lemma 1.2, i.e. ⟨X, γ˙⟩ is a linear function along γ, and the fact that a linear function has no more than one zero unless it is identically zero. □ ¶ Normal Jacobi fields on spaces with constant sectional curvature. Let (M, g) be a Riemannian manifold with constant sectional curvature k, i.e. R(X, Y )Z = −k(⟨X, Z⟩Y − ⟨Y, Z⟩X). Let γ be a normal geodesic in M, and X a normal Jacobi field along γ. Then R( ˙γ, X) ˙γ = −k(⟨γ, ˙ γ˙⟩X − ⟨X, γ˙⟩γ˙) = −kX. So the equation for a normal Jacobi field X along γ becomes ∇γ˙ ∇γ˙ X + kX = 0. Again we take an orthonormal frame {ei(t)} along γ so that • e1(t) = ˙γ(t), • each ei(t) is parallel along γ, as we did in the proof of Theorem 1.3, and write X = Xm i=2 X i (t)ei(t)

6LECTURE17:JACOBIFIELDSthentheequationforthecoefficientXi(t)becomesxi(t) + kxi(t) = 0,2 0,ctta.Xi(t) :if k = 0,ci sinh(V-kt)+d'cosh(v-kt),ifk0(cos(Vkt),k>0Vkk=0snk(t) =k=0 and cnz(t)=sn(t)=Nsinh(V-kt)cosh(V-kt),k<0k<0so that we can write Xi(t) = c'sn(t) + dcnk(t)I Jacobi fields with V(O) = 0.For simplicity let a = 0 for the defining interval [a, 6] of . In many applicationswe need geodesic variations that fix one end, i.e. with (O) =(O)for all s.Ofcourse the Jacobi field for such geodesic variations satisfies V(o) = 0. Conversely, ifV is a Jacobi field along with V(O) = 0, then in (2) we may take(s)=(0), T(s)=(0), W(s) =V(0) Vand get an explicit geodesic variation with one end fixed, whose variation field is V:Proposition 2.5.If V is a Jacobi field along geodesic withV(O)=0, thenf(t, s) = exp(0) (t((0) + sV(0) V)).is a geodesic variation of with s(O)=(O) and whose variation field is V.In particular, by calculating the variation field of the above variation via itsformula, we getCorollary 2.6. If V is a Jacobi field along geodesic with V(0) = 0, thenV(t) = fs(t, 0) = (d exp(o)t(0)(tV,V)

6 LECTURE 17: JACOBI FIELDS then the equation for the coefficient Xi (t) becomes X¨i (t) + kXi (t) = 0, 2 ≤ i ≤ m. The solution to this equation is X i (t) =    c i sin(√ kt) √ k + d i cos(√ kt), if k > 0, c i t + d i , if k = 0, c i sinh(√ −kt) √ −k + d i cosh(√ −kt), if k 0 t, k = 0 sinh(√ −kt) √ −k , k 0 1, k = 0 cosh(√ −kt), k < 0 so that we can write Xi (t) = c i snk(t) + d i cnk(t). ¶ Jacobi fields with V (0) = 0. For simplicity let a = 0 for the defining interval [a, b] of γ. In many applications we need geodesic variations that fix one end, i.e. with γs(0) = γ(0) for all s. Of course the Jacobi field for such geodesic variations satisfies V (0) = 0. Conversely, if V is a Jacobi field along γ with V (0) = 0, then in (2) we may take ξ(s) ≡ γ(0), T(s) ≡ γ˙(0), W(s) ≡ ∇γ˙ (0)V and get an explicit geodesic variation with one end fixed, whose variation field is V : Proposition 2.5. If V is a Jacobi field along geodesic γ with V (0) = 0, then f(t, s) = expγ(0)(t( ˙γ(0) + s∇γ˙ (0)V )). is a geodesic variation of γ with γs(0) = γ(0) and whose variation field is V . In particular, by calculating the variation field of the above variation via its formula, we get Corollary 2.6. If V is a Jacobi field along geodesic γ with V (0) = 0, then V (t) = fs(t, 0) = (d expγ(0))tγ˙ (0)(t∇γ˙ V )

LECTURE17:JACOBIFIELDS7I Taylor's expansion of the Jacobi field with V(O) = 0Now let V,W be Jacobi fields along a geodesic withV(0) = 0, V(o)V = X, E T,M and W(0) = 0, Vr(o)V= Y, E T,M.According to Corollary 2.6, we haveV(t) =(dexpp)t(o)(tXp) and W(t) =(dexPp)t(o)(tYp)Let f(t) = (V(t), W(t)). Then we havef(0) = (V(0), W(0)) = 0,f'(0) = (V(0) V, W(0)) + (V(0), V(0) W) = 0,f"(0) = (V(0) V,V, W(0)) + 2(V(0) V, V(0) W) + (V(0), V(0) V,W) = 2(Xp, Yp).To computemorederivatives, we notethat in view of V(o)=0V(0) V,V = R(%(0), V(0))(0) = 0,and similarly V(o) ,W = 0. So [We abbreviate the kth composition V,... , to (*)]((3-I)v, v()W)(0) = 0,f"(0) =>(V(4-I)v, V,W)(0) = 4(V()v, V,W)(0) + 4(,V, (3)W)(0)f"(0) =To calculate the third order derivative, we note that if we take the (k-2)th covariantderivative of the Jacobi field equation forV, thenv(k(k-2-I) R)(%, )V)= 0,where we used the factsVw(R(X,Y)Z)=(VwR)(X,Y)Z+R(VwX, Y)Z+R(X, VwY)Z+R(X,Y)VwZand V,= 0. Taking k = 3, we getV(3)V - (V,R)(6, V)- R(%, V,V)= 0.Evaluate at t = 0, and use V(0) = 0, we get (v(3)v)(0) = R((0), X,)%(0). Thusf""(0)=4(R(%(0),X,)(0),Yp)+4(Xp,R((0),Yp)%(0))=-8Rm((0),Xp, (0), Yp).So we get(V(t), W(t) = (Xp, Yp)t? - Rm(%(0), Xp, (0), Yp)t4 + O(t5).2In particular, if we take W = V and assume [Xpl =- 1, thenIV(t)2 = t2_ 1Rm(%(0), Xp, (0), X,)t* + O(t5)3

LECTURE 17: JACOBI FIELDS 7 ¶ Taylor’s expansion of the Jacobi field with V (0) = 0. Now let V, W be Jacobi fields along a geodesic γ with V (0) = 0, ∇γ˙ (0)V = Xp ∈ TpM and W(0) = 0, ∇γ˙ (0)V = Yp ∈ TpM. According to Corollary 2.6, we have V (t) = (d expp )tγ˙ (0)(tXp) and W(t) = (d expp )tγ˙ (0)(tYp). Let f(t) = ⟨V (t), W(t)⟩. Then we have f(0) = ⟨V (0), W(0)⟩ = 0, f ′ (0) = ⟨∇γ˙ (0)V, W(0)⟩ + ⟨V (0), ∇γ˙ (0)W⟩ = 0, f ′′(0) = ⟨∇γ˙ (0)∇γ˙ V, W(0)⟩ + 2⟨∇γ˙ (0)V, ∇γ˙ (0)W⟩ + ⟨V (0), ∇γ˙ (0)∇γ˙W⟩ = 2⟨Xp, Yp⟩. To compute more derivatives, we note that in view of V (0) = 0, ∇γ˙ (0)∇γ˙ V = R( ˙γ(0), V (0)) ˙γ(0) = 0, and similarly ∇γ˙ (0)∇γ˙W = 0. So [We abbreviate the k th composition ∇γ˙ · · · ∇γ˙ to ∇ (k) γ˙ ] f ′′′(0) = X 3 l=0  3 l  ⟨∇(3−l) γ˙ V, ∇ (l) γ˙ W⟩(0) = 0, f ′′′′(0) = X 4 l=0  4 l  ⟨∇(4−l) γ˙ V, ∇γ˙W⟩(0) = 4⟨∇(3) γ˙ V, ∇γ˙W⟩(0) + 4⟨∇γ˙ V, ∇ (3) γ˙ W⟩(0). To calculate the third order derivative, we note that if we take the (k−2)th covariant derivative of the Jacobi field equation for V , then ∇ (k) γ˙ V − X k−2 l=0  k − 2 l  (∇ (k−2−l) γ˙ R)( ˙γ, ∇ (l) γ˙ V ) ˙γ = 0, where we used the facts ∇W (R(X, Y )Z)= (∇W R)(X, Y )Z+R(∇W X, Y )Z+R(X, ∇W Y )Z+R(X, Y )∇W Z and ∇γ˙ γ˙ = 0. Taking k = 3, we get ∇ (3) γ˙ V − (∇γ˙R)( ˙γ, V ) ˙γ − R( ˙γ, ∇γ˙ V ) ˙γ = 0. Evaluate at t = 0, and use V (0) = 0, we get (∇ (3) γ˙ V )(0) = R( ˙γ(0), Xp) ˙γ(0). Thus f ′′′′(0)=4⟨R( ˙γ(0), Xp) ˙γ(0), Yp⟩+4⟨Xp, R( ˙γ(0), Yp) ˙γ(0)⟩=−8Rm( ˙γ(0), Xp, γ˙(0), Yp). So we get ⟨V (t), W(t)⟩ = ⟨Xp, Yp⟩t 2 − 1 3 Rm( ˙γ(0), Xp, γ˙(0), Yp)t 4 + O(t 5 ). In particular, if we take W = V and assume |Xp| = 1, then |V (t)| 2 = t 2 − 1 3 Rm( ˙γ(0), Xp, γ˙(0), Xp)t 4 + O(t 5 )