LECTURE19:CONJUGATE POINT AND APPLICATIONS1.CONJUGATE POINTSI The index form.We want to understand the mechanism for geodesics to fail to be length mini-mizing among nearby curves with the same endpoints. So we go back to the secondvariation formula for a proper variation f(t,s) = s(t) of a geodesic : [a,b] -→ Mwith variation fieldX,d?E(%) =((R(,X),X)+(V,X,V,X))dtds2If the second variation is positive, then is length minimizing among these s's fors small. Since any vector field can be realized as a variation field[See PSet 3], we areledtostudythequadraticformI(X, X) :=(R(,X),X)+(V,X,V,X))dtfor any vector field X along , and thus are led to study its polarizationI(X,Y) :=/((R(,X),Y)+(V,X,V,Y))dt(R(%,X)-V,V,X,Y)dt+(V,X,Y)In many applications one need to consider a variation whose variation field isonly continuous and piecewise smooth. So in this most general case, the inder formI =Iof thegeodesicis a symmetric bilinear form defined onV=V,={X is a continuous piecewise smooth vector field along)bytheformulaI(X,Y)=((R(,X),Y)+(V,X,V,Y))dt(1)[(R(,X)-VVX,dt+(V,X,)a-(V(tX-V(),Y),where a < ti < ... < ts <b are those points where X is not smooth, and V(t)Xmeans limt+tj+ Vs(t)X.1
LECTURE 19: CONJUGATE POINT AND APPLICATIONS 1. Conjugate points ¶ The index form. We want to understand the mechanism for geodesics to fail to be length minimizing among nearby curves with the same endpoints. So we go back to the second variation formula for a proper variation f(t, s) = γs(t) of a geodesic γ : [a, b] → M with variation field X, d 2 ds2 s=0 E(γs) =Z b a ⟨R( ˙γ,X) ˙γ,X⟩+⟨∇γ˙ X,∇γ˙ X⟩ dt. If the second variation is positive, then γ is length minimizing among these γs’s for s small. Since any vector field can be realized as a variation field[See PSet 3], we are led to study the quadratic form I(X, X) :=Z b a ⟨R( ˙γ,X) ˙γ,X⟩+⟨∇γ˙ X,∇γ˙ X⟩ dt for any vector field X along γ, and thus are led to study its polarization I(X, Y ) :=Z b a ⟨R( ˙γ,X) ˙γ,Y ⟩+⟨∇γ˙ X,∇γ˙ Y ⟩ dt = Z b a ⟨R( ˙γ, X) ˙γ−∇γ˙ ∇γ˙ X, Y ⟩dt+ ⟨∇γ˙ X, Y ⟩|b a . In many applications one need to consider a variation whose variation field is only continuous and piecewise smooth. So in this most general case, the index form I = I γ of the geodesic γ is a symmetric bilinear form defined on V = Vγ = {X is a continuous piecewise smooth vector field along γ} by the formula (1) I(X, Y )=Z b a ⟨R( ˙γ,X) ˙γ,Y ⟩+⟨∇γ˙ X,∇γ˙ Y ⟩ dt = Z b a ⟨R( ˙γ, X) ˙γ−∇γ˙∇γ˙X, Y⟩dt+ ⟨∇γ˙ X, Y ⟩|b a− X k j=1 ⟨∇γ˙ (t + j )X−∇γ˙ (t − j )X, Y ⟩, where a < t1 < · · · < tk < b are those points where X is not smooth, and ∇γ˙ (t + j )X means limt→tj+ ∇γ˙ (t)X. 1
2LECTURE19:CONJUGATEPOINTANDAPPLICATIONSI Index form v.s. Jacobi fields.Now werelate the index form I with Jacobi fields along :[a,b]→M.DenoteVO = V := [X e VI X(a) = 0, X(b) = 0).WehaveProposition 1.1. Let V V. Then V is a Jacobi field along if and only if forany X E vo, I(V, X) = 0.Proof. () According to (1), if X E vo and V is a Jacobi field (which has to besmooth) along , then I(V,X) = 0.() Conversely assume V eV satisfies I(V,X)=0 for all X evo, anda=to<ti<..<t<t+1=bis a subdivision of [a,b] so that V is smooth on each [t,,ti+i].. First take a smooth function f : [a,b] -→ R with f(ti) = 0 for all i andf(t) > 0 for all t g [to,ti, ...,th+i], and defineX = f(t)(R(,V)- V,V,V).Then X e vo and so0 = I(V,X) =f(t)IR(,V)-,,VdtIt follows that V is a Jacobi field on each (t,tij+1).. Next let's choose any X' vo withX(t) = V(tt)V- V()VThen0 = I(V,X) = -IVs(t)V- Vs(t)VI2.It follows that V is of class C1 at each ti口By uniqueness, Vis smooth.SoVis a Jacobifield.I Conjugate points.Inparticular,if thereis a nonzero Jacobifield Valongwith V(ta)=V(ta)=0,then I(V,V) = 0, where = liti,tal, and thus I is not positive definite. As a result,may fail to be length minimizing for the variation with variation field V.Definition 1.2. Let (M,g) be a Riemannian manifold, : [a, b] → M a geodesic,and ti + t2 e [a, b]j. If there exists a Jacobi field V along which is not identicallyzero, such that V(ti) = V(t2) = 0, then we say (t2) is conjugate to (ti) along
2 LECTURE 19: CONJUGATE POINT AND APPLICATIONS ¶ Index form v.s. Jacobi fields. Now we relate the index form I with Jacobi fields along γ : [a, b] → M. Denote V 0 = V 0 γ := {X ∈ V | X(a) = 0, X(b) = 0}. We have Proposition 1.1. Let V ∈ V. Then V is a Jacobi field along γ if and only if for any X ∈ V0 , I(V, X) = 0. Proof. (=⇒) According to (1), if X ∈ V0 and V is a Jacobi field (which has to be smooth) along γ, then I(V, X) = 0. (⇐=) Conversely assume V ∈ V satisfies I(V, X) = 0 for all X ∈ V0 , and a = t0 < t1 < · · · < tk < tk+1 = b is a subdivision of [a, b] so that V is smooth on each [tj , tj+1]. • First take a smooth function f : [a, b] → R with f(ti) = 0 for all i and f(t) > 0 for all t ̸∈ {t0, t1, · · · , tk+1}, and define X = f(t)(R( ˙γ, V ) ˙γ − ∇γ˙ ∇γ˙ V ). Then X ∈ V0 and so 0 = I(V, X) = Z b a f(t)|R( ˙γ, V ) ˙γ − ∇γ˙ ∇γ˙ V | 2 dt. It follows that V is a Jacobi field on each (tj , tj+1). • Next let’s choose any X′ ∈ V0 with X ′ (ti) = ∇γ˙ (t + i )V − ∇γ˙ (t − i )V. Then 0 = I(V, X′ ) = − X k i=1 |∇γ˙ (t + i )V − ∇γ˙ (t − i )V | 2 . It follows that V is of class C 1 at each ti . By uniqueness, V is smooth. So V is a Jacobi field. □ ¶ Conjugate points. In particular, if there is a nonzero Jacobi field V along γ with V (t1) = V (t2) = 0, then I γ¯ (V, V ) = 0, where ¯γ = γ|[t1,t2] , and thus I γ¯ is not positive definite. As a result, γ¯ may fail to be length minimizing for the variation with variation field V . Definition 1.2. Let (M, g) be a Riemannian manifold, γ : [a, b] → M a geodesic, and t1 ̸= t2 ∈ [a, b]. If there exists a Jacobi field V along γ which is not identically zero, such that V (t1) = V (t2) = 0, then we say γ(t2) is conjugate to γ(t1) along γ
3LECTURE19:CONJUGATEPOINTANDAPPLICATIONSNote that according to Corollary 2.4 in Lecture 17, any Jacobi field V alongsatisfying V(ti)=0 and V(t2)=0 (where ti+t2)must be a normal Jacobi field.So if q =(t2) is a conjugate point of p = (ti) along , thenJr,ti,t = [V / V is a Jacobi field along with V(ti) = V(t2) = 0)is a vector subspace of the space J+ of normal Jacobi fields along .Definition 1.3.If g=(t2) is a conjugate point of p=(ti)along ,we calln,t (t2) := dim Jr,ti,t2the multiplicity of the conjugate point q to p along .By definition, if q is conjugate to p along a geodesic , then p is conjugate to qalong the geodesic -,with the same multiplicity.We haveLemma 1.4. Suppose dimM = m, then n,t (t2)≤m -1.Proof. As we have seen in Lecture 17, a Jacobi field V is uniquely determined byV(ti) and V(ti)V. Moreover, V is normal implies V(t)V e ((ti))-. So J,ti,t2 isisomorphictoasubspaceof((0, u) / E(%(ti))) C[(u, v) / u, uET(t1)M)口and theconclusion follows.Erample. Consider the round sphere (Sm, ground) whose sectional curvature is 1. Let:[0,i] →M be a normal geodesic startingfrom anyp.Then in Lecture 17wehave seen that any normal Jacobi field along with V(O) =0 must be of the form1aV(t) =c'sin(t)e;(t),i=2where fe;(t)) is a parallel orthonormal frame along , with ei(t) =(t). It follows. if has length less than , then there is no conjugate point of p,. if the length of is between π and 2π, then the antipodal point () =-pis the only conjugate point to the north pole along any geodesic starting atp, and its multiplicity equals m-l.We may also repeat the same argument for (Rm,go) and (Hm,ghyperbotic), andarrive at the conclusion that there is no conjugate point at all. In fact the same resultholds for any Riemannian manifold whose sectional curvatures are non-positive:Proposition 1.5. Let (M,g) be a Riemannian manifold whose sectional curvatureis non-positive. Then any p e M has no conjugate point along any geodesic.Proof. Let be any geodesic from (o)=p and V any nonzero normal Jacobi fieldalong with V(0) = 0. Let f(t) = (V(t),V(t)). Thenf'(t) = 2(V(t) V, V)
LECTURE 19: CONJUGATE POINT AND APPLICATIONS 3 Note that according to Corollary 2.4 in Lecture 17, any Jacobi field V along γ satisfying V (t1) = 0 and V (t2) = 0 (where t1 ̸= t2) must be a normal Jacobi field. So if q = γ(t2) is a conjugate point of p = γ(t1) along γ, then Jγ,t1,t2 = {V | V is a Jacobi field along γ with V (t1) = V (t2) = 0} is a vector subspace of the space J ⊥ γ of normal Jacobi fields along γ. Definition 1.3. If q = γ(t2) is a conjugate point of p = γ(t1) along γ, we call nγ,t1 (t2) := dim Jγ,t1,t2 the multiplicity of the conjugate point q to p along γ. By definition, if q is conjugate to p along a geodesic γ, then p is conjugate to q along the geodesic −γ, with the same multiplicity. We have Lemma 1.4. Suppose dim M = m, then nγ,t1 (t2) ≤ m − 1. Proof. As we have seen in Lecture 17, a Jacobi field V is uniquely determined by V (t1) and ∇γ˙ (t1)V . Moreover, V is normal implies ∇γ˙ (t1)V ∈ ( ˙γ(t1))⊥. So Jγ,t1,t2 is isomorphic to a subspace of {(0, v) | v ∈ ( ˙γ(t1))⊥} ⊂ {(u, v) | u, v ∈ Tγ(t1)M} and the conclusion follows. □ Example. Consider the round sphere (S m, ground) whose sectional curvature is 1. Let γ : [0, l] → M be a normal geodesic starting from any p. Then in Lecture 17 we have seen that any normal Jacobi field along γ with V (0) = 0 must be of the form V (t) = Xm i=2 c i sin(t)ei(t), where {ei(t)} is a parallel orthonormal frame along γ, with e1(t) = ˙γ(t). It follows • if γ has length less than π, then there is no conjugate point of p, • if the length of γ is between π and 2π, then the antipodal point γ(π) = −p is the only conjugate point to the north pole along any geodesic starting at p, and its multiplicity equals m − 1. We may also repeat the same argument for (R m, g0) and (Hm, ghyperbolic), and arrive at the conclusion that there is no conjugate point at all. In fact the same result holds for any Riemannian manifold whose sectional curvatures are non-positive: Proposition 1.5. Let (M, g) be a Riemannian manifold whose sectional curvature is non-positive. Then any p ∈ M has no conjugate point along any geodesic. Proof. Let γ be any geodesic from γ(0) = p and V any nonzero normal Jacobi field along γ with V (0) = 0. Let f(t) = ⟨V (t), V (t)⟩. Then f ′ (t) = 2⟨∇γ˙ (t)V, V ⟩
4LECTURE19:CONJUGATEPOINTANDAPPLICATIONSand thus, in view of R(, V,, V) =-K(, V)//2|V2 ≥ 0,f"(t) = 2(V,V,V,V) +2/V,VI2 =2R(%,V,, V) +2/V,VI2 ≥ 0Since f(o) =0, f'(O) =0 and f"(0) >0, we conclude f(t) > 0 for t > 0. In other口words, V has no other zeroes along So p has no conjugate point along .Remark. In the proof we also get f"(t) ≥ -K(t)l/2 f(t), where K(t) = K((t), V(t)from which onemay derive better lower bounds of f.2.CONJUGATEPOINTSVIA CRITICALPOINTS OFTHE EXPONENTIALMAPI Conjugate points v.s. the exponential map.It turns out that conjugate points are exactly those points where the exponentialmap fails to be diffeomorphism. In fact, suppose : [o, I] → M is a geodesic, and Vis a Jacobi field along with V(O) = 0, then by Corollary 2.6 in Lecture 17V(t) = (d exPp)t(0)(tVs(o) V).Itfollowsq=(to) is conjugate top=(o)there is a nonzero Jacobi field V along so that V(0) = 0 and V(to) = 0(o)V 0 and 0 =V(to) = (dexPp)to(o)(toV(o)V)ker(dexPp)to(0)0.Moreover, (*) also implies that Jy,p,q is isomorphic to ker(dexp,)to(0).So we getthe following useful characterization of conjugate point:Theorem 2.1.Let:[o,1] -→M be a geodesic.Then q=(to) is a conjugate pointof p = (O) if and only if exPp is singular at to(O). Moreover, the multiplicityng,p(q) = dim ker(dexpp)to%(0)I The Cartan-Hadamard theorem.As an immediate application, we proveTheorem 2.2 (Cartan-Hadamard). Let (M,g) be a complete Riemannian manifoldwith non-positive sectional curvature, then(1) for any pE M, the erponential map exp, : T,M-→ M is a covering map.(2) if M is also simply connected, then exp, is a diffeomorphism.Proof.According to Proposition 1.5 and Theorem 2.1,exPp:TpM→Mis a local diffeomorphism everywhere. So (1) follows from Corollary 2.4 in Lecture 15.If M is simply connected, then any covering map to M must be a homeomorphism.口Since exp, is also a local diffeomorphism, it must be diffeomorphism
4 LECTURE 19: CONJUGATE POINT AND APPLICATIONS and thus, in view of R( ˙γ, V, γ, V ˙ ) = −K( ˙γ, V )|γ˙ | 2 |V | 2 ≥ 0, f ′′(t) = 2⟨∇γ˙ ∇γ˙ V, V ⟩ + 2|∇γ˙ V | 2 = 2R( ˙γ, V, γ, V ˙ ) + 2|∇γ˙ V | 2 ≥ 0. Since f(0) = 0, f′ (0) = 0 and f ′′(0) > 0, we conclude f(t) > 0 for t > 0. In other words, V has no other zeroes along γ. So p has no conjugate point along γ. □ Remark. In the proof we also get f ′′(t) ≥ −K(t)|γ˙ | 2 f(t), where K(t) = K( ˙γ(t), V (t)), from which one may derive better lower bounds of f. 2. Conjugate points via critical points of the exponential map ¶ Conjugate points v.s. the exponential map. It turns out that conjugate points are exactly those points where the exponential map fails to be diffeomorphism. In fact, suppose γ : [0, l] → M is a geodesic, and V is a Jacobi field along γ with V (0) = 0, then by Corollary 2.6 in Lecture 17, V (t) = (d expp )tγ˙ (0)(t∇γ˙ (0)V ). It follows q = γ(t0) is conjugate to p = γ(0) ⇐⇒there is a nonzero Jacobi field V along γ so that V (0) = 0 and V (t0) = 0 (∗) ⇐⇒∇γ˙ (0)V ̸= 0 and 0 = V (t0) = (d expp )t0γ˙ (0)(t0∇γ˙ (0)V ) ⇐⇒ker(d expp )t0γ˙ (0) ̸= 0. Moreover, (∗) also implies that Jγ,p,q is isomorphic to ker(d expp )t0γ˙ (0). So we get the following useful characterization of conjugate point: Theorem 2.1. Let γ : [0, l] → M be a geodesic. Then q = γ(t0) is a conjugate point of p = γ(0) if and only if expp is singular at t0γ˙(0). Moreover, the multiplicity nγ,p(q) = dim ker(d expp )t0γ˙ (0). ¶ The Cartan-Hadamard theorem. As an immediate application, we prove Theorem 2.2 (Cartan-Hadamard). Let (M, g) be a complete Riemannian manifold with non-positive sectional curvature, then (1) for any p ∈ M, the exponential map expp : TpM → M is a covering map. (2) if M is also simply connected, then expp is a diffeomorphism. Proof. According to Proposition 1.5 and Theorem 2.1, expp : TpM → M is a local diffeomorphism everywhere. So (1) follows from Corollary 2.4 in Lecture 15. If M is simply connected, then any covering map to M must be a homeomorphism. Since expp is also a local diffeomorphism, it must be diffeomorphism. □
LECTURE19:CONJUGATEPOINTANDAPPLICATIONS5Definition2.3.A complete simply-connected Riemannian manifold with non-positivecurvature is called a Cartan-Hadamard manifold, or an Hadamard manifold.Remark. Let (M, g) be a complete Riemannian manifold. We say p E M is a poleof (M,g) if expp : TpM → M is non-singular everywhere, i.e. p has no conjugatepoint along any geodesic.Repeating the proof of Cartan-Hadamard theorem wordby word, we can proveTheorem 2.4. If (M, g) has a pole p, then exPp : T,M -→ M is a smooth covering.I The Killing-Hopf theorem.AsanotherapplicationweproveTheorem 2.5 (Killing-Hopf). Let (M,g) be a complete Riemannian manifold ofconstant sectional curvature k, then the Riemannian universal cover of (M,g) is(a) (Sm, ground) if k >0,(b) (Rm, go) if k = 0,(c) (Hm, -ghyperbolic) if k < 0.Proof. It is enough to work on the Riemannian universal covering of (M,g) directlyi.e. prove that if (M,g) is also simply connected, then (M,g) is isometric to one ofthese model spaces above. It is also enough to prove the theorem for k = O, ±1Case 1: k = -1 or k = 0 Write (Sm, g) = (Hm, ghyperbolic) for k = -1, and (Sm,g) =(IR", go) for k = 0. Choose any point p e Sm and fix any linear isometryL : (T,Sm,gp) → (TpM,9p)and considerF= exp, oL o (exp,)-1 : (Sm,g) → (M,g).By Cartan's local isometry theorem, F is a local isometry. By Cartan-Hadamardtheorem, exp, : TpM -→ M is a diffeomorphism. So F is a diffeomorphism, and thusanisometry.Case 2:k-1Again we start with a point p e Sm and fix any linear isometryL : (TSm, gp) → (TpM, gp).SinceexPp : Bx(0) c Tpsm → Sm \{-P)is adiffeomorphism, byCartan's local isometrytheorem themapFi =exPpoL o (exPp)-1 : (Sm /[-p),ground)→ (M,g)is a local isometry.Similarly we start with q + ±p and get a local isometryF2 = exPFi(a) o(dFi)g o (expa)-1 : (Sm /{-q), ground) → (M, g).Note that by construction,F2(Q) = exPFi(@) (dFi)a 0 (exPg)-1(q) = exPFi(@) (dFi)a(0) = exPFi(@)(0) = Fi()
LECTURE 19: CONJUGATE POINT AND APPLICATIONS 5 Definition 2.3. A complete simply-connected Riemannian manifold with non-positive curvature is called a Cartan-Hadamard manifold, or an Hadamard manifold. Remark. Let (M, g) be a complete Riemannian manifold. We say p ∈ M is a pole of (M, g) if expp : TpM → M is non-singular everywhere, i.e. p has no conjugate point along any geodesic. Repeating the proof of Cartan-Hadamard theorem word by word, we can prove Theorem 2.4. If (M, g) has a pole p, then expp : TpM → M is a smooth covering. ¶ The Killing-Hopf theorem. As another application we prove Theorem 2.5 (Killing-Hopf). Let (M, g) be a complete Riemannian manifold of constant sectional curvature k, then the Riemannian universal cover of (M, g) is (a) (S m, 1 k ground) if k > 0, (b) (R m, g0) if k = 0, (c) (Hm, − 1 k ghyperbolic) if k < 0. Proof. It is enough to work on the Riemannian universal covering of (M, g) directly, i.e. prove that if (M, g) is also simply connected, then (M, g) is isometric to one of these model spaces above. It is also enough to prove the theorem for k = 0, ±1. Case 1: k = −1 or k = 0 Write (S m, g¯) = (Hm, ghyperbolic) for k = −1, and (S m, g¯) = (R m, g0) for k = 0. Choose any point ˜p ∈ S m and fix any linear isometry L : (Tp˜S m, gp˜) → (TpM, gp) and consider F = expp ◦L ◦ (expp˜ ) −1 : (S m, g¯) → (M, g). By Cartan’s local isometry theorem, F is a local isometry. By Cartan-Hadamard theorem, expp : TpM → M is a diffeomorphism. So F is a diffeomorphism, and thus an isometry. Case 2: k = 1 Again we start with a point ˜p ∈ S m and fix any linear isometry L : (Tp˜S m, gp˜) → (TpM, gp). Since expp˜ : Bπ(0) ⊂ Tp˜S m → S m \ {−p˜} is a diffeomorphism, by Cartan’s local isometry theorem the map F1 = expp ◦L ◦ (expp˜ ) −1 : (S m \ {−p˜}, ground) → (M, g) is a local isometry. Similarly we start with ˜q ̸= ±p˜ and get a local isometry F2 = expF1(˜q) ◦(dF1)q˜ ◦ (expq˜ ) −1 : (S m \ {−q˜}, ground) → (M, g). Note that by construction, F2(˜q) = expF1(˜q) ◦(dF1)q˜ ◦ (expq˜ ) −1 (q) = expF1(˜q) ◦(dF1)q˜(0) = expF1(˜q) (0) = F1(˜q)