LECTURE 20:THE INDEX FORM1.LENGTH MINIMIZING THROUGHINDEX FORMI Index form as Hessian.Lasttimes we defined the indexform I =Iof thegeodesic: [o,]→MI(X,Y)=((R(,X),Y)+(V,X,V,Y))dtdefinedonV- V,={V is a continuous piecewise smooth vector field along ]If o<ti<...<tk<l are those pointswhereV is not smooth,then(1)I(X,Y)=/(R(, )-VVx,)dt+(V,,)-(V(X-V(tX,)Recall that geodesics with (O) = p and = q are precisely the critical pointsof the energy functional defined on the space Co., of “piecewise smooth curves withfixed endpointsparametrized on [o, ]".TheindexformI=Ioriginates fromthesecond variation, namely for a proper variation s(t) of ,d2E(s) =((R(%,X),X)+(V,X,V,X))dt = I(X,X),ds2whereXevo is the variation field ofs.Sowhat aboutI(X,Y)?It is also secondvariationoftheenergyfunctionalEorthelengthfunctionalL.butwithrespectto"mixed directions": Let r,s(t) be a two-parameter variation of = o.o with fixedendpoints.Denote thevariationfields corresponding to thetwoparameterdirectionsbyX and Y.Obviously X,Y evo.Then as in PSet 3 one can prove02E(r.s) = I(X, Y).OrosSo the index form I of , restricted the subspace vo, can be regarded as the Hessianof the energy/length functional defined on co, at the critical pointRecall from calculus: at a critical point p of a multi-variable smooth function f,.if Hessp(f)ispositivedefinite,then pis an isolatedlocal minimum,.if Hessp(f)(u,u)<ofor some u, then p cannot be a local minimum,. if Hessp(f) is positive semi-definite but not positive definite, we can't drawa conclusion on the behavior of f near p.1
LECTURE 20: THE INDEX FORM 1. Length minimizing through index form ¶ Index form as Hessian. Last times we defined the index form I = I γ of the geodesic γ : [0, l] → M, I(X, Y )=Z b a ⟨R( ˙γ,X) ˙γ,Y ⟩+⟨∇γ˙ X,∇γ˙ Y ⟩ dt defined on V = Vγ = {V is a continuous piecewise smooth vector field along γ}. If 0 < t1 < · · · < tk < l are those points where V is not smooth, then (1) I(X, Y )=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 ⟩ Recall that geodesics γ with γ(0) = p and γ l = q are precisely the critical points of the energy functional defined on the space C 0,l p,q of “piecewise smooth curves with fixed endpoints parametrized on [0, l]”. The index form I = I γ originates from the second variation, namely for a proper variation γs(t) of γ, d 2 ds2 s=0 E(γs) =Z l 0 ⟨R( ˙γ,X) ˙γ,X⟩+⟨∇γ˙ X,∇γ˙ X⟩ dt = I(X, X), where X ∈ V0 is the variation field of γs. So what about I(X, Y )? It is also second variation of the energy functional E or the length functional L, but with respect to “mixed directions”: Let γr,s(t) be a two-parameter variation of γ = γ0,0 with fixed endpoints. Denote the variation fields corresponding to the two parameter directions by X and Y . Obviously X, Y ∈ V0 . Then as in PSet 3 one can prove ∂ 2 ∂r∂s r=s=0 E(γr,s) = I(X, Y ). So the index form I of γ, restricted the subspace V 0 , can be regarded as the Hessian of the energy/length functional defined on C 0,l p,q at the critical point γ. Recall from calculus: at a critical point p of a multi-variable smooth function f, • if Hessp(f) is positive definite, then p is an isolated local minimum, • if Hessp(f)(v, v) < 0 for some v, then p cannot be a local minimum, • if Hessp(f) is positive semi-definite but not positive definite, we can’t draw a conclusion on the behavior of f near p. 1
2LECTURE20:THEINDEXFORMIt turns out that the same phenomena happens for the energy/length functional:Theorem 1.1. Let : [0, 1] → M be a geodesic from p = (O) to q =(). Then(1) The inder form I is positive definite on vop has no conjugate point along ..Moreoverin this case is an “isolated" length minimizing among nearbycurves: there erists e > 0 so that for any piecewise smooth curve : [0, I] M from p to q satisfying dist((t),(t)) < e, we have L() ≥ L(), withequality hold if and only if is a re-parametrization of (2) There erists X vo with I(X,X) <0 there erists t<l such that q =(t)is conjugate to p along.Moreover in this case is not length minimizing among nearby curves:there is a proper variation of so that L(s)<L() for 0<s<e.(3)The inder formI is positive semi-definitebut not positive definite onvoqis the first conjugate point of p along .Remark. Note that we only claim that is minimizing among nearby curves. It ispossible that there exists other shorter geodesics from p to q. For example, for thecylinders there is no conjugate point (since the sectional curvature is O), but there areinfinitely many geodesics between any given two points: each is minimizing among"nearby curves", but only one of them is minimizing among all curves.As a corollary of part (1),we get the following important property:Corollary 1.2. Suppose p =(0) has no conjugate point along : [0, 1] → M. If Vis a Jacobi feld along , and X E V satisfies X(O) = V(O), X(1) = V(l), thenI(V, V) ≤I(X, X),with equality holds if and only if X-V.Proof. Since X is a Jacobi field and X(O) = V(O), X(U) = V(U), by the equation(1),I(V, V) = I(V,X).Itfollowsfrompart (1)ofTheorem1.1that0 ≤I(V - X, V - X) = I(V,V) - 2I(V,X)+I(X,X) = -I(V,V)+ I(X,X),口with equality holds if and only if V - X = 0.I Proof of Theorem 1.1, part (1).We need the following lemma which is essentially Theorem 2.6 in Lecture 13:Lemma 1.3. Suppose &, = EnT,M contains a line segment [0, ]X,. Let : [0, 1] -Ep be a piecewise smooth curve with (0) = 0, p(t) =lXp. Then for(t) := exPp(tXp)(0 ≤t ≤1)and(t) :=exp(sp(t))(0 ≤t≤1),we have L() ≥ L(). Moreover, if expp is non-singular along [0,1]Xp, then theequality L()= L() holds if and only if is a re-parametrization of
2 LECTURE 20: THE INDEX FORM It turns out that the same phenomena happens for the energy/length functional: Theorem 1.1. Let γ : [0, l] → M be a geodesic from p = γ(0) to q = γ(l). Then (1) The index form I is positive definite on V 0⇐⇒p has no conjugate point along γ. • Moreover in this case γ is an “isolated” length minimizing among nearby curves: there exists ε > 0 so that for any piecewise smooth curve γ¯ : [0, l] → M from p to q satisfying dist(γ(t), γ(t)) < ε, we have L(¯γ) ≥ L(γ), with equality hold if and only if γ¯ is a re-parametrization of γ. (2) There exists X ∈ V0 with I(X, X) < 0 ⇐⇒ there exists t < l ¯ such that q¯ = γ(t¯) is conjugate to p along γ. • Moreover in this case γ is not length minimizing among nearby curves: there is a proper variation of γ so that L(γs) < L(γ) for 0 < |s| < ε. (3) The index form I is positive semi-definite but not positive definite on V 0 ⇐⇒ q is the first conjugate point of p along γ. Remark. Note that we only claim that γ is minimizing among nearby curves. It is possible that there exists other shorter geodesics from p to q. For example, for the cylinders there is no conjugate point (since the sectional curvature is 0), but there are infinitely many geodesics between any given two points: each is minimizing among “nearby curves”, but only one of them is minimizing among all curves. As a corollary of part (1), we get the following important property: Corollary 1.2. Suppose p = γ(0) has no conjugate point along γ : [0, l] → M. If V is a Jacobi field along γ, and X ∈ V satisfies X(0) = V (0), X(l) = V (l), then I(V, V ) ≤ I(X, X), with equality holds if and only if X = V . Proof. Since X is a Jacobi field and X(0) = V (0), X(l) = V (l), by the equation(1), I(V, V ) = I(V, X). It follows from part (1) of Theorem 1.1 that 0 ≤ I(V − X, V − X) = I(V, V ) − 2I(V, X) + I(X, X) = −I(V, V ) + I(X, X), with equality holds if and only if V − X = 0. □ ¶ Proof of Theorem 1.1, part (1). We need the following lemma which is essentially Theorem 2.6 in Lecture 13: Lemma 1.3. Suppose Ep = E ∩TpM contains a line segment [0, l]Xp. Let φ : [0, l] → Ep be a piecewise smooth curve with φ(0) = 0, φ(l) = lXp. Then for γ(t) := expp (tXp)(0 ≤ t ≤ l) and γ(t) := expp (φ(t))(0 ≤ t ≤ l), we have L(γ) ≥ L(γ). Moreover, if expp is non-singular along [0, l]Xp, then the equality L(γ) = L(γ) holds if and only if γ is a re-parametrization of γ
3LECTURE20:THEINDEXFORMProof. WLOG, we may assume p(t) + 0 for all t E (0, I). Write p(t) = r(t)e(t),where r(t) = o(t)l and e(t) E SpM has unit length. Theng(t) =r(t)e(t) +r(t)e(t)and e(t) e(t). According to Gauss lemma,(t)/ = (d exP,)p(t)(t)| ≥ I(d exPp)p(t)(r(t)e(t)| = [r(t).Therefore,L()≥r(t)|dt ≥ Ir() - r(0)/ = I/p(0)/ - Ip(0)II = 1[Xp/ = L().If exp, is non-singular along [o, IX,, then for e small enough exp, is non-singularin the e-neighborhood of [o, i]Xp. Now suppose the equality holds, then we have(dexpp)(t)e(t) = 0,thus e(t) = 0 for all t. It follows that e(t) is a constant unit vector, i.e. (t) = r(t)efor some e E S,M. Obviously e=X,/X,/ is the direction vector of X,. Moreover口 cannot change sign. So (t) = expp(Xp) is a re-parametrization of Remark.There is no conflict with Theorem 1.1(2). Suppose contains a conjugatepoint of p. Since the exponential map exp, is not even a local diffeomorphism at aconjugate point of p, it is possible that a curve that is close to cannot be realizedas the image of a curve in T,M with the same endpoints under the exponential map.Nowweprovethe“moreover"andpart in (1)of Theorem1.1.(1)Moreover: Suppose p has no conjugate point along .Find a subdivision0=to<ti<...<t<t+1=land open neighborhoods V, 1 ≤ i ≤ k, of the line segment [ti,ti+i](0) in T,M sothat exp, is a diffeomorphism on each V. Denote U; = exp,(V). According to ourassumption on , for e small enough, ([t, ti+il) c U,. Now definep(t) = (expplv)-1((t),ti-1 ≤t≤t.Then p(t) is a piecewise smooth curve in T,M connecting O to l(0) with exp,(s(t) =(t). So the conclusion follows from Lemma 1.3.(1): According to the part (1)Moreover that we just proved, if p = (0) hasno conjugatepoint along ,then foranyX evo,I(X,X)≥0.(Otherwiseone canconstruct a variation with L(s)< L()) If I is not positive definite on vo, thenI(Y,Y)=for someY evo.Itfollows that foranyZ vo and anyAER,0 ≤ I(Y -^Z,Y-^Z) = -2)I(Y,Z) +^I(Z,Z)As a consequence, I(Y,Z) = o for any Z e vo. In other words, Y is a Jacobi fieldSince q = (I) is not a conjugate point of p, and Y(O) = 0, Y() = 0, we must haveY = 0. So I is positive definition on vo
LECTURE 20: THE INDEX FORM 3 Proof. WLOG, we may assume φ(t) ̸= 0 for all t ∈ (0, l). Write φ(t) = r(t)e(t), where r(t) = |φ(t)| and e(t) ∈ SpM has unit length. Then φ˙(t) = ˙r(t)e(t) + r(t) ˙e(t), and e(t) ⊥ e˙(t). According to Gauss lemma, |γ˙(t)| = |(d expp )φ(t)φ˙(t)| ≥ |(d expp )φ(t)( ˙r(t)e(t))| = |r˙(t)|. Therefore, L(γ) ≥ Z l 0 |r˙(t)|dt ≥ |r(l) − r(0)| = ||φ(l)| − |φ(0)|| = l|Xp| = L(γ). If expp is non-singular along [0, l]Xp, then for ε small enough expp is non-singular in the ε-neighborhood of [0, l]Xp. Now suppose the equality holds, then we have (d expp )φ(t)e˙(t) = 0, thus ˙e(t) = 0 for all t. It follows that e(t) is a constant unit vector, i.e. φ(t) = r(t)e for some e ∈ SpM. Obviously e = Xp/|Xp| is the direction vector of Xp. Moreover, r˙ cannot change sign. So γ(t) = expp ( r(t) |Xp|Xp) is a re-parametrization of γ. □ Remark. There is no conflict with Theorem 1.1(2). Suppose γ contains a conjugate point of p. Since the exponential map expp is not even a local diffeomorphism at a conjugate point of p, it is possible that a curve that is close to γ cannot be realized as the image of a curve in TpM with the same endpoints under the exponential map. Now we prove the “moreover” and ⇐= part in (1) of Theorem 1.1. (1)Moreover : Suppose p has no conjugate point along γ. Find a subdivision 0 = t0 < t1 < · · · < tk < tk+1 = l and open neighborhoods Vi , 1 ≤ i ≤ k, of the line segment [ti , ti+1] ˙γ(0) in TpM so that expp is a diffeomorphism on each Vi . Denote Ui = expp (Vi). According to our assumption on γ, for ε small enough, γ([ti , ti+1]) ⊂ Ui . Now define φ(t) = (expp Vi ) −1 (¯γ(t)), ti−1 ≤ t ≤ ti . Then φ(t) is a piecewise smooth curve in TpM connecting 0 to lγ˙(0) with expp (φ(t)) = γ(t). So the conclusion follows from Lemma 1.3. (1)⇐= : According to the part (1)Moreover that we just proved, if p = γ(0) has no conjugate point along γ, then for any X ∈ V0 , I(X, X) ≥ 0. (Otherwise one can construct a variation with L(γs) < L(γ).) If I is not positive definite on V 0 , then I(Y, Y ) = 0 for some Y ∈ V0 . It follows that for any Z ∈ V0 and any λ ∈ R, 0 ≤ I(Y − λZ, Y − λZ) = −2λI(Y, Z) + λ 2 I(Z, Z). As a consequence, I(Y, Z) = 0 for any Z ∈ V0 . In other words, Y is a Jacobi field. Since q = γ(l) is not a conjugate point of p, and Y (0) = 0, Y (l) = 0, we must have Y ≡ 0. So I is positive definition on V 0
4LECTURE20:THEINDEXFORMI Proof of Theorem 1.1, part (2) and (3).Lemma 1.4. Suppose q = (to) is NOT conjugate to p = (0) along a geodesic : [o,] - M. Then for any X, e T,M and X, e T,M, there erists a uniqueJacobi field V along so that V(O) = X, and V(to) = Xp.Proof. Let J,be the space of all Jacobi fields along .Define a mappinge : J-→ T,M × T,M, V -(V) = (V(0), V(to)Since is not a conjugate point of p, is injective. But is linear, and dim J,一dim(T,M × T,M) = 2m are of same dimension, so is an linear isomorphism.Nowweprovepart (2)and (3)of Theorem1.1.(2)Moreoverand(2):LetXbeanonzero Jacobifield alongwithX(O)0, X(t) = 0. Note that V@X + 0, otherwise X will be identically zero. Let Z bea smooth vector field along withZ(0) = 0, Z(U) = 0, Z(0) = -V()X.Denote = lo,t], 2 = l[t,], Z1 = Zlo,t] and Z2 = Z[(t, For any n >0 we putYl=X(t)+nZi(t),foro≤t<t,Y,(t) :Yz = nz2(t),fort<t<l.ThenI(, )) = -V(XP +n(Z1,ZI)and thusI(Yn, Yn) =-n/Vs(0)X/2 +nI1(Z1, Z1) +n12(Z2, Z2) < 0for n small enough. This proves the theorem.(3): Fix any c e (0, ). For any X e vo, we may write X = Xi(t)e;(t), where[e] are orthonormal and parallel along with ei =.LetX(t) =X(=)e;(t), 0 ≤t≤c.Then Xc is a vector field along := lo.g with Xe(0) = 0, Xc(c) = 0. Since p hasno conjugate point along lo.c, we get from (1) that I(Xc, X°) ≥ 0. It follows thatI(X,X) = lime-→a I(Xe, X) ≥ 0. So I is positively semi-definite on v0ObviouslyI is not positively definite on vo, since if X is a nonzero Jacobi fieldalong with X(a) = 0, X(b) = 0, then we have I(X, X) = 0.Thisfollows from (2) ()and (3) ().(1)(2)This follows from (1) (<) and (3) (<).(3)This follows from (1) () and (2) (←)
4 LECTURE 20: THE INDEX FORM ¶ Proof of Theorem 1.1, part (2) and (3). Lemma 1.4. Suppose q = γ(t0) is NOT conjugate to p = γ(0) along a geodesic γ : [0, l] → M. Then for any Xp ∈ TpM and Xq ∈ TqM, there exists a unique Jacobi field V along γ so that V (0) = Xp and V (t0) = Xp. Proof. Let Jγ be the space of all Jacobi fields along γ. Define a mapping Θ : Jγ → TpM × TqM, V 7→ Θ(V ) = (V (0), V (t0)). Since q is not a conjugate point of p, Θ is injective. But Θ is linear, and dim Jγ = dim(TpM × TqM) = 2m are of same dimension, so Θ is an linear isomorphism. □ Now we prove part (2) and (3) of Theorem 1.1. (2)Moreover and (2)⇐= : Let X be a nonzero Jacobi field along γ with X(0) = 0, X(t¯) = 0. Note that ∇γ˙ (t¯)X ̸= 0, otherwise X will be identically zero. Let Z be a smooth vector field along γ with Z(0) = 0, Z(l) = 0, Z(t¯) = −∇γ˙ (t¯)X. Denote γ1 = γ|[0,t¯] , γ2 = γ|[t,l ¯ ] , Z1 = Z|[0,t¯] and Z2 = Z|[t,l ¯ ] . For any η > 0 we put Yη(t) := Y 1 η = X(t) + ηZ1(t), for 0 ≤ t ≤ t,¯ Y 2 η = ηZ2(t), for t¯≤ t ≤ l. Then I γ1 (Y 1 η , Y 1 η ) = −η|∇γ˙ (t¯)X| 2 + η 2 I γ1 (Z1, Z1) and thus I γ (Yη, Yη) = −η|∇γ˙ (t¯)X| 2 + η 2 I γ1 (Z1, Z1) + η 2 I γ2 (Z2, Z2) < 0 for η small enough. This proves the theorem. (3)⇐= : Fix any c ∈ (0, l). For any X ∈ V0 , we may write X = Xi (t)ei(t), where {ei} are orthonormal and parallel along γ with e1 = ˙γ. Let X c (t) = X i ( lt c )ei(t), 0 ≤ t ≤ c. Then Xc is a vector field along γ c := γ|[0,c] with Xc (0) = 0, Xc (c) = 0. Since p has no conjugate point along γ|[0,c] , we get from (1) that I(Xc , Xc ) ≥ 0. It follows that I(X, X) = limc→a I(Xc , Xc ) ≥ 0. So I is positively semi-definite on V 0 . Obviously I is not positively definite on V 0 , since if X is a nonzero Jacobi field along γ with X(a) = 0, X(b) = 0, then we have I(X, X) = 0. (1)=⇒ : This follows from (2) (⇐=) and (3) (⇐=). (2)=⇒ : This follows from (1) (⇐=) and (3) (⇐=). (3)=⇒ : This follows from (1) (⇐=) and (2) (⇐=)
LECTURE20:THEINDEXFORM52.MORSEINDEXTHEOREMI Morse index of a geodesic.Let's continue our “finite dimension manifold" v.s.“infinite dimension spaceCo"analoguea bit further.Forafinite dimensionalmanifold M,there isaremarkabie theory,known as Morse theory,that relates the topology of M to the behaviorof critical points of a Morse function[the existence is guaranteed by Sard's theorem]:Suppose f e Co(M) is a Morse function,i.e. all critical points off are non-degenerateli.e. the Hessian Hessp(f) at any critical point p of f isnon-singularl, then M has the homotopy type of a CW-complex whoseA-cells are in one-to-one correspondence with critical points of indexA of f[the index of a critical point p is the number of negative eigenvaluesof Hessp(f).For example, if a compact manifold M admits a Morsefunctionwithonlytwo criticalpoints,thentheyhavetobemaximumand minimum, so their indexes have to be m and 0. In this case onecan prove that M is homeomorphic to Sm.What is the analogue in our setting? We already have. critical points w geodesics : [o, ] → M,.Hessian at the critical point the index form I of defined on v0So“is a non-degenerate critical point"should be translated to “"the index formI of defined on vo has trivial null space". On the other hand, we have seen lasttime that I(V, W) = O for all W E vo if and only if V is Jacobi field. SotheindexformI ofdefined onvohastrivial null space→there is no Jacobi field V along with V(O) = V(l) = 0q =(U) is not a conjugate point of p=(O) along Soweget. non-degenerate critical point geodesics : [0, ] → M so that g = (l) isnota conjugatepoint of p=(o)along,:the index of a non-degenerate critical point w the dimension of the maximalsubspace of vo on which I is negative definite.By the explanations above, we are naturally led to studyDefinition 2.1. Let (M,g) be a Riemannian manifold, and :[o,1] -→M a geo-desic.Wewill callind() = max dim(A c vo IL is negatively definite)the inder of , and callN() = dim(X vo I(X,Y)= O for all Y E VO)the nullity of
LECTURE 20: THE INDEX FORM 5 2. Morse Index Theorem ¶ Morse index of a geodesic. Let’s continue our “finite dimension manifold” v.s. “infinite dimension space C 0,l p,q” analogue a bit further. For a finite dimensional manifold M, there is a remarkable theory, known as Morse theory, that relates the topology of M to the behavior of critical points of a Morse function[the existence is guaranteed by Sard’s theorem]: Suppose f ∈ C ∞(M) is a Morse function, i.e. all critical points of f are ✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿ non-degenerate[i.e. the Hessian Hessp(f) at any critical point p of f is non-singular], then M has the homotopy type of a CW-complex whose λ-cells are in one-to-one correspondence with critical points of ✿✿✿✿✿ index λ of f[the index of a critical point p is the number of negative eigenvalues of Hessp(f)]. For example, if a compact manifold M admits a Morse function with only two critical points, then they have to be maximum and minimum, so their indexes have to be m and 0. In this case one can prove that M is homeomorphic to S m. What is the analogue in our setting? We already have • critical points ↭ geodesics γ : [0, l] → M, • Hessian at the critical point ↭ the index form I of γ defined on V 0 . So “γ is a non-degenerate critical point” should be translated to “the index form I of γ defined on V 0 has trivial null space”. On the other hand, we have seen last time that I(V, W) = 0 for all W ∈ V0 if and only if V is Jacobi field. So the index form I of γ defined on V 0 has trivial null space ⇐⇒there is no Jacobi field V along γ with V (0) = V (l) = 0 ⇐⇒q = γ(l) is not a conjugate point of p = γ(0) along γ. So we get • non-degenerate critical point ↭ geodesics γ : [0, l] → M so that q = γ(l) is not a conjugate point of p = γ(0) along γ, • the index of a non-degenerate critical point ↭ the dimension of the maximal subspace of V 0 on which I is negative definite. By the explanations above, we are naturally led to study Definition 2.1. Let (M, g) be a Riemannian manifold, and γ : [0, l] → M a geodesic. We will call ind(γ) = max dim{A ⊂ V0 | I|A is negatively definite} the index of γ, and call N(γ) = dim{X ∈ V0 | I(X, Y ) = 0 for all Y ∈ V0 } the nullity of γ