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
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
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
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) (⇐=)
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
6 LECTURE 20: THE INDEX FORM Remark. The geometric meanings of N(γ) and ind(γ) are clear: (1) The nullity N(γ) equals the multiplicity of q = γ(l) as a conjugate point of p = γ(0). In particular, N(γ) = 0 if q = γ(l) is not conjugate to p along γ. (2) The index ind(γ) is “the number of independent directions” towards which γ can be deformed to shorter curves with the same endpoints. ¶ Morse index theorem. It turns out that there is a well-developed infinite dimensional Morse theory: As in the finite dimensional case one can show that there exists p, q so that all geodesics connecting p and q are non-degenerate in the above sense. Again Cpq = C 0,l p,q has the homotopy type of a CW-complex whose λ-cells are in one-to-one correspondence with geodesics of ✿✿✿✿✿ index λ. This also gives the topological information of M, since Cpq is homotopy equivalent to Cpp, and πk+1(M) ∼= πk(Cpp). As an application we can explain Serre’s result we mentioned in the remark on page 3 in Lecture 14: by using algebraic topology Serre proved Hi(Cpq, R) ̸= 0 for infinitely many i, which implies that for any k, there exists a geodesic whose index is great than k. By Morse index theorem below, there must be infinitely many geodesics connecting p and q. So it is important to study the index of a geodesic. In what follows we will denote by γ t the geodesic γ|[0,t] , so that the corresponding index and nullity are ind(γ t ) and N(γ t ). The following theorem is fundamental in this direction: Theorem 2.2 (Morse Index Theorem). For any geodesic γ : [0, l] → M, we have ind(γ) = X 0<t<l N(γ t ) < ∞. As a consequence, we get a quantitative version of Theorem 1.1(2) “moreover”, Corollary 2.3. For any geodesic γ : [0, l] → M, γ(0) has only finitely many conjugate points along γ. If we denote these conjugate points (except possibly γ(b)) by γ(t1), · · · , γ(tk) (a < t1 < · · · < tk < b), then ind(γ) = X k j=1 nγ,0(tj ). ¶ Proof of Morse index theorem. To prove the theorem, the first step is to reduce the infinite dimension space V 0 to a finite dimensional subspace T1, so that the index and nullity of I|V0 is the same as I|T1 . As a consequence, ind(γ) is finite. To construct T1, let’s recall that near any point one can find a strongly convex neighborhood U, so that any two points in U can be connected by a unique minimal geodesic which is contained in U. Now we use finitely many such strongly convex
LECTURE 20: THE INDEX FORM 7 neighborhoods U0, U1, · · · , Uk to cover γ, and take 0 = t0 0. Thus I|T2 is positive definite. As a result, we get (2) ind(γ) = max dim{A ⊂ T1 | I|A is negatively definite} and thus ind(γ) ≤ dim T1 < +∞. To get the precise formula of ind(γ) in the theorem, we need the following lemmas which reveals the continuity property of the index ind(γ t ): Lemma 2.4. ind(γ t ) is non-decreasing in t. Lemma 2.5. ind(γ t ) = ind(γ t−ε ) for ε small enough. Lemma 2.6. ind(γ t+ε )=ind(γ t )+N(γ t ) for ε small enough. Proof of Moser Index Theorem. According to the previous lemmas, we have • ind(γ t ) is a non-decreasing and left-continuous step function. • For t small, ind(γ t ) = 0 (because γ t is a minimal geodesic for t small). • If γ(t) is not a conjugate point of γ(0), then ind(γ τ ) is constant for τ ∈ (t − ε, t + ε). • If γ(t) is a conjugate point of γ(0), then ind(γ τ ) is a step function in (t − ε, t + ε) with a “jump from right” of size N(γ t ) at τ = t The theorem follows from these facts. □
8 LECTURE 20: THE INDEX FORM ¶ Proof of Lemmas. Proof of Lemma 2.4. Let 0 < t < s. For any X ∈ V0 γ t , we can extend X to X′ ∈ V0 γ s by zero extension. Since I γ t (X, X) = I γ s (X′ , X′ ), the conclusion follows. □ Proof of Lemma 2.5. Suppose t ∈ (tj , tj+1), where tj is a subdivision of (0, l) as above. Then by (2), ind(γ t ) = {A ⊂ T γ t 1 | I|A is negatively definite}, where T γ t 1 is the set of piecewise smooth vector fields in V 0 γ t which are Jacobi fields on each (ti , ti+1). Under the isomorphism Φt : T γ t 1 → Tγ(t1)M ⊕ · · · ⊕ Tγ(tk)M the restriction of I on T γ t 1 becomes a bilinear form I t defined on Tγ(t1)M ⊕· · ·⊕Tγ(tk)M. It depends smoothly on t since for X, ⃗ Y⃗ ∈ Tγ(t1)M ⊕ · · · ⊕ Tγ(tk)M, by (1), I t (X, ⃗ Y⃗ )=X i<j I γi (X i , Y i ) + I γet (X t , Y t )=⟨∇γ˙ (tj+)X t , Y ⟩ + terms independent of t, where γ i = γ|[ti,ti+1] and Xi = (Φt ) −1 (X⃗ )|[ti,ti+1] for i < j, while γe t = γ|[tj ,t] and Xt = (Φt ) −1 (X⃗ )|[tj ,t] . So if I t is negative definite on a subspace, so is I t−ε for ε small, i.e. ind(γ t−ε ) ≥ ind(γ t ). The conclusion follows from Lemma 2.4. □ Proof of Lemma 2.6. As in the proof of Lemma 2.5, we identify the restriction of I on T γ t 1 with the bilinear form I t defined on Tγ(t1)M ⊕ · · · ⊕ Tγ(tk)M which depends smoothly on t. Then I t is positive definite on a subspace of dimension mk−ind(γ t )− N(γ t ). It follows that I t+ε is positive definite on this subspace for ε small enough. This implies ind(γ t+ε ) ≤ mk − (mk − ind(γ t ) − N(γ t )) = ind(γ t ) + N(γ t ). On the other hand, we have • By the same continuity argument, I t+ε is negative definite on the space where I t is negative definite. • I t+ε is negative definite on the null space of I t : If 0 ̸= X⃗ is in the null space of I t , then (Φt ) −1 (X⃗ ) is a Jacobi field along γ t and Xtj ̸= 0 (since there is no Jacobi field that vanishes at both tj and t). So if we denote by Xet the zero extension of Xt to [tj , t + ε], then by Corollary 1.2, I γet+ε (X t+ε , Xt+ε ) < Iγet+ε (Xet , Xet ) = I γet (X t , Xt ) and thus I t+ε (X, ⃗ X⃗ ) < It (X, ⃗ X⃗ ) = 0. So we get ind(γ t+ε ) ≥ ind(γ t ) + N(γ t ) which finishes the proof. □