Index lemma

Set-up and the initial-value family. [F1, F2, F3, given] Put $T:=\dot{\gamma }$; by hypothesis no $t\in (a,b]$ makes $\gamma (a)$ and $\gamma (t)$ conjugate along $\gamma {\mathrm{\mid }}_{[a,t]}$. For $v\in T_{\gamma (a)}M$ let $J_v$ be the unique smooth Jacobi field along $\gamma$ with $J_v(a)=0$ and $D_t{J}_v(a)=v$ (one-sided at $a$), which exists and is unique by [F1]. For $v,z\in T_{\gamma (a)}M$ and $\lambda ,\mu \in \mathbb{R}$ the field $\lambda J_v+\mu J_z$ is a smooth Jacobi field by [F2], and it has the initial data of $J_{\lambda v+\mu z}$; uniqueness in [F1] therefore gives $J_{\lambda v+\mu z}=\lambda J_v+\mu J_z\mathrm{.}$ In particular every map ${\mathrm{\Phi }}_t:T_{\gamma (a)}M\to T_{\gamma (t)}M$, ${\mathrm{\Phi }}_t(v):=J_v(t)$, is real-linear, and ${\mathrm{\Phi }}_a=0$.

The maps ${\mathrm{\Phi }}_t$ are isomorphisms for $t\in (a,b]$. [F3, F14, step 1.1] Let $t\in (a,b]$ and $v\in \ker {\mathrm{\Phi }}_t$, so that $J_v(t)=0$. If $v\mathrm{\ne }0$, then $J_v$ is not the zero field, because its initial derivative is $v\mathrm{\ne }0$; it is a Jacobi field along the nondegenerate segment $\gamma {\mathrm{\mid }}_{[a,t]}$ vanishing at both $\gamma (a)$ and $\gamma (t)$, so by [F3] those points would be conjugate along $\gamma {\mathrm{\mid }}_{[a,t]}$, contradicting the hypothesis. Hence $\ker {\mathrm{\Phi }}_t=\{0\}$. The spaces $T_{\gamma (a)}M$ and $T_{\gamma (t)}M$ both have dimension $n$ (the dimension of $M$), so rank-nullity gives $\mathrm{rank}{\mathrm{\Phi }}_t=n-0=n$ and ${\mathrm{\Phi }}_t$ is surjective, hence bijective [F14].

Existence and uniqueness of the Jacobi field with prescribed endpoint values. [F1, F2, step 2.1] By [F1] choose a smooth Jacobi field $K$ along $\gamma$ with $K(a)=u$ (for instance the field with $D_{t}K(a)=0$). By step 2.1 there is a unique $d\in T_{\gamma (a)}M$ with ${\mathrm{\Phi }}_b(d)=w-K(b)$. Then $J:=K+J_d$ is a smooth Jacobi field by [F2] with $J(a)=u+0=u,J(b)=K(b)+{\mathrm{\Phi }}_b(d)=w\mathrm{.}$ If $J^{\mathrm{\prime }}$ is a second smooth Jacobi field with these endpoint values, then $J-J^{\mathrm{\prime }}$ is a Jacobi field vanishing at $a$, so by the uniqueness clause of [F1] it equals $J_c$ with $c:=D_t(J-J^{\mathrm{\prime }})(a)$; from $0=(J-J^{\mathrm{\prime }})(b)={\mathrm{\Phi }}_b(c)$ and step 2.1 we get $c=0$ and $J^{\mathrm{\prime }}=J$.

Decomposition of an admissible field and cancellation against $J$. [F4, F5, step 2.1] Let $V\in {\mathcal{X}}_{\mathrm{p}\mathrm{w}}^1(\gamma )$ with $V(a)=u$, $V(b)=w$ and put $W:=V-J$. Then $W$ is continuous and piecewise $C^1$ with $W(a)=W(b)=0$, so $W\in {\mathcal{X}}_0(\gamma )$ [F4]. Bilinearity and symmetry of the index form [F4] give $I_{\gamma }(V,V)=I_{\gamma }(J,J)+2I_{\gamma }(J,W)+I_{\gamma }(W,W)\mathrm{.}$ The integration-by-parts identity [F5] applies to the smooth Jacobi field $J$ and the continuous piecewise $C^1$ field $W$: the jump terms vanish because $J$ is smooth, the interior term vanishes by the Jacobi equation, and both boundary terms vanish because $W(a)=W(b)=0$. Hence $I_{\gamma }(J,W)=0$ and $I_{\gamma }(V,V)=I_{\gamma }(J,J)+I_{\gamma }(W,W)\mathrm{.}$ It therefore suffices to prove $I_{\gamma }(W,W)\ge 0$ for every $W\in {\mathcal{X}}_0(\gamma )$, with equality only for $W=0$.

The Jacobi basis and the coefficient field. [F1, F7, F8, F9, F10, F14, step 2.1] Choose a parallel frame $E_1,\dots ,E_n$ along $\gamma$ [F9], started from a basis of $T_{\gamma (a)}M$; by [F7] the frame is orthonormal if its initial basis is. Put $J_i:=J_{E_i(a)}$, so that $J_i$ is a smooth Jacobi field with $J_i(a)=0$ and $D_t{J}_i(a)=E_i(a)$ [F1]. By steps 1.1 and 2.1, for every $t\in (a,b]$ the map ${\mathrm{\Phi }}_t$ is real-linear and bijective, hence carries the basis $E_1(a),\dots ,E_n(a)$ of $T_{\gamma (a)}M$ to a basis $J_1(t),\dots ,J_n(t)$ of $T_{\gamma (t)}M$ [F14] in which every field on $(a,b]$ has uniquely determined coefficients. Let $W\in {\mathcal{X}}_0(\gamma )$ and write $W(t)={\sum }_{i=1}^n{f}_i(t)J_i(t),t\in (a,b]\mathrm{.}$ In the parallel frame the fields $J_i$ have smooth coordinate columns $a_i(t)$ and $W$ has the continuous piecewise $C^1$ coordinate column $w(t)$ with $w(a)=w(b)=0$: for an orthonormal parallel frame the coefficients are the functions $t\mapsto g(W(t),E_i(t))$ and $t\mapsto g(J_i(t),E_k(t))$, which have the claimed regularity by the product rule [F7]; for a general parallel frame the dual frame is smooth by [F10]. The matrix $A(t)=(a_1(t)\mathrm{\mid }\cdots \mathrm{\mid }{a}_n(t))$ is invertible for $t\in (a,b]$ by step 2.1, and $f(t)=A(t{)}^{-1}w(t)$ is continuous on $(a,b]$ and $C^1$ on each piece by Cramer's rule [F10]. In the parallel frame $D_t$ acts on fields with coefficients by differentiation of the coefficients [F8], so $D_{t}W$ has coefficients $w^{\mathrm{\prime }}=A^{\mathrm{\prime }}f+A{f}^{\mathrm{\prime }}$ and the field $A{f}^{\mathrm{\prime }}$ (with coefficient column $A(t)f^{\mathrm{\prime }}(t)$) is well defined on $(a,b]$.

Pointwise identity for the integrand. [F2, F6, F15, step 3.3] Write $P:=A{f}^{\mathrm{\prime }}$ and $Q:=A^{\mathrm{\prime }}f$ for the fields whose coefficient columns are $A{f}^{\mathrm{\prime }}$ and $A^{\mathrm{\prime }}f$, so that $D_{t}W=P+Q$ by step 3.3 and $\mathrm{Rm}(W,T,T,W)$ refers to $W=Af$. Expanding, $⟨D_{t}W,D_{t}W⟩-\mathrm{Rm}(W,T,T,W)=⟨P,P⟩+2⟨P,Q⟩+⟨Q,Q⟩-\mathrm{Rm}(Af,T,T,Af)\mathrm{.}$ The columns of $A$ are Jacobi fields, so the Wronskian identity [F6] applied to the Jacobi fields with coefficient columns $g,h\in {\mathbb{R}}^n$ gives $⟨A^{\mathrm{\prime }}(t)g,A(t)h⟩=⟨A(t)g,A^{\mathrm{\prime }}(t)h⟩\text{for all }t\in [a,b],$ both sides being the Wronskian of two Jacobi fields evaluated at $t$, with value $0$ at $t=a$ because $A(a)=0$. Taking $g=f^{\mathrm{\prime }}$, $h=f$ turns $⟨P,Q⟩=⟨A{f}^{\mathrm{\prime }},A^{\mathrm{\prime }}f⟩$ into $⟨A^{\mathrm{\prime }}{f}^{\mathrm{\prime }},Af⟩$, the symmetric metric [F15] identifying the two mixed terms; the Jacobi equation for the columns, together with $\mathrm{Rm}(Af,T,T,Af)=⟨R(Af,T)T,Af⟩$ [F15], then gives $2⟨P,Q⟩+⟨Q,Q⟩-\mathrm{Rm}(W,T,T,W)=\frac{d}{dt}⟨A^{\mathrm{\prime }}f,Af⟩\mathrm{.}$ Indeed the derivative of $\phi (t):=⟨A^{\mathrm{\prime }}(t)f(t),A(t)f(t)⟩$ is $⟨A^{\mathrm{\prime }\mathrm{\prime }}f+A^{\mathrm{\prime }}{f}^{\mathrm{\prime }},Af⟩+⟨A^{\mathrm{\prime }}f,A^{\mathrm{\prime }}f+A{f}^{\mathrm{\prime }}⟩$, and $A^{\mathrm{\prime }\mathrm{\prime }}=-RA$ for the matrix of the curvature endomorphism, so the four terms match the left side term by term using the symmetry just proved. Therefore $⟨D_{t}W,D_{t}W⟩-\mathrm{Rm}(W,T,T,W)=⟨A{f}^{\mathrm{\prime }},A{f}^{\mathrm{\prime }}⟩+{\phi }^{\mathrm{\prime }}(t)$ on each piece of the subdivision.

Integration against a vanishing field. [F10, F11, F12, step 4.1] Integrating step 4.1 over each closed piece of the subdivision and summing, $I_{\gamma }(W,W)={\int }_a^b⟨A{f}^{\mathrm{\prime }},A{f}^{\mathrm{\prime }}⟩dt+\phi (b)-\phi (a^{+}),$ where each piece integral exists by [F11] (applied to the continuous function $\phi$ and its integrable derivative) and the sum over the pieces telescopes. At $t=b$ we have $W(b)=0$, so $\phi (b)=⟨A^{\mathrm{\prime }}(b)f(b),A(b)f(b)⟩=⟨A^{\mathrm{\prime }}(b)f(b),W(b)⟩=0$. As $t\downarrow a$, the matrix $A(t)$ satisfies $A(t)=(t-a)(I+o(1))$: its columns are $C^1$ with $A(a)=0$ and $A^{\mathrm{\prime }}(a)=I$ in the frame, and Newton–Leibniz [F11] applied to the entries gives $A(t)={\int }_a^t{A}^{\mathrm{\prime }}(s)ds=(t-a)(I+o(1))$. Since $w(a)=0$ and $w$ is $C^1$ on the first piece, likewise $w(t)=(t-a)(w^{\mathrm{\prime }}(a)+o(1))$, and Cramer's rule [F10] gives $(I+o(1){)}^{-1}=I+o(1)$; hence $f(t)=(I+o(1){)}^{-1}(w^{\mathrm{\prime }}(a)+o(1))=O(1)$ is bounded near $a$, while $A^{\mathrm{\prime }}$ is bounded on $[a,b]$ and $W(t)\to W(a)=0$. Therefore $\phi (t)=⟨A^{\mathrm{\prime }}(t)f(t),W(t)⟩\to 0$ as $t\downarrow a$, and $I_{\gamma }(W,W)={\int }_a^b⟨A(t)f^{\mathrm{\prime }}(t),A(t)f^{\mathrm{\prime }}(t)⟩dt\ge 0$ by [F12], the integrand being continuous on each piece and nonnegative.

Equality case and conclusion. [F13, step 3.2, step 5.1] Suppose $I_{\gamma }(W,W)=0$ for some $W\in {\mathcal{X}}_0(\gamma )$. By step 5.1 the integrand $⟨A{f}^{\mathrm{\prime }},A{f}^{\mathrm{\prime }}⟩$ is continuous and nonnegative on each closed piece and the sum of the piece integrals is $0$; since every piece integral is nonnegative [F12], each piece integral vanishes, and [F13] makes $⟨A(t)f^{\mathrm{\prime }}(t),A(t)f^{\mathrm{\prime }}(t)⟩=0$ for every $t$ of that piece. The metric is positive definite and $A(t)$ is invertible for $t\in (a,b]$ (step 2.1), so $f^{\mathrm{\prime }}(t)=0$ there; hence $f$ is constant on $(a,b)$ and, being continuous on $(a,b]$, constant on $(a,b]$ with value $c\in {\mathbb{R}}^n$. Then $W(t)=A(t)c$ on $(a,b]$ and, evaluating at $b$, $0=W(b)=A(b)c$ with $A(b)$ invertible, so $c=0$ and $W=0$. Therefore $I_{\gamma }(W,W)>0$ for every nonzero $W\in {\mathcal{X}}_0(\gamma )$, and step 3.2 gives $I_{\gamma }(V,V)=I_{\gamma }(J,J)+I_{\gamma }(W,W)\ge I_{\gamma }(J,J),$ with equality exactly when $W=0$, that is exactly when $V=J$. This proves both assertions.

Boundary and choice audit. [A1, F1, F2, F3, F4, F5, step 2.1, step 6.1] Every hypothesis is used at the point where it is needed: $a<b$ and the exclusion of conjugate instants on $(a,b]$ are exactly what make each ${\mathrm{\Phi }}_t$, $t\in (a,b]$, an isomorphism in step 2.1, and the exclusion at $t=b$ is what makes the endpoint-value problem in step 3.1 uniquely solvable and the constant $c$ in step 6.1 vanish. If $\gamma$ is constant, then $T=0$ and the conjugacy hypothesis is automatic: [F3] states that ${\mathcal{K}}_{\gamma }(a,b)=\{0\}$ for a constant geodesic, and [F2] makes the curvature term of the Jacobi equation and of the index-form integrand vanish (the Jacobi-field convention gives $D_t^{2}J=0$ when $\dot{\gamma }=0$, and $R$ is $C^{\mathrm{\infty }}$-linear in its second slot, so $R(\cdot ,0)=0$), so all steps above apply verbatim and the constant case is included; the same steps never divide by $T$ or by the speed. In dimension zero $T_{\gamma (a)}M=\{0\}$, so $u=w=0$, $V=0$, $J=0$ and both sides of the inequality are $0$. Dimension one and non-unit speed are unconstrained by any step. Included endpoints use the one-sided conventions of [F1], [F4] and [F5]. Exactly the declared ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through the index-form, Jacobi-field, Wronskian and integration-by-parts suppliers [A1]; the construction of the fields $J_v$ uses the uniqueness statement of [F1] rather than any selection, the parallel frame is fixed once by [F9], and the finite-dimensional linear algebra of steps 2.1, 3.3 and 6.1 makes no choice. No converse is claimed: the lemma asserts an inequality and its equality case, not that failure at some $t\in (a,b]$ produces a nonzero $W$ with $I_{\gamma }(W,W)=0$. $\mathrm{\square }$