The normal index form before the first conjugate instant is positive definite

(The Jacobi operator preserves the normal bundle and is selfadjoint.) Let $J,K$ be normal fields. For the bilinear form $\mathcal{R}(X,Y,Z,W):=g(R(X,Y)Z,W)$ the symmetries of [L2] read $\mathcal{R}(X,Y,Z,W)=-\mathcal{R}(Y,X,Z,W)=-\mathcal{R}(X,Y,W,Z)=\mathcal{R}(Z,W,X,Y)$. Hence $g(C_T(J),T)=\mathcal{R}(T,J,T,T)=-\mathcal{R}(T,J,T,T)=0$, so $C_T(J)$ is again normal by [L1]. For selfadjointness, $g(C_T(J),K)=\mathcal{R}(T,J,T,K)=-\mathcal{R}(T,J,K,T)=-\mathcal{R}(K,T,T,J)=\mathcal{R}(T,K,T,J)=g(C_T(K),J)$, using the last-pair skew, pair exchange and first-pair skew in that order; by symmetry of $g$ this is $g(J,C_T(K))$, so $C_T$ is $g$-selfadjoint.

(Matrix form in a parallel frame.) Let $E_1,E_2,E_3$ be a parallel orthonormal frame of $N_{\gamma }$ by [L4] and write $J={\sum }_i{J}^i{E}_i$, $K={\sum }_i{K}^i{E}_i$. Then $D_{s}J={\sum }_i(J^i{)}^{\mathrm{\prime }}{E}_i$ and $g(D_{s}J,D_{s}K)={\sum }_i(J^i{)}^{\mathrm{\prime }}(K^i{)}^{\mathrm{\prime }}$ by orthonormality; and $C_T(J)={\sum }_j{J}^j{C}_T(E_j)$, so $g(C_T(J),K)={\sum }_{i,j}{J}^j{C}^i{}_j{K}^i$ with $C^i{}_j=g(C_T(E_j),E_i)$, the coefficient functions being smooth along $\gamma$. By step 1.1 the matrix $C(s)$ is symmetric for every $s$.

(The fundamental solution.) By [L5] applied to the first-order system for $(A,A^{\mathrm{\prime }})$ with coefficient matrix $\left(\begin{array}{cc}0& I\\ C& 0\end{array}\right)$, the problem $A^{\mathrm{\prime }\mathrm{\prime }}=CA$, $A(a)=0$, $A^{\mathrm{\prime }}(a)=I_3$ has a unique solution on $[a,b]$. For $v\in {\mathbb{R}}^3$ the field $s\mapsto A(s)v$ in the frame is the Jacobi field along $\gamma$ with initial data $(0,v)$: its coefficient vector satisfies $(Av{)}^{\mathrm{\prime }\mathrm{\prime }}=CAv$ and the initial data are $(Av)(a)=0$, $(Av{)}^{\mathrm{\prime }}(a)=v$, which is the Jacobi initial-value problem of [L3] in the frame of step 2.1. Hence $A(s)v=0$ for a nonzero $v$ would exhibit a nonzero Jacobi field vanishing at $a$ and $s$, contradicting the absence of conjugate instants; so $A(s)$ is invertible for every $s\in (a,b]$.

(The Wronskian and $\mathrm{\Theta }$.) Define $W(s):=A(s{)}^T{A}^{\mathrm{\prime }}(s)-A^{\mathrm{\prime }}(s{)}^{T}A(s)$, a $3\times 3$ matrix function on $[a,b]$. Then $W^{\mathrm{\prime }}=A^{\mathrm{\prime }T}{A}^{\mathrm{\prime }}+A^T{A}^{\mathrm{\prime }\mathrm{\prime }}-A^{\mathrm{\prime }\mathrm{\prime }T}A-A^{\mathrm{\prime }T}{A}^{\mathrm{\prime }}=A^{T}CA-(CA{)}^{T}A=0$ because $C$ is symmetric by step 2.1, and $W(a)=0$; hence $W=0$ on $[a,b]$. On $(a,b]$ set $\mathrm{\Theta }:=A^{\mathrm{\prime }}{A}^{-1}$, which is well defined by step 3.1. The identity $W=0$ gives $A^T{A}^{\mathrm{\prime }}=A^{\mathrm{\prime }T}A$, and multiplying on the left by $(A^T{)}^{-1}$ and on the right by $A^{-1}$ gives ${\mathrm{\Theta }}^T=\mathrm{\Theta }$; differentiability of $A$ and $A^{-1}$ gives ${\mathrm{\Theta }}^{\mathrm{\prime }}=A^{\mathrm{\prime }\mathrm{\prime }}{A}^{-1}-A^{\mathrm{\prime }}{A}^{-1}{A}^{\mathrm{\prime }}{A}^{-1}=C-{\mathrm{\Theta }}^2$.

(The factorization identity.) Let $J$ be a continuous piecewise-$C^1$ normal field vanishing at $a$ and $b$, with coefficient vector $j(s)$ in the frame of step 2.1. On each smooth piece, $\mathrm{\mid }{j}^{\mathrm{\prime }}-\mathrm{\Theta }j{\mathrm{\mid }}^2=\mathrm{\mid }{j}^{\mathrm{\prime }}{\mathrm{\mid }}^2-2j^{\mathrm{\prime }T}\mathrm{\Theta }j+j^T{\mathrm{\Theta }}^{2}j$, because $\mathrm{\Theta }$ is symmetric. For $\epsilon >0$ sufficiently small, integrate by parts on the smooth pieces of $[a+\epsilon ,b]$; the interior endpoint terms cancel because $j$ and $\mathrm{\Theta }$ are continuous. The identity $(j^T\mathrm{\Theta }j{)}^{\mathrm{\prime }}=2j^{\mathrm{\prime }T}\mathrm{\Theta }j+j^T{\mathrm{\Theta }}^{\mathrm{\prime }}j$ gives ${\int }_{a+\epsilon }^b-2j^{\mathrm{\prime }T}\mathrm{\Theta }jds=-[j^T\mathrm{\Theta }j{]}_{a+\epsilon }^b+{\int }_{a+\epsilon }^b{j}^T{\mathrm{\Theta }}^{\mathrm{\prime }}jds\mathrm{.}$ At $b$ the boundary term vanishes because $j(b)=0$. At $a$, one-sided $C^1$ regularity gives $j(s)=(s-a)j^{\mathrm{\prime }}(a)+o(s-a)$ and $j^{\mathrm{\prime }}(s)=j^{\mathrm{\prime }}(a)+o(1)$. By [L6], $j^T\mathrm{\Theta }j=O(s-a)\to 0$ and $\mathrm{\Theta }(s)j(s)=j^{\mathrm{\prime }}(a)+o(1)$, so $j^{\mathrm{\prime }}(s)-\mathrm{\Theta }(s)j(s)\to 0$ on the first smooth piece as $s\downarrow a$. Define the factorized integrand at $a$ to be this limit squared, namely $0$; it is then piecewise continuous on $[a,b]$. Using ${\mathrm{\Theta }}^{\mathrm{\prime }}=C-{\mathrm{\Theta }}^2$ from step 4.1 and letting $\epsilon \downarrow 0$ in the integrated identity yields $I(J)={\int }_a^b(\mathrm{\mid }{j}^{\mathrm{\prime }}{\mathrm{\mid }}^2+j^{T}Cj)ds={\int }_a^b\mathrm{\mid }{j}^{\mathrm{\prime }}-\mathrm{\Theta }j{\mathrm{\mid }}^{2}ds\mathrm{.}$ This is the factorization in (c), with its right-hand integrand continuously extended at $a$.

(Positivity.) By step 5.1, $I(J)={\int }_a^b\mathrm{\mid }{J}^{\mathrm{\prime }}-\mathrm{\Theta }J{\mathrm{\mid }}^{2}ds\ge 0$. If $I(J)=0$, nonnegativity gives zero integral on every compact subinterval inside a smooth piece; [L7] makes the continuous integrand zero there, so $J^{\mathrm{\prime }}=\mathrm{\Theta }J$ on each open piece. Then $(A^{-1}J{)}^{\mathrm{\prime }}=A^{-1}{J}^{\mathrm{\prime }}-A^{-1}{A}^{\mathrm{\prime }}{A}^{-1}J=A^{-1}(J^{\mathrm{\prime }}-\mathrm{\Theta }J)=0$ on $(a,b]$, so $A^{-1}J$ is constant on each open piece; its continuity across the finitely many joints makes this one constant $v$ on $(a,b]$, giving $J(s)=A(s)v$; since $J(b)=0$ and $A(b)$ is invertible by step 3.1, $v=0$ and $J=0$. Conversely $J=0$ gives $I(J)=0$. This proves the positivity statement, and with steps 1.1, 2.1, 3.1 and 5.1 the lemma.