Existence and uniqueness of jacobi fields from initial data

If $M$ is empty there is no supplied geodesic. If $n=0$, then $T_{\gamma (a)}M=\{0\}$ and the only field along $\gamma$ is zero by [F1]; it is the unique Jacobi field with the only possible initial data. Assume $n>0$. The finite-dimensional inner-product space $T_{\gamma (a)}M$ has an orthonormal basis $e_1,\dots ,e_n$ by [F7].

For each $e_i$, [F2] gives a unique parallel field $E_i$ on all of $I$ with $E_i(a)=e_i$. By [F3], $D_t{E}_i=0$. For any $t\in I$, restrict to the compact segment between $a$ and $t$; [F4] shows the fields there remain orthonormal. Thus $E_1(t),\dots ,E_n(t)$ are an orthonormal basis of $T_{\gamma (t)}M$ for every $t$. Only the given finite basis is extended.

Put $T=\dot{\gamma }$ and define the smooth matrix $A(t)$ by $A_{ij}(t)=\mathrm{Rm}(E_j(t),T(t),T(t),E_i(t))\mathrm{.}$ Writing a field as $J(t)={\sum }_j{x}_j(t)E_j(t)$, [F3] gives $D_{t}J={\sum }_j{x}_j^{\mathrm{\prime }}{E}_j$ and $D_t^{2}J={\sum }_j{x}_j^{\mathrm{\prime }\mathrm{\prime }}{E}_j$. The Jacobi equation [F1] therefore becomes $x^{\mathrm{\prime }\mathrm{\prime }}+A(t)x=0$. With $y=(x,x^{\mathrm{\prime }})$ it is the first-order system $y^{\mathrm{\prime }}=C(t)y,C(t)=\left(\begin{array}{cc}0& I_n\\ -A(t)& 0\end{array}\right)\mathrm{.}$ The coefficient $C$ is smooth by [F5].

Let $\xi ,\eta \in {\mathbb{R}}^n$ be the coordinates of $v,w$ in the initial frame and put $y_0=(\xi ,\eta )$. For each $t\mathrm{\ne }a$, the closed segment $K_t$ between $a$ and $t$ is a compact interval contained in $I$. Apply [F6] to $C{\mathrm{\mid }}_{K_t}$ and the $2n\times 2n$ initial matrix whose first column is $y_0$ and whose other columns are zero. Its first column is the unique vector solution $y_t$ with value $y_0$ at $a$; uniqueness for vector solutions follows by placing any such solution in the first column of a matrix with other columns zero. If $s,t\in I$, their two segments lie in the compact interval spanned by $a,s,t$; uniqueness on that interval makes $y_s$ and $y_t$ agree wherever both are defined. These unique compact-interval solutions therefore glue to a well-defined solution $y$ on all of $I$. Around each interior time it agrees with a solution on a compact segment extending on both sides, and at included endpoints it has the corresponding one-sided derivatives. Smoothness follows from the smooth coefficient system.

Write $y=(x,p)$ and set $J(t)={\sum }_i{x}_i(t)E_i(t)$. The system gives $p=x^{\mathrm{\prime }}$ and $p^{\mathrm{\prime }}=-Ax$, so [F3] yields $D_{t}J={\sum }_i{p}_i{E}_i,D_t^{2}J=-{\sum }_i(Ax{)}_i{E}_i=-R(J,T)T\mathrm{.}$ Thus [F1] makes $J$ Jacobi on all of $I$. Its initial coordinates are $x(a)=\xi$ and $p(a)=\eta$, so $J(a)=v$ and $D_{t}J(a)=w$. This proves existence, including when $\gamma$ is constant, for which $A=0$.

If $\tilde{J}$ is another Jacobi field with the same initial data, its coordinates in the same parallel frame satisfy the same system and initial value. Uniqueness in [F6] on each compact segment between $a$ and $t$ gives $\tilde{J}(t)=J(t)$ for every $t\in I$, so $\tilde{J}=J$. Included endpoints use the one-sided convention from [F1]. The finite basis and the unique gluing use no choice axiom; the argument imposes no compactness or completeness condition on $I$ or $M$.