The dimension of Jacobi field

Question

The Jacobi field is defined as $$J^{''}(t)+R(\gamma^{'}(t),J(t))\gamma^{'}(t)=0$$ since it is a system of $2n$ order, spanned by $\{J,J^{'}\}$, so it is dimension of $2n$.

But I don't get if we add a condition of $J(0)=0$, then the dimension is $n$. ( I can't see why the system is immeddiately reduced to $n$ linear independent equations which is expained by our teaching assistant ).

Similarly, if we set the Jacobi field normal, that is $\langle J,\gamma ^{'}\rangle=0$, then the dimension of normal Jacobi field is $2(n-1)$.

It seems to be a trivial consequence, but I really don' see it. Can anyone else help explain it ?

Answer 1

A Jacobi field $J(t)$ determined by initial values of $J$ and $J'$ at a point. This proved by simply rewrite the Jacobi equation in terms of a parallel orthonormal frame along $\gamma$ and observe that the resulting equation is a system of linear second-order ODEs. For definiteness, here is a reference for this claim :

(Proposition 10.2 John Lee's Introduction to Riemannian Manifolds, 2nd ed) Let $M$ be a Riemannian or pseudo-Riemannian manifold. Suppose $I \subseteq \mathbb{R}$ is an interval, $\gamma : I \to M$ is a geodesic, $a \in I$ and $\gamma(a) = p$. For every pair of vectors $v,w \in T_pM$, there is a unique Jacobi field $J$ along $\gamma$ satisfying initial conditions $J(a)=v$ and $J'(a)=w.$

So if we denote $\mathfrak{X}(\gamma)$ as the space of smooth vector field along $\gamma$ and $\mathscr{J}(\gamma) \subseteq \mathfrak{X}(\gamma)$ be the set of Jacobi fields along $\gamma$. Then above proposition says that the map $ P : \mathscr{J}(\gamma) \to T_pM \oplus T_pM $ defined as
$$ J \mapsto (J(a), J'(a)). $$ is an isomorphism. Therefore it follows that $\text{dim }\mathscr{J}(\gamma) = 2n$.

Therefore the subspace of $\mathscr{J}(\gamma)$ that satisfy $J(a)=0$ (we can take $a=0$ from the beginning if we want) is just $P^{-1}(\{0\} \oplus T_pM)$ which is clearly have dimension $n$ since $P$ is an isomorphism.

Similarly for the subspace of normal Jacobi field, denoted by $\mathscr{J}(\gamma)^{\perp} \subseteq \mathscr{J}(\gamma)$ :

if $J \in \mathscr{J}(\gamma)^{\perp}$, then $\langle J(a), \gamma'(a)\rangle = 0$. Since $T_pM = \langle \gamma'(a)\rangle \oplus \langle \gamma'(a)\rangle^{\perp}$, this means $J(a) \in \langle \gamma'(a)\rangle^{\perp}$. Normal Jacobi field $J$ also satisfy $\langle J'(a), \gamma'(a) \rangle=0$. This also lead to $J'(a) \in \langle \gamma'(a)\rangle^{\perp}$. Hence $$ \mathscr{J}(\gamma)^{\perp} = P^{-1} \big( \langle \gamma'(a)\rangle^{\perp} \oplus \langle \gamma'(a)\rangle^{\perp} \big) $$
this implies $$ \text{dim }\mathscr{J}(\gamma)^{\perp} = \text{dim }\langle \gamma'(a)\rangle^{\perp} + \text{dim } \langle \gamma'(a)\rangle^{\perp} = 2(n-1). $$

Sorry for the late answer by the way.