Let $M$ be a riemannian manifold (let $\left\langle \cdot,\cdot \right\rangle_{p}$ be the scalar product on $T_{p}M$). Let $p \in M$ and $\xi \in T_{p}M$. We consider the geodesic $\gamma \, : \, t \, \mapsto \, \exp_{\gamma(0)}(t\xi)$ where $\gamma(0)=p$.

Let $\eta \in T_{p}M$. We can introduce a "small perturbation" of the geodesic $\gamma$ : $\gamma_{\varepsilon}(t) = \exp_{\gamma(0)}(t(\xi+\varepsilon \eta))$. Let $f \, : \, (t,\varepsilon) \, \mapsto \, \exp_{\gamma(0)}(t(\xi + \varepsilon \eta))$. We definie $J(t) = \frac{\partial f}{\partial \varepsilon}(t,0)$. Then, $\forall t, \, J(t) \in T_{\gamma(t)}M$. $J$ is a Jacobi vector field.

Let $P_{\gamma,0,t} \, : \, T_{p}M \, \rightarrow \, T_{\gamma(t)}M$ be the parallel transport along $\gamma$. I have read that we have the following approximation : for a "small" $\delta t$,

$$ \frac{J(\delta t)}{\delta t} - P_{\gamma, 0, \delta t}(\eta) = o(\delta t) $$

The Jacobi vector field $J$ gives a first order approximation of the parallel transport of $\eta$ along the geodesic $\gamma$. I would like to prove this approximation but I don't really know how to start...

Considering that $J(t) = \frac{\partial f}{\partial \varepsilon}(t,0)$, we can easily show that

$$ J(t) = \mathrm{D}_{t \xi} \left( \exp_{\gamma(0)} \right) \cdot (t \eta) $$

So that, $J(\delta t) = \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot (\delta t \eta) = \delta t \, \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta$. Then,

$$ \frac{J(\delta t)}{\delta t} = \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta$$

We want to prove that $\mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta - P_{\gamma,0,\delta t}(\eta)$ goes to $0$ as $\delta t \, \rightarrow \, 0$. One idea I had was to consider

$$ \Vert \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta - P_{\gamma,0,\delta t}(\eta) \Vert_{\gamma(\delta t)}$$ since the parallel transport is a linear isometry, $\Vert P_{\gamma,0,\delta t}(\eta) , P_{\gamma,0,\delta t}(\eta) \Vert_{\gamma(\delta t)} = \Vert \eta \Vert_{p}$. But it doesn't seem to be very helpful. Can anyone give me a hint ?

Thank you for your help.


This is to say $D_{\dot{\gamma}(0)} J = \eta$. To prove, in a small neighborhood U of p, use the geodesic normal coordinates: choose an orthonormal basis $\{e_1,\dots, e_n\}$ of $T_pM$, and $\{\alpha^1,\dots,\alpha^n\}$ the dual basis, then $x^i=\alpha^i(\exp_{\gamma(0)}^{-1})$ gives the geodesic normal coordinates in U. And we have: $$\langle \frac{\partial}{\partial{x^i}}(0), \frac{\partial}{\partial{x^j}}(0)\rangle=\delta_{ij}$$ $$D_{\frac{\partial}{\partial{x^i}}(0)}\frac{\partial}{\partial{x^j}}=0$$ Let $\xi=\xi^ie_i$ and $\eta=\eta^ie_i$, and let $\bar{\xi}=(\xi^1,\dots,\xi^n)$ and $\bar{\eta}=(\eta^1,\dots,\eta^n)$, we have $f(t,\varepsilon)=t(\bar{\xi}+\varepsilon\bar{\eta})$, and $J(t)=t\eta^i\frac{\partial}{\partial{x^i}}$. So $$D_{\dot\gamma(0)}J=\eta^i\frac{\partial}{\partial{x^i}}+tD_{\dot\gamma(0)}(\eta^i\frac{\partial}{\partial{x^i}})=\eta^i\frac{\partial}{\partial{x^i}}=\eta$$

  • 1
    $\begingroup$ Thanks for your answer ! I don't understand why it suffices to prove that $\mathrm{D}_{\dot{\gamma}(0)}J = \eta$... $\endgroup$ – jibounet Jul 22 '13 at 8:43
  • $\begingroup$ Just let $\delta t \rightarrow 0$ $\endgroup$ – Xipan Xiao Jul 22 '13 at 14:18
  • $\begingroup$ Xipan, I'm also not sure I understand why $D_{\dot{\gamma}(0)} J = \eta$is enough to prove the limit is $0$, i.e; for a "small" $\delta t$, $$ \frac{J(\delta t)}{\delta t} - P_{\gamma, 0, \delta t}(\eta) = o(\delta t) $$ Could you please be more explicit? $\endgroup$ – Mathmath Apr 28 '15 at 16:30
  • $\begingroup$ Note one can define the connection D by the transport P, see "Recovering the connection from the parallel transport" from en.wikipedia.org/wiki/Parallel_transport. Also note that J(0)=0. $\endgroup$ – Xipan Xiao Apr 28 '15 at 17:38
  • $\begingroup$ I'm familiar with both of the above, but sorry still don't see the how that means the limit = 0. I did it myself, I could only prove that the limit, if exists is bounded, i.e. $O(t)$, but couldn't prove $o(t)$. Would appreciate if you be more explicit. $\endgroup$ – Mathmath Apr 28 '15 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.