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.


1 Answer 1


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 $\mathrm{D}_{\dot{\gamma}(0)}J = \eta$... $\endgroup$
    – pitchounet
    Jul 22, 2013 at 8:43
  • $\begingroup$ Just let $\delta t \rightarrow 0$ $\endgroup$
    – Xipan Xiao
    Jul 22, 2013 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? $\endgroup$ Apr 28, 2015 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, 2015 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. $\endgroup$ Apr 28, 2015 at 20:25

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