# isometry and exponential map

I got stuck on the following questions. Can anyone give me idea how to proceed?

Suppose $M$ is a Riemannian manifold and $\phi: M \to M$ an isometry map. If $\phi(p)=p$ and $\phi(q)=q$ prove that $(d\phi)_p exp_p^{-1}(q)=exp_p^{-1}(q)$. You can assume that $exp_p$ is well defined on its inputs.

Thanks!

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What does $\phi$ do to geodesics between $p$ and $q$? –  Neal Nov 5 '12 at 4:00
It is identity on the geodesic. –  dmm Nov 5 '12 at 4:01

I think I figured it out. So as $\phi$ is isometry it fixes the geodesic through $p,q$. Now, let $v=exp_p^{-1}(q)$. We have that $exp_p(tv)$ is the geodesic through connecting $p$ and $q$. By differentiation of $\phi \circ exp_p (tv)=exp_p(tv)$ we get
$$(d\phi)_{exp_p(tv)}(d \: exp_p)_{tv}v=(d \: exp_p)_{tv} v.$$
At $t=0$ we know $(d \: exp_p)_{0}=Id$ and $exp_p(0)=p$, so we get $(d\phi)_pv=v$, which is the required equality.
What if there is more than one geodesic between $p$ and $q$? For a concrete example, let $M$ be a sphere, $p$ and $q$ be the north and south poles, and $\phi$ be a rotation fixing the north and south poles. –  Jason DeVito Nov 5 '12 at 4:18