Suppose $(M,g)$ is a Riemannian manifold, $d$ the induced metric (by geodesic lengths), and $f$ is an isometry on $M$ in the metric space sense. I want to prove $f$ is a diffeomorphism using the exponential map.
But the first thing is I want to prove smoothness. One fact I know is that $f$ sends geodesics to geodesics. So here I go.
For each $p$, consider some $\epsilon>0$ so that we have a geodesic ball $\exp_p(\bar B_\epsilon(0))$ around $p$ (in which $\bar B$ means closed balls in $T_pM$). Within this ball, I want to show $f$ is smooth. For each $q\in \exp_p(\bar B_\epsilon(0))$ there exists a unique $v_q\in T_pM$ so that $\|v_q\|\le\epsilon$ and that $\exp_p(v_q)=q$. Thus
$$f(q)=\exp_{F(p)}(d(f)_p v_q)$$
since $d(f)_p$ preserves lengths in tangent vectors. (Here $df$ denotes the differential of $f$, not to be confused with the metric $d$.)
However, I'm having trouble showing why $q\mapsto v_q\in T_pM$ is smooth. Is there any explicit expression of this map? (I know it is continuous of course (and indeed an isometry) since $d(p,q)=\|v_q\|$. But how to show the direction of $v_q$ also varies smoothly with $q$?)
Best Answer
There is a very explicit expression of the map $q\mapsto v_q$: it's the inverse of the exponential map $\exp_p^{-1}$ (well you write $\exp_p (v_q) = q$....). In particular, it is smooth as $\exp_p$ is a local diffeomorphism (I assume you know how to prove that).
Note that $q\mapsto v_q$ is indeed NOT an isometry. Your formula $d(p, q) = \| v_q\|$ holds true, but it works only for $p$. If $q_1, q_2$ are both not $p$, you do not have
$$ d(q_1, q_2) = \| v_{q_1} - v_{q_2}\|.$$
Indeed the metric $g_p$ defined on $B_\epsilon(0) \subset T_pM$ is flat, so in general $\exp_p$ cannot be an isometry.