Question: It is well known that if $\varphi:M\to \tilde{M}$ is an isometry between Riemannian manifolds, then $\varphi$ maps geodesics of $M$ to geodesics of $\tilde{M}$. I am wondering if it is sufficient that $\varphi$ is only a local isometry.
To prove the case of global isometries, one first prove that if $\gamma$ is a smooth curve on $M$ and $V$ is a vector field along $\gamma$ (denoted $V\in\mathscr{T}(\gamma)$), then
$$\varphi_\ast D_tV=\tilde{D}_t(\varphi_\ast V),$$
where $D_t$ and $\tilde{D}_t$ are the covariant derivatives of $\gamma$ and $\varphi\circ\gamma$, respectively. Does this formula also holds for local isometries?
I managed to reduce the problem to the following result (I think), but I am not sure how to prove it.
Let $\gamma:I\to M$ be a curve, $U\subseteq M$ and open set, and $J\subseteq I$ a subinterval such that $\gamma(J)\subseteq U$. Let $\bar{\gamma}=\gamma|_J$. Then, $\bar{\gamma}$ is a curve in the open submanifold $U$, so it has a covariant derivative $\bar{D}_t$ in $U$. Now, I want to show that if $V$ is a vector field along $\gamma$, and $\bar{V}:J\to TU$ its restriction as a vector field along $\bar{\gamma}$ in $U$, then
$$\iota_\ast(\bar{D}_t\bar{V}(t))=D_tV(t),\quad\forall t\in J,$$
where $\iota:U\to M$ is inclusion. How do I prove this?
If the above is true then we have the following proof.
Let $\gamma:I\to M$ be a geodesic, and let $\tilde{\gamma}=\varphi\circ\gamma$. We want to show that $\tilde{D}_t\dot{\tilde{\gamma}}(t_0)=0$ for all $t_0\in I$. Fix $t_0\in I$ and let $U\subseteq M$ be a neighbourhood of $\gamma(t_0)$ such that $\varphi$ restricts to an isometry $\bar{\varphi}:U\to \tilde{U}$, where $\tilde{U}\subseteq\tilde{M}$ is open. Now, $\gamma^{-1}(U)$ is an open set containing $t_0$, so let $J\subseteq I$ be its connected component containing $t_0$. We then have a curve $\bar{\gamma}:J\to U$ in $U$. Now, note that
$$\bar{\dot{\tilde{\gamma}}}=\overline{\varphi_\ast\dot{\gamma}}=\bar{\varphi}_\ast\dot{\bar{\gamma}},$$
so by the above result
$$
\begin{align}
\tilde{D}_t\dot{\tilde{\gamma}}(t_0) &= \iota_\ast(\bar{\tilde{D}}_t(\bar{\varphi}_\ast\dot{\bar{\gamma}})(t_0)) \\
&= \iota_\ast\bar{\varphi}_\ast\bar{D}_t\dot{\bar{\gamma}}(t_0)\\
&=0
\end{align}
$$
since
$$
\iota_\ast\bar{D}_t\dot{\bar{\gamma}}(t_0)=D_t\dot{\gamma}(t_0)=0,
$$
and $\iota_\ast$ is injective.
Best Answer
This is just an elaboration of the comment of @JohnMa . In fact all the concepts involved in you question are local, so they are compatible with restricting to an open subset. (You can restrict the metric and the Levi Civita connection and the restriction of the connection is the Levi-Civita connection of the restricted metric. This follows from uniqueness of the Levi-Civita connection, since the restricted connection is metric and torsion-free. This also is the argument needed for the missing step in your proof. Then you get that connections of the initial metric are connections for the restricted metric and so on.) Hence local isometries map geodesics to geodesics.