Let $f$ be an isometry (i.e a diffeomorphism which preserves the Riemannian metrics) between Riemannian manifolds $(M,g)$ and $(N,h).$
One can argue that $f$ also preserves the induced metrics $d_1, d_2$ on $M, N$ from $g, h$ resp. that is, $d_1(x,y)=d_2(f(x),f(y))$ for $x,y \in M.$ Then, it's easy to show that $f$ sends geodesics on $M$ to geodesics on $N,$ using the length minimizing property of geodesics and that $f$ is distance-preserving.
My question,
Is it possible to derive the result without using the distance-preserving property of isometries, by merely the definition?
What I have found so far;
Let $\gamma : I \to M$ be a geodesic on $M$, i.e. $\frac{D}{dt}(\frac{d\gamma}{dt})=0,$ where $\frac{D}{dt}$ is the covariant derivative and $ \frac{d\gamma}{dt}:=d\gamma(\frac{d}{dt}).$ Let $t_0 \in I,$ we have to show that $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt})=0$ at $t=t_0,$ or $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0.$
We also know that
$$ \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}= \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}.$$
Since $\frac{d}{dt} \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=2 \langle \frac{D}{dt}(\frac{d \gamma}{dt}|_{t=t_0}),\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=0,$ therefore
$$\frac{d}{dt} \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}=$$
$$2\langle \frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle _{f(\gamma(t_0))}=0.$$
How can I conclude from $\langle\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}) \rangle _{f(\gamma(t_0))}=0$ that $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0?$
Best Answer
Your calculation looks like an attempt to prove the naturality of the Levi-Civita connection, the fact that @Zhen Lin implicitly points to. In the settings of the question it can be stated as $$ \nabla^g_X{Y}=f^* \left( \nabla^{(f^{-1})^* g}_{\operatorname{d}f(X)} \operatorname{d}f(Y) \right) $$
Notice also that in fact you are using two different connections: one for vector fields along $\gamma \colon I \rightarrow M$ induced from $\nabla^g$ on $M$, and another one for vector fields along $f \circ \gamma \colon I \rightarrow N$ induced from $\nabla^h$ on $N$. Due to the naturality property they agree, and it may be helpful to distinguish $D^g_t:=\nabla^g_{\frac{d}{dt}{\gamma}}$ and $D^h_t:=\nabla^h_{\frac{d}{dt}(f \circ \gamma)}$ in the present calculation. Indeed, using $$\frac{\operatorname{d}}{\operatorname{d}t}(f\circ \gamma)=\operatorname{d}f(\frac{\operatorname{d}}{\operatorname{d}t}\gamma) $$ we get $$ D^h_t{\frac{\operatorname{d}(f\circ\gamma)}{\operatorname{d}t}} = f_* \left( D^g_t{\frac{\operatorname{d}\gamma}{\operatorname{d}t}} \right) =0 $$