Q) If $(M,g)$ is a Riemannian manifold, $p \in M$ and $f \, : \, M \, \rightarrow \, M$ an isometry such that $\mathrm{D}_{p}f \cdot v = v$ for some $v \in T_{p}M$ then, for the geodesic $\gamma \, : \, [a,b] \, \rightarrow \, M$ with $\gamma(0) =p$ and $\gamma'(0)=v$, then prove that we have $f \circ \gamma = \gamma$
My work so far: Once I have proven that $f \circ \gamma$ is a geodesic, then by uniqueness of geodesic it automatically proves that its equal to $\gamma$. I have shown that $f \circ \gamma$ is locally length minimizing because $\gamma$ is a geodesic and they have the same integrated length due to isometry (chosen over any small interval). So the I have proven the last part of the question, but need to see that $f \circ \gamma$ would indeed be a geodesic initialized at $p$
My question is this: how do I show that $f \circ \gamma$ has constant speed and initialzed at the same $p$ that $\gamma$ is?
Best Answer
First note that $D_pf$ is a linear mapping from $T_pM$ to $T_{f(p)}M$. Since $v\in T_pM$ and $D_pf$ maps $v$ to $v$, $(D_pf)v=v\in T_pM\cap T_{f(p)}M$, which implies $f(p)=p$. This proves $f\circ\gamma(0)=f(\gamma(0))=f(p)=p$.
Second, the initial velocity of $f\circ\gamma$ is $(f\circ\gamma)'(0)=(D_{\gamma(0)}f)\gamma'(0)=(D_pf)v=v$.