The question is asking to me prove that: Consider a fixed point set $F=\{x \in S : f(x)=x\}$ in a smooth Riemannian surface with $f:S \rightarrow S$ be an isometry. If $F$ is a smooth 1-manifold, then we can define a smooth curve (parametrized by arc length) that contained in $F$ by $\gamma :I \rightarrow F$ where $I$ is an open interval in $\mathbb{R}$. Than $\gamma$ is a geodesic.
What I have done so far is. Since geodesic is uniquely defined by initial position and velocity. So we can define a geodesic
$$g:I \rightarrow S \text{ with } g(0)=p \text{ and }g'(0)=\gamma'(p)$$
where $p \in F$.
And I know isometry of a geodesic is a geodesic, so $f(g)$ is a geodesic. Isn't the next step is to show $f(g) = \gamma$ ? or my approach isn't on the right track.
Best Answer
Yes, your approach is on the right track. A way to continue:
Since the restriction of $f$ to $F$ is the identity, for every $p\in F$ the restriction of the derivative $D_pf$ to $F$ is the identity. Therefore, $(f\circ g)'(0)=g'(0)$, hence $f\circ g=g$ by the uniqueness of geodesics. By definition of $F$, this gives $g\subset F$.
Both $\gamma$ and $g$ are open subsets of $F$, so locally they are the same thing. Since the property of being a geodesic is local, $\gamma$ is a geodesic.