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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.

SamC
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1 Answers1

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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.