Check that a curve is a geodesic.

Question

Suppose $M$ is a two-dimensional manifold. Let $\sigma:M \rightarrow M$ be an isometry such that $\sigma^2=1$. Suppose that the fixed point set $\gamma=\{x \in M| \sigma(x)=x\}$ is a connected one-dimensional submanifold of $M$. The question asks to show that $\gamma$ is the image of a geodesic.

Answer 1

Let $N=\{x\in M:\sigma(x)=x\}$ and fix $p\in N$.

Exercise 1: Prove that either $1$ or $-1$ is an eigenvalue of $d\sigma_p:T_p(M)\to T_p(M)$.

Exercise 2: Prove that if $v\in T_p(M)$ is an eigenvector of $d\sigma_p:T_p(M)\to T_p(M)$ of sufficiently small norm, then the unique geodesic $\gamma:I\to M$ for some open interval $I\subseteq \mathbb{R}$ such that $\gamma(0)=p$ and $\gamma'(0)=v$ has image contained in $N$. (Hint: isometries of Riemannian manifolds take geodesics to geodesics.)

I hope this helps!