# Isometry and geodesic

Let $F: M \rightarrow N$ an isometry and $M,N$ two riemannian manifold. How can I prove that the set of fixed points of F isometry (among riemannian manifold) is a geodesic? In general is it a curve?

• How can a map $M\to N$ have fixed points if $N$ is not a subset of $M$? Commented Mar 1, 2013 at 12:21
• Consider $M=N=S^n\subset\mathbb{R}^n$ the standard $n$-dimensional sphere. Then any isometry $F:M\to N$ is the rotation in $\mathbb{R}^n$, and the fixed point set of $F$ consists only two points ("north pole and south pole"). So in general it is not a curve. But it depends on what you mean by "in general".
– Paul
Commented Mar 1, 2013 at 13:03
• @Paul: For even dimensional spheres, all isometries fix (at least) two points, but on odd dimensional spheres, an isometry need not fix any points. Also note that it's easy to construct isometries of spheres having fixed point sets of any dimension. Commented Mar 1, 2013 at 13:07
• @JasonDeVito: Oh yeah! You are right. I was thinking about the $2$-dimensional sphere, without thinking deeply about the higher dimensional cases. I should be more careful. Thanks for pointing out the mistake.
– Paul
Commented Mar 1, 2013 at 13:22

Suppose $S \subset \textrm{Iso}(M,g)$ is a set of isometries. Then each connected component of the fixed point set is a totally geodesic submanifold $X \subset M$. Here totally geodesic means that the second fundamental form of $X$ in $M$ is identically zero.