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?
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asked Mar 1, 2013 at 12:02
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4$\begingroup$ How can a map $M\to N$ have fixed points if $N$ is not a subset of $M$? $\endgroup$– Emanuele PaoliniCommented Mar 1, 2013 at 12:21
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$\begingroup$ 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". $\endgroup$– PaulCommented Mar 1, 2013 at 13:03
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3$\begingroup$ @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. $\endgroup$– Jason DeVito - on hiatusCommented Mar 1, 2013 at 13:07
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$\begingroup$ @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. $\endgroup$– PaulCommented Mar 1, 2013 at 13:22
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1 Answer
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See Proposition 24 of Chapter 5, section 10 in Peter Petersen's book Riemannian Geometry, which states the following:
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.
Thus (as is basically indicated in the comments) it makes sense to look at collections of isometries instead of a single isometry.
In this case, if the fixed point set happens to be one dimensional, it will be a geodesic.
