# is every totally geodesic submanifold the set of fixed points of some isometries?

It is well known that the set of fixed points of an isometry $\phi:(M,g)\rightarrow (M,g)$ is a totally geodesic embedded submanifold. (e.g here ).

I ask whether the converse is true, i.e is every totally geodesic embedded submanifold $N \subset M$ can be realized as the set of fixed points of some isometries?

One trivial obstruction is that $N$ should be closed.

(Since the $Id,\phi$ are continuous and the diagonal in $M \times M$ is closed, the set of fixed points is always closed in $M$).

So, if we assume $M$ is connected, then of course we have to omit from our discussions open submanifolds (which I think are all totally geodesic but cannot be closed, hence cannot be a fixed-points-set).

Update: The answer is negative. A brief summary of the idea: take a small enough compact geodesic segement. Any isometry which fixes it, must "fix some more" of the whole geodesic the segment is a part of.

Now I wonder if every such submanifold must be the fixed point of some diffeomorphism(s)? (not necessarily an isometry).

• Open submanifolds certainly can be closed. For example, if $M$ has three or more connected components, all of which are isometric to each other, then there is an isometry whose fixed-point set is exactly one component, which is both open and closed. – Jack Lee Apr 15 '15 at 21:16
• I agree of course. That is why I mentioned that if we assume $M$ is connected then there are no open-and-closed submanifolds (except $M$ itself). Hence, since a fixed point set is readily seen to be closed, it cannot be open if we exclude $M$ itself as a trivial option. (Or equivalently the set of fixed points of the identity). – Asaf Shachar Apr 15 '15 at 21:43

This simply cannot be true. A generic Riemannian manifold does not have any isometries, but still the image of any geodesic is a totally geodesic submanifold.

i think it is unlikely that this is true. Take a dumbbell type shape with rotational symmetry. Make the bar bit pinch in. The smallest circle is a totally geodesic submanifold and the invariant under rotations. Now deform the rest of the barbell whilst leaving a small neighbourhood of the smallest circle unchanged. If you make it lumpy enough there won't be any isometries.

I will elaborate based on Andreas Cap's suggestion.

First I note that I meant only to discuss embedded submanifolds. (I have edited the question to include this).

Now, since in general a geodesic can intersects itself (even infinitely many times) it's image is not necessarily an embedded submanifold. (only an immersed one).

However, since every immersion is locally an embedding, we can take a "small enough segment" of any geodesic, which will be an embedded submanifold. In particular we can take the domain of our segment to be a compact interval, so the image will be compact, hence closed, as required by our counterexample.

This will be a totally geodesic closed embedded submanifold. (To see it is totally geodesic, we can just note that any geodesic in it will have to locally minimize length, hence coinside with the original sub-segment of the geodesic).

In particular we can take for a concrete example $\mathbb{S^2}$. Take $N$ any contiguous part of a great circle (which is not the entire circle). Let $\phi$ be an isometry which fixes $N$. Now since $\phi$ take geodesic to geodesic, it must take a segment of the great circle which intersects $N$ but not contained in it, to itself (by uniqueness of geodesics with a given starting point & velocity). Hence, $\phi$ fixes more than just $N$. (Actually it is easy to see $\phi$ must fix the entire great circle).

Actually this answer shows Andreas's idea can be applied to any manifold, that is: There are no Riemannian manifolds, whose every totally geodesic embedded closed submanifold is the set of fixed points of isometries.