Welcome to MathOverflow

Question

3

2

Let $M$ be a complete Riemannian manifold with bounded sectional curvature and $G$ a compact connected Lie group acts smoothly on $M$. Consider the fixed point set $F$, it is of course a submanifold of $M$ by the slice theorem. Let ${F_i}$ be the connected components of $F$. Then for each $i$, is there a sequence of Riemannian manifolds ${M_j},j\in\mathbb{N}$ with $M_0=M$ such that ${M_j}$ collapses to $F_i$ while keeping their sectional curvatures uniformly bounded?

If in general such a sequence does not exist, how about the case $G=T$? Here $T$ is a finite-dimensional torus.

edited Sep 20 at 6:21

asked Sep 19 at 17:42

Acky
227●6

1

Motivation? Could you point us to a good definition of `collapsing'? Is it true that each Euclidean space `collapses' to a point? [If not, take T = S^1 acting in standard linear fashion on the plane to get a counterexample. – Richard Montgomery Sep 19 at 17:58

1

Collapsing in the Gromov-Hausdorff sense. – Acky Sep 19 at 18:05

2

Do you require that all $M_j$ are diffeomorphic to $M$? If not, why not take $M_j=F_i\times $(small circle)? – Sergei Ivanov Sep 19 at 18:16

3

"...Consider the fixed point set F, it is of course a submanifold of M by the slice theorem". Note that it is really simpler than that; in geodesic coordinates at a point p of F, the fixed point set is locally the linear subspace left fixed by the linearized action at p. – Dick Palais Sep 19 at 18:22

1

I assume you give $F$ the induced metric from $M$. If the answer to your question is yes, then the curvature of $F$ is bounded from below, so in effect you hope that the fixed point set of any smooth compact group action has curvature bounded below. Why would that be true? – Igor Belegradek Sep 19 at 18:25

show 5 more comments

Anton Petrunin · Accepted Answer · 2012-09-20 17:20:36Z UTC

As it was noted in the comments you probably wanted to say that the action is isometric and $M_n$ is diffeomorphic to $M$ for all $n$. (Otherwise the question has no sense.)

In this case answer is NO. Consider $\mathbb S^1$ action on $\mathbb S^3$ with fixed point set $\mathbb S^1$ and note that simply connected spaces can not GH-converge to $\mathbb S^1$.

For the second part, it seems that you may only get a torus as the fixed set (?). In this case the answer is obviously YES.

Collapsing of Riemannian manifolds with a group action

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.