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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.

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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 '12 at 17:58
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Collapsing in the Gromov-Hausdorff sense. –  Acky Sep 19 '12 at 18:05
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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 '12 at 18:16
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"...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 '12 at 18:22
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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 '12 at 18:25
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up vote 2 down vote accepted

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.

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