Collapsing of Riemannian manifolds with a group action - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:35:27Z http://mathoverflow.net/feeds/question/107597 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107597/collapsing-of-riemannian-manifolds-with-a-group-action Collapsing of Riemannian manifolds with a group action Acky 2012-09-19T17:42:21Z 2012-09-20T17:20:36Z <p>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?</p> <p>If in general such a sequence does not exist, how about the case $G=T$? Here $T$ is a finite-dimensional torus.</p> http://mathoverflow.net/questions/107597/collapsing-of-riemannian-manifolds-with-a-group-action/107701#107701 Answer by Anton Petrunin for Collapsing of Riemannian manifolds with a group action Anton Petrunin 2012-09-20T17:20:36Z 2012-09-20T17:20:36Z <p>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.)</p> <p>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 $\mathbb S^1$.</p> <p>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.</p>