most recent 30 from http://mathoverflow.net 2013-05-26T02:29:27Z http://mathoverflow.net/feeds/question/116058 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116058/free-and-cellular-g-action-implies-free-g-complex thedef 2012-12-11T09:16:27Z 2012-12-11T20:36:39Z

<p>Recall that a CW-complex $X$ with an action of a group $G$ which permutes the cells (i.e., for any $g \in G$ and any cell $\sigma \subseteq X$, $g\sigma$ is a cell) is called a <em>$G$-complex</em>. If the action permutes the cells freely ($g\sigma = \sigma$ implies $g=1$), $X$ is a <em>free G-complex</em>. </p> <p>Clearly, if $X$ is a free $G$-complex, then the $G$-action on $X$ is free (i.e., for any $g \in G$ and any $x \in X$, $gx = x$ implies $g=1$). A question that pops to my mind every once in a while is the following: is a $G$-complex with a free $G$-action a free $G$-complex? I see that if $g\sigma = \sigma$ for some nontrivial $g \in G$ and a cell $\sigma$, then $g$ has infinite order (for a finite group cannot act freely on a contractible space), but this doesn't seem to get me anywhere.</p>

http://mathoverflow.net/questions/116058/free-and-cellular-g-action-implies-free-g-complex/116061#116061 Jeremy Rickard 2012-12-11T09:44:48Z 2012-12-11T20:36:39Z

<p>If $G$ acts freely on a CW-complex, permuting the cells, then the stabilizer of a cell must be finite (and therefore trivial, as pointed out in the question).</p> <p>This can be shown by induction on the dimension, the case of 0-cells being trivial. If $\sigma$ is an $n$-cell, with $n\geq 1$, let $H$ be the stabilizer of $\sigma$. Then $H$ permutes the set of cells with dimension less than $n$ in the closure of $\sigma$. But there are only finitely many such cells, and inductively each has finite (indeed, trivial) stabilizer. Thus $H$ is finite.</p>