Welcome to MathOverflow

Question

1

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 $G$-complex. If the action permutes the cells freely ($g\sigma = \sigma$ implies $g=1$), $X$ is a free G-complex.

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.

flag	asked Dec 11 at 9:16 thedef 11●1

	The usual definition of a $G$-CW-complex also requires that a group element fixing a cell setwise also fixes it pointwise. For example the unit interval with the $\mathbb{Z}/2$-action given by reflection at $1/2$ and the usual G-CW structure is not a G-CW-complex. I never came across the notion of a $G$-complex. – HenrikRüping Dec 11 at 13:33
	Maybe the difference is just one of based vs. unbased worlds? Both seem like reasonable definitions to me. – David White Dec 11 at 15:10

Jeremy Rickard · Answer 1 · 2012-12-11 20:36:39Z UTC

2

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).

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.

edited Dec 11 at 20:36

answered Dec 11 at 9:44

Jeremy Rickard
1,584●1●6●9

2

This argument applies when the cell has positive dimension, because then there is at least one lower-dimensional cell involved in the boundary. Fortunately a separate argument is available to show that an infinite group cannot act freely on a $0$-cell. – Tom Goodwillie Dec 11 at 14:04

A separate argument is also needed if the complex $X$ is not locally finite (in which case the number of lower-dimensional cells whose images intersect the boundary of the $n$-cell, could be infinite). – Misha Dec 11 at 16:46

@Misha: Doesn't the boundary of an $n$-cell $\sigma$ always involve only finitely many cells in the $(n-1)$-skeleton? Local finiteness means there are only finitely many higher dimensional cells whose boundaries involve $\sigma$, and I don't think my argument uses that. – Jeremy Rickard Dec 11 at 18:38

@Tom: True. I've slightly edited my answer so that I don't lie about 0-cells. :-) – Jeremy Rickard Dec 11 at 20:18

Free and cellular G-action implies free G-complex?

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.