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.

  • $\begingroup$ 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. $\endgroup$ Dec 11, 2012 at 13:33
  • $\begingroup$ I have a question, why can we use finite group cannot act freely on a contractible space? A CW-complex is not necessary to be contractible. $\endgroup$
    – 6666
    Apr 7, 2017 at 22:53
  • 1
    $\begingroup$ @6666 That fact is being applied to the action of the cyclic subgroup generated by $g$ on the single cell $\sigma$, which is contractible. $\endgroup$ Apr 8, 2017 at 8:28

1 Answer 1


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.

  • 4
    $\begingroup$ 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. $\endgroup$ Dec 11, 2012 at 14:04
  • $\begingroup$ 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). $\endgroup$
    – Misha
    Dec 11, 2012 at 16:46
  • $\begingroup$ @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. $\endgroup$ Dec 11, 2012 at 18:38
  • 1
    $\begingroup$ @Tom: True. I've slightly edited my answer so that I don't lie about 0-cells. :-) $\endgroup$ Dec 11, 2012 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.