Cocompact action with finite stabilizer implies locally finite

Given a discrete group $$G$$ acting cellularly and cocompactly on a cell-complex $$X$$ with finite stabilizers, I am struggling to show $$X$$ is locally finite.

I am trying to show any vertex $$x\in X$$ is contained in finitely many cells. Since $$X/G$$ is compact, thus locally finite, which implies there are only finitely many orbits of cells containing $$x$$. So it suffices to show each orbit of a cell containing $$x$$ has only finitely many cells (containing $$x$$).

I am trying to use the orbit-stabilizer theorem to deduce that. However, I think given a cell $$C$$ containing $$x$$ with $$g\cdot C$$ containing $$x$$ as well, $$g$$ might not always fix $$x$$. Thus $$g$$ might not be in the stabilizer of $$x$$. So I have no idea why each cell-orbit containing $$x$$ has only finitely many elements.

Suppose that $$X$$ is not locally finite.

Then there is some $$x$$ in the intersection of infinitely many cells. Moreover, since the quotient is compact, there must be some cell $$C$$, $$x\in C$$ such that the set $$H=\{g\in G|x\in C^g\}$$ is infinite.

Now note that $$x^{H^{-1}}\subset C$$. If this set is infinite, then our group action wasn't discrete, and if this set is finite, then by the pigeonhole principle, there is some infinite $$K\subset H$$ such that $$x^{K^{-1}}$$ is a singleton $$\{x'\}$$, and therefore $$KK^{-1}\subset \text{Stab(x')}$$ is infinite.

• My answer depends on a discrete cellular cocompact action action inducing discrete orbits of points, so this answer is modulo that fact.
– ZKe
Commented Dec 3, 2023 at 23:08
• Thanks for the answer, does $C^g$ mean the points in $C$ fixed by $g$?
– Kat
Commented Dec 4, 2023 at 18:34
• No, it is the image of the cell $C$ by the action of $g$.
– ZKe
Commented Dec 4, 2023 at 19:13

There is a standard lemma in the theory of CW complexes, namely Proposition A.1 in the Appendix of Hatcher's Topology:

Every compact subset of a CW complex $$X$$ is contained in a finite subcomplex of $$X$$.

For the action of $$G$$ on $$X$$ to be cocompact means, by definition, that there is a compact $$K \subset X$$ such that $$G \cdot K = X$$. Let $$L \subset X$$ be a finite subcomplex containing $$K$$, and so $$G \cdot L = X$$ (one sees from this argument that cocompactness of the action is equivalent to the existence of a finite subcomplex $$L$$ such that $$G \cdot L = X$$).

Arguing by contradiction, suppose there exists a $$0$$-cell $$x \in X$$ such that $$X$$ is not locally finite at $$x$$. By definition of local finiteness, this means that there are infinitely many (open) cells $$e \subset X$$ of the given CW structure such that $$x \in \overline e$$. Referring to $$\overline e$$ as a closed cell, this is equivalent to saying that there are infinitely many closed cells $$\overline e$$ containing $$x$$.

Note also that the number of closed cells contained in $$L$$ is finite (because the number of (open) cells contained in $$L$$ is finite). It follows that for each $$a \in G$$ the number of closed cells contained in $$a \cdot L$$ is finite. But there are infinitely many closed cells that contain $$x$$, and so by applying the pigeonhole principle it follows that the set $$A = \{a \in G \mid x \in a \cdot L\} = \{a \in G \mid a^{-1} \cdot x \in L\}$$ is infinite.

Knowing that $$x$$ is a $$0$$-cell, it follows that each $$a^{-1} \cdot x$$ is a $$0$$-cell. But $$L$$ has only finitely many $$0$$-cells. By another application of the pigeonhole principle, there exists an infinite subset $$B \subset A$$ such that $$b^{-1} \cdot x$$ is constant independent of $$b$$. Picking one $$b' \in B$$, it follows that $$b \cdot (b')^{-1}\cdot x = x$$ for all $$b \in B$$. This proves that $$x$$ has infinite stabilizer, contradicting the hypothesis.

• I understood this question to be about cell complexes and not CW complexes.
– ZKe
Commented Dec 4, 2023 at 17:04
• CW complexes are more general than cell complexes. Commented Dec 4, 2023 at 17:09
• This answer math.stackexchange.com/a/1244701/79927 has other conventions. I believe that it is a matter of the cell structures being different. And since OPs action is cellular, I think that matters? I am not certain. My intuition was that whichever point failed local finiteness might not actually be a zero cell in the complex.
– ZKe
Commented Dec 4, 2023 at 17:52
• Thanks for the answer, but why not locally finite at $x$ implies $G\cdot x\cap L$ has infinitely many elements?
– Kat
Commented Dec 4, 2023 at 18:15
• That's a consequence of the definition of the topology on a CW complex; I can add a more detailed answer to my post later, if you like. Commented Dec 4, 2023 at 18:50