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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.

Kat
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2 Answers2

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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.

ZKe
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  • My answer depends on a discrete cellular cocompact action action inducing discrete orbits of points, so this answer is modulo that fact. – ZKe Dec 03 '23 at 23:08
  • Thanks for the answer, does $C^g$ mean the points in $C$ fixed by $g$? – Kat Dec 04 '23 at 18:34
  • No, it is the image of the cell $C$ by the action of $g$. – ZKe Dec 04 '23 at 19:13
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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.

Lee Mosher
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  • I understood this question to be about cell complexes and not CW complexes. – ZKe Dec 04 '23 at 17:04
  • CW complexes are more general than cell complexes. – Lee Mosher Dec 04 '23 at 17:09
  • This answer https://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 Dec 04 '23 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 Dec 04 '23 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. – Lee Mosher Dec 04 '23 at 18:50
  • @Zackkenyon: I don't see that post as being particularly authoritative on the definition of a cell complex. In fact, the literature over many decades is rather fuzzy on the best definition of a cell complex. Nowadays the concept of a CW complex, which is very solidly established, has become the standard. In some older books, I have seen a cell complex defined as a CW complex such that the attaching map of each $n$-cell is a homeomorphism onto a spherical subcomplex of the $n-1$ skeleton. In reality, though, nowadays I think that "cell complex" is used synonymously with "CW-complex". – Lee Mosher Dec 04 '23 at 18:53
  • I would be very grateful if you could add a more detailed answer! Thanks! – Kat Dec 04 '23 at 18:57
  • @LeeMosher Sorry to be a bother, as it is not my question, but I guess I don't understand why OPs question bothers with the discreteness of $G$ if they are talking only about CW complexes. It is clear from your answer that in that case this fact is immaterial. – ZKe Dec 05 '23 at 00:10
  • @LeeMosher I wonder if you could add a more detailed answer about the consequence of the definition of the topology on a CW complex. I would be very grateful if you could help. – Kat Dec 06 '23 at 21:54
  • I added some details. – Lee Mosher Dec 07 '23 at 13:11
  • @LeeMosher Thank you so much! – Kat Dec 08 '23 at 05:24