4
$\begingroup$

I'm studying finiteness properties of groups and in my research I found the following theorem, which is quite useful:

$G$ is of type $F_n$ (i.e. there is a CW complex X with contractible universal covering and finite $n$-skeleton such that $\pi_1(X) = G$) if, and only if, $G$ acts freely, cocompactly, properly, faithfully and cellularly in a CW complex which is $(n-1)$-connected, that if, its homotopy groups of order less than $n$ are trivial.

The proof is not complicated, but I just realized that my argument does not work for $n = 1$. I've been trying to do this proof but I don't have any idea on how to proceed. The theorem, explicitly, is:

$G$ is finitely generated if, and only if, $G$ acts freely, cocompactly, faithfully, cellularly and properly in a connected CW complex.

Any help would be appreciated.

EDIT: as per C Squared and Moishe comments, I'll explain what's my problem in more detail.

So the proof for the general statement is the following:

Suppose $G$ is of type $F_n$ and take $X$ to be a CW complex with the prerequisites of the definition. Then, the universal covering $\tilde{X}$ is a CW-complex and we know, by construction of $\tilde{X}$ (which is far too long to post here, but can be foun in Geoghegan's book "Topological Methods in Group Theory", starting on page 87), that:

  • The action of $G$ on $\tilde{X}$ is free on cells (which is the definition of a free action in the context of CW complexes) and cellular, and it takes cells of $\tilde{X}$ to other cells with same dimension;
  • The action of $G$ in $\tilde{X}^n$ is cocompact, since $X^n = \tilde{X}^n/G$ is finite; The construction yields a covering map $\tilde{X}^n \to X^n$ and since $\tilde{X}^n$ is simply connected, the deck group of such a covering is $\pi_1(X) = G$, so $G$ acts properly in $\tilde{X}^n$.

So $\tilde{X}^n$ is an $(n-1)$-connected CW complex in which the action of $G$ is good enough.

Now, suppose $n \geq 2$. Then, we can prove the converse. If $X$ is an $(n-1)$-connected CW complex in which the action is good, then we know that:

  • $X \to X/G$ is a covering map, and since $X$ is simply connected (hence why we need $n \geq 2$) then $\pi_1(X/G) = G$;
  • Covering projections preserve homotopy groups of order bigger than $1$, so $\pi_k(X/G) = 1$ for all $1 < k \leq n-1$;
  • Since the action of $G$ is free and cellular, $X/G$ is a CW complex and its cells are images of cells of $X$ by the quotient map;
  • Since the action of $G$ is cocompact, then $X/G$ is compact, hence finite, so its $n$-skeleton is finite. Now, we can add cells of dimension $n+1$ or higher to $X/G$ to ensure that the higher homotopy groups are also all trivial. Since we are adding only cells of dimension $n+1$ or higher, the $n$-skeleton of the resulting space is also finite. Take this result space to be $Y$, and then $Y$ is a space that satisfies the definition of $G$ being of type $F_n$.

The thing is, I alredy figured out a proof for the converse in the case $n = 1$, and I'll be posting it below.

$\endgroup$
5
  • $\begingroup$ do you mind showing your argument and why it fails for $n = 1$? $\endgroup$ Commented Jun 23 at 14:13
  • $\begingroup$ Use the Cayley graph. $\endgroup$ Commented Jun 23 at 14:29
  • $\begingroup$ It's if and only if statement: which direction do you have trouble with? $\endgroup$ Commented Jun 24 at 0:51
  • $\begingroup$ @CSquared done. $\endgroup$ Commented Jun 24 at 9:39
  • $\begingroup$ @MoisheKohan done. $\endgroup$ Commented Jun 24 at 9:39

1 Answer 1

Reset to default
0
$\begingroup$

So, I figured it out with help from my amazing advisors.

What we actually want to prove is: if $G$ acts in a good enough way on a connected CW complex $X$, then $G$ is finitely generated.

Start by taking $Y = X/G$. We know, because the action is free on cells and cellular, that $Y$ is a CW complex with cells being the images, by the quotient map, of cells of $X$. Since the action is cocompact, $Y$ is finite, and in particular, $Y^1$ and $Y^0$ are finite.

Since the action preserver dimension of the cells, elements of some $v \in Y^0$ are all vertices of $X$. Since $Y^0$ is finite, we can take $V \subset X^0$ finite so that it contains a single representative for every $v \in Y^0$.

Take an orientation on $Y$ and induce an orientation on $X$ by orienting cells of $X$ with the corresponding orientation of ther images. Now, for every edge $\tilde{e} \in Y^1$, take $\tilde{u}_{\tilde{e}}$ and $\tilde{v}_{\tilde{e}}$ to be the initial and final points of $\tilde{e}$. They have representatives $u_{\tilde{e}}$ and $v_{\tilde{e}}$ in $V$. Take $e$ to be a lifting of $\tilde{e}$ with initial point in $u_{\tilde{e}}$. We know that the final point of $e$, denoted $v_e$, is in the orbit of $v_{\tilde{e}}$, so there is $g_e \in G$ with $g_e v_{\tilde{e}} = v_e$.

Since $Y^1$ is finite, the set $S = \{g_e \mid e \in Y^1\}$ is finite. We will now prove that $G = \langle S \rangle$. If $g \in G$, take $v \in V$ and a path $\tau$ from $v$ to $gv$. Project this path in $Y$ by the quotient map. Since the start and end points of $\tau$ are in the same $G$-orbit, the projection $\tilde{\tau}$ is a loop, and a finite one, since $Y^1$ is finite. By construction of the fundamental group, we have that $g$ is a product of the elements $g_e^{\pm 1}$ which correspond to the edges of $\tilde{\tau}$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.