Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Motivation: I'm an operator algebraist and I'm looking for an answer to the main question in order to build non-trivial spectral triples for a class (as large as possible) of discrete groups.
$\to$ It's noncommutative geometry "à la Alain Connes".

Definition : Let $G$ be a group, then a total space for $G$ is a weakly contractible topological space $X$ (i.e., every homotopy groups $\pi_{n}(X)$ are trivial), together with a free action of $G$ (i.e, $\forall x \in X$ and $\forall g \in G$, if $g.x = x$ then $g = e$). It is generally noted $EG$.

Definition : A group $\Gamma$ is finitely generated if there is a finite generating subset $S \subset \Gamma $.
In general such groups can be defined by generators $S$ and relations $R$, as $\Gamma := \langle S \hspace{0.1cm} \vert \hspace{0.1cm} R \rangle$.
The pair $(\Gamma , S)$ is called a marked group. We take $S=S^{-1}$ and the identity element $e \not \in S$.

Definition: Let $\Gamma = \langle S \hspace{0.1cm} \vert \hspace{0.1cm} R \rangle$ be a finitely generated marked group, and let $\Gamma_{n}$ be the sets of irreducible $n$-blocks, defined as follows :

  • $\Gamma_{0} := \Gamma$
  • $\Gamma_{1}:=\{ \{ g ,gs \} \hspace{0.1cm} \vert \hspace{0.1cm} g \in \Gamma, s \in S \}$
  • Let $n \geq 1$, $\Gamma_{n+1}$ is the set of irreducible $(n+1)$-blocks, defined by induction :
    A $(n+1)$-block is a nonempty finite set $\mathfrak{a}$ of irreducible $n$-blocks such that:

    $$ \forall \mathfrak{b} \in \mathfrak{a}, \forall \mathfrak{c} \in \mathfrak{b}, \exists ! \mathfrak{b}'\in \mathfrak{a} \text{ such that } \mathfrak{b} \cap \mathfrak{b}' = \{\mathfrak{c}\}$$ A $(n+1)$-block $\mathfrak{a}''$ is called irreducible if $\forall \mathfrak{a}, \mathfrak{a}'$ $(n+1)$-blocks:
    $$ \mathfrak{a}'' = \mathfrak{a} \triangle \mathfrak{a}' := (\mathfrak{a} \cup \mathfrak{a}') \backslash \mathfrak{a} \cap \mathfrak{a}' \Rightarrow max(\vert \mathfrak{a} \vert ,\vert \mathfrak{a}'\vert) \ge \vert \mathfrak{a}''\vert $$

Geometric realization: Let $\Gamma$ be a finitely generated marked group, and let $\Gamma_{n}$ its set of irreducible $n$-blocks. Its geometric realization $X$ is the topological space built by induction :
Each element $\mathfrak{a}$ of $\Gamma_{n}$ is related to a $n$-polytope $r(\mathfrak{a}) \subset X$ as follows :

  • $\mathfrak{a} \in \Gamma_{0}$ is related to a vertex.
  • $\mathfrak{a} \in \Gamma_{1}$ is related to an edge.
  • $\mathfrak{a} \in \Gamma_{2}$ is related to the regular polygon with $\vert \mathfrak{a} \vert$ edges.
  • $\mathfrak{a} \in \Gamma_{n+1}$ is related to the $(n+1)$-polytope, obtained as the (n+1)-dimensional filling of the gluing of the $\vert \mathfrak{a} \vert$ $n$-polytopes related to the elements of $\mathfrak{a}$, such that $\forall \mathfrak{b}, \mathfrak{b}' \in \mathfrak{a}$ then: $r(\mathfrak{b} \cap \mathfrak{b}') = r(\mathfrak{b}) \cap r(\mathfrak{b}')$.

Definition : Let $X_{n}$ be the regular CW complex obtained by gluing together all these $m$-polytopes $\forall m \leq n$ such that $\forall \mathfrak{a}, \mathfrak{a}' \in \Gamma_{m}$ then $r(\mathfrak{a} \cap \mathfrak{a}') = r(\mathfrak{a}) \cap r(\mathfrak{a}')$.

Note that the Cayley graph is $X_{1}$, and the Cayley complex is "close" to $X_{2}$.

Each $X_{n}$ admits a natural distance $d_{n}$, and $X_{n-1}$ naturally embeds into $X_{n}$.

Definition : Let $d$ be the distance on $ \bigcup{X_{n}}$, defined as follows : $d(x,y) = lim_{n \to \infty} d_{n}(x,y)$.

Remark : There is a small abuse in the previous definition because $d_{n}(x,y)$ is defined only for $x, y \in X_{n}$. But because we take $n \to \infty$, there is no problem.

Definition : Let $X:=\overline{\bigcup{X_{n}}}$, the complete metric space obtained as a completion of $\bigcup{X_{n}}$ for $d$. We call it the total Cayley space of the pair $(\Gamma, S)$.

Main question : Is the total Cayley space $X$ of a finitely generated marked group $(\Gamma, S)$, a total space $E \Gamma$, for $\Gamma$ torsion-free ?

Dimension :
- if $\Gamma_{n} \neq \emptyset$ and $\Gamma_{n+1} = \emptyset $, then $X$ is an $n$-dimensional topological space and $X \simeq X_{n}$.
- if $\forall n \in \mathbb{N}$ then $\Gamma_{n} \neq \emptyset$, then $X$ is an $\infty$-dimensional topological space.
$\to$ The Thompson group $F$ (see B. Steinberg comment below).

Sketch of proof (it needs to be completed)

Finite dimensional case:

  • The natural action of $\Gamma$ on $X$ is free.
    Let $x \in X$ and $g \in G$ such that $g.x = x$, then $\exists n \geq 0$ and $\mathfrak{a} \in \Gamma_{n}$ such that $x \in r(\mathfrak{a})$ and $g.\mathfrak{a} = \mathfrak{a}$. So a finite subset of $\Gamma$ is stable under the left multiplication by $g$, then $g$ is torsion, so $g=e$ by assumption. $\square$

  • $X$ is weakly contractible:
    By construction, an embedding of a sphere $\mathbb{S}^{n-1}$ is homotopic to the border of the realization $r(\mathfrak{a})$ of an irreducible $n$-block $\mathfrak{a}$. Now $r(\mathfrak{a})$ is a polytope, so it's homotopic to a point, its border also. $\square$

$\infty$-dimensional case:

  • The natural action of $\Gamma$ on $X$ is free (inspired by Andreas Thom) :
    Let $x \in X$ and $g \in G$ such that $g.x = x$. Now, $x = [(x_{n})]_{\sim}$ and $g.x = [(g.x_{n})]_{\sim}$, so $d_{n}(x_{n},g.x_{n}) \to 0$. Let $n$ with $d_{n}(x_{n},g.x_{n})$ sufficiently small, and let $\mathfrak{a}, \mathfrak{a'}\in \Gamma_{n}$ such that $x_{n} \in r(\mathfrak{a})$ and $g.x_{n} \in r(\mathfrak{a'})$. By construction, $r(\mathfrak{a}) \cap r(\mathfrak{a'}) \ne \emptyset$ and $\exists r \le n$, $\mathfrak{b} \in \Gamma_{r}$ such that $r(\mathfrak{b}) \subset r(\mathfrak{a}) \cap r(\mathfrak{a'})$ and $g.\mathfrak{b} = \mathfrak{b}$. The result follows as previously. $\square$
  • $X$ is weakly contractible ? See the generalized post here.

In the literature $E \Gamma$ is well known as the universal covering of the classifying space $B \Gamma$, which is (in the discrete groups case) the Eilenberg-MacLane space $K(\Gamma , 1)$. There is also the Milnor construction.
The construction here, specific to finitely generated groups, is very combinatorial.

Question : Is there a good reference containing this combinatorial construction ?

share|cite|improve this question
Have you read the paper of Baumslag-Bridson-Miller-Short on fibre products, non-positive curvature and decision problems? I think they cover some of this stuff in their 1-2-3 theorem, but I didn't quite understand all of your question. (It is not your fault, it is because I'm lazy. Sorry.) – user1729 Aug 29 '13 at 16:39
@user1729 : thank you very much for this reference. I will look at it right now. Note that I know Hamish Short, he was my professor of geometric group theory when I was student. Maybe I can ask him. – Sébastien Palcoux Aug 29 '13 at 16:53
A question on the geometric realization: How do you know the $(n+1)$ dimensional filling is unique, or indeed even exists? For example, what if one of your 3-blocks is a collection of 2-blocks that form a Klein bottle? – MartianInvader Sep 7 '13 at 21:45
I think you need more than orientability - it sounds like you want your $n$-blocks to become cells in a CW-complex, which means they need to not only be orientable, but be $(n-1)$ spheres. – MartianInvader Sep 9 '13 at 19:08
I can't come up with an example off the top of my head, and I suspect you would need to go beyond a $3$-block to find one. To prove it in 3 dimensions, I would try a strategy like the following: since a non-sphere surface will have nontrivial fundamental group, we can find a loop that's not nullhomotopic, which will necessarily be a 2-block on its own, and we can use such 2-blocks to subdivide the surface, showing it's not irreducible. This argument won't work in general, though, since higher-dimensional non-sphere manifolds can have trivial $(n-1)$-homotopy groups. – MartianInvader Sep 10 '13 at 20:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.