Are there familiar expressions for (finite) joins of finite groups?

Milnor construction of the classifying space of a topological group $G$ is given in terms of infinite joins of $G$. Schwarz then proved that the $k+1$ iterated self join of a group $G$ classifies $G$-principal bundles having (reduced) sectional category at most $k$.

My question is if there are familiar expressions for the topological spaces arising from such finite joins of topological groups (I am particularly interested in the case of discrete groups, see below).

For instance, (finite dimensional) spheres with antipodal action are the (finite) joins of $Z/2$ (thought of as a discrete group). Likewise, odd dimensional spheres with complex scalar multiplication are the (finite) joins of the unitary complex numbers. And the case of finite cyclic groups works similarly. (After the comment by Omar, I noted that the last assertion was too loose. Omar clarified the situation and essentially answered the original question.)

I would be particularly interested if anyone can point at a "familiar" topological description of the finite joins of $(Z/2)^s$ (also thought of as a discrete group), the elementary abelian 2-group of rank $s$.

• I've never heard about any notion of join of groups (there are notions of join of lattices [posets], and of join of two topological spaces). Could you provide a definition or a reference? – YCor Mar 19 '16 at 21:59
• Thanks for the clarification. I should have said that I meant to consider a finite group as a discrete topological space (so I am asking about joins as topological spaces). I am clarifying this point in my question above. – Jesus Gonzalez Mar 19 '16 at 22:11
• Hi Jesus! And the case of finite cyclic groups works similarly": of course, then $k$-fold join of $(\mathbb{Z}/2)^s$ will be homeomorphic to the $k$-fold join of $\mathbb{Z}/2^s$ (so possibly a sphere?) but with a different action. – Mark Grant Mar 21 '16 at 7:11
• You won't get (single) spheres from finite groups of order bigger than 2: I haven't thought about the actions, but the homotopy type of $G^{\ast n}$ depends only on the order of $|G|$ and is a wedge of $(|G|-1)^n$ copies of $S^{n-1}$. (When $|G|=2$ this is, of course, a single sphere.) – Omar Antolín-Camarena Mar 21 '16 at 23:19
• Omar, you are absolutely right. $G$ can be thought of as the wedge of $|G|-1$ copies of $S^0$, and then one can inductively use the fact that the join $X \ast Y$ has the homotopy type of the suspension of the smash product of $X$ and $Y$. Figuring out the action from here might now be accessible. Thanks! – Jesus Gonzalez Mar 22 '16 at 13:22

The $G$ action on $G^{\ast n}$ seems to be somewhat complicated: here's a description, but I hope someone can find a simpler one.

Let $G$ be a finite group. $G^{\ast n}$ can be described as a simplicial complex with vertex set $G \times [n]$ (where $[n] = \{1, 2, \ldots, n\}$), whose maximal simplices are sets of the form $\{(a_1, 1), (a_2, 2), \ldots, (a_n, n)\}$ for some $(a_1, \ldots, a_n) \in G^n$ ("pick one element of $G$ from each column of $G \times [n]$").

The union $T$ of the maximal simplices where at least one $a_i = e$ (where $e$ is the identity of $G$) is a contractible space --it can easily be retracted onto the simplex $\{(e,1), (e,2), \ldots, (e,n)\}$. The quotient $G^{\ast n}/T$ is then a wedge of copies of $S^{n-1}$, one for each of the $(|G|-1)^n$ maximal simplices that don't have any $a_i=e$. The $S^{n-1}$ corresponding to the maximal simplex $\{(a_1, 1), \ldots, (a_n, n)\}$ is $\{e, a_1\} \ast \{e, a_2\} \ast \cdots \ast \{e, a_n\} \subset G^{\ast n}$.

(I had first computed the homotopy type as Jesús did in this comment, but that didn't seem to make it easy to figure out the action.)

On $G^{\ast n}$ the $G$-action is straightforward, but let's see what it corresponds to in the quotient $G^{\ast n}/T$. For each $\mathbf{a} = (a_1, \ldots, a_n)$, let $\sigma(\mathbf{a})$ be the corresponding $(n-1)$-sphere in $G^{\ast n}/T$. The preimage of $\sigma(\mathbf{A})$ in $G^{\ast n}$ has maximal simplices $\{(a'_1, 1), \ldots, (a'_n,n)\}$ where each $a'_i$ is either $a_i$ or $1$. Applying the action of $g \in G$, this preimage goes to the $(n-1)$-sphere with faces $\{(ga'_1, 1), \ldots, (ga'_n,n)\}$. In $G^{\ast n}$ the faces with some $ga'_i=e$ get contracted to the basepoint, and the others will wrap around $\sigma(g \cdot \mathbf{a}')$ with degree $\pm 1$.

Some care is needed to figure out the signs, but I think they work out as follows: $$g|_{\sigma(\mathbf{a})} \simeq \sum_{I \subseteq [n]} (-1)^{|I|} \sigma\left(g \cdot (I \ominus \mathbf{a})\right),$$ where:

• $I \ominus \mathbf{a} \in G^n$ is $\mathbf{a}$ with all $a_i$ with $i \in I$ replaced by $e$,
• $g \cdot \mathbf{a}'=(ga'_1, \ldots ga'_n)$,
• the sum is to be interpreted as follows: the domain $\sigma(\mathbf{a})$ is first mapped to $\bigvee_{I \subset [n]} S^{n-1}$ by a (multiple-)pinch map, and then the $I$-th "wedgeand" maps to $\sigma(g \cdot I \ominus \mathbf{a})$ with degree $(-1)^{|I|}$, except that if any coordinate of $g \cdot I \ominus \mathbf{a}$ is $e$, the entire $I$-th wedgeand is sent to the basepoint of $G^{\ast n}/T$.

I'm not sure you can really strictify this to an honest $G$-action on $G^{\ast n}/T$, but at the very least, the above describes for each $g \in G$ an endomap on $G^{\ast n}/T$ such that the square $$\require{AMScd}\begin{CD} G^{\ast n} @>{g}>> G^{\ast n} \\ @V{\pi}VV @V{\pi}VV \\ G^{\ast n}/T @>{g}>> G^{\ast n}/T \end{CD}$$ commutes up to homotopy.

Notice that in the case $G$ is cyclic of order $2$, with generator $\tau$, there is only one $(n-1)$-sphere, namely $\sigma(\tau, \ldots, \tau)$ and in the expression for the action of $\tau$ only the term with $I = [n]$ is not crushed to the basepoint, which says that $\tau$ acts on $\sigma(\tau, \ldots, \tau)$ with degree $(-1)^n$ --which agrees, as expected, with the antipodal action.

ADDED LATER: Ben Williams figured out that the somewhat messy formula above with $I \ominus \mathbf{a}$ in it meant that $G$ acts on $H_{n-1}(G^{\ast n})$ by the $n$-th tensor power of the reduced regular representation and asked me if there was a space level explanation of that. Here's my explanation: if one is willing to settle for the homotopy type of the suspension of $G^{\ast n}$, we have that $G$-equivariantly $\Sigma G^{\ast n} \cong (SG)^{\wedge n}$, where $SG$ denotes the unreduced suspension of $G$; and $SG \cong \bigvee_{|G|-1} S^1$ with $H_1(SG)$ being the reduced regular representation.