# $G$-CW complex structure of universal a $\mathcal{F}$-space

Let $G$ be a finite group and $H$ be an abelian subgroup of $G$. Let $\mathcal{F}$ be a family of all subgroups of $H$ , i.e. $\mathcal{F}= \{K : K \leq H\}$ Define universal $\mathcal{F}$-space $E\mathcal{F}$ as in Peter May's Alaska Notes.

Question 1: What is $G$-CW complex structure of $E\mathcal{F}$ ?

Question 2: Is the $n$-th skeleton of $E\mathcal{F}$ is $G$-homotopy equivalence with $G/H \ast G/H \ast \cdots \ast G/H$ (n+1 topological join)?

Any hint or reference will be appreciated.

Thank you.

Firstly, your family $$\mathcal{F}$$ is not closed under conjugation if $$H$$ is not normal. Depending on what you want to do, this may not be an issue.
There are two references for the construction of $$E\mathcal{F}$$ that I know of (but I would be glad to hear of more). The first is tom Dieck's book Transformation groups, where in Chapter I.6 he states that a model for $$E\mathcal{F}$$ is given by the infinite join $$E\mathcal{F} = X \ast X \ast \cdots$$ where $$X=\bigsqcup G/H_a$$ is a disjoint union of orbit types. This is a $$G$$-CW complex in an obvious way, and suggests that the answer to your second question is no.
The second reference is Lück's Transformation groups and algebraic $$K$$-theory, available from the author's webpage (scroll down to Books). Apparently, using the results in Chapter 2 of that book, one can build a $$G$$-CW complex model for $$E\mathcal{F}$$ by an iterative process of attaching $$G$$-cells. The construction is not carried out explicitly, however, and I'm not sure of the details (perhaps this is the topic for a separate MO question).
• Actually I'm interested to calculate integer graded Bredon cohomology of $X = G/H \ast G/H \ast \cdots \ast G/H$ with constant coefficient system , which is equivalent to calculated the cohomology of $X/G.$ For this reason , I need $G$-CW complex structure of $X.$ How can I proceed ? Any hint? – Surojit Aug 21 '15 at 8:03
• The action of $G$ on your space $X$ is not free (its isotropy groups are all conjugates of $H$) so I don't see why the Bredon cohomology should reduce to the ordinary cohomology of the quotient. – Mark Grant Aug 21 '15 at 9:07
• Ah, OK. Maybe it would be better to ask a separate question about Bredon cohomology of joins. Before doing so, you could think about how this relates to the question mathoverflow.net/questions/211122/… given that $Y\ast Z\simeq \Sigma Y\wedge Z$. – Mark Grant Aug 25 '15 at 6:10