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).

  • $\begingroup$ 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? $\endgroup$ – Surojit Aug 21 '15 at 8:03
  • $\begingroup$ 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. $\endgroup$ – Mark Grant Aug 21 '15 at 9:07
  • $\begingroup$ Because I have taken constant coefficient system. $\endgroup$ – Surojit Aug 21 '15 at 9:09
  • $\begingroup$ 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 $Y\ast Z\simeq \Sigma Y\wedge Z$. $\endgroup$ – Mark Grant Aug 25 '15 at 6:10
  • $\begingroup$ I wasn't sure whether the post is worth bumping (and editing) just for this, but here is a working link: him.uni-bonn.de/lueck/publications.php (and a link directly to the book) $\endgroup$ – Martin Sleziak Aug 15 '19 at 6:54

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