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Let $\mathcal{S}$ be the $\infty$-category of spaces, and let $G$ be a finite group, and let $BG$ be the groupoid with one object and automorphisms given by $G$. Consider the $\infty$-category $\mathrm{Fun}(BG, \mathcal{S})$. (Note: This can be presented via a model category of $G$-spaces where the weak equivalences are the nonequivariant homotopy equivalences. In particular, this should not be confused with the category of $G$-equivariant spaces.) Some natural examples of objects in this $\infty$-category include the orbits $G/H$ for $H \subset G$ a subgroup (regarded as discrete spaces). Say that an object in this $\infty$-category is $G$-finite if it belongs to the smallest subcategory of $\mathrm{Fun}(BG, \mathcal{S})$ generated by finite colimits and retracts by the orbits $\{G/H\}$ for $H \subset G$. In other words, the object can be represented by a retract of a finite $G$-CW complex.

If an object of $\mathrm{Fun}(BG, \mathcal{S})$ is $G$-finite, then clearly the underlying space is a retract of a finite CW complex. I'd like to know if the converse is true. That is, is there a simple example of an object of this category (i.e., a space with a $G$-action) whose underlying space is finitely dominated (i.e., compact in $\mathcal{S}$) but which does not lie in the subcategory generated under finite colimits and retracts by the $\{G/H\}_{H \subset G}$?

It is also possible to ask the same question for the $\infty$-category $\mathrm{Sp}$ replacing $\mathcal{S}$: that is, one considers $\mathrm{Fun}(BG, \mathrm{Sp})$ and the unreduced suspension spectra of orbits, $\Sigma^\infty_+ (G/H)$, for $H \subset G$. These generate a thick subcategory of $\mathrm{Fun}(BG, \mathrm{Sp})$ which is contained in the subcategory $\mathrm{Fun}(BG, \mathrm{Sp}^\omega)$ spanned by those objects in $\mathrm{Fun}(BG, \mathrm{Sp})$ whose underlying spectrum is finite (if we write $\mathrm{Sp}^\omega$ for the $\infty$-category of finite spectra). Are these subcategories the same? That is, is $\mathrm{Fun}(BG, \mathrm{Sp}^\omega)$ generated as a thick subcategory by the $(G/H)_+$?

Remark: Suppose $X \in \mathrm{Fun}(B\mathbb{Z}/p, \mathrm{Sp}^\omega)$. Then, if $X$ belongs to the thick subcategory generated by the orbits, the $p$-completion of the Tate construction $X^{t \mathbb{Z}/p}$ will be the $p$-completion of a finite spectrum by the Segal conjecture. I'm curious if it is possible to construct an object $X$ as above such that $X^{t \mathbb{Z}/p}$ does not satisfy this. That would give a negative answer to the second question.

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    $\begingroup$ In Fun(BG,S), the orbits G/H generally won't be compact when H is nontrivial. But G itself generates the compact objects under finite colimits and retracts (this follows formally from the fact that it is a compact object which corepresents a conservative functor: in this case, the functor which forgets the G-action). $\endgroup$
    – Jacob Lurie
    Nov 28, 2014 at 19:17
  • $\begingroup$ @JacobLurie: I agree, but I'm looking for objects in $\mathrm{Fun}(BG, \mathcal{S})$ whose image in $\mathcal{S}$ is compact; they don't have to be compact in $\mathrm{Fun}(BG, \mathcal{S})$ itself. $\endgroup$
    – Akhil Mathew
    Nov 28, 2014 at 19:27
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    $\begingroup$ Ah, misunderstood you. I don't have very much helpful to suggest, then. One comment: in the special case G = Z/p, if a space X with an action of G is to have the form you describe, then it satisfies the Sullivan conjecture. In particular, the homotopy fixed points for the action of G on the p-adic completion X_p must itself arise as the p-adic completion of a finite space. Possibly this could give you a way to test if something is a counterexample? (A finite X w/G action such that (X_p)^hG is not the p-adic completion of anything finite?) $\endgroup$
    – Jacob Lurie
    Nov 28, 2014 at 20:15
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    $\begingroup$ Actually wait: I've misunderstood the question again, and what I said doesn't necessarily apply. (It applies when X comes from a finite object in the "genuine" equivariant category. But your finiteness condition looks like it might be weaker than that, even without allowing retracts.) $\endgroup$
    – Jacob Lurie
    Nov 28, 2014 at 21:25
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    $\begingroup$ For $\mathbb{Z}/p$-actions on finite spectra, then if a spectrum $X \in \mathrm{Fun}(B\mathbb{Z}/p, \mathrm{Sp})$ belongs to the thick subcategory generated by $S^0$ and $(\mathbb{Z}/p)_+$, its Tate construction needs to be the $p$-adic completion of a finite spectrum. I was hoping that it might be possible to argue along those lines. $\endgroup$
    – Akhil Mathew
    Nov 28, 2014 at 21:31

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