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Question: Let $G$ be a compact Lie group (you can assume that $G$ is finite if you like). Is the category of finite $G$-spectra idempotent complete?

Here, by "finite $G$-spectra", I mean those objects in the category of genuine $G$-spectra which can be constructed via a finite number of sums and cofiber sequences from orbit cells $S^n \wedge (G/H)_+$ where $H \subseteq G$ is a closed subgroup. This includes all representation spheres. By "category", I mean either the stable $\infty$-category or the triangulated category, whichever you prefer (in the stable case, a category is idempotent complete iff its homotopy category is, so it doesn't affect the question.)

When $G$ is trivial, the answer is yes, but I suspect the answer in general is no, -- maybe already for $G = C_2$?

By a theorem of Thomason, triangulated subcategories of a triangulated category $\mathcal T$ which generate $\mathcal T$ under splitting of idempotents are in bijection with subgroups of $K_0(\mathcal T)$. So another way to phrase the question is the following: let $\mathcal T$ be the category of compact $G$-spectra. Then is t$K_0(\mathcal T)$ generated as a group by the classes of $G$-orbits $\Sigma^n(G/H)_+$?

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    $\begingroup$ I suspect the answer is no, and you can get a counterexample from a finitely-dominated finite CW-complex with nontrivial Wall finiteness obstruction (this obstruction factors over taking stabilisation). $\endgroup$ Mar 29, 2021 at 21:01
  • $\begingroup$ @OscarRandal-Williams I'm confused -- when $G$ is trivial, I believe the answer is yes... I don't have a proof in front of me, but I think the idea is to use that any compact object in spectra must have finitely-generated homology, and to use this to construct a model of it with finitely many cells. $\endgroup$
    – Tim Campion
    Mar 29, 2021 at 21:05
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    $\begingroup$ I think the standard proof for spectra will work here as well (using Mackey functors valued homology) but it's too late here for me to try to write a proof :) $\endgroup$ Mar 29, 2021 at 21:17
  • $\begingroup$ @Tim Sure, not every group has nontrivial reduced projective class group, but some do, e.g. $G=\mathbb{Z}/23$. $\endgroup$ Mar 29, 2021 at 21:49
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    $\begingroup$ @Tim. Yes, that is what I mean. $\endgroup$ Mar 29, 2021 at 22:25

2 Answers 2

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I think the correct setting to look at this question is that of

W. Lück, "Transformation groups and algebraic K-theory". Lecture Notes in Mathematics, 1408. Mathematica Gottingensis. Springer-Verlag, Berlin, 1989.

Lück considers the $K$-theory of the category $\mathbb{Z}\mathrm{Or}(G)$ of functors $\mathrm{Or}(G)^{op} \to \mathbb{Z}\text{-mod}$, and in particular proves (Theorem 10.36) that there is a decomposition $$K_0(\mathbb{Z}\mathrm{Or}(G)) \cong \bigoplus_{\text{conjugacy classes of } H \leq G} K_0(\mathbb{Z}\pi_0(WH))$$ where $WH = N_G(H)/H$ is the Weyl group of $H \leq G$. (Stated in this way it is valid for $G$ a compact Lie group.) The image of a functor $M : \mathrm{Or}(G)^{op} \to \mathbb{Z}\text{-mod}$ in the $H$th component is $$M(G/H)/Im(M(\phi) \text{ for all nonisomorphisms } \phi : G/H \to G/H')$$ with its evident $\pi_0(WH)$-action.

By taking $G$-cellular chains, a finitely-dominated $G$-CW-complex or $G$-CW-spectrum $Y$ has an Euler characteristic $$\chi(Y) \in K_0(\mathbb{Z}\mathrm{Or}(G))$$ which satisfies inclusion-exclusion as usual. Under the splitting described above one calculates that $$\chi(S^n \wedge G/H_+) = (-1)^n [\mathbb{Z}\pi_0(WH)].$$ Thus any finite $G$-spectrum $Y$ in the sense of the question has $\chi(Y)$ given by a sum of such terms, and so the image $$Wall(Y) \in \bigoplus_{\text{conjugacy classes of } H \leq G} \widetilde{K}_0(\mathbb{Z}\pi_0(WH))$$ of $\chi(Y)$ in this quotient obstructs $Y$ being equivalent to a finite $G$-spectrum.

Lück also shows (Theorem 14.12) that all elements of $K_0(\mathbb{Z}\mathrm{Or}(G))$ arise from some finitely-dominated $G$-CW-complex.

Putting this together, I think it follows that finite $G$-spectra are idempotent-complete if and only if we have $\widetilde{K}_0(\mathbb{Z}\pi_0(WH))=0$ for all $H \leq G$.

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  • $\begingroup$ If $Wall(Y) \neq 0$, then this shows that $Y$ is not equivalent to a finite $G$-spectrum, but how do you see that $Wall(Y) = 0$ implies that $Y$ is equivalent to a finite $G$-spectrum? Could there be other obstructions? $\endgroup$ Mar 30, 2021 at 14:29
  • $\begingroup$ Thanks! I'm a bit confused about how to regard $\chi(Y)$ as a stable invariant. I can see that the inclusion-exclusion property implies that $\chi(\Sigma Y) = -\chi(Y)$. So assuming that the stabilization of compact $G$-spaces can be constructed via a Spanier-Whitehead type construction, I buy that $\chi$ is well-defined on the stabilization of $G$-spaces. But the stabilization of $G$-spaces is only naive $G$-spectra. In order to pass to genuine $G$-spectra, it seems we need to know how $\chi$ interacts with smashing with representation spheres. Perhaps that's also in Lück's book... $\endgroup$
    – Tim Campion
    Mar 30, 2021 at 14:54
  • $\begingroup$ @TimCampion: I think that for the approach you are suggesting you just need to show that $\chi(\text{regular representation sphere})$ is a unit in $K_0(\mathbb{Z}\mathrm{Or}(G))$, which I have not done but am sure it is. What I had in mind was instead taking the "Mackey-functor valued cellular chains" of the $G$-CW-complex $Y$, and letting $\chi(Y)$ be the underlying complex of presheaves on $\mathrm{Or}(G)$. $\endgroup$ Mar 30, 2021 at 15:16
  • $\begingroup$ @ChrisSchommer-Pries: Lück shows that for $G$-CW-complexes it is the only obstruction. I don't see any step that would not go through for $G$-spectra. $\endgroup$ Mar 30, 2021 at 15:21
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    $\begingroup$ This may be a bit late to the battle, but I think one doesn't need to go through spaces, iirc you can prove that $K(Sp_G) = \bigoplus_{[H]\leq G} K(\mathbb S[W_GH])$, which on $K_0$ immediately shows that your description of $K_0(\mathbb Z Or(G))$ is $K_0(Sp_G)$, so that any class in the RHS is realized as a $K$-theory class of a compact $G$-spectrum (in particular if one of them is nonzero, you get an obstruction to finiteness). Conversely, a theorem of Thomason shows that a dense subcategory is entirely determined by its image on $K$-theory, so this shows that it's the only obstruction $\endgroup$ May 30, 2021 at 15:41
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To complement Oscar's more systematic answer, let me expand my comment about the case $G = \mathbf{Z}/p\mathbf{Z}$ for a prime number $p$, where the answer is no when $\tilde{K}_0(\mathbf{Z}[G]) \neq 0$. Candidate non-finite retracts of finite $G$-spectra have already been presented in the comments, to see that they are not equivalent to finite genuine spectra we can use that the natural homomorphisms $$\tilde{K}_0(\mathbf{Z}[G]) \to \tilde{K}_0(\mathbf{Z}[\zeta_p]) \to \tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}])$$ are both isomorphisms. The first of these is an isomorphism by a theorem of Rim. To see that the second map is an isomorphism we use that $\mathbf{Z}[\zeta_p,p^{-1}] = \mathbf{Z}[\zeta_p,(\zeta_p-1)^{-1}]$ and that $(\zeta_p - 1) \subset \mathbf{Z}[\zeta_p]$ is a prime ideal (see e.g., the proofs of Lemma 1.3 and 1.4 on page 2 of Washington's book, in fact $p = (\zeta_p - 1)^{p-1} u$ for $u \in \mathbf{Z}[\zeta_p]^\times$), so exercise 3.8(b) in chapter I of Weibel's K-book applies.

For any genuine $G$-spectrum $X$, the fixed points $X^e$ for the trivial subgroup come with an action of $G$, so we can form $$C_* (X;\mathbf{Z}[\zeta_p,p^{-1}]) := C_* (X^e;\mathbf{Z}) \otimes_{\mathbf{Z}[G]} \mathbf{Z}[\zeta_p,p^{-1}].$$ This defines an exact functor $C_* (-;\mathbf{Z}[\zeta_p,p^{-1}])$ from the stable $\infty$-category of genuine $G$-spectra to $\mathcal{D} (\mathbf{Z}[\zeta_p,p^{-1}])$. This functor sends $\Sigma^\infty_+ (G/e) \mapsto \mathbf{Z}[\zeta_p,p^{-1}]$ and an easy exercise shows that it sends $\Sigma^\infty_+ (G/G) \mapsto 0$. Therefore a genuine $G$-spectrum which is finite in your sense will be sent to a compact object in $\mathcal{D} (\mathbf{Z}[\zeta_p,p^{-1}])$ representing $0 \in \tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}])$.

Any element of $\tilde{K}_0(\mathbf{Z}[\zeta_p,p^{-1}]) $ may be represented by the image under this functor of a retract of $ ( \Sigma^\infty_+ G ) ^{\oplus n} $ for some $n$. Some such retract then won't be finite when $\tilde{K}_0 (\mathbf{Z}[\zeta_p,p^{-1}]) \neq 0$.

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