Timeline for Are finite $G$-spectra idempotent complete?
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6 events
| when toggle format | what | by | license | comment | |
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| May 30, 2021 at 15:41 | comment | added | Maxime Ramzi | 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 | |
| Mar 30, 2021 at 15:21 | comment | added | Oscar Randal-Williams | @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. | |
| Mar 30, 2021 at 15:16 | comment | added | Oscar Randal-Williams | @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 $\mathrm{Or}(G)$. | |
| Mar 30, 2021 at 14:54 | comment | added | Tim Campion | 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... | |
| Mar 30, 2021 at 14:29 | comment | added | Chris Schommer-Pries | 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? | |
| Mar 30, 2021 at 8:42 | history | answered | Oscar Randal-Williams | CC BY-SA 4.0 |