Timeline for Are finite $G$-spectra idempotent complete?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 4, 2021 at 11:53 | answer | added | user171227 | timeline score: 5 | |
| Apr 1, 2021 at 21:07 | comment | added | Tim Campion | @DenisNardin The tension between your intuition and Oscar's almost-complete answer is killing me! | |
| Mar 30, 2021 at 8:42 | answer | added | Oscar Randal-Williams | timeline score: 9 | |
| Mar 29, 2021 at 23:24 | comment | added | user171227 | If $G$ is cyclic of prime order $p$ and with generator $g \in G$, you can invert $p(g-1) \in \pi_0(S[G])$. This process kills $S^n \wedge G/G_+$ completely and the chains of the resulting spectrum are naturally modules over $R = \mathbb{Z}[p^{-1},\zeta_p]$. For a finite spectrum $V$ the chains $C_*(V[(p(g-1))^{-1}];\mathbb{Z})$ will therefore admit a filtration where the associated graded is a direct sum of shifts of finitely generated free $R$-modules, which can presumably be ruled out in $K_0(R)$. | |
| Mar 29, 2021 at 22:28 | comment | added | Tim Campion | Interesting, thanks! I think I'd buy that either Oscar's suggestion or user171227's suggestion would probably give something which is not finite as a Borel $G$-spectrum, but because the notion of being a projective $\mathbb Z[G]$-module is closely tied up with freeness of the $G$-action, I wonder if such constructions might nevertheless be finite as genuine $G$-spectra? | |
| Mar 29, 2021 at 22:25 | comment | added | Oscar Randal-Williams | @Tim. Yes, that is what I mean. | |
| Mar 29, 2021 at 22:17 | comment | added | Tim Campion | Oh -- are you saying to take $X$ to be some finitely-dominated CW complex with nonvanishing Wall obstrutction, and then to regard the universal cover $\tilde X$ as a $G$-space for $G = \pi_1(X)$, and then think that maybe $\Sigma^\infty \tilde X$ might be a finitely-dominated but not finite $G$-spectrum (which happens to be Borel)? | |
| Mar 29, 2021 at 22:14 | comment | added | Tim Campion | @OscarRandal-Williams Now I'm more confused -- $K_0(\mathbb Z[\pi_1(X)])$ is where the Wall finiteness obstruction lives, but $\pi_1(X)$ has nothing to do with the $G$ in the question with respect to which things are equivariant. Moreover, the Wall finiteness obstruction vanishes for any 1-fold suspension space... Are you suggesting that the category of finite spectra is not idempotent complete? If so, then I think that's incorrect. If not, then how does equivariance play a role in your suggestion? Are you talking about some form of equivariant Wall finiteness obstruction? | |
| Mar 29, 2021 at 21:49 | comment | added | Oscar Randal-Williams | @Tim Sure, not every group has nontrivial reduced projective class group, but some do, e.g. $G=\mathbb{Z}/23$. | |
| Mar 29, 2021 at 21:17 | comment | added | Denis Nardin | 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 :) | |
| Mar 29, 2021 at 21:05 | comment | added | Tim Campion | @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. | |
| Mar 29, 2021 at 21:03 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Mar 29, 2021 at 21:01 | comment | added | Oscar Randal-Williams | 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). | |
| Mar 29, 2021 at 20:54 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Mar 29, 2021 at 20:42 | history | asked | Tim Campion | CC BY-SA 4.0 |