Timeline for Is there algebraic $K$-theory of a group independent of the base ring?
Current License: CC BY-SA 4.0
10 events
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| Apr 30, 2021 at 14:24 | comment | added | John Klein | There is a dumb, useless way to answer your question in the affirmative: take the coproduct of all $K(R[G])$ as $R$ ranges over all isomorphism classes of rings. | |
| Apr 19, 2021 at 22:15 | comment | added | Ben Wieland | My last comment is a little misleading. The Whitehead space depends not just on a group, but on a space $K(S)\wedge X\to K(S[\Omega X])\to Wh(X)$. Generalizing beyond the Wall obstruction and the Whitehead torsion requires moving from $Z$ to $S$, but it also requires more details about the space than just its fundamental group. | |
| Apr 16, 2021 at 18:43 | comment | added | Ben Wieland | @TimCampion The sphere is better than the integers. The cofiber on spectra of $K(S)\wedge BG\to K(SG)$ is called the Whitehead space. Its component group is where the Wall finiteness obstruction lives, its fundamental group is where the Whitehead torsion lives, and its higher homotopy groups, where the choice of the sphere makes a difference, are the correct generalization. | |
| Apr 16, 2021 at 18:32 | comment | added | Ben Wieland | Look up the Assembly Map. For torsion-free groups, the standard conjecture, closely related to the Borel conjecture, is that $K(ZG)=K(Z)\wedge BG$. The Farrell-Jones conjecture reduces the general case to the virtually cyclic case, the case of finite groups and finite groups extended by $Z$. What if $R=ZH$ is a group ring? Then $RG=Z[H\times G]$ is again a group ring. This is asking us to relate the $K$-theory of a product to $K$-theory of the factors. This turns out not to be so nice, even if one of the factors is $Z$ and the other is finite. | |
| Apr 16, 2021 at 18:29 | comment | added | David Corwin | @DavidHandelman Yes but then you get the K-theory of $\mathbb{Z}$, which is quite nontrivial (e.g. its torsion is related to deep conjectures in number theory), and I want this to be purely about the group. | |
| Apr 16, 2021 at 6:55 | comment | added | Denis Nardin | @TimCampion $K_0(\mathbb{Z}[G])=K_0(\mathbb{S}[G])$ since $\mathbb{S}[G]→\mathbb{Z}[G]$ is a map of connective ring spectra which is an isomorphism on $\pi_0$ (there are various ways of proving this but the quickest is probably just applying the weighty theorem of the heart, aka Gillet-Waldhausen) | |
| Apr 16, 2021 at 1:57 | comment | added | Tim Campion | Even better, you could take $K_0(\mathbb S[G])$, where $\mathbb S[G] = \Sigma^\infty_+ G$ is the group ring of $G$ over the sphere spectrum. Although $K_0(\mathbb Z[G])$ is where e.g. the Wall finiteness obstruction lives, so is arguably more useful. | |
| Apr 16, 2021 at 0:55 | comment | added | David Handelman | I'm sure you've considered K${}_0 ({\bf Z} G)$ (K$_0$ of the integral group ring); this is a K-theoretic invariant of $G$ that is independent of the choice of $R$, and there is a natural map $K_0 ({\bf Z}G) \to K_0 (R G)$, ... | |
| Apr 15, 2021 at 22:33 | history | edited | David Corwin | CC BY-SA 4.0 |
added 20 characters in body
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| Apr 15, 2021 at 20:26 | history | asked | David Corwin | CC BY-SA 4.0 |