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Given a ring R and a group $G$, I can consider the group ring $R[G]$ and then take the algebraic $K$-theory $K(R[G])$. This the $K$-theory of the category $\operatorname{Rep}_R(G)$. As a variant, one might start with a connected simply connected Lie group and take the universal enveloping algebra of its Lie algebra with coefficients in $R$. Then $K_0(R[G])$ is the representation ring, for example.

If $G$ is trivial, then this is just the $K$-theory of $R$, so it depends greatly on the ring $R$.

My question: is it possible to extract a $K$-theoretic invariant of the group $G$, independent of $R$?

Here's one possible approach: one could consider the map $K(R) \to K(R[G])$ and take the cofiber. Then one might ask whether this cofiber is independent of $R$ after some kind of completion (maybe $p$-completion?) or under certain conditions.

Another approach: could one find a spectrum $K(G)$ so that $K(R[G]) = K(R) \wedge K(G)$. Again, even if this is not true as stated, might it be true after completion or under some reasonable conditions?

Even if neither approach works, hopefully these two ideas explain what I mean by "$K$-theoretic invariant of the group $G$, independent of $R$". I'm interested in any other ideas or approaches to this question.

Let me add what one might expect for $K_0$. It would make sense to hope that for a reductive Lie group $G$, we have $K_0(G)=K_0(\mathbb{C}[G])$, possibly taking reduced $K_0$. But for higher $K$-groups, one of course can't just take $\mathbb{C}$-coefficients.

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    $\begingroup$ 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 $K_0 ({\bf Z}G) \to K_0 (R G)$, ... $\endgroup$
    – David Handelman
    Commented Apr 16, 2021 at 0:55
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    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Commented Apr 16, 2021 at 1:57
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    $\begingroup$ @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) $\endgroup$
    – Denis Nardin
    Commented Apr 16, 2021 at 6:55
  • $\begingroup$ @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. $\endgroup$
    – David Corwin
    Commented Apr 16, 2021 at 18:29
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    $\begingroup$ @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. $\endgroup$
    – Ben Wieland
    Commented Apr 16, 2021 at 18:43

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