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David Corwin
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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 $R$. Is

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.

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 $R$. 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.

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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David Corwin
  • 15.4k
  • 10
  • 83
  • 123

Is there algebraic $K$-theory of a group independent of the base ring?

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 $R$. 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.