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Let $k$ be a ring (resp. profinite ring), $G$ a group (resp. profinite group), and $k[G]$ the group algebra (resp. completed group algebra).

For any such $G$, we may associate to it the group of units $k[G]^\times$ of $k[G]$, and this association is clearly functorial. Has this functor been studied at all?

For example, given an exact sequence $$1\rightarrow G\rightarrow G'\rightarrow G''\rightarrow 1$$ we get a sequence $$1\rightarrow k[G]^\times/k^\times\rightarrow k[G']^\times/k^\times\rightarrow k[G'']^\times/k^\times\rightarrow 1$$ which is probably not exact, but at least the map $k[G]^\times/k^\times\rightarrow k[G']^\times/k^\times$ is injective, and so one might hope that the functor $G\mapsto k[G]^\times$ is left exact.

If it's left-exact, is this actually exact? If not, is anything known about its right-derived functors?

I suppose this is best posed under the assumption that $G$ is abelian, though I'm also very interested in the nonabelian case (or at least whatever still makes sense there in terms of cohomology)

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  • $\begingroup$ What do you mean by "Has this functor been studied" precisely? There has been done quite a lot in order to compute $k[G]^{\times}$; cf. Kaplansky's unit conjecture for group rings. These computations use functoriality all the time, because it is a basic feature. $\endgroup$
    – HeinrichD
    Mar 10, 2017 at 11:28
  • $\begingroup$ @HeinrichD do you have any references or keywords to google? I'm especially interested in the profinite situation. $\endgroup$
    – Will Chen
    Mar 10, 2017 at 13:01

1 Answer 1

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Forget about exactness; this functor isn't additive, which pretty much tanks any hope of doing homological algebra to it even if you restrict to abelian groups.

Here's a special case that's easy to understand. Suppose $G$ is finite and $k$ has characteristic not dividing $|G|$. Then $k[G]$ is semisimple, so we have a decomposition

$$k[G] \cong \prod_i M_{n_i}(D_i)$$

where the product is indexed over the irreducible representations $V_i$ of $G$ over $k$, with $\text{End}_G(V_i) = D_i$ division algebras and $n_i = \dim_{D_i} V_i$. Hence

$$k[G]^{\times} \cong \prod_i GL_{n_i}(D_i)$$

and quotienting by $k^{\times}$ just removes the trivial representation. The functoriality with respect to morphisms $f : G \to H$ comes from pulling back irreps of $H$ to $G$ and seeing how they decompose.

$k[G]^{\times}$ naturally occurs as the automorphism group of the forgetful functor from $G$-representations to $k$-vector spaces (regarded just as a functor, not as a monoidal functor).

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  • $\begingroup$ Would you mind elaborating on why this functor isn't additive? Also is it easy to see that $k[G]^\times$ is the automorphism group of the forgetful functor? (Do you have a reference perhaps?) $\endgroup$
    – Will Chen
    Mar 9, 2017 at 4:29
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    $\begingroup$ @oxeimon: the functor is not additive because the map on $k[G] \to k[H]$ induced by $f + g \colon G \to H$ does not agree with the product of the maps induced by $f$ and $g$. For example, consider $G = H = C_2 = \langle \sigma \rangle$, with $f = g = \operatorname{id}$. Note that $\sigma - 2 \in k[G]$ is a unit, since $(\sigma - 2)(\sigma + 2) = -3$. But multiplication by $2$ induces the $k$-linear map $\sigma \to 1$, which sends $\sigma - 2$ to $1-2 = -1$. This does not differ from $(\sigma - 2)^2 = - 4 \sigma+5$ by a scalar, so $G \mapsto k[G]^\times/k^\times$ is not an additive functor. $\endgroup$ Mar 9, 2017 at 5:00
  • $\begingroup$ @R.vanDobbendeBruyn Ah, great. I see. Thanks! $\endgroup$
    – Will Chen
    Mar 9, 2017 at 5:14
  • $\begingroup$ @Qiaochu: don't Dold and Puppe have some sort of formalism for derived functors of a non-additive functor? I believe the trick is to view everything simplicially, but besides that I don't really know how it works, nor what it actually gives you (I'm sure some of the usual statements have to be modified). $\endgroup$ Mar 9, 2017 at 5:15
  • $\begingroup$ @oxeimon: regarding the question about automorphisms of the forgetful functor, this follows from the Yoneda lemma. $\endgroup$ Mar 9, 2017 at 5:50

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