3
$\begingroup$

Let $R$ be a commutative ring with unity and let $G$ be a finite group. Let $\gamma \in Z^{2}(G,R^{\times})$ be a $2$-cocycle. The twisted group ring (of $G$ over $R$ with respect to $\gamma$) $R *_{\gamma} G$ is an $R$-algebra defined as follows: it is the free $R$-module with basis $\{ u_g \}_{g \in G}$ where the multiplication on the basis is defined via $u_{g}u_{h} = \gamma(g,h)u_{gh}$ for all $g,h \in G$ and $ru_g =u_gr$ for all $g \in G$ and $r \in R$. (Note that this also goes under different names such as crossed product order. Moreover, the term twisted group ring sometimes has a different but related meaning.)

The structure of $R *_{\gamma} G$ only depends on the class of $\gamma$ in $H^2(G,R^{\times})$. It would appear that this result is well-known to experts (see page 2 of this preprint, for example https://arxiv.org/abs/1611.00499). However, I can't seem to find a proof is this in the literature.

Is there a nice a reference (textbook or article) that gives an introduction to such twisted group rings and in particular proves the above fact on the relation with $H^2(G,R^{\times})$? Ideally, it should be for general $R$ rather than the special case that $R$ is a field.

Note that I don't need a proof (I know how to do this), but rather a reference to a proof, etc.

$\endgroup$
  • 1
    $\begingroup$ If $\gamma$ and $\gamma'$ differ by $\partial c$, then the map $u_g \mapsto c_g u_g$ is an isomorphism between the two group rings. $\endgroup$ – LSpice Jul 10 '18 at 11:29
  • 6
    $\begingroup$ (Meaning, it's probably too short to expect to find a reference.) $\endgroup$ – LSpice Jul 10 '18 at 11:36
  • $\begingroup$ @LSpice I agree that the proof is very short. But it would still be nice to have a reference that gives an introduction to twisted group rings and their basic properties. $\endgroup$ – Henri Johnston Jul 10 '18 at 12:06
  • 2
    $\begingroup$ Here is a more general construction I don't know where to find in the literature: $\gamma$ can actually be generalized to an action of $G$ on $R$ regarded as an $\text{Ab}$-enriched category with one object. Equivalently, to a homomorphism from $G$ to the automorphism $2$-group of $R$, whose $\pi_0$ is $\text{Out}(R)$ and whose $\pi_1$ is $Z(R)^{\times}$ ($R$ need not be commutative and $G$ need not be finite). The $2$-cocycle case is the special case where the induced map on $\pi_0$ is trivial. The twisted group ring is given by taking the homotopy quotient by this action. $\endgroup$ – Qiaochu Yuan Jul 10 '18 at 17:04
1
$\begingroup$

Computations of the sort you want are done in Pierce's Associative algebras, Chapter 14. It's not precisely your contextx, but the computations in that chapter definitely should prepare you properly to see that there is one obvious map that works.

Concretely, Pierce shows that if $E/F$ is a Galois extension with Galois group $G$, then there is a bijection between elements of the Brauer group $B(E/F)$, which you can think of as isomorphism classes of simple central $F$-algebras with $E$ as maximal subfield, and $H^2(G,E^\times)$, among other things.

The construction goes as you suggest: to every $2$-cocycle one can associate such an $F$-algebra by taking the group algebra $FG$ with a crossed multiplication to obtain a map $Z^2(G,E^\times)\to B(E/F)$, which one then checks is trivial on boundaries and descends to the desired isomorphism.

$\endgroup$
  • $\begingroup$ Thanks for this. I agree that the ideas are very similar, but unfortunately, it's not quite what I need (I actually know of a number of references that go over the same material as Chapter 14 in Pierce's book). $\endgroup$ – Henri Johnston Jul 12 '18 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.