7
$\begingroup$

Is there a known relation between the Hochschild cohomology of group algebras and cohomology of groupoids?

Clarification: It is known that 1-dimensional Hochschild cohomology of the Group algebra C[G] is isomorphic to so-called external derivations of C[G]. On the other hand, it is known that the space of derivations can be identified with so-called characters on a groupoid aG of adjoint actions of the group G.

Therefore 1-dimensional Hochschild cohomology can be identified with 1-dimensional cohomology of the Cayley complex of the groupoid aG in the case when the group G is a finite presented group. I would like to know if a similar identification is known for the Hochschild cohomologies of higher dimensions

$\endgroup$
4
  • 3
    $\begingroup$ This question is very vague. To begin, what do you mean with cohomology of groupoids? Cohomology of the classifying space? Or something different? And what kind of relation are you looking for? I'm temporarily voting to close as "unclear what you're asking" but I'll retract the vote if the question is edited to make it more focused. $\endgroup$ Commented Feb 12, 2018 at 14:31
  • 3
    $\begingroup$ Didn't the question just get changed to reflect the answer? The question now asks about only groups and how it relates to group cohomology which perhaps is a different question entirely than the OP had in mind. $\endgroup$ Commented Feb 12, 2018 at 19:04
  • 2
    $\begingroup$ @BenjaminSteinberg Agreed: I'm rolling it back -- I strongly dislike anything that comes across as putting one's words in another's mouth, even if that wasn't the intention $\endgroup$
    – Yemon Choi
    Commented Feb 12, 2018 at 20:17
  • 1
    $\begingroup$ @YemonChoi, my apologies. I was only trying to make the question more precise so that it will not be closed. In any case, I hope my answer below will be of some help to the OP. $\endgroup$ Commented Feb 12, 2018 at 21:39

1 Answer 1

6
$\begingroup$

For every $\mathbb{Z}G$-bimodule $M$ there exists a $G$-module $U(M)$ such that the Hochschild cohomology of $\mathbb{Z}G$ with coefficients in $M$ is naturally isomorphic to the group cohomology of $G$ with coefficients in $U(M)$.

To see this observe that there is an adjunction between $G$-modules and $\mathbb{Z}G$-bimodules defined as follows: the right adjoint $$U: \mathbb{Z}G-{\rm BiMod} \to G-{\rm Mod}$$ sends a $\mathbb{Z}G$-bimodule $M$ to the $G$-module $M$ with the action given by $(g,m) \mapsto gmg^{-1}$, and the left adjoint $L: G-{\rm Mod} \to \mathbb{Z}G-{\rm BiMod}$ sends a $G$-module $M$ to the $\mathbb{Z}G$-bimodule $$L(M) := M \otimes \mathbb{Z}G = \oplus_{g \in G}M\left<g\right>$$ where the right action is given by $(\sum_g m_g\left<g\right>,h) \mapsto \sum_g m_g\left<gh\right>$ and the left action is given by $(h,\sum_g m_g\left<g\right>) \mapsto \sum_gh(m_g)\left<hg\right>$. It then follows from general categorical considerations that for every $\mathbb{Z}G$-bimodule $M$ there is a natural isomorphism $$ {\rm HH}^n(\mathbb{Z}G,M) = {\rm Ext}_{\mathbb{Z} G-{\rm BiMod}}(\mathbb{Z}G,M) = {\rm Ext}_{\mathbb{Z} G-{\rm BiMod}}(L(\mathbb{Z}),M) \cong $$ $$ {\rm Ext}_{G-{\rm Mod}}(\mathbb{Z},U(M)) = {\rm H}^n(G,U(M)). $$ In particular, the Hochschild cohomology ${\rm HH}^n(\mathbb{Z}G) := {\rm HH}^n(\mathbb{Z}G,\mathbb{Z}G)$ is naturally isomorphic to the group cohomology of $G$ with coefficients in the $G$-module $\mathbb{Z}G$ equipped with the conjugation action.

When $G$ is finite the group cohomology $H^n(G, \mathbb{Z}G)$ is isomorphic to the cohomology $H^n(aG,\mathbb{Z})$ of the adjoint groupoid $aG$, and so the connection that you state holds in higher dimensions. I'm not sure what happens when $G$ is infinite.

Remark (of a somewhat unrelated nature):

The connection above comes from interpreting the group cohomology of $G$ as the Quillen cohomology of the ($\infty$-)-groupoid ${\bf B}G$ and the Hochschild cohomology of $\mathbb{Z}G$ as the Quillen cohomology the dg-category ${\bf B}(\mathbb{Z}G)$. The connection then arises from the natural adjunction between $\infty$-groupoids and dg-categories.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .