Connection between the classifications of group extensions and group-graded algebras in terms of non-abelian cohomology

Question

First, consider group extensions with non-abelian kernel $$1\to K\to G \to Q \to 1$$ It is well-known that these are classified by certain cohomological objects, specifically: Any such extension induces an outer action, i.e. a group homomorphism $\omega: Q\to\operatorname{Out}(K)$ (which turns $Z(K)$ into a $\mathbb{Z}Q$-module) and to any outer action we can associate an obstruction in $H^3(Q,Z(K))$ that vanishes iff any extension with the given outer action exists. And if they exist, they are in bijection with the non-abelian cohomology $H^2(Q,K,\omega)$.

Second, consider $Q$-graded, crossed $k$-algebras, i.e. $k$-algebras with a decomposition $A=\bigoplus_{q\in Q} A_q$ such that $1\in A_1$, $A_{q_1} \cdot A_{q_2} \subseteq A_{q_1 q_2}$, and all $A_q$ contain a unit. Again, this situation induces an outer action $\omega: Q\to\operatorname{Out}(A_1)$ (and $Z(A_1)$ becomes a $kQ$-module), for any such outer action there is an obstruction in $H^3(Q,Z(A_1)^\times)$ that vanishes iff any $Q$-graded, crossed $k$-algebra with the given outer action exists. And if they exists, they are classified by non-abelian cohomology $H^2(Q,A_1^\times,\omega)$.

These statements are so similar that it is natural to ask:

Question: What is a natural, common generalization of both statements?

There are natural constructions relating the two: Given a group extension, $A:=k[G]$ is $Q$-graded and crossed with $A_q:=k[qK]$. Conversely, given a crossed algebra, the group of homogeneous units fits into a natural extension $1\to A_1^\times \to (A^\times)_{homog.} \to Q\to 1$

But these constructions do not immediately give implications from one theorem to the other: The group extension version does not give you the algebra version, because $\operatorname{Out}(A_1)$ may be different from $\operatorname{Out}(A_1^\times)$ and $Z(A_1)^\times$ may have little to do with $Z(A_1^\times)$.

Conversely, the algebra version does not give you the group version, $A_1=k[K]$, again because $\operatorname{Out}(K)$ and $\operatorname{Out}(k[K])$ can be very different.

Answer 1

In the meantime I found the right nlab pages to read... The answer seems to be 2-groups! Specifically automorphism 2-groups. I will write up what I have come to understand so far (though I have not checked every last detail of this):

1. What's a 2-group? Like an ordinary group is a category with 1 object in which every morphism is invertible, a (coherent) 2-group is a 2-category with only one object in which every 1- and 2-morphism is invertible (and we have picked a weak inverse of every 1-morphism).

   • Generic Example: For every 2-category $\mathscr{C}$ and every object $\mathcal{C}\in\mathscr{C}$ there is a automorphism-2-group $\operatorname{AUT}_\mathscr{C}(\mathcal{C})$ which has the self-equivalences $\mathcal{C}\to\mathcal{C}$ as 1-morphisms and natural isomorphisms between those as 2-morphisms.

   • This includes both the case of groups by letting $\mathscr{C}$ be the 2-category of categories and the case of $k$-algebra by letting $\mathscr{C}$ be the 2-category of $k$-linear categories.

     A group $K$ is a category with one object and $\operatorname{AUT}_\mathsf{cat}(K)$ is a 2-group with $\operatorname{Aut}_{\mathsf{Grp}}(K)$ as its set of 1-morphisms and for every $h\in K$ a 2-morphism $\alpha \to \kappa_h\circ\alpha$ where $\kappa_h$ is the conjugation with $h$.

     A $k$-algebra is a $k$-linear category with one object and we get its automorphism-2-group $\operatorname{AUT}_{k\mathsf{-cat}}(A)$ which similary consists of $\operatorname{Aut}_{k\mathsf{-Alg}}(A)$ and $A^\times$ as its sets of 1- and 2-morphisms respectively.

   • Other example: Any ordinary group $Q$ can "upgraded" to a 2-group $\mathscr{Q}$ by letting 1-morphisms be elements of $Q$ just as usual and letting 2-morphisms be only identities.

2. Conversely, any 2-group $\mathscr{G}$ naturally defines an ordinary group $\pi_1(\mathscr{G})$ by considering 1-morphisms up to equivalence.

   In our examples

   • $\pi_1(\operatorname{AUT}_{\mathsf{cat}}(K)) = \operatorname{Out}(K)$,
   • $\pi_1(\operatorname{AUT}_{k\mathsf{-cat}}(A)) = \operatorname{Out}(A)$,
   • $\pi_1(\mathscr{Q}) = Q$ I will write $\operatorname{Out}_\mathscr{C}(\mathcal{C})$ as shorthand for $\pi_1(\operatorname{AUT}_\mathscr{C}(\mathcal{C}))$

3. In addition, any 2-group $\mathscr{G}$ also defines an abelian group $\pi_2(\mathscr{G})$ consisting of the 2-isomorphisms of the identity 1-morphism (which is abelian by an Eckmann-Hilton argument) and $\pi_1$ naturally acts on $\pi_2$ by conjugation.

   • In the generic example, $\pi_2(\operatorname{AUT}_\mathscr{C}(\mathcal{C})))$ is the unit group of the commutative monoid $Z(\mathcal{C})$ on which $\operatorname{Out}_\mathscr{C}(\mathcal{C})$ acts by "conjugation"
   • $\pi_2(\operatorname{AUT}_\mathsf{cat}(K)) = Z(K)$ with the natural action by $\operatorname{Out}(K)$.
   • Similary, $\pi_2(\operatorname{AUT}_{k\mathsf{-cat}}(A)) = Z(A)^\times$ with the natural action by $\operatorname{Out}(A)$.
   • And trivially $\pi_2(\mathscr{Q}) = 1$

4. Where do extensions and gradings come in? A common generalisation of both theorem in my question is the following:

Let $Q$ be a group,$(\mathcal{V},\oplus,\otimes)$ be a bimonoidal category and $\mathscr{C}:=\mathcal{V}\mathsf{-cat}$ be the 2-category of categories enriched in $(\mathcal{V},\otimes)$.

Then: For every group homomorphism $\omega: Q\to\operatorname{Out}_\mathscr{C}(\mathcal{C}_1)$, the non-abelian cohomology $H^2(Q,\operatorname{AUT}_\mathscr{C}(\mathcal{C}_1),\omega)$ is in canonical bijection with "crossed, $Q$-graded categories" $\mathcal{C}$ that have $\mathcal{C}_1$ as their degree-1-part and induced outer action $\omega$.

Furthermore: These two sets are non-empty iff a certain obstruction $o(\omega)\in H^3(Q,Z(\mathcal{C}_1)^\times)$ vanishes.

5. This begs the question: What's a "graded category"? A graded category is a category $\mathcal{C}$ enriched in $(\mathcal{V},\otimes)$ that is equipped with a decomposition $\mathcal{C}(x,y) = \bigoplus_{q\in Q} \mathcal{C}_q(x,y)$ such that $\operatorname{id}_x \in \mathcal{C}_1(x,x)$ and composition of morphisms decomposes accordingly as $\circ: \mathcal{C}_p \otimes \mathcal{C}_q \to \mathcal{C}_{pq}$.

   In particular: Considering only the morphisms in $\mathcal{C}_1$ gives us another category (with the same set of objects).

   A graded category is crossed if every $\mathcal{C}_q(x,x)$ contains an isomorphism.

   • Every group extension $1\to K\to G\to Q\to 1$ gives a $Q$-grading on the one-object category $G$ if we let $(\mathcal{V},\oplus,\otimes)$ be $(\mathsf{Set},\sqcup,\times)$, namely the decomposition into cosets of $K$. These gradings are always crossed.

   • A graded $k$-algebra $A=\bigoplus_{q\in Q}$ gives a $Q$-grading on the one-object category $A$ if we let $(\mathcal{V},\oplus,\otimes)$ be $(k\mathsf{-mod},\oplus,\otimes)$. These gradings are crossed iff they are crossed in the usual sense, i.e. if all $A_q$ contain at least one unit.

   A crossed $Q$-graded category induces an outer action $Q\to\operatorname{Out}_{\mathcal{V}\mathsf{-cat}}(\mathcal{C}_1)$ by choosing 1-isomorphisms $u_{q,x}\in \mathcal{C}_q(x,x)$ for every $x\in Ob(\mathcal{C})$ and $q\in Q$ and mapping $q$ to the equivalence class $[\alpha_q]\in\operatorname{Out}(\mathcal{C}_1)$ of the "conjugation"-functor $\alpha_q: \mathcal{C}_1\to\mathcal{C}_1, (x\xleftarrow{f}y) \mapsto (x \xleftarrow{u_{q,x} \circ f \circ u_{q,y}^{-1}} y)$.

   • For group extensions $1\to K\to G\to Q\to 1$ this is exactly the outer action $Q\to\operatorname{Out}(K)$ induced by conjugation.
   • For crossed, $Q$-graded algebras this is also the outer action $Q\to\operatorname{Out}(A_1)$ induced by conjugation.

6. Now, what precisely does the theorem do? For every fixed group homomorphism $\omega: Q \to \pi_1(\operatorname{AUT}_\mathscr{C}(\mathcal{C}_1))$ it establishes really two pairs bijection between the following three sets:

   • $Q$-graded, crossed categories extending $\mathcal{C}_1$ that induce $\omega$

     $$\uparrow \downarrow$$

   • 2-group morphisms $\mathscr{Q}\to\operatorname{AUT}_\mathscr{C}(\mathcal{C}_1)$ up to equivalence that induce $\omega$ on the $\pi_1$-groups, where $\mathscr{Q}$ is the 2-group upgrade of $Q$.

     $$\uparrow \downarrow$$

   • non-abelian cohomology $H^2(Q,\operatorname{AUT}_\mathscr{C}(\mathcal{C}_1),\omega)$, i.e. 2-cocycles $(\alpha,\chi)$ up to 2-coboundaries

     • $(\alpha,\chi)$ being a 2-cocycle means $\alpha: Q \to \{1\text{-morphisms}\}, \chi: Q^2 \to \{2\text{-morphisms}\}$ with $\chi_{p,q}: \alpha_{p} \alpha_{q} \Rightarrow \alpha_{pq}$, $\alpha_q$ is in the equivalence class $\omega(q)$, and the 2-coycle condition holds: $\chi_{xy,z}\circ\chi_{x,y} = \chi_{x,yz}\circ(\alpha_x(\chi_{y,z}))\circ\text{associator}$ as 2-morphisms $(\alpha_x\circ\alpha_y)\circ\alpha_z \Rightarrow \alpha_{xyz}$
     • A 2-coboundary $(\alpha,\chi) \to (\alpha',\chi')$ is map $\lambda: Q\to\{2\text{-morphisms}\}$ such that $\lambda_x: \alpha_x \Rightarrow \alpha'_x$ with $\lambda_{xy}\circ\chi_{x,y} = \chi'_{x,y}\circ(\lambda_x\lambda_y)$ as 2-morphism $\alpha_x \circ \alpha_x \Rightarrow \alpha'_{xy}$.

7. And finally: How does one prove all of that? If I have not made any huge mistakes, the classical proofs from Schreier theory all carry over if one replaces all equations by the appropriate commutative diagrams.

   There is even a regular action of the group $H^2(Q,Z(\mathcal{C}_1)^\times)$ on the set $H^2(Q,\operatorname{AUT}_\mathscr{C}(\mathcal{C}_1),\omega)$ if it is non-empty given by $[\zeta] . [\alpha,\chi] := [\alpha, \zeta.\!\chi]$ where $\zeta.\!\chi$ means the 2-morphism $\alpha_p\circ\alpha_q \overset{\chi_{p,q}}{\Longrightarrow} \alpha_{pq} \overset{1_{\alpha_{pq}} \ast \zeta_{p,q}}{\Longrightarrow} \alpha_{pq}$