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

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.

• What exactly are $N$ and $qN$? Aug 19, 2021 at 6:22
• Typos, that's what they are :-) Aug 19, 2021 at 6:59

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.

1. 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.
2. 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 $$\alpha_x \circ \alpha_x \Rightarrow \alpha'_{xy}$$.
3. 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 $$\alpha_p\circ\alpha_q \overset{\chi_{p,q}}{\Longrightarrow} \alpha_{pq} \overset{1_{\alpha_{pq}} \ast \zeta_{p,q}}{\Longrightarrow} \alpha_{pq}$$