Crossed modules in context of gerbes

Question

Question : How does Crossed modules comes into the set up of gerbes.

I am reading notes on 1- and 2-gerbes by Lawrence Breen. Once he defines torsors, he introduces notion of crossed modules. It was not clear how crossed modules come into picture of gerbes suddenly from nowhere.

Some search on internet gives an expository article which gives a proof of result that

Third cohomology group $H^3(G, M)$ of a group $G$ with coefficients in an abelian $G$-module $M$ is in bijection to the set $Ext^2(G, M)$ of equivalence classes of crossed module extensions of $G$ with $M$.

This gives some vague reason why crossed modules comes into the study of gerbes.

As gerbes are seen as elements of third cohomology class here, I am expecting that understanding about crossed modules in same setup as that of gerbes might give better idea of gerbes.

Can some one please tell me if this one of the reasons why crossed modules comes into picture when studying gerbes? Is there similar result as what I have quoted above in case of sheaf/Cech cohomology where gerbes are studied.

Any references will be useful.

Answer 1

This answer is meant to elaborate on my remark above concerning crossed modules as models for 2-groupoids. A crossed module consists of a group $G$ acting on a group $H$, together with a $G$-equivariant group homomorphism $f:H \to G$ (where we consider $G$ as acting on itself by conjugation). From this data we can construct a 2-groupoid $X$ as follows: $X$ has a single object $x_0 \in X$, the $1$-morphisms from $x_0$ to itself are the elements of $G$ (with composition given by the group structure of $G$), and for $g_0,g_1 \in G$, the $2$-morphisms from $g_0$ to $g_1$ are the elements $h \in H$ such that $f(h)g_0 = g_1$. Vertical composition of 2-morphisms is then given by the group structure of $H$, while the vertical composition uses both this and the action of $G$ on $H$. More precisely, if $h \in H$ is such that $f(h)g_0 = g_1$ and $h' \in H$ is such that $f(h')g_0'= g_1'$ then their horizontal composition is the element $h(h')^{g_0} \in H$ (where $(h')^{g_0}$ denotes the action of $g_0$ on $h'$), considered as a 2-morphism from $g_0g'_0$ to $g_1g_1'$.

It can then be shown that all non-empty connected 2-groupoids arise in this way, i.e., are equivalent (in the sense of 2-groupoids) to a 2-groupoid that comes from a crossed module. This is not completely obvious: in principle, given a connected 2-groupoid $X$ and an object $x_0 \in X$ we may be inclined to construct the corresponding crossed module by setting $G = {\rm Hom}_X(x_0,x_0)$ and letting $H$ be the set of pairs $(\alpha,h)$ where $\alpha: x_0 \to x_0$ is a 1-morphism from $x_0$ to itself and $h$ is a 2-morphism from the identify on $x_0$ to $\alpha$. This construction is in principle "correct", only that it may give $G$ and $H$ which are not groups but only monoids: indeed, $X$ being a 2-groupoid doesn't mean that every 1-morphism is invertible on the nose, only that it is invertible up to an invertible 2-morphism. Nonetheless, one can still show that there is a way to choose $G$ and $H$ such that the resulting 2-groupoid will be equivalent to $X$. There is of course also the issue that in order to reconstruct the cross module we had to choose a base point in $X$. This reflects the fact that cross modules don't model 2-groupoids, but rather pointed 2-groupoids, i.e., 2-groupoids equipped with a base point (also known as 2-groups).