A right crossed module is a homomorphism of groups $\partial\colon F\to G$ together with a right action of $G$ on $F$, written $(g,f)\mapsto f^g$, satisfying certain conditions.

The question is, whether a crossed module $F\to G$ is in some sense abelian.

Following
Norrie,
we set
$$
Z_F=\{f\in F\mid f^g=f \ \ \forall g\in G\},\ \text{ then }\ Z_F\subset Z(F);
$$
$$
Z_G=\{g\in Z(G)\mid f^g=f\ \ \forall f\in F\}.
$$
Here $Z(G)$ denotes the center of $G$.
One easily checks that $\partial(Z_F)\subset Z_G$. We write $\partial_Z$ for the induced homomorphism
$\partial_Z\colon Z_F\to Z_G.$
We say that the abelian crossed module $\partial_Z\colon Z_F\to Z_G$ is the *center*
of the crossed module $\partial\colon F\to G$.

The following definition is a version of a definition of González-Avilés.

*Definition.* A crossed module $F\to G$ is called *quasi-abelian*,
if the morphism of crossed modules
$$
(Z_F\to Z_G)\to (F\to G)
$$
is a quasi-isomorphism. This means that the induced homomorphisms
$$
\ker\partial_Z\to\ker\partial\ \ \text{ and }\ \ {\rm coker}\ \partial_Z\to{\rm coker}\ \partial
$$
are isomorphisms. In other words,

(i) $\partial(F)\cdot Z_G=G$ and

(ii) $\partial(Z_F)=Z_G\cap \partial(F)$.

We obtain an example of a quasi-abelian crossed module $G^{\rm sc}\to G$ from a connected reductive group $G$ over a field $k$ of characteristic 0. Here $G^{\rm sc}$ is the universal covering of the commutator subgroup $G^{\rm ss}:=[G,G]$ of our reductive group $G$. We have a differential $$ \partial\colon G^{\rm sc}\twoheadrightarrow G^{\rm ss}\hookrightarrow G. $$ By functoriality, $G$ acts on $G^{\rm sc}$ on the right. Let $\bar k$ denote an algebraic closure of $k$, then we obtain a quasi-abelian crossed module $G^{\rm sc}(\bar k)\to G(\bar k)$.

A *braiding* of a crossed module $F\to G$ is a map
$$
\{\ \} \colon G\times G \to F,\ g_1,g_2\mapsto \{g_1,g_2\}
$$
satisfying certain conditions, in particular,
$$
\partial\{g_1,g_2\}=g_1^{-1} g_2^{-1} g_1 g_2.
$$
A braiding is called symmetric if $\{g_1,g_2\}\{g_2,g_1\}=1$.
A braiding is called Picard if it is symmetric and also $\{g,g\}=1$.

We define a canonical braiding of a quasi-abelian crossed module as follows. Let $g_1,g_2\in G$. By (i) we can write $$ g_1=z_1\cdot\partial(f_1),\ g_2=z_1\cdot\partial(f_2), \ \text{ where } z_1,z_2\in Z_G. $$ Then we set $$ \{g_1,g_2\}=f_1^{-1}f_2^{-1}f_1 f_2. $$ Using (ii), one can prove that this braiding is well defined. It is symmetric and even Picard. We see that any quasi-abelian crossed module admits a Picard braiding.

**Question.** What are examples of a non-quasi-abelian crossed module admitting a braiding?
Admitting a symmetric braiding?