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](http://archive.numdam.org/article/BSMF_1990__118_2_129_0.pdf),
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](http://arxiv.org/abs/1110.4542).


*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?



Non-quasi-abelian braided crossed modules
Recall that free simplicial groups model all connected homotopy types. A simplicial group $G$ is free if each $G\_n$ is free and degeneracies are defined by maps between the bases.


A simplicial group is $n$-reduced if $G\_i=0$ for $i\leq n$. They model $(n+1)$-connected homotopy types.


A simplicial group $G$ is $n$-coskeletal if the Moore complex $N\_\*(G)$ vanishes in dimensions higher or equal than $n$, $N\_i(G)=0$, $i\geq n$. They model $n$-Postnikov pieces, i.e. spaces with no higher homotopy groups in dimensions $>n$.


The $n$-reduced $m$-coskeletal free simplicial groups model $(n+1)$-connected $m$-Postnikov pieces, so there are plenty of them. 


Let $G$ be a $0$-reduced $3$-coskeletal simplicial group. Conduché showed that its Moore complex, which reduces to $\partial\colon N\_2(G)\rightarrow G\_1$, is a braided crossed module. Since $G\_1$ is free, the center of $\partial$ is $\partial\_N \colon Z\_{N\_2(G)}\rightarrow 0$. Hence, $\partial$ is not quasi-abelian unless $\pi\_1G=0$. In this way you obtain tons of examples, at least one for each simply connected $3$-Postnikov piece $X$ with $\pi\_2X\neq 0$. For the symmetric case, simply start with a $1$-reduced $4$-coskeletal simplicial group.


There are also non-quasi-abelian braided and symmetric crossed modules where the bottom group has a center. Indeed, take $G$ as in the previous paragraph and let $\bar G$ be the quotient of $G$ by all triple commutators. Denote $\bar\partial$ the braided and symmetric crossed modules which is the Moore complex of $\bar G$. An important result of Curtis shows that the natural projection $G\twoheadrightarrow \bar G$ is a quasi-isomorphism, hence so is the induced morphism $\partial\rightarrow\partial'$. If $B$ is a basis of the free group $G\_1$ ($G\_2$ in the symmetric case), then $Z\_{\bar G\_1}=Z(\bar G\_1)\cong\wedge^2\mathbb{Z}[B]$ coincides with the commutator subgroup. Therefore $\bar \partial\_N$ is surjective, so $\bar\partial$ cannot be quasi-abelian unless $\pi\_1G=0$ ($\pi\_2G=0$ in the symmetric case).


I also want to remark that being quasi-abelian is not a nice notion from the homotopy or derived point of view. Any braided or symmetric crossed module is quasi-isomorphic to one of the examples above. Hence, there are quasi-isomorphisms $\partial\stackrel{\sim}\rightarrow\partial'$ such that $\partial'$ is quasi-abelian and $\partial'$ is not. In other words, the only braided or symmetric crossed modules which are derived quasi-abelian are $A\rightarrow 0$, with $A$ an abelian group.


Let me add an explicit example, which fits into the second family of examples pointed out above. Let $\partial \colon F\rightarrow G$ be the braided crossed module defined as follows. The group $G$ is the quotient of the free group on two generators $\{x,y\}$ by triple commutators. The group other group is $F=\otimes^2\mathbb{Z}[x,y]$, which is free abelian with basis $\{x\otimes x,x\otimes y,y\otimes x,y\otimes y\}$. The homomotphism $\partial$ sends $a\otimes b$ to the commutator of $a$ and $b$ and the braiding is defined by $\{x,y\}=x\otimes y$ (or the other way round, depending on your conventions). The cernter of $\partial$ is the surjective homomorphism $\partial\_Z\colon F\twoheadrightarrow \wedge^2\mathbb{Z}[x,y]\colon a\otimes b\mapsto a\wedge b$ with trivial braiding. The morphism $\partial\_Z\rightarrow\partial$ induces an isomorphism on kernels, but not on cokernes since $\partial\_Z$ is surjective but $\operatorname{coker} \partial=\mathbb{Z}[x,y]$.


If you want a symmetric example, replace $\otimes^2\mathbb{Z}[x,y]$ with the reduced tensor product $\hat\otimes^2\mathbb{Z}[x,y]$ which is the quotient by $a\otimes b+b\otimes a$, i.e. it is isomorphic to $\mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}$ with generators $x\hat \otimes x$, $y\hat \otimes y$, and $x\hat \otimes y=-y\hat \otimes x$.


