Non-quasi-abelian braided crossed modules

Question

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?

Answer 1

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 $x\hat \otimes y=-y\hat \otimes x$.