Groups lying horizontally in 2-groups

Question

Let $\mathcal{G}$ be a coherent 2-group. Following Baez:HDA5 $\mathcal{G}$ is uniquely determined up to 2-equivalence by the following data: The fundamental group $\pi_1(\mathcal{G})=G$, the second homotopy group $\pi_2(\mathcal{G})=H$, the action of $G$ on $H$ by group autos and a class in $a\in H^3(G,H)$. Now theres a third group associated to $\mathcal{G}$ which consists of the morphisms of $\mathcal{G}$ and where multiplication is given by tensor product (horizontal composition in the 2-category). Of course this is only a monoid a priori but becomes a group after quotienting out the cumbersome to state but rather obvious congruence relation defined below.

Let $\epsilon_X:I\rightarrow X\otimes\bar{X}$, $\eta_X:\bar{X}\otimes X\rightarrow I$ be the unit and counit adjoint equivalences. If $f:X\rightarrow Y$ is an arrow in $\mathcal{G}$, there's the so called mate of $f$ denoted $\hat{f}:\bar{X}\rightarrow\bar{Y}$ and defined as $$\hat{f}=(i_{\bar{X}}\otimes\epsilon_Y)(i_{\bar{X}}\otimes f^{-1}\otimes i_{\bar{Y}})(\eta_X\otimes i_{\bar{Y}})$$ Here $i$ is the identity and I use composition from left to right! It is a result of Laplaza that for any object $X$ there's at most one isomorphism $X\cong I$ built out of $\eta$ and $\epsilon$. If there is one, call $X$ simple. Now define $f:X\rightarrow Y$ and $g:A\rightarrow B$ equivalent iff both $X\otimes\bar{A}$ and $Y\otimes\bar{B}$ are simple and the composition $$I\rightarrow X\otimes\bar{A}\xrightarrow{f\otimes\hat{g}}{}Y\otimes\bar{B}\rightarrow I$$ is the identity.

Now from the above classification of 2-groups it should be possible to construct this group out of $G,H$ and $a$. Does anybody know how this operation on the data $G,H,a$ looks like? Is it a well known construction?

Answer 1

Let $\partial\colon C_1\rightarrow C_0$ be a crossed module, i.e. a group homomorphism together with a right action of $C_0$ on $C_1$, denoted exponentially $c_1^{c_0}$, satisfying the following laws: $$\partial(c_1^{c_0})=-c_0+c_1+c_0$$ $$c_1^{\partial(c_1')}=-c_1'+c_1+c_1'.$$ Here I denote the group laws additively, despite they are non abelian in general.

Crossed modules are equivalent to strict $2$-groups. The strict $2$-group $C_*$ associated to $\partial\colon C_1\rightarrow C_0$ has set of objects in $C_0$ and set of morphisms $C_0\ltimes C_1$, the semidirect product. I'll show that the group you define for $C_*$ is this semidirect product $C_0\ltimes C_1$. Hence it is not invariant under tensor equivalences.

The source and target of $(c_0,c_1)\in C_0\ltimes C_1$ are $$(c_0,c_1)\colon c_0+\partial(c_1)\longrightarrow c_0.$$ Composition is defined by $$(c_0,c_1)\circ(c_0+\partial(c_1),c_1')=(c_0,c_1+c_1').$$ The tensor product is the sum $+$ and the tensor unit is $0\in C_0$. In this case $\epsilon$ and $\eta$ are the identity map on the tensor unit, i.e. $(0,0)\in C_0\ltimes C_1$.

The mate of a morphism $f=(c_0,c_1)\in C_0\ltimes C_1$ is $\hat{f}=(-c_0,-c_1^{-c_0})\in C_0\ltimes C_1$. The only simple object is the tensor unit $0\in C_0$ and two morphisms $f=(c_0,c_1),g=(c_0',c_1')\in C_0\ltimes C_1$ are quivalent iff $$(0,0)=f+\hat{g}=(c_0,c_1)+(-c_0',-(c_1')^{-c_0'})=(c_0-c_0',c_1^{-c_0'}-(c_1')^{-c_0'}) =(c_0-c_0',(c_1-c_1')^{-c_0'})$$ i.e. iff $c_0=c_0'$ and $c_1=c_1'$, that is $f=g$. Therefore your group is simply the semidirect product $C_0\ltimes C_1$.

Answer 2

I have not checked that your construction in invariant under (2-)equivalence of (coherent) 2-group. If it is not invariant, then your question is ill-posed.

Assuming your question is well-posed and hence invariant under 2-equivalence, then it suffices to calculate what happens when the coherent 2-group is skeletal and the adjoint equivalences are trivial (every coherent 2-group is equivalent to one of these by the Baez-Lauda paper you cited). An easy calculation for this case shows that you get what you called $\pi_1(\mathcal{G}) = G$.

Answer 3

Contra Chris, I think the group you want is actually the semidirect product $G \ltimes H$, if I understand your construction correctly. We can identify $H = \operatorname{End}(I)$ with $\operatorname{End}(X)$ for any $X$ by tensoring on the right by $X$. The tensor product of $h \otimes X$ with $h' \otimes X'$ is then $h (X \rhd h') \otimes X \otimes X'$, where $X \rhd h'$ denotes the adjoint action of $X$ oh $h$. In the skeletal case, this is exactly $G \ltimes H$.

My concern about your construction (which I believe is responsible for the difference between my answer and Chris's) is that if $X$ and $A$ are equivalent, there's no canonical map $I \to X \otimes \overline{A}$; such maps are in bijection with equivalences between $X$ and $A$. And indeed, if you quotient out by all such maps, you're left with $G$. But in general, you would need to choose a contractible space of equivalences between $X$ and every object equivalent to $X$, which is essentially equivalent to passing to the skeletal case.