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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?

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3 Answers 3

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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 $C_0\ltimes C_1$.

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@ Fernando, Chris, Evan: Thank you for your answers. Oh yes, this question is ill-posed. Consider e.g. the free 2-group generated by an object A and an iso from I to A. Then pi_1 is trivial and pi_2 also, so it is 2-equivalent to the trivial 2-group. But in the first case, the group I defined is ZxZ and in the second case trivial. Can someone confirm this. But maybe I can rescue a bit of the question by restricting to 2-groups freely gen. by some objects and some isos between tensor expressions of these objects. Can something be said in this case? –  Werner Thumann Jan 31 '13 at 22:55
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@Werner, free crossed modules are a special case of what you propose, hence the computation above works as a counterexample too. BTW, I think that, in general, free objects are not better behaved for this king of question, because they are universal examples, hence anything that fails in general fails also for free ones. –  Fernando Muro Feb 1 '13 at 8:08
    
Fernando, thank you for your help, I will think over all this. –  Werner Thumann Feb 1 '13 at 8:45

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 $\pi_1(\mathcal{G}) = G$.

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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.

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