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Let G be a braided monoidal groupoid: it does no harm to suppose that the monoidal product on G is strictly associative, so I'll do that.

"With inverses" means that for every object $X$ of G, there is an object $Y$ and an isomorphism $X\otimes Y\approx 1$ in G, where $1$ is the unit object.

I'd like to assume, without loss of generality, that inverses exist "on the nose", so that for every object $X$, there is an object $X^{-1}$ such that $X\otimes X^{-1} = 1$. That is, the objects of G form a group.

First question: am I allowed to do this?

Now I can get a 2-category BG by "delooping" G (using the monoidal structure), so that BG has only one object *, and the category of morphisms BG(*,*) is exactly G. This is a 2-groupoid with one object, and it carries some sort of additional structure encoding the braiding.

A connected 2-groupoid is exactly the same thing as a crossed module, which consists of data $(H,F, d: H\to F, \phi: F\to \mathrm{Aut}(H))$, where $H$ and $F$ are groups and $d$ and $\phi$ are homomorphisms. In terms of G, F is the group of objects of G, while H is the set of 1-morphisms in G with unit object as domain.

Second question: what extra structure do I put on the crossed module to encode the braiding?

I want to understand such G which are free on some set S of objects. In the translation to crossed modules, the group F will have to be the free group on S.

Third question: how do you describe the group H in this crossed module? (That's what I mean by "explicit".)

There's an extensive literature on braided monoidal categories, so I bet someone has thought about this.

(Oh, and I can deloop one more time to get a weak 3-groupoid B2G, and this thing will model a homotopy 3-type X. If G is free, X is a wedge of 2-spheres. Because of this, I know things like the image and kernel of $d: H\to F$. But what is $H$ itself?)

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It's easy enough to describe the free G on one generator, so it amounts to the same thing to give an explicit description of a coproduct of braided monoidal groupoids with inverses. –  Charles Rezk Nov 11 '09 at 8:14
    
To clarify, you are working in the 1-category (not 2-category) of strict braided monoidal groupoids with strict inverses and (strict?) braided monoidal functors? Otherwise, I don't see why H and F would be uniquely determined. –  Reid Barton Nov 11 '09 at 18:13
    
Yes this is what I mean. –  Charles Rezk Nov 12 '09 at 20:02
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1 Answer 1

I think the answer to the first question is yes. Let $C$ be a monoidal category whose monoid of isomorphism classes is a group, so that every object has an inverse. For each object $X$ pick an object $X^{-1}$ and isomorphisms $l_X : X^{-1} \otimes X \to 1$, $r_X : X \otimes X^{-1} \to 1$ such that the composition $(r_X \otimes X)(X \otimes l_X) : X \to X \otimes X^{-1} \otimes X \to X$ is the identity. (This should be no problem; by assumption we can find $X^{-1}$, $l_X$, $r_X$ satisfying all but the last identity, and we can alter $r_X$ to make the last identity hold.) We also need to choose $1^{-1} = 1$, $l_1 = r_1 = \mathrm{id}$.

Now we build a strict monoidal category $D$ whose objects are the free group on the objects of $C$, where the monoidal structure $\cdot$ is strict and given by the multiplication in this group, and where the Hom sets are "pulled back" along the function $f : \mathrm{Ob}\,D \to \mathrm{Ob}\,C$ which sends a reduced word in objects of $C$ and their inverses to the tensor product of those objects and their chosen inverse objects. This induces a fully faithful functor $F : D \to C$ which is obviously also surjective and thus an equivalence. We need to make it a monoidal functor. To do so, note that $F(A \cdot B) = F(A) \otimes F(B)$ unless there is cancellation between the words $A$ and $B$ in the free group. Where there is cancellation, we use the isomorphisms $l_X$ and $r_X$ to build a map $F(A \cdot B) \to F(A) \otimes F(B)$. The monoidal functor conditions are satisfied because of the conditions on $l_X$ and $r_X$. Finally, the monoid of objects of $D$ is by definition a group.

As for the braiding, we should be able to pull it back along $F$ also, so that $F$ is braided monoidal.

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