Is there a “groupoid integral” with values in a groupoid?

Let $G = \{G_1 \rightrightarrows G_0\}$ be a finite groupoid, i.e. $G_1,G_0$ are both finite sets, and let $A$ be $\mathbb Q$-module. Regard $A$ as a discrete groupoid $A \rightrightarrows A$, and let $f: G\to A$ be a functor — equivalently, there is a set $G_0/G_1$ of isomorphism classes of $G$, and $f$ is an $A$-valued function on this set. Then Baez and Dolan define a groupoid integral: $$\int_G f = \sum_{x\in G_0/G_1} \frac{f(x)}{\lvert {\rm Aut}(x)\rvert}$$ where to make the above precise I'm using the fact that if $x,y \in G_0$ are isomorphic, then $\lvert {\rm Aut}(x)\rvert = \lvert {\rm Aut}(y)\rvert$, and choosing an isomorphism $x \to y$ in fact induces an isomorphism ${\rm Aut}(x) \to {\rm Aut}(y)$. Anyway, the point is that $\int_G f$ actually depends only on the "stack" $G_0 // G_1$, where for our purposes "stack" can mean "groupoid up to equivalence": if $G,G'$ are equivalent groupoids and $f':G' \to A$ is the functor corresponding to $f$ under the equivalence, then $\int_Gf = \int_{G'}f'$.

Suppose now that $A$ is not a $\mathbb Q$-module but some (possibly weak) version of a "$\mathbb Q$-module in groupoids". (If you want, I have no objection to you thinking of $A$ as a strict object — the weakened version would replace all the axioms for a $\mathbb Q$ vector space with natural isomorphisms that have their own coherency.) Then it still makes sense to talk about functors $f: G \to A$.

Question: Does there exists an extension of the groupoid integral above to integrals of the form $\int_Gf$ where $f: G \to A$ is a functor but $A$ is not discrete, and that plays well with equivalences of groupoids $G \to G'$ and $A \to A'$?

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$\int_Gf$ only depending on the 'stack' is trivial for the case of discrete $A$ (as in discrete as a groupoid), because $G_0/G_1 \simeq G'_0/G'_1$ for equivalent groupoids $G, G'$. I'd like to think of A as a groupoid with one object, so that the sum is the product in that groupoid, but then I don't know how to think of the functor $f$. –  David Roberts Apr 16 '10 at 1:05
For the more general case, an easier first case to consider is when $A$ is/comes from a crossed module $V_1 \to V_0$ in the category of $\mathbb{Q}$-modules. (On a technical note, since all such modules are free, the naive 2-category of such crossed modules is the 'right' one, you don't need to pass to a localisation, such as Noohi's butterflies/papillons). BTW these crossed modules are usually called Baez-Crans 2-vector spaces. My guess is that the desired integral will give an element of $V_1\times V_0$, which when one translates back to groupoids is an arrow of $A$. –  David Roberts Apr 16 '10 at 1:08

1 Answer

Yes. It seems you're only looking for an object up to isomorphism, so all you really need is divisibility and addition to yield objects that are well-defined up to isomorphism. Assuming your notion of rational vector space in groupoids involves a Picard groupoid (i.e., symmetric monoidal category with all objects invertible) with "multiplication by a/b" functors, and a zoo of compatibility isomorphisms under multiplication and addition, then you can use the Baez-Dolan formula.

This is a groupoid version of pushing forward a constant sheaf on a zero dimensional stack (which is well-known under suitable assumptions on stabilizers and the characteristic of the coefficients).

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So I'm confused. I have a groupoid $G$ and another groupoid $A$, and $A$ happens to be a $\mathbb Q$-vector space object. So if I use the Baez-Dolan formula, I can pick out an object of $A$ up to isomorphism. But I certainly cannot pick out an object of $A$, and I would like to, with their formula. Moreover, suppose that $g\in G$ and $a\in A$ are objects. Then $f$ gives a map of groups $f_g: {\rm Aut}(g) \to {\rm Aut}(a)$, and I would expect the integral, far from just dividing by $|{\rm Aut}(g)|$, to involve both groups (for example, only divide by $|{\rm ker}(f_g)|$). –  Theo Johnson-Freyd Apr 16 '10 at 2:47
I guess by "plays well with equivalences" I mean something like: if I have an equivalence $G \sim G'$, then I'd expect $\int_G f \cong \int_{G'}f'$ as objects in $A$, and I'd even expect to be able to compute this isomorphism from the equivalence. But I would like to be able to pick out a particular object of $A$. –  Theo Johnson-Freyd Apr 16 '10 at 2:49
I have heard that some category theorists frown upon picking out particular objects in a category. A better integral may involve an object of A together with some datum that satisfies a universal property. You won't get a unique object this way, but it will be unique up to unique isomorphism (or will lie in a contractible space of solutions, if you're working with higher categories). It is difficult to tell what precise properties you want, since you don't provide concrete motivation in your question. –  S. Carnahan Apr 16 '10 at 4:43