5
$\begingroup$

(I'd like to premise that I'm not an expert about these topics (just a student), so many of my doubts and perplexities are probably symptoms of my mathematical immaturity)

I'm currently studying differentiable stacks and I'm a little confused by the following statement:

Many classes of interesting geometric objects (orbit spaces, orbifolds, differentiable spaces,...) can be seen as differentiable stacks

This statement fascinated me, because I thought that its meaning was the following

Let $\mathscr{C}$ be a class of geometric objects (orbit spaces, orbifolds, differentiable spaces,...) and let's suppose that we have a natural notion of isomorphism of objects of $\mathscr{C}$. We associate to each object $C\in \mathscr{C}$ a Lie groupoid $\mathcal{G}(C)$ so that: $$C_1\cong C_2 \iff \mathcal{G}(C_1),\mathcal{G}(C_2) \text{ are Morita equivalent }$$ for any $C_1,C_2\in \mathscr{C}$.

This felt natural to me and (in good faith) I took for granted it was like this. But as I went on studying, I realized that probably I am wrong. Let's take the example of orbit spaces.

Let $M$ be a smooth manifold equipped with a left action of a Lie group $G$. Clearly the information of the action is completely encoded in the action groupoid $G\ltimes M$.

But why Morita equivalence should encode a natural notion of isomorphism?

I'm aware that the Morita equivalence of the action groupoids is equivalent to the homeomorphism of the orbit spaces and the isomorphism of the normal representations of the action groupoids.

But in a way this just moves the problem to why "homeomorphism of the orbit spaces and isomorphism of the normal representations of the action groupoids" is the correct notion of isomorphism between spaces of the type $M/G, N/H$.

It's like we are proceding backwards: first we presuppose that Morita invariance is the correct notion of equivalence and then whatever the Morita equivalence implies it's the correct notion of isomorphism between objects of $\mathscr{C}$.

This feels a bit unnatural to me. If I were forced to invent by myself a notion of isomorphism between $M/G,N/H$, I'd probably say that it's a homeomorphism $F:M/G\to N/H$ such that $F, F^{-1}$ both lift to smooth maps $M\to N,N\to M$. I'm NOT saying that this is the smartest approach: I'm just saying that this is what feels more natural to my (mathematically immature) mind.

Thank you in advance for your help!

$\endgroup$
1
  • 1
    $\begingroup$ The title is grammatically incorrect. Did you mean to say "What" rather than "Why"? $\endgroup$ Commented Apr 25 at 17:42

1 Answer 1

4
$\begingroup$

Stacks form an (∞,1)-category. The latter informal notion has many equivalent implementations: simplicial category, topological category, quasicategory (also known as ∞-category), Segal category, complete Segal space. For the purposes of this answer, the simplest model might be that of a relative category. A relative category is a category $C$ equipped with a subcategory $W$ of weak equivalences such that $C$ and $W$ have the same objects.

The relative category of stacks in groupoids on a site $S$ is defined as follows. Objects are functors $\def\op{{\sf op}}\def\Grpd{{\sf Grpd}}S^\op→\Grpd$. Morphisms are natural transformations of functors. Weak equivalences are local equivalences, which can be defined most easily if the site $S$ has enough points (true for differentiable stacks), in which case we can define them simply as maps whose stalks are equivalences of groupoids.

To connect the above description of local equivalences to Morita equivalences, we need to connect the model of relative categories to other models, like simplicial categories; in our case, categories enriched in groupoids are sufficient. Given a relative category $R=(C,W)$ and two objects $X,Y∈C$, the mapping simplicial set (or, in our case, mapping groupoid) can be computed using the general construction known as the hammock localization, which is somewhat difficult to work with in practice.

In our case, a much simpler computation is possible: the relative category of stacks admits a model structure, in which case the mapping groupoid can be computed more easily: take a cofibrant replacement $Q X→X$, a fibrant replacement $Y→R Y$, and compute the groupoid of maps $X→Y$ as the groupoid of natural transformations of functors $Q X → R Y$. (This procedure has the same origin as projective/injective resolutions in homological algebra.)

For the case of stacks in groupoids, a projective resolution $Q X$ of $X$ can be computed as the Čech nerve $Č(U)$ of some covering family $U$ (e.g., an open cover) of $X$. Furthermore, any stack $Y$ in groupoids is projectively fibrant, so we can take $R Y = Y$. Thus, the groupoid of morphisms $X→Y$ can be computed as the groupoid of natural transformations $Č(U)→Y$.

The canonical map $Č(U)→X$ is not an equivalence of Lie groupoids in the traditional sense, i.e., there is no morphism of Lie groupoids $X→Č(U)$ such that the composition $X→Č(U)→X$ is identity. However, the bundlization of the map $Č(U)→X$ is a left and right principal bibundle over the Lie groupoids $Č(U)$ and $X$. This motivates the notion of a Morita equivalence of Lie groupoids. More details can be found in Etendues and stacks as bicategories of fractions by Dorette A. Pronk, which proves a version of the equivalence stated above.

As a side remark, these days many researchers find the language of bibundles and Morita equivalences of Lie groupoids somewhat unwieldy to use. The equivalent language of presheaves, as presented in the first few paragraphs above, has the same expressive power, and is often easier to use in practice.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .