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I have often encountered definitions of the kind "stacks are equivalence classes of groupoids under Morita equivalence" in topological or differentiable context, with the notion of Morita equivalence appropriate for the context.

I have two questions:

1) can one define Deligne-Mumford or Artin stacks this way, as Morita equivalence classes of algebraic groupoids?

2) if yes, what is the connection between the corresponding notion of Morita equivalence and the one for the rings (i.e. rings are called Morita equivalent if they have equivalent categories of modules over them)?

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3 Answers

up vote 4 down vote accepted

Regarding your question relating Morita equivalence as defined for internal groupoids and as defined for rings:

Given an internal groupoid $G$ (say, Lie, topological or algebraic), it defines a presheaf of groupoids on the ambient category ($Diff$, $Top$ or $Sch$). The stackification of this presheaf is the category of (right, say) principal $G$-bundles. Thus two groupoids define equivalent stacks (isomorphic in the correct 2-categorical sense) whenever their categories of principal bundles are equivalent. This is the groupoid version of 'Morita equivalence' a la rings. But when these two groupoids present equivalent stacks, they are 'Morita equivalent' in the sense there is a span between them of 'essential equivalences' - fully faithful essentially surjective (-appropriately interpreted) functors. So the two notions coincide, at least when the ambient category has enough quotients of the right sort of reflexive coequalisers (as the examples I list do). In the more general case with no quotients, as I mention in my comment to the original question, I'm not sure what happens.


EDIT: It's been a while since I answered this, but now I know what happens in the case I mention in the final sentence above the line. One shouldn't take the category of principal bundles as being the stackification of the internal groupoid $G$, rather take the category of internal anafunctors with codomain $G$ and with domain a groupoid with no non-identity arrows. The arrows are a little tricky to describe, but one gets a stack over the base category with fibre over $X$ the hom-category $Hom(X,G)$ in the bicategory of internal groupoids, anafunctors and transformations. This will be covered in a forthcoming paper of mine.

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It is important not to define stacks as Morita-equivalence classes of groupoids. The correct statement is:

"isomorphism classes of stacks are the same as Morita-equivalence classes of groupoids"

where isomorphism is to be understood in the 2-category that stacks form.


Let's do a similar mistake in a more familiar context, just to see how wrong it then feels. Let's try to define the notion of finite dimensional vector space. What do you think of:

$$\text{"finite dimensional vector spaces are natural numbers" ?}$$

It's a pretty bad definition. Nevertheless, the statement "isomorphism classes of finite dimensional vector spaces are the same as natural numbers" is correct.

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Can we say that the category of stacks is equivalent to the category of groupoids whose morphisms are isomorphism classes of Morita equivalences? –  Dmitri Pavlov Sep 12 '10 at 15:17
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Groupoids form a natural 2-category, where the morphisms are functors and the 2-morphisms are natural transformations. This 2-category is not equivalent to stacks but you can enhance it by throwing in more morphisms. You can do this by adding the so-called "bibundles" or, equivalently, by formally inverting the Morita equivalences. Either way you get a new 2-category, equivalent to (presentable) stacks, whose objects are still groupoids. The equivalences in this new 2-cat are precisely the Morita equivalences. –  Chris Schommer-Pries Sep 12 '10 at 17:14
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The main literature reference that I know of is Dorette Pronk's "Entendues and stacks as bicategories of fractions". There this is proven for stacks on the site of smooth manifolds. I believe a similar proof works in most of the sites considered in algebraic geometry. This statement is not true over arbitrary sites, but it is true for quite a large class of interest. You need your site to admit quotients by certain nice simply transitive group actions. This allows you to define the composition of "bibundles" and hence a 2-category. ... (cont) –  Chris Schommer-Pries Sep 12 '10 at 18:39
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(cont) Once you have this 2-cat, there is an easy proof that it is equivalent to the full sub-2-category of presentable stacks. The proof is analogous to the proof that the 2-category of algebras, bimodules, and bimodule maps is equivalent to the full sub-2-category of abelian categories, right exact functors, and nat. trans spanned by those abelian categories of the form A-mod for an algebra A. –  Chris Schommer-Pries Sep 12 '10 at 18:43
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Actually one does not need quotients to invert Morita equivalences, but then one has to use anafunctors instead of bibundles. See my paper arxiv.org/abs/1101.2363 –  David Roberts Feb 7 '11 at 8:18
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1)I think that if there's a surjective representable map from a scheme X to stack M, then we can define the groupoid X\times_M X with base X. With different X, the corresponding groupoids are Morita equivalent. If the map from X to M is etale then it's DM stack. Therefore, we can define them that way.

2) if yes, what is the connection between the corresponding notion of Morita equivalence and the one for the rings (i.e. rings are called Morita equivalent if they have equivalent categories of modules over them)?

We can look at the groupoid C*-algebras of the corresponding groupoids. Then the Morita equivalent groupoids yield Morita equivalent algebras (but not in the opposite direction.)

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1) part of the question was what to mean by Morita equivalence of groupoids. I guess one can define groupoids to be M. eqvt when categories of groupoid representations are equivalent, is it what you mean? in that case I have another question: is there a concerete definition of Morita equivalence for groupoids in the spirit of M. equivalence for rings (the one using projective generator). Also, are two stacks represented by Morita equivalent algebraic groupoids isomorphic? –  Dima Sustretov Sep 11 '10 at 22:23
    
No, I mean by M.E of Groupoid $G_1/G_0$ and $H_1/H_0$ is there exist a bimodule P over G and H, such that groupoids G and H act on P ect.. You may take a look into Hilsum Skandalis Map and bibundle in arxiv.org/pdf/math/0506484 The modules of groupoid also can be found there or in a paper topological stack. Regarding to your question about being isomorphic, the answer is yes. Just play around with the diagram, it's just several line. –  Hanh Duc Do Sep 12 '10 at 2:59
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