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I have two naive questions about stacks.

1) Is it possible to define stacks in the Zariski topology?

Presuming you can:

2) If a stack has a coarse moduli, and the coarse moduli space is a scheme, then does that mean that your stack is a stack in the Zariski topology?

In general, I am trying to understand why a new notion of open cover is necessary if all I am interested in is remembering stabilizers. Certainly this is too simple a mind-set, so feel free to enlighten me.

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1.) It's possible to define stacks on ANY category equipped with a Grothendieck topology (such a category with a topology is called a site). In particular, this holds true for the Zariski site. Moreover, there is always a way to define an "Artin stack"- these are those stacks which arise as torsors for a groupoid object in your site. Outside of algebraic geometry, these give rise to notions of topological and differentiable stacks, for instance.

EDIT: As long as groupoid objects exist in your category.

2.) As in Harry's post, any stack which is a stack in a site which is finer than the Zariski topology is also a stack in the Zariski topology.

To address your general question as to "why a new notion of open cover is necessary if all I am interested in is remembering stabilizers", you should learn a bit about Grothendieck topologies. I'll make a couple remarks:

i) If all you cared about were stabilizers, then you wouldn't need to use any covers at all; ordinary fibered categories would do the trick!

Indeed, take a group object in your site acting an object, and take the action groupoid- it is a groupoid object. Look at the pseudo-functor which assigns each object of your site the groupoid of maps into this groupoid object (considering the object as a groupoid with all identity arrows). This remembers the stabilizers for this action.

ii) (subcanonical) Grothendieck topologies are a choice of a type of cover for your objects, in such a way that this object is the colimit of these covers, AND "this is important to remember". This is a little imprecise, so, allow me to elaborate via an example from topology:

Let $U_i$, $i\in I$ be an open cover of a space X. Then, continuous maps from X to another space Y are in bijection with with continuous maps $f_i:U_i \to Y$ which agree on their intersection. This is just saying that X is the colimit of this open cover. Instead, we can view this a property of the presheaf $Hom(blank,Y)$ represented by $Y$ on the category of topological spaces (for you set theorists, choose a Grothendieck universe).

For any $X$ and any open cover $U_i$, $i\in I$ of $X$, (let $Hom(blank,Y)=F$)

the natural map $F(X) \to \varprojlim \left[{\prod{F(U_i)}} \rightrightarrows {\prod{F(U_{ij})}}\right]$

is a bijection.

If $F$ is any presheaf, this is just saying $F$ is a sheaf. Since this is NOT true for an ARBITRARY presheaf $F$, X is no longer the colimit of its open covers in the full category of all presheaves. The same argument holds for all fibred categories- it's only true if we restrict to STACKS (and $X$ then becomes the weak colimit of this cover, but, never mind).

The reason you add the condition for descent for covers, is so that "all maps into your stack from a space are continuous". More precisely, and more generally, it's so that maps from a space, scheme, whatever you site is, into a stack can be determined by mapping out of elements of some covering of your object in a way that glues (for stacks, rather than sheaves, they don't need to AGREE on the intersection, but, agree up to an invertible 2-cell, plus some coherency conditions).

Combining these ideas, if you have a group acting on an object, the pseudo-functor produced by the action groupoid is rarely a stack with respect to your topology, but you can stackify it, and then it will become one and still remember all the stabilizers. I hope this helps!

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What you said about Artin stacks being definable over any site is not true. You need a "geometric context" in the sense of Toën. That is, you need a distinguished class of morphisms that commutes with finite sums, is squarable, has finite pullbacks, is compatible with the topology etc. Then an Artin stack is a a stack representable by a "geometric space" in a suitable sense. Your examples are true, but the total statement is not. It's in Cours 2. –  Harry Gindi Apr 16 '10 at 2:27
    
Well, I suppose that yes, I do need a class of groupoid objects to exist such that when I take their enriched simplicial nerve, each structure map admits setions with respect to the Grothendieck topology. –  David Carchedi Apr 16 '10 at 10:02
    
I don't know what you're talking about, but it sounds very cool. Where can I read about it? –  Harry Gindi Apr 16 '10 at 10:10
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If you have groupoid objects (which means that certain pullbacks need to exist), you can look at those stacks arising as torsors of groupoid objects (these will be the stackification of pseudo-functors of the form Hom( ,G)). Whether or not these deserve to be called Artin stacks in general may indeed be contentious, but this is why I used quotation marks. I agree that you will need additional axioms for these "Artin stacks" to be equivalent to a bicategory of groupoids and torsors, so maybe this is what you mean. What is "Cours 2"? I would be interested in looking at what these conditions are. –  David Carchedi Apr 16 '10 at 11:45
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@Harry, my other comment (about simplicial objects) is not quite what I meant to say. But, what I should have said is that for these stacks to behave nicely (other than what I said above), sometimes the structure maps of the groupoid are required to be some sort of "local fibration". To see this in the topological context, see: Noohi's Foundations of Topological Stacks I. –  David Carchedi Apr 16 '10 at 11:52
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1.) Yes, you can define a stack in on any site.

A stack is merely a fibered category satisfying descent along all covers (with respect to a specific Grothendieck topology). There's a particularly beautiful form of the descent condition given in Vistoli's notes. That is, a fibered category $F$ on $\mathcal{C}$ is a stack if given any object $S$ of $\mathcal{C}$, and any covering sieve $T$ of $S$, the induced functor $Hom(S,F)\to Hom(T,F)$ is an equivalence of categories. Here, of course, we replace $S$ and $T$ with their representable pseudofunctors/fibered categories.

If you meant an algebraic stack, they are étale stacks, so they are in particular Zariski stacks. There are also representability conditions (the diagonal is representable by an algebraic space), but those are not particularly important for this discussion.

2.) Since every stack in a finer Grothendieck topology is automatically a stack in the Zariski topology, then it follows that these two things are unrelated (in particular, the finer the topology, the more covering sieves you have, which makes it harder to meet the descent condition.)

I can't recommend the notes by Angelo Vistoli on Descent, Grothendieck Topologies, Fibered Categories, and Stacks enough. They're really top-notch (and avaiable on his website).

For your underlying question:

I'm sure that you've probably heard things said like "The Zariski Topology is the 'Wrong' topology for Algebraic Geometry". This is true for a number of reasons involving cohomology that I don't really understand. However, it turns out that if you live in the étale topology, a lot of things become much easier to work with in terms of geometry and definitions. In particular, a scheme is merely an étale sheaf of sets on $Aff$ satisfying a covering condition by smooth monomorphisms of representable functors. I don't know if you have ever seen the definition of the global Zariski topology, but it's really quite nasty. (It's in Demazure-Gabriel, if you're dead set on reading about it).

Anyway, I'm not the person to tell you about the wonderful geometric properties of the étale topology, but the rough idea is that the étale topology is similar to the analytic topology for complex schemes.

However, the nice geometric properties of the étale topology come at the price of having to prove a lot of foundational results from commutative algebra. In particular, many of the proofs characterizing (formally) smooth, unramified, and étale morphisms require Zariski's main theorem, which is actually very deep and notoriously hard to prove.

Edit: Toën's notes on stacks are wonderful as well, but they assume that you've got a pretty good grasp of descent. In particular, schemes are simply sheaves, while the classical presentation of a scheme as a locally ringed space is simply given as the "locally ringed space associated with the scheme". This viewpoint is quite useful to make the transition to thinking about (particularly) the small étale topos associated with a scheme, which looks almost like a locally ringed space but in fact gives us a lot of useful information not available in the classical presentation. The small étale topos is the perfect illustration of why a topos is a "generalization of the notion of a space".

Links:

Vistoli's Notes on Descent

Toën's Notes on Stacks

Edit 2: To address Brian's concerns, I'd like to note that the Zariski topology is an extremely important tool (especially for commutative algebra in my experience, however little that may be). I have problems with the global Zariski topology, which I maintain is "more trouble than it's worth".

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Harry, do not parrot a claim about erroneous nature of "most" published proofs of ZMT unless you verified it for yourself. Proofs in EGA and Hartshorne are fine. ZMT in no deeper than other things (e.g., Artin's work) needed to make algebraic spaces useful. Don't besmirch value of Zariski topology, lest you find yourself caught in circular reasoning (e.g., dubious you could get anywhere without having to prove that formally etale lfp maps are open for the Zariski topology; real work isn't avoided by making definitions). You lack the experience to pass judgement on "best" foundations. –  BCnrd Apr 16 '10 at 3:07
    
Brian, my emphasis in this post was on avoiding the definition of the global Zariski topology on the category of affines. The Zariski topology on a scheme is an invaluable tool for proving things in commutative algebra and algebraic geometry. I guess you could say that I was besmirching the global Zariski topology, but I have never seen it used as more than a proof of concept for the Grothendieck topologies. With respect to ZMT, I've removed the note because you're right, I have not checked them myself. That's not to say I don't trust Mel, but you make a good point that I should check myself. –  Harry Gindi Apr 16 '10 at 9:04
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The global Zariski topology has its uses, see: A. Hirschowitz: Cohérence et dualité sur le gros site de Zariski. Proceedings Trento 1988, Lect. Notes in Math.1389 (1989) 91-102. –  Leo Alonso Jan 20 '11 at 11:33
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