Summer stacks

This is the page for the 2012 Summer stacks reading group.

Resources

The book in progress of Behrend, Fulton, Kresch and other people is available here: [1]

Thanks to Sukhendu we have a copy of Champs algebriques" by Laumon and Moret-Bailly, currently in Ed's office.

The Stacks Project: [2]

Milestones

6/1 Finish Chapter 1

6/14 Finish Chapter 2

6/29 Finish Chapter 3

7/14 Finish Chapter 4

7/28 Finish Chapter 5

Comments, Questions and (hopefully) Answers

Introduction

Q: On page 5, the authors talk about the fundamental groupoid of a topological space. I'm not excellent with fiber products, so I'm having trouble seeing how the map m they exhibit really is a map m as in the definition of a groupoid. More precisely, why is it okay that it's only defined when we can concatenate the paths? I'm assuming that this is the whole point of the definition of groupoid, and I'm missing it... -Christelle

A: I figured it out myself :) The fiber product is along s (source) and t (target), which I assume means that the elements of the fiber product are pairs (f,g) such that target(f)=source(g). Thus it's okay for m to only be defined on those elements because that's all there is.

Q: I'm finding the [math]\displaystyle{ \tilde{S} }[/math] construction in the segment on moduli space of triangles pretty confusing. Is it just (non-canonically) isomorphic to a disjoint union of 6 copies of [math]\displaystyle{ S }[/math]? (I'm emphasizing the non-canonicity thing since despite the notation it looks as though [math]\displaystyle{ \tilde{S} }[/math] ought to depend on [math]\displaystyle{ T }[/math] as well as [math]\displaystyle{ S }[/math] but I don't quite grok how that works) Dewey 21:23, 26 May 2012 (UTC)

A: [math]\displaystyle{ \tilde{S} }[/math] might be not isomorphic to a disjoint union of 6 copies of [math]\displaystyle{ S }[/math] at all. Take the example in the previous page: [math]\displaystyle{ S }[/math] being a circle, and X being a family of equilateral triangles which rotates by 120 in one revolution around the circle. The corresponding [math]\displaystyle{ \tilde{S} }[/math] is a NON-TRIVIAL principal [math]\displaystyle{ \mathfrak{S}_3 }[/math] bundle over [math]\displaystyle{ S^1 }[/math]. Imagine that you start from the point [math]\displaystyle{ (e^{0i},123)\in\mathfrak{S}_3 }[/math] and walk along the base, when you return, you will arrive at [math]\displaystyle{ (e^{0i},231) }[/math]. Continue for another round, you get [math]\displaystyle{ (e^{0i},312) }[/math]. Yet another round you return to the starting point [math]\displaystyle{ (e^{0i},123) }[/math]. If you start with [math]\displaystyle{ (e^{0i},213) }[/math], then you get the other three points in the fiber of the principal bundle. So [math]\displaystyle{ \tilde{S} }[/math] has two connected component, each is a 3-cover of the base circle. This is also consistent with your feeling that [math]\displaystyle{ \tilde{S} }[/math] should depend on [math]\displaystyle{ T }[/math] and [math]\displaystyle{ S }[/math]. -Dongning 05:13, 29 May 2012 (UTC)

Q: This question is sort of tangential, but working on the "moduli space of triangles section" now and I noticed something kind of funny. Usually saying that [math]\displaystyle{ \tilde{T} }[/math] is the moduli space of ordered triangles would just mean that there is a natural isomorphism from the functor [math]\displaystyle{ S \to \{X \to S\} }[/math] where [math]\displaystyle{ X \to S }[/math] is a family of ordered triangles on S, to the functor [math]\displaystyle{ Hom(-,\tilde{T}) }[/math]. But here this is more structure. Since the morphisms in [math]\displaystyle{ \tilde{\mathfrak{T}} }[/math] are required to be isometries on each fiber there is actually a functor from [math]\displaystyle{ \tilde{\mathfrak{T}} }[/math] to the category [math]\displaystyle{ \tilde{T}-Top }[/math] of spaces over [math]\displaystyle{ \tilde{T} }[/math], that is, the objects spaces with a specified maps to [math]\displaystyle{ \tilde{T} }[/math] and the morphisms are commutative triangles. Is there some way to phrase this in a way that is more like the traditional definition of a moduli space? Like, maybe replace [math]\displaystyle{ Hom(-,\tilde{T}) }[/math] with the functor Top --> Cat sending [math]\displaystyle{ S }[/math] to the fully subcategory of [math]\displaystyle{ \tilde{T}-Top }[/math] consisting of morphisms [math]\displaystyle{ S \to \tilde{T} }[/math]? Dewey 22:42, 27 May 2012 (UTC)

Something similar is going on in the section on elliptic curves. Rather than look at a moduli functor we look at a category of families of elliptic curves. Interestingly the morphisms are all required to be cartesian commutative squares. This should somehow correspond to the fact that in the moduli functor setup, the map on hom-sets is defined using pullback. Maybe what is really giong on in the moduli space of triangles section is not that the map on fibers is required to be an isometry, but that this makes all the morphisms into Cartesian squares.

A: I've been thinking about the fact that the diagrams have to be Cartesian too. Here is an elliptic curve example: Say you have a morphism from [math]\displaystyle{ E \to Spec(K) }[/math] to [math]\displaystyle{ E' \to Spec(K') }[/math], by the Cartesian property this means that [math]\displaystyle{ E \times_{Spec(K)} Spec(K') }[/math] is isomorphic to [math]\displaystyle{ E' }[/math]. In other words, [math]\displaystyle{ E }[/math] and [math]\displaystyle{ E' }[/math] are the "same" elliptic curve, all we've done is a base change. It makes sense that in a moduli space there would be a map between an elliptic curve and the same elliptic curve over a field extension. In the case of triangles, requiring that the map be an isometry similarly makes sure that it's the "same" triangle.

Q: I think I'm confused about the groupoid coming from an atlas for an orbifold. On page 23 the book says that the map [math]\displaystyle{ R \to U \times U }[/math] is never injective unless X is just a manifold with its trivial orbifold structure. But consider the case [math]\displaystyle{ X = \mathbf{C} }[/math], with a single patch. In that case R is given by triples [math]\displaystyle{ (z_1,z_2,\phi) }[/math] with [math]\displaystyle{ \phi }[/math] a germ of a holomorphic map from a neighbourhood of the first point to a neighbourhood of the second. But complex discs have a whole [math]\displaystyle{ SL_2(\mathbf{R}) }[/math] of automorphisms, so the fiber of R over any point of [math]\displaystyle{ U \times U }[/math] should be a whole [math]\displaystyle{ SL_2(\mathbf{R}) }[/math], rather than a point. Dewey 16:58, 28 May 2012 (UTC)

A: Key point is that [math]\displaystyle{ \phi }[/math] is a map over X. In this case, where the coordinate patches are given by the identity [math]\displaystyle{ X \to X }[/math], [math]\displaystyle{ \phi }[/math] is necessarily the identity map.

Q: I (think I) solved problem 1.15, but to do it I needed to use the fact that the groups [math]\displaystyle{ G_\alpha }[/math] from the orbifold data are all finite. If you relax that condition and let them be infinite does anyone have a counterexample? I'm interested in this because supposedly orbifolds are analagous to DM stacks and when you relax the condition on the groups being infinit, but require them to be algebraic, you should get something like an Artin stack, so this infinite vs finite groups thing could be a big deal. Note: as far as the DM and Artin stacks thing goes, I have no idea what I am talking about.

C: I think a lot of the problems where you find a map of groupoids satisfying conditions i and ii boil down to the following statement: Let A and B be two groups acting on a space X such that the actions of A and B commute and B acts faithfully. Then A acts on Y, the quotient of X by B, and there is a map of groupoids [math]\displaystyle{ (A \times B \times X \rightrightarrows X) \to (A \times Y \rightrightarrows Y) }[/math] satisfying conditions i and ii

C: A note for those who have not messed around with the j-invariant before: at least over an algebraically closed field, the j invariant for a polynomial [math]\displaystyle{ y^2 = f(x)^3 }[/math] depends only on the roots of f and their multiplicities.

Q: On page 17 in the definition of a "modular family" the text says that the condition that any first order deformation of any fiber in one of the families C --> S is captured by a tangent vector on the base S, and goes on to say that this is like requiring that the map from S to the (nonexistent) moduli space of curves be etale. If the moduli space M existed, then a deformation of [math]\displaystyle{ C_s }[/math], the fiber of C over [math]\displaystyle{ s\in S }[/math], should correspond to a family [math]\displaystyle{ C' \to \mathrm{spec} k[\epsilon]/(\epsilon^2) }[/math] whose fiber over 0 is [math]\displaystyle{ C_s }[/math]. Saying that this deformation is captured by a tangent vector to S should be saying that this family over [math]\displaystyle{ k[\epsilon]/(\epsilon^2) }[/math] can be obtained by fibering [math]\displaystyle{ C \to S }[/math] with a map [math]\displaystyle{ \mathrm{spec}\, k[\epsilon]/(\epsilon^2) \to S }[/math]. On the other hand, a family over [math]\displaystyle{ k[\epsilon]/(\epsilon^2) }[/math] is the same thing as a tangent vector v in M, and saying that such a fiber can be captured by a tangent vector to S should mean that v is in the image of the map of tangent spaces induced by the map [math]\displaystyle{ S \to M }[/math] corresponding to the family [math]\displaystyle{ C \to S }[/math] (phew!). So: this claim sort of makes sense if asking for a map to be etale is like asking for it to induce a surjection on tangent spaces. Can someone who knows about etale maps say if that's right?

Chapter 1

Q:

A:

Summer plans

If you feel like telling us your general plans for the summer, so that we'll know when you are around Madison, please do so here:

Ed: Leaving June 2, back around August 1.

Jeff: Leaving June 17, back July 8.

Evan: Leaving May 22, back June 20.

Christelle: Leaving June 24, back July 20, leaving August 4.

David: Leaving June 17, back July 8. Leaving July 31, back August 14th.