# Difference between revisions of "Summer stacks"

(→Introduction) |
(→Introduction) |
||

Line 29: | Line 29: | ||

Q: I'm finding the <math>\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>S</math>? (I'm emphasizing the non-canonicity thing since despite the notation it looks as though <math>\tilde{S}</math> ought to depend on <math>T</math> as well as <math>S</math> but I don't quite grok how that works) [[User:Dewey|Dewey]] 21:23, 26 May 2012 (UTC) | Q: I'm finding the <math>\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>S</math>? (I'm emphasizing the non-canonicity thing since despite the notation it looks as though <math>\tilde{S}</math> ought to depend on <math>T</math> as well as <math>S</math> but I don't quite grok how that works) [[User:Dewey|Dewey]] 21:23, 26 May 2012 (UTC) | ||

+ | |||

+ | C: I' thinking about the moduli space of triangles right now, trying to understand why the category of families of unordered triangles is the same as <math>[\tilde{T}/S_3]</math>, and I noticed something kind of funny. Usually I'd think of the moduli functor as a functor from spaces to sets: it eats a space S and outputs the set of families over it. And saying that the moduli space is <math>\tilde{T}</math> means that this moduli functor <math>F</math> is isomorphic to <math>hom(-,\tilde{T})</math>. But in this case, since morphisms of families were defined to be isometries on each fiber, you get more structure. You get an improved moduli functor <math>G</math> from spaces to categories, which eats S and outputs the category of families over <math>S</math>. If you think of <math>Hom(-,\tilde{T})</math> as an "improved" hom-functor mapping to the category of morphisms to <math>\tilde{T}</math> (like the category of <math>\tilde{T}</math>-schemes) then G is isomorphic to <math>Hom(-,\tilde{T})</math>. I think this is important when you are proving that the correspondence <math>\mathfrak{T} \to [\tilde{T}/S_3]</math> is a functor, though for sure you don't need to dress it up this high-falutin' language to prove this. [[User:Dewey|Dewey]] 21:47, 27 May 2012 (UTC) | ||

=== Chapter 1 === | === Chapter 1 === |

## Revision as of 16:47, 27 May 2012

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

## Contents

## 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]\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]S[/math]? (I'm emphasizing the non-canonicity thing since despite the notation it looks as though [math]\tilde{S}[/math] ought to depend on [math]T[/math] as well as [math]S[/math] but I don't quite grok how that works) Dewey 21:23, 26 May 2012 (UTC)

C: I' thinking about the moduli space of triangles right now, trying to understand why the category of families of unordered triangles is the same as [math][\tilde{T}/S_3][/math], and I noticed something kind of funny. Usually I'd think of the moduli functor as a functor from spaces to sets: it eats a space S and outputs the set of families over it. And saying that the moduli space is [math]\tilde{T}[/math] means that this moduli functor [math]F[/math] is isomorphic to [math]hom(-,\tilde{T})[/math]. But in this case, since morphisms of families were defined to be isometries on each fiber, you get more structure. You get an improved moduli functor [math]G[/math] from spaces to categories, which eats S and outputs the category of families over [math]S[/math]. If you think of [math]Hom(-,\tilde{T})[/math] as an "improved" hom-functor mapping to the category of morphisms to [math]\tilde{T}[/math] (like the category of [math]\tilde{T}[/math]-schemes) then G is isomorphic to [math]Hom(-,\tilde{T})[/math]. I think this is important when you are proving that the correspondence [math]\mathfrak{T} \to [\tilde{T}/S_3][/math] is a functor, though for sure you don't need to dress it up this high-falutin' language to prove this. Dewey 21:47, 27 May 2012 (UTC)

### Chapter 1

Q:

A:

### Chapter 2

### Chapter 3

### Chapter 4

### Chapter 5

## 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.