## What’s a Stack? June 19, 2008

Posted by A.J. Tolland in Algebraic Geometry, mathematical physics.

Algebraic stacks are essential to my research.  This is more acceptable now than it was twenty years ago, but it still presents a bit of a language barrier.  Most mathematicians, I think, don’t know what a stack is in the way that they know what a manifold or a scheme is.  So I want to use this post to explain what stacks are, with an eye towards their appearance in mathematical physics.  I won’t quite define them (see Vistoli’s notes for that), but I’ll get you a lot closer than Harris & Morrison do (see p. 139), hopefully close enough to be comfortable that you know what’s going on when someone says “stack”.

Let’s start by saying that a space is what you get when you start with a set and then add some geometry. Maybe make the set into a manifold, maybe make it into a scheme; you can choose your favorite category.   The elements of the set become the points of your space.

A stack is what you get when you start with a groupoid instead of a set, and then add geometry. A groupoid, remember, is a category whose morphisms are all isomorphisms.  This means that the points of a stack aren’t just elements of some set; they also come equipped with a bunch of relations, telling you which points are isomorphic to each other.

So why would anyone try to make a groupoid into a geometry?

Well, one reason is that it would be nice if your category of spaces was closed under quotients.  This obviously isn’t true; for example, there’s no nice manifold or scheme structure on the orbit set of $\mathbb{C}P^1$ modulo $\mathbb{C}^{\times}$.   But it is true if we embed our category of spaces into a larger category of stacks.   Every space is a stack in a tautological way, because any set can be made into a groupoid by simply adding the identity morphisms.  And there’s a good way of taking quotients in the category of stacks:  If we have space $X$ acted on by a group $G$, then we can define the quotient $X/G$ to be the stack whose underlying groupoid is the action groupoid of $G$ on $X$, i.e., the category which has an object for every element $x \in X$ and a morphism from $x$ to $x'$ for every $g \in G$ such that $x' = gx$.  The quotient stack $X/G$ will be smooth if $X$ and the action of $G$ are.  And its dimension will be what you would naively expect:

$\qquad \qquad dim(X/G) = dim(X) - dim(G)$.

Obviously, I’m cheating here. I told you what the points of $X/G$ are, but I haven’t told you what the geometry is, how the points fit together.  But hopefully it’s clear that the action groupoid is not a bad substitute for the orbit set.   We’ve enriched the problem a little.  We can recover the orbit set from the action groupoid, if we want it, by taking equivalence classes, but we’ve also remembered  how the group acts. The orbit set of $G$ on $X$ doesn’t know how big the stabilizers of the points of $X$ are, but the action groupoid does.  This is roughly why quotient stacks can be smooth when orbit set quotients aren’t; the degrees of freedom only add up correctly if you remember to count the stabilizers.

Hopefully that’s enough motivation to make it worth suffering through some definitions.  We need to find a setting in which we can make sense of the idea that the points of a stack form a groupoid.

The usual way of doing this, at least in the world of algebraic geometry, is to use the functor of points.  If you have a scheme $X$, you can consider the functor $F_X = Hom(\bullet,X)$ from schemes to sets defined by $S \mapsto Hom(S,X)$.  They call this the functor of points, because if $S$ is a point, then $Hom(S,X)$ is just the set of points of $X$.  But the functor of points knows about more than points; it knows how all the points fit together, which of them lie on a given subscheme and so forth.

In fact, if you know the functor $F_X$ in some abstract terms, you can actually recover $X$.  For example, the functor which assigns to $S$ the set of units in the global sections of the structure sheaf $\mathcal{O}_S$ is actually the functor of points of the multiplicative group.  Morally speaking, you can recover $X$ from the functor $F_X$ because $F_X$ knows all there is to know about the open affine covers of $X$.  Which means that any geometry we can do with $X$ can also be done with the functor $F_X$.

Which means that we can shift our perspective a bit. We can forget about spaces as much as possible, and try to use functors as our geometric objects.

This actually works.  A little more precisely:  The functor of points has a pretty remarkable property.  It’s actually a sheaf on the category of schemes.   This means two things:  First, the category of schemes has something like a topology:  a (Zariski) open set in the category of schemes is an open embedding $S' \hookrightarrow S$. (You can check that this definition has all the properties you’d want in a topology.)  Second, it means that we can recover the functor of points on any given scheme $S$ by knowing the functor of points on open subschemes $S' \hookrightarrow S$; the sheaf condition means that we can glue the local data together.

So we can try to study the following objects:  sheaves $F$ for the Zariski topology on the category of schemes which have affine open covers, i.e., there is an affine scheme $A$ (the atlas) and a surjection $F_A \to F$, telling us how $F$ is glued together from the affine functors of points.

We don’t get anything new this way.  These sheaves are exactly the schemes.  But we can get new geometric objects if we choose a finer topology.  For example, we can give the category of schemes the “etale” topology, declaring the open sets to be etale morphisms, not all of which are Zariski embeddings.

It’s a big theorem that the functor of points of a scheme is a sheaf in the etale topology.  But not every etale sheaf with a cover by (the functors of points of) affine schemes arises from a scheme; there are new geometric objects here.  (In the literature, these etale sheaves go by the slightly awful name “algebraic spaces”.  I try to avoid this name, since I like to reserve the term “space” for when I’m being deliberately vague.)

In fact, you can also show that the functor of points is a sheaf for even finer topologies, such as when the open sets are smooth morphisms.  But I’m not going to dwell on this fact, because I wanted to tell you about stacks.

We get stacks by replacing the set-valued functor of points with a new functor valued in groupoids.  Doing this correctly — specifically, spelling out what it means to have a sheaf of groupoids — takes a little bit of care, and goes by the name “descent theory”.  There are two issues:  First, you need to know that you can patch together isomorphisms living over open sets $U$ and $V$ if they agree on the intersection $U \cap V$.  Second, you need to know that if you have two objects, one on $U$ and one on $V$, then you can glue them together if their restrictions to $U \cap V$ are isomorphic.

Algebraic geometers usually make this notion — the groupoid-valued sheaf — precise by introducing “categories fibered in groupoids, satisfying the descent conditions”.  But let’s just stick with the intuitive idea; this is a blog post!  (Look at Vistoli’s notes if you’re curiosu.)

A groupoid-valued sheaf $F$ is a stack if it has a cover by affine schemes, i.e., if

1) there is a morphism $f: F_A \to F$ to $F$ from (the stack associated to the functor of points of) an open scheme $A$, and

2) the fibers of $f$ are schemes, meaning that, for any morphism $g$ from a scheme $S$ to $F$, the fiber product $S \times_F F_A$ is a scheme.

Generally, people speak of Deligne-Mumford stacks when the category of schemes has the etale topology, and of Artin stacks when the category of schemes has the smooth topology.  (There are also some technical conditions I won’t get into.)

So what is all this good for?

I know of two good answers.  First, like I said, you can construct quotient stacks.  There’s some real fun to be had here.  Consider the quotient map $q: pt \to pt/G$.  The point is definitely an affine scheme.  Can we convince ourselves, using the idea that $pt/G$ is the action groupoid of $G$ on the point, that the morphism $q$ should have only schemes as fibers?   Well, consider the fiber of $q$ over another map $\phi: pt' \to pt/G$; this fiber will consist of a point $p \in pt$ (only one), a point $p' \in pt'$ (only one), and a morphism $\alpha: q(p) \to \phi(p')$ in the action groupoid.  Since such morphisms correspond to elements of $G$, we see that the fiber of $q$ over the embedded point $\phi: pt' \to pt/G$ is precisely a copy of $G$, although without a distinguished identity.  From this, we can make one amusing conclusion:  the dimension of the $pt/G$ is equal to minus the dimension of $G$.

What’s more, the fibers of $q$ fit together nicely; in fact, the morphism $q: pt \to pt/G$ is actually a principal $G$-bundle.  Crazier yet, the morphism $q: pt \to pt/G$ is actually the universal principal $G$-bundle.  (Liar’s proof:  The total space is contractible!)  This is actually taken to be a definition:  the stack $pt/G$ is the functor which assigns to a scheme $S$ the groupoid of principal $G$-bundles on $S$.  In particular, the stack $pt/\mathbb{C}^{\times}$ is the classifying stack of line bundles.

Note, however, that $pt/\mathbb{C}^{\times}$ is not the familiar classifying space $\mathbb{C}P^\infty$.  The former has dimension $-1$; the latter has dimension $\infty$.  The latter classifies line bundles; the former classifies line bundles together with their automorphisms.

The other thing stacks are good for is moduli problems.  In algebraic geometry, we often want to parametrize objects of a given kind, modulo isomorphism.  One of the best ways of doing this is to show that the set of isomorphism classes of objects forms a space, because then you can introduce parametrizations just by putting coordinate charts on this space.   But it often happens that these sets don’t have nice geometric structures; they might not have any structure at all, or they might be very singular.  Or they might be spaces, but not classifying spaces.  In fact, these sort of annoyances crop up precisely when the objects in question have non-trivial automorphisms.

But this is precisely the situation where stacks are useful.  (In fact, it’s what Deligne & Mumford invented them for. [Edit:  Jason Starr says that stacks predate D & M.  See his comment below.])  So what we do instead now, in moduli theory, is look at the groupoid valued functor which assigns to a scheme $S$ the groupoid of all families of relevant objects parametrized by $S$.  If we’re lucky, we can show that this functor is actually a stack, and then we can do geometry, thinking of the atlas $A$ as a collection of coordinate charts on the stack.

For example, there’s a moduli stack $\mathcal{M}_g$; this is the stack which assigns to a scheme $S$ the groupoid of all families over $S$ of smooth genus $g$ complex curves.  (A family of curves on $S$ is a flat morphism $f: C \to S$ whose fibers are curves.)  This is one of the nicest stacks there is.  It’s smooth, connected, and has constant dimension $3(g-1)$.  It’s of finite type, which means that it can be covered by finitely many coordinate charts, each of which is a smooth affine scheme of the appropriate dimension carrying a family of complex curves.

Even nicer, $\mathcal{M}_g$ carries a universal curve $\pi: C_g \to \mathcal{M}_g$.  This is the stack assigning to a scheme the groupoid of pairs $(C,x)$, where $C$ is a curve over $S$, and $x: S \to C$ is a morphism introducing a point onto each fiber of $f$.  The morphism $\pi$ forgets this marked point; thus the fiber of $\pi$ over any curve $C \in \mathcal{M}_g(pt)$ is a copy of $C$.  In fact, by definition, any family of curves $C \to S$ is tautologically the pullback of $C_g$ along some morphism $\phi: S \to \mathcal{M}_g$.  In other words, $\mathcal{M}_g$ lives up to its name; it’s the classifying stack for smooth complex curves of genus $g$!

Whew!  OK, did anyone actually read all that?

1. Charles - June 19, 2008

Excellent. Definitely the best intuition building thing I’ve seen on stacks. Far better than the “What is … a stack?” from the AMS.

2. A.J. Tolland - June 19, 2008

Thanks, Charles. I was hoping you’d like it. I’m flattered by the comparison to Edidin’s article, but it’s hardly fair to him. We don’t have word count limits on the internet, and I still didn’t manage to give a definition!

3. Scott Carnahan - June 19, 2008

I think the distinction between $\mathbb{C}P^\infty$ and $pt/\mathbb{C}^\times$ is more that complex line bundles are classified by some category, and it can be presented either as continuous maps to the first space with homotopies, or algebraic maps to the second with internal symmetries. The two actually share a lot of properties, e.g., I think Olsson and Laszlo have a computation of l-adic cohomology of $B\mathbb{G}_m$ that coincides with that of $\mathbb{P}^\infty$. In particular, an analytic continuation of the Lefschetz trace formula yields a point count of $\frac{1}{q-1}$ over the field of q elements, which is what you’d expect from quotienting by the multiplicative group.

I prefer to use the term “geometric object” instead of “space” when I’m being vague, because it already sounds super-vague, and there is a smaller chance of an overloaded term confusing people.

4. A.J. Tolland - June 19, 2008

Scott,

The two…err…geometric thingamagummies…do definitely share some equivariant properties. But it’s possible to tell the difference between them, even in topology-land; the K-theory of $\mathbb{C}P^\infty$ is the localization of the K-theory of $pt/\mathbb{C}^{\times}$ at the ideal of virtual representations of dimension zero.

I wasn’t aware of the Olsson-Laszlo computation; that’s pretty cool!

5. Allen Knutson - June 19, 2008

did anyone actually read all that?

Yeah, but the word “physics” didn’t reappear : -8 (
Followup awaited!

6. A.J. Tolland - June 19, 2008

Sorry, Allen! We’re getting there, but I need to talk about gauge theory a little first.

7. Charles - June 19, 2008

Possibly part of the difference is that I read that one much earlier in my education…and that I’m at a rather stacky conference right now. Either way, it’s helped me to put together a few things that have been floating in my head about stacks that I didn’t quite grasp before, which the Edidin article didn’t.

8. Alessandro - June 20, 2008

Good post!
What’s the relation with the point of view of a stack as something ”probed” by maps from any space to it?

9. Urs Schreiber - June 20, 2008

What’s the relation with the point of view of a stack as something “probed” by maps from any space to it?

It’s the same point of view!

A (pre)sheaf on an abstract site (i.e. one not necessarily of the form open subsets of some space) is a something of which you only know how to probe it by objects of your site. More precisely, the presheaf assigns to any object the “set of probes” by that object.

Allow the “set of probes” here to be a category of probes and arrive at (pre)stacks.

10. Alessandro - June 20, 2008

Useful as always, Urs! ;)

11. Urs Schreiber - June 20, 2008

Alessandro,

here is an illuminating example:

the stacky incarnation of the classifying space BG of G-bundles:

whatever the “space” BG is, by definition it is supposed to have the property that a map from a probe space X into it is a G-bundle over X, and that every G-bundle over X arises this way.

So, the universal G-bundle is supposed to be such that a probe of it by X is precisely a G-bundle over X.

But (using Yoneda once) that’s precisely the stacky definition of BG!

12. Alessandro - June 20, 2008

Ok, now, this is quite a good example.
Just one thing: with “the universal G-bundle” you meant BG, and not EG->BG, the… universal G-bundle, right?

13. A.J. Tolland - June 20, 2008

Allessandro,

Yes, Urs meant to say “[the base space BG of] the universal G-bundle is supposed to be such that a probe of it by X is precisely a G-bundle over X.”

In fact, BG = pt/G. I said this above without quite drawing attention to it. The quotient map q: pt -> pt/G is a principal G-bundle whose universal space is contractible, so pt = EG and pt/G = BG. (In Urs’ notation. I usually reserve BG for the topological space and use pt/G for the stack.)

What’s the relation with the point of view of a stack as something “probed” by maps from any space to it?

Urs basically answered this, but I want to elaborate a bit. Let $S$ be a scheme, and $\tilde{S}$ be the stack associated to this scheme. If we have another stack $F$, we can probe it by looking for functors $\tilde{S} \to F$; these are homomorphisms in the category of stacks. But if you trace out the definitions, you can show that the collection of such functors is equivalent to the groupoid $F(S)$ that the stack $F$ assigns to $S$.

14. Alessandro - June 20, 2008

I’ve asked this as you mentioned applications to mathematical physics, and that’s how one usually sees stacks arising in string theory in some cases. (For what I’ve read about it, that is)

15. Urs Schreiber - June 20, 2008

Urs meant to say “[the base space BG of] the universal G-bundle [...]

Thanks for catching that!

I usually reserve BG for the topological space and use pt/G for the stack.

I went through a period of lots of trouble with adjusting notation for the different incarnations of BG such that it would both make sense and make people see intuitively what is supposed to be meant.

There is also the groupoid incarnation of BG. That’s just the one-object groupoid with G as its space of morphisms. That groupoid has a somewhat canonical name as

pt//G

with two (!) slashes. In general, for V any space acted on by a group one “should” write (I suppose)

V//G

for the “weak quotient”, namely for the corresponding action groupoid.

Using that, an old theorem by Segal says that the realization of the nerve of the groupoid

G//G

(the action groupoid of G acting on itself from the left, say) is the universal G-bundle over BG, i.e. that the image of the sequence of groupoids

G -> G//G -> pt//G

under

|.| : Cat -> Top

is the universal G-bundle as a topological space

|G| -> |G//G| -> |pt//G|
=
G -> EG -> BG .

to emphasize this notationally I have settled on writing

${\bf E}G := G//G$

and

${\bf B}G = pt//G.$

This way realizing nerves just sends boldface to roman face. :-)

16. Urs Schreiber - June 20, 2008

and that’s how one usually sees stacks arising in string theory in some cases.

In many of the cases that people talk about stacks, string theory or elsewhere, I often feel that just talking about the corresponding Lie groupoids (if any) would help clarify the picture.

Well, it might just be me, but I often have the impression that people like to invoke stacks where one categorical dimension down they would not invoke sheaves (even though they could). Which is odd.

Well, I should either list examples or shut up. I think for the moment I’ll just up. ;-)

17. Alessandro - June 20, 2008

Fair enough ;)

18. Jason Starr - June 20, 2008

This post on stacks is nice. As a matter of historical reference,
stacks are much older than the article of Deligne and Mumford. Of course groupoids have been around “forever”. But “champ” in its current meaning goes back at least to Deligne’s Exp. XVIII in SGA 4. The definition of Deligne-Mumford stacks does seem to first arise in the article of Deligne and Mumford. For most of us, more important than where stacks were first defined is where they were profitably applied. And it is hard to think of a more profitable application of stacks than in the article of Deligne and Mumford.

Curiously, Mumford asserts in GIT that what we now call “Deligne-Mumford stable curves” were first invented by Mayer and Mumford years earlier. As with stacks, the marvelous application of stable maps to settle a conjecture in the paper of Deligne and Mumford is more memorable than the first definition of stable maps. This seems to be true of most definitions.

19. Chris Brav - June 20, 2008

Two basic questions about 15. by Urs:

1. In the sequence of groupoids G –> G//G –> pt//G,
is the groupoid G understood to have objects indexed by elements of the group G and only identity morphisms?

2. When G is some group with non-discrete topology, the sequence G –> G//G –> pt//G should presumably be understood to lie in the category of topological groupoids rather than in discrete groupoids. Is there some enriched nerve so that {Topological categories} –> {Simplicial sets} –> Top would produce EG and BG with the right topologies relative to the given topology on G?

20. Chris Schommer-Pries - June 21, 2008

@ Chris Brav

1. Yes. You view G as a groupoid with just identity morphisms.

2. Yes. If G has a topology, then the nerve construction doesn’t produce a simplicial set, but rather a simplicial space . The same formula that you use in the discrete case works here to give you a geometric realization.

For nice groups this works exactly as you would hope and really gives a model for the classifying space. If G is a bad group, say it fails to be locally contractible or worse, then things are more subtle. For example $EG \to BG$ may fail to be locally trivial. But for such bad groups you really have to think about what you mean by bundle and classifying space anyway.

All this is in one of Segal’s many great papers:
Classifying Spaces and Spectral Sequences

21. A.J. Tolland - June 21, 2008

Jason,

Thanks for the correction! I’ve put in a caveat above.

22. diogo - June 21, 2008

Hello
can you give me some of idea of what might be the use of this on a symplectic view point? you were talking about DM, moduli of curves etc… (on a just symplectic manifold)

thank you nice day

23. David Ben-Zvi - June 22, 2008

About $\mathbb{CP}^\infty$ vs BG_m: if we’re working
on the category of rings or schemes, let’s say over C, or
over the category of complex manifolds, there are actually
three different objects – infinite dim projective space,
BG_m and BS^1. The first classifies line subbundles of
a trivial infinite rank line bundle, the second classifies algebraic line
bundles, and the third classifies S^1 bundles.

(I think part of the confusion above is the distinction between
complex line bundles, which are topological objects, and
algebraic or holomorphic line bundles).

Here we think of S^1 = BZ as either a groupoid, or as a topological space/homotopy
or simplicial set/homotopy, which happens to be 1-truncated
- ie it has no homotopy groups above pi_1.
If we think of CP^infty as a homotopy type, rather than
as a scheme, then we get exactly this BS^1 = B(BZ).
One quick way to tell the difference between S^1 and G_m — or
equivalently between BS^1 and BG_m — is that the former is
a “constant” group/stack — it assigns the same groupoid
to everything (up to sheafification — so eg to any field
it will assign the same groupoid), while the latter
varies greatly with the ring that we input.

24. David Ben-Zvi - June 23, 2008

Hi – I posted a comment yesterday which seems to have
been swallowed up by the ether (or spam filter)..
just trying again to see what happens!

25. hilbertthm90 - June 25, 2008

Wow. This is utterly fascinating! As someone starting grad school and not sure what he wants to go into, I think I may have found a new topic to consider. Thanks!

26. Scott Carnahan - June 30, 2008

Grothendieck wrote a letter to Serre in 1959 complaining about problems arising when parametrizing objects with automorphisms (I learned this in Olsson’s stacks class last year). “Champ” is defined in Raynaud’s exp 13 in SGA 1, which was presumably built from Grothendieck’s lectures in 1960-61, but it references Giraud’s thesis (1966) and SGA 4 as Starr mentioned. I suppose this sort of circularity should be expected when it takes 10 years for your notes to get to published form.

David – Thanks for clarifying the algebraic/topological distinction. It seems to be a persistent point of confusion for me. I’ve heard there is a way in which P^n approximates BG_m in the algebraic category as n grows, but I don’t remember any precise functorial statements.

27. bb - July 5, 2008

There is a precise sense in which P^n approximates BG_m if one works in the A^1-homotopy category (obtained, roughly speaking, by contracting A^1 to a point in the category of sheaves on the lisse-Nisnevich site of a field). More specifically, if one defines the infinite dimensional affine space as X = colim_n A^n (with the usual embedding maps), then there’s a free G_m-action on X – 0. The quotient is simply P^oo = colim_n P^n. The natural map colim_n P^n = (X – 0)/G_m -> BG_m is an equivalence in the A^1-homotopy category by the same formal argument as in topology.

This means: the BG_m-valued points F(BG_m) of any A^1-homotopy invariant sheaf F (such as Quillen’s K-theory or Bloch’s higher Chow groups or etale cohomology with good coeffecients) can be computed as colim_n F(P^n) — this is at least one precise sense in which this is an approximation. If F is reasonably nice, then one can usually stop at a certain n.

One can similarly give algebraic approximations of the classifying spaces (in the etale topology) of any algebraic group (cf: papers by Totaro and Morel-Voevodsky).

28. Jason Starr - July 6, 2008

Scott,

Deligne also cites Giraud’s thesis for the notion of “champ”. So perhaps this is the origin (I don’t have “Correspondance Grothendieck-Serre” here to read Grothendieck’s letter).

By the way, you have to be careful when citing “Raynaud” as in your last post. Expose XIII of SGA 1 and other exposes in SGA 1 and SGA 2 are due to Madame Mich\’ele Raynaud and not to Monsieur Michel Raynaud. I often became confused over this
when I first read SGA 1 and SGA 2.

29. Jason Starr - July 6, 2008

Bhargav,

The algebraic approximations of the classifying space of a reductive group by Totaro have another very nice feature. The algebraic approximations are “close” to projective schemes, and thus BG is “nearly proper”. Apparently this idea goes back to Bogomolov although I haven’t been able to track down the precise reference. Johan and I used this recently in extending Johan’s proof of his “Period-Index Theorem” to characteristic 0, and also in our work on Serre’s “Conjecture II” (there are 2 articles about this “discriminant avoidance” on Johan’s webpage).

30. Not Even Wrong » Blog Archive » This and That - July 8, 2008

[...] the posts worth reading over at Secret Blogging Seminar, there’s a nice posting by A. J. Tolland explaining what a “stack” [...]

31. Jason Starr - July 9, 2008

Re 29 –

“to characteristic 0″ should be “from characteristic 0″.

32. Scott Carnahan - July 10, 2008

bb and Jason,

That’s pretty cool stuff. Thanks for pointing it out.

I’m a bit curious about what Mich\`ele Raynaud has been up to during the last 30 years (perhaps this sort of question is frowned upon). Google points to sundry government workers and a professional soprano sharing her name.

33. estraven - July 10, 2008

I think we should also know that there’s a difference between a stack which is in fact a sheaf of groupoids, and a DM (resp Artin) algebraic stack, which is a stack that locally in the ‘etale (resp fppf/smooth) topology looks like a scheme.

Another good way of thinking about stacks is that schemes have an underlying set; stacks have a set each of whose points has an automorphism group.

What makes algebraic geometers so hot about algebraic stacks is that…they’re geometric objects! You can do more or less everything you know from schemes – you have coherent sheaves, cotangent complex (it’s a generalization of the cotangent sheaf), Chow groups, etc. Artin stacks (those where the automorphism group of the points is possibly positive dimensional) tend to be a little bit more weird; e.g. BG has negative dimension (=-dim G).

But if you can get past being freaked out, they’re pretty useful.

34. estraven - July 10, 2008

BTW, around 2002 Michèle Raynaud
approved the electronic publication
of SGA1. So she was 1) alive and 2) somewhere where some mathematician could find her.

35. Scott Carnahan - July 10, 2008

It shouldn’t be too surprising that a mathematician could find her, since her husband is a mathematician.

[edit: That was a pretty stupid link to use. I should have just said that M. Raynaud is married to M. Raynaud, and their names differ by the letter e and an accent grave.]

36. estraven - July 11, 2008

I keep forgetting that some people know stacks, but not enough french to tell that Michel is a male name and Michèle a female one.
Are you sure that the two Raynauds are married to each other? I couldn’t find any evidence on the web.

37. Jason Starr - July 11, 2008

Estraven,

I was also told that Madame Mich\’ele Raynaud is married to Monsieur Michel Raynaud. But I cannot confirm if it is true.

38. Carnival of Mathematics « Rigorous Trivialities - July 11, 2008

[...] beyond cats, we get to learn about Stacks from A. J. Tolland at the Secret Blogging Seminar. Granted, he doesn’t actually DEFINE what a stack is, but [...]

39. Thomas Riepe - July 14, 2008

What’s about vanishing cycles/monodromy and stacks?

40. A.J. Tolland - July 14, 2008

Thomas,

It’s not entirely clear what you are asking.

41. Ben Webster - July 14, 2008

Thomas-

Just to clarify, your question is confusing in part because “what’s about X” isn’t really an established English turn-of-phrase, so it’s difficult to interpret what you mean by it (did you mean “what about X”?). Not to mention that “monodromy” is a rather big subject, so it’s kind of unclear what you want to know about it.

42. David Speyer - July 14, 2008

Thomas — I think you have two separate questions here. The first how to use vanishing cycles to compute/restrict monodromy. In this context, we are talking about a family $X$ of varieties over a punctured disc. Vanishing cycles, very roughly, are cycles in $X$ which become boundaries when we compactify $X$ “nicely” over the whole disc. The monodromy in question is that of the bundle-with-connection on the punctured disc where the fiber over the point $t$ is $H^i(X_t)$.
I know that there is supposed to be a connection between the two concepts (due to Deligne), but I don’t understand it. The second question is how this generalizes to stacks.

The first matter is already a very hard topic and I don’t think it makes sense to try to understand the stacky issues until you understand the situation for varieties. I don’t have that understanding yet, so I can’t help you. I will say that, based on what I have understood, I don’t see any reason to think that the stack perspective will make things easier.

If you, or any of our readers, do have a good understanding of how vanishing cycles and monodromy work, even in the case of families of ordinary smooth varieties, and can write it up at an expository level comparable to the Secret Blogging Seminar’s norm, I would be glad to link to it, or promote it as a guest post.

43. anon - July 14, 2008

here’s a guess at what it could mean in a specific situation: say we have a proper flat family $f:X \to S$ parametrized by the pointed unit disc (S,0) (more algebraically, the spectrum of a strict complete dvr). let $j:T \to S - 0$ be the universal cover (more algebraically, the map from the separable closure of the fraction field of the dvr to the dvr). then, in the derived category of sheaves on X, one has an adjunction map $R \to Rj_*(R)$ where R is the constant sheaf on some coefficient ring. the homotopy cokernel of this cospecialisation morphism is the sheaf of vanishing cycles, and its cohomology groups are the groups of vanishing cycles.

here’s why i think this is reasonable: by the proper base change theorem, $R\Gamma(R)$ computes the cohomology of the fibre over 0, while $R\Gamma(Rj_*(R))$ computes the cohomology of a general fibre. thus, the cokernel of the cospecialisation map is dual to a group that computes the homological cycles on the general fibre which vanish in the central fibre.

44. Thomas Riepe - July 16, 2008

Hi,

please excuse my crude use of your language – the simpleminded question was just: Do there exist extensions to stacks (and applications) of, say, SGA 7.2 Exp.s I, XIV and XIII, in which Deligne introduces (even drawing some pictures, which must have been very unfashionable then) the formalism of vanishing cycles and monodromy? My own understanding is of course only very rudimentary and I am just trying to dig through some articles which interest me in Asterisque 223. Illusie’s website contains several IMHO very interesting articles on related issues too, e.g. about vanishing cycles on higher dim. bases. Helpfull I found “Homology and Intersection Homology of Alg. Var.s” by Looijenga in IAS/Park City Math. Ser., vol.3, 1997 and “Notes on perverse sheaves and vanishing cycles” by Massey in the arxiv.

45. Moduli Spaces and Base Change « Rigorous Trivialities - July 23, 2008

[...] July 17, 2008 Last time we spoke of representable functors and talked about how to check if a functor is representable. The whole idea being that if we can first construct a functor that SHOULD be the functor of points for the scheme we want, and then we check representability and we’ve got it. Well, it doesn’t always happen like that. In fact, a lot of the time, the functors we get from moduli problems aren’t quite representable as schemes. That’s why people bother with algebraic spaces and with stacks. [...]

46. steve meagher - August 14, 2008

The notion of a stack appears first in Mumford’s paper
“Picard group of a moduli problem” which was published in 1963.

He defines what he calls a moduli topology for genus g curves; this is the stack of genus g curves. He then computes its Picard group in genus 1. This is probably the best introduction to stacks available.

It is not actually that big a theorem that a scheme is a sheaf for on the etale site. In fact it can easily be shown for the flat site. See for example Milne’s notes on etale cohomology(the notes on the web not the book from 1981) page Prop 6.8 p35.

47. CHL - December 6, 2008

[QUESTION on stacks]

Dear All;

In view of the high level discussions in this blog I just came across
more basic questions.
—————————

(1) I was reviewing the book Laumon and Moret-Bailly
[L-BL] “Champs algebriques” recently , trying to locate
the definition/notion of an ${\cal O}_{\cal X}$-module
${\cal M}$ on algebraic stacks ${\cal X}/{\cal Y}$
that is “flat” over ${\cal Y}$.
It is not there, (I hope I did not overlooked something
as it is written in French.)

Presumably, one can give a definition of the notion of
“flatness”
after pulling everything to the related schematic atlas
of the algebraic stacks in question,
(e.g. some description in the work of P.G. Goerss:
“Quasi-coherent sheaves on the moduli stack of
formal groups”, arXiv:0802.0996 [math.AT])
or following the similar treatment in [L-MB: (12.5), p.108]
to define first the notion of “stalk” of ${\cal M}$
at a “point” on ${\cal X}$ (cf. [L-MB: Chapter 11])
and then define the notion of “flatness” of ${\cal M}$
on ${\cal X}$ over ${\cal Y}$
similar to that for a ${\cal O}_X$-module on scheme
$X/Y$.

Q. I wonder if there are references in which a definition
of flatness of ${\cal M}$ on ${\cal X}$ over ${\cal Y}$
is explicitly given so that I can quote and put to use
instead of having to do things from scratches?

(2) I had expected this to be “everywhere” in literatures
on stacks but found myself wrong.

Q. Is that because it is too trivial to write it down
once one knows how to define it in the case
of schemes $X/Y$ or
because there are subtleties which takes effort
to explain and resolve and, hence, people just
leave it unless necessary?

(I think this notion for the case when ${\cal X}$ and
${\cal Y}$ are both Deligne-Mumford stacks are okay;
but I have to think more about the Artin stack case.)

——————————
I would appreciate some illumination from hosts/hostess
and readers of this wonderful blog
if any of you happen to know and get the time.

Pretty sorry for this bother.
Best thanks to all.

— Sincerely, CHL, 2008.11.06

48. A.J. Tolland - December 7, 2008

CHL,

Flatness can be checked in an atlas. I’m not sure what the correct reference on this is, but I don’t think there are any subtleties here. Most likely, the existing literature handles the definition of a flat morphism of stacks as follows:

1) List off a long collection of properties (flat, locally of finite presentation, etc) of morphisms which are local on the domain in various topologies. Reference EGA IV.

2) Observe that these properties also make sense for morphisms of stacks.

So you might not find anything of the form “Definition: A morphism of stacks is flat if …”. Look instead for a remark explaining that any property which is local on the domain in the appropriate topology can be defined for morphisms of stacks.

49. CHL - December 7, 2008

Dear AJ;

————————————–

(1) It seems that the notion of a flat morphism between
algebraic stacks is given more often in the literature
(after defining representable 1-morphisms and then,
by definition, reducing everything to schemes
after taking fibered-product,
e.g. [L-MB] or [Deligne-Mumford] classical paper
on irreducibility of moduli space of stable curves ).

But for (quasi-)coherent sheaves ${\cal M}$ on an algebraic
stack ${\cal X}$ over another algebraic stack ${\cal Y}$,
I as yet do not see an explicit formal definition for
“${\cal M}$ is flat over ${\cal Y}$”
that I can quote to make sure.
This surprises me a little bit.

(2) I will re-read EGA IV these two days to see if there is
it looks that I have better to give a terse/brief explanation
or definition of the flatness of a coherent sheaf on
${\cal X}/{\cal Y}$ when I use it.

———————————————
Best thanks to you.

— Sincerely, CHL, 2008.12.07

50. hyh - February 4, 2009

I have a question: suppose a curve over a scheme is given, how can one tell the corresponding morphism to M_g is etale/smooth or not? Thank you.

51. davidspeyer - February 4, 2009

Short answer: Check whether the map on tangent spaces is everywhere surjective. (And, if you want etale, whether your base has the same dimension as $M_g$.) Assuming that this short answer isn’t clear enough, here is a very long answer.

Suppose we have a family of curves over $D:=\mathrm{Spec\ } k[\epsilon]/\epsilon^2$. We’ll call the family $\tilde{C}$, and write $C$ for the fiber over $\mathrm{Spec\ } k$. For now, let’s also assume that $C$ is smooth. From such a family, we get a map from $D$ to $M_g$. As you may know, a map from $D$ to a variety $X$, with $\mathrm{Spec\ } k$ mapping to $x$, is equivalent to a (Zariski) tangent vector to $X$ at $x$. Also, as you may have heard, the tangent space to $M_g$ at $C$ is $H^1(C, T_C)$, where $T_C$ is the tangent bundle to $C$. So, from our family $\tilde{C} \to D$, we are supposed to get an element of $H^1(C, T_C)$. The simpler question is: how?

Let $U_i$ be an affine cover of $C$. Let $\tilde{U}_i$ be the open subschemes of $\tilde{U}_i$ supported on the same points as the $U_i$. It turns out that $\tilde{U}_i \cong U_i \times D$, although the isomorphism is not canonical. Choose such an isomorphism $\phi_i$ for each $i$. Exercise: Let $\tilde{U}$ be $\mathrm{Spec\ } \left( k[\epsilon]/\epsilon^2 \right)[x,y]/\langle \ y^2 - x(x-1)(x+1+\epsilon) \ \rangle$, so $U$ is $\mathrm{Spec\ } k[x,y]/\langle \ y^2-x(x-1)(x+1) \ \rangle$. Write down an isomorphism between $\tilde{U}$ and $U \times D$.

Writing $U_{ij}$ for $U_{i} \cap U_j$, and $\tilde{U}_{ij}$ for the corresponding open subscheme of $\tilde{C}$, we get TWO isomorphisms between $\tilde{U}_{ij}$ and $U_{ij} \times D$, one by restricting $\phi_i$ and one by restricting $\phi_j$. The difference, $\phi_j \circ \phi_i^{-1}$, is an automorphism of $U_{ij} \times D$. The corresponding map of rings is of the form $f + \epsilon g \mapsto f + \epsilon g +\epsilon \theta_{ij}(f)$, for a certain map $\theta$ from $\mathcal{O}(U_{ij})$ to itself. Exercise: Show that this formula gives a map of rings if and only if $\theta_{ij}$ is a derivation.

So, on each $U_{ij}$, we get a derivation $\theta_{ij}$ on the ring of functions on $U_{ij}$ — in other words, a section of $T_C$ over $U$. Moreover, the $\theta_{ij}$ form a cocycle. The class of this cocycle in $H^1(C, T_C)$ is the map $D \to M_g$.

Whew! That finishes the first question!

Now, what if we have a family of curves over $B$. Let $b$ be a smooth point of $B$, write $C$ for the curve over $b$, and let $T_b$ be the tangent space to $B$ at $b$. We are supposed to get a linear map $T_b \to H^1(C, T_C)$. What is it?

You could take a bunch of different maps $D \to B$, until you covered a basis for $T_b$. But it is nicer to do everything at once. Let $D$ be the second infinitesimal neighborhood of $b$. In other words, let $\mathcal{O}_b$ be the local ring at $b$ and $\mathfrak{m}_b$ the maximal ideal; then
$D = \mathrm{Spec\ }\mathcal{O}_b/\mathfrak{m}_b^2$. But, more explicitly, $D = \mathrm{Spec\ } k \oplus T_b$, where the product of any two elements in $T_b$ is zero. I’ll denote that ring as $k[T_b]/T_b^2$. Taking an affine cover as before, we get automorphisms of $\mathcal{O}(U_{ij}) \times k[T_b]/T_b^2$. Such an automorphism is of the form $f + \sum t_k g_k \mapsto f + \sum t_k g_k + \sum_k t_k \theta_{ij, k}(f)$, where $t_k$ is a basis of $T_b$. The $\theta_{ij,k}$ form the components of a homomorphism $T_b \to H^0(U_{ij}, T_b)$. Looking at all the $U_{ij}$ together, we have a cocycle, so we get a homorphism $T_b \to H^1(C, T_C)$ as desired.

So. That’s how to compute the map on tangent spaces. As I said at the beginning, once you have that, you know how to check etaleness/smoothness at a point.

I have a few more things to say, but I think they will go in a seperate comment.

52. davidspeyer - February 4, 2009

OK, the few more things.

(1) A very important case is where your family of curves is embedded in some fixed ambient space $P$, such as the family of degree $n$ curves in $\mathbb{P}^2$. In this case, the first order deformations of $C$ within $P$ are given by $H^0(C, N_{P/C})$, where $N_{P/C}$ is the normal bundle to $C$. Remember that we have a short exact sequence

$0 \mapsto T_C \mapsto T_P|_C \mapsto N_{P/C} \mapsto 0$

(where $T_P|_C$ is the restriction to $C$ of the tangent bundle to $P$).

The corresponding map $H^0(C, N_{P/C}) \to H^1(T_C, C)$ is just the boundary map in the corresponding long exact sequence. In particular, if $H^1(C, T_P|_C)=0$ then you automatically know that the map is smooth at $C$!. (Does anyone know a reference for this?)

(2) Since I’m asking for references, I’ll give one: a generally good reference for the things I’ve been writing is the first chapter of Hartshorne’s notes on deformation theory.

(3) So, what if the curve $C$ has nodes? Then the tangent bundle gets replaced by the logarithmic tangent bundle. To see what this means, let’s look at what a derivation $\theta$ of $k[x,y]/xy$ looks like. We must have $x \theta(y) + y \theta(x)=0$, so $x^2 \theta(y)=0$ and we deduce that $\theta(y) = g_1 y + g_2 y^2 + \cdots$. Similarly, $\theta(x) = f_1 x + f_2 x^2 + \cdots$. In the logarithmic tangent bundle, we require that $f_1+g_1=0$. I think of this as saying that “the coefficient of $xy$ in $x \theta(y) + y \theta(x)$ is zero”.

(4) So, what if the curve $C$ has automorphisms? This would be important if we wanted to know whether the map from $B$ to the coarse moduli space of $M_g$ was a smooth map. The Zariski tangent space to the coarse moduli space, at a point with automorphisms, is NOT $H^1(C, T_C)$. But I think that working with stacks means we get to ignore this issue. (Others, please correct me if I’m wrong.)

53. Deligne and Mumford on the Moduli of Curves « Rigorous Trivialities - February 18, 2009

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