Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Some background on GIT

Suppose G is a reductive group acting on a scheme X. We often want to understand the quotient X/G. For example, X might be some parameter space (like the space of possible coefficients of some polynomials which cut out things you're interested in), and the action of G on X might identify "isomorphic things", in which case X/G would be the "moduli space of isomorphism classes of those things."

It happens that the quotient X/G often doesn't exist, or is in some sense bad. For example, if the closures of two G-orbits intersect, then those orbits must get mapped to the same point in the quotient, but we'd really like to be able to tell those G-orbits apart. To remedy the situation, the idea is to somehow remove the "bad locus" where closures of orbits could intersect. For reasons I won't get into, this is done by means of choosing a G-linearized line bundle L on X (i.e. a line bundle L which has an action of G which is compatible with the action on X). Then we define the semi-stable and stable loci

Xss(L) = {x∈X|there is an invariant section s of some tensor power of L such that x∈Xs (the non-vanishing locus of s) and Xs is affine}
Xs(L) = {x∈Xss(L)| the induced action of G on Xs is closed (all the orbits are closed)}

Note that Xss(L) and Xs(L) are G-invariant. Then the basic result is Theorem 1.10 of Geometric Invariant Theory:

Theorem. There is a good quotient of Xss(L) by G (usually denoted X//LG, I believe), and there is a geometric (even better than good) quotient of Xs(L) by G. Moreover, L descends to an ample line bundle on these quotients, so the quotients are quasi-projective.


My Question

Are there some conditions you can put on G, X, L, and/or the action/linearization to ensure that the quotient X//LG or Xs(L)/G is projective?

Part of the appeal of the GIT machinery is that the quotient is automatically quasi-projective, so there is a natural choice of compactification (the projective closure). The problem is that you then have to find a modular interpretation of the compactification so that you can actually compute something. Is there some general setting where you know that the quotient will already be compact? If not, do you have to come up with a clever trick for showing that a moduli space is compact every time, or do people always use the same basic trick?

share|improve this question
    
Dumb comment: can you explain a bit more how "the quotient is automatically quasi-proj, so there is a natural choice of compactification" bit goes? I don't think quasi-proj means "I come with a given map to a projective space", I think it is something weaker. But maybe I'm missing something in this context. –  Kevin Buzzard Nov 20 '09 at 21:16
    
For what it's worth, with the moduli spaces I've considered in my mathemical lifetime (abelian varieties plus extra structure), compactness (properness) is the same as "given a family of ab vars + structure, can it degenerate into something worse?" and the technique is always "think about the Neron model". In other words, use the moduli problem itself, and not the GIT construction of it. I guess what I'm saying is that in the cases I know, properness, when it's true, comes from a study of the functor, not the construction of the representing object. –  Kevin Buzzard Nov 20 '09 at 21:19
    
Third unrelated comment: Anton---you do a very good job of laying out questions. Lots of fonts, links, boxes etc. Very clear. Can I "view source" somehow? –  Kevin Buzzard Nov 20 '09 at 21:21
1  
I'm not sure if asking here is the correct MO etiquette, but here goes: what is a "modular interpretation of the compactification"? –  Alberto García-Raboso Nov 20 '09 at 23:57
6  
Since Anton went to the trouble of giving some background on GIT, perhaps here is a good place to mention some notes on GIT (arxiv.org/abs/math/0512411) which are excellent but maybe less well-known than the standard references. I should also stress that they make a real effort to make a technical subject accessible to people without a strong algebro-geometric background. They may also be extra-useful for you Buzzard, but for a different reason: if you want to ask the author a question, you can knock on his door easily enough! –  Joel Fine Nov 21 '09 at 0:09

4 Answers 4

up vote 4 down vote accepted

I'm not sure if this is the sort of thing you are after but one can say the following.

Suppose we work over a base field $k$. If $X$ is proper over $k$ and the $G$-linearized invertible sheaf $L$ is ample on $X$ then the uniform categorical quotient of $X^{ss}(L)$ by $G$ is projective and so it gives a natural compactification of $X^s(L)/G$.

I believe this is mentioned in Mumford's book but no proof is given. It goes via considering $\mathrm{Proj}R_0$ where $R_0$ is the subring of invariant sections of $$\oplus_i H^0(X,L^{\otimes^i})$$ and showing that this is the quotient.

This is stated in Theorem 1.12 in these notes of Newstead. I haven't found a proof written down - but either showing it directly or doing it by reducing to the case of projective space shouldn't be too hard I don't think.

share|improve this answer
    
Yes. That's exactly the sort of thing I'm looking for. I felt like the assumption that X is projective should be enough, but since X^ss isn't proper, it's not clear to me why X^ss/G should be. A reference would be most appreciated. Of course, it would also be nice to have a general result which shows that projective space (viewed as a GIT quotient of X=A^n) is projective, but I'll take the hypothesis that X is projective if I can get it. –  Anton Geraschenko Nov 21 '09 at 1:00
1  
Maybe I'm missing something, but isn't this in Nick's notes? arxiv.org/abs/math.AG/0502366 –  Aaron Bergman Nov 21 '09 at 1:29
    
@Anton. I'm afraid I think about this like a differential geometer, but here's one way to see properness in certain cases. When you take the GIT quotient of something smooth and over C, you can often relate it to a symplectic reduction. You choose a Kahler metric on X invariant under a maximal compact subgroup K of G and ask for a moment map mu for the K action. In good cases, the GIT quotient is equal to the symplectic reduction mu^{-1}(0)/K. If mu^{-1}(0) is compact, so is the quotient. In the case of CP^n arising as the quotient of C^N+1-0, K is the circle and mu^{-1}(0) is the N-sphere. –  Joel Fine Nov 21 '09 at 11:27

If your desired moduli space can be interpreted as a moduli space of quiver representations, then you may be in luck. For moduli of $\theta$-(semi)stable quiver representations, there is a simple criterion equivalent to the stable and semistable loci being equal: the dimension vector $\alpha$ is indecomposable (ie, not a non-trivial sum of elements) in $\theta^\{\perp}\cap \mathbb{Z}^{Q_0}_{\geq 0}$, where $Q_0$ is your set of verticies, and $\theta$ is the fixed stability parameter.

To learn more about moduli of quiver representations, I recommend King's paper "Representations of finite dimensional algebras."

share|improve this answer
    
Can you confirm that I linked to the right paper? –  Anton Geraschenko Nov 21 '09 at 0:18
    
Thank you! Yes, that is correct. –  David Steinberg Nov 21 '09 at 0:25

Is the answer "when there are no non-constant invariant functions" too stupid? (By construction, functions on the GIT quotient are invariant global functions on $X$). Because that seems to show that $X$ projective implies $X^{ss}(L)/G$ projective.

share|improve this answer
    
How do you know that the map from X^{ss}(L) to Proj of the ring of invariant functions is surjective? You are giving a criterion for when that Proj is projective. –  David Speyer Nov 21 '09 at 4:42
    
That said, I feel like this is addressed somewhere in Mumford's book, and that there is some good criterion which lets you conclude the map is surjective. –  David Speyer Nov 21 '09 at 4:43
    
Rereading Anton's question, he asked two things: (1) When is X//G projective? (2) When is X^s/G projective? It seems to me that he should also have asked (3) When is the image of X^{ss} in X//G projective? I was thinking about (2) and (3); your answer is a correct answer to (1). –  David Speyer Nov 21 '09 at 5:31
    
Imposing the condition that there are no invariant sections of any power of L does tell you that X^ss(L)/G has trivial affine envelope, but it could still fail to be projective. If X is P^2 minus a point, and G is trivial, any information you could get from regular functions or sections of a line bundle would apply equally well to the case P^2/G, but one of the quotients is projective, and the other is not. –  Anton Geraschenko Nov 21 '09 at 18:29

As Greg Stevenson points out if X is projective then $X^{ss}//G$ is projective. Now if $X^s = X^{ss}$ then clearly $X^s//G$ is also projective. This much is true for general spaces. A proof can be found in Newstead's book "Introduction to Moduli problems and orbit spaces". In the quiver setting of course the space X is an affine variety. Here for any stability condition $\theta$, $X^{ss}//_{\theta}G$ is projective over $X//G = Spec(C[X]^G)$. Also even in the quiver variety case a criterion for stable = semistable seems hard and I'm not even sure if one is known.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.