When are GIT quotients projective? - MathOverflow

When are GIT quotients projective?

2009-11-20T20:59:10Z

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

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

Note that X^ss(L) and X^s(L) are G-invariant. Then the basic result is Theorem 1.10 of Geometric Invariant Theory:

Theorem. There is a good quotient of X^ss(L) by G (usually denoted X//_LG, I believe), and there is a geometric (even better than good) quotient of X^s(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 X^s(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?

Answer by David Steinberg for When are GIT quotients projective?

2009-11-20T22:49:02Z

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."

Answer by Greg Stevenson for When are GIT quotients projective?

2009-11-21T00:31:07Z

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.

Answer by Ben Webster for When are GIT quotients projective?

2009-11-21T04:04:38Z

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.

Answer by Avan Thiyagarajan for When are GIT quotients projective?

2009-11-21T04:50:34Z

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.