Unless there is some specific interest, I think this will be the final post in the series about taking quotients of varieties by actions of group schemes. Recall that everything is being done over an algebraically closed field, . I’ll discuss what happens in the projective case in this post. To cover such a topic completely is of course impossible, but hopefully this post will illlustrate some of the major ideas and results. There are also some more or less completely worked through, hands on, examples to show this subject is still rooted in polynomial computations!

In the following assume that is a reductive algebraic group scheme acting on an algebraic variety . Unless stated otherwise, a point will always mean a closed point. If is a linearized line bundle on , then we make the following definition:

**Definition**: i) A point is called semistable (with respect to ) if there exists a positive number m and (that is, an invariant global section of the m-th tensor power) such that is affine and contains .

ii) A point is called stable, if the condition in i) is satisfied, and in addition all orbits in are closed.

iii) If a point does not satisfy i), we say it is unstable.

(for those following along in GIT, Mumford also defines other stability notions, but is these which we will focus on).

Let me try to make some sense of this defintion. Assume that is very ample and so embeds equivariantly into projective space . Then where is the subspace of degree m polynomials vanishing on . At the level of invariants then, we have . An can be represented by a polynomial which is invariant modulo . This says exactly that will be constant on orbits of the action. For a point , we can choose representing it. We’ll have that if and only if . So the unstable points can be identified with the set . For an exercise using this description: show that the unstable points are exactly the ones represented by such that the origin is in the closure of the orbit .

Denote by and the collection of all semistable (respectively stable) points of . We are ready to state today’s main theorem:

**Theorem**: There is a categorical quotient such that is affine and is quasi projective. Furthermore, there is an open set of such that and the restricted map is a geometric quotient.

**Remark on Proof**: The proof is actually not so difficult. To show the existence of , simply remark that categorical quotients of the various (affine!) exist and can be glued together (by uniqueness of categorical quotients) to give a categorical quotient of . It is only slightly more difficult to show that the result is quasi-projective. To construct an ample line bundle on , first take a trivialization of on each peice . On the overlap , note that the function is invariant and so by construction is a function on . Not going into the details, these transition functions on define an ample line bundle. For the complete proof see Mumford’s GIT.

In fact, when is projective and is ample on , then is isomorphic to where . In particular, the quotient is projective.

Let’s work through some examples.

1. Let act on projective space by the formula where the are integers. I should have mentioned last time (when discussing the existence of linearizations) that there is an exact sequence where denotes the linearzed line bundles. This is actually a fairly easy exercise. Using this sequence then, the kernel becomes identified with the characters and in our example this is so that . This can be seen directly in the following way:

Any linearized line bundle must look like for some number m. This defined a equivariant embedding of into projective space (for the appropriate N) by the Veronese embedding with coordinates where . The action on has to be . Then a linearization is given by a representation of in lifting the action on projective space. It is defined by the formula for some integer . The pair indexes the linearized line bundles which we denote . Raising the bundle to the r-th power corresponds to replacing with . Replacing by a tensor power of itself does not change the semi-stable locus (exercise!) so we may take where by definition, only makes sense if q divides N. Invariance then means that for any non-zero values of t, we have . Assume now that . If or , we have from the above formula that for all possible values of N. This shows that the semi-stable locus is empty for in this range. When , then we have that . Then the semistable locus is exactly and the quotient is . As increases, for example, , there are more invariant polynomials and the categorical quotient changes. For example, if then the polynomial is invariant. One can show in general that if stays between two q’s then the quotient does not change, but otherwise certainly does.

2. In this example, and again . Suppose the action is given by . On affine space, all line bundles are trivial, so we may write (for the associated space) . Let be the -linearization defined by . A section of this bundle $s: X \rightarrow L_a$ may be written . The group acts on these global sections via . So a section defined by the polynomial is invariant if and only if . (Similarly for global sections of ). When , only the constant polynomials are invariant, so the semistable locus is all of but the quotient is . That is, the quotient is a single point. However, if , then the semistable locus is . This can be seen because the invariant functions are . And then . Recall that . If , this gives the standard projective space . However, if the q’s are not all one, but say, still positive, the result is called weighted projective space . Of course, unlike projective space, it need not be smooth. For example suppose acts on via . The quotient on the patch is given by . We’ll leave it as an easy exercise to see that this is not smooth.

3. Let’s work out one more special case where the weights of the action are not all positive. Suppose that acts on via . When (with notation as in example 2) , as above all points of are semi-stable. It is not difficult to see that the ring of invariants is which is isomorphic to . It is a quadric cone in . Even more interestingly, take . We’ll leave it as an exercise to show that the semistable locus is . This set is covered by the opens and . Then (similarly for the other open). I claim that the quotient of the semi-stable locus is a closed subvariety in given by where are homogeneous coordinates on . The last set of claims which is not difficult to verify are as follows: 1. The stable locus actually agrees with the semistable locus (as it did in the above examples!) so the quotient is a geometric one. 2. In this case it is smooth. 3. There is a map from the quotient when to the quotient when . 4. In fact, that morphism is a blowing up of at the conical point, that is, the origin.

There is more to the theory of taking and identifying quotients. In the above examples, identifying the semi-stable and stable locus was not so difficult, but as the examples become more complicated, this process becomes a bit harder. There are tools to do this (namely the numerical criterion), and it is fun and still hands on. Also as the above examples illustrated, varying the linearization changes the (semi) stable locus and so also the quotient. Is there a systematic description of how they are related, if indeed they are? The third example is supposed to be an illustration of this phenomenon. Also one can leave the category of schemes to take “finer” quotients, but that requires more foundational work. Unless there is overwhelming interest though, perhaps I’ll stop here in this subject for now.

September 15, 2008 at 6:30 pm

How does the semistable locus change under blowing up? If you start with a projective X for which the semistable locus is dense, then the rational map X – – > X//G can be “regularized” by blowing up X. Is there a way to see what to blow up by investigating the unstable locus?

September 17, 2008 at 9:18 am

That’s an interesting question — one I don’t know the answer to. I’ll think about it though.

I do know that in many cases, two quotients X//G (with respect to different linearizations), are birational and related by (often calculatable) flips. (This is work due to Michael Thaddeus: J. American Math Soc. 1996 no 3).

September 17, 2008 at 4:13 pm

I had one naive idea, but now I realize it is wrong. I thought that perhaps, after replacing X by a blowing up and replacing the original invertible sheaf by some new invertible sheaf, one could arrange for the semistable locus to equal all of X. But since the morphism X_{ss} –> X//G is affine, if the fiber dimension is positive, X_{ss} can never be projective (the fibers are closed subsets of X_{ss} which are affine).