Let be a projective variety (so that is an ample line bundle). Let be a coherent sheaf on . Define . The Hilbert function of is (this depends on the line bundle we have fixed). Alternatively, define this as . The idea is that for a sufficiently large , all non-zero cohomology of will vanish (because is ample). Define the Hilbert polynomial to be a polynomial in which agrees with for (this exists). Thanks to Riemann-Roch, the Hilbert polynomial depends only on the Chern classes of and of .
If is a vector bundle, then in fact,
We can consider the monic version of this polynomial.
Fix a coherent sheaf on . A quotient of is an exact sequence
of quotients with base
is a quotient of
is the projection, which is flat over
has the same Hilbert polynomial
, for all
We put an equivalence relation on quotients by identifying quotients with the same kernel.
Check that when this is just a quotient.
The Quot functor takes a scheme to the set of all families of quotients of with base . We can also fix a polynomial and restrict to quotients with that Hilbert polynomial.
If , then a quotient is just the structure sheaf of some closed subscheme , by considering the support of .
To see this, take an affine open set : . A quotient is a surjective map , so , for some ideal . Then . Let be the ideal sheaf of , i.e., on an open set , .
So in this case, the Quot functor takes a scheme to the set of all families of subschemes of with base . This is called the Hilbert functor. If it is representable, it is called the Hilbert scheme.
If we consider Example (MISSING), the Hilbert scheme is .
Set the Hilbert polynomial to be . Show that (as a scheme/functor of points - not just as a set).
To show that the Quot functor is representable, we will need the Grassmannian functor. Let be a field, and a finite dimensional vector space over . Let be an integer such that . Then is a functor from the opposite category of to Sets, such that
I will sketch how to show that this functor is representable. First, fix a of dimension . Consider the subset of such that the restriction of to gives an isomorphism . Let be the subfunctor defined by these subsets.
is representable. To see this, note that there is an isomorphism . Then we get a composition , which is the right inverse of
the inclusion So corresponds to morphisms which splits the inclusion. If such that , then corresponds to morphisms .
So is the functor , defined by
is representable by . To see this, recall that the universal property of the symmetric algebra is the following: given a -vector space and a -module , then there is a natural isomorphism . Thus,
The idea now is to show that the are open (we know that they are affine) subfunctors, that agree on intersections, and so they can be glued together to get . Fix and , and consider a quotient . Consider open subschemes of such that the restriction of to lies in the image of , and choose a maximal such , call it . Then this is the definition of being an open sub-functor. As varies, the form an open covering of , so the are an affine open covering of . After checking the cocycle condition, we can see that we can glue the together, and this scheme represents .
Finally, you can use the Plucker embedding to show that is projective (and indeed a smooth irreducible variety).
Let be a coherent sheaf over a projective variety . Then is representable.
First, let's fix a quotient . As is ample, there is an such that for all , . Moreover, for sufficiently large, is globally generated. This means that there is an exact sequence
, and taking the quotient defines
So the quotient is determined by the exact sequence of vector spaces
But this exact sequence is a point of
(recall that by definition
We assume that you can choose
We then need to generalize and do the above over a base , in order to get a subfunctor of the Grassmannian. The Grassmannian is representable, so this gives a subscheme of the Grassmannian, called Quot. For the details of the proof, see Huybrechts-Lehn.
Now we can give the outline of Simpson's construction of the moduli space of vector bundles over with given Hilbert polynomial .
- Fix a bundle on with Hilbert polynomial .
- Find such that all higher cohomology of vanishes, and is generated by its global sections. Again, this means that
- Fix an identification
- Under this identification, is a a quotient sheaf of , so
- Divide by the choice of identification in 3). That is, divide by the action of .
- Find a stability condition to do GIT for this quotient, and let Quot' denote the semi-stable points.
- Go back to step 2, and show that can be chosen universally for all bundles satisfying the stability condition.
- Check that is a moduli space of stable bundles/sheaves.
In step 4, we identified as an element of divided by the action.
In a previous exercise, we showed that the stability condition for a subspace , under the action on was for all proper subspaces ,
So as an element of
is stable by checking proper subspaces
. By unwinding
and evaluating sections,
gives rise to a subsheaf of
is stable if and only if for all proper subsheaves
This is called Gieseker stability. If we consider the leading coefficient as
, then we get slope stability
Slope stability implies Gieseker stability, and on a curve, these are equivalent.
Stability is a generic condition.
Show that both Gieseker stability and slope stability ( is the slope of ) satisfy the see-saw property. That is, for all exact sequences
and similarly when we consider Gieseker stability instead of slope stability.
Show that a stable vector bundle is simple.
Stability rules out the possibility of the 'jump phenomenon'.
If are families of stable sheaves parametrized by , and for all , , then .
By definition, are sheaves on , flat over with the same Hilbert polynomial on each fibre, and stable on each fibre.
Let be the projection. is a sheaf on (defined by ), so we can construct the direct image sheaf under , . By definition, for an open , . The fiber over is , as these are coherent sheaves.
Because are simple by the previous exercise, the fiber over is .
Show that is torsion free over . Recall that a sheaf over a scheme is torsion-free if for all , is a torsion-free - module.
If are families of vector bundles, then is a vector bundle, so as it is a line bundle over , it is a line bundle over . For sheaves, we need the previous exercise to show that it is a a line bundle over .
Pick a nonzero element of the fiber over . By definition, this is a non-zero morphism . So we can form two exact sequences:
Using the see-saw property of
and the stability of
, we have that
. By definition of
, it depends on the Hilbert polynomial of a sheaf, and so because
are families of fixed Hilbert polynomials,
. So we have a contradiction, unless
is an isomorphism.