So, today I discovered something rather nice, that I think could easily get more time in books and the like, but doesn’t. We’ll first have need of a theorem that I don’t want to prove:

**Theorem**: Let be a projective variety over and let be an invertible sheaf on . Then is ample if and only if for all coherent sheaves, there is a number , such that we have for all .

This is a very important theorem. What it does is it tells us that for large enough , we have . It is, in fact, a key point in the difference between Hilbert functions and Hilbert polynomials. Here’s a quick lemma (also without proof, because it isn’t the point)

**Lemma**: Let with ample. Then is a graded ring (multiplication by tensor product) and is isomorphic to the ring where is the ideal of the image of the map given by .

What this really says is that the function is the Hilbert function of its image. Now, we know from before that is eventually equal to a polynomial. By the theorem above, we know that is eventually equal to , so is eventually equal to the Hilbert polynomial. The story is actually better: it IS the Hilbert polynomial, in all degrees. We’ll first prove it in the case of curves:

Let be an ample divisor on a curve . Then the Riemann-Roch theorem for curves says that . But this is just , which is a linear polynomial, whose lead term is the degree of the image, and we even get the arithmetic genus right where we want it.

Note that the only reason for the ample condition is to make this the Hilbert polynomial of a curve in projective space. In general, we do get Hilbert polynomials of divisors.

Next we’ll look at the case of surfaces. We define to be the intersection number of and for divisors on our surface . This is defined carefully in many books, and will be here if I do some surface theory after I stop the toric geometry, but the gist is that it is the number of points of intersection of and , whenever that makes sense.

For surfaces, there’s a Riemann-Roch theorem, which says . If we take ample and then apply Riemann-Roch to , we get , which is precisely the Hilbert polynomial of a surface embedded in projective space, because is the degree of the embedding. Even better, this tells us what the coefficient of the middle term means.

This will work out in general. There’s the Grothendieck-Riemann-Roch Theorem, which we apply to a smooth projective variety , set to be a point, and take an ample divisor, it then gives us the equation where is the Chern character of the line bundle associated to and is the Todd class, which is a cohomology class associated to . So we apply it to and obtain . Now, is a map from to .

Now take . Then . The Todd class of is just another cohomology class on , and we multiply by it, through. Then takes the degree part of this element of and this will give us a polynomial in of degree , and it is precisely the Hilbert polynomial!

So what’s the point? The Riemann-Roch theorem makes it easy to see that the Hilbert function is eventually a polynomial. Even better, it tells you that the Hilbert function is a red herring, and that the real thing of interest is , the Euler characteristic, which is precisely the Hilbert polynomial. So among other things, this is why the Hilbert function, which looks like it should be the more fundamental thing, actually fails to capture low degree information, like the arithmetic genus of a projective variety. My favorite part is that this formulation tells you what happens on the middle terms, and no one ever seems to mention this. However, there is geometric information there. For instance, in the case of surfaces, the coefficient of is , which tells us a few things relating to the geometry of the canonical divisor, which is extremely important to things like the Minimal Model and Mori programs.

October 10, 2008 at 10:38 am

You probably need some more hypotheses in the theorem to make it bidirectional. As stated, it implies that the trivial bundle on P^n is ample, which is false. Non-triviality of the bundle isn’t enough either as the example of a non-torsion degree 0 on an abelian variety shows. Requiring that the higher cohomology of F(n) vanishes for any coherent sheaf F with n = n_F >> 0 is certainly necessary, and probably sufficient.

October 10, 2008 at 10:46 am

Ahh, yeah, that’s the hypothesis. Was working from flaky memory late last night. Will correct it.

October 10, 2008 at 5:10 pm

Sorry to nitpick, but the correct statement probably needs the n to depend on F. If there was a fixed N that worked for every F, then taking F = K_X(-N) would say H^top(K_X) = 0.

October 12, 2008 at 4:05 pm

Yeah, and that’d be me making a quick instead of a careful correction. Quantifiers rearranged.

October 17, 2013 at 6:58 pm

Nice post!