## The Hilbert Polynomial Explained

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 $X$ be a projective variety over $\mathbb{C}$ and let $\mathscr{L}$ be an invertible sheaf on $X$. Then $\mathscr{L}$ is ample if and only if for all $\mathscr{F}$ coherent sheaves, there is a number $n>>0$, such that we have $H^i(X,\mathscr{F}\otimes\mathscr{L}^{\otimes n})=0$ for all $i>0$.

This is a very important theorem. What it does is it tells us that for large enough $n$, we have $\chi(\mathscr{L})=\sum_{i=0}^\infty (-1)^i \dim H^0(X,\mathscr{L}^{\otimes n})=\dim H^0(X,\mathscr{L}^{\otimes n})$. 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 $R=\oplus_{n=0}^\infty H^0(X,L^{\otimes n})$ with $L$ ample. Then $R$ is a graded ring (multiplication by tensor product) and is isomorphic to the ring $\mathbb{C}[x_0,\ldots,x_n]/I$ where $I$ is the ideal of the image of the map given by $L$.

What this really says is that the function $f(n)=\dim H^0(X,L^{\otimes n})$ is the Hilbert function of its image. Now, we know from before that $f(n)$ is eventually equal to a polynomial. By the theorem above, we know that $f(n)$ is eventually equal to $\chi(L^{\otimes n})$, so $\chi(L^{\otimes n})$ 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 $D$ be an ample divisor on a curve $C$. Then the Riemann-Roch theorem for curves says that $\dim H^0(X,nD)-\dim H^1(X,nD)=\deg(nD)-g+1$. But this is just $\chi(nD)=n\deg D-g+1$, 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 $D\cdot D'$ to be the intersection number of $D$ and $D'$ for $D,D'$ divisors on our surface $S$. 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 $D$ and $D'$, whenever that makes sense.

For surfaces, there’s a Riemann-Roch theorem, which says $\chi(D)=\frac{1}{2}(D-K)\cdot D+\chi(0)$. If we take $D$ ample and then apply Riemann-Roch to $nD$, we get $\chi(nD)=\frac{1}{2}(nD-K)\cdot nD+\chi(0)=\frac{D\cdot D}{2}n^2-\frac{D\cdot K}{2}n+\chi(0)$, which is precisely the Hilbert polynomial of a surface embedded in projective space, because $D\cdot D$ 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 $X$, set $Y$ to be a point, and take $D$ an ample divisor, it then gives us the equation $\chi(D)=f_*(ch(D)td(X))$ where $ch(D)$ is the Chern character of the line bundle associated to $D$ and $td(X)$ is the Todd class, which is a cohomology class associated to $X$. So we apply it to $nD$ and obtain $\chi(nD)=f_*(ch(nD)td(X))=f_*(\exp(c_1(nD))td(X))$. Now, $f_*$ is a map from $H^*(X,\mathbb{Z})$ to $\mathbb{Z}$.

Now take $\dim X=m$. Then $\exp(c_1(nD))=\sum_{i=0}^m \frac{1}{i!}c_1(nD)^i=\sum_{i=0}^m \frac{n^i}{i!}c_1(D)^i$. The Todd class of $X$ is just another cohomology class on $X$, and we multiply by it, through. Then $f_*$ takes the degree $m$ part of this element of $H^*(X,\mathbb{Z})$ and this will give us a polynomial in $n$ of degree $m$, 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 $\chi(nD)$, 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 $n$ is $-\frac{D\cdot K}{2}$, 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. Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 5 Responses to The Hilbert Polynomial Explained

1. anon says:

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.

2. Charles Siegel says:

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

3. anon says:

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.

4. Charles Siegel says:

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

5. lucasbraune says:

Nice post!