# How to compute this Hilbert polynomial?

I am trying to solve Hartshorne's exercise V $$1.2$$.

Let $$H$$ be a very ample divisor on the surface $$X$$, corresponding to a projective embedding $$X\subseteq P^N$$. Then the Hilbert polynomial of $$X$$ can be written as $$P(z)=\frac{1}{2}az^2+bz+c$$, where $$a=H^2,b=H^2/2+\pi,c=1+p_a$$. Here the Hilbert polynomial is defined as $$P(n)=\mathcal{X}(\mathcal{F_H}(n))$$, where $$\mathcal{F}$$ is the invertible sheaf corresponds to divisor $$H$$, $$\mathcal{X}$$ means the Euler characteristic, and $$\mathcal{F_H}(n)=\mathcal{F_H}\otimes \mathcal{O}_X(n)$$. And $$H^2$$ denotes the self-intersection number of $$H$$, $$\pi$$ is the genus of a nonsingular curve representing $$H$$, and $$p_a$$ means the arithmetic genus.

First, we want to find $$c=P(0)$$, which should be $$1+p_a=\mathcal{X}(\mathcal{O}_X)$$. But $$P(0)=\mathcal{X}(\mathcal{F_H}(0))=\mathcal{X}(\mathcal{F_H})$$. Why does it equal to $$\mathcal{X}(\mathcal{O}_X)$$? Besides, if we can compute $$P(1)=\mathcal{X}(\mathcal{F_H}(1))$$ and $$P(-1)=\mathcal{X}(\mathcal{F_H}(-1))$$, then we can determine the polynomial. But none of these are simple to me, so I hope someone could help.

Also, Hartshorne states that if $$C$$ is any curve on $$X$$, then its degree in $$P^N$$ is just $$C.H$$. These should mean the Hilbert polynomial of $$C$$ is $$P(z)=(C.H)z+c$$, and $$c=\mathcal{X}(\mathcal{F_C}(0))$$. But I still have no idea on it.

By the way, I hope someone can tell me a systematic way to compute Hilbert polynomial. Thanks!

first your definition of Hilbert polynomial isn't correct:$$P(n)=\mathcal{X}(O_X(n))$$ where $$O_X(n)=O_X\otimes \mathcal{F}_H^n$$
now $$P(0)=\mathcal{X}(O_X)=1+p_a$$. In general from Riemann-Roch theorem we have $$P(n)=\mathcal{X}(\mathcal{F}_H^n)=\frac{1}{2}(nH.nH-nH.K)+1+p_a$$ so you must prove that $$H.K=H^2-(2\pi-2)$$ which is the Adjunction formula.
for the second part note that $$H\cap C$$ is a very ample divisor on C with degree $$C.H$$ then use the Riemann-Roch theorem on the curve C.
• I double checked the definition of Hilbert polynomial of a sheaf, which is the Euler characteristic of the sheaf with the tensor product of some power of the twisting sheaf $\mathcal{O}(1)$. Thus, it seems that my definition in the post is correct, or maybe I misunderstand $\mathcal{O}(n)$? But your answer really makes sense! – Yuyi Zhang May 10 at 4:04
• First note that $\mathcal{O}_X(1)$ is just $\mathcal{F}_H$.Second you want the the euler charactristic of structure sheaf $\mathcal{O}_X$ not $\mathcal{O}_X(1)=\mathcal{F}_H$ – ali May 10 at 7:28