# Hilbert polynomial as function of the Segre classes

Let $X\subset\mathbb{P} ^ N$ be a smooth irreducible complex projective variety of dimension 3 (or better yet, dimension $n$). Is it possible to express the Hilbert polynomial of $X$ as a function of the degrees of Segre classes of $X$? If so, how?

Thanks.

• Chern classes can always be written in terms of Segre classes (for example $c_1(E)=−s_1(E)$, $c_2(E)=s_1(E)^2−s_2(E)$...), so the answer is yes! Commented Mar 21, 2012 at 16:33
• diverietti, how does your comment relate to gio's question? Commented Mar 21, 2012 at 16:55
• Maybe I misunderstand gio's question! The coefficients of the Hilbert polynomial are intersection products of Chern classes of $X$ with the hyperplane class. I just thought he wanted to express them as intersection products of Segre classes of $X$ with the hyperplane class... Commented Mar 21, 2012 at 17:01
• I interpreted the degrees that gio mentions as products of the hyperplane class with various Segre classes... Commented Mar 21, 2012 at 17:06
• @gio:Perhaps a way of seeing this is the following: If you let $Y$ be the blow up of $\mathbb{P}^n$ along $X$, then one can define the Segre classes as $s_p=\pi_*(E^p)$. Also, the hilbert polynomial of the ideal sheaf of $X$ is nothing but the function $$\chi(O_Y(m\pi^*H-E)$$ for $m$ large where $H$ is a hyperplane and $E$ is the exceptional divisor. But by Riemann-roch, $\chi (O_Y(m\pi^* H-E))$ can be expressed as a polynomial in the numbers $\pi^*(H)^i\cdot E^{n-i}$, which are the the same as the degrees of the Segre classes. Commented Mar 22, 2012 at 8:51

You should say what bundle you are taking Chern/Segre classes of. (As diverietti points out, knowledge of the two is equivalent.) Some people use "Chern class of $$X$$" to mean "Chern class of the tangent bundle to $$X$$." If this is what you mean, then no, because you haven't incorporated the data of the embedding.
Alternatively, you could mean the classes of $$\mathcal{O}(1)$$ restricted from projective space. In this case, there is only one nonzero Chern class, $$c_1$$, and the Segre classes are $$s_k = (-c_1)^k$$. So all you can see is $$\int_X (c_1)^{\dim X}$$, better known as $$\deg X$$. This is equivalent to knowing the first term of the Hilbert polynomial, but no more.
If you know $$c_1(\mathcal{O}(1))$$ and you know the Chern (or Segre) classes of $$T_X$$, then yes! This is the Hirzebruch-Riemann-Roch formula.
ADDED We have a short exact sequence of vector bundles on $$X$$: $$0 \to T_X \to T_{\mathbb{P}^n}|_X \to N_{X/\mathbb{P}^n} \to 0$$. So $$c(T_X) c(N_{X/\mathbb{P}^n}) = c(T_{\mathbb{P}^n})$$ (where $$c$$ is the total Chern class). Remembering that $$s(E) c(E) = 1$$, with $$s$$ the total Segre class, we have $$c(T_X) = c( T_{\mathbb{P}^n}) s(N_{X/\mathbb{P}^n}).$$ Writing $$H$$ for the hyperplane class $$c_1(\mathcal{O}(1))$$, we have $$c(T_{\mathbb{P}^n}) = (1+H)^{n+1} - H^{n+1}$$. So knowing the degree of $$X$$ and the Segre class of the normal bundle determines the Chern classes of the tangent bundle. There may be some further simplification that can be performed, but I don't see it right now.
• If I have learned well, the Chern classes of $X$ are the Chern class of the tangent sheaf $\mathcal{T}_X$, while the Segre classes of $X$ are the Segre classes of the normal bundle $\mathcal{N}_{X,\mathbb{P}^N}$. The Hilbert polynomial of $X$ can be express as function of the Chern classes of $X$, but how do I get the Chern class of $X$ from the Segre classes of $X$?