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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.

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  • $\begingroup$ 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! $\endgroup$
    – diverietti
    Commented Mar 21, 2012 at 16:33
  • $\begingroup$ diverietti, how does your comment relate to gio's question? $\endgroup$
    – J.C. Ottem
    Commented Mar 21, 2012 at 16:55
  • $\begingroup$ 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... $\endgroup$
    – diverietti
    Commented Mar 21, 2012 at 17:01
  • $\begingroup$ I interpreted the degrees that gio mentions as products of the hyperplane class with various Segre classes... $\endgroup$
    – diverietti
    Commented Mar 21, 2012 at 17:06
  • $\begingroup$ @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. $\endgroup$
    – J.C. Ottem
    Commented Mar 22, 2012 at 8:51

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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.

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  • $\begingroup$ 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 $X$? $\endgroup$
    – gio
    Commented Mar 21, 2012 at 17:10

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