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Why does $\dim \Gamma(X)_n=\chi(\mathcal{O}_X(n))$ (probably, for sufficiently large $n$), where $\Gamma(X)_n$ is $n$-dimensional component of homogeneous coordinate ring of $X$ and $\mathcal{O}_X(n)$ is $n$-twisting sheaf on $X$?

On the one hand, Hilbert polynomial of projective variety $X \subset \mathbb{P}^m$ is defined as $H_X(n)=\dim \Gamma(X)_n$.

On the other, it can be defined for any sheaf as $H_L(n)=\chi (L^{\otimes n})$. I'd like to see the connection.

VividD
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evgeny
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1 Answers1

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There's a slight conflict here with the terminology I'm used to, but I think the basic fact being used is Serre vanishing: for $X$ projective over a Noetherian ring $A$ and $\mathscr{F}$ coherent on $X$ we have $H^i(X, \mathscr{F}(m)) = 0$ for $i > 0$ and $m \gg 0$ [see Hartshorne III.5.2 or Vakil 18.1.4]. In particular, for large $m$ we have $\chi(X, \mathscr{F}(m)) = h^0(X, \mathscr{F}(m))$.

Hoot
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    My apologies if I'm underestimating your question. There is also something to say depending on how you defined the coordinate ring -- as one learns early on in Hartshorne, $\mathscr{O}(1)$ produces the "best" ring, but it may be different from the original one. – Hoot Jun 09 '14 at 18:19
  • Thank you for answer, I just did not know about this vanishing. – evgeny Jun 09 '14 at 18:23
  • I see. I added some references. – Hoot Jun 09 '14 at 18:26