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Given a smooth projective variety $X \subset \Bbb{CP}^k$ why is it true that global sections of $O(l)|_X, l >> 0 $ are just global sections of $O(l)$ on $\Bbb{CP}^k$ restricted to $X$?

Here $O(l)$ is just an appropriate power of the anti-tautological line bundle on $\Bbb{CP}^k$.

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Powers of the CANONICAL bundle on $CP^k$ have no sections. You should say "dual tautological" instead of "canonical". – Sasha Nov 2 '13 at 7:17
Perhaps you wanted to say that for any given $X$ there exists an embedding for which this is true? Or that this is true for $l\gg 0$? – Sándor Kovács Nov 2 '13 at 8:41
Edited the question. Of course I meant the dual tautological line bundle and sufficiently high powers. – The Common Crane Nov 2 '13 at 14:02
up vote 2 down vote accepted

This is an answer to the edited question

Consider the short exact sequence $$ 0 \to \mathscr I_X(l) \to \mathscr O_{\mathbb P^k}(l) \to \mathscr O_X(l) \to 0, $$ and the long exact sequence of cohomology it induces: $$ 0 \to H^0(\mathbb P^k, \mathscr I_X(l)) \to H^0(\mathbb P^k,\mathscr O_{\mathbb P^k}(l)) \to H^0(X,\mathscr O_X(l)) \to H^1(\mathbb P^k, \mathscr I_X(l)) \to\dots . $$

By Serre's theorem $H^1(\mathbb P^k, \mathscr I_X(l))=0$ for $l\gg 0$ and hence the previous map $H^0(\mathbb P^k,\mathscr O_{\mathbb P^k}(l)) \to H^0(X,\mathscr O_X(l))$ is surjective.

You will not be able to get injectivity. In fact, the larger the $l$, the "less" injective that map is. $\dim H^0(\mathbb P^k,\mathscr O_{\mathbb P^k}(l))$ grows at the order of $l^k$ while $\dim H^0(X,\mathscr O_X(l))$ grows at the order of $l^{\dim X}$, so there is nothing you can do to guarantee injectivity. Quite the opposite, the only time you might have injectivity is for small $l$'s.

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Thanks for your answer. As it is quite apparent, I lack serious background in AG. One more small thing, which theorem of Serre are you exactly referring to? – The Common Crane Nov 4 '13 at 3:41
It's usually called "Serre's vanishing theorem" and can be found in almost any AG textbook, look in Hartshorne or in volume 1 of Lazarsfeld's positivity book. – Dan Petersen Nov 4 '13 at 7:40

This is not true. The restriction map $H^0(\Bbb{CP}^k, \mathcal{O}(l))\rightarrow H^0(X,\mathcal{O}_X(l))$ is neither injective nor surjective in general. The kernel gives the space of hypersurfaces of degree $l$ containing $X$. Surjectivity for all $l$ means that $X$ is projectively normal, a property which does not hold in general.

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You are correct. The question I meant was: why is this possible for high enough l? Hopefully this is indeed possible. – The Common Crane Nov 2 '13 at 15:47
if $l >>0$ then $H^1(P^k, I_{X}(l) ) = 0 $ – aginensky Nov 2 '13 at 21:15

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