Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\Bbb{CP}^k$.

share|improve this question
2  
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
2  
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

2 Answers 2

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.

share|improve this answer
    
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
1  
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.