Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to prove that for an algebraic surface $X$ (under some extra assumptions that are probably not important) there the space $H^0(X,\Omega_X^1)$ is trivial, i.e. that there exist no globally defined differential forms with not poles. Since this is part of an assignment, I am certainly not asking for the reason/proof of why this is true. I would however like to ask for a reference.

  1. I am aware of a number of (apparently) related vanishing theorems. The most relevant one Kodaira-Akizuki-Nakano vanishing theorem, which (for the purposes of the problem) says that: $$ H^q(X,\Omega_X^p \otimes \mathcal{L}^{-1}) = 0 $$ for an ample invertible sheaf $\mathcal{L}$ and $p+q < \operatorname{dim} X = 2$. (see http://en.wikipedia.org/wiki/Akizuki%E2%80%93Nakano_vanishing_theorem) Does there exist a similar general result, which does not include a line bundle, and which is accessible (say, at a graduate level)? If so, a reference would be highly appreciated.

  2. I am aware that there exist some results that immediately solve the problem, but appear to be a few orders of magnitude too strong. For instance, there are papers that prove that $H^0(X,S^m \Omega_X^1) = 0$ as a very special case of the theory, listed below. However, the proofs are rather inaccessible. I would be extremely grateful for a reference to some weaker/special versions of such theorems.

  3. I would appreciate a reference to any piece of literature that might be helpful in learning how to solve this problem, or the problem of computing the Hodge numbers $\dim H^p(X,\Omega^q_X)$ in a more general setup.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

The condition of embedding in $\P^3$ is very strong! These surfaces are extremely special. One thing that's often helpful in cases like these is the conormal exact sequence (the same thing you use to prove the adjunction formula). This says that if $Y$ is a smooth subvariety of $X$ (included via $i : Y \to X$), you get $0 \to \mathcal I_Y/\mathcal I_Y^2 \to i^\ast \Omega_X \to \Omega_Y \to 0$. In the case that the codimension of $Y$ is one, the first term is nothing but the conormal bundle, and in the $\mathbb P^3$ case you can just compute the things in the long exact sequence.

As to (1), there is no vanishing $H^q(X,\Omega^p) = 0$ unless something special is going on (some extra added "positivity", often). There are a handful of cases where you can, for instance on Fano varieties ($-K_X$ is ample) -- see if you can derive the statement here. There are some related theorems (eg you can relax to $\mathcal L$ big and nef) -- Lazarsfeld's "Positivity in Algebraic Geometry" describes many such statements and their connections. (2) I doubt these things actually solve the problem immediately, since you really do have to use the assumption that it's a hypersurface. (3) I don't know much about Hodge theory, so I don't have any good suggestions except to read one of the standard texts on the subject, eg Griffiths-Harris or the vol I of Voisin's. You might be able to find notes that work out more examples.

share|improve this answer
    
Thank you for pointing out how strong the condition of being embedded in $\mathbb{P}^3$ is! My initial intuition was that there should be a lot of freedom in choice of such surface, and that one should look for a general result. –  Feanor Jan 19 '13 at 12:45
add comment

I suspect that the extra assumptions probably are important -- otherwise, can you just take e.g. an abelian surface, even the product of two elliptic curves. This has $H^0(X,\Omega^1)$ 2-dimensional, spanned by $dz$ and $dz^\prime$.

share|improve this answer
    
Well, the assumption is that $X$ is given by a homogenous equation in $\mathbb{P}^3$. This did not strike me as a very strong condition, though. –  Feanor Jan 18 '13 at 19:38
add comment

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.