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

I'm looking for a reference for Hard Lefschetz theorem in algebraic De Rham cohomology. By this I mean the statement that

If $i: Y \hookrightarrow X$ is a smooth hyperplane section of a smooth projective algebraic variety $X$ of dimension $n$ over a field $k$ of characteristic zero and $\omega \in H^2_{DR}(X)$ is its image under the cycle class map, then the operator

$L^j: H^{n-j}_{DR}(X) \to H^{n+j} _{DR}(X) $

sending $x$ to $x \cdot \omega^j$ is an isomorphism for $j=1, \ldots, n$.

Is a proof of this written somewhere?

Weak Lefschetz is proved in Hartshorne's paper "On the de Rham cohomology..." chapter III, section 7 but I was completely unable to find a reference for the former.

Help please! ;)

share|improve this question
2  
Griffiths-Harris, "Principles of Algebraic Geometry", p. 122 –  Francesco Polizzi Jun 21 '13 at 18:14
    
Thanks Francesco. Is that a proof for algebraic de Rham cohomology or the analytic one? I guess if you know how to do that for analytic you can deduce it for algebraic but you need some small argument when the base field is not $\CC$, don't you? –  vicban Jun 21 '13 at 18:50
2  
To expand Francesco's comment: after a field reduction followed an extension, you can assume that the ground field is $\mathbb{C}$. Apply Grothendieck's algebraic de Rham theorem to see that algebraic de Rham cohomology is isomorphic to singular cohomology. After this, the standard proof applies. However, if you are asking for a completely self contained elementary proof, then there isn't one. –  Donu Arapura Jun 21 '13 at 18:57
1  
What about the Hard Lefschetz theorem in étale cohomology? Do you also need to pass by Betti cohomology to prove it? Is Hard Lefschetz proved in this way for all Weil cohomologies known to satisfy it? –  vicban Jun 21 '13 at 19:20
    
No, Deligne's proof for etale cohomology is entirely different and is based on a strong form of the Weil conjectures. This is what I meant by saying that there isn't an elementary proof. –  Donu Arapura Jun 21 '13 at 20:53

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.