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

Sommese's theorem is a natural generalization of the Weak Lefschetz; for a smooth projective (connected) $X$, an ample vector bundle $E/X$ of rank $e$, and a section $s:X\to E$ it states that the cohomology of $X$ is isomorphic to the cohomology of the zero locus $Z$ of $s$ ($Z=\{ x\in X:\ s(x)=0 \}$) up to degree $\dim X-e-1$. Sommese himself stated this theorem for singular cohomology. Yet it seems that the proof of the statement given in Lazarsfeld's 'Positivity in Algebraic Geometry' book works over an algebraically closed base field of any characteristic (and etale cohomology). The argument is described at the very beggining of Chapter 7; see http://books.google.ru/books?id=rd4sIp0f79cC&pg=PA65&hl=ru&source=gbs_toc_r&cad=4#v=onepage&q&f=false Lazarsfeld verifies that there is projection $p:P(E)\setminus Z^*\to X\setminus Z$ for a certain zero locus $Z^\ast$ on $P(E)$.

So my questions are:

  1. Is there anything that prevents applying Lazarsfeld's argument over a positive characteristic algebraically closed base field $K$? It seems an easy exercise to verify that $p$ is Zariski-locally a vector bundle. In any case, is Sommese's theorem known for $char K>0$?

  2. Why the space of one-dimensional quotients of $E$ considered by Lazarsfeld is the projectivization of $E$ and not of its dual?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

1. The argument seems to work fine in positive characteristic.

2. In Grothendieck's convention, the projectivization of vector bundle is defined with 1-dimensional quotients, and not 1-dimensional subspaces. The reason is that quotients work for arbitrary coherent sheaves, while 1-dimensional subspaces don't.

share|improve this answer
    
Thank you very much! Please tell me if you will have any doubts in 1 (later).:) –  Mikhail Bondarko Oct 14 '12 at 17:54

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.