Let $P$ be a smooth connected projective variety (say, over complex numbers); $H$ is its smooth hyperplane section. What can be said about the Zariski cohomology of $H$ with constant coefficients? It is certainly zero in positive degrees if $H$ is irreducible. Does it vanish in lower degrees (those that are smaller than $\operatorname{dim}P-1$) in general?

Upd. It seems that with rational coefficients the answer is positive. Indeed, the cohomology groups in question would be exactly the weight zero part of the singular cohomology of $H$ (considered as a sequence of mixed Hodge structures), whereas for $i<\operatorname{dim}P-1$ the Weak Lefschetz theorem yields that $H^i_{sing}(H)$ is pure of weight $i$. Yet for torsion (or integral) coefficients the answer is not clear to me. Could the Zariski cohomology in question have torsion? Is there an upper bound on the exponent of this torsion (that depends on $\operatorname{dim}P$)?

P.S. Cf. Does Artin's vanishing hold for '$E_2$-weight pieces' for (torsion) cohomology of affine varieties?

  • $\begingroup$ I am confused by the fact that you ask about Zariski cohomology of the constant sheaf, and you then talk about Weak Lefschetz. The Lefschetz theorems are about singular cohomology, which is not closely related to Zariski cohomology of the constant sheaf. Are you sure you are asking the question you want to ask? $\endgroup$ – David E Speyer Dec 5 '11 at 20:01
  • $\begingroup$ Yes, I do.:) Certainly, Zariski cohomology is quite distinct from singular or etale one; yet there are some relations between them. $\endgroup$ – Mikhail Bondarko Feb 15 '12 at 11:43

It does not necessarily vanish. For example, if $Y$ is the union of two irreducible curves meeting at more than one point and $A$ is a non-zero abelian group, $H^1(Y, A)$ is not zero.

  • $\begingroup$ Could this happen in higher dimensions? I want the number of the non-vanishing group to be smaller than $dim P-2$. $\endgroup$ – Mikhail Bondarko Dec 2 '11 at 7:01
  • $\begingroup$ Besides, I would certianly like to understand which $Y$ could be presented as $H_1\cap H_2$. $\endgroup$ – Mikhail Bondarko Dec 2 '11 at 7:03

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