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

Ok, so this is where I reveal my ignorance as an algebraic geometer: I had previously asked about pushforward of line bundles from the smooth locus of a variety to the whole thing. I think I understand basically how that picture works now, but I have a variation thereof that I would like to ask about.

Let $X$ is a normal variety (let's say irreducible quasi-projective of finite type over $\mathbb{C}$; you can even assume terminal and $\mathbb{Q}$-factorial if you like), $Y$ its smooth locus and $Z$ its singular locus. In fact, let's say the dimension of $Z$ is at most $\dim X - 4$ just for good measure.

What can be said about the local cohomology $H^i_Z(\mathcal{O}_X)$ which fills in the exact sequence $$\cdots \to H^i_Z(\mathcal{O}_X)\to H^i(X;\mathcal{O}_X)\to H^i(Y;\mathcal{O}_Y)\to \cdots?$$

What I'd love to say is that this is 0 in degrees $\leq 3$, but obviously, I'd accept other answers if they happen to be true.

share|improve this question
add comment

1 Answer

This is essentially answered in my answer to a question raised after my answer to an MO question here.

Basically the point is that what you want is true provided your variety is $S_4$. (See the above link for a proof). In particular, if your variety is Cohen-Macaulay (which terminal and even klt singularities are, but not log canonical) then you're in business.

In general, this vanishing is essentially equivalent to that your variety be $S_4$. If you consider a $4$-dimensional variety $X$ with a single singular point $z\in X$ that is $S_2$, but not $S_4$, then $X$ is normal but $H^i_z(X,\mathscr O_X)\neq 0$.

For an explicit example of such a singularity consider a cone over an abelian threefold. This is a non-klt log canonical singularity. The fact that this is such an example follows from the condition that tells you what $S_m$ a cone is based on the cohomology groups of the scheme it is a cone over. For a complete proof of that criterion in a rather general case see Lemma 3.1 of this paper of Patakfalvi.

You might also need the cohomological interpretation of depth, which is (one of) Grothendieck's vanishing theorem. See for example in Bruns-Hezog.

share|improve this answer
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.