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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 this 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.
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Funny you should ask. :)

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 factparticular, it 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 abelian threefold. This is a non-klt log canonical singularity. The fact that this is 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 this 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.
show/hide this revision's text 2 edited body; added 400 characters in body

Funny you should ask. :) 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 fact, it is essentially equivalent to that. If you consider a $4$-dimensional variety $X$ with a single singular point $z\in X$ that is $S_2$, but not $S_3$, 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 abelian threefold. The fact that this is 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 this criterion in a rather general case see Lemma 3.1 of this paper of Patakfalvi.
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