# Vanishing of local cohomology groups

Let $k$ be a field and let $X$ be a smooth separated $k$-variety. Let $T$ be a closed integral subscheme of $X$ of generic point $\eta$. The object of interest here is the local cohomology group $$H^{q}_{\eta}(\Omega^r_{X/k})\; \text{ for some }\;q,r>0,$$ defined, for example, as colimit over $\eta \in U$ for $U$ open in $X$ of $$H^q_{\overline{\eta}\cap U}(U, ({\Omega^{r}_{X/k}})_{|U}).$$ Is it true that for $q<codim_{X}(T)$ those groups are $0$ for all $r$? If not, is this true for some special value of $r$ (better, what is the right range of values for $r$)?

A small comment: when one says "local cohomology" then the automatic reaction is to look at duality statements. Well, if $A=\mathcal{O}_{X,\eta}$ with maximal ideal $\mathfrak{m}$, then Grothendieck tells us that $$H^q_{\mathfrak{m}}(\Omega^r_{X/k}) \xrightarrow{\sim} \text{Hom}( \text{Ext}_A^{a-q}(\Omega^r_{X/k, \eta}, A), I)$$ where $I$ is a dualizing module for $A$ and $a=codim_{X}(T)= \dim A$. Since $A$ is local regular, we can use $I= H^a_\mathfrak{m}(A)$.

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You almost answered this yourself. Because $\Omega^r_{X/k}$ is locally free , the question is just whether $H^q_\mathfrak m(A)$ is zero for $q = 0, \ldots, \dim(A) - 1$. Since you know about Matlis duality, my guess is you know how to compute $H^q_\mathfrak m(A)$. Good luck.
OK, no sorry, my fault! Well, using your duality thing, you have to prove $\text{Ext}^i_A(A, A)$ is zero for $i$ not equal to zero. Does this help? – Count Dracula Apr 26 '14 at 16:10
Not entirely... You reduce to $Ext^i(A,A)$ because you make the Ext commute with the direct sum? This does not work in general, right? Anyway, is it obvious that $Ext^i(A,A)=0$ - by Koszul computation, for example? – FedeB Apr 27 '14 at 8:10