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Let $X$ denote the flag variety of a semi-simple group $G$ (in characteristic 0) and let $T_X$ denote its tangent bundle. I would like to ask the following question(s):

1) Is it true that for any $n\geq 0$ we have $H^i(X,T_X^{\otimes n})=0$ for $i>0$?

2) More generally, let $\lambda$ be a dominant weight of $G$ and let $\mathcal O(\lambda)$ be the corresponding line bundle on $X$. Is it true that $H^i(X, T_X^{\otimes n}\otimes \mathcal O(\lambda))=0$ for $i>0$?

When tensor powers of $T_X$ are replaced by symmetric powers, this is known to be true (for example it is proved in a paper of Kumar, Lauritzen and Thomsen).

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It's been almost 9 years, so maybe you are no longer interested in this question, but in a recent(-ish) preprint with Maxim Smirnov we have shown that the answer to the first question is no in general, even when restricted to the exterior powers of the tangent bundle (which describes the Hochschild cohomology of $X$).

In Hochschild cohomology of generalised Grassmannians we have mostly focused on the case of $G/P$ where $P$ is a maximal parabolic subgroup, and discuss when there is such vanishing for exterior powers. But we also give an explicit family of examples (namely for the symplectic Grassmannian $\operatorname{SGr}(3,2n)$) which has an $\mathrm{H}^1$, which is what you are looking for.

If you care about the full flag variety, rather than generalised Grassmannians, then there are also counterexamples, including in type A, starting from $\mathrm{SL}_6$. This one was obtained using Macaulay2 by Allen Knutson.

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I don't have a complete answer, but let me just note that your question 2) would be true if there is a Frobenius splitting of the cotangent bundle $T^*_{X^n}$ of $X^n$ which compatibly splits the diagonal copy $\Delta$ of $T^*_X$, the cotangent bundle of $X$. Since $T^*_{X^n}$ is the cotangent bundle of the flag variety of the semisimple algebraic group $G^n$, there is a candidate for this splitting, namely the splitting of Kumar, Lauritzen, and Thomsen that you mention. I don't know, though, if their splitting compatibly splits $\Delta$; that is an interesting question.

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  • $\begingroup$ Thanks; I understand this argument, but I don't know if the above-mentioned splitting is indeed compatible with the diagonal. $\endgroup$
    – Alexander Braverman
    Commented Nov 30, 2011 at 21:53
  • $\begingroup$ Kumar, Lauritzen, and Thomsen give a more explicit version of their splitting in the type A case and mention that it's the same as a splitting due to Mehta and van der Kallen. In the paper of Mehta and van der Kallen they show that various subvarieties of the cotangent bundle are compatibly split in type A. They don't consider the diagonal case, but it may follow from their construction, which has the benefit of being more hands-on. You might want to contact one of them and ask. (Also, van der Kallen is active on MathOverflow). $\endgroup$
    – Chuck Hague
    Commented Dec 4, 2011 at 17:08

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