When relative Lie algebra cohomology of a parabolic subalgebra is trivial?

${\bf Question}$: List parabolic subalgebras whose relative Lie algebra cohomology with respect to its semi-simple subalgebra are trivial. In other words, list all pairs of a Dynkin graph $\Gamma$ and its subgraph $S$ such that the relative Lie algebga cohomology of a parabolic subalgebra $p_S$ of the Kac-Moody Lie algebra $\mathfrak{g}_{\Gamma}$ assigned with the graph $\Gamma$ relative to the Kac--Moody subalgebra assigned to a subgraph $S$ with trivial coefficients is trivial: $H^{>0}(p_s, g_S;\mathbb{C}) = 0$ and $H^{0}( p_S, \mathfrak{g}_S;\mathbb{C})=\mathbb{C}$.

Note that by a parabolic subalgebra $p_S$ I mean the subalgebra generated by $e_{\alpha}$ with $\alpha\in \Gamma$ and $f_{\alpha}$ with $\alpha\in S$. Where $\langle e_{\alpha},h_{\alpha}, f_{\alpha}\rangle$ is an $sl_2$ triple assigned with a simple root $\alpha$. What means that parabolic Lie subalgebra does not contain the entire Cartan subalgebra. I.e. $h_\alpha\in p_S$ iff $\alpha\in S$.

${\bf \text{Equivalent formulation using BGG resolution:}}$ list the cases of $\Gamma\supset S$ when $\forall\omega\in W_{\Gamma} / W_{S}$ (all elements of the quotient of the Weyl group of $\Gamma$ by the Weyl group of $S$) there exists $\alpha=\alpha(\omega)\in S$ such that the corresponding scalar product $\langle\omega(\rho)-\rho, \alpha\rangle\neq 0$ is different from zero.

In particular, there are no examples when $\Gamma=A_n$. However, relative cohomology vanishes when $\mathfrak{g}_{\Gamma}$ is affine Lie algebra and $\mathfrak{g}_S$ is the corresponding finite-dimensional semisimple Lie algebra.

Are there any other examples?

• So your "parabolic" subalgebra $p_S$ is, in fact, the commutator $[\mathfrak{p}_S, \mathfrak{p}_S]$ of the true parabolic subalgebra of $\mathfrak{g}_S$ that is typically denoted $\mathfrak{p}_S$. – Victor Protsak Jun 9 '13 at 1:01