How to compute the dimension of the ideals in the lower central series of a nilradical of a parabolic subalgebra?

Let $\mathfrak g$ be simple complex Lie algebra, $\mathfrak p$ a parabolic subalgebra associated to $\Pi_0 \subset \Pi\subset \triangle$, where $\Pi$ (resp. $\triangle$) is the set of simple roots (resp. positive roots) with respect to some Cartan subalgebra $\mathfrak h$. Denote Let $\mathfrak n$ the nilradical of $\mathfrak p$.

It is known that $\gamma \in\triangle$ is a root "corresponding to" $\mathfrak n$ if and only if $coord_{\alpha}(\gamma)>0$ for some $\alpha\in\Pi_0$. Let $o(\gamma)=\sum_{\alpha \in \Pi_0}coord_{\alpha}(\gamma)$. The lower central series of $\mathfrak n$ is $\mathfrak n=C^0(\mathfrak n)\supset C^1(\mathfrak n)\supset C^2(\mathfrak n)\supset \cdots \supset C^{k-1}(\mathfrak n) \supset C^k(\mathfrak n)=0$, so that $k$ is the nilpotency index of $\mathfrak n$.

It is possible to describe $C^j(\mathfrak n)$ in terms of the root system: $$C^j(\mathfrak n)=\bigoplus_{\stackrel{\gamma\in\triangle}{o(\gamma)=j+1}} X_{\gamma},$$ were $X_\gamma$ is the root vector associated to $\gamma$.

My question is: How can I compute $\dim C^j(\mathfrak n)$, depending on $\Pi_0$ and $\mathfrak g$? (using root systems, Cartan matrices, and so on) Mainly, the question is: Is it possible to find, given $t\in\mathbb N$, the cardinal of $\{\gamma\in\triangle: o(\gamma)=t\}$? How??

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    $\begingroup$ Maybe you want to have a look at: H. Azad, M. Barry and G. Seitz: On the structure of parabolic subgroups. Comm. Alg. 18 (1990), 551-562. They determine the structure of the unipotent radical of a parabolic subgroup of a split Chevalley group. I guess the result transfers, as you are interested in the Lie algebra of the unipotent radical of a parabolic in a complex Lie group. $\endgroup$ – Matthias Wendt Mar 21 '15 at 21:07

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