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Fix a $\mathbb Z_+^n$-graded Lie algebra ${\frak a}=\oplus_{r \in\mathbb Z_+^n}^{} {\frak a}[r]$ such that ${\frak g}:={\frak a}[0]$ is a finite-dimensional semisimple Lie algebra over the complex numbers and ${\frak a}[r]$ is a finite-dimensional $\frak g$-module for all $r\in\mathbb Z_+^n$.

We have a natural ideal given by ${\frak a_+}=\oplus_{r \in\mathbb Z_+^n, r\ne 0}^{} {\frak a}[r]$. Denote by $U(\frak a_+)$ its universal enveloping algebra. Notice that $U(\frak a_+)$ inherits a $\mathbb Z_+^n$-gradation from $\frak a$.

How to describe $U({\frak a}_+)[k]$ as a $\frak g$-module using the PBW Theorem ? (the graded pieces of $U({\frak a}_+))$

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The symmetrization map (which is a bijection by the PBW theorem) $S(\mathfrak{a}_+)\to U(\mathfrak{a}_+)$ respects the degree and the $\mathfrak{g}$-module structure. – user8268 Jan 21 '12 at 22:18
But how should I proceed to write explicitly the decomposition of each graded piece as a sum of tensor products of symmetric powers? – Lucas Jan 21 '12 at 23:13
@user8268: I think that he wants a decomposition of each piece in terms of tensor products of symmetric powers of ${\frak a}[r_i]$ for suitable choice of $a[r_i]$. I don't know how to do either. – Binai Jan 22 '12 at 17:55
$(S\mathfrak{a}_+)[k]$ is the direct sum of $S^{a_1}(\mathfrak{a}[b_1])\otimes\dots\otimes S^{a_m}(\mathfrak{a}[b_m])$ over all $a_1 b_1+\dots+a_m b_m=k$, $b_1<\dots<b_m$. – user8268 Jan 22 '12 at 19:54
@user8268 The gradation is in $\mathbb Z_+^n$, then something more is necessary. There is no meaning for $b_1<\cdots< b_m$ in this case. Do you know where there exists a proof in the case of $\mathbb Z_+$-gradation? I think that it is possible to extend it... Otherwise, you could write one here to help! – Binai Jan 22 '12 at 21:05

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