# PBW-Theorem and multigraded Lie algebras

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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It would be better to make an effort to improve the question rather than replacing it with a short sentence. –  S. Carnahan Jan 25 '12 at 5:39

As $\mathfrak{a}[0]$-modules we have: $$U(\mathfrak{a}_+)[k]=\bigoplus_{\substack{l\geq1 \\ c_1r_1+\cdots+c_lr_l=k}}S^{c_1}(\mathfrak{a}_+[r_1])\otimes\cdots\otimes S^{c_l}(\mathfrak{a}_+[r_l])$$
PS: I assume you work over a field of characteristic zero and I use the isomorphism of graded $\mathfrak a$-modules $S(\mathfrak a)\to U(\mathfrak a)$ given by the symmetrization map.
Dear Lucas. I used the fact that $S(V\oplus W)=S(V)\otimes S(W)$. Then from the PBW theorem you get that $U(\mathfrak a_+)[k]$ is isomorphic to $S(\mathfrak a_+)[k]$ as a $\mathfrak a[0]$-module. –  DamienC Jan 25 '12 at 17:11
Well. If $V$ is homogeneous of degree $k$ then $S^c(V)$ is homogeneous of degree $ck$. That's all what you need to know. –  DamienC Jan 25 '12 at 20:30