0
$\begingroup$

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 $U({\frak a}_+))$

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ It would be better to make an effort to improve the question rather than replacing it with a short sentence. $\endgroup$
    – S. Carnahan ♦
    Commented Jan 25, 2012 at 5:39

1 Answer 1

Reset to default
0
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Could you give me details or reference? I don't know how to get this expression! $\endgroup$
    – Lucas
    Commented Jan 25, 2012 at 14:32
  • $\begingroup$ 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. $\endgroup$
    – DamienC
    Commented Jan 25, 2012 at 17:11
  • $\begingroup$ Ok. I can see that you are using this fact. But I can't see how you are adjusting the powers of these symmetric algebras. $\endgroup$
    – Lucas
    Commented Jan 25, 2012 at 18:53
  • $\begingroup$ 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. $\endgroup$
    – DamienC
    Commented Jan 25, 2012 at 20:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .