Decomposition of universal enveloping algebra of the general linear Lie algebra

Question

Hi all I have a question, let $\mathfrak g = \mathfrak{gl}(n)$ be the general linear Lie algebra, and $U$ the enveloping algebra of $\mathfrak g$. Now we regard $U$ as a $\mathfrak g$-module $U^{\text{ad}}$ through the adjoint action of $\mathfrak g$, that is, $x\cdot u = xu-ux$, for all $x\in \mathfrak g$ and $u\in U$.

$\bf My ~Question$: What is the $\mathfrak g$-decomposition of $U^{\text{ad}}$ into direct sum of irreducible $\mathfrak g$-modules? Thanks!

Answer 1

For any reductive Lie algebra $\mathfrak{g}$, the enveloping algebra $U=U(\mathfrak{g})$ is isomorphic to $\mathrm{Sym}(\mathfrak{g})$ as a $\mathfrak{g}$-module. This fact is essentially the PBW theorem: the $\mathrm{ad}(\mathfrak{g})$-stable filtration $$U^{\leq d}=\mathbf{C} \{ x_1 \cdots x_e \ | \ x_i \in \mathfrak{g}, \ e \leq d \}$$ has associated graded algebra $\mathrm{Sym}(\mathfrak{g})$, and by complete reducibility $$U^{\leq d} \cong \bigoplus_{e \leq d} U^{\leq e}/U^{\leq e-1}$$ as $\mathfrak{g}$-modules.

Answer 2

Yes. Write $\mathfrak{g} := \mathfrak{gl}_n$.

$\mathfrak{U}(\mathfrak{g})$ is a sum of finite dimensional $\mathfrak{g}$-modules. This follows from the following identity: $$\text{ad}x. (x_1 x_2 \cdots x_m) = \sum_{i = 1}^m x_1 \cdots [x, x_i] \cdots x_m.$$ Note the bracket doesn't change the length of the 'word' $x$ acts on. This means the subspace spanned by the PBW basis vectors with length no greater than $m$ is a submodule of the adjoint action. Using PBW theorem again to get the claim. Notice this claim is true for any Lie algebra.

Each submodule has the property that the center of $\mathfrak{g}$, i.e. scalar matrices, act as numbers, hence semisimply. It is not hard to see that these modules are completely reducible. Therefore so is $\mathfrak{U}(\mathfrak{g} = \mathfrak{gl}_n)$.