Decomposing the universal enveloping algebra into a direct sum of irreducibles

Question

In the case of $\mathfrak{g} = \mathfrak{su}_2$ (the Lie algebra of $SU(2)$), the universal enveloping algebra can be decomposed into irreducibles as $$ U(\mathfrak{g}) = \bigoplus_{j=0}^\infty V_{2j+1}, $$ where $V_d$ is an irreducible representation of $SU(2)$ of dimension $d$ (e.g., see here).

Can we make a similar statement for $SU(n)$ in general? That is, which irreducibles does $U(\mathfrak{g})$ decompose into (and with what multiplicity) when $\mathfrak{g} = \mathfrak{su}_n$?

I would guess there is some explicit map/algorithm to compute this? For example, when $\mathfrak{g} = \mathfrak{su}_2$ then \begin{align*} \mathfrak{g}^{\otimes 0} &= V_1 \\ \mathfrak{g}^{\otimes 1} &= V_3 \\ \mathfrak{g}^{\otimes 2} &= V_1 \oplus V_3 \oplus V_5 \\ \mathfrak{g}^{\otimes 3} &= V_1 \oplus V_3 \oplus V_3 \oplus V_3 \oplus V_5 \oplus V_5 \oplus V_7. \\ \end{align*} So it seems like $V_{2j+1}$ can be roughly related to $\mathfrak{g}^{\otimes j}$.

Answer 1

First of all, it is unclear to me why you compute $\mathfrak{g}^{\otimes j}$. The universal enveloping algebra is a quotient of the tensor algebra, so $T^j(\mathfrak{g}) = \mathfrak{g}^{\otimes j}$ will not share the irreducible decomposition of $\mathcal{U}(\mathfrak{g})$.

Secondly, the decomposition of $\mathfrak{su}(2)$ seems wrong: recall that $\mathcal{U}(\mathfrak{g}) \cong \mathcal{S}(\mathfrak{g})$ (the symmetric algebra, which can be interpreted as polynomials in $\mathfrak{g}$) as $\mathfrak{g}$-modules. Then clearly $\mathcal{S}^2(\mathfrak{g})$ has dimension 6, generated by polynomials $e^2, f^2, h^2, eh, fe, fh$. At the same time, $e^2$ is a maximal vector of weight 4, so it generates an irreducible $\mathfrak{g}$-module isomorphic to $V_5$. It follows that the decomposition of $\mathcal{U}(\mathfrak{g})$ is $V_1 \oplus V_5$. But we also have $\mathcal{U}^0(\mathfrak{g}) \cong V_1$ (ie the scalars). Thus we see that $V_1$ appears multiple times in the decomposition, and not just once.

So let's think this through from the start. First of all, what is the decomposition of $\mathfrak{g} = \mathfrak{su}(2)$ as a $\mathfrak{g}$-module into simple modules? Well, we know $\mathfrak{su}(n)$ is simple, and a submodule of the adjoint representation is the same as an ideal. Therefore, the adjoint representation must be simple. For $\mathfrak{su}(2)$, we have simply $\mathfrak{su}(2) \cong V_3$ as $\mathfrak{su}(2)$-modules. Next, we need a way to compute the decomposition of symmetric powers of $V_3$. This sort of problem (with tensor, symmetrix or exterior powers) is called $\textbf{plethysm}$. It is a hard problem in general but many specific cases have been worked out. For the tensor product case, there is Steinberg's formula. The symmetric power case of $\mathfrak{su}(2)$ it is not too bad, see for instance this post.

For $\mathfrak{su}(3)$, we can see that $\mathfrak{su}(3) \cong L(\alpha_1 + \alpha_2)$. This is the irreducible module with highest weight vector having weight $\alpha_1 + \alpha_2$. Next we would need to find the decomposition of $\mathcal{S}^j(L(\alpha_1 + \alpha_2))$. Unfortunately I didn't find anything that would help with this through a cursory online search.

Answer 2

Question: "Can we make a similar statement for $SU(n)$ in general? That is, which irreducibles does $U(\mathfrak{g})$ decompose into (and with what multiplicity) when $\mathfrak{g} = \mathfrak{su}_n$?"

Answer: If $V$ is any irreducible $SU(n)$-module, let $0 \neq v \in V$ be any non zero vector. Since $SU(n)$ is semi simple, the surjective map

$$\phi_v: U(\mathfrak{su}(n)) \rightarrow V$$

defined by $\phi_v(z):=zv$, splits. Hence $U(\mathfrak{su}(n)) \cong W \oplus V$. Hence any irreducible module $V$ is a direct summand of $U(\mathfrak{su}(n))$.

Note: The above argument proves that an irreducible $SU(n)$-module $W$ is either finite dimensional or countably infinite dimensional, since by the PBW theorem it follows $U(\mathfrak{su}(n))$ is countably infinite dimensional.