Derivative operator representation of U(sl3)

Hi, I'm Akaghef. I calculated derivative operator representation of $U(\mathfrak{sl}_3)$.(I don't have enough time to calculate $U_q$.)

0. Review: the $\mathfrak{sl}_2$ derivative-operator model

We recall the standard realization of $\mathfrak{sl}_2$ by differential operators,
which serves as the guiding prototype.
Let $\mathfrak{sl}_2=\langle E,F,H\rangle$ with relations
$$ [H,E]=2E,\qquad [H,F]=-2F,\qquad [E,F]=H. $$
Consider the polynomial ring $\mathbb C[x,y]$ and define
$$ E := x\,\partial_y,\qquad F := y\,\partial_x,\qquad H := x\,\partial_x - y\,\partial_y. $$
A direct computation shows that these operators satisfy the $\mathfrak{sl}_2$
relations, hence define a Lie algebra homomorphism
$$ \mathfrak{sl}_2 \longrightarrow \mathrm{Der}\big(\mathbb C[x,y]\big) \subset A_2, $$
and therefore a representation of $U(\mathfrak{sl}_2)$ on $\mathbb C[x,y]$.
The polynomial ring decomposes by total degree:
$$ \mathbb C[x,y]=\bigoplus_{n\ge0}\mathbb C[x,y]_n, $$
and each homogeneous component $\mathbb C[x,y]_n$ is stable under the action.
Moreover, $\mathbb C[x,y]_n$ is irreducible, with highest-weight vector $x^n$,
and is isomorphic to the $(n+1)$-dimensional irreducible representation $L(n)$.
Thus the derivative-operator representation of $\mathfrak{sl}_2$ “packages” all
finite-dimensional irreducible representations:
$$ \mathbb C[x,y] \;\cong\; \bigoplus_{n\ge0} L(n), $$
where each irreducible appears exactly once as a homogeneous component.
This clean direct-sum decomposition by degree is the feature we aim to generalize
to $\mathfrak{sl}_3$.

1. Differential-operator realization of $\mathfrak{sl}_3$

We considered the Weyl algebra $A_6$ generated by
$$ M_{ij} := x_i \partial_{x_j} \qquad (1 \le i,j \le 6), $$
which satisfy the $\mathfrak{gl}_6$ relations
$$ [M_{ij}, M_{kl}] = \delta_{jk} M_{il} - \delta_{il} M_{kj}. $$
Using this, we defined a Lie algebra homomorphism
$$ \rho : \mathfrak{sl}_3 \longrightarrow A_6 $$
by interpreting $(x_1,x_2,x_3)$ as coordinates of $V \cong \mathbb C^3$ and
$(x_4,x_5,x_6)$ as coordinates of the dual space $V^*$, with the correct
contragredient action.
Explicitly,
$$ \begin{aligned} E_1 &= M_{12} - M_{54}, &\quad F_1 &= M_{21} - M_{45}, \\ E_2 &= M_{23} - M_{65}, &\quad F_2 &= M_{32} - M_{56}, \end{aligned} $$
and $H_i = [E_i,F_i]$.
By direct computation using the $\mathfrak{gl}_6$ commutation relations,
we verified that these operators satisfy all defining relations of
$\mathfrak{sl}_3$ (Cartan relations, Chevalley relations, and Serre relations).
Hence this gives a well-defined representation of $U(\mathfrak{sl}_3)$ by
differential operators.

2. Occurrence of all finite-dimensional irreducibles

On the polynomial ring $\mathbb C[x_1,\dots,x_6]$, the action decomposes according
to bidegree
$$ \mathcal A_{A,B} := \mathbb C[x_1,x_2,x_3]_A \otimes \mathbb C[x_4,x_5,x_6]_B. $$
For each $(a,b)$, the vector $x_1^a x_4^b$ is a highest-weight vector of weight
$a\omega_1 + b\omega_2$, and the $U(\mathfrak{sl}_3)$-submodule it generates is
isomorphic to the irreducible module $L(a,b)$. Thus every finite-dimensional
irreducible $L(a,b)$ occurs inside this Weyl-algebra representation.
Using the standard proposition:
Proposition (separation by finite-dimensional irreducibles).
For a complex semisimple Lie algebra $\mathfrak g$,
$$ \bigcap_{V\ \mathrm{finite\text{-}dim.\ irrep}} \mathrm{Ann}_{U(\mathfrak g)}(V) = \{0\}. $$
we concluded that, since all $L(a,b)$ appear, the induced algebra homomorphism
$$ U(\mathfrak{sl}_3) \longrightarrow A_6 $$
is injective.
From a representation-theoretic viewpoint, every finite-dimensional irreducible
$\mathfrak{sl}_3$-module is of the form $L(a,b)$ with highest weight
$a\omega_1+b\omega_2$, and can be realized inside
$$\mathrm{Sym}^a(V)\otimes \mathrm{Sym}^b(\wedge^2 V) $$
where $V=\mathbb C^3$ is the defining representation and $\wedge^2 V\simeq V^*$ for $\mathfrak{sl}_3$.
The differential-operator model is simply a concrete incarnation of this fact,
obtained by identifying $\mathrm{Sym}(V)$ and $\mathrm{Sym}(V^*)$ with polynomial
rings and letting $\mathfrak{sl}_3$ act by derivations.
The invariant pairing $V\otimes V^*\to\mathbb C$ explains the appearance of the
polynomial $p$, and quotienting by $p$ isolates the top irreducible $L(a,b)$ in
each bidegree.

3. Invariant polynomial and quotient by $p=0$

We identified the polynomial
$$ p := x_1 x_4 + x_2 x_5 + x_3 x_6 $$
as $\mathfrak{sl}_3$-invariant under the $V \oplus V^*$ action.
Consequently, the ideal $(p)$ is $\mathfrak{sl}_3$-stable, and the differential
operators descend to the quotient algebra
$$ \mathbb C[x_1,\dots,x_6]/(p). $$
In each bidegree $(A,B)$, multiplication by powers of $p$ accounts for the chain
$$ L(A,B),\ L(A-1,B-1),\ \dots, $$
inside $\mathcal A_{A,B}$. After imposing the relation $p=0$, all lower terms are
killed, and each bidegree component becomes exactly one irreducible module:
$$ \mathbb C[x_1,\dots,x_6]/(p) \;\cong\; \bigoplus_{a,b \ge 0} L(a,b). $$
Thus, the quotient by $p=0$ provides a precise analogue of the $\mathfrak{sl}_2$
situation, where homogeneous components directly realize irreducible
representations.