Motivation of this model of irreducible representation of $\mathfrak{sl}(2,\mathbb{F})$ of highest weight $m$ In Humphrey's book "Introduction to Lie Algebras and Representation Theory", Exercise 7.4 introduces a model of the irreducible representation of $L := \mathfrak{sl}(2,\mathbb{F})$ as follows:
Let $X, Y$ be a basis for the two dimensional vector space $\mathbb{F}^2$, on which $L$ acts as usual. Let $\mathcal{R} := \mathbb{F}[X,Y]$ be the polynomial algebra in two variables, and extend the action of $L$ to $\mathcal{R}$ by the derivation rule:
$$
z \cdot (fg) = (z \cdot f) g + f(z \cdot g),
$$
for $z \in L, f,g \in \mathcal{R}$. Then

*

*This extension is well-defined and that $\mathcal{R}$ becomes an $L$-module,

*The subspace of homogeneous polynomials of degree $m$, $\mathcal{R}_m$, with basis $X^m, X^{m-1}Y, \ldots, XY^{m-1}, Y^m$, is invariant under $L$ and irreducible of highest weight $m$.

I've checked the above two points. My question is: How do people come up such a model for the highest weight representation $V(m)$? Are there any hidden backgrounds behind this construction? For example, using Lie groups to solve PDEs or something like that? (Since this action is quite similar to Leibniz rule, so maybe "taking derivations" is involved?)
My "attempts": When I was doing this exercise, I just blindly checked everything but not knew what was really going on. So I'm hoping for some references or explanations.
(I have read some basics on Lie groups in the scope of John Lee's "Introduction to Smooth Manifolds" before.)
Thank you all so much for your answers and comments :)
 A: Question: "My question is: How do people come up such a model for the highest weight representation $V(m)$? Are there any hidden backgrounds behind this construction?"
Answer: If you let $V:=\mathbb{F}\{e_1,e_2\}$ be a two dimensional vector space over $\mathbb{F}$ and $L:=\mathfrak{sl}(V)$, it follows all symmetric powers $S^d(V)$ and $S^d(V^*)$ are canonically $L$-modules. The symmetric power $S^d(V^*) \cong \mathbb{F}[x_1,x_2]_d$ is isomorphic to the vector space of degree d homogeneous polynomials in $x_i:=e_i^*$ - the dual basis. Most irreducible $SL(\mathbb{F}^n)$-modules can be constructed using tensor and exterior powers of the standard module $W:=\mathbb{F}^n$. There is an explicit construction of all irreducible finite dimensional $SL(W)$-modules as submodules of
$$S^{d_1}(W)\otimes S^{d_2}(\wedge^2 W) \otimes \cdots \otimes S^{d_{n-1}}(\wedge^{n-1} W).$$
Example: for $n=2$ you get a canonical map
$$W \otimes W \rightarrow \wedge^2 W \cong T$$
where $T$ is the trivial $SL(W)$-module ($\wedge^2 W$ is trivial since for all $g\in SL(W)$ we have $det(g)=1$). It follows
$$W \cong W^* \otimes T \cong W^*$$
Hence $S^d(W) \cong S^d(W^*)$ are isomorphic as $SL(W)$-modules for all $d \geq 1$ and as you have noted above: This gives all irreducible finite dimensional modules in the case when $n=2$.
More generally $\wedge^{n-1}W \cong W^*$ and $S^d(W^*) \cong S^d(\wedge^{n-1}W)$ are isomorphic as $SL(W)$-modules. You may view $W, \wedge^{n-1} W$ as a $GL(W)$ module and there is no isomorphism
$$W \cong W^*$$
as $GL(W)$-modules, since the determinant $T$ is non-trivial for $GL(W)$.
