As is known to all, the Lie algebra $\frak{sl}_2$ admits a very nice representation on $$ \mathbb{K}[X,Y] $$ the polynomials in two variables, given by $$ E \mapsto X\frac{\partial }{\partial Y}, ~~ F \mapsto Y\frac{\partial }{\partial X}, ~~ H \mapsto X\frac{\partial }{\partial X}  Y\frac{\partial }{\partial Y}. $$ Does an analogous representation exist for general $\frak{sl}_n$?

3$\begingroup$ These can be thought as vector field on $\mathbb P^1_{\mathbb K}$, and it indeed generalise. For a semisimple algebraic group $G$, global sections on the flag variety $G/B$ of the tangent sheaf are isomorphic to $\mathfrak g$ as Lie algebra ($B$ is a Borel subgroup). $\endgroup$– Nicolas HemelsoetJul 11, 2022 at 15:32

5$\begingroup$ Mapping $E_{ij}\mapsto X_i\partial_j$ yields an embedding of $\mathfrak{gl}_n$. $\endgroup$– YCorJul 11, 2022 at 15:36

$\begingroup$ @YCor: Nice! Thanks a lot. But to be sure, this is an embedding of $\frak{gl}_n$ into the linear operatos on the polynomials in $n$variables? $\endgroup$– Jake WetlockJul 11, 2022 at 15:45

1$\begingroup$ @Sergey: Fixed! Thanks for pointing this out! $\endgroup$– Jake WetlockJul 11, 2022 at 15:54

1$\begingroup$ @TimCromby yes, $X_i\partial_j$ is viewed as derivation of $k[X_1,\dots,X_n]$. $\endgroup$– YCorJul 11, 2022 at 16:05
1 Answer
The group $GL(V)$ acts on a vector space $V$ by linear automorphisms, and this induces the action of its Lie algebra $\mathfrak{gl}(V)$, i.e. a homomorphism to the Lie algebra of vector fields (differentiations). Explicitly $$e_{ij} \mapsto X_i \partial_j.$$ The composition of two differentiations is a second order differential operator, and the commutator is another differentiation. All these polynomial differential operators naturally act on the ring $\mathbb{K}[V]$ of polynomial functions.
For $\mathfrak{sl}(V) \subset \mathfrak{gl}(V)$ just restrict to a subalgebra. For explicit operators like $E,F,H$ choose a basis of $\mathfrak{sl}(V)$, e.g. $e_{ij}$ for $i\neq j$ and $e_{ii}  e_{i+1,i+1}$.

$\begingroup$ Great! Thanks a lot! To check I have it correctly, the weight of the monomial $X_i$ is $(0, \dots, 0, 1,1, 0, \dots, 0)$? $\endgroup$ Jul 11, 2022 at 16:10

1$\begingroup$ In order to have the name here, which you certainly know, your suggested basis for $\mathfrak{sl}(V)$ is an example of a Chevalley basis (for an arbitrary semisimple Lie algebra). $\endgroup$– LSpiceJul 11, 2022 at 16:24

$\begingroup$ P.S. I guess every representation can be realized as a subrepresentation of this representation? $\endgroup$ Jul 11, 2022 at 16:54

2$\begingroup$ @TimCromby No for $n\neq 2$ this does not capture all representations. Indeed what we have found is equivalent to the symmetric tensor algebra of $V$ so this splits into irreducible pieces of the form $S^k(V)$ (i.e. the degree $k$ homogeneous polynomials in $V$). So, for example, we are missing $\bigwedge^k V$ the exterior powers. To capture all representations we need to take the full tensor powers of $V$ rather than just its symmetric powers. $\endgroup$– CallumJul 11, 2022 at 17:19

$\begingroup$ @Callum: Yes it's clear. Thanks for the answer! $\endgroup$ Jul 11, 2022 at 17:28