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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 $\frak{sl}_n$?

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    $\begingroup$ 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). $\endgroup$ Jul 11, 2022 at 15:32
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    $\begingroup$ Mapping $E_{ij}\mapsto X_i\partial_j$ yields an embedding of $\mathfrak{gl}_n$. $\endgroup$
    – YCor
    Jul 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? $\endgroup$ Jul 11, 2022 at 15:45
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    $\begingroup$ @Sergey: Fixed! Thanks for pointing this out! $\endgroup$ Jul 11, 2022 at 15:54
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    $\begingroup$ @TimCromby yes, $X_i\partial_j$ is viewed as derivation of $k[X_1,\dots,X_n]$. $\endgroup$
    – YCor
    Jul 11, 2022 at 16:05

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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 $e_{ii} - e_{i+1,i+1}$.

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  • $\begingroup$ 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
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    $\begingroup$ 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). $\endgroup$
    – LSpice
    Jul 11, 2022 at 16:24
  • $\begingroup$ P.S. I guess every representation can be realized as a sub-representation of this representation? $\endgroup$ Jul 11, 2022 at 16:54
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    $\begingroup$ @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. $\endgroup$
    – Callum
    Jul 11, 2022 at 17:19
  • $\begingroup$ @Callum: Yes it's clear. Thanks for the answer! $\endgroup$ Jul 11, 2022 at 17:28

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