1
$\begingroup$

Let $k$ be an algebraic closed field of characteristic 0. We write

$$X=\left( \begin{array}{ccc} 0 & 1\\ 0 & 0\\ \end{array} \right),~~ Y=\left( \begin{array}{ccc} 0 & 0\\ 1 & 0\\ \end{array} \right),~~ H=\left( \begin{array}{ccc} 1 & 0\\ 0 & -1\\ \end{array} \right)$$

Consider the polynomial representation of $sl_{2}(k)$, $$\rho:sl_{2}(k)\rightarrow End(k[x,y]),~~ \rho(X)=x\frac{\partial}{\partial y},~~\rho(Y)=y\frac{\partial}{\partial x},~~\rho(H)=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}$$. Then:

  1. The universal enveloping algebra $U(sl_{2}(k))$ is generated by the set {$\rho(X),~\rho(Y),~\rho(H)$} as an algebra.

  2. We write $L=sl_{2}(k)$.

$k[x,y]$ can be written as the direct sum of finite dimensional simple modules:

$k[x,y]=\displaystyle{\bigoplus_{n=0}^{\infty} k[x,y]_{n}}$.

Moreover, if $M$ is a finite dimensional irreducible representation of $sl_{2}(k)$ with $dim(M)=n+1$, then $M\cong k[x,y]_{n}$ as $L$-modules.
Can we extend this result to some kinds of Lie algebras?

Specifically, I try to find a class $\mathfrak{L}$ of Lie algebras, such that $\forall L\in\mathfrak{L}$, there is an $L$-module $V$ ("polynomial representation of $L$") possessing the following properties:

  1. The universal enveloping algebra $U(L)$ is generated by some "differential operators of $V$" as an algebra.

  2. $V$ can be written as a direct sum of simple modules: $V=\displaystyle{\bigoplus_{i\in \Lambda} V_{i}}$. Moreover, if $M$ is a simple $L$-module, there exists $i\in \Lambda$ such that $M\cong V_{i}$ as $L$-modules.

$\endgroup$
2
  • $\begingroup$ I get into trouble when I type the $2\times 2$ matrices $X$, $Y$ and $H$. Can somebody help me to edit it? $\endgroup$
    – sife
    Commented May 1, 2013 at 17:22
  • $\begingroup$ @John Wiltshire-Gordon: Thanks you! $\endgroup$
    – sife
    Commented May 1, 2013 at 17:32

0

You must log in to answer this question.