Showing that any irreducible $\mathfrak{sl}(2)$-representation has a $1$-dimensional kernel for its nilpotent elements.

Question

I want to prove, that for any irreducible $\mathfrak{sl}(2)$-representation $(\rho, V)$ in the standart basis $\{e,f,h\}$ we have that $dim( ker(\rho(e))=1$, and same for $f$.

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My idea:

First we recall some definitions: Recall that $\mathfrak{sl}(n,\mathbb{C})$ has the standart basis $\{e,f,h\}$ with relations $$[h,e]=2e, [h,f]=-2f, [e,f]=h.$$ We are assuming that $V$ finite dimensional. Also, since $h$ is semisimple, it acts diagonally on $V$ and since $\mathbb{C}$ is algebraically closed, this yields a decomposition of $V$ as direct sum of eigenspaces $V_{\lambda} = \{v\in V \mid h.v = \lambda v\}$ with $\lambda \in \mathbb{C}$. Whenever we have that $V_{\lambda} \neq 0$, we call $\lambda$ a weight of $h$ in $V$ and $V_{\lambda}$ a weight space.

We now have, that $V$ is a irreducible $\mathfrak{sl}(2,\mathbb{C})$-module. Choose a maximal vector, say $v_0$ and set $v_{-1}=0, v_i = (1/i!)f^i.v_0$. According to Humphrey´s Lemma (page 32), we have the follwing formulas: $$h.v_i = (\lambda - 2i)v_i,$$ $$f.v_i =(i+1)v_{i+1},$$ $$e.v_i = (\lambda - i +1)v_{i-1}$$ for $i \geq 0$. In particular for $i=0$ it yields $$e.v_0 = (\lambda +1)v_{-1}=0$$ My idea was to use this, in order to show that $$ker(\rho(e)) = span\{v_i \mid i=0 \} = \mathbb{C}v_0.$$ Unfortunately it didn´t work out.

Answer 1

The $n$-dimensional irreducible representation $V=\langle v_0,\ldots ,v_{n-1}\rangle$ can be given by the formulas $\rho(h)v_i=(n-1-2i)v_i$, $\rho(e)v_i=(n-i)v_{i-1}$ and $\rho(f)v_i=(i+1)v_{i+1}$. Then the matrices of $\rho(e)$, $\rho(f),\rho(h)$ are given by $$ \rho(e)=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & n-1 \\ 0 & 0 & \cdots & 0 & 0 \end{pmatrix}, $$

$$ \rho(f)=\begin{pmatrix} 0 & 0 & \cdots & 0 & 0 \\ n-1 & 0 & \ddots & 0 & 0 \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & 2 & 0 & 0 \\ 0 & \cdots & 0 & 1 & 0 \end{pmatrix} $$ and $$ \rho(h)=\begin{pmatrix} n-1 & 0 & \cdots & 0 & 0 \\ 0 & n-3 & \cdots & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \ddots & 3-n & 0 \\ 0 & 0 & \cdots & 0 & 1-n \end{pmatrix} $$ Then $\ker(\rho(e))={\rm span}(v_0)$ is $1$-dimensional, as well as $\ker(\rho(f))={\rm span}(v_{n-1})$.