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

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 $$f$$.

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.

• Why are you assuming that $V$ is finite-dimensional? Besides, I don't see how you justify that $\ker(\rho(e))=\operatorname{span}\{v_0\}$. Commented Jul 6, 2022 at 7:55

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 $$\ker(\rho(f))={\rm span}(v_{n-1})$$.