Calculations in a $sl_\alpha$-module

Question

I want to understand the proof of the following theorem:

For $\lambda\in H^*$ we have: $\dim M_\lambda<\infty \Leftrightarrow \lambda \in \Lambda^+$

($H$ is a Cartan subalgebra of a Lie algebra $L$ and $M_\lambda:=V_\lambda/S_{max}$ where $V_\lambda$ is the Verma-module of $L$ and $S_{max}$ is the unique maximal proper submodule of $V_\lambda$)

At the beginning of the proof we consider $M_\lambda$ as a finite-dimensional $sl_\alpha$-module ($sl_\alpha$ is a Lie subalgebra of $L$, which is generated by $\{e_\alpha,f_\alpha, h_\alpha\}$ and these three elements correspondend to the generators $\{e,f,h\}$ of $sl(2,\mathbb{C})$) Now they say, if we have an irreducible, finite dimensional $sl_\alpha$-module $V$ and $e.v=0$ for some $v\in V$, then we get $h.v=mv$ with $m\in\mathbb{Z}_{+}$. Can you give me an explanation for that? Later in the proof they say: If $V$ is a finite-dimensional $sl_\alpha$-module and $x\in V\backslash\{0\}$ satisfies $h.x=mx$, then $f^m.x\neq0$ if $m\geq 0$ and $e^{-m}.x\neq 0$ if $m<0$. How can I prove that?

I know that any irreducible finite dimensional $sl(2,\mathbb{C})$-module $N$ is isomorphic to $M_\lambda$ with $\lambda=\dim(N)-1$. ($M_\lambda=V_\lambda/S_\lambda$ and $V_\lambda$ is the Verma-module of $sl(2,\mathbb{C})$ and $S_\lambda$ is the unique maximal proper submodule of $V_\lambda$ for $\lambda\in \mathbb{Z}_{+}$).

Answer 1

You posted two questions.

1) That's because of the classification of the irreducible $\mathfrak{sl}(2,\mathbb{C})$-modules. For each $m\in\mathbb{N}$ there is (up to isomorphism) one and only one such module and, for that module, the eigenspace with eigenvalue $0$ of the action of $e$ on that module is one-dimensional and if $v$ belongs to that space, $h.v=(m-1)v$. Since $m\in\mathbb N$, $m-1\in\mathbb{Z}_+$.

2) Again, this follows from the classification of the irreducible $\mathfrak{sl}(2,\mathbb{C})$-modules. I am assuming here that there is a typo in your question and that, when you wrote $e^m.x$ what you really meant was $e^{-m}.x$.

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More details for 1) Take $m\in\mathbb N$. There is one and, up to isomorphism, only one irreducible $m$-dimensional representation $V_m$ of $\mathfrak{sl}(2,\mathbb{C})$. For that representation, the eigenvalues of the action of $h$ are $m-1,m-3,m-5,\dots,-(m-1)$. Furthermore, if $v$ is an eigenvector of the action of $h$ with eigenvalue $m-1$, then $e.v=0$. On the other hand, the kernel of the action of $e$ is precisely $\mathbb{C}v$. Indeed, there is a basis $(v_{m-1},v_{m-3},\ldots,v_{-(m-1)})$ of $V_m$ such that, if $k\in\{m-1,m-3,\ldots,-(m-1)\}$, then

1. $h.v_k=kv_k$;
2. $e.v_k=\begin{cases}v_{k+2}&\text{ if }k<m-1\\0&\text{ otherwise.}\end{cases}$

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More details for 2) It turns out that the action of $f$ on the previous base is given by$$f.v_k=\begin{cases}0&\text{ if }k=-(m-1)\\\eta_kv_{k-2}&\text{ otherwise,}\end{cases}$$where the $\eta_k$'s are non-zero integers. Suppose now that $h.v=kv$ for some integer $k$; suppose furthermore that $k\geqslant0$. Could we have $f^k.v=0$? No, because the kernel of the action of $f$ on this module is $\mathbb{C}v_{-(m-1)}$, $h.v_{-(m-1)}=-(m-1)v_{-(m-1)}$, and $-(m-1)<0$. For a similar reason, if $k\leqslant0$, then $e^{-k}.v\neq0$.