irreducible module has a nonzero weight space if and only if $u\left( H\right) \cdot v$ is finite dimensional for all $v\in V$

Question

Question:

Throughout this question, $L$ will denote a semisimple Lie algebra(over the algebraically closed field $F$ of characteristic 0), $H$ a fixed CSA of $L$, $u(H)$ universal enveloping algebra of $H$.

Let $V$ be an irreducible $L-$module.Then $V$ has a (nonzero) weight space if and only if $u\left( H\right) \cdot v$ is finite-dimensional for all $v\in V$.

This question comes from J.E. Humphreys's book Introduction to Lie algebras and Representation Theory. Exercises p111 2.(b)

Attempt:

$\Rightarrow $ Since irreducible and $V$ has a (nonzero) weight space, so $V$ is the direct sum of its weight space $V=\sum V_{\lambda }$. For $v\in V$, $v=\sum^{s}_{i=1} v_{i}$,$v_{i}\neq 0 $, then we have $$h\cdot v=h\cdot \sum v_{i}=\sum \lambda_{i} \left( h\right) v_{i}\in <v_{1},...,v_{s}>$$ for all $h\in u\left( H\right)$,so $u\left( H\right) \cdot v$ is finite-dimensional.

Can we prove $u\left( H\right) \cdot v=<v_{1},...,v_{s}>$?

$\Leftarrow$ I don't have a good idea.

Answer 1

$\Leftarrow$ Denote the representation by $\pi$. Let $v\in V$, $v\ne 0$. Let $W=\pi(u(H))v$. Then by assumption $W$ is finite dimensional. Since $H\cdot u(H)\subseteq u(H)$ then $$\pi(H)(W)=\pi(H)\pi(u(H))v=\pi(H\cdot u(H))v\subseteq\pi(u(H))v=W$$

Since $H$ is Abelian, so is $\pi(H)\rvert_W\subseteq \mathfrak{gl}(W)$. An Abelian Lie algebra is solvable. By Lie's Theorem (Section 4.1 in the book), $W$ contains a common eigenvector $w\in W$, $w\ne 0$ of $\pi(H)\rvert_W$ and then $\operatorname{span}(\{w\})$ is a non-zero weight space.

Notes:

• $\pi:u(L)\to\operatorname{End}_F(V)$ is a homomorphism of associative algebras. $u(H)$ is an associative subalgebra of $u(L)$. $H$ is a Lie subalgebra of $u(H)$ whose bracket derived from the associative multiplication in $u(H)$ coincides with its intrinsic Lie bracket. $\pi$, as an associative homomorphism, is also a Lie homomorphism on the Lie bracket derived from the associative multiplication.
• The weight space $\operatorname{span}(\{w\})$ is non-zero, but the associated weight may be zero. This is not a problem as other weight spaces are derived as in section 20.1 of the book.