Showing highest weight vectors are weight vectors

Question

Let $\frak{g}$ be a simple complex Lie algebra with a choice of maximal torus. We denote the associated Cartan--Weyl presentation by $E_\alpha, F_\alpha, H_\alpha$, for $\alpha$ a simple root. For $V$ a representation of $\frak{g}$, we define a highest weight vector $v \in V$ to be one such that $$ E_{\alpha}v = 0, \text{ for all simple roots } \alpha $$ What is the easiest way to see that such a vector is a weight vector, which is to say a common eigenvector of the elements $H_\alpha$?

Answer 1

First, I believe you mean to refer to $v$ as a maximal vector. Secondly, it is not true in general that, for an arbitrary representation, a maximal vector is a weight vector. Take $V=L(\lambda)\oplus L(\mu)$ to be the direct sum of two irreducible highest weight modules. Then, if $v_\lambda\in L(\lambda)$ and $v_\mu\in L(\mu)$ are highest weight vectors, $v=v_\lambda+v_\mu$ is a maximal vector that is not a weight vector.

A highest weight vector is defined to be a weight vector that is annihilated by all $E_\alpha$. For an irreducible finite dimensional module, highest weight vectors exist:

Assume $V$ is an irreducible $\mathfrak{g}$-module, then $V$ has a basis of weight vectors $\beta=\{v_\lambda\mid \lambda\in \mathfrak{h}^*\}$. Since $\beta$ is finite, we may choose $v_\lambda$ with $\lambda$ maximal. Since the weight of $E_\alpha v_\lambda$ is $\lambda+\alpha>\lambda$, we must have $E_\alpha v_\lambda=0$ for all $\alpha$. QED.

(Note that when passing to infinite dimensional modules, it is standard practice to work inside the category $\mathcal{O}$. Among the properties of modules in category $\mathcal{O}$ are the conditions that these modules have bases of weight vectors, and the weights of these modules are bounded above.)