Product of root vectors transform weight vector into another weight vector.

Question

Let $\mathfrak h$ a Cartan subalgebra of the complex semisimple Lie algebra $\mathfrak g$ and let $V$ be a representation of $\mathfrak g$. Let $v \in V$ be a weight vector of weight $\lambda$, that is, $h v = \lambda(h) v$ for all $h \in \mathfrak h$ and let $n_\alpha \in \mathfrak g_\alpha$ be a root vector for the root $\alpha \in \mathfrak h^\ast$, that is, $[h,n_\alpha] = \alpha(h) n_\alpha$ for all $h \in \mathfrak h$. It is easy to calculate that $n_\alpha v$ (action of $n_\alpha \in U(\mathfrak g)$ on $v$) then is a weight vector of weight $\lambda + \alpha$. Now if $n_\alpha \in \mathfrak g_\alpha, n_\beta \in \mathfrak g_\beta$ are considered as elements in the universal enveloping algebra $U(\mathfrak g)$, then I want to deduce that $n_\beta n_\alpha v$ is a weight vector of weight $\lambda + \alpha + \beta$. How is this calculation done? (note: $n_\beta n_\alpha$ denotes multiplication in $U(\mathfrak g)$) Do I have to invoke the PBW-theorem and how?

Answer 1

You have a result that says for all weights $\mu$ and elements $u \in V_\mu$ (the $\mu$-weight space of $V$) and all roots $\gamma$ and all $n_\gamma \in \mathfrak{g}_\gamma$ we have $n_\gamma u \in V_{\mu+\gamma}$.

Apply this once with $u=v, \mu = \lambda, \gamma = \alpha$: you get $n_\alpha v \in V_{\lambda + \alpha}$. Apply it again with $u=n_\alpha v, \mu = \lambda + \alpha, \gamma = \beta$ and you get $n_\beta (n_\alpha v) \in V_{\lambda + \alpha + \beta}$. Isn't this what you want?