Let $\frak{g}$ be a complex simple Lie algebra of rank $l$. For $\frak{h}$ a choice of Cartan subalgebra, let $\alpha_1, \cdots, \alpha_r$ be the corresponding choice of simple roots, $X_{\alpha_i}, H_{\alpha_i}, X_{\alpha_i}$ the CartanWeyl basis, and $\pi_1, \cdots, \pi_l$ the fundamental weights. For the irreducible $\frak{g}$module $V(\pi_k)$ let $v \in V(\pi_k)$ be a highest weight vector, i.e. $$ X_{\alpha_i}v = 0 ~~ \forall i=1, \dots, r. $$ How will the elements $X_{\alpha_i}$ act on $v$? Initial experiments suggest that $$ X_{\alpha_i}v = 0, ~~ \forall i \neq k. $$ However I cannot seem to see a general proof. Any help is very appreciated.

$\begingroup$ Should $F_{\alpha_i}$ be $X_{\alpha_i}$? $\endgroup$– LSpiceAug 30, 2021 at 1:23

1$\begingroup$ Are you sure this is true in Type $B_2$? $\endgroup$– Sam HopkinsAug 30, 2021 at 1:35

2$\begingroup$ As pointed out by @LSpice you seem to be conflating two of the (three) standard notations for the generators: $X_{\alpha}, H_{\alpha}, X_{\alpha}$ and $E_{\alpha},H_{\alpha}, F_{\alpha}$. $\endgroup$– F ZaldivarAug 30, 2021 at 2:56

$\begingroup$ Sorry for the typo, it has been fixed. $\endgroup$– johhnyelgertonAug 30, 2021 at 17:14
1 Answer
Let α be a simple root that is not α_{k}. Associated to this simple root is a subalgebra isomorphic to $\mathfrak{sl}_2$. For this subalgebra, an element of weight ω_{k} has weight zero. In the fundamental representation you are considering, such an element is highest weight by assumption and therefore by the classificaition of representations of $\mathfrak{sl}_2$ is annihilated by F_{α}.

$\begingroup$ Nice answer, thanks a lot! So it seems that there is nothing special about the $V(\pi_k)$case, that is to say, for any highest weight vector in $V(\lambda)$, where $\lambda = \sum_i a_i\pi_i$, we have $X_{\alpha_i}v = 0$ whenever $a_i = 0$? $\endgroup$ Aug 30, 2021 at 15:06