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 Cartan--Weyl 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.
$\begingroup$
$\endgroup$
4
-
$\begingroup$ 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
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Let α be a simple root that is not αk. Associated to this simple root is a subalgebra isomorphic to $\mathfrak{sl}_2$. For this subalgebra, an element of weight ωk 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 Fα.
answered Aug 30, 2021 at 6:37
-
$\begingroup$ 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