# Total number of extremal weights in irreps corresponding to fundamental weights equals total number of positive roots?

Let $\mathfrak g$ be a semisimple Lie algebra and $\mathfrak g = \mathfrak n \oplus \mathfrak h \oplus \mathfrak n^-$ be the triangular decomposition. A weight of a finite dimensional irreducible representation of $\mathfrak g$ is called extremal if it is in the Weyl group orbit of the highest weight.

For a fundamental weight $\omega_i$, write $n_i$ for the number of extremal weights in the finite dimensional irreducible $\mathfrak g$-representation $M(\omega_i)$ corresponding to $\omega_i$. Am I correct that $\sum_i n_i$ equals the number of positive roots in the root system for $(\mathfrak g, \mathfrak h)$? If the answer is yes, how does one prove it, and is there a natural bijection between $\bigcup_i \{\text{extremal weights for } M(\omega_i)\}$ and the set of positive roots?

Thank you very much in advance!

• Unless I messed up my calculations, this fails for both type $A_1$ and $A_2$, where we get more elements in the orbits of the fundamental weights than positive roots (note that this is really just a root system question, as the representation itself does not actually play any role). I get $2$ extremal weights for $A_1$, namely $\omega_1$ and $w_0(\omega_1) = -\omega_1$. For $A_2$ I just need $w_0$ again to get too many, since this gives me $2$ for each $\omega_i$ and there are only $3$ positive roots. – Tobias Kildetoft Aug 3 '18 at 7:30
• You are completely right. Thanks! And sorry for such a stupid question. – user312073 Aug 3 '18 at 18:36