Show that given an irreducible module of heighest weight $0$ over a lie algebra has dimension $1$

Question

Given $L$ a semisimple Lie algebra over an algebraically closed field of characteristic $0$ and given $V(0)$ the (standard cyclic) irreducible $L$-module of heighest weight $0$, then I want to show that the set of weights $\Pi(0)=\{ 0 \}$

(Here the notation is the one in Humpreys).

Given this I can automatically say that $dim(V(0))=1$ because in general $m_\lambda(\lambda)=1$. Where $m_\lambda(\lambda)$ stands for the dimensione of the $\lambda$-weightspace of $V(\lambda)$.

For now, i know only that $\Pi(0)$ is saturated, but then?

Answer 1

There is a unique irreducible highest weight representation with any given highest weight. The trivial representation is an irreducible representation of highest weight 0.