For a simple complex Lie algebra $\frak{g}$, let $V$ be an irreducible $\frak{g}$-module. Is it true that the weights of the non-zero weight vectors in $V$ are less than the highest weight vector and greater than the lowest weight vector with respect to the partial order on weights? If not, what is a simple counterexample?

## 1 Answer

One nice way to see this is using the PBW theorem. Write $\mathfrak{g} = \mathfrak{n}_- \oplus \mathfrak{h} \oplus \mathfrak{n}_+$ in the usual way. Take an ordered basis of $\mathfrak{g}$ consisting of, first, a basis for $\mathfrak{n}_-$, then a basis for $\mathfrak{h}$ and then a basis for $\mathfrak{n}_+$. Every PBW monomial thus factors as a product $x_- x_0 x_+$ where $x_- \in U(\mathfrak{n}_-)$, $x_0 \in U(\mathfrak{h})$ and $x_+ \in U(\mathfrak{n}_+)$.

Now, let $v_0$ be a highest weight vector in $V$. Then $U(\mathfrak{g}) v_0$ is a subrep of $V$ which, since $V$ is simple, must equal $V$. So $V$ is spanned by $x_- x_0 x_+ v$ for $x_-$, $x_0$ and $x_+$ as above.

Now, if $x_+$ is a positive degree monomial in $\mathfrak{n}_+$, then $x_+ v_0=0$ since $v_0$ is highest weight, so we can consider just the span of $x_- x_0 v$. And $v$ is an eigenvector for every $x_0 \in U(\mathfrak{h})$, so we can consider just the span of $x_- v_0$ for $x_- \in U(\mathfrak{n}_-)$. In short, we have proven that $V = U(\mathfrak{n}_-) v_0$. But it is clear that acting on $v_0$ by anything in $\mathfrak{n}_-$ lowers the weight.

Incidentally, we have only used the easy part of the PBW theorem here, which is that the PBW monomials span $U(\mathfrak{g})$; we didn't need the hard part, which is that they are linearly independent.