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?



Do weight vectors live between the highest and lowest weights?
One nice way to see this is using the [PBW theorem](https://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Birkhoff%E2%80%93Witt_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.


