# Do weight vectors live between the highest and lowest weights?

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?

• Yes, this is true. In fact the weights live in the convex hull of the $W$ orbit of the highest weight. Oct 27, 2021 at 23:13

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.