There is the well-known fact that if $\lambda$ is the highest weight of $V$ a finite-dimensional irreducible $\mathfrak{g}$-module, then $w_0 \lambda$ is the lowest weight (here $\mathfrak{g}$ is any complex semisimple Lie algebra), where $w_0$ is the longest element of the Weyl group.
Why is this true?
Related questions have been asked on MSE, but not answering why, only giving the statement.
It has almost been answered in this MO post but not concretely for me.
Specifically, this is the proof I am reading:
The set of weights of $V$, denote by $\chi(V)$, are invariant under the Weyl group action.
Let $Q_+$ be the semigroup generated by $\Phi^+$.
Since $w_0 \cdot Q_+ = -Q_+$, we have $w_0 (\lambda) \prec \mu$ for all $\mu \in \chi(V)$, which implies the assertion.
Why is the bolded statement correct? I'm not even sure it is, just graphically looking at the picture of the weights of the representation of $\mathfrak{sl}_3$ with highest weight $\lambda = \varpi_1 + 2 \varpi_2$ below.
The weight marked with the open circle at $2\pi/3$ representing $s_2 s_1 (\lambda – \alpha_2)$ does not graphically appear to be a higher weight than $w_0(\lambda)$.
What am I misunderstanding here/how do I reason the above statement algebraically?
Best Answer
Perhaps a picture says more than words here.
If $x = s_2 s_1(\lambda - \alpha_2)$, then the green region contains all weights which are $\preceq x$. The green region is the region $x - \mathbb{R}_{\geq 0}\alpha_1 - \mathbb{R}_{\geq 0}\alpha_2$: every point that can be written as $x$ plus some strictly negative linear combination of simple roots, or equivalently $x$ plus some strictly negative linear combination of simple roots. (Beware: this order is usually defined by $y \preceq x$ if and only if $y \in x - \mathbb{N}\alpha_1 - \mathbb{N} \alpha_2$, so even though it looks like $0 \preceq \lambda$ from the region-description, actually $0$ and $\lambda$ are incomparable).
Algebraically, we have that $y \preceq x$ if and only if $y + q = x$ for some $x \in Q^+$. Applying $w_0$ to both sides gives that $w_0 y = -w_0 q + w_0 x$, and since $- w_0 q \in Q^+$ we have $w_0 x \preceq w_0 y$.