This is an question about an exercise in Humphreys book on Lie algebras. First of all a bunch of definitions and notation, see ยง13 in Humphreys for details.
Let $\Phi$ be a root system, $\Delta$ a base for $\Phi$. The set of weights of $\Phi$ is the set $\Lambda = \{\lambda : \langle \lambda, \alpha \rangle \in \mathbb{Z} \text{ for all $\alpha \in \Phi$} \}$ where $\langle x, y \rangle$ is the notation for Cartan integers. The set of dominant weights (wrt the base $\Delta$) is the set
$$\Lambda^+ = \{\lambda \in \Lambda: \langle \lambda, \alpha \rangle \geq 0 \text{ for all $\alpha \in \Delta$}\}$$
For weights we have the partial order $\prec$ wrt $\Delta$ ($\mu \prec \lambda$ if and only if $\lambda – \mu$ is a finite sum of positive roots).
Now we call a subset $\Pi$ of $\Lambda$ saturated if for all $\lambda \in \Pi$, $\alpha \in \Phi$ and integer $i$ between $0$ and $\langle \lambda, \alpha \rangle$, we have $\lambda – i\alpha \in \Pi$. We say that $\Pi$ has highest weight $\lambda$ if $\lambda \in \Lambda^+$ and $\mu \prec \lambda$ for all $\mu \in \Pi$.
It is immediate that $\Pi$ is closed under the action of the Weyl group. What I want to prove is that for any $\lambda \in \Lambda^+$, there is a unique saturated set $\Pi$ with highest weight $\lambda$.
Some ideas: Everything follows once we can prove that the following set is saturated:
$$\Pi = \{\sigma \mu: \mu \in \Lambda^+, \mu \prec \lambda, \sigma \in W\}$$
where $W$ is the Weyl group.
It is enough to prove that for all $\mu \in \Lambda^+$, $\mu \prec \lambda$ the following holds:
For all $\alpha \in \Phi$ and $i$ between $0$ and $\langle \mu, \alpha \rangle$, the element $\mu – i\alpha$ is conjugate under $W$ to some $\mu' \in \Lambda^+$, $\mu' \prec \lambda$.
I can see how this holds in the case where $\alpha \in \Delta$. Now $\langle \lambda, \alpha \rangle \geq 0$. Let $0 \leq i \leq \langle \lambda, \alpha \rangle$ and $\mu = \lambda – i \alpha$. Consider two cases:
If $0 \leq i \leq \langle \lambda, \alpha \rangle / 2$: then $\mu$ is dominant, because $\langle \mu, \alpha \rangle \geq 0$ by the condition on $i$ and because $\langle \alpha, \beta \rangle \leq 0$ for $\beta \in \Delta$, $\beta \neq \alpha$.
If $\langle \lambda, \alpha \rangle / 2 \leq i \leq \langle \lambda, \alpha \rangle$: Now using the reflection $\sigma_\alpha$ with respect to $\alpha$, we have $\sigma_\alpha(\mu) = \lambda – j \alpha$, where $0 \leq j \leq \langle \lambda, \alpha \rangle / 2$. So we are done by the first case.
How to prove this in the general case? I think it should be enough to do this for positive roots, but I have no idea how to generalize from the case $\alpha \in \Delta$.
Best Answer
From the definition of the set $\Pi$ it is immediate that $\sigma(\Pi)=\Pi$ for all $\sigma\in W$. By Lemma 13.2A we have $\sigma\mu\prec\mu$ for all dominant weights $\mu$ and all $\sigma\in W$. This allows us to redescribe the set $\Pi$ as $$ \Pi=\{\mu\in\Lambda\mid \sigma\mu\prec\lambda\ \text{for all $\sigma\in W$}\}.\qquad(*) $$ Let then $\mu\in\Pi$ and $\alpha\in\Phi$ be arbitrary. Consider the $\alpha$-chain of weights connecting $\mu$ and $\mu-m\alpha$, $m=\langle\mu,\alpha\rangle$. Let us fix an integer $i$ between $0$ and $m$, and study the weight $\mu'=\mu-i\alpha$.
The claim is that $\mu'\in \Pi$. We are going to apply description $(*)$, so let $\sigma\in W$ be arbitrary. The upshot is that $\sigma(\mu')$ is on the $\sigma(\alpha)$-chain from $\sigma\mu$ to $\sigma(\mu-m\alpha)=\sigma(\mu)-m\sigma(\alpha)$. So according to the sign of $m$ and the positivity of $\sigma(\alpha)$ we have either $$ \sigma\mu\prec\sigma\mu'\prec\sigma\mu-m\sigma(\alpha) $$ or $$ \sigma\mu-m\sigma(\alpha)\prec\sigma\mu'\prec\sigma\mu. $$
But here $\sigma\mu$ and $\sigma(\mu-m\alpha)=\sigma s_\alpha\mu$ are both in the $W$-orbit of $\mu$, so according to $(*)$ they are both $\prec\lambda$. So irrespective of which alternative holds we get that $\sigma(\mu')\prec\lambda$.
So $\sigma(\mu')\prec\lambda$ for all $\sigma\in W$, and thus by $(*)$ we have that $\mu'\in\Pi$. Q.E.D.