# Proving that there exists a saturated set with given highest weight

This is an question about an exercise in Humphreys book on Lie algebras. First of all a bunch of definitions and notation, see §13 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 $\alpha \in \Delta$.

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.

• Sorry about being tardy in typing this up. My copy of Humphreys (1972 edition) was in my office, I'm on vacation, and I needed to check how the key result of Lemma 13.2A related to this exercise in the order of Humphreys' presentation. – Jyrki Lahtonen Jul 12 '14 at 7:33
• Thanks a lot, I think this works! – spin Jul 16 '14 at 6:30
• @Jyrki: The end terms $\sigma(\mu)$ and $\sigma(\mu-m\alpha)$ are in the $W$-orbit of $\mu$; this is clear to me. But the next step, why both are less than $\lambda$ is not clear to me. – Beginner Feb 2 '17 at 11:41
• @Beginner: If we are discussing the same step, then we are working under the assumption that $\mu\in\Pi$. Therefore $\sigma\mu\prec\lambda$ for all $\sigma\in W$. – Jyrki Lahtonen Feb 2 '17 at 11:56
• @Jyrki: Thanks for quick clarification. So in short, if $x$ is a weight less than $\lambda$, then each member of $W$-orbit of $x$ is less than $\lambda$ which you have pointed in first two lines; got it. I noticed this later, and was just writing arguments step-by-step. – Beginner Feb 2 '17 at 12:00