# The Unapologetic Mathematician

## Lengths of Weyl Group Elements

With our theorem from last time about the Weyl group action, and the lemmas from earlier about simple roots and reflections, we can define a few notions that make discussing Weyl groups easier. Any Weyl group element $\sigma\in\mathcal{W}$ can be written as a composition of simple reflections

$\displaystyle\sigma=\sigma_{\alpha_1}\dots\sigma_{\alpha_t}$

where all $\alpha_k\in\Delta$ are simple roots for some choice of a base $\Delta\subseteq\Phi$. In general we can do this in many ways, and some will have larger values for $t$ than others. But there must be some minimal number of simple reflections it takes to make $\sigma$ — some smallest possible value of $t$. This number we call the “length” $l(\sigma)$ of the root $\sigma$ relative to $\Delta$, and an expression that uses this minimal number of reflections is called “reduced”. By definition we set $l(1)=0$ for the identity element, since we can write it with no reflections at all.

Now, we also have another characterization of the length of a root. Let $n(\sigma)$ be the number of positive roots $\alpha\in\Phi^+$ for which $\sigma(\alpha)\prec0$ — the number of roots that $\sigma$ moves from $\Phi^+$ to $\Phi^-$. I say that $l(\sigma)=n(\sigma)$ for all $\sigma\in\mathcal{W}$, and I’ll proceed by induction on $l(\sigma)$. Indeed, the base case is obvious, since the only element of $\mathcal{W}$ with length zero is the identity, and it sends no positive roots to negative roots.

If this assertion is true for all $\tau\in\mathcal{W}$ with $l(\tau), then we write $\sigma$ in a reduced form as $\sigma_{\alpha_1}\dots\sigma_{\alpha_t}$ and set $\alpha=\alpha_t$. By one of our lemmas, we see that $\sigma(\alpha)\prec0$. By another of our lemmas we know that $\sigma_\alpha$ merely permutes the positive roots other than $\alpha$, and so $n(\sigma\sigma_\alpha)=n(\sigma)-1$. On the other hand, $l(\sigma\sigma_\alpha)=l(\sigma)-1, and our inductive hypothesis allows us to conclude that $l(\sigma\sigma_\alpha)=n(\sigma\sigma_\alpha)$, and thus that $l(\sigma)=n(\sigma)$.