The Unapologetic Mathematician

Mathematics for the interested outsider

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)<l(\sigma), 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<l(\sigma), and our inductive hypothesis allows us to conclude that l(\sigma\sigma_\alpha)=n(\sigma\sigma_\alpha), and thus that l(\sigma)=n(\sigma).

February 8, 2010 Posted by | Geometry, Root Systems | 2 Comments

   

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