The Unapologetic Mathematician

Mathematics for the interested outsider

Properties of Irreducible Root Systems III

Today we conclude with our series of lemmas on irreducible root systems.

If \Phi is irreducible, then roots in \Phi have at most two different lengths. Here I mean actual geometric lengths, as measured by the inner product, not the “length” of a Weyl group element. Further, any two roots of the same length can be sent to each other by the action of the Weyl group.

Let \alpha and \beta be two roots. We just saw that the \mathcal{W}-orbit of \alpha spans V, and so not all the \sigma(\alpha) can be perpendicular to \beta. From what we discovered about pairs of roots, we know that if \langle\alpha,\beta\rangle\neq0, then the possible ratios of squared lengths \frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2} are limited. Indeed, this ratio must be one of \frac{1}{3}, \frac{1}{2}, {1}, {2}, or {3}.

If there are three distinct root-lengths, let \alpha, \beta, and \gamma be samples of each length in increasing order. We must then have \frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}=2 and \frac{\lVert\gamma\rVert^2}{\lVert\alpha\rVert^2}=3, and so \frac{\lVert\gamma\rVert^2}{\lVert\beta\rVert^2}=\frac{3}{2}, which clearly violates our conditions. Thus there can be at most two root lengths, as asserted. We call those of the smaller length “short roots”, and the others “long roots”. If there is only one length, we call all the roots long, by convention.

Now let \alpha and \beta have the same length. By using the Weyl group as above, we may assume that these roots are non-orthogonal. We may also assume that they’re distinct, or else we’re already done! By the same data as before, we conclude that \alpha\rtimes\beta=\beta\rtimes\alpha=\pm1. We can replace one root by its negative, if need be, and assume that \alpha\rtimes\beta=1. Then we may calculate:

\displaystyle[\sigma_\alpha\sigma_\beta\sigma_\alpha](\beta)=[\sigma_\alpha\sigma_\beta](\beta-\alpha)=\sigma_\alpha(-\beta-\alpha+\beta)=\alpha.

We may note, in passing, that the unique maximal root \beta is long. Indeed, it suffices to show that \langle\beta,\beta\rangle\geq\langle\alpha,\alpha\rangle for all \alpha\in\Phi. We may, without loss of generality, assume \alpha is in the fundamental doman \overline{\mathfrak{C}(\Delta)}. Since \beta-\alpha\succ0, we must have \langle\gamma,\beta-\alpha\rangle\geq0 for any other \gamma\in\overline{\mathfrak{C}(\Delta)}. In particular, we have \langle\beta,\beta-\alpha\rangle\geq0 and \langle\alpha,\beta-\alpha\rangle\geq0. Putting these together, we conclude

\displaystyle\langle\beta,\beta\rangle\geq\langle\alpha,\beta\rangle\geq\langle\alpha,\alpha\rangle

and so \beta must be a long root.

February 12, 2010 Posted by | Geometry, Root Systems | 4 Comments

   

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