# The Unapologetic Mathematician

## 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

1. Typo: gamma squared over alpha squared should be three not two.

Comment by Gilbert Bernstein | February 14, 2010 | Reply

2. thanks, fixed

Comment by John Armstrong | February 14, 2010 | Reply

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