The Unapologetic Mathematician

Mathematics for the interested outsider

Root Strings

First, a lemma: given a root system \Phi, let \alpha and \beta be nonproportional roots. That is, \beta\neq\pm\alpha. Then if \langle\alpha,\beta\rangle>0 — if the angle between the vectors is strictly acute — then \alpha-\beta is also a root. On the other hand, if \langle\alpha,\beta\rangle<0 then \alpha+\beta is also a root. We know that -\beta is a root. If we replace \beta with -\beta we can see that the second of these assertions follows from the first, so we just need to prove that one.

We know that \langle\alpha,\beta\rangle is positive if and only if \alpha\rtimes\beta is. So let’s look at the table we worked up last time. We see that when \alpha\rtimes\beta is positive, then either that or \beta\rtimes\alpha equals {1}. If it’s \alpha\rtimes\beta, then

\sigma_\beta(\alpha)=\alpha-(\alpha\rtimes\beta)\beta=\alpha-\beta

is a root. On the other hand, if \beta\rtimes\alpha=1 then \alpha-\beta=-\sigma_\alpha(\beta) is a root.

So given two nonproportional and nonorthogonal roots \alpha and \beta we’re guaranteed to have more than one vector of the form \beta+k\alpha for some integer k in the root system \Phi. We call the collection of all such vectors in \Phi the \alpha-string through \beta.

Let r be the largest integer so that \beta-r\alpha\in\Phi, and let q be the largest integer so that \beta+q\alpha\in\Phi. I say that the root string is unbroken. That is, \beta+k\alpha\in\Phi for all -r\leq k\leq q.

Indeed, if there’s some integer k so that \beta+k\alpha\notin\Phi, then we can find p and s with -r\leq p<s\leq q so that

\displaystyle\begin{aligned}\beta+p\alpha&\in\Phi\\\beta+(p+1)\alpha&\notin\Phi\\\beta+(s-1)\alpha&\notin\Phi\\\beta+s\alpha&\in\Phi\end{aligned}

But then the above lemma tells us that \langle\alpha,\beta+p\alpha\rangle\geq0, while \langle\alpha,\beta+s\alpha\rangle\leq0. Subtracting, we find

\displaystyle\begin{aligned}\langle\alpha,\beta+p\alpha\rangle-\langle\alpha,\beta+s\alpha\rangle&=\langle\alpha,(\beta+p\alpha)-(\beta+s\alpha)\rangle\\&=\langle\alpha,(p-s)\alpha\\&=(p-s)\langle\alpha,\alpha\rangle\end{aligned}

The two inequalities tell us that this difference should be positive, but \langle\alpha,\alpha\rangle is positive and p-s is negative. Thus we have a contradiction, and the root string must be unbroken from \beta-r\alpha to \beta+q\alpha.

We can also tell that \sigma_\alpha just adds a positive or negative multiple of \alpha to any root. Then it’s clear from the geometry that \sigma_\alpha just reverses a root string end to end. That is, \sigma_\alpha(\beta+q\alpha)=\beta-r\alpha. But we can also calculate

\displaystyle\begin{aligned}\sigma_\alpha(\beta+q\alpha)&=\sigma_\alpha(\beta)+q\sigma_\alpha(\alpha)\\&=\beta-(\beta\rtimes\alpha)\alpha-q\alpha\\&=\beta-((\beta\rtimes\alpha)+q)\alpha\\&=\beta-r\alpha\end{aligned}

Thus r-q=\beta\rtimes\alpha. And so the length of the \alpha string through \beta can be no more than 4.

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January 29, 2010 - Posted by | Geometry, Root Systems

6 Comments »

  1. Cool! Some of this was new to me. Thank you!

    Comment by Jonathan Vos Post | January 30, 2010 | Reply

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  6. this is great thanks – helping me finally understand some of this root system stuff, just in time for my exams!

    Comment by twentyfifteen | May 21, 2012 | Reply


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