# The Unapologetic Mathematician

## 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 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 $4$.

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

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