# 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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5. […] But beyond that, we don’t know which Cartan matrices actually arise, and that’s the whole point of our project. For now, though, we will assume that our matrix does in fact arise from a real root system, and see how to use it to construct a root system whose Cartan matrix is the given one. And our method will hinge on considering root strings. […]

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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