# The Unapologetic Mathematician

## Properties of Irreducible Root Systems I

Now we can turn towards the project of classifying irreducible root systems up to isomorphism. And we start with some properties of irreducible root systems.

First off, remember a root system $\Phi$ is reducible if we can write it as the disjoint union of two collections of roots $\Phi=\Psi\uplus\Psi'$ so that each root in $\Psi$ is perpendicular to each one in $\Psi'$. I assert that, for any base $\Delta\subseteq\Phi$, $\Phi$ is reducible if and only if $\Delta$ can itself be broken into two collections $\Delta=\Gamma\uplus\Gamma'$ in just the same way. One direction is easy: if we can decompose $\Phi$, then the roots in $\Delta$ are either in $\Psi$ or $\Psi'$ and we can define $\Gamma=\Delta\cap\Psi$ and $\Gamma'=\Delta\cap\Psi'$.

On the other hand, if we write $\Delta=\Gamma\uplus\Gamma'$ with each simple root in $\Gamma$ perpendicular to each one in $\Gamma'$, then we will find a similar decomposition of $\Phi$. But we know from our study of the Weyl group that every root in $\Phi$ can be sent by the Weyl group to some simple root in $\Delta$. So we define $\Psi$ to be the collection of roots whose orbit includes a point of $\Gamma$, and $\Psi'$ to be the collection of roots whose orbit includes a point of $\Gamma'$. So a vector in $\Psi$ is of the form $\sigma(\alpha)$ for some Weyl group element $\sigma\in\mathcal{W}$ and some simple root $\alpha\in\Gamma$. The Weyl group element $\sigma$ can be written as a sequence of simple reflections. A simple reflection corresponding to a root in $\Gamma$ adds some multiple of that root to the vector, and thus leaves the subspace spanned by $\Gamma$ invariant; while a simple reflection corresponding to a root in $\Gamma'$ leaves the subspace spanned by $\Gamma$ fixed point-by-point. Thus any root in $\Psi$ must lie in the subspace spanned by $\Gamma$, and similarly any root in $\Psi'$ must lie in the subspace spanned by $\Gamma'$. This shows that $\Phi=\Psi\uplus\Psi'$ is exactly the sort of decomposition we’re looking for.

If $\Phi$ is irreducible, with base $\Delta$, then there is a unique maximal root $\beta\in\Phi$ relative to the partial ordering $\prec$ on roots. In particular, the height of $\beta$ is greater than the height of any other root in $\Phi$, $\langle\beta,\alpha\rangle\geq0$ for all simple roots $\alpha\in\Delta$, and in the unique expression

$\displaystyle\beta=\sum\limits_{\alpha\in\Delta}k_\alpha\alpha$

all of the coefficients $k_\alpha$ are strictly positive.

First of all, if $\beta$ is maximal then it’s clearly positive, and so each $k_\alpha$ is either positive or zero. Let $\Gamma$ be the collection of simple roots $\alpha$ so that $k_\alpha>0$, and $\Gamma'$ be the collection of simple roots $\alpha$ so that $k_\alpha=0$. Then $\Delta=\Gamma\uplus\Gamma'$ is a partition of the base. From here we’ll assume that $\Gamma'$ is nonempty and derive a contradiction.

If $\alpha'\in\Gamma'$, then $\langle\alpha,\alpha'\rangle\leq0$ for any other simple root $\alpha\in\Delta$. From this, we can calculate

$\displaystyle\langle\beta,\alpha'\rangle=\left\langle\sum\limits_{\substack{\alpha\in\Delta\\\alpha\neq\alpha'}}k_\alpha\alpha,\alpha'\right\rangle=\sum\limits_{\substack{\alpha\in\Delta\\\alpha\neq\alpha'}}k_\alpha\langle\alpha,\alpha'\rangle\leq0$

Since $\Phi$ is irreducible, there must be at least one $\alpha\in\Gamma$ and $\alpha'\in\Gamma'$ so that $\langle\alpha,\alpha'\rangle<0$, and so we must have $\langle\beta,\alpha'\rangle<0$ for this $\alpha'$. This proves that $\beta+\alpha'$ must also be a root, which contradicts the maximality of $\beta$. And so we conclude that $\Gamma'$ is empty, and all the coefficients $k_\alpha>0$. In passing, we can use the same fact to show that $\langle\beta,\alpha\rangle\geq0$ for all $\alpha\in\Delta$, or else $\beta$ wouldn’t be maximal.

This same argument applies to any other maximal root $\beta'$, giving $\langle\alpha,\beta'\rangle$ for all $\alpha\in\Delta$ with the inequality strict for at least one $\alpha$. We calculate

$\displaystyle\langle\beta,\beta'\rangle=\sum\limits_{\alpha\in\Delta}k_\alpha\langle\alpha,\beta'\rangle>0$

which tells us that $\beta-\beta'$ is a root unless $\beta=\beta'$. But if $\beta-\beta'$ is a root, then either $\beta\prec\beta'$ or $\beta\succ\beta'$, contradicting the assumption that both are maximal. Thus $\beta$ must be unique.

February 10, 2010 - Posted by | Geometry, Root Systems

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