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.

4 Comments »

[…] of Irreducible Root Systems II We continue with our series of lemmas on irreducible root […]

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[…] of Irreducible Root Systems III Today we conlude with our series of lemmas on irreducible root […]

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[…] root system is reducible, but the other three are irreducible. For each of these, we can see that there is a unique maximal root. However, doesn’t; both and are […]

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[…] where after adding each term we have one of the positive roots. In fact, this path hits all but one of the six positive roots on its way to the unique maximal root. […]

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