The Unapologetic Mathematician

Mathematics for the interested outsider

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 | 4 Comments

   

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