# The Unapologetic Mathematician

## Irreducible Root Systems

Given a root system $\Phi$ in an inner product space $U$ we may be able to partition it into two collections $\Phi=\Psi\uplus\Psi'$ so that each root in $\Psi$ is orthogonal to every root in $\Psi'$ and vice versa. In this case, the subspace $V$ spanned by $\Psi$ and the subspace $V'$ spanned by $\Psi'$ are orthogonal to each other, and together they span the whole of $U$. Thus we can write $U=V\oplus V'$, and thus $\Phi$ is the coproduct $\Psi\amalg\Psi'$.

In this situation, we say that $\Phi$ is a “reducible” root system. On the other hand, if there is no way of writing $\Phi$ as the coproduct of two other root systems we say it’s “irreducible”. We will show that every root system is the coproduct of a finite number of irreducible root systems in a unique way (up to reordering the irreducibles).

Okay, first off we can show that there’s at least one decomposition into irreducibles, and we’ll proceed by induction on the dimension of the root system. To start with, any one-dimensional root system is of the form $\{\alpha,-\alpha\}$, and this is definitely irreducible. On the other hand, given an $n$-dimensional root system $\Phi$, either it’s irreducible or not. If it is, we’re done. But if not, then we can break it into $\Psi$ and $\Psi'$, each of which has dimension strictly less than $n$. By the inductive hypothesis, each of these can be written as the coproduct of a bunch of irreducibles, and so we just take the coproduct of these two collections.

On the other hand, how do we know that the decomposition is essentially unique? This essentially comes down to what the guys at the Secret Blogging Seminar call the diamond lemma: if we have two different ways of partitioning a root system, then there is some common way of partitioning each of those, thus “completing the diamond”. Thus, any choices we may have made in showing that a decomposition exists ultimately don’t matter.

So, let’s say we have a root system $\Phi\subseteq U$. Suppose that we have two different decompositions, $U=V_1\oplus\dots\oplus V_m$ and $U=V'_1\oplus\dots\oplus V'_n$. In each case, every root in $\Phi$ is actually in exactly one of the subspaces $V_i$ and exactly one of the subspaces $V'_j$, thus partitioning the root system on the one hand as $\Phi=\Psi_1\amalg\dots\amalg\Psi_m$ and $\Phi=\Psi'_1\amalg\dots\amalg\Psi'_n$.

All we have to do is define $W_{ij}=V_i\cap V'_j$. From here we can see that

\displaystyle\begin{aligned}V_i&=\bigoplus\limits_{j=1}^nW_{ij}\\V'_j&=\bigoplus\limits_{i=1}^mW_{ij}\\U&=\bigoplus\limits_{\substack{1\leq i\leq m\\1\leq j\leq n}}W_{ij}\end{aligned}

Further, each root in $\Psi$ is in exactly one of the $W_{ij}$, so we have a decomposition of the root system.

This result does a lot towards advancing our goal. Any root system can be written as the coproduct of a bunch of irreducible root systems. Now all we have to do is classify the irreducible root systems (up to isomorphism) and we’re done!

January 27, 2010 - Posted by | Geometry, Root Systems

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