## Irreducible Root Systems

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

In this situation, we say that is a “reducible” root system. On the other hand, if there is no way of writing 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 , and this is definitely irreducible. On the other hand, given an -dimensional root system , either it’s irreducible or not. If it is, we’re done. But if not, then we can break it into and , each of which has dimension strictly less than . 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 . Suppose that we have two different decompositions, and . In each case, every root in is actually in exactly one of the subspaces and exactly one of the subspaces , thus partitioning the root system on the one hand as and .

All we have to do is define . From here we can see that

Further, each root in is in exactly one of the , 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!

[...] 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 [...]

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

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

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[...] This breaks the base up into a bunch of mutually-perpendicular subsets, which give rise to irreducible components of the root system [...]

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[...] Classification of (Possible) Root Systems At long last, we can state the classification of irreducible root systems up to isomorphism. We’ve shown that for each such root system we can construct a [...]

Pingback by The Classification of (Possible) Root Systems « The Unapologetic Mathematician | February 19, 2010 |

[...] of root systems. In particular, this leads us to the idea of decomposing a root system into irreducible root systems. If we can classify these pieces, any other root system will be built from [...]

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