## Properties of Irreducible Root Systems II

We continue with our series of lemmas on irreducible root systems.

If is irreducible, then the Weyl group acts irreducibly on . That is, we cannot decompose the representation of on as the direct sum of two other representations. Even more explicitly, we cannot write for two nontrivial subspaces and with each one of these subspaces invariant under . If is an invariant subspace, then the orthogonal complement will also be invariant. This is a basic fact about the representation theory of finite groups, which I will simply quote for now, since I haven’t covered that in detail. Thus my assertion is that if is an invariant subspace under , then it is either trivial or the whole of .

For any root , either or . Indeed, since , we must have . As a reflection, breaks into a one-dimensional eigenspace with eigenvalue and another complementary eigenspace with eigenvalue . If contains the -eigenspace, then . If not, then is perpendicular to or couldn’t be invariant under , and in this case .

So then if isn’t in then it must be in the orthogonal complement . Thus every root is either in or in , and this gives us an orthogonal decomposition of the root system. But since is irreducible, one or the other of these collections must be empty, and thus must be either trivial or the whole of .

Even better, the span of the -orbit of any root spans . Indeed, the subspace spanned by roots of the form is invariant under the action of , and so since is irreducible it must be either trivial (clearly impossible) or the whole of .

Are we ready for Dynkin diagrams?

Comment by Jonathan Vos Post | February 11, 2010 |

Not yet.

Comment by John Armstrong | February 11, 2010 |

[…] and be two roots. We just saw that the -orbit of spans , and so not all the can be perpendicular to . From what we discovered […]

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[…] also see that the Weyl orbit of a root spans the plane in the irreducible cases. But, again, in the Weyl orbits of and only span their […]

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