# The Unapologetic Mathematician

## Properties of Irreducible Root Systems II

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

If $\Phi$ is irreducible, then the Weyl group $\mathcal{W}$ acts irreducibly on $V$. That is, we cannot decompose the representation of $\mathcal{W}$ on $V$ as the direct sum of two other representations. Even more explicitly, we cannot write $V=W\oplus W'$ for two nontrivial subspaces $W$ and $W'$ with each one of these subspaces invariant under $\mathcal{W}$. If $W$ is an invariant subspace, then the orthogonal complement $W'$ 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 $W$ is an invariant subspace under $\mathcal{W}$, then it is either trivial or the whole of $V$.

For any root $\alpha\in\Phi$, either $\alpha\in W$ or $W\subseteq P_\alpha$. Indeed, since $\sigma_\alpha\in\mathcal{W}$, we must have $\sigma_\alpha(W)=W$. As a reflection, $\sigma_\alpha$ breaks $V$ into a one-dimensional eigenspace with eigenvalue $-1$ and another complementary eigenspace with eigenvalue ${1}$. If $W$ contains the $-1$-eigenspace, then $\alpha\in W$. If not, then $\alpha$ is perpendicular to $W$ or $W$ couldn’t be invariant under $\sigma_\alpha$, and in this case $W\subseteq P_\alpha$.

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

Even better, the span of the $\mathcal{W}$-orbit of any root $\alpha\in\Phi$ spans $V$. Indeed, the subspace spanned by roots of the form $\sigma(\alpha)$ is invariant under the action of $\mathcal{W}$, and so since $V$ is irreducible it must be either trivial (clearly impossible) or the whole of $V$.

February 11, 2010 - Posted by | Geometry, Root Systems

1. Are we ready for Dynkin diagrams?

Comment by Jonathan Vos Post | February 11, 2010 | Reply

2. Not yet.

Comment by John Armstrong | February 11, 2010 | Reply

3. […] 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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4. […] 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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