# The Unapologetic Mathematician

## The Fundamental Weyl Chamber

When we first discussed Weyl chambers, we defined the fundamental Weyl chamber $\mathfrak{C}(\Delta)$ associated to a base $\Delta$ as the collection of all the vectors $\lambda\in V$ satisfying $\langle\lambda,\alpha\rangle>0$ for all simple roots $\alpha\in\Delta$. Today, I want to discuss the closure $\overline{\mathfrak{C}(\Delta)}$ of this set — allowing $\langle\lambda,\alpha\rangle=0$ — and show that it’s a fundamental domain for the action of the Weyl group $\mathcal{W}$.

To be more explicit, saying that the fundamental Weyl chamber $\overline{\mathfrak{C}(\Delta)}$ is a fundamental domain means that each vector $\mu\in V$ is in the orbit of exactly one vector in $\overline{\mathfrak{C}(\Delta)}$. That is, there is a unique $\lambda\in\overline{\mathfrak{C}(\Delta)}$ so that $\lambda=\sigma(\mu)$ for some $\sigma\in\mathcal{W}$.

First, to existence. Given a vector $\mu\in V$, we consider its orbit — the collection of all the $\sigma(\mu)$ as $\sigma$ runs over all elements of $\mathcal{W}$. We have to find a vector in this orbit which lies in the fundamental Weyl chamber $\overline{\mathfrak{C}(\Delta)}$. To do this, we’ll temporarily extend our partial order to all of $V$ by saying that $\mu\prec\nu$ if $\nu-\mu$ is a nonnegative $\mathbb{R}$-linear combination of simple roots. Relative to this order, pick a maximal vector $\lambda=\sigma(\mu)$; that is, one so that for any $\nu=\tau(\mu)$ we never have $\nu\succ\lambda$. There may well be more than one such maximal vector, given what we’ve said so far, but there will always be at least one.

I say that this $\lambda=\sigma(\mu)$ is actually in the fundamental Weyl chamber. Indeed, if it weren’t then there would be some simple root $\alpha\in\Delta$ so that $\langle\lambda,\alpha,\rangle<0$. But then we could look at the vector $\sigma_\alpha(\lambda)=[\sigma_\alpha\sigma](\mu)\in\mathcal{W}\mu$. We calculate

$\displaystyle\sigma_\alpha(\lambda)-\lambda=-2\frac{\langle\lambda,\alpha\rangle}{\langle\alpha,\alpha,\rangle}\alpha$

which is a positive $\mathbb{R}$-linear combination of simple roots. Thus $\sigma_\alpha(\lambda)\succ\lambda$, which is impossible by assumption. In fact, this gives us a method for constructing a maximal vector in the orbit. Just start with $\mu$ and form its inner product with all the simple roots. If we find one for which the inner product is negative, reflect the vector through the plane perpendicular to that simple root. Eventually, you’ll end up with a vector in the fundamental Weyl chamber!

Now for uniqueness: if there are two vectors $\lambda_1$ and $\lambda_2$ in the orbit $\mathcal{W}\mu$ that lie within the fundamental Weyl chamber, then we must have $\lambda_1=\sigma(\lambda_2)$ for some $\sigma\in\mathcal{W}$. What I’ll show is that if we have $\lambda_1=\sigma(\lambda_2)$ for two vectors in the fundamental Weyl chamber, then $\sigma$ must be the product of simple reflections which leave $\lambda_2$ fixed, and thus $\lambda_1=\lambda_2$.

We’ll prove this by induction on the length of the Weyl group element $\sigma$. If $l(\sigma)=0$, then $\sigma$ is the identity and the statement is obvious. If $l(\sigma)>0$ then (by the result we proved last time) $\sigma$ must send some positive root to a negative root. In particular, $\sigma$ cannot send all simple roots to positive roots. So let’s say that $\alpha\in\Delta$ is a simple root for which $\sigma(\alpha)\prec0$. Then we observe

$\displaystyle0\geq\langle\lambda_1,\sigma(\alpha)\rangle=\langle\sigma^{-1}(\lambda_1),\alpha\rangle=\langle\lambda_2,\alpha\rangle\geq0$

since $\lambda_1$ and $\lambda_2$ are both in the fundamental Weyl domain. Thus it is forced that $\langle\lambda_2,\alpha\rangle=0$, that $\sigma_\alpha(\lambda_2)=\lambda_2$, and then that $[\sigma\sigma_\alpha](\lambda_2)=\lambda_1$. But $\sigma\sigma_\alpha$ sends fewer positive roots to negative ones than $\sigma$ does, so $l(\sigma\sigma_\alpha) and we can invoke the inductive hypothesis to finish the job.

The upshot of all this is that we know what the space of orbits of $\mathcal{W}$ looks like! It has one point for each vector $\lambda\in\overline{\mathfrak{C}(\Delta)}$. If $\lambda$ is in the interior of this fundamental domain, then the orbit looks just like a copy of $\mathcal{W}$. On the other hand, if $\lambda$ lies on one of the boundary hyperplanes the orbit looks like “half” of the Weyl group. That is, if $\langle\lambda,\alpha\rangle=0$ then $\sigma(\lambda)=[\sigma\sigma_\alpha](\lambda)$, so both of the corresponding group elements “collapse” into one point in this orbit. As $\lambda$ lies on more and more of the boundary hyperplanes, more and more of the orbit “folds up”, until finally at $\lambda=0$ we have an orbit consisting of exactly one point.

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February 9, 2010 - Posted by | Geometry, Root Systems

## 2 Comments »

1. […] Indeed, it suffices to show that for all . We may, without loss of generality, assume is in the fundamental doman . Since , we must have for any other . In particular, we have and . Putting these together, we […]

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2. […] vector and drag it around, watching how the orbit changes. No matter where you place , notice that there is exactly one image in the fundamental domain, as we […]

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