The Unapologetic Mathematician

Mathematics for the interested outsider

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)<l(\sigma) 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.

About these ads

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 [...]

    Pingback by Properties of Irreducible Root Systems III « The Unapologetic Mathematician | February 12, 2010 | Reply

  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 [...]

    Pingback by Some Root Systems and Weyl Orbits « The Unapologetic Mathematician | February 15, 2010 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 388 other followers

%d bloggers like this: