The Fundamental Weyl Chamber
When we first discussed Weyl chambers, we defined the fundamental Weyl chamber associated to a base
as the collection of all the vectors
satisfying
for all simple roots
. Today, I want to discuss the closure
of this set — allowing
— and show that it’s a fundamental domain for the action of the Weyl group
.
To be more explicit, saying that the fundamental Weyl chamber is a fundamental domain means that each vector
is in the orbit of exactly one vector in
. That is, there is a unique
so that
for some
.
First, to existence. Given a vector , we consider its orbit — the collection of all the
as
runs over all elements of
. We have to find a vector in this orbit which lies in the fundamental Weyl chamber
. To do this, we’ll temporarily extend our partial order to all of
by saying that
if
is a nonnegative
-linear combination of simple roots. Relative to this order, pick a maximal vector
; that is, one so that for any
we never have
. 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 is actually in the fundamental Weyl chamber. Indeed, if it weren’t then there would be some simple root
so that
. But then we could look at the vector
. We calculate
which is a positive -linear combination of simple roots. Thus
, which is impossible by assumption. In fact, this gives us a method for constructing a maximal vector in the orbit. Just start with
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 and
in the orbit
that lie within the fundamental Weyl chamber, then we must have
for some
. What I’ll show is that if we have
for two vectors in the fundamental Weyl chamber, then
must be the product of simple reflections which leave
fixed, and thus
.
We’ll prove this by induction on the length of the Weyl group element . If
, then
is the identity and the statement is obvious. If
then (by the result we proved last time)
must send some positive root to a negative root. In particular,
cannot send all simple roots to positive roots. So let’s say that
is a simple root for which
. Then we observe
since and
are both in the fundamental Weyl domain. Thus it is forced that
, that
, and then that
. But
sends fewer positive roots to negative ones than
does, so
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 looks like! It has one point for each vector
. If
is in the interior of this fundamental domain, then the orbit looks just like a copy of
. On the other hand, if
lies on one of the boundary hyperplanes the orbit looks like “half” of the Weyl group. That is, if
then
, so both of the corresponding group elements “collapse” into one point in this orbit. As
lies on more and more of the boundary hyperplanes, more and more of the orbit “folds up”, until finally at
we have an orbit consisting of exactly one point.
[…] 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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[…] 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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