The Weyl Group of a Root System
Let’s take a root system 
in the inner product space 
. Each vector 
in gives rise to a reflection in 
, the group of transformations preserving the inner product on . So what sorts of transformations can we build up from these reflections? The subgroup of 
generated by the reflections 
for all 
is called the Weyl group 
of the root system. It’s pronounced “vile”, but we don’t mean that as any sort of value judgement.
Anyway, we can also realize as a subgroup of the group of permutations on the vectors in . Indeed, by definition each sends each vector in back to another vector in , and so shuffles them around. So if has 
vectors, the Weyl group can be realized as a subgroup of 
.
In particular, is a finite group, as a subgroup of another finite group. In fact, we even know that the number of transformations in divides 
. It may well (and usually does) have elements which are not of the form , but there are still only a finite number of them.
The first thing we want to take note of is how certain transformations in 
act on by conjugation. Specifically, if 
leaves invariant, then it induces an automorphism on that sends the generator to 
— which (it turns out) is the generator 
— for all . Further, it turns out that 
for all 
.
Indeed, we can calculate
)=\tau(\sigma_\alpha(\beta))=\tau(\beta-(\beta\rtimes\alpha)\alpha)=\tau(\beta)-(\beta\rtimes\alpha)\tau(\alpha)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%5B%5Ctau%5Csigma_%5Calpha%5Ctau%5E%7B-1%7D%5Cright%5D%28%5Ctau%28%5Cbeta%29%29%3D%5Ctau%28%5Csigma_%5Calpha%28%5Cbeta%29%29%3D%5Ctau%28%5Cbeta-%28%5Cbeta%5Crtimes%5Calpha%29%5Calpha%29%3D%5Ctau%28%5Cbeta%29-%28%5Cbeta%5Crtimes%5Calpha%29%5Ctau%28%5Calpha%29&bg=e6e6e6&fg=333333&s=0 "\displaystyle\left[\tau\sigma_\alpha\tau^{-1}\right](\tau(\beta))=\tau(\sigma_\alpha(\beta))=\tau(\beta-(\beta\rtimes\alpha)\alpha)=\tau(\beta)-(\beta\rtimes\alpha)\tau(\alpha)")
Now, every vector in is of the form 
for some 
, and so sends it to the vector 
, which is again in , so it leaves invariant. The transformation also fixes every vector in the hyperplane 
, for if is orthogonal to , then the above formula shows that is left unchanged by the transformation. Finally, sends 
to 
.
This is all the data we need to invoke our lemma, and conclude that is actually equal to . Specifying the action on the generators of is enough to determine the whole automorphism. Of course, we can also just let 
act on each element of by conjugation, but it’s useful to know that the generating reflections are sent to each other exactly as their corresponding vectors are.
Now we can calculate from the definition of a reflection
)=\sigma_{\tau(\alpha)}(\tau(\beta))=\tau(\beta)-(\tau(\beta)\rtimes\tau(\alpha))\tau(\alpha)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%5B%5Ctau%5Csigma_%5Calpha%5Ctau%5E%7B-1%7D%5Cright%5D%28%5Ctau%28%5Cbeta%29%29%3D%5Csigma_%7B%5Ctau%28%5Calpha%29%7D%28%5Ctau%28%5Cbeta%29%29%3D%5Ctau%28%5Cbeta%29-%28%5Ctau%28%5Cbeta%29%5Crtimes%5Ctau%28%5Calpha%29%29%5Ctau%28%5Calpha%29&bg=e6e6e6&fg=333333&s=0 "\displaystyle\left[\tau\sigma_\alpha\tau^{-1}\right](\tau(\beta))=\sigma_{\tau(\alpha)}(\tau(\beta))=\tau(\beta)-(\tau(\beta)\rtimes\tau(\alpha))\tau(\alpha)")
Comparing this with the equation above, we find that 
, as asserted.
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January 21, 2010 -
Posted by John Armstrong |
Algebra, Geometry, Group Actions, Group theory, Root Systems
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You write that:
It’s clear that this defines a homomorphism from the Weyl group into
. But it doesn’t seem trivially obvious that this is injective — a priori, couldn’t there be two funny products of 
‘s which aren’t equal but agree on elements of 
?
That’s a good question, Peter. But there can’t be, because
spans the whole vector space. So if any two linear transformations agree on 
, they must agree on the whole vector space, and thus must be the same transformation.
On the other hand, you might be thinking that some linear transformation in
can be written as a product of the 
in two different ways. This is possible, but it leads to a relation in the Weyl group. Notice that I never said that 
was freely-generated (beyond the obvious fact that each generator is an involution).
Does that make sense?
Ah! I’d forgotten that
spans the space. OK, it’s obvious 
(It was indeed your first interpretation that I meant, not your second.)
In fact, I guess even if we don’t assume that
spans the space, they agree on the span of 
as you say, and they also act trivially on the orthogonal complement of 
(since each 
does), so still agree on the whole space.
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“pronounced “vile”, but we don’t mean that as any sort of value judgement”, lovely really enjoyed reading this blog.