## The Action of the Weyl Group on Weyl Chambers

With our latest lemmas in hand, we’re ready to describe the action of the Weyl group of a root system on the set of its Weyl chambers. Specifically, the action is “simply transitive”, and the group itself is generated by the reflections corresponding to the simple roots in any given base .

To be a bit more explicit, let be any fixed base of . Then a number of things happen:

- If is any regular vector, then there is some so that for all . That is, sends the Weyl chamber to the fundamental Weyl chamber .
- If is another base, then there is some so that . That is, sends to . We say that the action of the Weyl group is “transitive” on bases and their corresponding Weyl chambers.
- If is any root, then there is some so that .
- The Weyl group is generated by the for .
- If for some , then is the identity transformation. That is, the only transformation in the Weyl group that sends a base back to itself is the trivial one. We say that the action of the Weyl group is “simple” on bases and their corresponding Weyl chambers.

What we’ll do is let be the group generated by the for , as in the fourth assertion. We’ll show that this group satisfies the first three assertions, and then show that .

Let be a regular vector and write for the half-sum of the positive roots

Choose some so that is as large as possible. If is simple, then is in too, so we find

which forces for all . None of these inner products can actually equal zero, because if one was then we would have and wouldn’t be regular. Therefore lies in the fundamental Weyl chamber, as desired.

For the second assertion, we know that there must be some regular in the positive half-space for each root , and the first assertion then applies to send to .

For the third assertion, we can invoke the second assertion as long as we know that every root lies in *some* base . We can find some that’s in no other hyperplane perpendicular to another root (other than ). Then pick some close enough so that , but also for all . The root must then belong to the base .

Okay, now let’s show that . We just need to show that each reflection for (all of which together generate ) is an element of . But using our third assertion we can find some so that . Then

and so .

Finally, suppose that is some non-identity element of so that . Thanks to our fourth assertion we can write as a string of basic reflections, and we can assume that is as small as possible. Then we must have by our final lemma from last time, but we also must have , which gives us a contradiction.

Mathworld has a cute animation,

http://mathworld.wolfram.com/WeylGroup.html

and begins their text with:

Let L be a finite-dimensional split semisimple Lie algebra over a field of field characteristic 0, H a splitting Cartan subalgebra, and Lambda a weight of H in a representation of L. Then

Lambda’ = LambdaS_alpha = lambda-(2(Lambda,alpha))/((alpha,alpha))(alpha)

is also a weight. Furthermore, the reflections S_alpha with alpha a root, generate a group of linear transformations in H_0^* called the Weyl group W of L relative to H, where H^* is the conjugate space of H and H_0^* is the Q-space spanned by the roots (Jacobson 1979, pp. 112, 117, and 119).

The Weyl group acts on the roots of a semisimple Lie algebra, and it is a finite group. The animations above illustrate this action for Weyl Group acting on the roots of a homotopy from one Weyl matrix to the next one (i.e., it slides the arrows from g to h) in the first two figures, while the third figure shows the Weyl Group acting on the roots of the Cartan matrix of the infinite family of semisimple lie algebras A_3 (cf. Dynkin diagram), which is the special linear Lie algebra, sl_4.

SEE ALSO: Cartan Matrix, Dynkin Diagram, Lie Algebra, Lie Algebra Root, Lie Group, Macdonald’s Constant-Term Conjecture, Root System, Root Lattice, Semisimple Lie Algebra, Weight Lattice, Weyl Chamber

Portions of this entry contributed by Todd Rowland

REFERENCES:

Jacobson, N. Lie Algebras. New York: Dover, 1979.

Comment by Jonathan Vos Post | February 6, 2010 |

It’s cute, but it goes so fast that if you don’t already know what it’s supposed to be, you’ll never understand it.

If you want more visuals, there are some coming.

Comment by John Armstrong | February 6, 2010 |

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