Seminar in Algebra and Number Theory Reflection Groups and Hecke Algebras Fall 2005 P. Achar Summary: Seminar in Algebra and Number Theory Reflection Groups and Hecke Algebras Fall 2005 P. Achar Problem Set 2a Hints Due: September 27, 2005 1. Given a parabolic subgroup W (generated by, say, I S), let V = (span ) , where is the set of simple roots corresponding to I. It's clear that W stabilizes V ; it remains to be shown that nothing outside of W does, so that W is equal to the stabilizer of V . First show that it's enough to consider w W W possessing a reduced expression ending with a generator not in I. If w ends in t, then by choosing v V such that v, s > 0 for all s / I, show that w-1 v, t < 0, so w-1 v = v, and w does not stabilize V . Next, given a subspace V V with stabilizer W , let 1 be the set of roots orthogonal to V . This is itself a root system. Choose positive and simple systems 1, 1 for it, and let W1 be the subgroup of W generated by the reflections in 1 (or, equivalently, 1). There are two things to prove: (1) W1 is a parabolic subgroup, or, equivalently, that 1 can be extended to a simple system for the whole group; and (2) W1 coincides with the stabilizer W . Once we prove the first part, the second part is easy--we did the same thing in the previous paragraph. To prove (1), we have to choose a positive system for W that contains 1 (that's easy) with the additional property that no root in 1 can be Collections: Mathematics