Lately in Lie Groups

Lie Groups 11/3/09

Claim from last time; The notation: R = roots of a semisimple lie algebra.  To reduce typing write a for \alpha; W is the Weyl group generated by reflections W_a

If you look at the complement of \cup_{a \in R} a^\perp you divide your vector space into Weyl chambers.  The first part of class was showing the Weyl groups acts simply and transitively ( = free and transitive?) on the chambers.

Next, let \theta_{ab} be the angle between the roots a,b \in R.  Then set  n_ab = 2(b, a)/(a, a) using (a,b) = |a||b|cos(\theta_{ab}) this can be re-written as

n_{ab} = 2cos(\theta_{ab})|b|/|a|.

Hence n_{ab}n_{ba} = 4 cos^2(\theta_{ab}).  So n_ab, n_ba have same sign if \theta_{ab} \ne \pm \pi/2.  Have n_{ab}n_{ba} = 4 iff \cos \theta_{ab} = \pm 1  iff a = \pm b.  

So assume this is not the case, then  0 \le n_ab*n_ba \le 3 (recall that n_ab have to be integers, essentially because reps of sl(2,C) only have integer eigen values.  Assume |b|\ge |a|, equivalently 3 \ge |n_{ab}| \ge |n_{ba}| Get table

\begin{array}{cccccccc} n_{ab}: & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ n_{ba}: & -1 & -1 & -1 & 0 & 1 & 1 & 1 \\ \cos(\theta_{ab}): & -\sqrt 3/2 & -\sqrt 2/2 & -1/2 & 0 & 1/2 & \sqrt 2/2 & \sqrt 3/2 \\ \theta_{ab}: & 5\pi/6 & 3\pi/4 & 2\pi/3 & \pi/2 & \pi/3 & \pi/4 & \pi/6 \end{array}

some of the possibilities 

rank 1: only have  A_1  corresponds to sl_2(C)

rank 2:  get A_1 \times A_1 and A_2  for angles pi/2, pi/3.  For angle pi/4 get B_2 which is  among other things corresponds to so(5,C).  You also get something for \pi/6.  This is said more explicitly in pg 257 of Hall’s book.

Prop (this is also done somewhere in Humphreys): If \mathfrak{g} is simple then h^* is an irrep of W, the Weyl group iff the root system is ireddcuible.

proof(sketch): Roughly, if the root system decomposes R = R' \cup R'' then the subalgebra generated by g_a for a \in R' will end being a proper ideal of \mathfrak{g}, contradicting that its simple.  More details are in pg. 73 of Humphreys.  The equivalence h^* irrep iff R irreducible, I didn’t write down.  But roughly seems to be if the root system is irreducible, then the action of the Weyl group will be transitive, meaning no sub invariant subspaces, and vice versa. 

Now examples in rank 3

…, didn’t write down.

About triangles on spheres.

a,b,c \in \{2,3,4,6\}, area of spherical triangle = pi/a + pi/b + pi/c - pi > 0.  Gave a proof that I didn’t write down.

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