# 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.

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