Lately in Lie Groups

This material is coming from chapter 21 of Fulton and Harris.  Below is mostly a regurgitation with a few clarifying remarks.

Recall axioms for a root system R,E, (,)

  1. R spans E
  2. a \in R implies (ka \in R iff k = \pm 1)
  3. The reflection W_a(b) = b - 2(b,a)/(a,a) a maps R to itself
  4. 2(b,a)/(a,a) is an integer.

##5(Clarifying point) Now b \ne \pm a.  Consider the longest string

b - pa, b - (p-1)a , ..., b - a, b, b+a, ..., b+q

by axiom 3 it follows that this string is sent to itself under W_a; in particular, 

W_a(b + ka) = W_a(b)-ka

so comparing where b + ka and b+(k+1)a get sent, it follows that the highest element b+qa gets mapped to the lowest element b-pa.  From this you get

b - pa = b - (n_{ba}+q)a

So n_{ba} = p - q. etc. etc.

##6 (clarifying point) a,b are as above and (b,a) > 0 then W_a(b) = b - ka is a root.  If k = 1 then b - a is a root.  If k > 1, it should still hold that b-a is a root.  

The point is \mathfrak{s}_a, the subalgebra generated by g_a, g_{-a} that’s isomorphic to \mathfrak{sl}(2,C) acts on \oplus g_{b + ka} and it breaks up into a direct sum of irreducible representations that have the property that if b+pa and b + (p+l)a are roots, then everything in between is a root.  You might also be able to prove this playing around with reflections W_a, W_b.

in any case, it follows that (a,b) > 0 implies a-b is a root and (a,b) <0 implies a+b is a root.

 If (a,b) = 0 and if a+b is a root then so is W_a(a+b) = b - a.  This shows a+b, a-b are either both roots or both not roots.

##7 if a,b are simple distinct roots, then \pm (a - b) cannot be roots, otherwise a = (a -b ) + b, contradicting a simple.

##8 From 6,7 it follows that (a,b) \le 0 so the angle between them satisfies \cos(\theta_{ab}) \le 0, meaning \pi/2 \le \theta_{ab} \le 3\pi/2 (i.e. not accute)

##9 an excercise in Fulton and Harris (that has hints) shows the simple roots are linearly independent.  Basically assume you can write a vector v in two different ways using disjoint subsets of the simple roots (i.e. you have a relation), take (v,v), show its zero to get a contradiction.

##10 As R spans E there must be exactly \dim E simple roots with consequently form a basis.  And its not hard to show any positive root can be written as a nonnegative combination of the simple roots.  

The rest of the lecture was about makin Dynkin diagrams and then a big theorem saying all the possible Dynkin diagrams are A_n, B_n, ... with 5 exceptions. 







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