# 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.