[u, v], u ∈ Lp , v ∈ Lq , p + q = n.

X X

We claim that

¦(u) = nu ∀ u ∈ Ln . (1.26)

X

For n = 1 this is immediate from the de¬nition of ¦. So by induction it is

enough to verify this on elements of the form [u, v] as above. We have

¦(uv ’ vu)

¦([u, v]) =

˜(u)¦(v) ’ ˜(v)¦(u)

=

q˜(u)v ’ p˜(v)u by induction

=

q[u, v] ’ p[v, u]

=

since ˜(w) = ad(w) for w ∈ LX

= (p + q)[u, v] QED.

We can now write down an explicit formula for the n’th term in the

Campbell-Baker-Hausdor¬ expansion. Consider the case where X consists of

two elements X = {x, y}, x = y. Let us write

∞

z ∈ LX , z =

z = log ((exp x)(exp y)) zn (x, y).

1

We want an explicit expression for zn (x, y). We know that

1

zn = ¦(zn )

n

34 CHAPTER 1. THE CAMPBELL BAKER HAUSDORFF FORMULA

and zn is a sum of non-commutative monomials of degree n in x and y. Now

∞ ∞

xp yq

(exp x)(exp y) =

p! q!

p=0 q=0

xp y q

= 1+ so

p!q!

p+q≥1

z = log((exp x)(exp y))

m

«

∞ m+1 pq

(’1) xy

=

m p!q!

m=1 p+q≥1

(’1)m+1 xp1 y q1 xp2 y q2 · · · xpm y qm

= .

p1 !q1 ! · · · pm !qm !

m

pi +qi ≥1

1

We want to apply n ¦ to the terms in this last expression which are of total

degree n so as to obtain zn . So let us examine what happens when we apply ¦

to an expression occurring in the numerator: If qm ≥ 2 we get 0 since we will

have ad(y)(y) = 0. Similarly we will get 0 if qm = 0, pm ≥ 2. Hence the only

terms which survive are those with qm = 1 or qm = 0, pm = 1. Accordingly we

decompose zn into these two types:

1

zn = (z + zp,q ), (1.27)

n p+q=n p,q

where

(’1)m+1 ad(x)p1 ad(y)q1 · · · ad(x)pm y

zp,q = summed over all

p1 !q1 ! · · · pm !

m

p1 + · · · + pm = p, q1 + · · · + qm’1 = q ’ 1, qi + pi ≥ 1, pm ≥ 1

and

(’1)m+1 ad(x)p1 ad(y)q1 · · · ad(y)qm’1 (x)

zp,q = summed over

p1 !q1 ! · · · qm’1 !

m

p1 + · · · + pm’1 = p ’ 1, q1 + · · · + qm’1 = q,

pi + qi ≥ 1 (i = 1, . . . , m ’ 1) qm’1 ≥ 1.

The ¬rst four terms are:

z1 (x, y) = x+y

1

z2 (x, y) = [x, y]

2

1 1

z3 (x, y) = [x, [x, y]] + [y, [y, x]]

12 12

1

z4 (x, y) = [x, [y, [x, y]]].

24

Chapter 2

sl(2) and its

Representations.

In this chapter (and in most of the succeeding chapters) all Lie algebras and

vector spaces are over the complex numbers.

2.1 Low dimensional Lie algebras.

Any one dimensional Lie algebra must be commutative, since [X, X] = 0 in any

Lie algebra.

If g is a two dimensional Lie algebra, say with basis X, Y then [aX +bY, cX +

dY ] = (ad ’ bc)[X, Y ], so that there are two possibilities: [X, Y ] = 0 in which

case g is commutative, or [X, Y ] = 0, call it B, and the Lie bracket of any

two elements of g is a multiple of B. So if C is not a multiple of B, we have

[C, B] = cB for some c = 0, and setting A = c’1 C we get a basis A, B of g with

the bracket relations

[A, B] = B.

This is an interesting Lie algebra; it is the Lie algebra of the group of all a¬ne

transformations of the line, i.e. all transformations of the form

x ’ ax + b, a = 0.

For this reason it is sometimes called the “ax + b group”. Since

ab x ax + b

=

01 1 1

we can realize the group of a¬ne transformations of the line as a group of two

by two matrices. Writing

a = exp tA, b = tB

35

36 CHAPTER 2. SL(2) AND ITS REPRESENTATIONS.

so that

a0 A0 1 b 0B