11.2 Chevalley™s theorem. . . . . . . . . . . . . . . . . . . . . . . . . . 189

11.2.1 Transcendence degrees. . . . . . . . . . . . . . . . . . . . 189

11.2.2 Symmetric polynomials. . . . . . . . . . . . . . . . . . . . 190

11.2.3 Fixed ¬elds. . . . . . . . . . . . . . . . . . . . . . . . . . . 192

11.2.4 Invariants of ¬nite groups. . . . . . . . . . . . . . . . . . . 193

11.2.5 The Hilbert basis theorem. . . . . . . . . . . . . . . . . . 195

11.2.6 Proof of Chevalley™s theorem. . . . . . . . . . . . . . . . . 196

Chapter 1

The Campbell Baker

Hausdor¬ Formula

1.1 The problem.

Recall the power series:

1 1 1 1

exp X = 1 + X + X 2 + X 3 + · · · , log(1 + X) = X ’ X 2 + X 3 + · · · .

2 3! 2 3

We want to study these series in a ring where convergence makes sense; for ex-

ample in the ring of n—n matrices. The exponential series converges everywhere,

and the series for the logarithm converges in a small enough neighborhood of

the origin. Of course,

log(exp X) = X; exp(log(1 + X)) = 1 + X

where these series converge, or as formal power series.

In particular, if A and B are two elements which are close enough to 0 we

can study the convergent series

log[(exp A)(exp B)]

which will yield an element C such that exp C = (exp A)(exp B). The problem

is that A and B need not commute. For example, if we retain only the linear

and constant terms in the series we ¬nd

log[(1 + A + · · · )(1 + B + · · · )] = log(1 + A + B + · · · ) = A + B + · · · .

On the other hand, if we go out to terms second order, the non-commutativity

begins to enter:

1 1

log[(1 + A + A2 + · · · )(1 + B + B 2 + · · · )] =

2 2

7

8 CHAPTER 1. THE CAMPBELL BAKER HAUSDORFF FORMULA

1 1 1

A + B + A2 + AB + B 2 ’ (A + B + · · · )2

2 2 2

1

= A + B + [A, B] + · · ·

2

where

[A, B] := AB ’ BA (1.1)

is the commutator of A and B, also known as the Lie bracket of A and B.

Collecting the terms of degree three we get, after some computation,

1 1 1

A2 B + AB 2 + B 2 A + BA2 ’ 2ABA ’ 2BAB] = [A, [A, B]]+ [B, [B, A]].

12 12 12

This suggests that the series for log[(exp A)(exp B)] can be expressed entirely

in terms of successive Lie brackets of A and B. This is so, and is the content of

the Campbell-Baker-Hausdor¬ formula.

One of the important consequences of the mere existence of this formula is

the following. Suppose that g is the Lie algebra of a Lie group G. Then the local

structure of G near the identity, i.e. the rule for the product of two elements of

G su¬ciently closed to the identity is determined by its Lie algebra g. Indeed,

the exponential map is locally a di¬eomorphism from a neighborhood of the

origin in g onto a neighborhood W of the identity, and if U ‚ W is a (possibly

smaller) neighborhood of the identity such that U · U ‚ W , the the product of

a = exp ξ and b = exp ·, with a ∈ U and b ∈ U is then completely expressed in

terms of successive Lie brackets of ξ and ·.

We will give two proofs of this important theorem. One will be geometric -

the explicit formula for the series for log[(exp A)(exp B)] will involve integration,

and so makes sense over the real or complex numbers. We will derive the formula

from the “Maurer-Cartan equations” which we will explain in the course of our

discussion. Our second version will be more algebraic. It will involve such ideas

as the universal enveloping algebra, comultiplication and the Poincar´-Birkho¬-

e

Witt theorem. In both proofs, many of the key ideas are at least as important

as the theorem itself.

1.2 The geometric version of the CBH formula.

To state this formula we introduce some notation. Let ad A denote the operation

of bracketing on the left by A, so

adA(B) := [A, B].

De¬ne the function ψ by

z log z

ψ(z) =

z’1

which is de¬ned as a convergent power series around the point z = 1 so

u u2 u u2

log(1 + u)

= (1 + u)(1 ’ + +···) = 1+ ’ +··· .

ψ(1 + u) = (1 + u)

u 2 3 2 6

1.2. THE GEOMETRIC VERSION OF THE CBH FORMULA. 9

In fact, we will also take this as a de¬nition of the formal power series for ψ in

terms of u. The Campbell-Baker-Hausdor¬ formula says that

1

log((exp A)(exp B)) = A + ψ ((exp ad A)(exp tad B)) Bdt. (1.2)

0

Remarks.

1. The formula says that we are to substitute

u = (exp ad A)(exp tad B) ’ 1

into the de¬nition of ψ, apply this operator to the element B and then integrate.

In carrying out this computation we can ignore all terms in the expansion of ψ

in terms of ad A and ad B where a factor of ad B occurs on the right, since

(ad B)B = 0. For example, to obtain the expansion through terms of degree

three in the Campbell-Baker-Hausdor¬ formula, we need only retain quadratic

and lower order terms in u, and so

t2

1

ad A + (ad A)2 + tad B + (ad B)2 + · · ·

u =

2 2

u2 2

(ad A) + t(ad B)(ad A) + · · ·

=

1

u u2 1 1 1

1 + ad A + (ad A)2 ’ (ad B)(ad A) + · · · ,

1+ ’ dt =

2 6 2 12 12

0

where the dots indicate either higher order terms or terms with ad B occurring

on the right. So up through degree three (1.2) gives

1 1 1

log(exp A)(exp B) = A + B + [A, B] + [A, [A, B]] ’ [B, [A, B]] + · · ·

2 12 12

agreeing with our preceding computation.

2. The meaning of the exponential function on the left hand side of the

Campbell-Baker-Hausdor¬ formula di¬ers from its meaning on the right. On

the right hand side, exponentiation takes place in the algebra of endomorphisms

of the ring in question. In fact, we will want to make a fundamental reinter-

pretation of the formula. We want to think of A, B, etc. as elements of a Lie

algebra, g. Then the exponentiations on the right hand side of (1.2) are still

taking place in End(g). On the other hand, if g is the Lie algebra of a Lie group

G, then there is an exponential map: exp: g ’ G, and this is what is meant by

the exponentials on the left of (1.2). This exponential map is a di¬eomorphism

on some neighborhood of the origin in g, and its inverse, log, is de¬ned in some

neighborhood of the identity in G. This is the meaning we will attach to the