AJ = JA.

Then

dAJ = JdA = 0.

and so

A’1 dAJ = A’1 JdA = JA’1 dA.

1.3. THE MAURER-CARTAN EQUATIONS. 13

We return to general considerations: Let us take the exterior derivative of

the de¬ning equation θ = A’1 dA. For this we need to compute d(A’1 ): Since

d(AA’1 ) = 0

we have

dA · A’1 + Ad(A’1 ) = 0

or

d(A’1 ) = ’A’1 dA · A’1 .

This is the generalization to matrices of the formula in elementary calculus for

the derivative of 1/x. Using this formula we get

dθ = d(A’1 dA) = ’(A’1 dA · A’1 ) § dA = ’A’1 dA § A’1 dA

or the Maurer-Cartan equation

dθ + θ § θ = 0. (1.8)

If we use commutator instead of multiplication we would write this as

1

dθ + [θ, θ] = 0. (1.9)

2

The Maurer-Cartan equation is of central importance in geometry and physics,

far more important than the Campbell-Baker-Hausdor¬ formula itself.

Suppose we have a map g : R2 ’ G, with s, t coordinates on the plane. Pull

θ back to the plane, so

‚g ‚g

g — θ = g ’1 ds + g ’1 dt

‚s ‚t

De¬ne

‚g

± = ±(s, t) := g ’1

‚s

and

‚g

β := β(s, t) = g ’1

‚t

so that

g — θ = ±ds + βdt.

Then collecting the coe¬cient of ds § dt in the Maurer Cartan equation gives

‚β ‚±

’ + [±, β] = 0. (1.10)

‚s ‚t

This is the version of the Maurer Cartan equation we shall use in our proof

of the Campbell Baker Hausdor¬ formula. Of course this version is completely

equivalent to the general version, since a two form is determined by its restriction

to all two dimensional surfaces.

14 CHAPTER 1. THE CAMPBELL BAKER HAUSDORFF FORMULA

1.4 Proof of CBH from Maurer-Cartan.

Let C(t) be a curve in the Lie algebra g and let us apply (1.10) to

g(s, t) := exp[sC(t)]

so that

‚g

±(s, t) = g ’1

‚s

= exp[’sC(t)] exp[sC(t)]C(t)

= C(t)

‚g

g ’1

β(s, t) =

‚t

‚

= exp[’sC(t)] exp[sC(t)] so by (1.10)

‚t

‚β

’ C (t) + [C(t), β] = 0.

‚s

For ¬xed t consider the last equation as the di¬erential equation (in s)

dβ

= ’(ad C)β + C , β(0) = 0

ds

where C := C(t), C := C (t).

If we expand β(s, t) as a formal power series in s (for ¬xed t):

β(s, t) = a1 s + a2 s2 + a3 s3 + · · ·

and compare coe¬cients in the di¬erential equation we obtain a1 = C , and

nan = ’(ad C)an’1

or

1 1

β(s, t) = sC (t) + s(’ad C(t))C (t) + · · · + sn (’ad C(t))n’1 C (t) + · · · .

2 n!

If we de¬ne

ez ’ 1 1 1

= 1 + z + z2 + · · ·

φ(z) :=

z 2! 3!

and set s = 1 in the expression we derived above for β(s, t) we get

d

exp(’C(t)) exp(C(t)) = φ(’ad C(t))C (t). (1.11)

dt

Now to the proof of the Campbell-Baker-Hausdor¬ formula. Suppose that

A and B are chosen su¬ciently near the origin so that

“ = “(t) = “(t, A, B) := log((exp A)(exp tB))

1.5. THE DIFFERENTIAL OF THE EXPONENTIAL AND ITS INVERSE.15

is de¬ned for all |t| ¤ 1. Then, as we remarked,

exp “ = exp A exp tB

so exp ad “ = (exp ad A)(exp t ad B) and hence

ad “ = log ((exp ad A)(exp t ad B)) .

We have

d d

exp “(t) = exp A exp tB

dt dt

= exp A(exp tB)B

= (exp “(t)B so

d

(exp ’“(t)) exp “(t) = B and therefore

dt

φ(’ad “(t))“ (t) = B by (1.11) so

φ(’ log ((exp ad A)(exp t ad B)))“ (t) = B.

Now for |z ’ 1| < 1

e’ log z ’ 1

φ(’ log z) =

’ log z

z ’1 ’ 1

=

’ log z

z’1

= so

z log z

z log z

ψ(z)φ(’ log z) ≡ 1 where ψ(z) := so

z’1

“ (t) = ψ ((exp ad A)(exp tad B)) B.