This proves (1.4) and integrating from 0 to 1 proves (1.2).

1.5 The di¬erential of the exponential and its

inverse.

Once again, equation (1.11), which we derived from the Maurer-Cartan equa-

tion, is of signi¬cant importance in its own right, perhaps more than the use we

made of it - to prove the Campbell-Baker-Hausdor¬ theorem. We will rewrite

this equation in terms of more familiar geometric operations, but ¬rst some

preliminaries:

The exponential map exp sends the Lie algebra g into the corresponding Lie

group, and is a di¬erentiable map. If ξ ∈ g we can consider the di¬erential of

exp at the point ξ:

d(exp)ξ : g = T gξ ’ T Gexp ξ

16 CHAPTER 1. THE CAMPBELL BAKER HAUSDORFF FORMULA

where we have identi¬ed g with its tangent space at ξ which is possible since g

is a vector space. In other words, d(exp)ξ maps the tangent space to g at the

point ξ into the tangent space to G at the point exp(ξ). At ξ = 0 we have

d(exp)0 = id

and hence, by the implicit function theorem, d(exp)ξ is invertible for su¬-

ciently small ξ. Now the Maurer-Cartan form, evaluated at the point exp ξ

sends T Gexp ξ back to g:

θexp ξ : T Gexp ξ ’ g.

Hence

θexp ξ —¦ d(exp)ξ : g ’ g

and is invertible for su¬ciently small ξ. We claim that

„ (ad ξ) —¦ θexp ξ —¦ d(expξ ) = id (1.12)

where „ is as de¬ned above in (1.3). Indeed, we claim that (1.12) is an immediate

consequence of (1.11).

Recall the de¬nition (1.3) of the function „ as „ (z) = 1/φ(’z). Multiply

both sides of (1.11) by „ (ad C(t)) to obtain

d

„ (ad C(t)) exp(’C(t)) exp(C(t)) = C (t). (1.13)

dt

Choose the curve C so that ξ = C(0) and · = C (0). Then the chain rule says

that

d

exp(C(t))|t=0 = d(exp)ξ (·).

dt

Thus

d

exp(’C(t)) exp(C(t)) = θexp ξ d(exp)ξ ·,

dt |t=0

the result of applying the Maurer-Cartan form θ (at the point exp(ξ)) to the

image of · under the di¬erential of exponential map at ξ ∈ g. Then (1.13) at

t = 0 translates into (1.12). QED

1.6 The averaging method.

In this section we will give another important application of (1.10): For ¬xed

ξ ∈ g, the di¬erential of the exponential map is a linear map from g = Tξ (g) to

Texp ξ G. The (di¬erential of) left translation by exp ξ carries Texp ξ (G) back to

Te G = g. Let us denote this composite by exp’1 d(exp)ξ . So

ξ

θexp ξ —¦ d(exp)ξ = d exp’1 d(exp)ξ : g’g

ξ

is a linear map. We claim that for any · ∈ g

1

exp’1 d(exp)ξ (·) = Adexp(’sξ) ·ds. (1.14)

ξ

0

1.6. THE AVERAGING METHOD. 17

We will prove this by applying(1.10) to

g(s, t) = exp (t(ξ + s·)) .

Indeed,

‚g

β(s, t) := g(s, t)’1 = ξ + s·

‚t

so

‚β

≡·

‚s

and

β(0, t) ≡ ξ.

The left hand side of (1.14) is ±(0, 1) where

‚g

±(s, t) := g(s, t)’1

‚s

so we may use (1.10) to get an ordinary di¬erential equation for ±(0, t). De¬ning

γ(t) := ±(0, t),

(1.10) becomes

dγ

= · + [γ, ξ]. (1.15)

dt

For any ζ ∈ g,

d

Adexp ’tξ ζ = Adexp ’tξ [ζ, ξ]

dt

= [Adexp ’tξ ζ, ξ].

So for constant ζ ∈ g,

Adexp ’tξ ζ

is a solution of the homogeneous equation corresponding to (1.15). So, by

Lagrange™s method of variation of constants, we look for a solution of (1.15) of

the form

γ(t) = Adexp ’tξ ζ(t)

and (1.15) becomes

ζ (t) = Adexp tξ ·

or

t

γ(t) = Adexp ’tξ Adexp sξ ·ds

0

is the solution of (1.15) with γ(0) = 0. Setting s = 1 gives

1

γ(1) = Adexp ’ξ Adexp sξ ds

0

and replacing s by 1 ’ s in the integral gives (1.14).

18 CHAPTER 1. THE CAMPBELL BAKER HAUSDORFF FORMULA

1.7 The Euler MacLaurin Formula.

We pause to remind the reader of a di¬erent role that the „ function plays in

mathematics. We have seen in (1.12) that „ enters into the inverse of the

exponential map. In a sense, this formula is taking into account the non-

commutativity of the group multiplication, so „ is helping to relate the non-

commutative to the commutative.

But much earlier in mathematical history, „ was introduced to relate the

discrete to the continuous: Let D denote the di¬erentiation operator in one

variable. Then if we think of D as the one dimensional vector ¬eld ‚/‚h it

generates the one parameter group exp hD which consists of translation by h.

In particular, taking h = 1 we have

eD f (x) = f (x + 1).

This equation is equally valid in a purely algebraic sense, taking f to be a

polynomial and

1 1

eD = 1 + D + D2 + D3 + · · · .

2 3!

This series is in¬nite. But if p is a polynomial of degree d, then Dk p = 0 for

k > D so when applied to any polynomial, the above sum is really ¬nite. Since

Dk eah = ak eah

it follows that if F is any formal power series in one variable, we have

F (D)eah = F (a)eah (1.16)

in the ring of power series in two variables. Of course, under suitable convergence

conditions this is an equality of functions of h.

For example, the function „ (z) = z/(1 ’ e’z ) converges for |z| < 2π since