The problem
Recall the power series: expX = 1 +X+1
We aim to analyze series within a ring where convergence is applicable, such as the ring of n×n matrices The exponential series converges universally, while the logarithmic series converges within a sufficiently small neighborhood around the origin Notably, we have the relationships log(exp(X)) = X and exp(log(1 + X)) = 1 + X, valid in the regions where these series converge or as formal power series.
When A and B are elements close to zero, we can analyze the convergent series log[(expA)(expB)], resulting in an element C such that expC = (expA)(expB) However, A and B may not commute By considering only the linear and constant terms in the series, we derive 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: log[(1 +A+1
[A, B] :−BA (1.1) is thecommutatorofAandB, also known as theLie bracketofAandB. Collecting the terms of degree three we get, after some computation,
The series for log[(expA)(expB)] can be fully represented through the successive Lie brackets of A and B, which is a key aspect of the Campbell-Baker-Hausdorff formula.
The existence of this formula has significant implications, particularly regarding the Lie algebra \( g \) of a Lie group \( G \) It provides insights into the local structure of \( G \) near the identity element, specifically detailing the product rule for two elements within the group.
The identity of a Lie group is closely related to its Lie algebra, as the exponential map serves as a local diffeomorphism from a neighborhood around the origin to a neighborhood of the identity For a smaller neighborhood U within W, the product of elements a = exp(ξ) and b = exp(η), both belonging to U, can be fully expressed using successive Lie brackets of ξ and η.
This article presents two proofs of a significant theorem: a geometric proof utilizing the explicit formula for the series log[(expA)(expB)], which incorporates integration and is applicable to real or complex numbers, derived from the Maurer-Cartan equations; and an algebraic proof that explores concepts such as the universal enveloping algebra, comultiplication, and the Poincaré-Birkhoff-Witt theorem Both proofs emphasize that the underlying ideas are as crucial as the theorem itself.
The geometric version of the CBH formula
To state this formula we introduce some notation Let adAdenote the operation of bracketing on the left byA, so adA(B) := [A, B].
Define the functionψby ψ(z) = zlogz z−1 which is defined as a convergent power series around the pointz= 1 so ψ(1 +u) = (1 +u)log(1 +u) u = (1 +u)(1ưu
1.2 THE GEOMETRIC VERSION OF THE CBH FORMULA 9
In fact, we will also take this as adefinition of the formal power series forψin terms ofu The Campbell-Baker-Hausdorff formula says that log((expA)(expB)) =A+
1 The formula says that we are to substitute u= (exp adA)(exptadB)−1 into the definition ofψ, apply this operator to the elementBand then integrate.
In our computation, we can disregard terms in the expansion of ψ involving ad A and ad B where ad B appears on the right, as (adB)B equals zero For instance, when deriving the expansion up to degree three in the Campbell-Baker-Hausdorff formula, we only need to consider quadratic and lower-order terms in u, resulting in u being equal to adA+1.
12(adB)(adA) +ã ã ã , where the dots indicate either higher order terms or terms with adB occurring on the right So up through degree three (1.2) gives log(expA)(expB) =A+B+1
12[B,[A, B]] +ã ã ã agreeing with our preceding computation.
The exponential function in the Campbell-Baker-Hausdorff formula has distinct meanings on its left and right sides On the right, it operates within the algebra of endomorphisms related to the specific ring We propose a fundamental reinterpretation of the formula, considering A, B, and similar elements as part of a Lie algebra, denoted as g Thus, the exponentiations on the right side still occur within End(g) Conversely, if g represents the Lie algebra of a Lie group, the interpretation shifts accordingly.
The exponential map, denoted as exp: g→G, represents the concept of exponentials referenced in equation (1.2) This map acts as a diffeomorphism within a neighborhood surrounding the origin, while its inverse, the logarithm, is defined in a region close to the identity in G This interpretation clarifies the use of the logarithm mentioned on the left side of equation (1.2).
The Campbell-Baker-Hausdorff formula reveals that the local structure of a Lie group \( G \), specifically the multiplication law for elements close to the identity, is entirely defined by its Lie algebra.
4 For example, we see from the right hand side of (1.2) that group multi- plication and group inverse are analytic if we use exponential coordinates.
5 Consider the functionτ defined by τ(w) := w
The function discussed is integral to analysis and is featured in the Euler-Maclaurin formula Specifically, it serves as the exponential generating function for the sequence (−1) k b k, where the b k represents the Bernoulli numbers Consequently, the relationship can be expressed as ψ(z) = τ(log z).
The formula is incorrectly attributed to mathematicians Campbell, Baker, and Hausdorff; however, Friedrich Schur's earlier paper, "Neue Begruendung der Theorie der endlichen Transformationsgruppen," published in Mathematische Annalen, predates their contributions significantly.
In 1890, Schur presented the composition law for a Lie group using convergent power series expressed in "canonical coordinates," which refer to linear coordinates on the Lie algebra He established recursive relations for the coefficients, leading to a refined version of the formulas discussed in this article Acknowledgments are given to Prof Schmid for providing this reference.
Our approach to proving equation (1.2) involves demonstrating a differential version, expressed as d/dt log((expA)(exptB)) = ψ((exp adA)(exptadB))B Notably, when t = 0, log(expA(exptB)) simplifies to A By integrating this differential equation from 0 to 1, we can establish the validity of (1.2) We define Γ = Γ(t) = Γ(t, A, B) as Γ = log((expA)(exptB)).
Then exp Γ = expAexptB and so d dtexp Γ(t) = expAd dtexptB
We will prove (1.4) by finding a general expression for exp(−C(t))d dtexp(C(t)) whereC=C(t) is a curve in the Lie algebra,g, see (1.11) below.
The Maurer-Cartan equations
In our derivation of equation (1.4) from equation (1.11), we will utilize a key property of the adjoint representation Specifically, for any element g in the group G, we define the associated linear transformation.
In geometric terms, the differential of a left multiplication by an element \( g \) maps \( TI(G) \) into the tangent space \( Tg(G) \) at the point \( g \) Conversely, right multiplication by \( g^{-1} \) brings this tangent space back to \( g \) The combined operation forms a linear map from \( g \) into itself, known as \( \text{Ad}_g \) It is important to note that \( \text{Ad} \) serves as a representation in this context.
In particular, for any A∈g, we have the one parameter family of linear trans- formations Ad(exptA) and d dtAd (exptA)X = (exptA)AX(exp−tA) + (exptA)X(−A)(exp−tA)
= (exptA)[A, X](exp−tA) so d dtAd exptA = Ad(exptA)◦adA.
The linear transformation adA operates on g, leading to the solution of the differential equation d/dt M(t) = M(t) adA, with the initial condition M(0) = I, which results in M(t) = exp(t adA) Consequently, we have the significant relationship Ad(exp(tA)) = exp(t adA) By setting t = 1, we derive a crucial formula.
As an application, consider the Γ introduced above We have exp(ad Γ) = Ad (exp Γ)
= (exp adA)(exp adtB) hence ad Γ = log((exp ad A)(exp adtB)) (1.7)
If G is a Lie group and γ = γ(t) is a curve on G with γ(0) = A ∈ G, then
A −1 γis a curve which passes through the identity att= 0 HenceA −1 γ 0 (0) is a tangent vector at the identity, i.e an element ofg, the Lie algebra ofG.
We have established a linear differential form θ on G, taking values in g If G is a subgroup of the group of all invertible n × n matrices over the real numbers, this form can be expressed as θ = A^(-1) dA.
The matrix A can be viewed as a set of n² real-valued functions defined on the group G, with dA representing the matrix of differentials of these functions The equation for θ is derived through matrix multiplication For clarity, we will focus on this specific case, although the main theorem presented in equation (1.8) is applicable to any Lie group and is widely recognized in the field.
The definitions of the groups we are considering amount to constraints on
A, and then differentiating these constraints show that A −1 dAtakes values in g, and gives a description ofg It is best to explain this by examples:
−I 0 and let Sp(n) consist of all matrices satisfying
Then dAJ a † +AJ dA † = 0 or
The equationBJ+J B † = 0 defines the Lie algebra sp(n).
• LetJ be as above and define Gl(n,C) to consist of all invertible matrices satisfying
Then dAJ =J dA= 0. and so
In our analysis, we revisit the defining equation θ = A^(-1) dA and proceed to take its exterior derivative To do this, we first compute d(A^(-1)) Given that d(AA^(-1)) = 0, we can derive that dA A^(-1) + A d(A^(-1)) = 0, leading to the result d(A^(-1)) = -A^(-1) dA A^(-1).
The generalization of the derivative formula for 1/x in elementary calculus applies to matrices, leading to the expression dθ=d(A −1 dA) =−(A −1 dAãA −1 )∧dA This simplifies to the Maurer-Cartan equation, represented as dθ+θ∧θ= 0.
If we use commutator instead of multiplication we would write this as dθ+1
The Maurer-Cartan equation is of central importance in geometry and physics, far more important than the Campbell-Baker-Hausdorff formula itself.
Suppose we have a mapg:R 2 →G, withs, tcoordinates on the plane Pull θ back to the plane, so g ∗ θ=g −1 ∂g
Then collecting the coefficient ofds∧dtin the Maurer Cartan equation gives
In our proof of the Campbell-Baker-Hausdorff formula, we will utilize a specific version of the Maurer-Cartan equation, which is entirely equivalent to the general formulation, as a two-form is uniquely defined by its restriction to all two-dimensional surfaces.
Proof of CBH from Maurer-Cartan
LetC(t) be a curve in the Lie algebragand let us apply (1.10) to g(s, t) := exp[sC(t)] so that α(s, t) = g −1 ∂g
For fixedtconsider the last equation as the differential equation (in s) dβ ds =−(adC)β+C 0 , β(0) = 0 whereC:=C(t), C 0 :=C 0 (t).
If we expandβ(s, t) as a formal power series ins(for fixedt): β(s, t) =a 1 s+a 2 s 2 +a 3 s 3 +ã ã ã and compare coefficients in the differential equation we obtaina1=C 0 , and nan=−(adC)an−1 or β(s, t) =sC 0 (t) +1
2!z+ 1 3!z 2 +ã ã ã and sets= 1 in the expression we derived above forβ(s, t) we get exp(−C(t))d dtexp(C(t)) =φ(−adC(t))C 0 (t) (1.11)
Now to the proof of the Campbell-Baker-Hausdorff formula Suppose that
AandB are chosen sufficiently near the origin so that Γ = Γ(t) = Γ(t, A, B) := log((expA)(exptB))
The differential of the exponential and its inverse
is defined for all|t| ≤1 Then, as we remarked, exp Γ = expAexptB so exp ad Γ = (exp adA)(exptadB) and hence ad Γ = log ((exp adA)(expt adB)).
We have d dtexp Γ(t) = expAd dtexptB
(exp−Γ(t))d dtexp Γ(t) = B and therefore φ(−ad Γ(t))Γ 0 (t) = B by (1.11) so φ(−log ((exp adA)(exptadB)))Γ 0 (t) = B.
= z−1 zlogz so ψ(z)φ(−logz) ≡ 1 whereψ(z) := zlogz z−1 so Γ 0 (t) = ψ((exp adA)(exptadB))B.
This proves (1.4) and integrating from 0 to 1 proves (1.2).
1.5 The differential of the exponential and its inverse.
The importance of equation (1.11), derived from the Maurer-Cartan equation, extends beyond its application in proving the Campbell-Baker-Hausdorff theorem To enhance understanding, we will express this equation using more familiar geometric operations, but first, we will address some preliminary concepts.
The exponential map, denoted as exp, transforms the Lie algebra into its corresponding Lie group and is a differentiable function For any element ξ in the Lie algebra g, we can analyze the differential of exp at ξ, expressed as d(exp)ξ, which maps the tangent space at ξ to the tangent space of the group G at the point exp(ξ) This mapping is possible because g is a vector space Notably, at the origin ξ = 0, the differential d(exp)0 acts as the identity map, and according to the implicit function theorem, d(exp)ξ remains invertible for sufficiently small values of ξ Additionally, the Maurer-Cartan form, evaluated at exp(ξ), provides a mapping from the tangent space of G at exp(ξ) back to the Lie algebra g.
Hence θ exp ξ ◦d(exp) ξ :g→g and is invertible for sufficiently smallξ We claim that τ(adξ)◦ θ exp ξ ◦d(exp ξ )
= id (1.12) whereτis as defined above in (1.3) Indeed, we claim that (1.12) is an immediate consequence of (1.11).
Recall the definition (1.3) of the function τ as τ(z) = 1/φ(−z) Multiply both sides of (1.11) byτ(adC(t)) to obtain τ(adC(t)) exp(−C(t))d dtexp(C(t)) =C 0 (t) (1.13)
Choose the curveC so thatξ=C(0) andη =C 0 (0) Then the chain rule says that d dtexp(C(t)) |t=0 =d(exp)ξ(η).
The application of the Maurer-Cartan form θ at the point exp(ξ) to the image of η, as influenced by the differential of the exponential map at ξ ∈ g, yields the expression =θexp ξd(exp)ξη Consequently, when evaluating equation (1.13) at t=0, it simplifies to equation (1.12) QED.
The averaging method
In this section we will give another important application of (1.10): For fixed ξ∈g, the differential of the exponential map is a linear map fromg=T ξ (g) to
T exp ξ G The (differential of) left translation by expξcarriesT exp ξ (G) back to
T e G=g Let us denote this composite by exp −1 ξ d(exp) ξ So θexp ξ◦d(exp)ξ p −1 ξ d(exp)ξ: g→g is a linear map We claim that for anyη∈g exp −1 ξ d(exp) ξ (η) Z 1 0
We will prove this by applying(1.10) to g(s, t) = exp (t(ξ+sη)).
The left hand side of (1.14) is α(0,1) where α(s, t) :=g(s, t) −1 ∂g
∂s so we may use (1.10) to get an ordinary differential equation forα(0, t) Defining γ(t) :=α(0, t),
Adexp −tξζ is a solution to the homogeneous equation related to (1.15) Utilizing Lagrange’s method of variation of constants, we seek a solution for (1.15) in the form of γ(t) = Ad exp −tξ ζ(t) Consequently, (1.15) is transformed into ζ 0 (t) = Adexp tξη, leading to the expression γ(t) = Adexp −tξ.
Adexp sξηds is the solution of (1.15) withγ(0) = 0 Settings= 1 gives γ(1) = Adexp −ξ
Adexp sξds and replacingsby 1−sin the integral gives (1.14).
The Euler MacLaurin Formula
The τ function in mathematics serves a crucial role, particularly in its involvement with the inverse of the exponential map, as highlighted in equation (1.12) This relationship acknowledges the non-commutativity of group multiplication, with τ acting as a bridge between non-commutative and commutative structures.
In early mathematical history, τ was introduced to connect discrete and continuous concepts Let D represent the differentiation operator in a single variable; when viewed as the one-dimensional vector field ∂/∂h, it generates the one-parameter group exphD, which facilitates translation by h.
In particular, takingh= 1 we have e D f (x) =f(x+ 1).
This equation is equally valid in a purely algebraic sense, taking f to be a polynomial and e D = 1 +D+1
This series is infinite But if pis a polynomial of degreed, then D k p= 0 for k > Dso when applied to any polynomial, the above sum is really finite Since
D k e ah =a k e ah it follows that ifF is any formal power series in one variable, we have
F(D)e ah =F(a)e ah (1.16) in the ring of power series in two variables Of course, under suitable convergence conditions this is an equality of functions ofh.
For example, the function τ(z) =z/(1−e −z ) converges for |z|