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Local convergence of the Baker–Campbell–Hausdorff series

Statement

Let g be a finite-dimensional real Lie algebra equipped with any norm ∥⋅∥. There is an ε>0 such that Dynkin's series

BCH⁡(X,Y)=∑N≥1HN(X,Y)

converges absolutely whenever ∥X∥+∥Y∥<ε. Moreover, for every 0<r0<ε, its partial sums converge uniformly on

Dr0:={(X,Y):∥X∥+∥Y∥≤r0}.

No assertion here identifies this convergent sum with log⁡(exp⁡Xexp⁡Y); that is the content of the following BCH theorem.

Facts & Assumptions

Given: A finite-dimensional real Lie algebra g with a norm.

[F1]

The degree-N Dynkin polynomial is the stated finite sum of right-nested commutators, and BCH is its formal degree-indexed series. Baker–Campbell–Hausdorff series.

[F2]

The Lie bracket is bilinear. Finite-dimensional Lie algebra.

[F3]

A chosen finite basis gives a bounded coordinate isomorphism for any norm. A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space.

[F4]

Every finite-dimensional normed real vector space is complete. Every finite-dimensional normed space is Banach.

[F5]

The real exponential series converges absolutely everywhere and obeys exp⁡(a+b)=exp⁡(a)exp⁡(b). The exponential series converges absolutely for every real argument. The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y).

[F7]

Proof

technique · direct
1.1F2F3

If g=0, every HN vanishes and the conclusion holds for every positive ε. Otherwise choose one finite basis. By [F2]--[F3], its finitely many structure constants and the bounded coordinate maps give a constant C≥0 such that ∥[U,V]∥≤C∥U∥∥V∥ for all U,V∈g. This chooses one finite witness, not a family.

2.1F1step 1.1induction

Induction on word length now gives ∥[Z1⋯ZN]R∥≤CN−1∏j=1N∥Zj∥ for every right-nested commutator in [F1].

2.2F1step 1.1cases

If C=0, all brackets vanish, so [F1] gives H1(X,Y)=X+Y and HN(X,Y)=0 for N≥2; take ε=1. Hence suppose C>0 and put ε=1/(4C).

3.1F5F6F7step 2.2

Fix 0<r0<ε and put R=Cr0<1/4. For a=C∥X∥, b=C∥Y∥ and (X,Y)∈Dr0, [F6] shows that the sum of the scalar weights ambn/(m!n!) over a block of positive total degree is q(a,b)=exp⁡(a+b)−1, and its degree-d part is (a+b)d/d!≤Rd/d!. Thus these coefficients are bounded degree by degree by those of q0:=∑d≥1Rd/d!. Since R<1/4 and d!≥1, [F5]--[F7] give 0≤q0≤∑d≥1(1/4)d=1/3<1.

4.1F1F7step 2.1step 3.1algebra

Apply step 2.1 to the formula in [F1]. For a k-block summand of total degree N, absorb CN into its scalar letter weights and retain the factor 1/C. Since 1/N≤1, the sum of the norms of all homogeneous terms, uniformly for (X,Y)∈Dr0, is bounded by the nonnegative degree expansion of 1C∑k≥1q0k/k≤1C∑k≥1q0k, which converges by [F7]. In particular, the resulting degree majorants MN satisfy ∥HN(X,Y)∥≤MN on Dr0 and ∑NMN<∞.

5.1F4step 4.1

For each (X,Y)∈Dr0, step 4.1 makes the partial sums Cauchy by the triangle inequality, and [F4] supplies their limit. Moreover, the norm of every tail is bounded by the corresponding scalar tail ∑N>mMN, independently of (X,Y); that tail tends to zero. Hence the convergence is absolute and uniform on Dr0.

6.1F1F2step 1.1step 2.2step 5.1∎

Any pair with ∥X∥+∥Y∥<ε lies in some Dr0 with ∥X∥+∥Y∥<r0<ε, so step 5.1 proves both claims. The zero and one-dimensional cases are included; in dimension one the bracket is zero. Degenerate brackets are allowed. There is no interval or endpoint, no metric beyond the arbitrary norm in the statement, no choice principle beyond choosing one finite basis, and no biconditional.

Depends on

Dependency tree · 2 levels

Sources