The BCH group of a nilpotent Lie algebra

If $\mathfrak{n}$ has class $c$, every Lie monomial of bracket length greater than $c$ vanishes. Thus the BCH expression supplied globally by [L3] is a finite polynomial. Its universal identities give $\mathrm{BCH}(x,0)=x=\mathrm{BCH}(0,x)$ and $\mathrm{BCH}(x,-x)=0$.

Let $N$ be the connected simply connected integration supplied by [L2]. By [L3], $\exp :\mathfrak{n}\to N$ is a diffeomorphism and the transported global product $x\ast y=\log (\exp x\exp y)$ is the truncated BCH polynomial; this agrees with the local formula in [L1]. Associativity, identity $0$, and inverse $-x$ follow from the laws of $N$.

The underlying manifold is $\mathfrak{n}\cong {\mathbb{R}}^{\dim \mathfrak{n}}$, hence is connected and simply connected, including dimension zero. The antisymmetric part of the quadratic BCH term is $[x,y]$, so differentiating the commutator recovers the original bracket.

In the Heisenberg algebra, $(ae+bf+cz)\ast (a^{\mathrm{\prime }}e+b^{\mathrm{\prime }}f+c^{\mathrm{\prime }}z)=(a+a^{\mathrm{\prime }})e+(b+b^{\mathrm{\prime }})f+(c+c^{\mathrm{\prime }}+\frac{1}{2}(a{b}^{\mathrm{\prime }}-b{a}^{\mathrm{\prime }}))z,$ because brackets of length three vanish. This is a concrete nonabelian instance. The declared ${\mathsf{A}\mathsf{C}}_{\omega }$ is exactly that inherited through [L2]–[L3]; the finite calculation adds no choice.