Baker–Campbell–Hausdorff theorem

By [F1], choose a BCH convergence ball. The smooth map $\mathrm{\Phi }(t,X,Y)={\exp }_G(tX){\exp }_G(tY)$ sends $[0,1]\times \{(0,0)\}$ to $e\in U$. Apply the tube lemma to ${\mathrm{\Phi }}^{-1}(U)$ and intersect the resulting neighborhood of $(0,0)$ with a sufficiently small sum-norm ball. For $(X,Y)$ in this neighborhood, put $g(t)=\mathrm{\Phi }(t,X,Y)$ and $H(t)={\log }_G(g(t))$; then $H$ is smooth, $H(0)=0$, and ${\exp }_G(H(t))=g(t)$ for all $0\le t\le 1$.

Right-trivializing the derivative of the product and using the two one-parameter-subgroup equations gives $d(R_{g(t{)}^{-1}}{)}_{g(t)}{g}^{\mathrm{\prime }}(t)=X+{\mathrm{Ad}}_{{\exp }_G(tX)}Y=X+e^{t{\mathrm{ad}}_X}Y$. Applying [F2] to $g(t)={\exp }_G(H(t))$ therefore gives $D(A(t))H^{\mathrm{\prime }}(t)=X+e^{t{\mathrm{ad}}_X}Y$, where $A(t)={\mathrm{ad}}_{H(t)}$ and $D(A)={\sum }_{n\ge 0}{A}^n\mathrm{/}(n+1)!$.

Since $\mathrm{Ad}$ is a representation, [F3] and step 1.1 give $e^{A(t)}={\mathrm{Ad}}_{\exp H(t)}=e^{t{\mathrm{ad}}_X}{e}^{t{\mathrm{ad}}_Y}$. Put $P(t)=e^{t{\mathrm{ad}}_X}{e}^{t{\mathrm{ad}}_Y}-I$. Continuity and the tube lemma allow a further shrinking, uniform in $t\in [0,1]$, so that $\parallel P(t)\parallel <q<1$ and $H(t)$ remains in a fixed small coordinate ball.

Define $Q(P)={\sum }_{j\ge 0}(-1{)}^j{P}^j\mathrm{/}(j+1)$. It converges absolutely for $\parallel P\parallel <1$ by [F6]. In the commutative formal subalgebra generated by one indeterminate $z$, [F5] gives $\log (e^z)=z$, hence $D(z)Q(e^z-1)=1$. Absolute operator convergence permits substitution $z=A(t)$ and coefficientwise multiplication, so $Q(P(t))$ is the two-sided inverse of $D(A(t))$. Thus step 2.1 becomes $H^{\mathrm{\prime }}(t)={\sum }_{j\ge 0}(-1{)}^{j}P(t{)}^j(X+e^{t{\mathrm{ad}}_X}Y)\mathrm{/}(j+1)$.

Expand $P(t)$ by [F2]: $P(t)={\sum }_{m,n\ge 0,m+n>0}{t}^{m+n}{\mathrm{ad}}_X^m{\mathrm{ad}}_Y^n\mathrm{/}(m!n!)$. The bound $\parallel P(t)\parallel \le q<1$, together with exponential scalar majorants after one further shrinking, makes the expansions in step 3.1 jointly absolutely and uniformly convergent on $[0,1]$. In finite coordinates [F7] therefore permits termwise multiplication, regrouping, and integration.

A term with $k-1$ positive blocks from $P(t{)}^{k-1}$ followed by the terminal $X$ has word degree $N$, coefficient $(-1{)}^{k-1}\mathrm{/}k$, and power $t^{N-1}$; a term followed by $e^{t{\mathrm{ad}}_X}Y$ has the same description, with final block $X^{m}Y$ and again power $t^{N-1}$. Every other possible final block in Dynkin's formula has at least two terminal equal letters and its right-nested commutator is zero. Hence integration from zero to one contributes the factor $1\mathrm{/}N$ and gives exactly the full degree-$N$ Dynkin polynomial $H_N(X,Y)$.

By vector-valued FTC and step 1.1, $H(1)={\int }_0^1{H}^{\mathrm{\prime }}(t)dt$. Steps 4.1–5.1 and the uniform convergence in [F1] identify this integral with ${\sum }_{N\ge 1}{H}_N(X,Y)=\mathrm{BCH}(X,Y)$. Since $H(1)={\log }_G({\exp }_{G}X{\exp }_{G}Y)$, the logarithmic identity follows; applying ${\exp }_G$ and using [F1] gives the asserted product identity.

A Lie group contains its identity. In dimension zero the identities are the unique identities, and in dimension one the bracket vanishes so BCH is $X+Y$ and the local product is additive in exponential coordinates. No adjoint endomorphism is assumed invertible: step 3.1 inverts $D(A)$ by a convergent series and never divides by $A$. Both endpoints of $[0,1]$ occur in steps 1.1 and 6.1. The only choice principle is ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited exactly through the local logarithm, exponential, and adjoint-exponential suppliers in [F1]–[F4]; all shrinkings select finitely many single witnesses. No metric independence beyond the arbitrary auxiliary norm and no biconditional is asserted.