Vtyjkyuk

I will use homological grading.

Let $mathfrakg$ be a dg Lie algebra. Then the set of elements of degree zero $mathfrakg_0$ acts on the set of Maurer-Cartan elements
$$mathrmMC(mathfrakg):=xinmathfrakg_-1mid dx+tfrac12[x,x] = 0$$
by what is called the gauge action. I will denote by $lambdacdot x$ the gauge action of an element $lambda$ of degree $0$ on a Maurer-Cartan element $x$. There is a well known Baker-Campbell-Hausdorff formula
$$mathrmBCH:mathfrakg_0timesmathfrakg_0longrightarrowmathfrakg_0$$
which makes $mathfrakg_0$ into a group and which is such that
$$mathrmBCH(lambda_1,lambda_2)cdot x = lambda_1cdot(lambda_2cdot x) .$$
It starts with
$$mathrmBCH(lambda_1,lambda_2) = lambda_1 + lambda_2 +frac12[lambda_1,lambda_2]+cdots$$
My question is:

Is it possible to completely determine the Baker-Campbell-Hausdorff formula by its properties alone?

More precisely, I would like to have a reference or proof for the following statement:

The Baker-Campbell-Hausdorff formula is the only associative binary operation $mathrmBCH:mathfrakg_0timesmathfrakg_0longrightarrowmathfrakg_0$ such that $mathrmBCH(lambda_1,lambda_2)cdot x = lambda_1cdot(lambda_2cdot x)$.

I've got an answer to my question. I will write it in the context of ordinary (non dg) Lie algebras, but it easily generalizes.

First of all, notice that we want a formula that is the same for any Lie algebra. Therefore, we want it to come from a BCH formula on the free Lie algebra on two generators $mathrmLie(x,y)$, i.e. free bracketings of the elements $x$ and $y$ modulo the Jacobi relation and antisymmetry of the bracket. Then we obtain the BCH formula for an arbitrary Lie algebra $mathfrakg$ by considering the map
$$phi:mathrmLie(x,y)longrightarrowmathfrakg$$
sending $xmapstolambda_1$ and $ytolambda_2$. We have
$$mathrmBCH(lambda_1,lambda_2):=phi(mathrmBCH(x,y)) .$$
Thus, the characterization of the BCH formula reduces to proving the uniqueness of such a formula for the free Lie algebra on two generators. It must satisfy
$$e^ad_mathrmBCH(x,y) = e^ad_xe^ad_y.$$
Suppose that there were a second possibility $widetildemathrmBCH(x,y) = mathrmBCH(x,y) + z$. Then for any $uinmathrmLie(x,y)$ we must have
$$sum_nge0frac1n!left(ad_mathrmBCH(x,y)right)^n(u) = e^ad_mathrmBCH(x,y)(u) = e^ad_widetildemathrmBCH(x,y)(u) = sum_nge0frac1n!left(ad_mathrmBCH(x,y) + ad_zright)^n(u)tag$star$$$
We will now prove that we must have $zin Z(mathrmLie(x,y))$, which implies that $z=0$ since the free Lie algebra on two generators has trivial center. We write
$$z = z_1 + z_2 + cdots$$, where $z_n$ is the part of $z$ consisting of bracketings of $n$ elements, and similarly for the other elements. So for example we have
$$x_1 = x ,quad x_n = 0quadtextfor nge2 , textandquadmathrmBCH(x,y)_2 = frac12[x,y] .$$

Lemma: We have $zin Z(mathrmLie(x,y))$ if and only if $[x,z] = 0$ and $[y,z] = 0$.

Proof: One direction is obvious. The other is done by induction using the Jacobi rule, where the $n$th step consists in proving that $z_n$ is in the center if $z$ commutes with $x$ and $y$.

Proposition: The element $z$ is zero.

Proof: Using the Lemma, we only consider $u=x$ and $u=y$.

This is also done by induction. The equation $(star)_1$ is automatically satisfied, since both sides equal $u_1 = u$. For $(star)_2$, we get the constraint
$$[mathrmBCH(x,y)_1,u] = [mathrmBCH(x,y)_1,u] + [z_1,u] ,$$
which gives $[z_1,u] = 0$ and implies that $z_1$ is central by the Lemma. Thus, $z_1=0$.

Proceeding this way, from $(star)_n+1$ we obtain the equation $[z_n,u] = 0$, which similarly implies $z_n=0$. Therefore, we conclude that $z=0$, which finishes the proof.