Obtaining higher order terms of the Baker-Campbell-Hausdorff formula as higher order commutators Let $G$ be a Lie group, then $(X,Y) \mapsto \log(\exp(X)\exp(Y))$ locally defines a group law on a neighbourhood of  $0 \in T_eG$. This local group law is analytic, and the first term is given by $X + Y$ and the second by $\frac{1}{2}[X,Y]$. According to Tits' textbook on Lie groups the Lie bracket may be obtained as the second order derivative of $G \times G \to G, \; (x,y) \mapsto xyx^{-1}y^{-1}$. (This is well-defined, because the first derivative of $(x,y) \mapsto xyx^{-1}y^{-1}$ vanishes.) My question is then: 

Are there maps $\pi_r: G \times G \to G$ for every $r \in \mathbb{N}$ such that $d^k \pi_r$ vanishes for all $k < r$ (so that we obtain a well defined $r$-linear form $d^r \pi_r: T_eG \times \underbrace{\dots}_{2 r\, \times} \times T_eG \to T_eG$), such that the $r$-th term in the Baker-Campbell-Hausdorff formula is given by $\frac{1}{r!}d^r \pi_r$?
  Furthermore, can the maps $\pi_r$ be constructed from the multiplication and inverse maps on $G$?

 A: Thm 3.16 in Peter Michor's book "Topics in differential geometry" describes how on any smooth manifold iterated Lie brackets of vector fields can be computed from higher order derivatives of commutators of flows of the vector fields on the manifold.
In the case of Lie groups, the flow $\Phi_X:G\times\mathbb{R}\to G$ of a left-invariant vector field starting from a point $p\in G$ is given simply by $\Phi_X(p,t)=p\exp(tX)$. So in this case the commutators of flows from the identity $e\in G$ can be written using the group commutators $(x,y)=xyx^{-1}y^{-1}$ as
$$\Phi_Y^{-1}\circ\Phi_X^{-1}\circ\Phi_Y\circ\Phi_X(e,t) = \exp(tX)\exp(tY)\exp(-tX)\exp(-tY) = (\exp(tX),\exp(tY)).$$
The individual iterated brackets involved in the Baker-Campbell-Hausdorff formula are then given by higher order derivatives of these. For example a bracket of length three can be obtained from
\begin{align}
[X,[X,Y]] &= \frac{1}{6}\frac{d^3}{dt^3}(x,(x,y))\Big\vert_{t=0}\\
&=\frac{1}6\frac{d^3}{dt^3}x(x,y)x^{-1}(x,y)^{-1}\Big\vert_{t=0}\\
&=\frac{1}6\frac{d^3}{dt^3}xxyx^{-1}y^{-1}x^{-1}yxy^{-1}x^{-1}\Big\vert_{t=0},
\end{align}
where $x=\exp(tX)$ and $y=\exp(tY)$.
In order to get the entire $r$-th term of the BCH formula from a single map, you can then take products of these group commutators. For instance if $\alpha(t)$ and $\beta(t)$ are commutators with $\frac{d^3}{dt^3}\alpha(t)\Big\vert_{t=0}=[X,[X,Y]]$ and $\frac{d^3}{dt^3}\beta(t)\Big\vert_{t=0}=[Y,[Y,X]]$ and all their lower order derivatives vanish, then also all the lower order derivatives of $\alpha\beta$ vanish, and
$$\frac{d^3}{dt^3}\alpha(t)\beta(t)\Big\vert_{t=0} = \frac{d^3}{dt^3}\alpha(t)\Big\vert_{t=0}+\frac{d^3}{dt^3}\beta(t)\Big\vert_{t=0} = [X,[X,Y]]+[Y,[Y,X]].$$
