The Baker-Campbell-Hausdorff series is composed by a sum in the form $C(X,Y) = X + Y + \sum_n C_n(X,Y),$ where $C_n(X,Y)$ are terms formed by commutators of $X$ and $Y$ in a $n-$dimensional Lie algebra $\mathfrak{g}$. This series converges if $|X|, |Y| < \rho$, for some $\rho>0$ small enough. My question is: consider $\mathfrak{g}$ a solvable Lie algebra endowed with a Riemannian metric. If we take $X,Y,Z \in \mathfrak{g}$ in which $|X|,|Y|,|Z|<\rho$ such that $C(X,Y)$ and $C(Z,Y)$ are convergent, is the series $C(X+Y,Z)$ still convergent?
I believe it is, given the following arguments: the original BCH series is given by \begin{equation} C(X,Y) = \sum_{k \geq 0} \frac{(-1)^{k+1}}{k} \left(\sum_{n \geq 0} \sum_j^{n}\frac{1}{j!(n-j)!}X^{n-j}Y^j\right)^k, \end{equation}
If we consider $C(X+Y,Z)$, for some $X,Y,Z \in \mathfrak{g}$ such that $|X|,|Y|,|Z| < \rho$, we get \begin{equation*} C(X+Y,Z) = \sum_{k \geq 0} \frac{(-1)^{k+1}}{k} \left(\sum_{n \geq 0} \sum_j^{n}\frac{1}{j!(n-j)!}(X+Y)^{n-j}Z^j\right)^k, \end{equation*}
The expression $(X+Y)^{n-j}$ is a homogeneous polynomial of degree $n-j$. Then $(X+Y)^{n-j} = \sum_{i} P_i(X,Y)$, where each $P_i$ is a polynomial in the form $k_{i,j} X^{i} Y^{n-j-i}$ or $k_{i,j} Y^{i} X^{n-j-i}$ where $k_{i,j} = \binom{n-j}{i}$, given the non-commutative property. Thus \begin{equation*} C(X+Y,Z) = \sum_{k \geq 0} \frac{(-1)^{k+1}}{k} \left(\sum_{n \geq 0} \sum_j^{n} \sum_{i}\frac{1}{j!(n-j)!} P_i(X,Y)Z^j\right)^k, \end{equation*} which (I believe) converges if $|X|,|Y|,|Z| < \rho$. More preciselly, we can use the expression provided by Wyss (ref https://arxiv.org/abs/1707.03861 Theorem 1) which splits $(X+Y)^{n-j}$ in the form. \begin{equation*} (X+Y)^{n-j} = \sum_{i}^{n-j} \binom{n-j}{i}\left( (X + ad_Y)^i Y^{n-j-i}\right). \end{equation*}
I know that if $|X+Y|< \rho$, it works perfectly.
Other question: and for a $C(X_1 + ... + X_m, Y)$, with $|X_i|, |Y|< \rho$?
If there is any hypothesis I need to add, please let me know. Any help is appreciated. Thank you.