Let $\mathfrak{g}$ be a finite-dimensional nilpotent Lie algebra over an algebraically closed field $k$ of characteristic zero. Throughout, let $x,y$ and $z$ be elements of $\mathfrak{g}$. The Baker-Campbell-Hausdorff series
$$ H(x,y):= x+y +\tfrac{1}{2}[x,y] + \text{(summands with iterated brackets)}$$
is finite. It defines a unipotent group structure $(x,y) \mapsto x \star y := H(x,y)$ on $\mathfrak{g}$ with neutral element $0$ and inverse $x^{-1}:=-x$. If $\mathfrak{g}=\mathrm{Lie}(G)$ for a unipotent group $G$, then the exponential map becomes an isomorphism of algebraic groups (with respect to $\star$), since $\exp(x) \exp(y)=\exp(H(x,y))=\exp(x\star y)$.
Having a group structure on $\mathfrak{g}$, we can define commutators $\langle x,y \rangle:= x\star y \star x^{-1} \star y^{-1} = H(H(x,y),H(-x,-y))$ and ask what is their relation to the bracket $[x,y]$. The Lie bracket is trivial for all $x$ and $y$ if and only if the commutator is trivial for all $x$ and $y$ (the group $G$ is commutative if and only if the Lie algebra $\mathrm{Lie}(G)$ is). It is also easy to prove that if the iterated brackets vanish (meaning that $[x,[y,z]]=0$ for all $x,y,z \in \mathfrak{g}$), then actually $\langle x,y \rangle = [x,y]$ for all $x,y$ and, in particular, also the iterated commutators vanish. I would like to know if the converse is also true, i.e., if vanishing of the iterated commutators implies vanishing of the iterated brackets. Computing this explicitly seems like a nightmare (I do not know of a compact description for $H(x,y)$), so I do not know how to go about this. Actually, are the three following conditions equivalent?
- $[x,[y,z]] = 0$ for all $x,y,z$
- $[x,y]=\langle x,y \rangle$ for all $x,y$
- $\langle x, \langle y,z \rangle \rangle= 0$ for all $x,y,z$
