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 $\langle x,y \rangle = [x,y]$ 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?

1. $[x,[y,z]] = 0$ for all $x,y,z$
2. $[x,y]=\langle x,y \rangle$ for all $x,y$
3. $\langle x, \langle y,z \rangle \rangle= 0$ for all $x,y,z$

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?

1. $[x,[y,z]] = 0$ for all $x,y,z$
2. $[x,y]=\langle x,y \rangle$ for all $x,y$
3. $\langle x, \langle y,z \rangle \rangle= 0$ for all $x,y,z$

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 bracket is trivial if and only if the commutator is trivial (since the group structure is unipotent if and only if the Lie algebra structure is nilpotent). 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 $\langle x,y \rangle = [x,y]$ 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?

1. $[x,[y,z]] = 0$ for all $x,y,z$
2. $[x,y]=\langle x,y \rangle$ for all $x,y$
3. $\langle x, \langle y,z \rangle \rangle= 0$ for all $x,y,z$

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 $\langle x,y \rangle = [x,y]$ 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?

1. $[x,[y,z]] = 0$ for all $x,y,z$
2. $[x,y]=\langle x,y \rangle$ for all $x,y$
3. $\langle x, \langle y,z \rangle \rangle= 0$ for all $x,y,z$