Criterion to construct a $\mathbb{Z}$-basis of a free $\mathbb{Z}$-Lie algebra

Let $$L(\mathbb{Z},n)$$ (resp. $$L(\mathbb{Q},n)$$) be the free Lie algebra over $$\mathbb{Z}$$ (resp. over $$\mathbb{Q}$$) with generating set $$\{x_1,\dots,x_n\}$$.

Let $$\mathcal B$$ be a $$\mathbb{Q}$$-basis of $$L(\mathbb{Q},n)$$ consisting of elements of the form $$[x_{i_1},[x_{i_2},\dots,[x_{i_{r-1}},x_{i_r}]\dots]]$$ for some $$i_1,\dots,i_r\in\{1,\dots,n\}$$. Is $$\mathcal B$$ automatically a $$\mathbb{Z}$$-basis of $$L(\mathbb{Z},n)$$?

• I'd avoid writing $n$ as index, since it often denotes the degree $n$ subspace of the Lie algebra. Rather write $L(A,n)$? then $L_d(A,n)$ is generated by $d$-fold brackets. Then the question should be checkable for small $d$: $d=2,3$ work. $d=4$ should be doable by hand, and for higher $d$ computer check might help.
– YCor
Aug 7 '19 at 14:03

Take for instance $$n=3$$, and attribute the degree $$\alpha_i$$ to $$x_i$$, so that $$L(\mathbb{Q},3)$$ becomes graded by the free $$\mathbb{Z}$$-module $$\oplus_{i=1}^3\mathbb{Z}\alpha_i$$. Then the homogeneous component of degree $$2\alpha_1+\alpha_2+\alpha_3$$ of $$L(\mathbb{Q},3)$$ has dimension $$3$$, and admits the $$\mathbb{Q}$$-basis $$\mathcal B=\{[x_1,[x_1,[x_2,x_3]]], [x_2,[x_1,[x_1,x_3]]],[x_3,[x_1,[x_1,x_2]]]\}$$. However, $$[x_1,[x_3,[x_1,x_2]]]=-\frac{1}{2}[x_1,[x_1,[x_2,x_3]]]+\frac{1}{2}[x_2,[x_1,[x_1,x_3]]]+\frac{1}{2}[x_3,[x_1,[x_1,x_2]]].$$
• OK, so indeed the counterexample comes as soon as $d=4$ (as soon as $n\ge 3$).