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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 $L(\mathbb{Z},n)$?

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  • $\begingroup$ 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. $\endgroup$
    – YCor
    Aug 7 '19 at 14:03
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It turns out my intuition was wrong and the answer is "no".

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, $$[x_1,[x_3,[x_1,x_2]]]=-\frac{1}{2}[x_1,[x_1,[x_2,x_3]]]+\frac{1}{2}[x_2,[x_1,[x_1,x_3]]]+\frac{1}{2}[x_3,[x_1,[x_1,x_2]]].$$

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  • $\begingroup$ OK, so indeed the counterexample comes as soon as $d=4$ (as soon as $n\ge 3$). $\endgroup$
    – YCor
    Aug 7 '19 at 19:12

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