Let $L$ be a free Lie algebra (over $\mathbb{Q}$) on generators $x\_1, x\_2, \ldots, x\_n$, and let $V\_k$ be the subspace spanned by the $k$-fold brackets. Let $U\_1 = \mathrm{span}\{ x\_i | i< n\}\subseteq V\_1$ and let $W\_n \subseteq V\_n$ be the span of the $n$-fold brackets that involve $x\_n$ at least once.


The bracket defines a function $V\_k\otimes V\_\ell \to V\_{k+\ell}$ (which is what I think of as bracketing and rewriting in terms of a chosen basis, but actually it's just bracketing). 


Question: Is the composition 
$$
U\_1 \otimes W\_m \to V\_1\otimes V\_m \to V\_{m+1}
$$
injective? ($m$ has no particular relation to $n$)


I have asked MAPLE to compute the dimensions of these vector spaces and it seems that the dimension of the target is generally large enough to accommodate such an injection (though I haven't proved a dimension inequality yet).



Injectivity of Rewrite Rule in a Free Lie Algebra
Yes. Embed the free Lie algebra in the free associative algebra, then consider the monomial with $x\_n$ as far right as possible (i.e., with the largest number of other $\{x\_i,i\leq n\}$ before it): it won't be cancelled. Let me now give details.


Consider $F$ the free (associative but non-commutative) algebra (over $\mathbb{Q}$) on generators $x\_1,\ldots,x\_n$. Let $[P,Q]\_F = P Q - Q P \in F$ be the usual commutator for elements $P,Q\in F$. Consider $L\_F$ the smallest subset of $F$ containing $x\_1,\ldots,x\_n$ which is closed under commutators and linear combinations. Then $L\_F$ endowed with the Lie bracket $[\cdot,\cdot]\_F$ is isomorphic to the free Lie algebra $L$.


Under this isomorphism, $U\_1$ is still the span of $x\_1,\ldots,x\_{n-1}$, and $W\_m$ is included in the vector space $W^F\_m$ of polynomials in $x\_1,\ldots,x\_n$ with no monomial independent of $x\_n$. An element of $U\_1\otimes W^F\_m$ is $\sum\_{i=1}^{n-1} x\_i \otimes P\_i$ for some $P\_i\in W^F\_m$. Its image under the composition you care about is $S = \sum\_{i=1}^{n-1} (x\_i P\_i - P\_i x\_i)$. Since everything is linear, we just need to show that $S \neq 0$ assuming that not all $P\_i = 0$.


Among all monomials of all $P\_i$, consider $m = x\_{a\_1} \cdots x\_{a\_k} x\_n x\_{b\_1} \cdots x\_{b\_l}$ with $k$ maximal. Say it is in $P\_j$. Then $S$ contains the monomial $x\_j m = x\_j x\_{a\_1} \cdots x\_{a\_k} x\_n x\_{b\_1} \cdots x\_{b\_l}$ with some non-zero coefficient coming from $x\_j P\_j$, and this monomial does not appear in any other term of $S$. Indeed, the monomial cannot appear in $x\_i P\_i$ for $i\neq j$ since it starts with $x\_j$ and not $x\_i$, and the monomial cannot appear in $P\_i x\_i$ for any $i$ since this would require the presence in $P\_i$ of a monomial $m'$ starting with $x\_j x\_{a\_1} \cdots x\_{a\_k} x\_n$, which contradicts the maximality of $k$.


It is probably possible to extend the proof to show that $U\_k \otimes W\_m \to V\_{k + m}$ is injective, where $U\_k$ is the span of $(k-1)$-fold brackets of $\{x\_i, i<n\}$.


