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Let $L=L(x,y)$ be the free Lie algebra generated by letters $x,y.$ For a vector subspace $V\leq L$ we denote by $[V,L]$ the vector space spanned by brackets $[v,l],v\in V,l\in L.$ A vector subspace $V\leq L$ is an ideal of $L$ if and only if $$V=V+[V,L].$$ Consider the following increasing sequence of vector spaces which starts from the ${\rm span}(x):$ $$V_0={\rm span}(x),$$ $$V_1=V_0+[V_0,L],$$ $$\dots$$ $$V_{n+1}=V_n+[V_n,L].$$ Then $$\bigcup_{n} V_n=(x),$$ where $(x)$ is the ideal of $L$ generated by $x.$

Question: How to prove that $$(x)\ne V_n$$ for any $n$? For example, I believe that the left-normed commutator $[x,\underbrace{y,\dots,y}_{n+1}]$ is not an element of $V_n.$ But I can't prove this.

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    $\begingroup$ Please don't use $\operatorname{id}$ for anything other than the identity map -- it's just too confusing. $\endgroup$ Mar 4 '20 at 15:09
  • $\begingroup$ Anyway, is $V = \operatorname{span}\left(x\right)$ here? Then you can just argue that $V_n$ lives in homogeneous degree $n$ and thus cannot cover the whole ideal. (Of course, you'd have to verify that the ideal has a nontrivial component in all positive degrees; but this is clear from the embedding of the free Lie algebra into the tensor algebra.) $\endgroup$ Mar 4 '20 at 15:11
  • $\begingroup$ No, $V_n$ lives in all homogeneous degrees for $n\geq 1$. For example, $V_1$ has the following element $[x,[x,y,...,y]].$ $\endgroup$ Mar 4 '20 at 15:37
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    $\begingroup$ Ah, I see -- it bounds the number of $x$'s, not the total number of factors. Then it's a less trivial question than I expected. Still, any explicit basis of the free Lie algebra (e.g., using Lyndon words) should do the trick. $\endgroup$ Mar 4 '20 at 15:42
  • $\begingroup$ I tried to use this explicit basis but I didn't succeed. $\endgroup$ Mar 4 '20 at 15:47
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Now I know the answer on my question. First of all we note that the ideal $(x)$ is freely generated by the elements $$x,[x,y],[x,y,y],\dots$$ as a Lie algebra. Indeed, it is easy to check that they generate the Lie algebra $(x)$ because any element of the Lyndon basis except $y$ lies in the algebra generated by these elements. The fact that the sequence is free follows from Theorem 2.9 of Section "Free sets of Lie polynomials" of Reutenauer's book. Indeed, these elements can be presented in $K\langle x,y \rangle$ as $$[x,\underbrace{y,\dots,y}_n]=\sum_{i=0}^n (-1)^{i} \binom{n}{i} y^ixy^{n-i}.$$ It is easy to check that they are right linearly independent over $K\langle x,y \rangle$.

Further, we consider the quotient $$M=(x)/[(x),(x)]$$ as a module over $L.$ Then the elements $$z_i=[x,\underbrace{y\dots,y}_i]+[(x),(x)]$$ form a basis of the vector space $M.$ The action of $L$ on $M$ is given by $[z_i,x]=0$ and $[z_i,y]=z_{i+1}.$ Then, if we denote by $V'_n$ the image of $V_n$ in $M,$ we obtain $$V'_n={\rm span}(z_0,\dots,z_n).$$

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