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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.

UPD: I hope that the ideal $(x)$ is freely generated as a Lie algebra by the elements $$x,[x,y], [x,y,y],\dots.$$
In this case I know how to prove this. We can 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 $L$ 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 $$V'_n={\rm span}(z_0,\dots,z_n).$$

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.

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.

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.

UPD: I hope that the ideal $(x)$ is freely generated as a Lie algebra by the elements $$x,[x,y], [x,y,y],\dots.$$
In this case I know how to prove this. We can 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 $L$ 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 $$V'_n={\rm span}(z_0,\dots,z_n).$$

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 $$\bigcup_{n} V_n={\rm id}(x),$$ where ${\rm id}(x)$ is the ideal of $L$ generated by $x.$

Question: How to prove that $${\rm id}(x)\ne V_n$$ 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.

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.