Symmetric polynomials in free center-by-metabelian Lie algebras of rank 2

1. Introduction and Notation

Let K be a field of characteristic zero. We write K-vector (sub)space instead of vector (sub)space over K. By “Lie algebra” we mean a Lie algebra over K. Let L be a Lie algebra. For $a,b\in L$, we write $[a,b]$ for the Lie commutator of a and b. For $c\ge 3$ and for elements $a_1,\dots ,a_c$ of L, we define the left-normed Lie commutator $[a_1,\dots ,a_{c-1},a_c]=[[a_1,\dots ,a_{c-1}],a_c]$. For a non-negative integer m and for $a,b\in L$, we write $[a,{}_{m}b]=[a,b,\dots ,b]$, where b is repeated m-times (for $m=0$, we write $[a,{}_{0}b]=a$). For K-vector subspaces A and B of L, we write $[A,B]$ for the K-vector subspace of L spanned by all the elements $[a,b]$, with $a\in A$ and $b\in B$. We denote $L^{\prime }={\gamma }_2(L)$, that is, $L^{\prime }$ is the derived algebra of L, and $L^{\prime \prime }$ $=$ ${(L^{\prime })}^{\prime }$. For a non-empty subset X of L, we write $L(X)$ for the Lie subalgebra of L generated by the set X.

Let L be a finitely generated Lie algebra with generators $a_1,\dots ,a_n$, $n\ge 2$. We say that an element $f(a_1,\dots ,a_n)\in L$ is a symmetric polynomial in L if $f(a_1,\dots ,a_n)=f(a_{\sigma (1)},\dots ,a_{\sigma (n)})$ for all $\sigma \in S_n$. We write $P_L$ for the set of all symmetric polynomials in L. Since for $f_1(a_1,\dots ,a_n),f_2(a_1,\dots ,a_n)\in P_L$, $[f_1(a_1,\dots ,a_n),f_2(a_1,\dots ,a_n)]\in P_L$, $P_L$ is a Lie subalgebra of L. We call $P_L$ the Lie subalgebra of symmetric polynomials in L.

For a positive integer n, with $n\ge 2$, let $L_n$ be the free Lie algebra of rank n freely generated by the set $\{{\ell }_1,\dots ,{\ell }_n\}$. It has been proved in [2] that $P_{L_n}$ is infinitely generated. The generating series for the number of free generators of the Lie subalgebra of invariants of the free Lie algebra under the action of a finite group is found in [10] (and in the case of prime characteristic, the growth of this series under the action of a homogeneous finite group of automorphisms has been studied in [11]). Let $M_n=L_n∕L_n^{\prime \prime }$, that is, $M_n$ is the free metabelian Lie algebra, freely generated by the set $\{y_1,\dots ,y_n\}$, with $y_i={\ell }_i+L_n^{\prime \prime }$, $i=1,\dots ,n$. It has been proved in [4, Theorem 2.3] that $P_{M_n}$ is infinitely generated (the analogous result for the Lie subalgebra of symmetric polynomials in the free metabelian Lie algebra over a field of prime characteristic has been proved in [3]). Furthermore, generating sets for $P_{M_n}$ are given in [6] for $n=2$ and in [5] for $n\ge 3$. In particular, in [6], it has been proved that the set

is a basis of the Lie algebra of symmetric polynomials in the free metabelian Lie algebra of rank 2 $P_{M_2}$ and the set is a minimal set of generators for this Lie algebra.

Our aim is to extend the results of [6] in the case of free center-by-metabelian Lie algebras of rank 2. The free center-by-metabelian Lie algebra $C_2$ of rank 2 is the quotient $C_2=L_2∕[L_2^{\prime \prime },L_2]$ freely generated by the set $\{x_1,x_2\}$, where $x_1={\ell }_1+[L_2^{\prime \prime },L_2]$ and $x_2={\ell }_2+[L_2^{\prime \prime },L_2]\}$. Since $P_{C_2}$ may be identified with the Lie algebra of invariants of the ${\mathbb{Z}}_2$-action permuting $x_1$ and $x_2$, it follows from [3] that $P_{C_2}$ is infinitely generated. Throughout the text, for non-negative integers m and n, we use the following notation :

and In Sec. 3, we prove that the set is a basis of the Lie algebra of symmetric polynomials in the free center-by-metabelian Lie algebra of rank 2 $P_{C_2}$ and in Sec. 4.2, we show that the set is a minimal generating set for this Lie algebra.

2. The Lie Algebra $C_2$

The proof of the next result is elementary. We point out that Lemma 1 is a special case of [7, Lemma 2.2].

Lemma 1. Let $C_2$ be the free center-by-metabelian Lie algebra of rank 2 freely generated by the set $\{x_1,x_2\}$.

(1)

Let $v\in C_2^{\prime }$ and $u_1,\dots ,u_m\in C_2,$ with $m\ge 3.$ Then, for $\sigma \in S_{m-1},$ $$[v,u_1,\dots ,u_{m-1},u_m]=[v,u_{\sigma (1)},\dots ,u_{\sigma (m-1)},u_m].$$

(2)

For $\sigma \in S_r,$ $v_1,v_2\in C_2^{\prime }$ and $z_1,\dots ,z_r\in \{x_1,x_2\},$ $$[v_1,z_1,\dots ,z_r,v_2]={(-1)}^{r+1}[v_2,z_{\sigma (1)},\dots ,z_{\sigma (r)},v_1].$$ In particular, if $v_1=v_2$ and r is even, then $[v_1,z_1,\dots ,z_r,v_1]=0$. Furthermore, for $(\sigma ,\tau )\in S_r\times S_t,$ $v_1^{\prime },v_2^{\prime }\in C_2^{\prime }$ and $z_1,\dots ,z_r,w_1,\dots ,w_t\in \{x_1,x_2\},$ $$\begin{array}{ccc}\hfill [v_1^{\prime },z_1,\dots ,z_r,[v_2^{\prime },w_1,\dots ,w_t]]& \hfill =\hfill & {(-1)}^{r+1}[v_2^{\prime },z_{\sigma (1)},\dots ,z_{\sigma (r)},w_{\tau (1)},\dots ,w_{\tau (t)},v_1^{\prime }]\hfill \\ \hfill & \hfill =\hfill & {(-1)}^t[v_1^{\prime },z_{\sigma (1)},\dots ,z_{\sigma (r)},w_{\tau (1)},\dots ,w_{\tau (t)},v_2^{\prime }].\hfill \end{array}$$

For non-negative integers a and $b,$ with $a\ge b,$ we write $\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right)=\frac{a!}{b!(a-b)!},$ where $0!=1.$ For $a\ge b-1\ge 0,$ we point out that $\left(\begin{array}{c}\hfill a\hfill \\ \hfill b-1\hfill \end{array}\right)+\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right)=\left(\begin{array}{c}\hfill a+1\hfill \\ \hfill b\hfill \end{array}\right)$.

Lemma 2. Let $C_2$ be the free center-by-metabelian Lie algebra of rank 2 freely generated by the set $\{x_1,x_2\}.$

(1)	For positive integers m and n, $$[x_2,x_1,{}_m{x}_1,{}_n{x}_2]=[x_2,x_1,{}_n{x}_2,{}_m{x}_1]-[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,[x_2,x_1]].$$ In particular, for $m+n$ even, $$[x_2,x_1,{}_m{x}_1,{}_n{x}_2]=[x_2,x_1,{}_n{x}_2,{}_m{x}_1].$$
(2)	For positive integers m and n, with $n\ge 2,$ $$\begin{array}{ccc}\hfill [x_2,x_1,{}_m{x}_1{,}_n{x}_1+x_2]& \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1{,}_{m+n-k}{x}_1{,}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k-1}{x}_1,{}_{k-1}{x}_2,[x_2,x_1]].\hfill \end{array}$$ In particular, for $m+n$ even, $$[x_2,x_1,{}_m{x}_1{,}_n{x}_1+x_2]=\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1{,}_{m+n-k}{x}_1{,}_k{x}_2].$$
(3)	For positive integers m and n, with $n\ge 2,$ $$\begin{array}{ccc}\hfill [x_2,x_1,{}_m{x}_2,{}_n{x}_1+x_2]& \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k}{x}_2,{}_k{x}_1]\hfill \\ \hfill & \hfill \hfill & -\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{k-1}{x}_1,{}_{m+n-k-1}{x}_2,[x_2,x_1]].\hfill \end{array}$$ In particular, for $m+n$ even, $$[x_2,x_1,{}_m{x}_2{,}_n{x}_1+x_2]=\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1{,}_{m+n-k}{x}_2{,}_k{x}_1].$$

Proof.

(1)	By Lemma 1(1), we get $$[x_2,x_1,{}_m{x}_1,{}_n{x}_2]=[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,x_1,x_2].$$ Using the Jacobi identity in the form $[a,b,c]=[a,c,b]+[a,[b,c]]$, we get $$\begin{array}{ccc}\hfill & \hfill \hfill & [x_2,x_1,{}_m{x}_1,{}_n{x}_2]=[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,x_2,x_1]\hfill \\ \hfill & \hfill \hfill & +[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,[x_1,x_2]]=[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,x_2,x_1]\hfill \\ \hfill & \hfill \hfill & -[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,[x_2,x_1]]=[x_2,x_1,{}_{m-1}{x}_1,{}_n{x}_2,x_1]\hfill \\ \hfill & \hfill \hfill & -[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,[x_2,x_1]].\hfill \end{array}$$ Again, by Lemma 1(1), $$[x_2,x_1,{}_{m-1}{x}_1,{}_n{x}_2,x_1]=[x_2,x_1,{}_n{x}_2,{}_{m-1}{x}_1,x_1]=[x_2,x_1,{}_n{x}_2,{}_m{x}_1]$$ and hence, by the above equation, $$[x_2,x_1,{}_m{x}_1,{}_n{x}_2]=[x_2,x_1,{}_n{x}_2,{}_m{x}_1]-[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,[x_2,x_1]].$$ In the case $m+n$ is even, by Lemma 1(2), $$[x_2,x_1,{}_{m-1}{x}_1,{}_{n-1}{x}_2,[x_2,x_1]]=0$$ and hence, by the above equation, $$[x_2,x_1,{}_m{x}_1,{}_n{x}_2]=[x_2,x_1,{}_n{x}_2,{}_m{x}_1].$$
(2)	Throughout the proof of Lemma 2(2), we write $u=[x_2,x_1]$. We use induction on n. For $n=2$, $$\begin{array}{ccc}\hfill [u,{}_m{x}_1,{}_2{x}_1+x_2]& \hfill =\hfill & [u,{}_{m+2}{x}_1]+[u,{}_{m+1}{x}_1,x_2]+[u,{}_m{x}_1,x_2,x_1]\hfill \\ \hfill & \hfill \hfill & +[u,{}_m{x}_1,{}_2{x}_2].\hfill \end{array}$$ Since, by the Jacobi identity in the form $[a,b,c]=[a,c,b]+[a,[b,c]]$, $$[u,{}_m{x}_1,x_2,x_1]=[u,{}_{m+1}{x}_1,x_2]+[u,{}_m{x}_1,u]$$ it follows from the above equation that $$\begin{array}{ccc}\hfill [u,{}_m{x}_1,{}_2{x}_1+x_2]& \hfill =\hfill & [u,{}_{m+2}{x}_1]+2[u,{}_{m+1}{x}_1,x_2]\hfill \\ \hfill & \hfill \hfill & +[u,{}_m{x}_1,{}_2{x}_2]+[u,{}_m{x}_1,u]\hfill \end{array}$$ that is, $$[u,{}_m{x}_1,{}_2{x}_1+x_2]=\sum _{k=0}^2\left(\begin{array}{c}\hfill 2\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+2-k}{x}_1,{}_k{x}_2]+[u,{}_m{x}_1,u]$$ and hence, the result is true for $n=2$. Assume that the result is true for some $n\ge 2$, that is, assume that $$\begin{array}{ccc}\hfill [u,{}_m{x}_1,{}_n{x}_1+x_2]& \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k-1}{x}_1,{}_{k-1}{x}_2,u].\hfill \end{array}$$ Then, by our inductive argument, $$\begin{array}{ccc}\hfill & \hfill \hfill & [u,{}_m{x}_1,{}_{n+1}{x}_1+x_2]=[u,{}_m{x}_1,{}_n{x}_1+x_2,x_1+x_2]\hfill \\ \hfill & \hfill \hfill & =\left[\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_k{x}_2]\right.\hfill \\ \hfill & \hfill \hfill & \left.+\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k-1}{x}_1,{}_{k-1}{x}_2,u],x_1+x_2\right].\hfill \end{array}$$ Since $[C_2^{\prime \prime },C_2]=\{0\}$, we have $$[[u,{}_{m+n-k-1}{x}_1,{}_{k-1}{x}_2,u],x_1+x_2]=0$$ and hence, by the above equation, we get $$\begin{array}{ccc}\hfill [u,{}_m{x}_1,{}_{n+1}{x}_1+x_2]& \hfill =\hfill & \left[\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_k{x}_2],x_1+x_2\right]\hfill \\ \hfill & \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_k{x}_2,x_1+x_2].\hfill \end{array}$$ Hence, $$\begin{array}{ccc}\hfill & \hfill \hfill & [u,{}_m{x}_1,{}_{n+1}{x}_1+x_2]=[u,{}_{m+n}{x}_1,x_1+x_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_k{x}_2,x_1+x_2]\hfill \\ \hfill & \hfill \hfill & =[u,{}_{m+n+1}{x}_1]+[u,{}_{m+n}{x}_1,x_2]+\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_k{x}_2,x_1]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k+1}{x}_2].\hfill \end{array}$$ Since, by Lemma 2(1), $$[u,{}_{m+n-k}{x}_1,{}_k{x}_2,x_1]=[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]+[u,{}_{m+n-k}{x}_1,{}_{k-1}{x}_2,u]$$ it follows from the above equation that $$\begin{array}{ccc}\hfill & \hfill \hfill & [u,{}_m{x}_1,{}_{n+1}{x}_1+x_2]\hfill \\ \hfill & \hfill \hfill & =[u,{}_{m+n+1}{x}_1]+[u,{}_{m+n}{x}_1,x_2]+\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k-1}{x}_2,u]+\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k+1}{x}_2]\hfill \\ \hfill & \hfill \hfill & =[u,{}_{m+n+1}{x}_1]+[u,{}_{m+n}{x}_1,x_2]+\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k-1}{x}_2,u]+\sum _{k=2}^{n+1}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k-1\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2].\hfill \end{array}$$ But, $$\begin{array}{ccc}\hfill & \hfill \hfill & \sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]+\sum _{k=2}^{n+1}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k-1\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & =n[u,{}_{m+n}{x}_1,x_2]+\sum _{k=2}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=2}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k-1\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]+[u,{}_m{x}_1,{}_{n+1}{x}_2]=n[u,{}_{m+n}{x}_1,x_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=2}^n\left\{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)+\left(\begin{array}{c}\hfill n\hfill \\ \hfill k-1\hfill \end{array}\right)\right\}[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]+[u,{}_m{x}_1,{}_{n+1}{x}_2]\hfill \\ \hfill & \hfill \hfill & =n[u,{}_{m+n}{x}_1,x_2]+\sum _{k=2}^n\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]+[u,{}_m{x}_1,{}_{n+1}{x}_2].\hfill \end{array}$$ Therefore, $$\begin{array}{ccc}\hfill & \hfill \hfill & [u,{}_m{x}_1,{}_{n+1}{x}_1+x_2]=[u,{}_{m+n+1}{x}_1]+[u,{}_{m+n}{x}_1,x_2]+n[u,{}_{m+n}{x}_1,x_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=2}^n\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k+1}{x}_1,{}_k{x}_2]+[u,{}_m{x}_1,{}_{n+1}{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k-1}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & =[u,{}_{m+n+1}{x}_1]+(n+1)[u,{}_{m+n}{x}_1,x_2]+\sum _{k=2}^n\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)\hfill \\ \hfill & \hfill \hfill & \times [u,{}_{m+n+1-k}{x}_1,{}_k{x}_2]+[u,{}_m{x}_1,{}_{n+1}{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k-1}{x}_2,u].\hfill \end{array}$$ Since $$\begin{array}{ccc}\hfill & \hfill \hfill & [u,{}_{m+n+1}{x}_1]+(n+1)[u,{}_{m+n}{x}_1,x_2]+\sum _{k=2}^n\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n+1-k}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +[u,{}_m{x}_1,{}_{n+1}{x}_2]=\sum _{k=0}^{n+1}\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n+1-k}{x}_1,{}_k{x}_2]\hfill \end{array}$$ it follows that $$\begin{array}{ccc}\hfill [u,{}_m{x}_1,{}_{n+1}{x}_1+x_2]& \hfill =\hfill & \sum _{k=0}^{n+1}\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n+1-k}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{m+n-k}{x}_1,{}_{k-1}{x}_2,u].\hfill \end{array}$$ Hence, Lemma 2 is also true for $n+1$ and hence, we get the required result. In the case $m+n$ is even, $m+n-k-1+k-1=m+n-2$ is also even. Hence, by Lemma 1(2) $[u,{}_{m+n-k-1}{x}_1,{}_{k-1}{x}_2,u]=0$ and hence, by the above equation, $$[x_2,x_1,{}_m{x}_1,{}_n{x}_1+x_2]=\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k}{x}_1,{}_k{x}_2].$$
(3)	Let $\sigma$ be the automorphism of $C_2$ defined by $\sigma (x_1)=x_2$ and $\sigma (x_2)=x_1$. Applying $\sigma$ on the equation $$\begin{array}{ccc}\hfill [x_2,x_1,{}_m{x}_1,{}_n{x}_1+x_2]& \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k-1}{x}_1,{}_{k-1}{x}_2,[x_2,x_1]],\hfill \end{array}$$ which has been proved in Lemma 2(2), we get $$\begin{array}{ccc}\hfill [x_1,x_2,{}_m{x}_2,{}_n{x}_2+x_1]& \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_1,x_2,{}_{m+n-k}{x}_2,{}_k{x}_1]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[x_1,x_2,{}_{m+n-k-1}{x}_2,{}_{k-1}{x}_1,[x_1,x_2]].\hfill \end{array}$$ Hence, $$\begin{array}{ccc}\hfill -[x_2,x_1,{}_m{x}_2,{}_n{x}_1+x_2]& \hfill =\hfill & -\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k}{x}_2,{}_k{x}_1]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k-1}{x}_2,{}_{k-1}{x}_1,[x_2,x_1]]\hfill \end{array}$$ that is, $$\begin{array}{ccc}\hfill [x_2,x_1,{}_m{x}_2,{}_n{x}_1+x_2]& \hfill =\hfill & \sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k}{x}_2,{}_k{x}_1]\hfill \\ \hfill & \hfill \hfill & -\sum _{k=1}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k-1}{x}_2,{}_{k-1}{x}_1,[x_2,x_1]].\hfill \end{array}$$ In the case $m+n$ is even, by Lemma 2(2), $$[x_2,x_1,{}_m{x}_1,{}_n{x}_1+x_2]=\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)[x_2,x_1,{}_{m+n-k}{x}_1,{}_k{x}_2].$$ Applying $\sigma$ on this equation, we may deduce the result.

 □

In the following, we give a basis of the free center-by-metabelian Lie algebra of rank 2 $C_2$. We point out that a basis of the free center-by-metabelian Lie algebra of finite rank was first determined in [8]. It was discovered afterwards that this paper is not completely correct and corrections have been given in [9]. In particular, in [9], it is determined a basis of the free center-by-metabelian Lie ring of finite rank over $\mathbb{Z}$.

Since $L_2∕L_2^{\prime }\cong C_2∕C_2^{\prime }$ in a natural way, the set $\{x_1+C_2^{\prime },x_2+C_2^{\prime }\}$ is a basis of the quotient $C_2∕C_2^{\prime }$. The derived algebra $L_2^{\prime }$ is a free Lie algebra, and the Lie commutators $[{\ell }_2,{\ell }_1,{}_m{\ell }_1,{}_n{\ell }_2]$, with $m,n\ge 0$, form a free generating set for $L_2^{\prime }$ (see, for example, [1, Sec. 2.4.2, p. 55]. We point out that in [2], it has been used the right-normed convention for Lie commutators). It follows that the set $\{[{\ell }_2,{\ell }_1,{}_m{\ell }_1,{}_n{\ell }_2]+L_2^{\prime \prime }:m,n\ge 0\}$ is a basis of the quotient $L_2^{\prime }∕L_2^{\prime \prime }$. Since $L_2^{\prime }∕L_2^{\prime \prime }\cong C_2^{\prime }∕C_2^{\prime \prime }$ in a natural way, the set $\{[x_2,x_1,{}_m{x}_1,{}_n{x}_2]+C_2^{\prime \prime }:m,n\ge 0\}$ is a basis of the quotient $C_2^{\prime }∕C_2^{\prime \prime }$. For a positive integer c, with $c\ge 5$, we write ${(C_2^{\prime \prime })}^{(c)}$ for the subspace of $C_2^{\prime \prime }$ spanned by all Lie commutators $[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]$, with $m+n=c-4$. Since ${(C_2^{\prime \prime })}^{(c)}$ is a subspace of $C_2^{\prime \prime }$ and, by Lemma 1(2), ${(C_2^{\prime \prime })}^{(c)}=\{0\}$ for c even, we obtain $C_2^{\prime \prime }={\oplus }_{\begin{array}{c}d\ge 5\\ d\mathrm{\text{odd}}\end{array}}{(C_2^{\prime \prime })}^{(d)}$. By Lemma 1(2), for an odd integer d, with $d\ge 5$, ${(C_2^{\prime \prime })}^{(d)}$ is spanned by all Lie commutators $[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]$, with $m,n\ge 0$ and $m+n=d-4$. It follows from [9, Theorems 7.1 and 7.2] that the set $\{[x_2,x_1,{}_{m_1}{x}_1,{}_{m_2}{x}_2,[x_2,x_1]],m_1+m_2=d-4\}$ is a basis of ${(C_2^{\prime \prime })}^{(d)}$.

In the following proposition, we summarize the aforementioned results. We point out that $C_2$ is a graded Lie algebra, $C_2={\oplus }_{c\ge 1}{C}_2^{(c)}$, where $C_2^{(c)}$ is the K-vector subspace of $C_2$ spanned by all Lie commutators of $C_2$ of homogeneous degree c.

Proposition 1. Let $C_2$ be the free center-by-metabelian Lie algebra of rank 2 freely generated by the set $\{x_1,x_2\}.$ Then

(1)	The set $\{x_1,x_2\}$ is a basis of the K-vector subspace $C_2^{(1)}.$
(2)	The set $\{[x_1,x_2]\}$ is a basis of the K-vector subspace $C_2^{(2)}.$
(3)	The set $\{[x_2,x_1,x_1],[x_2,x_1,x_2]\}$ is a basis of the K-vector subspace $C_2^{(3)}.$
(4)	For d even, with $d\ge 4,$ the set $\{[x_2,x_1,{}_m{x}_1{,}_n{x}_2]:m,n\ge 0,m+n=d-2\}$ is a basis of the K-vector subspace $C_2^{(d)}.$
(5)	For d odd, with $d\ge 5,$ the set $$\begin{array}{ccc}\hfill & \hfill \hfill & \{[x_2,x_1,{}_{m_1}{x}_1,{}_{n_1}{x}_2],[x_2,x_1,{}_{m_2}{x}_1,{}_{n_2}{x}_2,[x_2,x_1]]:\hfill \\ \hfill & \hfill \hfill & m_1,n_1,m_2,n_2\ge 0,m_1+n_1=d-2,m_2+n_2=d-4\}\hfill \end{array}$$ is a basis of the K-vector subspace $C_2^{(d)}$.
(6)	The set $$\begin{array}{ccc}\hfill & \hfill \hfill & \{x_1,x_2,[x_2,x_1,{}_{m_1}{x}_1,{}_{n_1}{x}_2],[x_2,x_1,{}_{m_2}{x}_1,{}_{n_2}{x}_2,[x_2,x_1]]:m_1,n_1,m_2,\hfill \\ \hfill & \hfill \hfill & \times n_2\ge 0,m_2+n_2\mathrm{\text{odd}}\}\hfill \end{array}$$ is a basis of $C_2$. In particular, the set $\{x_1+C_2^{\prime },x_2+C_2^{\prime }\}$ is a basis of the quotient $C_2∕C_2^{\prime },$ the set $\{[x_2,x_1,{}_m{x}_1,{}_n{x}_2]+C_2^{\prime \prime }:m,n\ge 0\}$ is a basis of the quotient $C_2^{\prime }∕C_2^{\prime \prime }$ and the set $\{[x_2,x_1{,}_{m_1}{x}_1,{}_{m_2}{x}_2,[x_2,x_1]]:m_1,m_2\ge 0,m_1+m_2\mathrm{\text{odd}}\}$ is a basis of $C_2^{\prime \prime }.$

3. A Basis of $P_{C_2}$

We show at first that there is a Lie algebra epimorphism from $P_{C_2}$ onto $P_{M_2}$. This is a special case of the following more general result.

Proposition 2. Let $L_2$ be the free Lie algebra of rank 2 freely generated by the set $\{l_1,l_2\},$ let $V(L_2)$ be a fully invariant ideal of $L_2$ such that $V(L_2)\subseteq L_2^{\prime \prime }$ and let $G_2=L_2∕V(L_2)$ be a relatively free Lie algebra freely generated by the set $\{b_1,b_2\},$ where $b_i={\ell }_i+V(L_2),$ $i=1,2.$ Let ${\pi }_{G_2}$ be the natural Lie algebra epimorphism from $G_2$ onto $M_2$ defined by ${\pi }_{G_2}(b_i)=y_i,$ $i=1,2.$ Then, the restriction of ${\pi }_{G_2}$ on the Lie algebra $P_{G_2}$ of symmetric polynomials in $L_2$ is a Lie algebra epimorphism from $P_{G_2}$ onto $P_{M_2}$.

Proof. It suffices to show that ${\pi }_{G_2}(P_{G_2})=P_{M_2}$. We show at first that ${\pi }_{G_2}(P_{G_2})\subseteq P_{M_2}$. Let ${\pi }_{G_2}(v(b_1,b_2))\in {\pi }_{G_2}(P_{G_2}),$ where $v(b_1,b_2)\in P_{G_2}.$ Therefore$,$ $v(b_1,b_2)=v(b_2,b_1)$ and hence$,$ ${\pi }_{G_2}(v(b_1,b_2))={\pi }_{G_2}(v(b_2,b_1)),$ that is$,$ $v(y_1,y_2)=v(y_2,y_1).$ Thus$,$ $v(y_1,y_2)\in P_{M_2}$ and so$,$ ${\pi }_{G_2}(v(b_1,b_2))=v(y_1,y_2)\in P_{M_2}.$ Hence$,$ ${\pi }_{G_2}(P_{G_2})\subseteq P_{M_2}$.

We show now that $P_{M_2}\subseteq {\pi }_{G_2}(P_{G_2}).$ Since$,$ by [6, Corollary 3.2], the set

is a basis of $P_{M_2}$, it suffices to show that $y_1+y_2\in {\pi }_{G_2}(P_{G_2})$ and $[y_2,y_1,{}_m{y}_1,{}_n{y}_2]-[y_2,y_1,{}_n{y}_1,{}_m{y}_2]\in {\pi }_{G_2}(P_{G_2})$ for all positive integers m and n, with $m>n\ge 0$. It is trivially true that for the element $b_1+b_2$ of $P_{G_2}$, ${\pi }_{G_2}(b_1+b_2)=y_1+y_2$ and hence, $y_1+y_2\in {\pi }_{G_2}(P_{G_2})$. Fix some positive integers m and n, with $m>n\ge 0$, and let $u=[b_2,b_1,{}_m{b}_1,{}_n{b}_2]-[b_2,b_1,{}_m{b}_2,{}_n{b}_1]$. We point out that $u\in P_{G_2}$. Furthermore, Since $M_2$ is a metabelian Lie algebra, it may be easily verified that and hence, by the above equation, Hence, $[y_2,y_1,{}_m{y}_1,{}_n{y}_2]-[y_2,y_1,{}_n{y}_1,{}_m{y}_2]\in {\pi }_{G_2}(P_{G_2})$. Thus, it follows that $P_{M_2}\subseteq {\pi }_{G_2}(P_{G_2})$. Therefore, we obtain the required result. □

From now on, we write $\pi$ instead of ${\pi }_{C_2}$ for the natural Lie algebra epimorphism from $C_2$ onto $M_2$ defined by $\pi (x_i)=y_i$, $i=1,2$. Since $[L_2^{\prime \prime },L_2]\subseteq L_2^{\prime \prime }$, the next result follows directly from Proposition 2.

Corollary 1. The restriction of $\pi$ on $P_{C_2}$ is a Lie algebra epimorphism from $P_{C_2}$ onto $P_{M_2}.$

Let $m,n\ge 0.$ It is straightforward to verify that $z_{m,n}=[x_2,x_1,{}_m{x}_1,{}_n{x}_2]-[x_2,x_1,{}_m{x}_2,{}_n{x}_1]\in P_{C_2}.$ Furthermore$,$ by Lemma 1(2)$,$

and hence, we get $w_{m,n}\in P_{C_2}$.

Proposition 3. Let $C_2$ be the free center-by-metabelian Lie algebra of rank 2 freely generated by the set $\{x_1,x_2\}.$

(1)	Let $f(x_1,x_2)$ be a symmetric polynomial in $C_2^{\prime \prime }.$ Then, $f(x_1,x_2)$ has the form $$f(x_1,x_2)=\sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}{w}_{m,n},$$ where $c_{m,n}\in K.$
(2)	Let $f(x_1,x_2)$ be a symmetric polynomial in $C_2^{\prime }.$ Then, $f(x_1,x_2)$ has the form $$f(x_1,x_2)=\sum _{m>n\ge 0}{b}_{m,n}{z}_{m,n}+\sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}{w}_{m,n},$$ where $b_{m,n},c_{m,n}\in K.$

Proof.

(1)

Since $f(x_1,x_2)\in C_2^{\prime \prime },$ by Proposition 1(6) we may write $$f(x_1,x_2)=\sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]],$$ where $c_{m,n}\in K$. Hence$,$ $$\begin{array}{ccc}\hfill f(x_2,x_1)& \hfill =\hfill & \sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_1,x_2,{}_m{x}_2,{}_n{x}_1,[x_1,x_2]]\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_2,{}_n{x}_1,[x_2,x_1]]\hfill \end{array}$$ and hence, by Lemma 1(1), $$\begin{array}{ccc}\hfill f(x_2,x_1)& \hfill =\hfill & \sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_n{x}_1,{}_m{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{n,m}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]].\hfill \end{array}$$ Since $f(x_1,x_2)=f(x_2,x_1)$, we get $$\begin{array}{ccc}\hfill & \hfill \hfill & \sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill \hfill & =\sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{n,m}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \end{array}$$ and since, by Proposition 1(6), the set $$\{[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]:m,n\ge 0,m+n\mathrm{\text{odd}}\}$$ is a basis of $C_2^{\prime \prime }$, it follows from the above equation that $c_{m,n}=c_{n,m}$ for all $m,n\ge 0$, with $m+n$ odd. Hence, $$\begin{array}{ccc}\hfill f(x_1,x_2)& \hfill =\hfill & \sum _{\begin{array}{c}m,n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill \hfill & +\sum _{\begin{array}{c}0\le m<n\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill \hfill & +\sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{n,m}[x_2,x_1,{}_n{x}_1,{}_m{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill \hfill & +\sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}[x_2,x_1,{}_n{x}_1,{}_m{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}([x_2,x_1,{}_m{x}_1,{}_n{x}_2,[x_2,x_1]]\hfill \\ \hfill & \hfill \hfill & +[x_2,x_1,{}_n{x}_1,{}_m{x}_2,[x_2,x_1]])\hfill \\ \hfill & \hfill =\hfill & \sum _{\begin{array}{c}m>n\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}{w}_{m,n}.\hfill \end{array}$$

(2)

Let $f(x_1,x_2)\in P_{C_2}\cap C_2^{\prime }$. Then, $\pi (f(x_1,x_2))=f(y_1,y_2)\in P_{M_2}$ and hence, by [6, Lemma 3.1], $$\pi (f(x_1,x_2)=\sum _{m>n\ge 0}{b}_{m,n}([y_2,y_1,{}_m{y}_1,{}_n{y}_2]-[y_2,y_1,{}_n{y}_1,{}_m{y}_2]).$$ But it is easily verified that $[y_2,y_1,{}_n{y}_1,{}_m{y}_2]=[y_2,y_1,{}_m{y}_2,{}_n{y}_1]$ and hence, by the above equation, $$\begin{array}{ccc}\hfill \pi (f(x_1,x_2)& \hfill =\hfill & \sum _{m>n\ge 0}{b}_{m,n}([y_2,y_1,{}_m{y}_1,{}_n{y}_2]-[y_2,y_1,{}_m{y}_2,{}_n{y}_1])\hfill \\ \hfill & \hfill =\hfill & \pi \left(\sum _{m>n\ge 0}{b}_{m,n}{z}_{m,n}\right).\hfill \end{array}$$ Hence, $$f(x_1,x_2)=\sum _{m>n\ge 0}{b}_{m,n}{z}_{m,n}+u(x_1,x_2),$$ where $b_{m,n}\in K$ and $u(x_1,x_2)\in \mathrm{\text{Ker}}(\pi )=C_2^{\prime \prime }$. Since $f(x_1,x_2),z_{m,n}\in P_{C_2}$, it follows that $u(x_1,x_2)\in P_{C_2}$ and hence, $u(x_1,x_2)\in P_{C_2}\cap C_2^{\prime \prime }$. Therefore, by Proposition 3(1), $$u(x_1,x_2)=\sum _{\begin{array}{c}m>n\ge 0\\ m+n\mathrm{\text{odd}}\end{array}}{c}_{m,n}{w}_{m,n}$$ and hence, we obtain the required result.

 □

For a positive integer d, we write $P_{C_2}^{(d)}$ for the K-vector subspace of $P_{C_2}$ spanned by all Lie commutators of $P_{C_2}$ of homogeneous degree d. Since $P_{C_2}$ is a Lie subalgebra of $C_2$, $P_{C_2}$ is a graded Lie algebra and hence, we may write $P_{C_2}={\oplus }_{d\ge 1}{P}_{C_2}^{(d)}$. It is straightforward to verify that $P_{C_2}^{(1)}$ is spanned by the element $x_1+x_2$ and $P_{C_2}^{(2)}=\{0\}$.

Theorem 1. Let $C_2$ be the free center-by-metabelian Lie algebra over a field K of characteristic 0 freely generated by the set $\{x_1,x_2\}$ and let $P_{C_2}$ be the Lie subalgebra of symmetric polynomials in $C_2.$ For non-negative integers m and n, let

andThen, the setis a basis for the Lie algebra $P_{C_2}$.

Proof. Since $P_{C_2}^{(1)}=\{a(x_1+x_2):a\in K\}$, $P_{C_2}^{(2)}=\{0\}$ and by Proposition 3(2), it follows that the set $X_1$ generates the Lie algebra $P_{C_2}$. Let

where $a,b_{m,n},c_{m,n}\in K$. To complete the proof, it suffices to show that $a=0$, $b_{m,n}=0$ for all m and n, with $m>n\ge 0$, and $c_{m,n}=0$ for all m and n, with $m>n\ge 0$ and $m+n$ odd. Since the set $\{x_1+C_2^{\prime },x_2+C_2^{\prime }\}$ is a basis of the quotient $C_2∕C_2^{\prime }$, working modulo $C_2^{\prime }$ in Eq. (3.1), we get $a=0$. By Lemma 2(1), Hence, since $a=0$, working modulo $C_2^{\prime \prime }$ in Eq. (3.1) we get Since, by Proposition 1(6), the set $\{[x_2,x_1,{}_k{x}_1,{}_l{x}_2]+C_2^{\prime \prime }:k,l\ge 0\}$ is a basis of the quotient $C_2^{\prime }∕C_2^{\prime \prime }$, it follows from the above equation that $b_{m,n}=0$ for all m and n, with $m>n\ge 0$. Since $a=0$ and $b_{m,n}=0$ for all m and n, with $m>n\ge 0$, it follows from Eq. (3.1) that Since, by Proposition 1(6), the set $\{[x_2,x_1,{}_k{x}_1,{}_l{x}_2,[x_2,x_1]]:k,l\ge 0,k+l\mathrm{\text{odd}}\}$ is a basis of $C_2^{\prime \prime }$, it follows from the above equation that $c_{m,n}=0$ for all $m>n\ge 0$, with $m+n$ odd. Hence, the set $X_1$ is a basis of $P_{C_2}$. □

Corollary 2.

(1)	The set $\{x_1+x_2\}$ is a basis of the K-vector subspace $P_{C_2}^{(1)}.$
(2)	$P_{C_2}^{(2)}=\{0\}.$
(3)	The set $\{z_{1,0}\}$ is a basis of the K-vector subspace $P_{C_2}^{(3)}.$
(4)	For d even, with $d\ge 4,$ the set $\{z_{m,n}:m>n\ge 0,m+n=d-2\}$ is a basis of the K-vector subspace $P_{C_2}^{(d)}.$
(5)	For d odd, with $d\ge 5,$ the set $$\begin{array}{ccc}\hfill & \hfill \hfill & \{z_{m_1,n_1},w_{m_2,n_2}:m_1>n_1\ge 0,m_1+n_1=d-2,m_2>n_2\ge 0,\hfill \\ \hfill & \hfill \hfill & m_2+n_2=d-4\}\hfill \end{array}$$ is a basis of the K-vector subspace $P_{C_2}^{(d)}.$

4. A Minimal Generating Set of $P_{C_2}$

4.1. Relations in $P_{C_2}$

In this section, our main aim is to show certain relations among the elements $x_1+x_2$, $z_{2r-1,0}$ and $w_{2s-1,0}$, where r and s are positive integers, which we need for the proof of Theorem 2.

Lemma 3. Let $C_2$ be the free center-by-metabelian Lie algebra freely generated by the set $\{x_1,x_2\}.$ Then

(1)	For non-negative integers m and n, with $m+n\ge 1,$ $w_{m,n}=w_{n,m}.$
(2)	For non-negative integers m and n, with $m+n\ge 1$ and $m+n$ even, $w_{m,n}=0.$
(3)	For non-negative integers $m_1,$ $m_2,$ $n_1$ and $n_2,$ $$[z_{m_1,n_1},z_{m_2,n_2}]={(-1)}^{m_2+n_2}{w}_{m_1+m_2,n_1+n_2}+{(-1)}^{m_2+n_2+1}{w}_{m_1+n_2,n_1+m_2}.$$
(4)	(a) For positive integers r and s, $$[z_{2r-1,0},{}_{2s-1}{x}_1+x_2]=\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)z_{2r+2s-k-2,k}.$$ (b) For positive integers r and s, $$\begin{array}{ccc}\hfill & \hfill \hfill & [z_{2r-1,0},{}_{2s}{x}_1+x_2]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill k\hfill \end{array}\right)z_{2r+2s-k-1,k}+\sum _{k=1}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)w_{2r+2s-k-2,k-1}.\hfill \end{array}$$
(5)	For positive integers r, s and t, $$\begin{array}{ccc}\hfill & \hfill \hfill & [z_{2r-1,0},{}_{2s-1}{x}_1+x_2,z_{2t-1,0}]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)(-w_{2r+2s+2t-k-3,k}+w_{2r+2s-k-2,2t+k-1}).\hfill \end{array}$$

Proof. Throughout the proof, we write $u=[x_1,x_2]$ and $a=x_1+x_2$.

(1)	Since $w_{m,n}=[u,{}_m{x}_1,{}_n{x}_2,u]+[u,{}_n{x}_1,{}_m{x}_2,u]$, the result is trivially true.
(2)	Since $w_{m,n}=[u,{}_m{x}_1,{}_n{x}_2,u]+[u,{}_n{x}_1,{}_m{x}_2,u]$ and $m+n$ is even, the result follows from Lemma 1(2).
(3)	We have $$\begin{array}{ccc}\hfill & \hfill \hfill & [z_{m_1,n_1},z_{m_2,n_2}]\hfill \\ \hfill & \hfill \hfill & =[[u,{}_{m_1}{x}_1,{}_{n_1}{x}_2]-[u,{}_{m_1}{x}_2,{}_{n_1}{x}_1],[u,{}_{m_2}{x}_1,{}_{n_2}{x}_2]-[u,{}_{m_2}{x}_2,{}_{n_2}{x}_1]]\hfill \\ \hfill & \hfill \hfill & =[[u,{}_{m_1}{x}_1,{}_{n_1}{x}_2],[u,{}_{m_2}{x}_1,{}_{n_2}{x}_2]]-[[u,{}_{m_1}{x}_1,{}_{n_1}{x}_2],[u,{}_{m_2}{x}_2,{}_{n_2}{x}_1]]\hfill \\ \hfill & \hfill \hfill & -[[u,{}_{m_1}{x}_2,{}_{n_1}{x}_1],[u,{}_{m_2}{x}_2,{}_{n_2}{x}_1]]+[[u,{}_{m_1}{x}_2,{}_{n_1}{x}_1],[u,{}_{m_2}{x}_2,{}_{n_2}{x}_1]]\hfill \end{array}$$ and by Lemma 1(2), $$\begin{array}{ccc}\hfill [[u,{}_{m_1}{x}_1,{}_{n_1}{x}_2],[u,{}_{m_2}{x}_1,{}_{n_2}{x}_2]]& \hfill =\hfill & {(-1)}^{m_2+n_2}[u,{}_{m_1+m_2}{x}_1,{}_{n_1+n_2}{x}_2,u],\hfill \\ \hfill [[u,{}_{m_1}{x}_1,{}_{n_1}{x}_2],[u,{}_{m_2}{x}_2,{}_{n_2}{x}_1]]& \hfill =\hfill & {(-1)}^{m_2+n_2}[u,{}_{m_1+n_2}{x}_1,{}_{n_1+m_2}{x}_2,u],\hfill \\ \hfill [[u,{}_{m_1}{x}_2,{}_{n_1}{x}_1],[u,{}_{m_2}{x}_1,{}_{n_2}{x}_2]]& \hfill =\hfill & {(-1)}^{m_2+n_2}[u,{}_{n_1+m_2}{x}_1,{}_{m_1+n_2}{x}_2,u],\hfill \\ \hfill [[u,{}_{m_1}{x}_2,{}_{n_1}{x}_1],[u,{}_{m_2}{x}_2,{}_{n_2}{x}_1]]& \hfill =\hfill & {(-1)}^{m_2+n_2}[u,{}_{n_1+n_2}{x}_1,{}_{m_1+m_2}{x}_2,u].\hfill \end{array}$$ Hence, $$\begin{array}{ccc}\hfill [z_{m_1,n_1},z_{m_2,n_2}]& \hfill =\hfill & {(-1)}^{m_2+n_2}[u,{}_{m_1+m_2}{x}_1,{}_{n_1+n_2}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & -{(-1)}^{m_2+n_2}[u,{}_{m_1+n_2}{x}_1,{}_{n_1+m_2}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & -{(-1)}^{m_2+n_2}[u,{}_{n_1+m_2}{x}_1,{}_{m_1+n_2}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & +{(-1)}^{m_2+n_2}[u,{}_{n_1+n_2}{x}_1,{}_{m_1+m_2}{x}_2,u]\hfill \\ \hfill & \hfill =\hfill & {(-1)}^{m_2+n_2}([u,{}_{m_1+m_2}{x}_1,{}_{n_1+n_2}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & +[u,{}_{n_1+n_2}{x}_1,{}_{m_1+m_2}{x}_2,u])\hfill \\ \hfill & \hfill \hfill & +{(-1)}^{m_2+n_2+1}([u,{}_{m_1+n_2}{x}_1,{}_{n_1+m_2}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & +[u,{}_{n_1+m_2}{x}_1,{}_{m_1+n_2}{x}_2,u])\hfill \\ \hfill & \hfill =\hfill & {(-1)}^{m_2+n_2}{w}_{m_1+m_2,n_1+n_2}+{(-1)}^{m_2+n_2+1}{w}_{m_1+n_2,n_1+m_2}.\hfill \end{array}$$
(4)	For a positive integer t, we have $$\begin{array}{ccc}\hfill [z_{2r-1,0},{}_t{x}_1+x_2]& \hfill =\hfill & [[u,{}_{2r-1}{x}_1]-[u,{}_{2r-1}{x}_2],{}_{t}a]\hfill \\ \hfill & \hfill =\hfill & [u,{}_{2r-1}{x}_1,{}_{t}a]-[u,{}_{2r-1}{x}_2,{}_{t}a].\hfill \end{array}$$ (a) By the above equation (for $t=2s-1$) and Lemma 2(2)–(3), we get $$\begin{array}{ccc}\hfill & \hfill \hfill & [z_{2r-1,0},{}_{2s-1}{x}_1+x_2]=[u,{}_{2r-1}{x}_1,{}_{2s-1}a]-[u,{}_{2r-1}{x}_2,{}_{2s-1}a]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{2r+2s-2-k}{x}_1,{}_k{x}_2]\hfill \\ \hfill & \hfill \hfill & -\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{2r+2s-2-k}{x}_2,{}_k{x}_1]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)([u,{}_{2r+2s-k-2}{x}_1,{}_k{x}_2]-[u,{}_{2r+2s-k-2}{x}_2,{}_k{x}_1])\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)z_{2r+2s-k-2,k}.\hfill \end{array}$$ (b) Similarly, by the above equation (for $t=2s$) and Lemma 2(2)–(3), we get $$\begin{array}{ccc}\hfill & \hfill \hfill & [z_{2r-1,0},{}_{2s}{x}_1+x_2]=[u,{}_{2r-1}{x}_1,{}_{2s}a]-[u,{}_{2r-1}{x}_2,{}_{2s}a]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{2r+2s-1-k}{x}_1,{}_k{x}_2]+\sum _{k=1}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{2r+2s-2-k}{x}_1,{}_{k-1}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & -\sum _{k=0}^{2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{2r+2s-1-k}{x}_2,{}_k{x}_1]\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)[u,{}_{k-1}{x}_1,{}_{2r+2s-2-k}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill k\hfill \end{array}\right)([u,{}_{2r+2s-k-1}{x}_1,{}_k{x}_2]-[u,{}_{2r+2s-k-1}{x}_2,{}_k{x}_1])\hfill \\ \hfill & \hfill \hfill & +\sum _{k=1}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)([u,{}_{2r+2s-k-2}{x}_1,{}_{k-1}{x}_2,u]\hfill \\ \hfill & \hfill \hfill & +[u,{}_{k-1}{x}_1,{}_{2r+2s-k-2}{x}_2,u])\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s}\left(\begin{array}{c}\hfill 2s\hfill \\ \hfill k\hfill \end{array}\right)z_{2r+2s-k-1,k}+\sum _{k=1}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)w_{2r+2s-k-2,k-1}.\hfill \end{array}$$
(5)	Since, by Lemma 3(4)(a), $$[z_{2r-1,0},{}_{2s-1}{x}_1+x_2]=\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)z_{2r+2s-k-2,k},$$ we get $$[z_{2r-1,0},{}_{2s-1}{x}_1+x_2,z_{2t-1,0}]=\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)[z_{2r+2s-k-2,k},z_{2t-1,0}].$$ Hence, by Lemma 3(3), $$\begin{array}{ccc}\hfill & \hfill \hfill & [z_{2r-1,0},{}_{2s-1}{x}_1+x_2,z_{2t-1,0}]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)[{(-1)}^{2t-1}{w}_{2r+2s+2t-k-3,k}+{(-1)}^{2t}{w}_{2r+2s-k-2,k+2t-1}]\hfill \\ \hfill & \hfill \hfill & =\sum _{k=0}^{2s-1}\left(\begin{array}{c}\hfill 2s-1\hfill \\ \hfill k\hfill \end{array}\right)(-w_{2r+2s+2t-k-3,k}+w_{2r+2s-k-2,k+2t-1}).\hfill \end{array}$$

 □

Lemma 4. Let $C_2$ be the free center-by-metabelian Lie algebra freely generated by the set $\{x_1,x_2\}$ and let $P_{C_2}$ be the Lie subalgebra of symmetric polynomials in $C_2$. Then, the following relations hold in $P_{C_2}$:

(1)	$[w_{2i-1,0},u]=0$ for all $i\ge 1$ and $u\in \{x_1+x_2,z_{2n-1,0},w_{2n-1,0}:n\ge 1\}$.
(2)	$[z_{2r-1,0},{}_{2s}{x}_1+x_2,z_{2t-1,0}]=0$ for all $r,t\ge 1$ and $s\ge 0$.
(3)	$[z_{2i-1,0},{}_k{x}_1+x_2,[z_{2j-1,0}{,}_{\lambda }{x}_1+x_2]]={(-1)}^{\lambda }[z_{2i-1,0},{}_{k+\lambda }{x}_1+x_2,z_{2j-1,0}]$ for all $i,j,k,\lambda \ge 1$.
$(4)$	$[z_{2i-1,0},{}_{2k-1}{x}_1+x_2,z_{2j-1,0},u]=0$ for all $i,k,j\ge 1$ and $u\in \{x_1+x_2,z_{2n-1,0}\}$.

Proof. (1) Since $[C_2^{\prime \prime },C_2]=\{0\}$ and $w_{2j-1,0}\in C_2^{\prime \prime }$, the result follows directly.

(2) Assume at first that $s=0$. Then, by Lemma 3(3), $[z_{2r-1,0},z_{2t-1,0}]=-w_{2r+2t-2,0}+w_{2r-1,2t-1}$. Since, by Lemma 3(2), $w_{2r+2t-2,0}=w_{2r-1,2t-1}=0$, it follows that $[z_{2r-1,0},z_{2t-1,0}]=0$. Assume now that $s>0$. Then, by Lemma 3(4)(b),

But, by Lemma 3(3), and hence, by Lemma 3(2), $[z_{2r+2s-k-1,k},z_{2t-1,0}]=0$ for all $k=0,\dots ,2s$. Furthermore, by Lemma 4(1), $[w_{2r+2s-k-2,k-1},z_{2t-1,0}]=0$ for all $k=1,\dots ,2s-1$. Thus, by the above equation, we obtain the required result.

(3) The result follows directly from Lemma 1(2).

(4) Since, $[z_{2i-1,0},{}_{2k-1}{x}_1+x_2,z_{2j-1,0}]\in C_2^{\prime \prime }$ and $[C_2^{\prime \prime },C_2]=\{0\}$, we get the result. □

4.2. The main theorem

Recall that $M_2$ is the free metabelian Lie algebra freely generated by the set $\{y_1,y_2\}$ and $P_{M_2}$ is the Lie subalgebra of symmetric polynomials in $M_2$. For a positive integer d, we write $P_{M_2}^{(d)}$ for the K-vector subspace of $P_{M_2}$ spanned by all Lie commutators of $P_{M_2}$ of homogeneous degree d. Since $P_{M_2}$ is a Lie subalgebra of $M_2$, $P_{M_2}$ is a graded Lie algebra and hence, we may write $P_{M_2}={\oplus }_{d\ge 1}{P}_{M_2}^{(d)}$. It has been proved in [6, Theorem 3.4] that $Y=\{y_1+y_2,[y_2,y_1,{}_{2m-1}{y}_1]-[y_2,y_1,{}_{2m-1}{y}_2]:m\ge 1\}$ is a set of generators for $P_{M_2}$.

Lemma 5. Let $C_2$ be the free center-by-metabelian Lie algebra freely generated by the set $\{x_1,x_2\}$ and let $P_{C_2}$ be the Lie subalgebra of symmetric polynomials in $C_2.$ Let $X_2=\{x_1+x_2,z_{2n-1,0},w_{2n-1,0}:n\ge 1\}$ and let $L(X_2)$ be the Lie subalgebra of $P_{C_2}$ generated by the set $X_2$. Then

(1)	$z_{m,n}\in L(X_2)$ for all m and n, with $m>n\ge 0$ and $m+n$ even.
(2)	$w_{m,n}\in L(X_2)$ for all m and n, with $m>n\ge 0$ and $m+n$ odd.
(3)	$z_{m,n}\in L(X_2)$ for all m and n, with $m>n\ge 0$ and $m+n$ odd.

Proof. Recall that $\pi$ is the natural Lie algebra epimorphism from $C_2$ onto $M_2$ defined by $\pi (x_i)=y_i$, $i=1,2$, and by Corollary 1, $\pi (P_{C_2})=P_{M_2}$. We point out that $\pi (x_1+x_2)=y_1+y_2$ and for non-negative integers m and n, with $m>n\ge 0$,

We point out that, in particular, $\pi (X_2)=Y$.

(1)	Fix some m and n, with $m>n\ge 0$ and $m+n$ even. Then, $z_{m,n}\in P_{C_2}^{(m+n+2)}$ and hence, $\pi (z_{m,n})\in P_{M_2}^{(m+n+2)}$. Since, as aforementioned, $\pi (P_{C_2})=P_{M_2}$ and $\pi (X_2)=Y$, it follows from [6, proof of Lemma 3.3], that $\pi (z_{m,n})=\pi (u_1)$, where $u_1\in P_{C_2}^{(m+n+2)}\cap L(X_2)$. Hence, we may write $z_{m,n}=u_1+u_2$, where $u_2\in \mathrm{\text{Ker}}(\pi )=C_2^{\prime \prime }$. Since $z_{m,n},u_1\in P_{C_2}^{(m+n+2)}$, it follows that $u_2\in P_{C_2}^{(m+n+2)}$ and hence, $u_2\in C_2^{\prime \prime }\cap P_{C_2}^{(m+n+2)}$. Since $m+n+2$ is even, it follows from Corollary 2(4) that $u_2=0$. Thus $z_{m,n}=u_1\in L(X_2)$ and hence, we get the required result.
(2)	Fix some m and n, with $m>n\ge 0$ and $m+n$ odd. Since for $n=0$, it is trivially true that $w_{m,0}\in L(X_2)$, we may assume that $m>n>0$. By Lemma 3(1), $$[z_{m,0},z_{n,0}]={(-1)}^n{w}_{m+n,0}+{(-1)}^{n+1}{w}_{m,n}.$$ Since $m+n$ is odd, by the definition of $X_2$ and by Lemma 5(1) we have $z_{m,0},z_{n,0}\in L(X_2)$. Therefore, $[z_{m,0},z_{n,0}]\in L(X_2)$ and hence, by the above equation, ${(-1)}^n{w}_{m+n,0}+{(-1)}^{n+1}{w}_{m,n}\in L(X_2)$. Since, by the definition of $X_2$, $w_{m+n,0}\in L(X_2)$, it follows that $w_{m,n}\in L(X_2)$ and hence, we get the required result.
(3)	Fix some m and n, with $m>n\ge 0$ and $m+n$ odd. Then, $z_{m,n}\in P_{C_2}^{(m+n+2)}$ and hence, $\pi (z_{m,n})\in P_{M_2}^{(m+n+2)}$. Since $\pi (P_{C_2})=P_{M_2}$ and $\pi (X_2)=Y$, it follows from [6, proof of Theorem 3.4] that $\pi (z_{m,n})=\pi (v_1)$, where $v_1\in P_{C_2}^{(m+n+2)}\cap L(X_2)$. Hence, by similar arguments as in the proof of Lemma 5(1), we may write $z_{m,n}=v_1+v_2$, where $v_1\in L(X_2)\cap P_{C_2}^{(m+n+2)}$ and $v_2\in C_2^{\prime \prime }\cap P_{C_2}^{(m+n+2)}$. Hence, by Proposition 3(1), $v_2$ has the form ${\sum }_{\begin{array}{c}\alpha >\beta \ge 0\\ \alpha +\beta =m+n+2\end{array}}{c}_{\alpha ,\beta }{w}_{\alpha ,\beta }$, where $c_{\alpha ,\beta }\in K$. Since, by Lemma 5(2), $w_{\alpha ,\beta }\in L(X_2)$ for all $\alpha$ and $\beta$, with $\alpha >\beta \ge 0$ and $a+b=m+n+2$, it follows that $v_2\in L(X_2)$. Since $v_1,v_2\in L(X_2)$, it follows that $z_{m,n}\in L(X_2)$ and hence, we get the required result.

 □

Theorem 2. Let $C_2$ be the free center-by-metabelian Lie algebra over a field K of characteristic 0 freely generated by the set $\{x_1,x_2\}$ and let $P_{C_2}$ be the Lie subalgebra of symmetric polynomials in $C_2.$ For a positive integer $n,$ let

and letThen, $X_2$ is a minimal generating set for the Lie algebra $P_{C_2}.$

Proof. We show at first that the set $X_2$ is a generating set for $P_{C_2}$. Let $L(X_2)$ be the Lie subalgebra of $P_{C_2}$ generated by the set $X_2$. Since, by Theorem 1, the set

is a basis for the Lie algebra $P_{C_2}$, and by Lemma 5, $z_{m,n}\in L(X_2)$ for all $m>n\ge 0$ and $w_{k,l}\in L(X_2)$ for all $k>l\ge 0$, with $k+l$ odd, it follows that $P_{C_2}\subseteq L(X_2)$ and hence, $P_{C_2}=L(X_2)$, that is, $P_{C_2}$ is generated by the set $X_2$.

To complete the proof, we have to show that $X_2$ is a minimal generating set for $P_{C_2}$. It suffices to show that $z_{2i-1,0}\notin L(X_2\setminus \{z_{2i-1,0}\})$ and $w_{2i-1,0}\notin L(X_2\setminus \{w_{2i-1,0}\})$ for all $i\ge 1$. Furthermore, since $P_{C_2}={\oplus }_{m\ge 1}{P}_{C_2}^{(m)}$, it suffices to show that $z_{2i-1,0}\notin L(X_2\setminus \{z_{2i-1,0}\})\cap P_{C_2}^{(2i+1)}$ and $w_{2i-1,0}\notin L(X_2\setminus \{w_{2i-1,0}\})\cap P_{C_2}^{(2i+3)}\cap C_2^{\prime \prime }$ for all $i\ge 1$. Clearly, we may assume that $i\ge 2$.

We show at first that $z_{2i-1,0}\notin L(X_2\setminus \{z_{2i-1,0}\})\cap P_{C_2}^{(2i+1)}$. Suppose on the contrary that $z_{2i-1,0}\in L(X_2\setminus \{z_{2i-1,0}\})\cap P_{C_2}^{(2i+1)}$. Hence, we may write

where $u\in P_{C_2}^{(2i+1)}\cap C_2^{\prime \prime }$. By the Lie algebra epimorphism $\pi$ (defined in Sec. 3) we get But, by [6, Theorem 3.5], there are no non-trivial relations among the elements $y_1+y_2$ and $[y_2,y_1{,}_{2i-2\lambda -1}{y}_1]-[y_2,y_1{,}_{2i-2\lambda -1}{y}_2]$, $\lambda =1,\dots ,i-1$ and hence, we get the required contradiction.

We show now that $w_{2i-1,0}\notin L(X_2\setminus \{w_{2i-1,0}\})\cap P_{C_2}^{(2i+3)}\cap C_2^{\prime \prime }$. Suppose on the contrary that $w_{2i-1,0}\in L(X_2\setminus \{w_{2i-1,0}\})\cap P_{C_2}^{(2i+3)}\cap C_2^{\prime \prime }$. But, by the relations in Lemma 4, the only non-trivial elements in $L(X_2\setminus \{w_{2i-1,0}\})\cap P_{C_2}^{(2i+3)}\cap C_2^{\prime \prime }$ are the elements of the form $[z_{2r-1,0},{}_{2s-1}{x}_1+x_2,z_{2t-1,0}]$, with $r+s+t=i+1$. Hence, $w_{2i-1,0}$ may be written as a K-linear combination of elements of the form $[z_{2r-1,0},{}_{2s-1}{x}_1+x_2,z_{2t-1,0}]$, with $r+s+t=i+1$. But, by Lemma 3(5),

Furthermore, by Lemma 3(1), $w_{\alpha ,\beta }=w_{\beta ,\alpha }$ for all $\alpha ,\beta \ge 1$. Hence, by the above equation, $[z_{2r-1,0},{}_{2s-1}{x}_1+x_2,z_{2t-1,0}]$ is a K-linear combination of elements of the form $-w_{m_1,n_1}+w_{m_2,n_2}$, where $m_1>n_1$, $m_2>n_2$ and $m_1+n_1=m_2+n_2=2i-1$. But then, $w_{2i-1,0}$ may be written as a K-linear combination of elements of the above form. Since, by Corollary 2(5), the set $\{w_{2i-1-k,k}:k=0,\dots ,i-1\}$ is a basis of the K-vector subspace ${(C_2^{\prime \prime })}^{(2i+3)}$, the elements $w_{2i-1,0}$ and $-w_{m_1,n_1}+w_{m_2,n_2}$, with $m_1+n_1=m_2+n_2=2i-1$, are linearly independent over K and hence, we obtain the required contradiction. □

Acknowledgment

The author would like to thank the referee for valuable comments and suggestions during the preparation of this paper.

ORCID

C. E. Kofinas  https://orcid.org/0000-0002-4007-6144