IA-automorphisms and Lie Algebras related to the McCool group

TL;DR

This paper studies a subgroup of the McCool group, proving it has a solvable conjugacy problem, analyzing its associated Lie algebra, and establishing its embedding into the IA automorphisms' Lie algebra.

Contribution

It introduces and analyzes the subgroup $I_n$ of the McCool group, providing its Lie algebra presentation and embedding into the Andreadakis-Johnson Lie algebra.

Findings

$I_n$ has a solvable conjugacy problem

Presented a Lie algebra for $I_n$

Embedded gr($I_n$) into the IA automorphisms' Lie algebra

Abstract

In the present work we investigate a subgroup $I_n$ of the McCool group $M_n$. We show that $I_n$ has solvable conjugacy problem. Next, we investigate its Lie Algebra gr($I_n$) and we find a presentation for it. Finally we show that gr($I_n$) is naturally embedded into the Andreadakis-Johnson Lie Algebra of the IA automorphisms of the free group $F_n$.

Equations

$$\chi_{ij}(x_{k})=\left\{\begin{array}[]{lr}x_{k}&\mbox{if}\ \ k\neq i\\ x^{-1}_{j}x_{i}x_{j}&\mbox{if}\ \ k=i.\end{array}\right.$$

$$[{\chi }_{ij},{\chi }_{kj}]=[{\chi }_{ij},{\chi }_{kl}]=[{\chi }_{ij}{\chi }_{kj},{\chi }_{ik}]=1,$$

$$y_{jk}^{-1}y_{il}y_{jk}=\left\{\begin{array}[]{lr}y_{il}&k=l\\ y_{il}&l>j\\ y^{-1}_{ik}y_{il}y_{ik}&k\neq l,l\leq j.\end{array}\right.$$

$$\{1\}=G_{n+1}⊲G_n⊲G_{n-1}⊲\dots ⊲G_2=I_n$$

$$g_n{g}_{n-1}\dots g_3{g}_2{w}_n{w}_{n-1}\dots w_3{w}_2{g}_2^{-1}{g}_3^{-1}\dots g_{n-1}^{-1}{g}_n^{-1}=y$$

$$g_n(g_{n-1}\dots g_2{w}_n{g}_2^{-1}\dots g_{n-1}^{-1})g_{n-1}(g_{n-2}\dots g_3{g}_2{w}_{n-1}{g}_2^{-1}\dots g_{n-2}^{-1})\dots$$

$$g_4(g_3{g}_2{w}_4{g}_2^{-1}{g}_3^{-1})g_3(g_2{w}_3{g}_2^{-1})(g_2{w}_2{g}_2^{-1})g_3^{-1}\dots g_n^{-1}=y.$$

$$a_n=g_{n-1}\dots g_2{w}_n{g}_2^{-1}\dots g_{n-1}^{-1}$$

$$\begin{array}[]{rll}b_{n}&=&g_{n-1}(g_{n-2}\ldots g_{2}w_{n-1}g_{2}^{-1}\ldots g_{n-2}^{-1})\\ &&g_{n-2}(g_{n-3}\ldots g_{2}w_{n-2}g_{2}^{-1}\ldots g_{n-3}^{-1})g_{n-3}\ldots\\ &&g_{4}(g_{3}g_{2}w_{4}g_{2}^{-1}g_{3}^{-1})g_{3}(g_{2}w_{3}g_{2}^{-1})(g_{2}w_{2}g^{-1}_{2})g_{3}^{-1}\ldots g_{n-1}^{-1}.\end{array}$$

$$g_n{a}_n{b}_n{g}_n^{-1}=y$$

$$(g_n{a}_n{b}_n{g}_n^{-1}{b}_n^{-1})b_n=y.$$

$$a_{n-1}=g_{n-2}\dots g_2{w}_{n-1}{g}_2^{-1}\dots g_{n-2}^{-1}\in H_{n-1}$$

$$\begin{array}[]{rll}b_{n-1}&=&g_{n-2}(g_{n-3}\ldots g_{2}w_{n-2}g_{2}^{-1}\ldots g_{n-3}^{-1})g_{n-3}\\ &&\ldots g_{4}(g_{3}g_{2}w_{4}g_{2}^{-1}g_{3}^{-1})g_{3}(g_{2}w_{3}g_{2}^{-1})(g_{2}w_{2}g^{-1}_{2})g_{3}^{-1}\ldots g_{n-2}^{-1}\in I_{n-2}.\end{array}$$

$$(g_n{a}_n{b}_n{g}_n^{-1}{b}_n^{-1})(g_{n-1}{a}_{n-1}{b}_{n-1}{g}_{n-1}^{-1}{b}_{n-1}^{-1})b_{n-1}=y,$$

$$(g_n{a}_n{b}_n{g}_n^{-1}{b}_n^{-1})(g_{n-1}{a}_{n-1}{b}_{n-1}{g}_{n-1}^{-1}{b}_{n-1}^{-1})\dots (g_3{a}_3{b}_3{g}_3^{-1}{b}_3^{-1})(g_2{w}_2{g}_2^{-1})=y,$$

$$g_3{a}_3{b}_3{g}_3^{-1}{b}_3^{-1}=z_3$$

$$g_3{a}_3(g_2{w}_2{g}_2^{-1})g_3^{-1}(g_2{w}_2^{-1}{g}_2^{-1})=z_3.$$

$$g_3{a}_3\varphi (g_3^{-1})=z_3.$$

$$\begin{array}[]{lll}L&=&L(U\oplus V)\\ &=&L(U)\oplus L(V\wr U)\\ &=&L(U)\oplus L(V)\oplus L(W),\end{array}$$

$${\mathcal{X}}_{V,U}=\mathcal{V}\cup (\bigcup _{a\ge 1}[\mathcal{V},{}_a\mathcal{U}]),$$

$$\mathcal{W}=[\mathcal{V},\mathcal{U}]\cup (\bigcup _{\genfrac{}{}{0ex}{}{a+b\ge 1}{a,b\ge 0}}[\mathcal{V},\mathcal{U},{}_a\mathcal{U},{}_b\mathcal{V}]),$$

$${\psi }_a([v,z,z_1,\dots ,z_{a-2}])={\varphi }_a([v,z,z_1,\dots ,z_{a-2}])+w_a,$$

$$\begin{array}[]{ccl}\psi_{a}([v,z,z_{1},\ldots,z_{a-2}])&=&[\psi_{2}([v,z]),z_{1},\ldots,z_{a-2}]\\ &=&[\phi([v,z]),z_{1},\ldots,z_{a-2}]+[w_{v,z},z_{1},\ldots,z_{a-2}]\\ &=&\phi_{a}([v,z,z_{1},\ldots,z_{a-2}])+[w_{v,z},z_{1},\ldots,z_{a-2}],\end{array}$$

$$[w_{v,z},z_1,\dots ,z_{a-2}]\in L(\mathcal{V}\cup (\bigcup _{1\le \beta \le a-2}[\mathcal{V},{}_{\beta }\mathcal{U}])).$$

$${\psi }_a([v,z,z_1,\dots ,z_{a-2}])={\varphi }_a([v,z,z_1,\dots ,z_{a-2}])+w_a,$$

$$\begin{array}[]{rll}L(V\wr U)&=&\Psi(L(V\wr U))\\ ({\rm By~{}Corollary}~{}\ref{c1})&=&\Psi(L(V)\oplus L(W))\\ (\Psi~{}{\rm automorphism})&=&\Psi(L(V))\oplus\Psi(L(W))\\ &=&L(V)\oplus L(\Psi(W))\end{array}$$

$$\mathcal{C}=\Psi (\mathcal{W})=\Psi ([\mathcal{V},\mathcal{U}])\cup (\bigcup _{\genfrac{}{}{0ex}{}{a+b\ge 1}{a,b\ge 0}}[\Psi ([\mathcal{V},\mathcal{U}]),{}_a\mathcal{U},{}_b\mathcal{V}]).$$

$$W_{m,\Psi }=\underset{\genfrac{}{}{0ex}{}{a,b\ge 0}{a+b=m-2}}{⨁}[\Psi ([V,U]),{}_{a}U,{}_{b}V].$$

$${\mathcal{W}}_{m,\Psi }=\bigcup _{\genfrac{}{}{0ex}{}{a,b\ge 0}{a+b=m-2}}[\Psi ([\mathcal{V},\mathcal{U}]),{}_a\mathcal{U},{}_b\mathcal{V}]$$

$$L(\Psi (W))=\underset{m\ge 2}{⨁}{L}_{\mathrm{grad}}^m(\Psi (W)).$$

Full text

1. Introduction and notation

For a positive integer $n$, with $n\geq 2$, let $F_{n}$ be a free group of rank $n$ generated by the set $\{x_{1},\ldots,x_{n}\}$ and $M_{n}$ be the McCool group (the basis-conjugating group), that is, the subgroup of Aut$(F_{n})$ generated by the automorphism $\chi_{ij}$, $1\leq i\neq j\leq n$, with

[TABLE]

For any group $G$ and $a,b\in G$, we write $[a,b]=a^{-1}b^{-1}ab$. As shown by McCool in [13], a complete set of relations for the group $M_{n}$ is the following

[TABLE]

where different letters correspond to different numbers. It is shown by Savushkina [16] (see, also, [2, Theorem 1] and [8, Theorem 1.2]) that $M_{n}$ is structured as a semi-direct product $M_{n}=D_{n-1}\rtimes M_{n-1}$, where $D_{n-1}$ is the group generated by the set $\{\chi_{1n},\ldots,\chi_{n-1,n},\chi_{n1},\ldots,$ $\chi_{n,n-1}\}.$ The group structure of $D_{n}$ at the moment is a mystery, although we know that the subgroup of $D_{n}$ generated by the set $\{\chi_{1n},\ldots,\chi_{n-1,n}\}$ is free abelian and the subgroup of $D_{n}$ generated by the set $\{\chi_{n1},\ldots,\chi_{n,n-1}\}$ is free.

In the present work, we investigate a subgroup of $M_{n}$, that arose in [3], the partial inner automorphisms group $I_{n}$, consisting of automorphisms acting on initial segments of the basis of $F_{n}$ by conjugations. In Section 2, we show that the conjugacy problem for $I_{n}$ is solvable, thus answering a question of Bardakov and Neshchadim in [3]. In Section 3, we develop an inductive technique that helps us to decompose and understand the graded Lie algebra $\rm gr(I_{n})$ of $I_{n}$ which leads us to give a presentation for $\rm gr(I_{n})$. Finally, in Section 4, by using standard arguments, we prove that $\rm gr(I_{n})$ naturally embeds into the Andreadakis-Johnson Lie algebra of the IA-automorphisms of the free group $F_{n}$.

2. Some group theoretic results

Bardakov and Neshchadim in [3] have shown that in $M_{n}$, we may define the subgroup $H_{n}$ generated by the elements $y_{ni}=\chi_{1i}\ldots\chi_{ni}$, $i=1,\ldots,n$, with the assumption that $\chi_{ii}=1$. It is easy to see that $y_{ni}$ is an inner automorphism of $F_{n}$, conjugating each generator by $x_{i}$. Obviously, the subgroup $H_{n}$ of $M_{n}$ generated by the set $\{y_{ni},i=1,\ldots,n\}$ is equal to Inn$(F_{n})$, the inner automorphisms of $F_{n}$. Similarly, for $k=2,\ldots,n-1$, they define $H_{k}$ to be the subgroup of $M_{n}$ generated by the set $\{y_{k1},\ldots,y_{kk}\}$, the inner automorphisms of $F_{k}$ considered as a subgroup of $M_{n}$. Finally, they define $I_{n}$ to be the subgroup of $M_{n}$ generated by the set $\{y_{\mu j}:2\leq\mu\leq n,1\leq j\leq\mu\}$. Now, one can easily verify that for $i>j$ the group $H_{j}$ normalizes $H_{i}$, namely, since $H_{i}$ is generated by the set $\{y_{i1},\ldots,y_{ii}\}$ and $H_{j}$ is generated by the set $\{y_{j1},\ldots,y_{jj}\}$, we have

[TABLE]

Furthermore, it is easily shown that $I_{n}$ is a poly-free group (see also [3]). Indeed, if $G_{i}=H_{n}\rtimes H_{n-1}\rtimes\ldots\rtimes H_{i}$, with $i\in\{2,\ldots,n\}$, then there is the tower of normal subgroups

[TABLE]

such that each $G_{i}/G_{i+1}\cong H_{i}$ is free. Thus, $I_{n}=G_{3}\rtimes H_{2}$ and so, any element of $I_{n}$ is uniquely written as $g_{3}w_{2}$, where $g_{3}\in G_{3}$ and $w_{2}\in H_{2}$. It is shown in [3] that a complete set of defining relations for $I_{n}$ is the following

1. (1)

$[y_{mi},y_{ri}]=1,~{}~{}~{}~{}~{}~{}2\leq r<m\leq n,$ 2. (2)

$[y_{mi},y_{rj}]=1,\ \ 2\leq r<i\leq m\leq n,~{}~{}1\leq j\leq r,$ 3. (3)

$[y_{mi},y_{rj}]=[y_{mi},y_{mj}],\ \ 2\leq r<m\leq n,~{}~{}1\leq i\leq r<m\leq n,~{}~{}1\leq j\leq r,j\neq i.$

Since the group $H_{j}$ normalizes $H_{i}$ for $i>j$, the elements of $I_{n}$ have a normal form. That is, every element of $I_{n}$ may be uniquely written in the form $w_{n}w_{n-1}\ldots w_{2}$, where each $w_{i}\in H_{i}={\rm Inn}(F_{i})\cong F_{i}$. This implies that the word problem in $I_{n}$ is solvable. Indeed, an element $w_{n}w_{n-1}\ldots w_{2}=1$ if and only if each $w_{i}=1$ and that is decidable in a free group. Moreover, the above allow us to answer positively a question of Bardakov and Neshchadim in [3].

Theorem 1**.**

The conjugacy problem in $I_{n}$ is solvable.

*Proof. *Let $x=w_{n}\ldots w_{3}w_{2}$, $y=z_{n}\ldots z_{3}z_{2}$ and $g=g_{n}\ldots g_{3}g_{2}$ $\in I_{n}$, where $w_{i},z_{i},g_{i}\in H_{i}$, $i=2,\ldots,n$, such that $gxg^{-1}=y$. The equation can be rewritten as

[TABLE]

or equivalently,

[TABLE]

[TABLE]

Write

[TABLE]

and

[TABLE]

Clearly, $a_{n}\in H_{n}$ and $b_{n}\in I_{n-1}$. Now, the initial equation can be written as

[TABLE]

or

[TABLE]

Since $H_{n}$ is a normal subgroup of $I_{n}$, we get $g_{n}a_{n}b_{n}g_{n}^{-1}b_{n}^{-1}\in H_{n}$. Now, we may repeat the above decomposition as follows. Write

[TABLE]

and

[TABLE]

So, our equation has the form

[TABLE]

with $g_{n}a_{n}b_{n}g_{n}^{-1}b_{n}^{-1}\in H_{n}$ and $g_{n-1}a_{n-1}b_{n-1}g_{n-1}^{-1}b_{n-1}^{-1}\in H_{n-1}$. Repeating the above we will have, in finitely many steps,

[TABLE]

where $g_{i}a_{i}b_{i}g_{i}^{-1}b_{i}^{-1}\in H_{i}$ for all $i\in\{3,\ldots,n\}$ and $g_{2}w_{2}g_{2}^{-1}\in H_{2}$. So, the word in the left is in normal form. Thus, for the two words to be equal we must have $g_{2}w_{2}g_{2}^{-1}=z_{2}$. But that is the conjugacy problem in $I_{2}={\rm Inn}(F_{2})\cong F_{2}$, which is well known to be decidable (see, for example, [11, Proposition 2.14]). If such a $g_{2}$ doesn’t exist, then $x,y$ are not conjugate and we are done. If such a $g_{2}$ exists, then we must also have

[TABLE]

or equivalently,

[TABLE]

Since $g_{2}w_{2}g_{2}^{-1}\in H_{2}$ acts as an automorphism, say $\phi$, on $H_{3}$, we have

[TABLE]

But this is the twisted conjugacy problem in $H_{3}={\rm Inn}(F_{3})\cong F_{3}$, which is decidable (see, for example, [4]). Hence, if such a $g_{3}$ does not exist, then $x,y$ are not conjugate. If it exists, we continue the same way. So, in a finite number of steps, either we fail to find some $g_{i}$ and so, $x,y$ are not conjugate, or we complete the algorithm by demonstrating the appropriate $g$. ∎

In fact, the above proof shows the following more general result.

Corollary 1**.**

Let $G$ be the iterated semi-direct product $G=(\ldots(G_{n}\rtimes G_{n-1})\rtimes\ldots\rtimes G_{3})\rtimes G_{2}$ such that the group $G_{j}$ normalizes $G_{i}$ for all $i>j$, $i,j\in\{2,\ldots,n\}$. Then, the conjugacy problem is decidable in $G$ if the twisted conjugacy problem is decidable in $G_{i}$ for all $i=2,\ldots,n$.

3. The Lie algebra of $I_{n}$

Throughout this paper, by “Lie algebra”, we mean a Lie algebra over the ring of integers $\mathbb{Z}$ and we use the left-normed convention for group commutators and Lie commutators. Let $G$ be a group. For a positive integer $c$, let $\gamma_{c}(G)$ be the $c$-th term of the lower central series of $G$. We point out that $\gamma_{2}(G)=G^{\prime}$; that is, the commutator subgroup of $G$. Write ${\rm gr}_{c}(G)=\gamma_{c}(G)/\gamma_{c+1}(G)$ for $c\geq 1$. The (restricted) direct sum of the quotients ${\rm gr}_{c}(G)$ is the associated graded Lie algebra of $G$, ${\rm gr}(G)=\bigoplus_{c\geq 1}{\rm gr}_{c}(G)$. The Lie bracket multiplication in $\rm gr(G)$ is $[a\gamma_{c+1}(G),b\gamma_{d+1}(G)]=[a,b]\gamma_{c+d+1}(G)$, where $a\gamma_{c+1}(G)$ and $b\gamma_{d+1}(G)$ are the images of the elements $a\in\gamma_{c}(G)$ and $b\in\gamma_{d}(G)$ in the quotient groups ${\rm gr}_{c}(G)$ and ${\rm gr}_{d}(G)$, respectively, and $[a,b]\gamma_{c+d+1}(G)$ is the image of the group commutator $[a,b]$ in the quotient group ${\rm gr}_{c+d}(G)$. Multiplication is then extended to ${\rm gr}(G)$ by linearity.

In the present section, we investigate the Lie algebra ${\rm gr}(I_{n})$ of $I_{n}$ and we give a presentation for it. For $\mathbb{Z}$-submodules $A$ and $B$ of any Lie algebra, let $[A,B]$ be the $\mathbb{Z}$-submodule spanned by $[a,b]$ where $a\in A$ and $b\in B$. Furthermore, $B\wr A$ denotes the $\mathbb{Z}$-submodule defined by $B\wr A=B+[B,A]+[B,A,A]+\cdots$. Also, for any non-empty subsets ${\mathcal{U}}$ and ${\mathcal{V}}$ of any Lie algebra, we denote by $[{\mathcal{V}},{\mathcal{U}},\underbrace{{\mathcal{U}},\ldots,{\mathcal{U}}}_{a},\underbrace{{\mathcal{V}},\ldots,{\mathcal{V}}}_{b}]$ the set of (Lie) commutators where the first element belongs to ${\mathcal{V}}$, the second to ${\mathcal{U}}$ and then $a$ consecutive elements belong to ${\mathcal{U}}$ and the last $b$ consecutive elements belong to ${\mathcal{V}}$. For brevity, this is also denoted by $[{\mathcal{V}},{\mathcal{U}},~{}_{a}{\mathcal{U}},~{}_{b}{\mathcal{V}}]$. For a free $\mathbb{Z}$-module $A$, let $L(A)$ be the free Lie algebra on $A$, that is, the free Lie algebra on $\mathcal{A}$, where $\mathcal{A}$ is an arbitrary $\mathbb{Z}$-basis of $A$. Thus, we may write $L(A)=L({\mathcal{A}})$. For a positive integer $c$, let $L^{c}(A)$ denote the $c$th homogeneous component of $L(A)$. It is well-known that $L(A)=\bigoplus_{c\geq 1}L^{c}(A)$.

The following result is a version of Lazard’s ”Elimination Theorem” (see [5, Chapter 2, Section 2.9, Proposition 10]). In the form written here, it is a special case of [6, Lemma 2.2] (see, also, [14, Section 2.2]).

Lemma 1** (Elimination Theorem).**

Let $U$ and $V$ be free $\mathbb{Z}$-modules with $U\cap V=\{0\}$, and consider the free Lie algebra $L(U\oplus V)$. Then, $U$ and $V\wr U$ freely generate Lie subalgebras $L(U)$ and $L(V\wr U)$, and there is a $\mathbb{Z}$-module decomposition $L(U\oplus V)=L(U)\oplus L(V\wr U)$. Furthermore, $V\wr U=V\oplus[V,U]\oplus[V,U,U]\oplus\cdots$ and, for each $n\geq 0$, there is a $\mathbb{Z}$-module isomorphism $[V,~{}_{n}U]\cong V\otimes\underbrace{U\otimes\cdots\otimes U}_{n}$ under which $[v,u_{1},\ldots,u_{n}]$ corresponds to $v\otimes u_{1}\otimes\cdots\otimes u_{n}$ for all $v\in V$ and all $u_{1},\ldots u_{n}\in U$.

As a consequence of Lemma 1, we have the following result.

Corollary 1**.**

For free $\mathbb{Z}$-modules $U$ and $V$, we write $L(U\oplus V)$ for the free Lie algebra on $U\oplus V$. Then, there is a $\mathbb{Z}$-module decomposition $L(U\oplus V)=L(U)\oplus L(V)\oplus L(W)$, where $W=W_{2}\oplus W_{3}\oplus\cdots$ such that, for all $m\geq 2$, $W_{m}$ is the direct sum of submodules $[V,U,~{}_{a}U,~{}_{b}V]$ with $a+b=m-2$ and $a,b\geq 0$. Each $[V,U,~{}_{a}U,~{}_{b}V]$ is isomorphic to $V\otimes U\otimes\underbrace{U\otimes\cdots\otimes U}_{a}\otimes\underbrace{V\otimes\cdots\otimes V}_{b}$ as $\mathbb{Z}$-module. Furthermore, $L(W)$ is the ideal of $L(U\oplus V)$ generated by the module $[V,U]$.

3.1. A decomposition of the free Lie algebra

Let $L=L({\mathcal{Y}})$ be the free Lie algebra on the finite set $\mathcal{Y}$ and decompose ${\mathcal{Y}}$ into a disjoint union ${\mathcal{Y}}={\mathcal{U}}\cup{\mathcal{V}}$ of finite non-empty subsets ${\mathcal{U}}=\{u_{1},\ldots,u_{\kappa}\}$ and ${\mathcal{V}}=\{v_{1},\ldots,v_{\lambda}\}$. Also, define $u_{1}<\cdots<u_{\kappa}<v_{1}<\cdots<v_{\lambda}$. Let $U$ and $V$ denote the free ${\mathbb{Z}}$-modules with bases ${\mathcal{U}}$ and ${\mathcal{V}}$, respectively. Since the $\mathbb{Z}$-module generated by ${\mathcal{Y}}$ is equal to $U\oplus V$ and $L({\mathcal{Y}})$ is free on the set ${\mathcal{Y}}$, $L$ is free on the set ${\mathcal{U}}\cup{\mathcal{V}}$ and we have $L=L(U\oplus V)$. By Lemma 1 and Corollary 1, we have

[TABLE]

where $W=W_{2}\oplus W_{3}\oplus\cdots$ such that, for all $m\geq 2$, $W_{m}=\bigoplus_{a+b=m-2}[V,U,~{}_{a}U,~{}_{b}V]$. Furthermore, $L(V\wr U)$ and $L(W)$ are the ideals in $L$ generated by the modules $V$ and $[V,U]$, respectively. In particular, $L(W)$ is the ideal in $L$ generated by the natural $\mathbb{Z}$-basis $[{\mathcal{V}},{\mathcal{U}}]=\{[v_{i},u_{j}]:1\leq i\leq\lambda,1\leq j\leq\kappa\}$ of $[V,U]$. Let ${\mathcal{X}}_{V,U}$ be the natural $\mathbb{Z}$-basis of $V\wr U$. That is,

[TABLE]

where $[{\mathcal{V}},~{}_{a}{\mathcal{U}}]=\{[v,z_{1},\ldots,z_{a}]:v\in{\mathcal{V}},z_{1},\ldots,z_{a}\in{\mathcal{U}}\}$ is the natural $\mathbb{Z}$-basis of the module $[V,~{}_{a}U]$. We point out that the set

[TABLE]

where $[{\mathcal{V}},{\mathcal{U}},~{}_{a}{\mathcal{U}},~{}_{b}{\mathcal{V}}]=\{[v,z,z_{1},\ldots,z_{a},v_{1},\ldots,v_{b}]:v,v_{1},\ldots,v_{b}\in{\mathcal{V}},z,z_{1},\ldots,z_{a}\in{\mathcal{U}}\}$, is the natural $\mathbb{Z}$-basis of $W$.

Lemma 2**.**

With the above notation, let $\phi$ be any automorphism of the free $\mathbb{Z}$-module $[V,U]$ and let $\psi_{2}:[{\mathcal{V}},{\mathcal{U}}]\rightarrow L(V\wr U)$ be the map given by $\psi_{2}([v_{i},u_{j}])=\phi([v_{i},u_{j}])+w_{i,j}$, where $w_{i,j}\in L^{2}(V)$ for all $1\leq i\leq\lambda$ and $1\leq j\leq\kappa$. Furthermore, for every $a\geq 3$, let $\psi_{a}$ be the mapping from $[{\mathcal{V}},{\mathcal{U}},~{}_{(a-2)}{\mathcal{U}}]$ into $L(V\wr U)$ satisfying the conditions $\psi_{a}([v,z,z_{1},\ldots,z_{a-2}])=[\psi_{2}([v,z]),z_{1},\ldots,z_{a-2}]$ for all $v\in{\mathcal{V}}$ and $z,z_{1},\ldots,z_{a-2}\in{\mathcal{U}}$. Define $\Psi:{\mathcal{X}}_{V,U}\rightarrow L(V\wr U)$ to be the induced map with $\Psi(v)=v$ for all $v\in{\mathcal{V}}$ and, for $a\geq 2$, $\Psi(v)=\psi_{a}(v)$ for all $v\in[{\mathcal{V}},{\mathcal{U}},~{}_{(a-2)}{\mathcal{U}}]$. Then,

1. (1)

$\psi_{2}$* extends to a $\mathbb{Z}$-linear monomorphism from $[V,U]$ into $L(V\wr U)$.* 2. (2)

For every $a\geq 3$, there exists an automorphism $\phi_{a}$ of the free $\mathbb{Z}$-module $[V,U,~{}_{(a-2)}U]$ such that, for all $v\in{\mathcal{V}}$ and $z,z_{1},\ldots,z_{a-2}\in{\mathcal{U}}$,

[TABLE]

where $w_{a}\in L({\mathcal{V}}\cup(\bigcup_{1\leq\beta\leq a-2}[{\mathcal{V}},~{}_{\beta}{\mathcal{U}}]))$. 3. (3)

$\Psi$* extends to a (Lie algebra) automorphism of $L(V\wr U)$.*

Proof.

1. (1)

Since $[{\mathcal{V}},{\mathcal{U}}]$ is a $\mathbb{Z}$-basis of $[V,U]$, $\psi_{2}$ extends to a $\mathbb{Z}$-linear mapping, denoted $\psi_{2}$ again, from $[V,U]$ into $L(V\wr U)$. We point out that $\psi_{2}([V,U])\subseteq L^{2}(V)\oplus[V,U]$. Since $L^{2}(V)\cap[V,U]=\{0\}$, $\phi$ is an automorphism of $[V,U]$ and the set $[{\mathcal{V}},{\mathcal{U}}]$ is a $\mathbb{Z}$-basis of $[V,U]$, we have $\psi_{2}$ is a $\mathbb{Z}$-linear monomorphism of $[V,U]$ into $L(V\wr U)$. 2. (2)

Let $a\geq 3$. Since $[{\mathcal{V}},{\mathcal{U}},~{}_{(a-2)}{\mathcal{U}}]$ is a $\mathbb{Z}$-basis for the $\mathbb{Z}$-module $[V,U,~{}_{(a-2)}U]$ and $\psi_{2}$ is a $\mathbb{Z}$-linear mapping, we have $\psi_{a}$ extends to a $\mathbb{Z}$-linear mapping from $[V,U,~{}_{(a-2)}U]$ into $L(V\wr U)$. For $a\geq 3$, let $\phi_{a}$ be the mapping from $[{\mathcal{V}},{\mathcal{U}},~{}_{(a-2)}{\mathcal{U}}]$ into $[V,U,~{}_{(a-2)}U]$ satisfying the conditions $\phi_{a}([v,z,z_{1},\ldots,$ $z_{a-2}])=[\phi([v,z]),z_{1},\ldots,z_{a-2}]$ for all $v\in{\mathcal{V}}$ and $z,z_{1},z_{a-2}\in{\mathcal{U}}$. Since $\phi$ is a $\mathbb{Z}$-linear mapping, $\phi_{a}$ extends to a $\mathbb{Z}$-linear mapping of $[V,U,~{}_{(a-2)}U]$. Let $\chi_{a}$ be the mapping from $[{\mathcal{V}},{\mathcal{U}},~{}_{(a-2)}{\mathcal{U}}]$ into $[V,U,~{}_{(a-2)}U]$ satisfying the conditions $\chi_{a}([v,z,z_{1},\ldots,z_{a-2}])=[\phi^{-1}([v,z]),z_{1},\ldots,z_{a-2}]$ for all $v\in{\mathcal{V}}$, $z,z_{1},\ldots,z_{a-2}\in{\mathcal{U}}$, where $\phi^{-1}$ is the inverse of $\phi$. We point out that the elements $z_{1},\ldots,z_{a-2}\in{\mathcal{U}}$ are invariant under $\phi_{a}$ and $\chi_{a}$ in the above expressions. Since $\phi^{-1}$ is the inverse of $\phi$, $z_{1},\ldots,z_{a-2}$ are invariant under $\phi_{a}$ and $\chi_{a}$ and $[{\mathcal{V}},{\mathcal{U}},~{}_{(a-2)}{\mathcal{U}}]$ is a $\mathbb{Z}$-basis of $[V,U,~{}_{(a-2)}U]$, we have $\chi_{a}$ is the inverse of $\phi_{a}$. Therefore, $\phi_{a}$ is an automorphism of $[V,U,~{}_{(a-2)}U]$. Now, for $v\in{\mathcal{V}}$, $z,z_{1},\ldots,z_{a-2}\in{\mathcal{U}}$,

[TABLE]

where $w_{v,z}\in L^{2}(V)$. By using the Jacobi identity in the form $[x,y,z]=[x,z,y]+[x,[y,z]]$, we may show that

[TABLE]

Therefore, for all $v\in{\mathcal{V}}$ and $z,z_{1},\ldots,z_{a-2}\in{\mathcal{U}}$,

[TABLE]

where $w_{a}\in L({\mathcal{V}}\cup(\bigcup_{1\leq\beta\leq a-2}[{\mathcal{V}},~{}_{\beta}{\mathcal{U}}]))$. 3. (3)

Since ${\mathcal{X}}_{V,U}$ is a free generating set of $L(V\wr U)$, $\Psi$ extends to an endomorphism of $L(V\wr U)$. Since $\Psi(v)=v$ for all $v\in{\mathcal{V}}$, we obtain from Lemma 2 (1)-(2) and [7, Lemma 2.1] that $\Psi$ is an automorphism of $L(V\wr U)$. ∎

Since $L(W)$ is a free Lie subalgebra of $L(V\wr U)$ on $W$ and $\Psi$ is an automorphism of $L(V\wr U)$, we have $\Psi(L(W))$ is a free Lie subalgebra of $L(V\wr U)$. In fact, $\Psi(L(W))=L(\Psi(W))$, that is, $\Psi(L(W))$ is a free Lie algebra on $\Psi(W)$. Furthermore, since $\Psi$ is an automorphism of $L(V\wr U)$ and $L(W)$ is an ideal of $L(V\wr U)$, we obtain $\Psi(L(W))=L(\Psi(W))$ is an ideal in $L(V\wr U)$. We point out that

[TABLE]

and so, \begin{array}[]{lll}L&=&L(U)\oplus L(V)\oplus L(\Psi(W)).\end{array} Write

[TABLE]

Since $\mathcal{W}$ is a $\mathbb{Z}$-basis for $W$ and $\Psi$ is an automorphism of $L(V\wr U)$, we have $\mathcal{C}$ is a $\mathbb{Z}$-basis of $\Psi(W)$ and so, $L(\Psi(W))=L({\mathcal{C}})$. For a positive integer $m$, with $m\geq 2$, let

[TABLE]

In particular, for $m\geq 2$, the set

[TABLE]

is a natural $\mathbb{Z}$-basis for $W_{m,\Psi}$ and so, $\Psi(W)=\bigoplus_{m\geq 2}W_{m,\Psi}$. Obviously, ${\mathcal{C}}=\bigcup_{m\geq 2}{\mathcal{W}}_{m,\Psi}$. Furthermore, for $m\geq 2$, we write $L^{m}_{\rm grad}(\Psi(W))$ for $L^{m}\cap L(\Psi(W))$. That is, $L^{m}_{\rm grad}(\Psi(W))$ is the $\mathbb{Z}$-submodule of $L^{m}$ spanned by all Lie commutators of the form $[w_{1},\ldots,w_{\kappa}]$, with $\kappa\geq 1$, $w_{i}\in{\mathcal{W}}_{m(i),\Psi}$ and $m(1)+\cdots+m(\kappa)=m$. It is easily checked that

[TABLE]

Proposition 1**.**

With the above notation, $L(\Psi(W))$ is an ideal in $L$.

*Proof. *To show that the Lie subalgebra $L(\Psi(W))$ of $L$ is an ideal in $L$, it is enough to show that $[w,u]\in L(\Psi(W))$ for all $w\in L(\Psi(W))$ and $u\in L$. Since $L(\Psi(W))$ is an ideal in $L(V\wr U)$ and by Lemma 1, it is enough to show that $[w,u]\in L(\Psi(W))$ for all $w\in L(\Psi(W))$ and $u\in L(U)$. Since any $u\in L(U)$ is written as a $\mathbb{Z}$-linear combination of Lie commutators $[u_{i_{1}},\ldots,u_{i_{\kappa}}]$, with $\kappa\geq 1$ and $u_{i_{1}},\ldots,u_{i_{\kappa}}\in{\mathcal{U}}$, and by using the Jacobi identity in the form $[x,[y,z]]=[x,y,z]-[x,z,y]$, it is enough to show that $[w,u_{i_{1}},\ldots,u_{i_{k}}]\in L(\Psi(W))$ for all $w\in L(\Psi(W))$ and $u_{i_{1}},\ldots,u_{i_{k}}\in{\mathcal{U}}$. Since $\mathcal{C}$ is a (free) generating set of $L(\Psi(W))=L({\mathcal{C}})$, any $w\in L(\Psi(W))$ is written as a $\mathbb{Z}$-linear combination of Lie commutators $[w_{1},\ldots,w_{\lambda}]$ with $w_{1},\ldots,w_{\lambda}\in{\mathcal{C}}$. By the linearity of the Lie bracket, it is enough to show that $[w_{1},\ldots,w_{\lambda},u_{i_{1}},\ldots,u_{i_{\kappa}}]\in L(\Psi(W))$ for all $w_{1},\ldots,w_{\lambda}\in{\mathcal{C}}$ and $u_{i_{1}},\ldots,u_{i_{\kappa}}\in{\mathcal{U}}$. By using induction on $\lambda$, the Jacobi identity in the form $[x,y,z]=[x,z,y]+[x,[y,z]]$ and the fact that $L(\Psi(W))$ is a Lie algebra, it suffices to show that $[w,u_{i_{1}},\ldots,u_{i_{\kappa}}]\in L(\Psi(W))$ for all $w\in{\mathcal{C}}$ and $u_{i_{1}},\ldots,u_{i_{\kappa}}\in{\mathcal{U}}$.

Notice that if $w\in\Psi([{\mathcal{V}},{\mathcal{U}}])$, then $[w,u_{i_{1}},\ldots,u_{i_{\kappa}}]\in{\mathcal{C}}$ for all $u_{i_{1}},\ldots,u_{i_{\kappa}}\in{\mathcal{U}}$. Thus, we assume that $w\in[\Psi([{\mathcal{V}},{\mathcal{U}}]),~{}_{a}{\mathcal{U}},~{}_{b}{\mathcal{V}}]$ for some integers $a,b\geq 0$ with $a+b\geq 1$ and $b\geq 1$. For elements $u,v_{1},\ldots,v_{m}\in L$, we denote

[TABLE]

Rewrite $w=[u,v_{i_{1}},\ldots,v_{i_{b}}]$, where $u\in[\Psi([{\mathcal{V}},{\mathcal{U}}]),~{}_{a}{\mathcal{U}}]$, $v_{i_{1}},\ldots,v_{i_{b}}\in{\mathcal{V}}$, $a,b$ non-negative integers with $a+b\geq 1$ and $b\geq 1$. By applying repeatedly the Jacobi identity in the form $[x,y,z]=[x,z,y]+[x,[y,z]]$, we get $w=[u,~{}_{\mu_{1}}v_{1},\ldots,~{}_{\mu_{\lambda}}v_{\lambda}]+\tilde{w}$, where $\mu_{1}+\cdots+\mu_{\lambda}=b$ and $\tilde{w}\in L(V\wr U)$. In particular, $\tilde{w}$ is a summand of elements of the form $[u,~{}_{\mu^{\prime}_{1}}v_{1},\ldots,~{}_{\mu^{\prime}_{\beta}}v_{\beta},z_{i_{1}},\ldots,z_{i_{\gamma}}]$, where $\mu^{\prime}_{1},\ldots,\mu^{\prime}_{\beta}\geq 0$, $\mu^{\prime}_{1}+\cdots+\mu^{\prime}_{\beta}\leq b$ and $z_{i_{1}},\ldots,z_{i_{\gamma}}\in L^{\prime}(V)$. Since $L(\Psi(W))$ is an ideal in $L(V\wr U)$, it suffices to consider the case where $w=[u,~{}_{\mu_{1}}v_{1},\ldots,~{}_{\mu_{\lambda}}v_{\lambda}]$, with $u\in[\Psi([{\mathcal{V}},{\mathcal{U}}]),~{}_{a}{\mathcal{U}}]$, $\mu_{1},\ldots,\mu_{\lambda}\geq 0$, $\mu_{1}+\cdots+\mu_{\lambda}=b\geq 1$ and $v_{1},\ldots,v_{\lambda}\in{\mathcal{V}}$. Since $\phi$ is an automorphism of $[V,U]$, we may write, for $i\in\{1,\ldots,\lambda\}$, $j\in\{1,\ldots,\kappa\}$,

[TABLE]

where $b_{i,j,r,s}\in\mathbb{Z}$ and $\tilde{w}_{i,j}\in L^{2}(V)$. By using the Jacobi identity in the expression $[u,~{}_{\mu_{1}}v_{1},\ldots,~{}_{\mu_{m}}v_{m},u_{i_{1}},\ldots,u_{i_{k}}]$, where $u\in[\Psi([{\mathcal{V}},{\mathcal{U}}]),~{}_{a}{\mathcal{U}}]$, $\mu_{1},\ldots,\mu_{m}\geq 0$, $\mu_{1}+\cdots+\mu_{m}=b\geq 1$, $v_{1},\ldots,v_{m}\in{\mathcal{V}}$ and $u_{i_{1}},\ldots,u_{i_{k}}\in{\mathcal{U}}$, replacing each $[v_{i},u_{j}]$ by the equation $(1)$ as many times as it is needed and since $L(\Psi(W))$ is an ideal in $L(V\wr U)$, we may show that $[u,~{}_{\mu_{1}}v_{1},\ldots,~{}_{\mu_{m}}v_{m},u_{i_{1}},\ldots,u_{i_{k}}]\in L(\Psi(W))$ (see [15, Example 1] for an example of the above procedure). Therefore, $L(\Psi(W))$ is an ideal in $L$. ∎

Since

[TABLE]

we have the following result.

Corollary 2**.**

Let $\Psi$ be the Lie algebra automorphism of $L(V\wr U)$ defined above. Then, for every positive integer $m$, with $m\geq 2$, $L^{m}=L^{m}(U)\oplus L^{m}(V)\oplus L^{m}_{\rm grad}(\Psi(W))$.

3.2. A further decomposition

Let $L({\mathcal{Y}})$ be the free Lie algebra on the finite set $\mathcal{Y}$ and decompose $\mathcal{Y}$ into a disjoint union ${\mathcal{Y}}={\mathcal{Y}}_{2}\cup\cdots\cup{\mathcal{Y}}_{n}$, with $n\geq 3$, of finite non-empty subsets of the form ${\mathcal{Y}}_{i}=\{y_{i1},y_{i2},\ldots,y_{ik_{i}}\}$, $2\leq i\leq n$ and $k_{i}\geq 1$. Moreover, define $y_{i1}<\cdots<y_{ik_{i}}$ and $y_{ik_{i}}<y_{(i+1)1}$ for all $i\in\{2,\ldots,n-1\}$. For $m\in\{2,\ldots,n\}$, let $Y_{m}$ be the free $\mathbb{Z}$-module with $\mathbb{Z}$-basis ${\mathcal{Y}}_{m}$. Thus,

[TABLE]

Then,

[TABLE]

Write $W_{(2,\ldots,n),2}=\bigoplus_{2\leq\kappa<\rho\leq n}[Y_{\rho},Y_{\kappa}]$. For $r\in\{2,\ldots,n-1\}$, let $U_{r}=Y_{r}\oplus\cdots\oplus Y_{n}$. We point out that ${\mathcal{U}}_{r}=\bigcup_{\kappa=r}^{n}{\mathcal{Y}}_{\kappa}$ is a natural $\mathbb{Z}$-basis of $U_{r}$. It is easily verified that

[TABLE]

For $r\in\{2,\ldots,n-1\}$, $[U_{r+1},Y_{r}]$ is generated by the following standard free generating set $[{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}]=\{[y_{\mu\nu},y_{rl}]:r+1\leq\mu\leq n,\nu=1,\ldots,k_{\mu},l=1,\ldots,k_{r}\}$. Furthermore, $W_{(2,\ldots,n),2}$ is generated by the set $K=\bigcup_{r=2}^{n-1}[{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}]$, which is a natural $\mathbb{Z}$-basis of $W_{(2,\ldots,n),2}$. Let $\phi_{r}$ be any automorphism of the free $\mathbb{Z}$-module $[U_{r+1},Y_{r}]$ and let $\psi_{2,r}:[{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}]\rightarrow[U_{r+1},Y_{r}]\oplus L^{2}(U_{r+1})$ be the map given by $\psi_{2,r}([y_{\mu\nu},y_{rl}])=\phi_{r}([y_{\mu\nu},y_{rl}])+w_{\mu,\nu,r,l}$, where $w_{\mu,\nu,r,l}\in L^{2}(U_{r+1})$, $\mu=r+1,\ldots,n$, $\nu=1,\ldots,k_{\mu}$ and $l=1,\ldots,r$. We point out that

[TABLE]

Since $\phi_{r}$ is an automorphism and $[U_{r+1},Y_{r}]\cap L^{2}(U_{r+1})=\{0\}$, we get $\psi_{2,r}$ is $1-1$. For $a\geq 3$, let $\psi_{a,r}$ be the mapping from $[{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r},~{}_{(a-2)}{\mathcal{Y}}_{r}]$ into $L(U_{r+1}\wr Y_{r})$ satisfying the conditions

[TABLE]

for all $u\in{\mathcal{U}}_{r+1}$ and $z,z_{1},\ldots,z_{a-2}\in{\mathcal{Y}}_{r}$. Let ${\mathcal{X}}_{U_{r+1},Y_{r}}$ be the natural $\mathbb{Z}$-basis of $U_{r+1}\wr Y_{r}$. That is,

[TABLE]

where $[{\mathcal{U}}_{r+1},~{}_{a}{\mathcal{Y}}_{r}]=\{[u,z_{1},\ldots,z_{a}]:u\in{\mathcal{U}}_{r+1},z_{1},\ldots,z_{a}\in{\mathcal{Y}}_{r}\}$ is the natural $\mathbb{Z}$-basis of the module $[U_{r+1},~{}_{a}Y_{r}]$. Furthermore, we define $\Psi_{r}:{\mathcal{X}}_{U_{r+1},Y_{r}}\rightarrow L(U_{r+1}\wr Y_{r})$ to be the map with $\Psi_{r}(u)=u$ for all $u\in{\mathcal{U}}_{r+1}$ and, for $a\geq 2$, $\Psi_{r}(u)=\psi_{a,r}(u)$ for all $u\in[{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r},~{}_{(a-2)}{\mathcal{Y}}_{r}]$. By Lemma 2 (3), $\Psi_{r}$ extends to a (Lie algebra) automorphism of $L(U_{r+1}\wr Y_{r})$. By applying Lemma 1 on $L(U_{r})$, we have

[TABLE]

where $W_{(r)}=W_{(r),2}\oplus W_{(r),3}\oplus\cdots$ such that, for all $\kappa\geq 2$,

[TABLE]

Furthermore, $L(U_{r+1}\wr Y_{r})$ and $L(W_{(r)})$ are the ideals in $L(U_{r})$ generated by the modules $U_{r+1}$ and $[U_{r+1},Y_{r}]$, respectively. By Proposition 1, $T_{r}=L(\Psi_{r}(W_{(r)}))$ is an ideal of $L(U_{r})$. Thus, for $r\in\{2,\ldots,n-1\}$,

[TABLE]

Furthermore, for $r\in\{2,\ldots,n-1\}$, let

[TABLE]

where, for $\kappa\geq 2$,

[TABLE]

and ${\mathcal{U}}_{r+1}={\mathcal{Y}}_{r+1}\cup\cdots\cup{\mathcal{Y}}_{n}$. Thus, for $r\in\{2,\ldots,n-1\}$, $T_{r}=L(\Psi_{r}(W_{(r)}))=L({\mathcal{C}}_{(r)})$ and $L(U_{r})=L(Y_{r})\oplus L(U_{r+1})\oplus T_{r}$. We define, for $r\in\{2,\ldots,n-1\}$,

[TABLE]

We point out that, for $r\geq 3$, $\tilde{T}_{r}=T_{r}\oplus\tilde{T}_{r-1}$.

Proposition 2**.**

With the above notation, for every $r\in\{3,\ldots,n-1\}$,

1. (1)

$L=\left(\bigoplus_{j=2}^{r-1}L(Y_{j})\right)\oplus L(U_{r})\oplus\tilde{T}_{r-1}$. In particular, for $r=n-1$, we have $L=\left(\bigoplus_{i=2}^{n}L(Y_{i})\right)\oplus\tilde{T}_{n-1}$. 2. (2)

$[T_{r},L]\subseteq\left(\sum_{j=2}^{r-1}[T_{r},L(Y_{j})]\right)+T_{r}+[T_{r},\tilde{T}_{r-1}]$.

*Proof. *(1) We induct on $r$. Let $r=3$. By Lemma 1, $L=L(Y_{2})\oplus L(U_{3}\wr Y_{2})$. By Lemma 2 (3), $\Psi_{2}$ extends to a Lie automorphism of $L(U_{3}\wr Y_{2})$ and so,

[TABLE]

We assume that our claim is valid for $\kappa-1$, with $4\leq\kappa\leq n-1$. Thus,

[TABLE]

By Lemma 1, applied on $L(U_{\kappa-1})$, we have $L(U_{\kappa-1})=L(Y_{\kappa-1})\oplus L(U_{\kappa}\wr Y_{\kappa-1}).$ As before, by Lemma 2 (3), $\Psi_{\kappa-1}$ extends to a Lie automorphism of $L(U_{\kappa}\wr Y_{\kappa-1})$ and so, by Proposition 1, $T_{\kappa-1}=L(\Psi_{\kappa-1}(W_{(\kappa-1)}))$ is an ideal of $L(U_{\kappa-1})$. But

[TABLE]

Thus, we have the following decomposition of $L$,

[TABLE]

and so, we obtain the required result.

(2) By Proposition 2 (1), for $r\in\{3,\ldots,n-1\}$,

[TABLE]

For any $r\in\{3,\ldots,n-1\}$ and by using the equations (2) and (3), we get the required result. ∎

By Corollary 2, for any $r\in\{3,\ldots,n-1\}$ and $m\geq 2$,

[TABLE]

Since $\tilde{T}_{n-1}=\bigoplus_{s=2}^{n-1}L_{\rm grad}(\Psi_{s}(W_{(s)}))$, we have, for any $m\geq 2$,

[TABLE]

By Proposition 2 (1) and the above equation, we have the following result.

Corollary 3**.**

For any positive integer $m$, with $m\geq 2$, $L^{m}=\left(\bigoplus_{i=2}^{n}L^{m}(Y_{i})\right)\oplus\tilde{T}_{n-1}^{m}$.

Let $J$ be the ideal of $L$ generated by the set ${\mathcal{J}}$ of all elements of the form $\psi_{2,r}([y_{\mu\nu},y_{rl}])$, $r=2,\ldots,n-1$, $\mu=r+1,\ldots,n-1$, $\nu=1,\ldots,k_{\mu}$ and $l=1,\ldots,r$. Since $J$ is a homogeneous ideal of $L$, we have $J=\bigoplus_{m\geq 2}J^{m}$, where $J^{m}=J\cap L^{m}$.

Lemma 3**.**

With the above notation, if $\tilde{T}_{n-1}$ is an ideal of $L$, then $\tilde{T}_{n-1}=J$ and so, $L=(\bigoplus_{i=2}^{n}L(Y_{i}))\oplus J$. In particular, for any positive integer $m$, with $m\geq 2$, $J^{m}=\tilde{T}_{n-1}^{m}=\bigoplus_{s=2}^{n-1}L^{m}_{\rm grad}(\Psi_{s}(W_{(s)}))$ and $L^{m}=\left(\bigoplus_{i=2}^{n}L^{m}(Y_{i})\right)\oplus J^{m}$.

*Proof. *Let $\tilde{T}_{n-1}$ be an ideal of $L$. We point out that ${\mathcal{J}}\subseteq\tilde{T}_{n-1}\subseteq J$. Since $\tilde{T}_{n-1}$ is an ideal in $L$, we obtain $J\subseteq\tilde{T}_{n-1}$, and so, $J=\tilde{T}_{n-1}$. Hence, by Corollary 3 and since $J$ is homogeneous, we obtain the required result. ∎

3.3. The case of $I_{n}$

Let us now consider the case of the subgroup $I_{n}$ of $M_{n}$, with $n\geq 3$. Let $L({\mathcal{Y}})$ be the free Lie algebra on the set ${\mathcal{Y}}$ which is decomposed as ${\mathcal{Y}}=\bigcup_{i=2}^{n}{\mathcal{Y}_{i}}$ with ${\mathcal{Y}}_{i}\cap{\mathcal{Y}}_{j}=\emptyset$ for $i\neq j$ and ${\mathcal{Y}}_{m}=\{y_{m1},\ldots,y_{mm}\}$, $2\leq m\leq n$, and define $y_{m1}<\cdots<y_{mm}$ and $y_{mm}<y_{(m+1)1}$ for all $m\in\{2,\ldots,n-1\}$. For $m\in\{2,\ldots,n\}$, let $Y_{m}$ be the free $\mathbb{Z}$-module generated by ${\mathcal{Y}}_{m}$ and thus $L=L({\mathcal{Y}})=L(Y_{2}\oplus\ldots\oplus Y_{n})$. For $r\in\{2,\ldots,n-1\}$, define $U_{r},T_{r}$ and $\tilde{T}_{r}$ as in the previous subsection. Namely, $U_{r}=Y_{r}\oplus\cdots\oplus Y_{n}$ with ${\mathcal{U}}_{r}=\bigcup_{\kappa=r}^{n}{\mathcal{Y}}_{\kappa}$ is a natural $\mathbb{Z}$-basis of $U_{r}$, $T_{r}=L(\Psi_{r}(W_{(r)}))$ and $\tilde{T}_{r}=\bigoplus_{s=2}^{r}T_{s}$. Furthermore, we define the map $\psi_{2,r}:[{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}]\rightarrow[U_{r+1},Y_{r}]\oplus L^{2}(U_{r+1})$ as follows:

$(F1)~{}\psi_{2,r}([y_{mi},y_{ri}])=[y_{mi},y_{ri}],~{}~{}~{}~{}~{}~{}2\leq r<m\leq n,$

$(F2)~{}\psi_{2,r}([y_{mi},y_{rj}])=[y_{mi},y_{rj}],\ \ 2\leq r<i\leq m\leq n,~{}~{}1\leq j\leq r,$

$(F3)~{}\psi_{2,r}([y_{mi},y_{rj}])=[y_{mi},y_{rj}]-[y_{mi},y_{mj}],\ \ 2\leq r<m\leq n,~{}~{}1\leq i\leq r<m\leq n,~{}~{}1\leq j\leq r,j\neq i.$

Lemma 4**.**

For $n\geq 4$, $r\in\{3,\ldots,n-1\}$ and $j\in\{2,\ldots,r-1\}$,

[TABLE]

*Proof. *Let $n\geq 4$ and fix $r\in\{3,\ldots,n-1\}$ and $j\in\{2,\ldots,r-1\}$. Let

[TABLE]

Note that $k\in\{1,\ldots,m\}$, $l\in\{1,\ldots,r\}$ and $t\in\{1,\ldots,j\}$. In what follows, we make extensive use of the definition of $\psi_{2,\mu}$ ($(F1),(F2)~{}{\rm and}~{}(F3)$) with $\mu\in\{2,\ldots,n-1\}$, the Jacobi identity of the form $[x,y,z]=[x,z,y]-[y,z,x]$ and the fact that $T_{\nu}$ is an ideal of $L(U_{\nu})$ for all $\nu=2,\ldots,n-1$, a consequence of Proposition 1. We separate various cases. We give full details for the proof of the first case. Similar arguments may be applied to the remaining cases.

1. (1)

$k,l\leq j\leq r-1$.

1. (a)

$k\neq l$. Then, we separate the following cases.

1. (i)

$t\neq k\leq j$ and $l\neq t\leq j$. Then, since $1\leq k\neq l\leq r-1$, we have, by using $(F3)$,

[TABLE]

By using the Jacobi identity in the form $[x,y,z]=[x,z,y]-[y,z,x]$,

[TABLE]

Since $1\leq t\neq k\leq j$ and $1\leq l\neq t\leq j$, we get, by using $(F3)$,

[TABLE]

Write

[TABLE]

By using the Jacobi identity, we have

[TABLE]

Since $1\leq k\neq l<r$, $1\leq t\neq l<r$ and $1\leq k\neq t<r$, we obtain, by using $(F3)$,

[TABLE]

Write

[TABLE]

We point out that

[TABLE]

and so,

[TABLE]

By using the Jacobi identity,

[TABLE]

Since $1\leq k\neq l<r$, $1\leq t\neq l<r$ and $1\leq k\neq t<r$, we obtain by using $(F3)$,

[TABLE]

Finally, write

[TABLE]

Since $[y_{mk},y_{mt},y_{ml}]=[y_{mk},y_{ml},y_{mt}]-[y_{mt},y_{ml},y_{mk}]$, we obtain

[TABLE] 2. (ii)

$1\leq t\neq k\leq j$ and $1\leq l=t\leq j$. Then, since $1\leq k\neq t\leq j<r$ and $l=t$, we have, by using $(F3)$,

[TABLE]

By using the Jacobi identity,

[TABLE]

Since $1\leq t\neq k\leq j$, we get, by using $(F3)$ and $(F1)$,

[TABLE]

Write

[TABLE]

and by using the Jacobi identity, we have

[TABLE]

Since $1\leq t\neq k\leq j<r$, we obtain, by using $(F3)$,

[TABLE]

Thus,

[TABLE] 3. (iii)

$1\leq t=k\leq j$ and $1\leq l\neq t\leq j$. Then, since $1\leq l\neq t\leq j<r$ and $k=t$, we have, by using $(F3)$,

[TABLE]

By using the Jacobi identity,

[TABLE]

Since $1\leq l\neq k\leq j$, we get, by using $(F1)$ and $(F3)$,

[TABLE]

Write

[TABLE]

Now,

[TABLE]

By using the Jacobi identity,

[TABLE]

Since $1\leq l\neq k\leq j<r$, we obtain, by using $(F3)$,

[TABLE]

Since $[y_{mk},y_{ml},y_{rk}]=[y_{mk},y_{rk},y_{ml}]-[y_{ml},y_{rk},y_{mk}]$ and $1\leq l\neq k\leq j<r$, we obtain, by using $(F3)$,

[TABLE]

Thus,

[TABLE] 2. (b)

$k=l$. Then,

[TABLE]

By using the Jacobi identity,

[TABLE]

Note that $1\leq k,t\leq j$ and so, we separate two cases.

1. (i)

$k=t$. Then,

[TABLE] 2. (ii)

$k\neq t\leq j$. Then, by using $(F3)$,

[TABLE]

Write

[TABLE]

By using the Jacobi identity, $(F1)$ and $(F3)$, we get

[TABLE]

Write

[TABLE]

and so,

[TABLE]

Thus,

[TABLE] 2. (2)

$k\leq j\leq r-1$ and $j+1\leq l\leq r$. Then,

[TABLE] 3. (3)

$j+1\leq k\leq r$ and $l\leq j\leq r$. Then,

[TABLE] 4. (4)

$j+1\leq k\leq r$ and $j+1\leq l\leq r$. Then,

[TABLE] 5. (5)

$r+1\leq k$

1. (a)

$j+1\leq l$. Then,

[TABLE] 2. (b)

$l\leq j$. Then,

[TABLE]

Remark 1**.**

By the proof of Lemma 4, we have, for $n\geq 4$, $r\in\{3,\ldots,n-1\}$ and $j\in\{2,\ldots,r-1\}$,

[TABLE]

Lemma 5**.**

For $n\geq 4$, $r\in\{3,\ldots,n-1\}$, $j\in\{2,\ldots,r-1\}$ and $a,b\geq 0$, $[\psi_{2,r}([U_{r+1},Y_{r}]),~{}_{a}Y_{r},~{}_{b}U_{r+1},Y_{j}]\subseteq\tilde{T}_{r}$.

*Proof. *For $a=b=0$, our claim follows from Lemma 4. Thus, we assume that $a,b\geq 0$ with $a+b\geq 1$. Let $w_{a,b}=[u,y_{1},\ldots,y_{a},y^{\prime}_{1},\ldots,y^{\prime}_{b}]$ with $u\in\psi_{2,r}([{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}])$, $y_{1},\ldots,y_{a}\in{\mathcal{Y}}_{r}$, $y^{\prime}_{1},\ldots,y^{\prime}_{b}\in{\mathcal{U}}_{r+1}$.

Suppose that $a+b=1$ and let $w=[u,y,z]$, where $u\in\psi_{2,r}([{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}])$, $y\in{\mathcal{Y}}_{s}$, $s\geq r$ and $z\in{\mathcal{Y}}_{j}$. By the Jacobi identity, $w=[u,z,y]+[u,[y,z]]$. Since $j<r$, either $[y,z]=\psi_{2,j}([y,z])$ or $[y,z]=\psi_{2,j}([y,z])+[y,y_{1}]$, with $y,y_{1}\in{\mathcal{Y}}_{s}$, $r\leq s$. Then, either

[TABLE]

or

[TABLE]

By Remark 1 and using the technique of Lemma 4 we get $w\in\tilde{T}_{r}$, since $\tilde{T}_{r}$ is a Lie subalgebra.

So, we assume that $a+b\geq 2$. Then, by using repeatedly the Jacobi identity, we get

[TABLE]

As before, we obtain $[w_{a,b},z]\in\tilde{T}_{r}$. ∎

Lemma 6**.**

For $n\geq 4$, $r\in\{3,\ldots,n-1\}$, $j\in\{2,\ldots,r-1\}$, $a,b\geq 0$ and $s\geq 1$, $[\psi_{2,r}([U_{r+1},Y_{r}]),~{}_{a}Y_{r},~{}_{b}U_{r+1},~{}_{s}Y_{j}]\subseteq\tilde{T}_{r}$.

*Proof. *For $a,b\geq 0$ and $s\geq 1$, we write

[TABLE]

where $u\in\psi_{2,r}([{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}])$, $y_{1},\ldots,y_{a}\in{\mathcal{Y}}_{r}$, $y^{\prime}_{1},\ldots,y^{\prime}_{b}\in{\mathcal{U}}_{r+1}$ and $z_{1},\ldots,z_{s}\in{\mathcal{Y}}_{j}$. It is enough to show that $w(a,b,s)\in\tilde{T}_{r}$. We use induction on $s$. By Lemma 5, $w(a,b,1)\in\tilde{T}_{r}$. Suppose that $w(a,b,s-1)\in\tilde{T}_{r}$ and write

[TABLE]

where $u_{(j)}\in T_{j}$, $j=2,\ldots,r$. Thus,

[TABLE]

Since $T_{2}$ is an ideal in $L$ and each $T_{j}$, $j\in\{2,\ldots,r\}$, is a free Lie algebra with a free generating set ${\mathcal{C}}_{(j)}$, we obtain from Lemma 5, $w(a,b,s)\in\tilde{T}_{r}$ and so, we get the required result. ∎

Let $J$ be the ideal of $L$ generated by the set ${\mathcal{J}}$ of all possible elements of the form

1. (1)

$[y_{mi},y_{ri}],~{}~{}~{}~{}~{}~{}2\leq r<m\leq n,$ 2. (2)

$[y_{mi},y_{rj}],\ \ 2\leq r<i\leq m\leq n,~{}~{}1\leq j\leq r,$ 3. (3)

$[y_{mi},y_{rj}]-[y_{mi},y_{mj}],\ \ 2\leq r<m\leq n,~{}~{}1\leq i\leq r<m\leq n,~{}~{}1\leq j\leq r,j\neq i.$

Proposition 3**.**

1. (1)

For $n\geq 3$, $r\in\{2,\ldots,n-1\}$ and $j\in\{1,\ldots,r-1\}$, $[T_{r},L(Y_{j})]\subseteq\tilde{T}_{r}$. 2. (2)

For all $r\in\{2,\ldots,n-1\}$ with $n\geq 3$, $\tilde{T}_{r}$ is an ideal in $L$. Furthermore, $J=\tilde{T}_{n-1}$.

Proof.

1. (1)

For $n=3$, our claim is valid, since $T_{2}$ is an ideal in $L$. Thus, we may assume that $n\geq 4$. For $r\in\{3,\ldots,n-1\}$, $T_{r}=L({\mathcal{C}}_{(r)})$, where

[TABLE]

Hence, for $r\in\{3,\ldots,n-1\}$ and $j\in\{1,\ldots,r-1\}$, it is enough to show that, for $a,b\geq 0$ and $s\geq 1$,

[TABLE]

which is an immediate consequence of Lemma 6. Therefore,

[TABLE]

for $r\in\{2,\ldots,n-1\}$ and $j\in\{1,\ldots,r-1\}$. 2. (2)

Let $r\in\{2,\ldots,n-1\}$ with $n\geq 3$. To show that $\tilde{T}_{r}$ is an ideal in $L$, we induct on $r$. For $r=2$, $\tilde{T}_{2}=T_{2}$, which is an ideal in $L$, and so, we suppose that our claim is valid for $k$ with $2\leq k\leq n-2$ and $n\geq 4$. That is, $\tilde{T}_{k}$ is an ideal in $L$. By Proposition 2 (2),

[TABLE]

Since $\tilde{T}_{k+1}=T_{k+1}\oplus\tilde{T}_{k}$ and $\tilde{T}_{k}$ is an ideal in $L$, it is enough to show that $[T_{k+1},L(Y_{j})]\subseteq\tilde{T}_{k+1}$ for all $j\in\{2,\ldots,\kappa\}$. By the equation (4) (for $r=k+1$), we get $\tilde{T}_{r}$ is an ideal of $L$. Since ${\mathcal{J}}\subseteq\tilde{T}_{n-1}$, we obtain $J\subseteq\tilde{T}_{n-1}$. Since, for all $r\in\{2,\ldots,n-1\}$, $T_{r}=L({\mathcal{C}}_{(r)})$ and ${\mathcal{C}}_{(r)}\subseteq J$, we have $T_{r}\subseteq J$ for all $r$. Therefore, $\tilde{T}_{n-1}\subseteq J$ and so, $J=\tilde{T}_{n-1}$. ∎

Theorem 2**.**

$L=(\bigoplus_{i=2}^{n}L(Y_{i}))\oplus J$.

*Proof. *By Proposition 2, we have $L=(\bigoplus_{i=2}^{n}L(Y_{i}))\oplus\tilde{T}_{n-1}$ and, by Proposition 3, we obtain the required result. ∎

Recall that for any $r\in\{2,\ldots,n-1\}$, $T_{r}=L(\Psi_{r}(W_{(r)}))$ and $L(U_{r})=L(Y_{r})\oplus L(U_{r+1})\oplus T_{r}$. By Corollary 2, for any $m\geq 2$,

[TABLE]

By Proposition 2 (1) and the equation (5), we have the following result.

Corollary 4**.**

With the above notation, for any $m\geq 2$, $L^{m}=(\bigoplus_{i=2}^{n}L^{m}(Y_{i}))\oplus(\bigoplus_{j=2}^{n-1}L^{m}_{\rm grad}(\Psi_{j}(W_{(j)})))$. In particular, for any $m\geq 2$, $J^{m}=\bigoplus_{j=2}^{n-1}L^{m}_{\rm grad}(\Psi_{j}(W_{(j)}))$.

3.4. A presentation of ${\rm gr}(I_{n})$

Let $F=F({\mathcal{Y}})$ be the free group on the finite set ${\mathcal{Y}}$ such that ${\mathcal{Y}}$ is decomposed into a disjoint union ${\mathcal{Y}}={\mathcal{Y}}_{2}\cup{\mathcal{Y}}_{3}\cup\ldots\cup{\mathcal{Y}}_{n}$ with $n\geq 3$, of subsets of the form ${\mathcal{Y}}_{m}=\{y_{m1},\ldots,y_{mm}\}$, with $2\leq m\leq n$, and such that

[TABLE]

It is well known that $\rm gr(F)$ is a free Lie algebra of finite rank $|{\mathcal{Y}}|$ with a free generating set $\{y_{ij}F^{\prime}\mid y_{ij}\in{\mathcal{Y}}\}$. Both $L$ and $\rm gr(F)$ are free Lie algebras of the same rank and they are isomorphic to each other as Lie algebras in a natural way, and from now on, we identify $L$ and $\rm gr(F)$. For all $c\geq 1$, we have $L^{c}=\gamma_{c}(F)/\gamma_{c+1}(F)$.

Let ${\mathcal{R}}$ be the subset of $\gamma_{2}(F)=F^{\prime}$ consisting of all possible elements of the form

1. (1)

$[y_{mi},y_{ri}],~{}~{}~{}~{}~{}~{}2\leq r<m\leq n,$ 2. (2)

$[y_{mi},y_{rj}],\ \ 2\leq r<i\leq m\leq n,~{}~{}1\leq j\leq r,$ 3. (3)

$[y_{mi},y_{rj}][y_{mi},y_{mj}]^{-1},\ \ 2\leq r<m\leq n,~{}~{}1\leq i\leq r<m\leq n,~{}~{}1\leq j\leq r,j\neq i.$

Let $N={\mathcal{R}}^{F}$ be the normal closure of ${\mathcal{R}}$ in $F$. Thus, $N$ is generated by the set $\{r^{g}=g^{-1}rg:r\in{\mathcal{R}},g\in F\}$. By using the presentation of $I_{n}$, the group $I_{n}$ may be identified with the quotient group $F/N$. Since $r\in\gamma_{2}(F)\setminus\gamma_{3}(F)$ for all $r\in{\mathcal{R}}$, we have $N\subseteq F^{\prime}$ and so, $I_{n}/I^{\prime}_{n}\cong F/F^{\prime}$. Let $\widetilde{{\mathcal{R}}}=\{[r,g]:r\in{\mathcal{R}},g\in F\setminus\{1\}\}\subseteq\gamma_{3}(F)$. Since $N\gamma_{3}(F)/\gamma_{3}(F)$ is generated by the set $\{r\gamma_{3}(F):r\in{\mathcal{R}}\}$, we have $N\gamma_{3}(F)/\gamma_{3}(F)=J^{2}$, where $J^{2}=\bigoplus_{j=2}^{n-1}\Psi_{j}(W_{(j),2})$. Since $L^{2}=(\bigoplus_{i=2}^{n}L^{2}(Y_{i}))\oplus J^{2}$, we get the set $\{r\gamma_{3}(F):r\in{\mathcal{R}}\}$ is $\mathbb{Z}$-linear independent and so, ${\mathcal{R}}\cap\widetilde{\mathcal{R}}=\emptyset$. Therefore, $N$ is generated by the disjoint union ${\mathcal{R}}\cup\widetilde{\mathcal{R}}$.

For a positive integer $d$, let $N_{d}=N\cap\gamma_{d}(F)$. Note that for $d\leq 2$, we have $N_{d}=N$. Also, for $d\geq 2$,

[TABLE]

Since $[g_{1},\ldots,g_{\kappa}]^{f}=[g_{1}^{f},\ldots,g_{\kappa}^{f}]$ for all $f\in F$, and $N$ is normal, we obtain $\{N_{d}\}_{d\geq 2}$ is a normal (descending) series of $N$. Clearly, each $N_{d}$ is normal in $F$. Since $[N_{\kappa},N_{\ell}]\subseteq N_{\kappa+\ell}$ for all $\kappa,\ell\geq 2$, we have $\{N_{d}\}_{d\geq 2}$ is a central series of $N$. Define ${\mathcal{I}}_{d}(N)=N_{d}\gamma_{d+1}(F)/\gamma_{d+1}(F)\leq\gamma_{d}(F)/\gamma_{d+1}(F)$. Since $N\subseteq F^{\prime}$, we get ${\mathcal{I}}_{1}(N)=\{F^{\prime}\}$. Since $N$ is a normal subgroup of $F$, we have the Lie subalgebra ${\mathcal{I}}(N)=\bigoplus_{d\geq 2}{\mathcal{I}}_{d}(N)$ of ${\rm gr}(F)$ is an ideal of ${\rm gr}(F)$ (see [10]). It is easily verified that ${\mathcal{I}}_{d}(N)\cong N_{d}/N_{d+1}$ as $\mathbb{Z}$-modules. By our definitions, identifications and the above discussion, ${\mathcal{I}}_{2}(N)=N\gamma_{3}(F)/\gamma_{3}(F)=J^{2}$. Since ${\mathcal{I}}_{2}(N)=J^{2}$, ${\mathcal{I}}(N)$ is an ideal of ${\rm gr}(F)=L$ and $J$ is generated by ${\mathcal{J}}$, we obtain $J\subseteq{\mathcal{I}}(N)$. Since $J$ is a homogeneous ideal, we get, for all $d\geq 2$, $J^{d}\subseteq{\mathcal{I}}_{d}(N)$. For a positive integer $m$, with $m\geq 2$, we write $(L^{m})^{*}=\bigoplus_{i=2}^{n}L^{m}(Y_{i})$. Denote ${\mathcal{V}}={\mathcal{J}}=\bigcup_{r=2}^{n-1}\psi_{2,r}([{\mathcal{U}}_{r+1},{\mathcal{Y}}_{r}])$ and ${\mathcal{V}}^{*}=\bigcup_{m=2}^{n}\{[y_{m\kappa},y_{m\ell}]:1\leq\ell<\kappa\leq m\}$. We point out that ${\mathcal{V}}^{*}$ is a natural $\mathbb{Z}$-basis of $(L^{2})^{*}$ and ${\mathcal{R}}\gamma_{3}(F)/\gamma_{3}(F)={\mathcal{V}}={\mathcal{J}}$, which is a natural $\mathbb{Z}$-basis of $J^{2}$.

Proposition 4**.**

For all positive integers $c$, ${\mathcal{I}}_{c+2}(N)=J^{c+2}$.

*Proof. *Let $c=1$. We denote by $N_{2,1}$ the subgroup of $N_{2}=N$ generated by the set $\mathcal{R}$ and let $N_{2,2}$ be the normal closure of $\widetilde{\mathcal{R}}$ in $F$. Since $N_{2}=N=N_{2,1}N_{2,2}$ and $N_{2,2}\subseteq\gamma_{3}(F)$, by using the modular law, we get

[TABLE]

An element $w$ of $N_{2,1}$ is written as a product of elements in ${\mathcal{R}}$ (in an abelian form) and an element $u\in N^{\prime}_{2,1}$. Since ${\mathcal{I}}_{2}(N)=J^{2}$ and ${\mathcal{R}}\gamma_{3}(F)/\gamma_{3}(F)$ is a $\mathbb{Z}$-basis of ${\mathcal{I}}_{2}(N)$, we have $N_{2,1}\cap\gamma_{3}(F)$ is generated by elements which belong to $N_{2,2}\cap\gamma_{4}(F)$. That is, $N_{2,1}\cap\gamma_{3}(F)\subseteq N_{2,2}$. Therefore, $N_{3}=N_{2,2}$ is generated by the set $\{[r,g]^{f}:r\in{\mathcal{R}},g\in F\setminus\{1\},f\in F\}$ and so, ${\mathcal{I}}_{3}(N)=J^{3}$.

Suppose that $c\geq 2$. Now, $I_{n}=H_{n}\rtimes H_{n-1}\rtimes\ldots\rtimes H_{2}$ and so, by induction on $n$ and a result of Falk and Randell [9, Theorem (3.1)], we have, for all $d\geq 3$, $\gamma_{d}(I_{n})/\gamma_{d+1}(I_{n})\cong\bigoplus_{i=2}^{n}\gamma_{d}(H_{i})/\gamma_{d+1}(H_{i})$. Since $H_{i}$ is a free group of rank $i$, $i\in\{2,\ldots,n\}$, we have $\gamma_{d}(H_{i})/\gamma_{d+1}(H_{i})\cong L^{d}(Y_{i})$. By Proposition 3 (2), Lemma 3 and Theorem 2,

[TABLE]

Hence, ${\rm rank}(\gamma_{d}(I_{n})/\gamma_{d+1}(I_{n}))={\rm rank}(L^{d})^{*}$. Now,

[TABLE]

Since $\gamma_{d+1}(F)\subseteq\gamma_{d}(F)$, we have, by the modular law,

[TABLE]

But, $\gamma_{d}(F)/(\gamma_{d+1}(F)N_{d})\cong(\gamma_{d}(F)/\gamma_{d+1}(F))/{\mathcal{I}}_{d}(N)$. Therefore, for all $d\geq 3$, $\gamma_{d}(I_{n})/\gamma_{d+1}(I_{n})\cong(\gamma_{d}(F)/\gamma_{d+1}(F))/{\mathcal{I}}_{d}(N)$ and so,

[TABLE]

Since $J^{c}\subseteq{\mathcal{I}}_{c}(N)$ for all $c\geq 2$, $(L^{d})^{*}$ is torsion-free for all $d\geq 3$ and by our definitions and identifications, we conclude from $(6)$ that ${\mathcal{I}}_{d}(N)=J^{d}$. Thus, we obtain the required result. ∎

Corollary 5**.**

${\mathcal{I}}(N)=J$.

*Proof. *Since $J=\bigoplus_{d\geq 2}J^{d}$ and ${\mathcal{I}}_{2}(N)=J^{2}$, we have from Proposition 4 that ${\mathcal{I}}(N)=J$. ∎

So, we are now able to prove the following.

Theorem 3**.**

$L/J\cong{\rm gr}(I_{n})$* as Lie algebras.*

*Proof. *Recall that ${\rm gr}(I_{n})=\bigoplus_{c\geq 1}\gamma_{c}(I_{n})/\gamma_{c}(I_{n+1})$. Since $I_{n}/I_{n}^{\prime}\cong F/NF^{\prime}=F/F^{\prime}\cong\mathbb{Z}^{k}$ with $k=(n-1)(n+2)/2$, we have ${\rm gr}(I_{n})$ is generated as a Lie algebra by the set $\{\alpha_{ij}:2\leq i\leq n,1\leq j\leq i\}$ with $\alpha_{ij}=y_{ij}I^{\prime}_{n}$. Since $L$ is a free Lie algebra of rank $k$ with a free generating set $\{y_{ij}:2\leq i\leq n,1\leq j\leq i\}$, the map $\xi:L\rightarrow{\rm gr}(I_{n})$ with $\xi(y_{ij})=\alpha_{ij}$ extends uniquely to a Lie algebra epimorphism. Hence $L/\ker\xi\cong{\rm gr}(I_{n})$ as Lie algebras. By definition, $J\subseteq\ker\xi$ and so, $\xi$ induces a Lie algebra epimorphism $\overline{\xi}:L/J\rightarrow{\rm gr}(I_{n})$. In particular, $\overline{\xi}(y_{ij}+J)=\alpha_{ij}$ for all $i,j$. Moreover, $\overline{\xi}$ induces a $\mathbb{Z}$-linear mapping $\overline{\xi}_{c}:(L^{c}+J)/J\rightarrow\gamma_{c}(I_{n})/\gamma_{c+1}(I_{n})$. For $c\geq 2$, and in view of the proof of Proposition 4, we have ${\rm rank}(\gamma_{c}(I_{n})/\gamma_{c+1}(I_{n}))={\rm rank}(L^{c})^{*}$. Since $J=\bigoplus_{m\geq 2}J^{m}$, we have $(L^{c}+J)/J\cong L^{c}/(L^{c}\cap J)=L^{c}/J^{c}\cong(L^{c})^{*}$ (by Theorem 2 and Lemma 3) and so, we obtain ${\rm ker}\overline{\xi}_{c}$ is torsion-free. Since ${\rm rank}(\gamma_{c}(I_{n})/\gamma_{c+1}(I_{n}))={\rm rank}(L^{c})^{*}$, we have ${\rm ker}\overline{\xi}_{c}=\{1\}$ and so, $\overline{\xi}_{c}$ is an isomorphism. Since $\overline{\xi}$ is an epimorphism and each $\overline{\xi}_{c}$ is isomorphism, we have $\overline{\xi}$ is an isomorphism. Hence, $L/J\cong{\rm gr}(I_{n})$ as Lie algebras. ∎

Corollary 6**.**

A presentation of $\rm gr(I_{n})$ is given by generators $y_{ij}$, with $i\in\{2,\ldots,n\}$, $j\in\{1,\ldots,n\}$ and $i\geq j$, together with the following relations:

1. (1)

$[y_{mi},y_{ri}],~{}~{}~{}~{}~{}~{}2\leq r<m\leq n,$** 2. (2)

$[y_{mi},y_{rj}],\ \ 2\leq r<i\leq m\leq n,~{}~{}1\leq j\leq r,$** 3. (3)

$[y_{mi},y_{rj}]-[y_{mi},y_{mj}],\ \ 2\leq r<m\leq n,~{}~{}1\leq i\leq r<m\leq n,~{}~{}1\leq j\leq r,j\neq i.$**

4. A natural embedding

For positive integers $n$ and $c$, with $n,c\geq 2$, let $F_{n}$ be a free group of rank $n$ and $F_{n,c-1}=F_{n}/\gamma_{c}(F_{n})$ be the free nilpotent group of rank $n$ and class $c-1$. The natural epimorphism from $F_{n}$ onto $F_{n,c-1}$ induces a group homomorphism $\pi_{n,c-1}:\rm Aut(F_{n})\rightarrow\rm Aut(F_{n,c-1})$ with kernel ${\rm I}_{c}{\rm A}(F_{n})$. For $c=2$ we write ${\rm I}_{2}{\rm A}(F_{n})={\rm IA}(F_{n})$ for the $IA$-automorphisms of $F_{n}$. It is well known that, for $t,s\geq 2$, $[{\rm I}_{t}{\rm A}(F_{n}),{\rm I}_{s}{\rm A}(F_{n})]\subseteq{\rm I}_{t+s-1}{\rm A}(F_{n})$. Since $F_{n}$ is residually nilpotent, we have $\bigcap_{c\geq 2}{\rm I}_{c}{\rm A}(F_{n})=\{1\}$. For $r\geq 2$, we write ${\mathcal{L}}^{r}({\rm I}{\rm A}(F_{n}))={\rm I}_{r}{\rm A}(F_{n})/{\rm I}_{r+1}{\rm A}(F_{n})$. It is known that ${\mathcal{L}}^{r}({\rm I}{\rm A}(F_{n}))$ is a free abelian group for all $r\geq 2$ (see, for example, [15, Section 1]). Form the (restricted) direct sum of the free abelian groups ${\mathcal{L}}^{r}({\rm I}{\rm A}(F_{n}))$ and denote it by ${\mathcal{L}}({\rm IA}(F_{n}))=\bigoplus_{r\geq 2}{\mathcal{L}}^{r}({\rm I}{\rm A}(F_{n}))$. It has the structure of a graded Lie algebra with ${\mathcal{L}}^{r}({\rm IA}(F_{n}))$ as component of degree $r-1$ in the grading and Lie multiplication given by $[\phi{\rm I}_{j+1}{\rm A}(F_{n}),\psi{\rm I}_{\kappa+1}{\rm A}(F_{n})]=(\phi^{-1}\psi^{-1}\phi\psi){\rm I}_{j+\kappa}{\rm A}(F_{n})$, for all $\phi\in{\rm I}_{j}{\rm A}(F_{n})$, $\psi\in{\rm I}_{\kappa}{\rm A}(F_{n})$ with $j,\kappa\geq 2$. Multiplication is then extended to ${\mathcal{L}}({\rm IA}(F_{n}))$ by linearity. The above Lie algebra is usually called the Andreadakis-Johnson Lie algebra of ${\rm IA}(F_{n})$. We point out that, for any positive integer $c$, $\gamma_{c}({\rm IA}(F_{n}))\subseteq{\rm I}_{c+1}{\rm A}(F_{n})$.

Let $G$ be a finitely generated subgroup of ${\rm IA}(F_{n})$ with $G/G^{\prime}$ torsion-free. For a positive integer $q$, let ${\mathcal{L}}^{q}_{1}(G)=\gamma_{q}(G)({\rm I}_{q+2}{\rm A}(F_{n}))/{\rm I}_{q+2}{\rm A}(F_{n})$. Form the (restricted) direct sum of free abelian groups ${\mathcal{L}}_{1}(G)=\bigoplus_{q\geq 1}{\mathcal{L}}^{q}_{1}(G)$. It is easily verified that ${\mathcal{L}}_{1}(G)$ is a Lie subalgebra of ${\mathcal{L}}({\rm IA}(F_{n}))$. Furthermore, if $\{y_{1}G^{\prime},\ldots,y_{m}G^{\prime}\}$ is a $\mathbb{Z}$-basis for $G/G^{\prime}$, then ${\mathcal{L}}_{1}(G)$ is generated as Lie algebra by the set

[TABLE]

By a natural embedding of ${\rm gr}(G)$ into ${\mathcal{L}}({\rm IA}(F_{n}))$, we mean that there exists a Lie algebra isomorphism $\zeta$ from ${\rm gr}(G)$ onto ${\mathcal{L}}_{1}(G)$ satisfying the conditions $\zeta(y_{i}G^{\prime})=y_{i}({\rm I}_{3}{\rm A}(F_{n}))$, $i=1,\ldots,m$. In this case, we also say that ${\rm gr}(G)$ is naturally isomorphic to ${\mathcal{L}}_{1}(G)$.

In this section we prove that there is a natural embedding of ${\rm gr}(I_{n})$ into ${\mathcal{L}}({\rm IA}(F_{n}))$.

Theorem 4**.**

For a positive integer $n\geq 2$, ${\rm gr}(I_{n})$ is naturally isomorphic to ${\mathcal{L}}_{1}(I_{n})$ as Lie algebra.

*Proof. *Let $n=2$. Since $I_{2}=H_{2}={\rm Inn}(F_{2})\cong F_{2}$, our claim follows from [15, Proof of Proposition 3]. Thus, we assume that $n\geq 3$. Recall that $I_{n}=(\ldots(H_{n}\rtimes H_{n-1})\rtimes\ldots)\rtimes H_{2}$, where each $H_{i}$ is a free group of finite rank $i$ with a free generating set $\{y_{i1},\ldots,y_{ii}\}$. Thus, ${\rm gr}(H_{i})\cong{\rm gr}(F_{i})$ as Lie algebras. By [15, Proof of Proposition 3], we have ${\rm gr}(H_{i})\cong{\mathcal{L}}_{1}(H_{i})$ by a Lie algebra isomorphism $\zeta_{i}$ satisfying the conditions $\zeta_{i}(y_{ik}H^{\prime}_{i})=y_{ik}({\rm I}_{3}{\rm A}(F_{i}))$, with $k=1,\ldots,i$, for all $i=2,\ldots,n$. Hence, ${\mathcal{L}}_{1}(H_{i})$ is a free Lie algebra of rank $i$. Since $y_{i1}\notin{\rm I}_{3}{\rm A}(F_{i})$, we have ${\mathcal{L}}_{1}(H_{i})$ is a non-trivial subalgebra of ${\mathcal{L}}({\rm IA}(F_{i}))$. In fact, ${\mathcal{L}}_{1}(H_{i})$ is generated by the set $\{y_{ik}({\rm I}_{3}{\rm A}(F_{i})):k=1,\ldots,i\}$. For $h\in F_{i}$, we write $\tau_{h}$ for the inner automorphism of $F_{i}$ defined by $\tau_{h}(x)=hxh^{-1}$ for all $x\in F_{i}$. Since $[\tau_{g},\phi]=\tau_{g^{-1}\phi^{-1}(g)}$ for all $\phi\in{\rm Aut}(F_{i})$ and $g\in F_{i}$, we have ${\mathcal{L}}_{1}(H_{i})$ is an ideal in ${\mathcal{L}}({\rm IA}(F_{i}))$ for all $i\in\{2,\ldots,n\}$.

Next, we claim that ${\mathcal{L}}_{1}(I_{n})=\bigoplus_{i=2}^{n}{\mathcal{L}}_{1}(H_{i})$. Fix $m\in\{3,\ldots,n\}$. First, we observe that $I_{m}=H_{m}\rtimes I_{m-1}$ and claim that ${\mathcal{L}}_{1}(I_{m})$ is additively equal to the direct sum of ${\mathcal{L}}_{1}(I_{m-1})$ and ${\mathcal{L}}_{1}(H_{m})$. Since ${\mathcal{L}}_{1}(H_{m})$ is an ideal in ${\mathcal{L}}({\rm IA}(F_{m}))$, we have ${\mathcal{L}}_{1}(I_{m-1})+{\mathcal{L}}_{1}(H_{m})$ is a Lie subalgebra of ${\mathcal{L}}_{1}(I_{m})$. Let $w\in{\mathcal{L}}_{1}(I_{m-1})\cap{\mathcal{L}}_{1}(H_{m})$. In the next few lines, we adopt the arguments given in the proof of [15, Proposition 4]. Since both ${\mathcal{L}}_{1}(I_{m-1})$ and ${\mathcal{L}}_{1}(H_{m})$ are graded Lie algebras, we may assume that $w\in{\mathcal{L}}^{d}_{1}(I_{m-1})\cap{\mathcal{L}}^{d}_{1}(H_{m})$ for some $d$. Thus, there are $u\in\gamma_{d}(I_{m-1})$ and $v\in\gamma_{d}(H_{m})$ such that $w=u{\rm I}_{d+2}{\rm A}(F_{m})=v{\rm I}_{d+2}{\rm A}(F_{m})$. To get a contradiction, we assume that $u,v\notin{\rm I}_{d+2}{\rm A}(F_{m})$. Therefore, $v\in\gamma_{d}(H_{m})\setminus\gamma_{d+1}(H_{m})$ and so, there exists $g\in\gamma_{d}(F_{m})\setminus\gamma_{d+1}(F_{m})$ such that $v=\tau_{g}\rho$, where $\rho\in\gamma_{d+1}(H_{m})$. Then, we have $u{\rm I}_{d+2}{\rm A}(F_{m})=\tau_{g}\rho{\rm I}_{d+2}(F_{m})=\tau_{g}{\rm I}_{d+2}(F_{m})$, since $\rho\in\gamma_{d+1}(H_{m})\subseteq{\rm I}_{d+2}{\rm A}(F_{m})$. So, $u^{-1}\tau_{g}\in{\rm I}_{d+2}{\rm A}(F_{m})$. Hence, $u^{-1}\tau_{g}(x_{m})=x_{m}h$, where $h\in\gamma_{d+2}(F_{m})$, by the definition of ${\rm I}_{d+2}{\rm A}(F_{m})$. Since $u^{-1}$ fixes $x_{m}$, we have

[TABLE]

Since $I_{m-1}\subseteq{\rm IA}(F_{m})$, we obtain $\gamma_{d}(I_{m-1})\subseteq\gamma_{d}({\rm IA}(F_{m}))\subseteq{\rm I}_{d+1}{\rm A}(F_{m})$ and so, $u^{-1}(x_{j})=x_{j}z_{j}$, where $z_{j}\in\gamma_{d+1}(F_{m})$. Since $\gamma_{d}(F_{m})$ is a fully invariant subgroup of $F_{m}$, we get $u^{-1}(g^{-1})=g^{-1}g_{1}$ with $g_{1}\in\gamma_{d+1}(F_{m})$. Moreover, since

[TABLE]

and $[x_{m},g^{-1}]=([x_{m},g]^{-1})^{g^{-1}}$, we have $[g,x_{m}]\in\gamma_{d+2}(F_{m})$. Since ${\rm gr}(F_{m})$ is a free Lie algebra of rank $m$ with a free generating set $\{x_{i}F^{\prime}_{m}:i=1,\ldots,m\}$ and $\gamma_{d}(F_{m})/\gamma_{d+1}(F_{m})$ is the $d$th homogeneous component of ${\rm gr}(F_{m})$, we obtain $[g,x_{m}]\in\gamma_{d+1}(F_{m})\setminus\gamma_{d+2}(F_{m})$ (see, for example, [12, Theorem 5.10]) which is a contradiction. Therefore, ${\mathcal{L}}_{1}(I_{m-1})\cap{\mathcal{L}}_{1}(H_{m})=\{0\}$ for all $m\in\{3,\ldots,n\}$. Since $I_{m}=H_{m}\rtimes I_{m-1}$ and the action of $I_{m-1}$ on $H_{m}$ is by conjugation, we have, by a result of Falk and Randell [9, Proof of Theorem (3.1)], $\gamma_{q}(I_{m})=\gamma_{q}(H_{m})\rtimes\gamma_{q}(I_{m-1})$ for all $q$. Hence, ${\mathcal{L}}^{q}_{1}(I_{m})={\mathcal{L}}^{q}_{1}(I_{m-1})+{\mathcal{L}}^{q}_{1}(H_{m})$ and so, ${\mathcal{L}}^{q}_{1}(I_{m})={\mathcal{L}}^{q}_{1}(I_{m-1})\oplus{\mathcal{L}}^{q}_{1}(H_{m})$ for all $q$. Therefore, ${\mathcal{L}}_{1}(I_{m})={\mathcal{L}}_{1}(I_{m-1})\oplus{\mathcal{L}}_{1}(H_{m})$ for all $m\in\{3,\ldots,n\}$ and so,

[TABLE]

Recall that, for $m\in\{2,\ldots,n\}$, we have ${\rm gr}(H_{m})\cong{\mathcal{L}}_{1}(H_{m})$ by an isomorphism $\zeta_{m}$ satisfying the conditions $\zeta_{m}(y_{mk}H^{\prime}_{m})=y_{mk}{\rm I}_{3}{\rm A}(F_{m})$ for all $k\in\{1,\ldots,m\}$. By a result of Andreadakis [1, Section 6], $\gamma_{c}(H_{m})=H_{m}\cap{\rm I}_{c+1}{\rm A}(F_{m})$ for all $c$ and $m\in\{2,\ldots,n\}$. Since

[TABLE]

and since any finitely generated residually nilpotent group is Hopfian (see, [12, Theorem 5.5]), we get the restriction of $\zeta_{m}$, say $\zeta_{m,c}$, on ${\rm gr}_{c}(H_{m})$ is a $\mathbb{Z}$-module isomorphism for all $c$ and $m\in\{2,\ldots,n\}$. For all $c$, let $\mu_{c}$ be the map from ${\rm gr}_{c}(H_{2})\oplus\ldots\oplus{\rm gr}_{c}(H_{n})$ into ${\mathcal{L}}^{c}_{1}(I_{n})$ defined by $\mu_{c}((u_{2}\gamma_{c+1}(H_{2}),\ldots,u_{n}\gamma_{c+1}(H_{n}))=\zeta_{2,c}(u_{2}{\rm I}_{c+2}{\rm A}(F_{n}))+\cdots+\zeta_{n,c}(u_{n}{\rm I}_{c+2}{\rm A}(F_{n}))$ for $u_{m}\in\gamma_{c}(H_{m})$ for all $m\in\{2,\ldots,n\}$. It is easily checked that $\mu_{c}$ is a $\mathbb{Z}$-module isomorphism. By a result of Falk and Randell [9, Theorem (3.1)], ${\rm gr}_{c}(I_{n})=\bigoplus_{i=2}^{n}{\rm gr}_{c}(H_{i})$ and so, ${\rm gr}_{c}(I_{n})$ is isomorphic to ${\mathcal{L}}_{1}^{c}(I_{n})$ (as $\mathbb{Z}$-modules) by means of $\mu_{c}$ for all $c$. Therefore, ${\rm rank}({\mathcal{L}}^{c}_{1}(I_{n}))={\rm rank}({\rm gr}_{c}(I_{n}))={\rm rank}(L^{c})^{*}$. By Theorem 3, $L/J\cong{\rm gr}(I_{n})$ by a Lie algebra isomorphism $\overline{\xi}$ satisfying the conditions $\overline{\xi}(y_{ij}+J)=y_{ij}{\rm I}^{\prime}_{n}$ for all $2\leq i\leq n$ and $1\leq j\leq i$. Let $\zeta$ be the map from $L$ into ${\mathcal{L}}_{1}(I_{n})$ such that $\zeta(y_{ij})=y_{ij}{\rm I}_{3}{\rm A}(F_{n})$ for all $2\leq i\leq n$ and $1\leq j\leq i$. Since $L$ is a free Lie algebra on the set $\{y_{ij}:2\leq i\leq n,1\leq j\leq i\}$, the map $\zeta$ extends uniquely to a Lie algebra epimorphism and so, $L/{\rm ker}\zeta\cong{\mathcal{L}}_{1}(I_{n})$. Clearly, $J\subseteq{\rm ker}\zeta$. Thus, $\zeta$ induces a Lie algebra epimorphism $\overline{\zeta}:L/J\rightarrow{\mathcal{L}}_{1}(I_{n})$ such that $\overline{\zeta}(y_{ij}+J)=y_{ij}{\rm I}_{3}{\rm A}(F_{n})$ for all $2\leq i\leq n$ and $1\leq j\leq i$. Moreover, $\overline{\zeta}$ induces a $\mathbb{Z}$-linear mapping $\overline{\zeta}_{c}:(L^{c}+J)/J\rightarrow{\mathcal{L}}^{c}_{1}(I_{n})$. By using similar arguments as in the proof of Theorem 3, we have $\overline{\zeta}$ is an isomorphism. Thus, $\overline{\zeta}(\overline{\xi})^{-1}$ is the required Lie algebra isomorphism from ${\rm gr}(I_{n})$ onto ${\mathcal{L}}_{1}(I_{n})$. ∎

Remark 2**.**

By the proof of Theorem 4, we obtain ${\mathcal{L}}_{1}^{c}(I_{n})={\mathcal{L}}_{1}^{c}(H_{2})\oplus\ldots\oplus{\mathcal{L}}^{c}(H_{n})$. Since ${\mathcal{L}}_{1}^{c}(H_{i})\cong\gamma_{c}(H_{i})/\gamma_{c+1}(H_{i})$ for all $i=2,\ldots,n$, we get ${\rm rank}({\mathcal{L}}_{1}(H_{i}))=\frac{1}{c}{\displaystyle\sum_{d|c}\mu(d)i^{\frac{c}{d}}}$. Thus, for any $c$, $\frac{1}{c}{\displaystyle\sum_{d|c}\mu(d)\{2^{\frac{c}{d}}+\ldots+n^{\frac{c}{d}}\}}\leq{\rm rank}({\mathcal{L}}^{c+1}({\rm IA}(F_{n}))$.