# On Schur Algebras and Derivations of Free Lie Algebras

Frederick Cohen  Department of Mathematics, University of Rochester,
RochesterHylan Bldg, 140 Trustee Rd, Rochester, NY 14627, U.S.A.
<EMAIL_ADDRESS>,  Mohamed Elhamdadi  Department of Mathematics,
University of South Florida, Tampa, FL 33620, U.S.A<EMAIL_ADDRESS>,
Tao Jin  PBC School of Finance, Tsinghua University, 43 Chengfu Road, Haidian
District, Beijing 100083, China<EMAIL_ADDRESS>and  Minghui Liu
Mathematics and Science Department, Florida College, Temple Terrace, FL 33617,
U.S.A<EMAIL_ADDRESS>
###### Abstract.

We investigate the action of Schur algebra on the Lie algebras of derivations
of free Lie algebras and operad structures constructed from it. We also show
that the Lie algebra of derivations is generated by quadratic derivations
together with the action of the Schur operad. Applications to certain
subgroups of the automorphism group of a finitely generated free group are
given as well.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Calculation with the Derivations
  4. 4 The Schur Algebra on $V_{n}(K)^{\otimes q}$
  5. 5 The Associativity and Anti-commutativity of the Schur Algebra
  6. 6 Action of the Schur Algebra on Derivations and Operads
  7. 7 A Short Exact Sequence of Lie Algebras

##  1\. Introduction

Let $F_{n}$ be the free group of rank $n$ and ${IA}_{n}$ be the subgroup of
$\mathrm{Aut}(F_{n})$ which induces the identity automorphism on
$\mathbb{Z}^{n}$, the abelianization of $F_{n}$. These groups have been
studied by Nielsen in 1917 [Nielsen], but their structures remained not well-
understood. In 1997, Krstić and McCool [KM] proved that $IA_{3}$ is not
finitely presentable and $H^{2}(IA_{3})$ is not finitely generated. Some
properties of the second rational cohomology of $IA_{n}$ were studied by A.
Pettet [P] in 2005. In 2007, M. Bestvina, K.-U. Bux and D. Margalit proved
some properties about a certain subgroup of $\mathrm{Out}\left(F_{n}\right)$,
the outer automorphism group of $F_{n}$, and its homology [BBM]. In 2011, S.
Galatius [G] proved that there is a homology equivalence

$\mathbb{Z}\times
B\left(\underset{{n\to\infty}}{\mathrm{colim}}\,\mathrm{Aut}(F_{n})\right)\to\underset{{n\to\infty}}{\mathrm{colim}}\,\Omega^{n}S^{n},$

where the right hand side is the infinite loop space, and $B$ stands for the
classifying space.

In order to study the groups $IA_{n}$ and $\mathrm{Aut}(F_{n})$, the Johnson
filtration

$\mathrm{Aut}(F_{n})=\mathcal{A}_{n}(0)\supset\mathcal{A}_{n}(1)\supset\cdots$

was first introduced by S. Andreadakis in 1965 [A]. The Johnson Lie algebra
$\mathrm{gr}_{\ast}^{J}(IA_{n})$ is the Lie algebra structure associated with
the Johnson filtration of the group $IA_{n}$. The group $IA_{n}$ is related to
$\mathrm{Der}_{\ast}^{\triangle}(L_{\mathbb{Z}}(X_{n}))$, the Lie algebra of
derivations of the free Lie algebra $L_{\mathbb{Z}}(X_{n})$, via the so-called
Johnson homomorphism, which was studied by N. Kawazumi [K], where
$X_{n}=\\{1,\ldots,n\\}$. There is a second descending filtration
$\mathcal{A}_{n}(k)^{\prime}$ of $\mathrm{Aut}(F(X_{n}))$ induced by the lower
central series of $IA_{n}$. S. Andreadakis conjectured that
$\mathcal{A}_{n}(k)=\mathcal{A}_{n}(k)^{\prime}$ for all $k$ and $n$; which is
shown to be false in general by L. Bartholdi in [B0, B1], though some special
cases were studied and proved by S. Andreadakis [A] and some other authors
such as K. Satoh [Sa].

In this article which is based on the thesis of T. Jin [Jin], we introduce a
new additional structure for the Lie algebra of derivations of a free Lie
algebra, which arises from actions of a classical algebra known as the Schur
algebra, which was first studied by I. Schur in the early twentieth century to
explore the representation theory of the symmetric group $\Sigma_{r}$ and the
general linear group $\mathrm{GL}(m,\mathbb{C})$ [Sch]. We also show that the
Lie algebra of derivations is generated by quadratic derivations together with
the action of the Schur operad. Applications to certain subgroups of the
automorphism group of a finitely generated free group are given as well. The
first main result of this paper is the following proposition:

######  Proposition.

The Schur algebras assemble to give an operad, denoted $\mathit{Schur}$.

See Proposition 6.8 for details. The Lie algebra of derivations is a “module”
over the operad $\mathit{Schur}$, which gives us the second main result of
this paper (See Theorem 6.9 for details):

######  Theorem.

The Lie algebra of derivations
$\mathrm{Der}_{*}=\displaystyle\bigoplus_{p=2}^{\infty}\operatorname{Hom}_{K}\left(V_{n},L^{p}(X_{n})\right)$
is a “module” over the operad $\mathit{Schur}$ which is generated by $2$
elements.

The paper is mainly based on the thesis of Jin Tao [Jin]. The article is
organized as follows: In Section 2, we recall the necessary ingredients needed
for the rest of the paper. In particular, the Lie algebra associated to the
lower central series of a group $G$ and the group of so-called
$\mathrm{IA}$-automorphisms of $G$. Section 3 deals with the calculation
involving the derivations of a free Lie algebra in order to explore the
structure of certain subgroups of the automorphism group of a finitely
generated free group. We prove the main theorem stating that any two distinct
elements in $\mathrm{IA}_{n}$ either generate a free group or a free abelian
group. In Section 4, we recall the definition of Schur algebra and we prove a
few results regarding the Schur algebra on the tensor product of a finitely
generated free $K$-module. In Section 5, we study some properties of Schur
algebras and their action on the Lie algebra of derivations of a free Lie
algebra. By using the inter-degree multiplication induced from a transfer-like
map, it is shown that the Schur algebra forms an associative and anti-
commutative graded K-algebra. In Section 6, the Schur-operad structure is
constructed. We show that the Lie algebra of derivations can be generated by
quadratic derivations together with the action of the Schur operad. We
introduce a short exact sequence of Lie algebras which is obtained from
certain group filtrations in Section 7.

##  2\. Preliminaries

In this section, we recall the necessary ingredients needed later in this
paper including the Lie algebra associated to the lower central series of a
group $G$ and the group of $\mathrm{IA}$-automorphism of $G$. Let
$X_{n}=\\{1,\ldots,n\\}$ and let the free Lie algebra generated by $X_{n}$
over a commutative ring with identity $K$ be denoted by $L_{K}(X_{n})$, or
simply $L(X_{n})$. The free Lie algebra $L_{K}(X_{n})$ is naturally a graded
$K$-module and

$L_{K}(X_{n})=\bigoplus_{p=1}^{\infty}L_{K}^{p}(X_{n}),$

where each component $L_{K}^{p}(X_{n})$ is a free $K$\- module. Let $G$ be a
group and $\\{\Gamma^{m}G\\}_{m\in\mathbb{Z}^{+}}$ be the lower central series
of $G$. There is a naturally induced homomorphism

$\mathrm{Aut}(G)\to\mathrm{Aut}(G/\Gamma^{m}G)$

for every positive integer $m$. Let the kernel of this induced homomorphism be
denoted by $J^{m}\mathrm{Aut}(G)$.

######  Definition 2.1.

A group $G$ is called residually nilpotent if
$\displaystyle\bigcap_{m=1}^{\infty}\Gamma^{m}G$ is the trivial group.

Recall that for an algebra $A$ over a field $K$, a derivation $D:A\to A$ is a
$K$-linear map such that

$D(xy)=(Dx)y+x(Dy).$

The set $\mathrm{Der}(A)$ of all derivations is a Lie algebra with the product

$[D,D^{\prime}]=DD^{\prime}-D^{\prime}D.$

The derivations of the free Lie algebra $L(X_{n})$ is a naturally graded
$K$-module

$\mathrm{Der}_{\ast}^{\triangle}\left(L(X_{n})\right)=\bigoplus_{p=1}^{\infty}\mathrm{Der}_{p}\left(L(X_{n})\right),$

where $\mathrm{Der}_{p}\left(L(X_{n}\right))$ is the component of degree $2p$;
that is,

$\mathrm{Der}_{p}\left(L(X_{n})\right)=\\{f\in\mathrm{Der}_{\ast}^{\triangle}\left(L(X_{n})\right)\mid
f(x_{i})\in L^{p}(X_{n}),1\leq i\leq n\\}.$

######  Remark 2.2.

Notice that for notation in this paper, we adopt the convention that an
element in $V_{n}(K)^{\otimes q}$ has degree $2q$; that is, we define

$T\left(V_{n}(K)\right)=\bigoplus_{q=0}^{\infty}T^{q}\left(V_{n}(K)\right),$

where

$T^{2q}\left(V_{n}(K)\right)=V_{n}(K)^{\otimes q}$

and

$T^{2q+1}\left(V_{n}(K)\right)=0.$

In particular, the grading of $Der_{p}(L(X_{n}))$ is the component of degree
$2p$.

Since every $f\in\mathrm{Der}_{\ast}^{\triangle}\left(L(X_{n})\right)$ is
uniquely determined by the restriction of $f$ on $X_{n}$, we have

$\mathrm{Der}_{p}\left(L(X_{n})\right)\cong\operatorname{Hom}_{K}\left(V_{n},L^{p}(X_{n})\right)$

and

$\mathrm{Der}_{\ast}^{\triangle}\left(L(X_{n})\right)\cong\bigoplus_{p=1}^{\infty}\operatorname{Hom}_{K}\left(V_{n},L^{p}(X_{n})\right).$

The sub-Lie algebra $\mathrm{Der}_{\ast}\left(L(X_{n})\right)$ which is
defined as

$\mathrm{Der}_{\ast}\left(L(X_{n})\right)=\bigoplus_{p=2}^{\infty}\operatorname{Hom}_{K}\left(V_{n},L^{p}(X_{n})\right)$

is called the derivations of the free Lie algebra $L(X_{n})$.

Let $V_{n}(K)$ (or simply $V_{n}$) be the free $K$-module generated by
$X_{n}=\\{x_{1},\cdots,x_{n}\\}$ and let $\Sigma_{q}$ be the symmetric group
on $q$ letters. The symmetric group $\Sigma_{q}$ acts on $V_{n}(K)^{\otimes
q}$ from the right by

$(y_{1}\otimes\cdots\otimes
y_{q})\cdot\sigma=y_{\sigma^{-1}(1)}\otimes\cdots\otimes y_{\sigma^{-1}(q)},$

where $\sigma\in\Sigma_{q}$ and $y_{1}\otimes\cdots\otimes y_{q}\in
V_{n}^{\otimes q}$.

Recall that for a given group $G$, an integral filtration of $G$ is a sequence
$\\{F^{m}G\\}_{m\in\mathbb{Z}^{+}}$ of subgroups of $G$ such that (1)
$F^{1}G=G$ and (2) $[F^{m}G,F^{n}G]\subseteq F^{m+n}G$. Given an integral
filtration $\\{F^{m}G\\}_{m\in\mathbb{Z}^{+}}$, there is an associated graded
Lie algebra structure

$\mathrm{gr}_{\ast}G=\bigoplus_{m=1}^{\infty}\mathrm{gr}_{m}G,$

where $\mathrm{gr}_{\ast}G=F^{m}G/F^{m+1}G$ [S].

Let $G$ be a group and $F^{m}G=\Gamma^{m}G$ be the $m$-th term of the lower
central series of $G$. We then have the following definition, which is a
special case of associated graded Lie algebra:

######  Definition 2.3.

The associated graded Lie algebra is called the Lie algebra associated to the
lower central series of $G$ and is denoted by
$\mathrm{gr}^{\mathrm{LCS}}_{\ast}G$.

######  Remark 2.4.

the Lie algebra associated to the lower central series of $G$ is a Lie
algebra, and graded, but it is not a "graded Lie algebra" as defined by
Milnor-Moore [MM]. For example, in a graded Lie algebra it may be the case
that $[x,x]$ is not equal to $0$. In the graded case

$[x,y]=(-1)^{|x||y|}[y,x].$

Thus if $x=y$ are of degree 1, then

$[x,x]=2x^{2}$

in the universal enveloping algebra (over a field).

To avoid this issue, we may double the grading of each element in the Lie
algebra for the descending central series where $[x,x]=0$.

######  Lemma 2.5.

Let $f$ be a homomorphism from a residually nilpotent group $G$ to a group
$H$. If the induced Lie algebra homomorphism

$f_{\ast}:\mathrm{gr}^{\mathrm{LCS}}_{\ast}G\to\mathrm{gr}^{\mathrm{LCS}}_{\ast}H$

is a monomorphism, then so is $f$.

###### Proof.

Suppose $f$ is not a monomorphism, then there exists some non-identity element
$g\in G$ such that $f(g)=1$. Since $G$ is residually nilpotent,
$g\in\Gamma^{r}G\setminus\Gamma^{r+1}G$ for some positive integer $r$. So

$f\left(g\cdot\Gamma^{r+1}G\right)\subseteq\Gamma^{r+1}H.$

But since $f_{\ast}$ is one-to-one, this implies $g\cdot\Gamma^{r+1}G$ is the
identity element in $\Gamma^{r}G/\Gamma^{r+1}G$, which implies
$g\in\Gamma_{r+1}G$ and this is a contradiction. ∎

######  Definition 2.6.

Let $G$ be a group. The group of IA-automorphism of $G$ is defined to be

$J^{2}\mathrm{Aut}(G):=\mathrm{ker}\left(\mathrm{Aut}(G)\to\mathrm{Aut}\left(G/[G,G]\right)\right).$

That is, each IA-automorphism of $G$ is defined as an automorphism of $G$
which induces the identity automorphism on $G/[G,G]$. Furthermore, we write
$J^{2}\mathrm{Aut}\left(F(X_{n})\right)$ as $\mathrm{IA}_{n}$.

Let $J^{m}\mathrm{Aut}(F(X_{n}))$ be the group of automorphisms of $F(X_{n})$
which induces identity on the quotient group $F(X_{n})/\Gamma_{m+1}F(X_{n})$.
The Johnson filtration on $IA_{n}$ is defined as the filtration

$IA_{n}=J^{2}\mathrm{Aut}(F(X_{n}))\supseteq
J^{3}\mathrm{Aut}(F(X_{n}))\supseteq\cdots.$

Clearly the kernel of the natural quotient

$\operatorname{Aut}(F(X_{n}))\to\mathrm{GL}(n,\mathbb{Z})$

is precisely $IA_{n}$. A finite set of generators of $IA_{n}$ is given in
[MR2109550] as follows:

$M_{n}=\\{\chi_{i,j}\mid 1\leq i,j\leq n,i\neq
j\\}\bigcup\\{\theta_{i,[x_{s},x_{t}]}\mid i\notin\\{s,t\\},1\leq i\leq
n,1\leq s<t\leq n\\},$

where

$\chi_{i,j}(x_{k})=\begin{cases}x_{k},\quad\quad\text{if }k\neq i,\\\
x_{j}^{-1}x_{k}x_{j},\text{if }k=i.\end{cases}$

and

$\theta_{i,[x_{s},x_{t}]}(x_{k})=\begin{cases}x_{k},\quad\quad\text{if }k\neq
i,\\\ x_{i}[x_{s},x_{t}],\text{if }k=i.\end{cases}$

We define two types of derivations of $L(X_{n})$ as follows:

$\widetilde{M}_{n}=\\{\widetilde{\chi}_{i,j}\mid 1\leq i,j\leq n,i\neq
j\\}\bigcup\\{\widetilde{\theta}_{i,[x_{s},x_{t}]}\mid i\notin\\{s,t\\},1\leq
i\leq n,1\leq s<t\leq n\\},$

where

$\widetilde{\chi}_{i,j}(x_{k})=\begin{cases}0,\quad\quad\text{if }k\neq i,\\\
[x_{i},x_{j}],\text{if }k=i.\end{cases}$

and

$\widetilde{\theta}_{i,[x_{s},x_{t}]}(x_{k})=\begin{cases}0,\quad\quad\text{if
}k\neq i,\\\ [x_{s},x_{t}],\text{if }k=i.\end{cases}$

##  3\. Calculation with the Derivations

This section deals with the calculation involving the derivations of a free
Lie algebra in order to explore the structure of certain subgroups of the
automorphism group of a finitely generated free group. We prove the main
theorem stating that any two distinct elements in $\mathrm{IA}_{n}$ either
generate a free group or a free abelian group. Let
$u=[\widetilde{\chi}_{i,j_{1}},\ldots,\widetilde{\chi}_{i,j_{r}}]$ be a Lie
monomial of degree $2r$ in
$\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right)$ ($n\geq 3$) and
$\check{u}$ be the Lie monomial obtained from $u$ by replacing
$\widetilde{\chi}_{i,j_{k}}$ in $u$ with $x_{j_{k}}$ for every $k=1,\ldots,r$.

The following lemma can be easily proved by induction on the degree of simple
Lie monomials:

######  Lemma 3.1.

We have

$u(x_{k})=\begin{cases}[x_{i},\check{u}],\text{ if }k=i;\\\ 0,\text{
otherwise.}\end{cases}$

The following proposition can be found in [CP]:

######  Proposition 3.2.

Let $L[S]$ be the free Lie algebra over the integers generated by a set $S$,
and let $a$ be an element of $S$ with $S$ of cardinality at least $2$. Then
the centralizer of $a$ in $L[S]$ is the linear span of $a$.

######  Lemma 3.3.

The set
$\widetilde{C}_{r}:=\\{\widetilde{\chi}_{i,j_{1}},\ldots,\widetilde{\chi}_{i,j_{r}}\\}$,
$1\leq i\leq n$ generates a free sub-Lie algebra
$S(\widetilde{C}_{r},\mathbb{Z})$ in
$\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right)$.

###### Proof.

Let $Y_{r}=\\{y_{1},\ldots,y_{r}\\}$ be a set of $r$ elements and let
$\widetilde{f}_{r}:L_{\mathbb{Z}}(Y_{r})\to S(\widetilde{C}_{r},\mathbb{Z})$
be the Lie epimorphism extended from the map $f_{r}$ which maps every $y_{k}$
to $\widetilde{\chi}_{i,j_{k}}$.

Let $u=\sum a_{k}u_{k}$ be a non-zero element in $L_{\mathbb{Z}}(Y_{r})$,
where each $u_{k}$ is a Lie monomial. For every Lie monomial $v$, let
$v^{\\#}$ be the lie monomial in
$\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right)$ obtained by replacing
every $y_{k}$ by $\widetilde{\chi}_{i,j_{k}}$. By Lemma 3.1,

$\widetilde{f}_{r}(u)(x_{k})=\sum_{k}a^{k}u_{k}^{\\#}(x_{k})=\begin{cases}[x_{i},\sum_{k}a_{k}\check{u}^{\\#}],$
if $k=i\\\ 0,$ if $k\neq i.\end{cases}$

Since $\sum_{k}a_{k}\check{u}^{\\#}$ is non-zero, so is
$[x_{i},\sum_{k}a_{k}\check{u}^{\\#}]$ by Proposition 3.2. Thus
$\widetilde{f}_{r}(u)$ is non-zero and therefore, $\widetilde{f}_{r}$ is an
isomorphism. ∎

Let $N$ be a fully invariant subgroup of $F(X_{n})$. The quotient group
$F(X_{n})/N$ is called the _universal variety group_ associated to $N$.

######  Definition 3.4.

Let $G$ be a universal variety group, then the Lie algebra associated with the
Johnson filtration on $\mathrm{IA}_{n}(G)$ is called the Johnson Lie algebra
associated with group $G$ and is denoted by
$\mathrm{gr}_{\ast}^{\mathrm{J}}\left(\mathrm{IA}_{n}(G)\right)$.

Let $A$ be a subgroup of a group $B$ and an integral filtration $\\{B_{m}\\}$
naturally induces a filtration $\\{A_{m}\\}$ on $A$. One easily proves the
following lemma:

######  Lemma 3.5.

The inclusion $i:A\hookrightarrow B$ induces a monomorphism of Lie algebras

$i_{\ast}:\mathrm{gr}_{\ast}A\to\mathrm{gr}_{\ast}B.$

As a special case, for a subgroup $G$ of $\mathrm{IA}_{n}$, the inclusion
$i_{G}:G\hookrightarrow\mathrm{IA}_{n}$ induces a monomorphism of Lie algebra

$i_{G,\ast}:\mathrm{gr}_{\ast}^{J}(G)\to\mathrm{gr}_{\ast}^{J}(\mathrm{IA}_{n}).$

The following lemma follows from Lemma 3.5 and the fact that
$\widetilde{J}_{n,\ast}$ is a monomorphism of Lie algebras:

######  Lemma 3.6.

For every subgroup $G$ of $\mathrm{IA}_{n}$, $\widetilde{J}_{G,\ast}$ is a
monomorphism of Lie algebras.

The following lemma is proven in [S]:

######  Lemma 3.7.

Let $F(X)$ be the free group generated by a set $X$. The canonical map
$X\to\mathrm{gr}_{1}^{\mathrm{LCS}}\left(F(X)\right)=F(X)/\left[F(X),F(X)\right]$
induces a natural isomorphism

$\phi_{X,\ast}:L_{\mathbb{Z}}(X)\to\mathrm{gr}_{\ast}^{\mathrm{LCS}}\left(F(X)\right).$

The following lemma given in [Jin] will be used to prove Theorem 3.11:

######  Lemma 3.8.

Let $n\geq r+1$ and let $G$ be the subgroup of $\mathrm{IA}_{n}$ generated by
$C_{r}:=\\{\chi_{i,j_{1}},\ldots,\chi_{i,j_{r}}\\}$, then $G$ is a free group
and the natural homomorphism

$\psi_{G,\ast}:\mathrm{gr}_{\ast}^{\mathrm{LCS}}(G)\to\mathrm{gr}_{\ast}^{\mathrm{J}}(G)$

is a monomorphism of Lie algebras. Let

$\widetilde{J}_{G,\ast}:\mathrm{gr}_{\ast}^{\mathrm{J}}(G)\to\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right)$

be the Johnson homomorphism of $G$. Then
$\widetilde{J}_{G,\ast}\left(\psi_{G,\ast}\left(\mathrm{gr}_{\ast}^{\mathrm{LCS}}(G)\right)\right)$
is exactly the free sub-Lie algebra $S(\widetilde{C}_{r},\mathbb{Z})$ in
$\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right)$.

Let $L_{n}$ be a collection of $n$ unknotted, unlinked circles in
$\mathbb{R}^{3}$, and let $P\Sigma_{n}$ be the group of motions of $L_{n}$
where each circle ends up back at its original position. The group
$P\Sigma_{n}$ can be represented as a group of free group automorphisms; that
is, $P\Sigma_{n}$ is the subgroup of $\mathrm{Aut}\left(F(X_{n})\right)$
defined by

$P\Sigma_{n}:=\\{\alpha\in\mathrm{Aut}\left(F(X_{n})\right)\mid\alpha(x_{i})\text{
is conjugate to }x_{i},i=1,\ldots,n\\};$

see [JMM] for details.

A presentation of $P\Sigma_{n}$ given in the following theorem appeared in
[McCool]:

######  Theorem 3.9.

A presentation of $P\Sigma_{n}$ is given by generators $\chi_{k,j}(k\neq j)$
with the following relations.

  1. (1)

For distinct $i,j,k$,
$\chi_{i,j}\cdot\chi_{k,j}\cdot\chi_{i,k}=\chi_{i,k}\cdot\chi_{i,j}\cdot\chi_{k,j}$.

  2. (2)

If $\\{j,k\\}\cap\\{s,t\\}=\emptyset$, then $[\chi_{k,j},\chi_{s,t}]=1$,

  3. (3)

$[\chi_{k,j},\chi_{s,j}]=1$,

  4. (4)

For distinct $i,j,k$, $[\chi_{i,k},\chi_{i,j}]=[\chi_{i,k},\chi_{k,j}^{-1}]$.

The following lemma can be proved by induction on the degree of Lie monomials
using Jacobi identity, while the base case is clear from Theorem 3.9:

######  Lemma 3.10.

Let $u$ be a Lie monomial consisting of $\widetilde{\chi}_{i,j}$ and
$\widetilde{\chi}_{i,j^{\prime}}$ for distinct $i,j,j^{\prime}$ in
$\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right)$ $(n\geq 3)$ and
$\hat{u}$ be the Lie monomial obtained by replacing all
$\widetilde{\chi}_{i,j^{\prime}}$ with $-\widetilde{\chi}_{j,j^{\prime}}$ in
the expression of $u$. Then when $\mathrm{deg}(u)\geq 2$,

$u=\hat{u}.$

######  Theorem 3.11.

Any two distinct elements in ${IA}_{n}$ either generate a free group or a free
abelian group. More precisely, let $\chi_{i,j}$ and
$\chi_{i^{\prime},j^{\prime}}$ be two distinct elements in ${IA}_{n}$. Then
$\chi_{i,j}$ and $\chi_{i^{\prime},j^{\prime}}$ generate

  1. (1)

a free abelian group, when
$\\{i,j\\}\cap\\{i^{\prime},j^{\prime}\\}=\emptyset$, or $j=j^{\prime}$ and
$i\neq i^{\prime}$.

  2. (2)

a free group, otherwise.

###### Proof.

When $\\{i,j\\}\cap\\{i^{\prime},j^{\prime}\\}=\emptyset$, or $j=j^{\prime}$
and $i\neq i^{\prime}$, the conclusion is straightforward. Otherwise, there
are a few possibilities:

  1. (1)

When $i=j^{\prime}$ and $j=i^{\prime}$, this case is essentially the same as
$\chi_{1,2}$ and $\chi_{2,1}$ in $IA_{2}$. Nielsen [Nielsen] proved that
$IA_{2}$ is a free group generated by $\chi_{1,2}$ and $\chi_{2,1}$ and the
result follows.

  2. (2)

When $i=i^{\prime}$ and $j\neq j^{\prime}$. When
$i>\mathrm{max}\\{j,j^{\prime}\\}$, it can be proved using upper triangular
McCool group [CPVW].

  3. (3)

When $j=i^{\prime}$ and $i\neq j^{\prime}$, let $H$ and $G$ be the subgroups
of $IA_{n}$ generated by $\\{\chi_{i,j},\chi_{j,j^{\prime}}\\}$ and
$\\{\chi_{i,j},\chi_{i,j^{\prime}}\\}$, respectively. Let $f:G\to H$ be the
group epimorphism defined by $f(\chi_{i,j})=\chi_{i,j}$ and
$f(\chi_{i,j^{\prime}})=\chi_{j,j^{\prime}}^{-1}$.

Consider the following diagram, which is commutative for $\ast\geq 2$ by Lemma
3.10, where $\psi_{G,\ast}$ and $\psi_{H,\ast}$ are the natural homomorphisms
from the LCS Lie algebras of $G$ and $H$ to the Johnson Lie algebras of $G$
and $H$, respectively; $\widetilde{J}_{G,\ast}$ and $\widetilde{J}_{H,\ast}$
are the Johnson homomorphism of $G$ and $H$, respectively.

${gr^{LCS}_{\ast}(H)}$${gr^{LCS}_{\ast}(G)}$${gr^{J}_{\ast}(H)}$${gr_{\ast}^{J}(G)}$${\mathrm{Der}\left(L_{\mathbb{Z}}(X_{n})\right)}$$\scriptstyle{\psi_{H,\ast}}$$\scriptstyle{f_{\ast}^{LCS}}$$\scriptstyle{\psi_{G,\ast}}$$\scriptstyle{\widetilde{J}_{H,\ast}}$$\scriptstyle{\widetilde{J}_{G,\ast}}$

By the lemmas 3.6 and 3.8, the map $\widetilde{J}_{G,\ast}\circ\psi_{G,\ast}$
is a monomorphism of Lie algebras, which implies that
$f_{\ast}^{\mathrm{LCS}}$ is a monomorphism when $\ast\geq 2$ while the case
$\ast=1$ is obvious.

  4. (4)

When $i=j^{\prime}$ and $j\neq i^{\prime}$, the proof is essentially the same
as the previous case.

Since
$f_{\ast}^{\mathrm{LCS}}:\mathrm{gr}_{\ast}^{\mathrm{LCS}}(G)\to\mathrm{gr}_{\ast}^{\mathrm{LCS}}(H)$
is a monomorphism, so is $f$ and the proof is finished.

∎

Let $G$ be a universal variety group and $m\geq 2$, there is a monomorphism

$J^{m}\mathrm{Aut}(G)/J^{m+1}\mathrm{Aut}(G)\to\operatorname{Hom}_{\mathbb{Z}}(G/[G,G],\Gamma^{m}G/\Gamma_{m+1}G).$

When $G=F(X_{n})$, this monomorphism gives a $\mathbb{Z}$-linear injection

$\widetilde{J}_{n,m}:\mathrm{gr}_{m}^{\mathrm{J}}({IA}_{n})\to\mathrm{Der}_{m+1}(L_{\mathbb{Z}}(X_{n})).$

Therefore, one obtains a $\mathbb{Z}$-module monomorphism, called the Johnson
homomorphism of $\mathrm{IA}_{n}$:

$\widetilde{J}_{n,\ast}:\mathrm{gr}_{\ast}^{\mathrm{J}}({IA}_{n})\to\bigoplus_{m=2}^{\infty}\mathrm{Der}_{m}\left(L_{\mathbb{Z}}(X_{n})\right).$

##  4\. The Schur Algebra on $V_{n}(K)^{\otimes q}$

In this section we recall the definition of Schur algebra and we prove a few
results regarding the Schur algebra on the tensor product of a finitely
generated free $K$-module.

######  Definition 4.1.

The Schur algebra on $V_{n}^{\otimes q}$ is the subalgebra of the algebra of
endomorphisms on $V_{n}^{\otimes q}$ consisting of those $K$-linear maps $f$
which make the following diagram commute for all $\sigma\in\Sigma_{q}$,

${V_{n}^{\otimes q}}$${V_{n}^{\otimes q}}$${V_{n}^{\otimes
q}}$${V_{n}^{\otimes
q}}$$\scriptstyle{f}$$\scriptstyle{\sigma}$$\scriptstyle{\sigma}$$\scriptstyle{f}$

We denote the Schur algebra on $V_{n}^{\otimes q}$ by $\mathrm{Schur}(q,n,K)$.
By convention we define $V_{n}^{\otimes 0}=K$.

The Schur algebra on $T(V_{n})$ will be denoted by
$\displaystyle{Schur(n,K)=\bigoplus_{q=0}^{\infty}\mathrm{Schur}(q,n,K)}$,
where $T(V_{n})=\displaystyle{\bigoplus_{q=0}^{\infty}V_{n}^{\otimes q}}$ is
the tensor algebra. A natural basis of $V_{n}^{\otimes q}$ is

$B_{n,q}:=\\{x_{i_{1}}\otimes\cdots\otimes x_{i_{q}},\quad 1\leq
i_{1},\ldots,i_{q}\leq n\\}.$

We consider the following proper subset $S_{n,q}$ of $B_{n,q}$ given by

$S_{n,q}:=\\{x_{i_{1}}\otimes\cdots\otimes x_{i_{q}},\quad 1\leq
i_{1}\leq\ldots\leq i_{q}\leq n\\}.$

Let $\Sigma_{q}$ be the symmetric group on $q$ letters and for the
$\Sigma_{q}$-action on $B_{n,q}$, the corresponding left $\Sigma_{q}$-orbit is
given by

$\mathscr{O}_{n,q}:=\\{\sigma\cdot u\mid\sigma\in\Sigma_{q}\text{ and }u\in
B_{n,q}\\}=\\{\sigma\cdot u\mid\sigma\in\Sigma_{q}\text{ and }u\in
S_{n,q}\\}.$

######  Definition 4.2.

For a Lie monomial $u\in L_{X_{n}}^{p}$, let $\mathrm{deg}_{x_{i}}(u)$ be the
number of occurrence of the letter $x_{i}$ in the expression of $u$. The
multi-degree of $u$ is defined as

$\mathrm{mdeg}(u)=\left(\mathrm{deg}_{x_{1}}(u),\ldots,\mathrm{deg}_{x_{n}}(u)\right).$

Define $T^{\triangle}\left(V_{n}(K)\right)$ as the two-sided ideal generated
by $V_{n}(K)$; that is,

$T^{\triangle}\left(V_{n}(K)\right)=\bigoplus_{q=1}^{\infty}T^{q}\left(V_{n}(K)\right).$

Let

$\mathfrak{b}_{n}:T^{\triangle}\left(V_{n}(K)\right)\to L_{K}(X_{n})$

by

$\mathfrak{b}_{n,2p}(x_{i_{1}}\otimes\cdots\otimes
x_{i_{p}})=\mathrm{ad}(x_{i_{1}})\cdots\mathrm{ad}(x_{i_{p-1}})(x_{i_{p}})\text{,}$

where $\mathfrak{b}_{n,2p}$ is the degree $2p$ component of the map
$\mathfrak{b}_{n}$.

######  Remark 4.3.

The map $\mathfrak{b}_{n}$ is essentially the standard map defined by Specht
and Wever; see page 169 of [Jacobson] and page 1197 of [Cohen95] for details.
In particular, the classical Specht-Wever relation can be written as

$b_{n,2p}\circ b_{n,2p}=pb_{n,2p}.$

The proof of the following proposition is straightforward:

######  Proposition 4.4.

Let $u$ and $v$ be elements in $B_{n,q}$. Then

$\\{\sigma\cdot u\mid\sigma\in\Sigma_{q}\\}=\\{\sigma\cdot
v\mid\sigma\in\Sigma_{q}\\}$

if and only if

$\mathrm{mdeg}\left(\mathfrak{b}_{n,2q}(u)\right)=\mathrm{mdeg}\left(\mathfrak{b}_{n,2q}(v)\right).$

For $u\in B_{n,q}$, let $\mathcal{I}_{u}$ be the stabilizer subgroup of $u$
under the action of $\Sigma_{q}$; that is,

$\mathcal{I}_{u}=\\{g\in\Sigma_{q}\mid g\cdot u=u\\}.$

######  Theorem 4.5.

For any $v\in B_{n,q}\setminus S_{n,q}$, there is a unique $u\in S_{n,q}$ and
some $\sigma\in\Sigma_{q}$ such that $v=\sigma(u)$. If $f\in Sch(q,n,K)$ then
$f$ satisfies the following two conditions

  1. (1)

$f(v)=\sigma(f(u))$ for any $\sigma\in\Sigma_{q}$ satisfying $v=\sigma(u),v\in
B_{n,q}\setminus S_{n,q}$ and $u\in S_{n,q}$;

  2. (2)

Let $f(u)=\displaystyle{\sum_{w\in B_{n,q}}a_{w}w}$ for $u\in S_{n,q}$, If
$\mathcal{I}_{u}\cdot w=\mathcal{I}_{u}\cdot w^{\prime}$, then
$a_{w}=a_{w^{\prime}}$. Or equivalently, the following is true for every $u\in
S_{n,q}$:

$f(u)=\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(a_{\mathcal{I}_{u}\cdot w}\cdot\sum_{v\in
B_{n,q},\mathcal{I}_{u}\cdot v=\mathcal{I}_{u}\cdot w}v\right),$

where $a_{\mathcal{I}_{u}\cdot w}\in K$.

Conversely, if $f\in\mathrm{End}\left(V_{n}^{\otimes q}\right)$ satisfies the
above two conditions, then $f\in\mathrm{Schur}(q,n,K)$.

###### Proof.

When $f\in Sch(q,n,K)$ then $f$, then $f$ satisfies the first condition by
Definition 4.1 and we shall prove the second condition. Assuming
$\mathcal{I}_{u}\cdot w=\mathcal{I}_{u}\cdot w^{\prime}$, let $\sigma$ be an
element in $\mathcal{I}_{u}$ such that $\sigma\cdot w=w^{\prime}$. Then

$\displaystyle\sum_{w\in B_{n,q}}a_{w}w$ $\displaystyle=f(u)$
$\displaystyle=f(\sigma\cdot u)$ $\displaystyle=\sigma\cdot f(u)$
$\displaystyle=\sigma\cdot\sum_{w\in B_{n,q}}a_{w}w$ $\displaystyle=\sum_{w\in
B_{n,q}}a_{w}\sigma\cdot w$

Since $\sigma\cdot w=w^{\prime}$, we conclude that $a_{w}=a_{w^{\prime}}$.

Now we prove condition (1) and (2) imply $f\in\mathrm{Sch}(q,n,K)$. It
suffices to prove

$f\left(\sigma(w)\right)=\sigma\left(f(w)\right)$

for every $w\in B_{n,q}$ and $\sigma\in\Sigma_{q}$.

#### Case 1

When $\sigma(w)=w$ and $w\in S_{n,q}$, condition (2) implies
$f\left(\sigma(w)\right)=\sigma\left(f(w)\right)$. We write $w=\sum_{\theta\in
B_{n,q}}a_{\theta}\theta$. Therefore, $\sigma\cdot f(w)=\sum_{\theta\in
B_{n,q}}a_{\theta}\sigma\cdot\theta$.

Since $\mathcal{I}_{w}\cdot(\sigma\cdot\theta)=\mathcal{I}_{w}\cdot\theta$
(Notice if $h\in I_{w}$, so is $h\cdot\sigma$),
$a_{\sigma\cdot\theta}=a_{\theta}$, so $\sigma\cdot f(w)=\sum_{\theta\in
B_{n,q}}a_{\theta}\sigma\cdot\theta=\sum_{\theta\in B_{n,q}}a_{\theta}\theta$.

#### Case 2

When $\sigma(w)=w$ and $w\notin S_{n,q}$, condition (1) implies
$f\left(\sigma(w)\right)=\sigma\left(f(w)\right)$.

#### Case 3

When $\sigma(w)\neq w$ and $w\in S_{n,q}$. In this case $\sigma\cdot w\in
B_{n,q}\setminus S_{n,q}$ and the conclusion is clear by condition (1).

#### Case 4

When $\sigma(w)\neq w$ and $w\notin S_{n,q}$. therefore exists a unique $u\in
S_{n,q}$ and some $\sigma^{\prime}\in\Sigma_{q}$ such that
$w=\sigma^{\prime}(u)$.

  1. (1)

When $\sigma(w)\in S_{n,q}$, then condition (1) implies

$f(w)=\sigma^{-1}(f(\sigma(w))),$

and the result follows.

  2. (2)

When $\sigma(w)\in B_{n,q}\setminus S_{n,q}$, then there are $u\in S_{n,q}$
and $\sigma^{\prime}\in\Sigma_{q}$ such that $w=\sigma^{\prime}(u)$. we obtain
from condition (1) that $f(w)=\sigma^{\prime}\left(f(u)\right)$ and a direct
calculation shows that
$f\left(\sigma(w)\right)=f(\sigma(\sigma^{\prime}(u)))=\sigma(\sigma^{\prime}(f(u)))\sigma\left(f(w)\right)$.

∎ The following theorem gives a necessary and sufficient condition when a map
from $S_{n,q}$ to $V_{n}^{\otimes q}$ can be extended to a $\Sigma_{q}$-linear
map.

######  Theorem 4.6.

A map $g$ from $S_{n,q}$ to $V_{n}^{\otimes q}$ can be extended to a
$\Sigma_{q}$-linear map if and only if $g$ satisfies condition (2) in Theorem
4.5.

###### Proof.

By Theorem 4.5, we only need to prove the case when $g$ satisfies condition
(2) in Theorem 4.5, it can be extended to a $\Sigma_{q}$-linear map.

We extend $g$ to $B_{n,q}\setminus S_{n,q}$ in the following manner: for each
$v\in B_{n,q}\setminus S_{n,q}$, there exists a unique $u\in S_{n,q}$ and some
$\sigma\in\Sigma_{q}$ such that $v=\sigma(u)$. Define
$f(v)=\sigma\left(g(u)\right)$. Condition (2) in Theorem 4.5 guarantees that
$f$ is well-defined. When $u\in S_{n,q}$, define $f(u)=g(u)$. So we have a
well-defined map $f:B_{n,q}\to B_{n,q}$ which extends $g$. Therefore we have
an extension (still denoted by $f$) of $g$ from $V_{n}^{\otimes q}$ to itself.
By the construction of $f$, it satisfies the two conditions in Theorem 4.5 and
is therefore $\Sigma_{q}$-linear. ∎

When the extension Theorem 4.6 exists, it is clearly unique. The following
corollary is proved in essentially the same manner.

######  Corollary 4.7.

A function from $R_{n,q}$, a set of representatives of the $\Sigma_{q}$-orbits
in $B_{n,q}$, to $V_{n}^{\otimes q}$ can be extended to a $\Sigma_{q}$-linear
map if and only if for every $u\in R_{n,q}$,

$g(u)=\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(a_{\mathcal{I}_{u}\cdot w}\cdot\sum_{v\in
B_{n,q},\mathcal{I}_{u}\cdot v=\mathcal{I}_{u}\cdot w}v\right),$

where $a_{\mathcal{I}_{u}\cdot w}\in K$.

Consider the canonical embedding

$I_{n}:L(X_{n})\to T(V_{n})$

and its restriction

$I_{n,q}:L^{q}(X_{n})\to T^{q}(V_{n}).$

######  Definition 4.8.

Let $\mathrm{Br}_{q}$ be the set of Lie bracketing of $Lie$ monomials of
degree $2q$, where the degree is doubled as in Remark 2.2. The the embedding
$I_{n,q}$ of $L^{q}(X_{n})$ into $T^{q}(V_{n})$ induces a mapping, called the
$q$-th bracketing function, $\mathrm{br}_{q}:\mathrm{Br}_{q}\to K(\Sigma_{q})$
such that for every Lie monomial $[x_{i_{1}},\cdots,x_{i_{q}}]$,

$I_{n,q}\left([x_{i_{1}},\cdots,x_{i_{q}}]\right)=\mathrm{br}_{q}\left(\llbracket[x_{i_{1}},\cdots,x_{i_{q}}]\rrbracket\right)(x_{i_{1}}\otimes\cdots\otimes
x_{i_{q}}).$

######  Example 4.9.

$\displaystyle I_{n,3}\left(\left[[x_{i},x_{j}],x_{k}\right]\right)=$
$\displaystyle[x_{i}\otimes x_{j}-x_{j}\otimes x_{i},x_{k}]$ $\displaystyle=$
$\displaystyle(x_{i}\otimes x_{j}-x_{j}\otimes x_{i})\otimes
x_{k}-x_{k}\otimes(x_{i}\otimes x_{j}-x_{j}\otimes x_{i})$ $\displaystyle=$
$\displaystyle x_{i}\otimes x_{j}\otimes x_{k}-x_{j}\otimes x_{i}\otimes
x_{k}-x_{k}\otimes x_{i}\otimes x_{j}+x_{k}\otimes x_{j}\otimes x_{i}$
$\displaystyle=$ $\displaystyle 1(x_{i}\otimes x_{j}\otimes
x_{k})-(12)(x_{i}\otimes x_{j}\otimes x_{k})$
$\displaystyle-(123)(x_{i}\otimes x_{j}\otimes x_{k})+(13)(x_{i}\otimes
x_{j}\otimes x_{k})$ $\displaystyle=$
$\displaystyle\left(1-(12)-(123)+(13)\right)(x_{i}\otimes x_{j}\otimes x_{k})$

Therefore,

$\mathrm{br}_{3}\left(\left[[,],\right]\right)=1-(12)-(123)+(13)\in
K(\sigma_{3}).$

Similarly,

$\mathrm{br}_{3}\left(\left[,[,]\right]\right)=1-(23)-(132)+(13)\in
K(\sigma_{3}).$

Every element $f\in\mathrm{Schur}(q,n,K)$ sends Lie elements to Lie elements,
as is shown in the following theorem.

######  Theorem 4.10.

For any $f\in\mathrm{Schur}(q,n,K)$ and Lie monomial
$[x_{i_{1}},\cdot,x_{i_{q}}]$, we have

$f\left([x_{i_{1}},\cdot,x_{i_{q}}]\right)=\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(a_{\mathcal{I}_{u}\cdot w}\cdot\sum_{v\in
B_{n,q},\mathcal{I}_{u}\cdot v=\mathcal{I}_{u}\cdot w}[v]\right),$

where $u=x_{i_{1}}\otimes\cdots\otimes x_{i_{q}}$, $a_{\mathcal{I}_{u}\cdot
w}\in K$ and all the Lie monomials have the same bracketing.

Let the cardinality of $\mathcal{O}_{u}$ be denoted by $d_{u}$. Fix $u$ and
let $f$ run through $\mathrm{Schur}(q,n,K)$, then the $d_{u}$-tuple
$(a_{\mathrm{I}_{u}\cdot w})_{\mathrm{I}_{u}\cdot w\in\mathcal{O}_{u}}$ runs
through $K^{d_{u}}$.

###### Proof.

From the definition of the $q$-th bracketing function $\mathrm{br}_{q}$ and
the $\Sigma_{q}$ linearity of $f$, we conclude that

$f\left([x_{i_{1}},\cdots,x_{i_{q}}]\right)=\mathrm{br}_{q}\left(\llbracket[x_{i_{1}},\cdots,x_{i_{q}}]\rrbracket\right)\left(f(u)\right).$

Since the orbits of $\mathcal{I}_{u}$’s action on $B_{n,q}$ form a partition
of $B_{n,q}$, we may write

$f(u)=\sum_{v\in B_{n,q}}a_{v}\cdot v=\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(\sum_{v\in B_{n,q},\mathcal{I}_{u}\cdot
v=\mathcal{I}_{u}\cdot w}a_{v}\cdot v\right).$

By the second condition in 4.5, we conclude that

$f\left([x_{i_{1}},\cdots,x_{i_{q}}]\right)=\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(a_{\mathcal{I}_{u}\cdot w}\sum_{v\in
B_{n,q},\mathcal{I}_{u}\cdot v=\mathcal{I}_{u}\cdot
w}\mathrm{br}_{q}\left(\llbracket[x_{i_{1}},\cdots,x_{i_{q}}]\rrbracket\right)(v)\right).$

From the definition of $\mathrm{br}_{q}$,

$\mathrm{br}_{q}\left(\llbracket[x_{i_{1}},\cdots,x_{i_{q}}]\rrbracket\right)(v)=[v],$

where $[v]$ is a Lie monomial whose bracketing is the same as the bracketing
of $[x_{i_{1}},\cdots,x_{i_{q}}]$. So

$f\left([x_{i_{1}},\cdot,x_{i_{q}}]\right)=\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(a_{\mathcal{I}_{u}\cdot w}\cdot\sum_{v\in
B_{n,q},\mathcal{I}_{u}\cdot v=\mathcal{I}_{u}\cdot w}[v]\right).$

For $u\in B_{n,q}$ there exists a unique $u^{\prime}\in S_{n,q}$ such that
$u=\sigma(u^{\prime})$ for some $\sigma\in\Sigma_{q}$. Let
$R_{n,q}=\left(S_{n,q}\cup\\{u\\}\right)\setminus\\{u^{\prime}\\}$. For every
$(a_{\mathrm{I}_{u}\cdot w})_{\mathrm{I}_{u}\cdot w\in\mathcal{O}_{u}}\in
K^{d_{u}}$, define $g:R_{n,q}\to V_{n}^{\otimes q}$ as follows:

$g(r)=\begin{cases*}\sum_{\mathcal{I}_{u}\cdot
w\in\mathcal{O}_{u}}\left(a_{\mathcal{I}_{u}\cdot w}\cdot\sum_{v\in
B_{n,q},\mathcal{I}_{u}\cdot v=\mathcal{I}_{u}\cdot w}v\right)&if $r=u$,\\\
0&if $r\neq u$.\end{cases*}$

By Corollary 4.7, $g$ can be uniquely extended to some
$f\in\mathrm{Schur}(q,n,K)$. So the coefficients of $(a_{\mathrm{I}_{u}\cdot
w})_{\mathrm{I}_{u}\cdot w\in\mathcal{O}_{u}}$ can be any element in
$K^{d_{u}}$ and this completes the proof. ∎

######  Proposition 4.11.

The map $\mathfrak{b}_{n}$ commutes with Schur algebra elements. That is, for
all $q\geq 1$ and $f\in\mathrm{Schur}(q,n,K)$,

$f\circ\mathfrak{b}_{n,2q}=\mathfrak{b}_{n,2q}\circ f.$

###### Proof.

Similar to the proof of Theorem 4.10, we obtain

$f\circ\mathfrak{b}_{n,2q}(x_{i_{1}}\otimes\cdots\otimes
x_{i_{q}})=\mathfrak{b}_{n,2q}\circ f(x_{i_{1}}\otimes\cdots\otimes
x_{i_{q}})$

for all $x_{i_{1}}\otimes\cdots\otimes x_{i_{q}}\in V_{n}(K)^{\otimes q}$. ∎

######  Proposition 4.12.

The Schur algebra elements send Lie elements to Lie elements. That is, for all
$f\in\mathrm{Schur}(q,n,K)$ and $u\in L_{K}^{q}(X_{n})$,

$f(u)\in L_{K}^{q}(X_{n}).$

###### Proof.

Consider the simple Lie monomial

$v=[x_{i_{1}},\ldots,x_{i_{q}}]\in V_{n}(K)^{\otimes q}.$

By the definition of $\mathfrak{b}_{n,2q}$,

$\mathfrak{b}_{n,2q}\left(f(x_{i_{1}}\otimes\cdots\otimes x_{i_{q}})\right)\in
L_{K}^{q}(X_{n}).$

That is,

$f\left(\mathfrak{b}_{n,2q}(x_{i_{1}}\otimes\cdots\otimes x_{i_{q}})\right)\in
L_{K}^{q}(X_{n}).$

Since every Lie element can be written as a linear combination of simple Lie
monomials, the result follows. ∎

Now we discuss the action of the Schur Algebra on the Lie algebra of
derivations of a free Lie algebra.

######  Definition 4.13.

For any map $h:X_{n}\to L^{q}(X_{n})$, the unique
$g\in\mathrm{Der}_{q}\left(L(X_{n})\right)$ satisfying $h=g|_{X_{n}}$ is
called the extension of $h$ and is denoted by

$g=h^{\mathrm{Ext}}$

######  Definition 4.14.

The action of $f\in\mathrm{Schur}(q,n,K)$ on
$\mathrm{Der}_{q}\left(L_{K}(X_{n})\right)$ is defined as the map

$\Phi_{f}:\mathrm{Der}_{q}\left(L_{K}(X_{n})\right)\to\mathrm{Der}_{q}\left(L_{K}(X_{n})\right)$

with

$\Phi_{f}(\theta)=\left(f\circ
I_{n,q}\circ\theta|_{X_{n}}\right)^{\mathrm{Ext}}.$

######  Proposition 4.15.

The action of the Schur algebra $\mathrm{Schur}(q,n,K)$ on $V_{n}^{\otimes q}$
induces an action on $\mathrm{Der}_{q}\left(L_{K}(X_{n})\right)$ and therefore
an action on $\mathrm{Der}_{q}\left(L_{K}(X_{n})\right)$

$\displaystyle\Phi_{n,q}:\mathrm{Schur}(q,n,K)\times\mathrm{Der}_{q}(L(X_{n}))$
$\displaystyle\to\mathrm{Der}_{q}(L(X_{n}))$ $\displaystyle(f,\theta)$
$\displaystyle\mapsto\Phi_{f}(\theta).$

###### Proof.

By the definition of $\mathrm{Schur}(q,n,K)$, $V_{n}^{\otimes q}$ is a left
$\mathrm{Schur}(q,n,K)$ module. Since every Schur algebra element preserves
Lie elements, the result follows. ∎

The following theorem can be derived directly from Proposition 4.15.

######  Theorem 4.16.

The Lie algebra of derivations $\mathrm{Der}_{*}(L_{K}(X_{n}))$ and
$Der_{*}^{\Delta}(L_{K}(X_{n}))$ are graded left-modules over the graded Schur
algebra $Sch(n,K)$. The action of the Schur algebra $Sch(n,K)$ on
$\mathrm{Der}_{*}(L_{K}(X_{n}))$ and $Der_{*}^{\Delta}(L_{K}(X_{n}))$ is given
by Proposition 4.15.

##  5\. The Associativity and Anti-commutativity of the Schur Algebra

Here we study some properties of Schur algebras and their action on the Lie
algebra of derivations of a free Lie algebra. By using the inter-degree
multiplication induced from a transfer-like map, it is shown that the Schur
algebra forms an associative and anti-commutative graded K-algebra, and the
so-called Schur operad is also constructed. The proof to the following lemma
is clear:

######  Lemma 5.1.

Given groups $A\leq B\leq G$, let $L_{A,B}$ and $L_{B,G}$ be complete sets of
representatives of the left cosets of $A$ in $B$ and of $B$ in $G$,
respectively. Then the following hold:

  1. (1)

If $(g,h)$ and $(g^{\prime},h^{\prime})$ are distinct elements in
$L_{B,G}\times L_{A,B}$, then $gh\neq g^{\prime}h^{\prime}$.

  2. (2)

The set

$\\{gh\mid(gh)\in L_{B,G}\times L_{A,B}\\}$

is a complete set of representatives of the left cosets of $A$ in $G$.

Let $\lambda$ be a partition of a positive integer $d$; $d=a_{1}+\cdots+a_{k}$
(the order matters). We define the Young subgroup of $\Sigma_{d}$, denoted by
$\Sigma_{\lambda}$ as the direct product
$\Sigma_{a_{1}}\times\cdots\times\Sigma_{a_{k}}$. Let $L_{\lambda}$ a complete
set of representatives of the left cosets of $\Sigma_{\lambda}$ in
$\Sigma_{d}$. Define the “transfer-like” map

$M_{\lambda}:\mathrm{Schur}(a_{1},n,K)\otimes\cdots\otimes\mathrm{Schur}(a_{k},n,K)\to\mathrm{Schur}(d,n,K)$

by

$\left(M_{\lambda}(f_{1}\otimes\cdots\otimes
f_{k})\right)(x_{i_{1}}\otimes\cdots\otimes x_{i_{d}})=\sum_{\sigma\in
L_{\lambda}}\sigma\left((f_{1}\otimes\cdots\otimes
f_{k})\left(\sigma^{-1}(x_{i_{1}}\otimes\cdots\otimes
x_{i_{d}})\right)\right).$

One easily checks that $M_{\lambda}$ is well-defined and is independent of the
choice of $L_{\lambda}$.

We define the inter-degree product

$\boxtimes:\mathrm{Schur}(0,n,K)\times\mathrm{Schur}(q,n,K)\to\mathrm{Schur}(q,n,K)$

by $f\boxtimes g:=f\cdot g$, where the right hand side makes sense as $f$ is
in $\mathrm{Schur}(0,n,K)=K$.

To simplify the notation, for $\lambda=(a_{1},\ldots,a_{k})$, when
specification of the partition is unnecessary, we denote
$M_{\lambda}(f_{1}\otimes\cdots\otimes f_{k})$ by
$\displaystyle\star_{i=1}^{k}f_{i}$ and when $k=2$, we also write it as
$f_{1}\star f_{2}$. We shall show this product is associative and commutative.

######  Lemma 5.2.

For $f_{i}\in\mathrm{Schur}(a_{i},n,K)$ $i=1,2,3$,

$\star_{i=1}^{3}f_{i}=(f_{1}\star f_{2})\star f_{3}=f_{1}\star(f_{2}\star
f_{3}).$

###### Proof.

Let $\lambda^{\prime}$ be the partition $(a_{1},a_{2})$ of $a_{1}+a_{2}$ and
let $L_{\lambda^{\prime}}$ be a complete set of representatives of the left
cosets of $\Sigma_{a_{1}}\times\Sigma_{a_{2}}$ in $\Sigma_{a_{1}+a_{2}}$. The
natural embedding of groups
$\Sigma_{a_{1}}\times\Sigma_{a_{2}}\hookrightarrow\Sigma_{a_{1}+a_{2}}$ gives
the embedding

$\Sigma_{a_{1}}\times\Sigma_{a_{2}}\times\Sigma_{a_{3}}\hookrightarrow\Sigma_{a_{1}+a_{2}}\times\Sigma_{a_{3}}.$

So $L_{\lambda^{\prime}}$ can also be regarded as a complete set of
representatives of the left cosets of
$\Sigma_{a_{1}}\times\Sigma_{a_{2}}\times\Sigma_{a_{3}}$ in
$\Sigma_{a_{1}+a_{2}}\times\Sigma_{a_{3}}$. Consider the groups

$\Sigma_{a_{1}}\times\Sigma_{a_{2}}\times\Sigma_{a_{3}}\leq\Sigma_{a_{1}+a_{2}}\times\Sigma_{a_{3}}\leq\Sigma_{a_{1}+a_{2}+a_{3}}.$

Let $L_{\alpha}$ be a complete set of representatives of the left cosets of
$\Sigma_{a_{1}+a_{2}}\times\Sigma_{a_{3}}$ in $\Sigma_{a_{1}+a_{2}+a_{3}}$. So
by Lemma 5.1, the set

$L_{\beta}:=\\{\tau\sigma\mid(\tau,\sigma)\in L_{\alpha}\times
L_{\lambda^{\prime}}\\}$

is a complete set of representatives of the left cosets of
$\Sigma_{a_{1}}\times\Sigma_{a_{2}}\times\Sigma_{a_{3}}$ in
$\Sigma_{a_{1}+a_{2}+a_{3}}$. From the definitions of $M_{\lambda}$, we
conclude that

$\star_{i=1}^{3}f_{i}=f_{1}\star(f_{2}\star f_{3}).$

The proof to

$\star_{i=1}^{3}f_{i}=(f_{1}\star f_{2})\star f_{3}$

is similar.

∎

######  Lemma 5.3.

For any partition $\lambda=(a_{1},a_{2})$ of a positive integer $d$,

$M_{\lambda}(f_{1}\otimes f_{2})=M_{\lambda^{\prime}}(f_{2}\otimes f_{1}),$

where $\lambda^{\prime}$ is the partition $(a_{2},a_{1})$ of $d$.

###### Proof.

Write $x_{i_{1}}\otimes\cdots\otimes x_{i_{n}}$ as the tensor product

$x_{i_{1}}\otimes\cdots\otimes x_{i_{n}}=b_{1}\otimes b_{2},$

where $b_{i}\in V_{n}^{\otimes a_{i}}$ for $i=1$ or $2$. Let
$\tau=(12)\in\Sigma_{2}$ and let $\sigma_{\tau}$ be the unique element in
$\Sigma_{d}$ such that

$\sigma_{\tau}(x_{i_{1}}\otimes\cdots\otimes x_{i_{n}})=b_{2}\otimes b_{1}.$

A direct calculation gives

$f_{1}\otimes f_{2}=\sigma_{\tau}^{-1}\circ(f_{2}\otimes
f_{1})\circ\sigma_{\tau},$

which implies

$M_{\lambda}(f_{1}\otimes f_{2})=\sum_{\sigma\in
L_{\lambda}}(\sigma\sigma_{\tau}^{-1})\circ(f_{2}\otimes
f_{1})\circ(\sigma\sigma_{\tau}^{-1})^{-1}.$

One easily shows that the right hand side of the equation above is precisely

$M_{\lambda^{\prime}}(f_{2}\otimes f_{1}).$

∎

The same ideas can be used to prove the case when the partition of $d$ is
$\lambda=(a_{1},\ldots,a_{k})$. More precisely, we have the following
proposition:

######  Proposition 5.4.

Let $\lambda=(a_{1},\ldots,a_{k})$ be a partition of a positive integer $d$
and let $\tau\in\Sigma_{k}$. Let $\lambda_{\tau}$ be the partition
$(a_{\tau(1)},\ldots,a_{\tau(k)})$. Then

$M_{\lambda}(f_{1}\otimes\cdots
f_{k})=M_{\lambda_{\tau}}(f_{\tau(1)}\otimes\cdots f_{\tau(k)}).$

The following lemma follows directly from the definition of $M_{\lambda}$.

######  Lemma 5.5.

If $f,f_{1},f_{2}\in\mathrm{Schur}(0,n,K)=K$ and
$g,g_{1},g_{2}\in\mathrm{Schur}(q,n,K)$,

$f\boxtimes(g_{1}+g_{2})=f\boxtimes g_{1}+f\boxtimes g_{2}$

and

$(f_{1}+f_{2})\boxtimes g=f_{1}\boxtimes g+f_{2}\boxtimes g.$

Putting these together, we have the following

######  Theorem 5.6.

The inter-degree multiplication $\boxtimes$ in the Schur algebra
$\mathrm{Schur}(n;K)$ is associative and commutative and satisfies the
distributive laws and thus the Schur algebra $\mathrm{Schur}(n;K)$ is an
associative and anti-commutative graded $K$-algebra.

##  6\. Action of the Schur Algebra on Derivations and Operads

In this section, the Schur-operad structure is constructed. We show that the
Lie algebra of derivations can be generated by quadratic derivations together
with the action of the Schur operad.

Let $\overline{S}(\widetilde{M}_{n},K)$ denote the smallest sub-Lie algebra of
$\mathrm{Der}_{*}L_{K}(X_{n})$ which contains $\widetilde{M}_{n}$ and is
invariant under the map $\Phi_{f}$ for any $f\in\mathrm{Schur}(q,n,K)$ for
some $q\geq 2$, where $\Phi_{f}$ is defined in Definition 4.14. Let

$\overline{S}_{q}(\widetilde{M}_{n},K)=\overline{S}(\widetilde{M}_{n},K)\bigcap\mathrm{Der}_{q}(L_{K}(X_{n})).$

######  Proposition 6.1.

Suppose $n,p\geq 2$. Then
$\overline{S}_{q}(\widetilde{M}_{n},K)=\mathrm{Der}_{q}(L_{K}(X_{n}))$ if and
only if the set

$\mathcal{B}(p,n,K)=\\{f_{i,[x_{j},u]}\in\mathrm{Der}_{q}(L_{K}(X_{n}))\mid
1\leq i,j\leq u\\}\subseteq\overline{S}_{q}({M_{n}},K),$

where $u$ is a simple Lie monomial of degree $2(p-1)$.

###### Proof.

We only prove if $\mathcal{B}(p,n,K)\subseteq\overline{S}_{q}({M_{n}},K)$,
then $\overline{S}_{q}({M_{n}},K)=\mathrm{Der}_{q}(L_{K}(X_{n}))$ as the
converse is trivial. Define

$f_{i,w}(x_{s})=\begin{cases}0,\text{if }s\neq i,\\\ w,\text{if
}s=i.\end{cases}$

It is clear that for any $p,n\geq 2$,

$\\{f_{i,w}\in\mathrm{Der}_{p}\left(L_{K}(X_{n})\right)\mid 1\leq i\leq n,w\in
L_{K}^{p}(X_{n})\\}$

is a generating set of $\mathrm{Der}_{q}(L_{K}(X_{n}))$. Since every Lie
monomial and be written as a linear combination of simple Lie monomials,
$\mathcal{B}(p,n,K)$ is a generating set of $\mathrm{Der}_{q}(L_{K}(X_{n}))$.
Thus $\mathrm{Der}_{q}(L_{K}(X_{n}))\subseteq\overline{S}_{q}({M_{n}},K)$. ∎

A direct computation gives the following proposition:

######  Proposition 6.2.

For any $i,j$ in $\\{1,\ldots,n\\}$ with $i\neq j$ and any Lie monomial $u$ of
degree $2k$,’

$[{\chi_{i,j}},f_{j,u}]=-f_{i,[x_{i},u]}+f_{j,{\chi_{i,j}}(u)},$

where $f_{j,u}$, $f_{i,[x_{i},u]}$ and $f_{j,{\chi_{i,j}}(u)}$ are defined in
the proof of Proposition 6.1.

Let $u$ be a Lie monomial of degree $2k$, then ${\chi_{i,j}}(u)$ is equal to
the sum of Lie monomials where each monomial is obtained from $u$ by replacing
each appearance of $x_{i}$ in $u$ by $[x_{i},x_{j}]$.

######  Example 6.3.

We assume that $i\neq k$.

$\displaystyle{\chi_{i,j}}\left(\left[[x_{i},x_{k}],x_{i}\right]\right)$
$\displaystyle=\left[{\chi_{i,j}}\left([x_{i},x_{k}]\right),x_{i}\right]+\left[[x_{i},x_{k}],{\chi_{i,j}}(x_{i})\right]$
$\displaystyle=\left[\left[{\chi_{i,j}}(x_{i}),x_{k}\right],x_{i}\right]+\left[\left[x_{i},{\chi_{i,j}}(x_{k})\right],x_{i}\right]+\left[[x_{i},x_{k}],[x_{i},x_{k}]\right]$
$\displaystyle=\left[\left[[x_{i},x_{k}],x_{k}\right],x_{i}\right]+\left[\left[x_{i},0\right],x_{i}\right]+\left[[x_{i},x_{k}],[x_{i},x_{k}]\right]$
$\displaystyle=\left[\left[[x_{i},x_{k}],x_{k}\right],x_{i}\right]+\left[[x_{i},x_{k}],[x_{i},x_{k}]\right]$

The proof of the following lemma is found on pages 65-67 of [Jin].

######  Lemma 6.4.

For every Lie monomial $u$ of degree $2k$ ($k\geq 2$), there exists
$h\in\mathrm{Schur}(k+1,n,K)$ such that

$h\left({\chi_{i,j}}(u)\right)=0$

and

$h\left(-[x_{i},u]=[x_{i},u]\right),$

which implies

$\Phi_{h}\left([{\chi_{i,j}},f_{j,u}]\right)=f_{i,[x_{i},u]},$

where $\Phi_{h}$ is defined as in Definition 4.14.

######  Theorem 6.5.

For every $n$,
$\overline{S}({M_{n}},K)=\mathrm{Der}_{*}\left(L_{K}(X_{n})\right)$.

###### Proof.

The case $n=1$ is trivial and we assume $n\geq 2$ and it suffices to prove

$\overline{S}_{p}({M_{n}},K)=\mathrm{Der}_{p}\left(L_{K}(X_{n})\right)$

and we shall prove by induction on $p$. Since ${M_{n}}$ is a generating set of
$\mathrm{Der}_{2}\left(L_{K}(X_{n})\right)$, the result is true for $p=2$.
Assume

$\overline{S}_{p}({M_{n}},K)=\mathrm{Der}_{p}\left(L_{K}(X_{n})\right)$

for $p=2,\ldots,k$.

Given any $1\leq i\leq n$ and any Lie monomial $u$ of degree $2k$. Let $1\leq
j\leq n$ and $j\neq i$. Then by Prop 6.2,

$[{\chi_{i,j}},f_{j,u}]=-f_{i,[x_{i},u]}+f_{j,{\chi_{i,j}}(u)}.$

By Lemma 6.4, there exists $h\in\mathrm{Schur}(k+1,n,K)$ such that

$\Phi_{h}\left([{\chi_{i,j}},f_{j,u}]\right)=f_{i,[x_{i},u]},$

which implies $f_{i,[x_{i},u]}\in\overline{S}_{k+1}({M_{n}},K)$ for any $i$
and any Lie monomial $u$ of degree $2k$.

For a given $\sigma\in\Sigma_{n}$ and $t\geq 1$, let $\zeta_{t,\sigma}$ be the
linear map $V_{n}^{\otimes t}\to V_{n}^{\otimes t}$ defined by

$\zeta_{t,\sigma}(x_{j_{1}}\otimes\cdots\otimes
x_{x_{j_{t}}})=x_{\sigma(j_{1})}\otimes\cdots\otimes x_{\sigma(j_{t})}.$

A direct calculation shows that

$\zeta_{k+1,(i,j)}\left([x_{i},\zeta_{k,(i,j)}(u)]\right)=[x_{j},u].$

Therefore,

$\Phi_{\zeta_{k+1,(i,j)}}\left(f_{i,[x_{i},v]}\right)=f_{i,[x_{j},u]};$

that is, $f_{i,[x_{j},u]}\in\overline{S}_{k+1}({M_{n}},K)$ for any $1\leq
i\neq j\leq n$ and any Lie monomial $u$ of degree $2k$ and by Proposition 6.1
the proof is finished. ∎

Let ${v}_{1},\ldots,{v}_{n}$ be elements in ${M_{n}}$ such that either each
${v}_{i}$ is ${\chi_{i,j}}$ with $i\neq j$ or each ${v}_{i}$ is
${\theta}_{i,[x_{s},x_{t}]}$ for distinct $i,s$ and $t$. Then it is easy to
check that every element in ${M_{n}}$ can be obtained from
${\gamma_{n}}:=\\{{v}_{1},\ldots,{v}_{n}\\}$ via the action of
$\mathrm{Schur}(2,n,K)$. Thus we have the following corollary:

######  Corollary 6.6.

For every $n$,
$\overline{S}({\Gamma_{n}},K)=\mathrm{Der}_{*}\left(L_{K}(X_{n})\right)$.

######  Definition 6.7.

An operad consists of

  1. (1)

a sequence $\left(P(m)\right)_{m\in\mathbb{Z}_{+}}$ of sets whose elements
$\theta$ are called the $m$-ary operation of $P$.

  2. (2)

an element $1\in P(1)$ called the identity.

  3. (3)

for all positive integers $m,k_{1},\ldots,k_{m}$, a composition function

$\displaystyle\circ:P(m)\times P(k_{1})\times\cdots\times P(k_{m})$
$\displaystyle\to P(k_{1}+\cdots k_{m})$
$\displaystyle(\theta,\theta_{1},\ldots,\theta_{m})$
$\displaystyle\mapsto\theta\circ(\theta_{1},\ldots,\theta_{m})$

satisfying the following coherence axioms:

  * •

identity: $\theta\circ(1,\ldots,1)=\theta=1\circ\theta$.

  * •

associativity:

$\displaystyle\theta\circ(\theta_{1}\circ(\theta_{1,1},\ldots,\theta_{1,k_{1}}),\ldots,\theta_{n}\circ(\theta_{n,1},\ldots,\theta_{n,k_{n}}))$
$\displaystyle={}$
$\displaystyle(\theta\circ(\theta_{1},\ldots,\theta_{n}))\circ(\theta_{1,1},\ldots,\theta_{1,k_{1}},\ldots,\theta_{n,1},\ldots,\theta_{n,k_{n}}),$

where the number of arguments correspond to the arities of the operations.

A morphism $f:P\to Q$ of operads is naturally defined a sequence

$\left(f_{m}:P(m)\to Q(m)\right)_{m\in\mathbb{Z}_{+}}$

of maps which is compatible with identity and composition.

The following theorem, whose proof is obvious, constructs the operad of Schur
Algebras.

######  Proposition 6.8.

Let
$\mathit{Schur}=\left(\mathrm{Schur}(m-1,n,K)\right)_{m\in\mathbb{Z}_{+}}$.
For each $m,k_{1},\ldots,k_{m}\in\mathbb{Z}_{+}$, define the operad
composition by

$\displaystyle\circ:P(m)\times P(k_{1})\times P(k_{m})$ $\displaystyle\to
P(k_{1}+\cdots k_{m})$ $\displaystyle(\theta,\theta_{1},\ldots,\theta_{m})$
$\displaystyle\mapsto\theta\boxtimes\theta_{1}\boxtimes\cdots\boxtimes\theta_{m}$

and the identity of $\mathscr{S}chur$ is defined as the identity element of
$\mathrm{Schur}(0,n,K)=K=P(1)$. This defines $\mathit{Schur}$ as an operad.

Proposition 4.15 and Corollary 6.6 imply the following result:

######  Theorem 6.9.

The Schur Operad $\mathscr{S}ch$ acts on $T\left(V_{n}(K)\right)$,
$L_{K}(X_{n})$, $\mathrm{Der}_{*}^{\triangle}\left(L_{K}(X_{n})\right)$ and
$\mathrm{Der}_{*}\left(L_{K}(X_{n})\right)$. Furthermore,
$\mathrm{Der}_{*}\left(L_{K}(X_{n})\right)$ is generated by
$\\{{v}_{1},\ldots,{v}_{n}\\}\subseteq{M_{n}}$ where either each ${v}_{i}$ is
${\chi_{i,j}}$ with $i\neq j$ or each ${v}_{i}$ is
${\theta}_{i,[x_{s},x_{t}]}$ for distinct $i,s$ and $t$.

##  7\. A Short Exact Sequence of Lie Algebras

We introduce a short exact sequence of Lie algebras which is obtained from
certain group filtrations.

Define the group epimorphism $p_{n}$ by $p_{n}:P\Sigma_{n}\to P\Sigma_{n-1}$
by

$p_{n}(\chi_{i,j})=\begin{cases}\chi_{i,j}\text{, if }n\in\\{i,j\\}\\\
1\text{, otherwise.}\end{cases}$

Then there is a short exact sequence of groups [CPVW]

$1\to\mathfrak{K}_{n}\to P\Sigma_{n}\xrightarrow{p_{n}}P\Sigma_{n-1}\to 1.$

The kernel $\mathfrak{K}_{n}$ is generated by the set
$P_{n}:=\\{\chi_{i,j}\mid n\in\\{i,j\\}\\}$ in the group $P\Sigma_{n}$ [CPVW].

Define $q_{n}:\mathfrak{K}_{n}\to\mathbb{Z}^{n-1}$ defined by

$q_{n}(\chi_{i,n})=\chi_{i,n}$

and

$q_{n}(w)=1\text{, if }w\text{ is a product of }\chi_{i,j}\text{'s other than
}\chi_{i,n}\text{'s}.$

Then there is a short exact sequence [CPVW]

$1\to\mathfrak{W}_{n}\to\mathfrak{K}_{n}\xrightarrow{q_{n}}\mathbb{Z}_{n-1}\to
1.$

######  Lemma 7.1.

Under the notation of the short exact sequence above, we have

$\mathrm{gr}_{2}^{J}(\mathbb{Z}^{n-1})=1,$

where $\mathbb{Z}^{n-1}$ is the free abelian group generated by
$\\{\chi_{i,n}\mid 1\leq i<n\\}$.

Recall for $h\in J^{m}IA_{n}$, $\widetilde{J}_{n,m}(h\cdot J^{m+1}IA_{n})$ is
the unique derivation of degree $2(m+1)$ given by the conditions

$\widetilde{J}_{n,m}\left(h\cdot
J^{m+1}IA_{n}\right)(x_{i})=\phi_{n,m+1}^{-1}\left(w_{i,m+1\cdot\Gamma^{m+2}F(X_{n})}\right)$
(1)

for $i=1,\ldots,n$; see page 29 of [Jin] for details. Recall also the morphism
of Lie algebras

$\psi_{n,\ast}:\mathrm{gr}_{\ast}^{\mathrm{LCS}}(IA_{n})\to\mathrm{gr}_{\ast}^{\mathrm{J}}(IA_{n})$

induced by the quotient map

$IA_{n}^{m}/IA_{n}^{m+1}\to J^{m}IA_{n}/J^{m+1}IA_{n}.$

Composing with $\widetilde{J}_{n,\ast}$, we obtain the morphism of Lie
algebras

$\widetilde{J}_{n,\ast}\circ\psi_{n,\ast}:\mathrm{gr}_{\ast}^{\mathrm{LCS}}(IA_{n})\to\mathrm{Der}_{\ast}\left(L_{\mathbb{Z}}(X_{n})\right).$
(2)

###### Proof.

We write every non-trivial element in $\mathbb{Z}^{n-1}$ as

$\chi_{i_{1},n}^{e_{1}}\cdot\chi_{i_{2},n}^{e_{2}}\cdots\chi_{i_{r},n}^{e_{r}}$

for some $r\geq 1$, $1\leq i_{1}<i_{2}<\cdots i_{r}<n$ and
$e_{1},e_{2},\ldots,e_{r}\neq 0$. By the formula (1), one gets

$\chi_{i_{1},n}^{e_{1}}\cdot\chi_{i_{2},n}^{e_{2}}\cdots\chi_{i_{r},n}^{e_{r}}(x_{i_{1}})=[x_{i_{1}},x_{j_{1}}]^{e_{1}}\cdot
u,$

where $u\in\Gamma^{3}F(X_{n})$. Thus

$\mathbb{Z}^{n-1}\cap J^{2}IA_{n}=1$

and the proof is completed. ∎

Now we prove a general version of the Falk-Randell Theorem [Falk].

######  Theorem 7.2.

Let

$1\to A\to B\to C\to 1$

be a split short exact sequence of groups thus allowing $A$ and $C$ be
identified as subgroups of $B$. let $\\{B_{m}\\}_{m\geq 1}$ be a filtration of
$B$. Let $A_{m}=B_{m}\cap A$ and $C_{m}=B_{m}\cap C$ and therefore,
$\\{A_{m}\\}$ and $\\{C_{m}\\}$ are filtrations of $A$ and $C$, respectively.
If $B_{m}=A_{m}C_{m}=\\{ac\mid a\in A_{m}\text{ and }c\in C_{m}\\}$ for each
$m$, then there is a short exact sequence of Lie algebras

$0\to\mathrm{gr}(A)\to\mathrm{gr}(B)\to\mathrm{gr}(C)\to 0$

which splits as a sequence of abelian groups.

###### Proof.

For each $m$, the split short exact sequence

$1\to A_{m}\to B_{m}\to C_{m}\to 1$

gives the following commutative diagram, the rows of which splits.

${1}$${1}$${1}$${1}$${A_{m+1}}$${B_{m+1}}$${C_{m+1}}$${1}$${1}$${A_{m}}$${B_{m}}$${C_{m}}$${1}$${0}$${\mathrm{gr}_{m}(A)}$${\mathrm{gr}_{m}(B)}$${\mathrm{gr}_{m}(C)}$${0}$${1}$${1}$${1}$

∎

######  Theorem 7.3.

Let

$1\to A\to B\to C\to 1$

be a split short exact sequence of groups and let $\\{B_{m}\\}_{m\geq 1}$ be a
filtration of $B$. We identify $A$ and $C$ as subgroups of $B$ according to
the corresponding embedding and cross section. Let $A_{m}=B_{m}\cap A$ and
$C_{m}=B_{m}\cap C$ and therefore, $\\{A_{m}\\}$ and $\\{C_{m}\\}$ are
filtrations of $A$ and $C$, respectively. Then there is a short exact sequence
of Johnson Lie algebras

$0\to\mathrm{gr}_{\ast}^{J}(\mathfrak{W}_{n})\to\mathrm{gr}_{\ast}^{J}(\mathfrak{K}_{n})\to\mathbb{Z}^{n-1}\to
0$

which splits as a sequence as abelian groups and therefore,

$\mathrm{gr}_{\ast}^{J}(\mathfrak{K}_{n})\cong\mathbb{Z}^{n-1}\oplus\mathrm{gr}_{\ast}^{J}(\mathfrak{W}_{n}).$

###### Proof.

Since the group $\mathfrak{K}_{n}$ is generated by the $w^{-1}\chi_{i,j}w$’s,
where $\chi_{i,j}\in P_{n}$, $w$ is a (possibly empty) product of
$\chi_{i^{\prime},j^{\prime}}^{\prime}s$. Thus every element
$a\in\mathfrak{K}_{n}$ can be written as

$a=\chi_{i_{1},j_{1}}^{e_{1}}\chi_{i_{2},j_{2}}^{e^{2}}\cdots\chi_{i_{r},j_{r}}^{e_{r}}\cdot
u$

for $e_{k}\in\mathbb{Z}$, $k=1,\ldots,r$ and some
$u\in\Gamma^{2}\mathfrak{K}_{n}$. Then

$\displaystyle q_{n}(a)$ $\displaystyle=\sum_{k=1}^{r}e_{k}\cdot
q_{n}(\chi_{i_{k},j_{k}})$ $\displaystyle=\sum_{1\leq k\leq
r,j_{k}=n}e_{k}\cdot\chi_{i_{k},j_{k}}.$

Suppose $a\in\mathfrak{W}_{n}$ and
$c=\displaystyle\sum_{l=1}^{n-1}d_{i}\chi_{l,n}$ is a non-trivial element in
$\mathbb{Z}^{n-1}$. Then

$ac=\chi_{i_{1},j_{1}}^{e_{1}}\chi_{i_{2},j_{2}}^{e^{2}}\cdots\chi_{i_{r},j_{r}}^{e_{r}}\cdot
u\cdot\chi_{1,n}^{d_{1}}\chi_{2,n}^{d_{2}}\cdots\chi_{n-1,n}^{d_{n-1}}.$

Since $u\in\Gamma^{2}\mathfrak{K}_{n}\subset J^{2}\mathfrak{K}_{n}$,

$ac\dot{J}^{2}\mathfrak{K}_{n}=\chi_{i_{1},j_{1}}^{e_{1}}\chi_{i_{2},j_{2}}^{e^{2}}\cdots\chi_{i_{r},j_{r}}^{e_{r}}\chi_{1,n}^{d_{1}}\chi_{2,n}^{d_{2}}\cdots\chi_{n-1,n}^{d_{n-1}}J^{2}\mathfrak{K}_{n}.$

A straightforward computation gives

$\widetilde{J}_{1,\mathfrak{K}_{n}}(ac\cdot J^{2}\mathfrak{K}_{n})=\sum_{1\leq
k\leq r,j_{k}\neq
n}e_{k}\cdot\widetilde{\chi}_{i_{k},j_{k}}+\sum_{l=1}^{n-1}d_{l}\cdot\widetilde{\chi}_{l,n}\neq
0$

since not all $d_{l}$’s are zeros. That is, for every $a\in\mathfrak{W}_{n}$
and $c\neq 1$, $ac\notin J^{2}\mathfrak{K}_{n}$. By Lemma 7.1 and Theorem 7.3,
there is a short exact sequence of Johnson Lie algebras

$0\to\mathrm{gr}_{\ast}^{J}(\mathfrak{W}_{n})\to\mathrm{gr}_{\ast}^{J}(\mathfrak{K}_{n})\to\mathbb{Z}^{n-1}\to
0$

which splits as a sequence as abelian groups and therefore,

$\mathrm{gr}_{\ast}^{J}(\mathfrak{K}_{n})\cong\mathbb{Z}^{n-1}\oplus\mathrm{gr}_{\ast}^{J}(\mathfrak{W}_{n}).$

## Acknowledgement

Mohamed Elhamdadi was partially supported by Simons Foundation collaboration
grant 712462.

### In Memoriam

Frederick Cohen passed away on january 16, 2022 at the age of 76 and we all
initiated the present work and Fred fully contributed to the whole paper.

∎

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