# Automorphisms of free metabelian Lie algebras

###### Abstract.

We show that every automorphism of a free metabelian Lie algebra $M_{n}$ of
rank $n\geq 4$ over an arbitrary field $K$ is almost tame, that is, it is a
product of so-called Chein automorphisms (or one-row transformations).
Moreover, we show that the group of all automorphisms $\mathrm{Aut}(M_{n})$ of
$M_{n}$ of rank $n\geq 4$ over a field $K$ of characteristic $\neq 3$ is
generated by all linear automorphisms, as well as one quadratic and one cubic
automorphism. The same result holds for fields of any characteristic if $n\geq
5$.

We also show that all Chein automorphisms of lower degree $\geq 4$ and all
exponential automorphisms of lower degree $\geq 5$ are tame, which contradicts
the results of [5, 40].

Ualbai Umirbaev111Department of Mathematics, Wayne State University, Detroit,
MI 48202, USA; Department of Mathematics, Al-Farabi Kazakh National
University, Almaty, 050040, Kazakhstan; and Institute of Mathematics and
Mathematical Modeling, Almaty, 050010, Kazakhstan, e-mail<EMAIL_ADDRESS>

Mathematics Subject Classification (2020): 17B40, 17B01, 17B30.

Key words: Automorphism, derivation, Lie algebra, free metabelian Lie algebra.

###### Contents

  1. 1 Introduction
  2. 2 Automorphisms of $M_{n}$
  3. 3 Jacobian matrices
  4. 4 Matrix notations
  5. 5 Tame automorphisms
  6. 6 Almost tame automorphisms
  7. 7 Special matrices
  8. 8 Inductive lemma
  9. 9 Exceptional automorphism
  10. 10 Elimination of a variable
  11. 11 Main results

##  1\. Introduction

The well known Jung–van der Kulk Theorem [18, 22] states that every
automorphism of the polynomial algebra $K[x,y]$ in two variables $x,y$ over an
arbitrary field $K$ is tame. An analogue of this result for free associative
algebras in two variables was proven by Makar-Limanov and Czerniakiewicz [14,
26]. The same result is true for free Poisson algebras in two variables over a
field of characteristic zero [28] (see also [27, 68, 29]). Moreover, the
automorphism groups of polynomial algebras, free associative algebras, and
free Poisson algebras in two variables are isomorphic. The tameness of
automorphisms of free right-symmetric algebras in two variables over any field
was proven in [21].

For any free algebra $A$ in the free variables $x_{1},\ldots,x_{n}$ denote by
$\phi=(f_{1},\ldots,f_{n})$ the endomorphism of $A$ determined by
$\phi(x_{i})=f_{i}$ for all $1\leq i\leq n$.

The automorphism groups of commutative and associative algebras generated by
three elements are much more complicated. The well-known Nagata automorphism
[33]

$\displaystyle(x+2y(zx-y^{2})+z(zx-y^{2})^{2},y+z(zx-y^{2}),z)$

of the polynomial algebra $K[x,y,z]$ over a field $K$ of characteristic $0$ is
proven to be wild [51, 64]. The well-known Anick automorphism (see [13, p.
398])

$\displaystyle(x+z(xz-zy),\,y+(xz-zy)z,\,z)$

of the free associative algebra $K\langle x,y,z\rangle$ over a field $K$ of
characteristic $0$ is also proven to be wild [65, 66]. The Nagata automorphism
gives an example of a wild automorphism of free Poisson algebras in three
variables. In 2006 I.P. Shestakov (unpublished) constructed an analogue of the
Anick automorphism

$\displaystyle(x+\\{xz-\\{y,z\\},z\\},y+z(xz-\\{y,z\\}),z)$

for free Poisson algebras in three variables.

It is well known [58] that the Nagata and Anick automorphisms are stably tame,
that is, they become tame after adding one more variable.

In 1981 Shafarevich [49] and in 1983 Anick [1] independently proved that every
automorphism of the polynomial algebra $K[x_{1},x_{2},\ldots,x_{n}]$ over a
field $K$ of characteristic zero can be approximated by tame automorphisms
with respect to two different but similar topologies. This can be considered
as a weaker generalization of the Jung-van der Kulk Theorem for polynomial
algebras of rank $n\geq 3$. In particular, the Nagata automorphism is wild but
can be approximated by tame automorphisms.

A classical theorem of combinatorial group theory asserts that every subgroup
of a free group is free. Nielsen established this remarkable result in 1921
for finitely generated subgroups [38]; in 1927, it was extended to all
subgroups by Schreier [55]. A variety of universal algebras is called Nielsen-
Schreier, if any subalgebra of a free algebra of this variety is free. In 1953
Shirshiv and in 1956 Witt proved that [52, 70] the varieties of Lie algebras
and Lie $p$-algebras over a field are Nielsen-Schreier. The varieties of all
non-associative algebras [23], commutative and anti-commutative algebras [53],
Lie superalgebras [30, 57], and Lie $p$-superalgebras [31] over a field are
also Nielsen-Schreier. It was recently shown [15] that the varieties of pre-
Lie (also known as right-symmetric) algebras and Lie-admissible algebras over
a field of characteristic zero are Nielsen-Schreier. Some other examples of
Nielsen-Schreier varieties can be found in [10, 32, 50, 61, 63].

In 1924, Nielsen also proved [39] that the automorphism group of a finitely
generated free group is generated by Nielsen transformations, and represented
this group as a finitely defined group. In 1964 P. Cohn proved [11] that all
automorphisms of finitely generated free Lie algebras over a field are tame.
Later this result was extended to free algebras of Nielsen-Schreier varieties
[24]. Defining relations between the elementary automorphisms of free algebras
of Nielsen-Schreier varieties are given in [67].

One of the classical and well-studied varieties of groups is the variety of
metabelian groups. In 1965 Bachmuth [2] proved that all automorphisms of the
free metabelian group of rank $2$ are tame. The first example of a non-tame
automorphism of the free metabelian group of rank $3$ was constructed by Chein
[9]. Bachmuth-Mochizuki proved [3] that the group of automorphisms of the free
metabelian group of rank $3$ is not even finitely generated. Strengthening
these results, Roman’kov proved [45] that the free metabelian group of rank
$3$ contains non-tame primitive elements and gave a method of constructing of
such elements. The well known Bachmuth-Mochizuki-Roman’kov Theorem [4, 44]
states that every automorphism of the free metabelian group of rank $n\geq 4$
is tame. More information about automorphisms of free metabelian groups can be
found in [48].

The variety of metabelian Lie algebras is also a classical and well-studied
variety of Lie algebras. In this paper we denote by $M_{n}$ a free metabelian
Lie algebra of rank $n$ over a field $K$. It is well known that any nontrivial
exponential automorphism of $M_{2}$ is wild (see, for example [54, Proposition
4],[42, Theorem 3]). In 1992, Drensky [16] proved that the exponential
automorphism $\mathrm{exp}(\mathrm{ad}[x_{1},x_{2}])$ of $M_{3}$ is wild. In
2008, Roman’kov [47] proved that the group of automorphisms of $M_{3}$ cannot
be finitely generated modulo the subgroup generated by all exponential and
tame automorphisms. Kabanov and Roman’kov [19] constructed strictly nontame
primitive elements of $M_{3}$.

An analogue of the Anick-Shafarevich Theorem [1, 49] for free metabelian Lie
algebras was proven in 1993 by Bryant and Drensky [8]. They proved that the
group of tame automorphisms is dense in the group of all automorphisms of
$M_{n}$ if $n\geq 4$. A sharpened version of this result can be found in [20].

In 2015, Nauryzbaev [37] proved that the group of tame automorphisms of
$M_{n}$ is generated by all linear automorphisms and the quadratic
automorphism

(1) $\displaystyle(x_{1}+[x_{2},x_{3}],x_{2},\ldots,x_{n})$

if either $n\geq 4$ and $K$ is a field of characteristic $\neq 3$, or $n\geq
5$ and $K$ is a field of arbitrary characteristic.

In 1992 V. Drensky [16], in 1993 Papistas [41], and in 1995 Bahturin and
Shpilrain [6] proved that free algebras of rank $n\geq 2$ of any polynilpotent
variety of Lie algebras that is not abelian or metabelian have wild
automorphisms. This result also makes the case of free metabelian Lie algebras
more important.

In 1992 Yu. Bahturin and S. Nabiyev [5] claimed that every nontrivial
exponential automorphism of the free metabelian Lie algebra $M_{n}$ of any
rank $n\geq 2$ over a field of characteristic zero is wild. More examples of
wild automorphisms of $M_{n}$ of rank $n\geq 4$ were given in 2008 by Z.
Özcurt and N. Ekici [40]. For a long time, the area specialists believed these
results were true. Unfortunately, both articles have recently been shown to
contain fatal errors [69].

Inner automorphisms of groups play a crucial role in the proofs of Bachmuth-
Mochizuki-Roman’kov Theorem [4, 44]. The group automorphism

$\displaystyle(x_{2}^{-1}x_{1}x_{2},x_{2},\ldots,x_{n})$

is extreemly important in their proofs and there is no analogue of this
automorphism in the case of free metabelian Lie algebras. This makes the case
of algebras more complicated.

One property of the free metabelian Lie algebras is that they admit non-
elementary automorphisms that change only one variable and fix the rest. All
automorphisms of $M_{n}$ of this type (including elementary ones) are called
Chein automorphisms or one-row transformations [17, 46](see definitions in
Section 2). The first example of the Chein automorphism is

(2) $\displaystyle(x_{1}+[[x_{2},x_{3}],x_{1}],x_{2},\ldots,x_{n})$

and no one could answer whether it is tame or not. For this reason, the
specialists in the area easily accepted the wrong results [5, 40]. This
automorphism is wild if $n=3$ [34], and is the first candidate for wildness
for all $n\geq 4$.

Chein automorphisms were accepted as almost elementary automorphisms. An
automorphism of $M_{n}$ is called almost tame if it is a product of Chein
automorphisms [34, 36].

In this paper we show that every automorphism of $M_{n}$ of rank $n\geq 4$ is
almost tame. Moreover, $K$ is a field of characteristic $\neq 3$ then the
group of all automorphisms $\mathrm{Aut}(M_{n})$ of $M_{n}$ of rank $n\geq 4$
is generated by all linear automorphisms, as well as a quadratic automorphism
(1) and a cubic automorphism (2). If $n\geq 5$ then the same result holds for
fields of any characteristic.

One of the most important points of the proof is that we notice that all Chein
automorphisms of lower degree $\geq 4$ and all exponential automorphisms of
lower degree $\geq 5$ are tame. These results contradict the results of [5,
40].

We follow the proof scheme given in [4], and in several places, we repeat the
proofs for completeness. The real difficulties arise in those places where
inner automorphisms were used in [4]. But in algebra, we can apply
linearization [71]. Thus the question about the existence of wild
automorphisms for a free metabelian Lie algebra $M_{n}$ of rank $n\geq 4$ is
still open as well as in the case polynomial and free associative algebras.
But in this case it is reduced to a concrete automorphism.

######  Problem 1.

Is the Chein automorphism (2) of the free metabelian algebra Lie algebra
$M_{n}$ of rank $n\geq 4$ tame?

The paper is organized as follows. In Section 2 we define some important
automorphisms and some subgroups of the automorphism group
$\mathrm{Aut}(M_{n})$. In Section 3, using a representation of $M_{n}$ via
Magnus embedding, we give the definition of the Fox derivatives and the
Jacobian matrix of an endomorphism, and determine some of their properties.
Using some representations of automorphisms from this section, in the
following sections we make calculations in the language of the Jacobian
matrices instead of automorphisms. For this, in Section 4 we clearly determine
some matrix notations. Section 5 is devoted to the study of Chein and
exponential automorphisms. Section 6 is devoted to the study of some special
class of matrices that includes the Jacobian matrices of an important class of
automorphisms. In Section 7 we prove a lemma on eliminating a variable in some
cases. Section 8 is devoted to the study of one exceptional automorphism

(3)
$\displaystyle(x_{1}+[[x_{1},x_{2}],x_{1}],x_{2}+[[x_{1},x_{2}],x_{2}],x_{3},\ldots,x_{n})$

of $M_{n}$. In Section 9 we show how to eliminate a variable in the general
case. In Section 10 the main result and some its corollaries are formulated.

##  2\. Automorphisms of $M_{n}$

Let $L_{n}$ be a free Lie algebra over a field $K$ freely generated by
$z_{1},z_{2},\ldots,z_{n}$. Then $Y_{n}=L_{n}/[L_{n},L_{n}]$ is an abelian Lie
algebra with a linear basis $y_{1},y_{2},\ldots,y_{n}$ and
$M_{n}=L_{n}/[[L_{n},L_{n}],[L_{n},L_{n}]]$ is a free metabelian Lie algebra
with free generators $x_{1},\ldots,x_{n}$, where $y_{i}=z_{i}+[L_{n},L_{n}]$
and $x_{i}=z_{i}+[[L_{n},L_{n}],[L_{n},L_{n}]]$ for all $1\leq i\leq n$.

Thus $M_{n}^{\prime}=[M_{n},M_{n}]$ is abelian and the set of all right normed
(nonassociative) words of the form

(4) $\displaystyle[\ldots[[x_{i},x_{i_{1}}],x_{i_{2}}],\ldots,x_{i_{k}}],\
i>i_{1}\leq i_{2}\leq\ldots\leq i_{k},$

is a linear basis of $M_{n}^{\prime}$ (see, for example [7, p. 139]).

Denote by $\phi=(f_{1},\ldots,f_{n})$ the endomorphism of $M_{n}$ such that
$\phi(x_{i})=f_{i}$ for all $i$. In particular,
$\mathrm{id}=(x_{1},\ldots,x_{n})$ is the identity automorphism. An
automorphism of $M_{n}$ of the form

$\displaystyle(x_{1},\ldots,x_{i-1},\alpha x_{i}+f,x_{i+1},\ldots,x_{n}),$

where $0\neq\alpha\in K$ and $f$ does not depend on $x_{i}$, is called
elementary. The subgroup $\mathrm{TAut}(M_{n})$ of $\mathrm{Aut}(M_{n})$
generated by all elementary automorphisms is called the tame automorphism
group, and the elements of this subgroup are called tame automorphisms of
$M_{n}$. Non-tame automorphisms of $M_{n}$ are called wild.

Free metabelian Lie algebras admit automorphisms of the form

$\displaystyle(x_{1},\ldots,x_{i-1},f_{i},x_{i+1},\ldots,x_{n}),\ 1\leq i\leq
n,$

which include all elementary automorphisms but not necessarily elementary
ones. They are called Chein automorphisms or one-row transformations [17].
These automorphisms move only one variable and fix the others. The subgroup
$\mathrm{ATAut}(M_{n})$ of $\mathrm{Aut}(M_{n})$ generated by all Chein
automorphisms will be called the almost tame automorphism group, and the
elements of this subgroup will be called called almost tame automorphisms of
$M_{n}$ [34]. Thus

$\displaystyle\mathrm{TAut}(M_{n})\subseteq\mathrm{ATAut}(M_{n})\subseteq\mathrm{Aut}(M_{n}).$

If $n=3$ then every non-elementary Chein automorphism is wild [34, 47] and the
automorphism (3) is not almost tame [34]. The case $=3$ is well studied and
defining relations for elementary automorphisms [35] and for Chein
automorphisms [36] are given.

As usual, the group of all linear automorphisms of $M_{n}$ will be identified
with the group of all invertible matrices $GL_{n}(K)$. Recall that the
symmetric group $S_{n}$ is embedded into $GL_{n}(K)$ and any permutation
$\pi\in S_{n}$ represents the automorphism
$\pi=(x_{\pi(1)},\ldots,x_{\pi(n)})$. For example, the transposition $(23)$
represents the automorphism $(23)=(x_{1},x_{3},x_{2},x_{4},\ldots,x_{n})$.

Obviously, the group of almost tame automorphisms $\mathrm{ATAut}(M_{n})$ is
generated by all linear automorphisms $GL_{n}(K)$ and all Chein automorphisms
of the form

(5) $\displaystyle(x_{1}+f,x_{2},\ldots,x_{n}),\ f\in M_{n}^{2}.$

Chein automorphisms of this form are described in Section 3.

The group of tame automorphisms $\mathrm{TAut}(M_{n})$ is generated by all
linear automorphisms $GL_{n}(K)$ and all automorphisms of the form (5) where
$f\in M_{n}^{2}$ does not depend on $x_{1}$. Moreover, if $n\geq 4$ and $K$ is
a field of characteristic $\neq 3$, then $\mathrm{TAut}(M_{n})$ is generated
[37] by all linear automorphisms $GL_{n}(K)$ and the quadratic elementary
automorphism (1). If $n\geq 5$ then the same result holds over fields of any
characteristic.

Let $0\neq z\in[M_{n},M_{n}]$. Then $\mathrm{ad}\,z:M_{n}\to
M_{n}(m\mapsto[z,m])$ is a locally nilpotent derivation and
$(\mathrm{ad}\,z)^{2}=0$. The exponential automorphism $\exp(\mathrm{ad}\,z)$
of $M_{n}$ can be written as

$\displaystyle\exp(\mathrm{ad}\,z)=\mathrm{id}+\mathrm{ad}\,z=(y_{1}+[z,y_{1}],\ldots,y_{n}+[z,y_{n}]).$

It is known that the exponential automorphism
$\exp(\mathrm{ad}\,[x_{1},x_{2}])$ is wild if $n=3$ [16, 34, 47].

Let

$\displaystyle M_{n}=T_{1}\oplus T_{2}\oplus\ldots\oplus T_{k}\oplus\ldots$

be the standard degree grading of $M_{n}$, i.e., $T_{k}$ is the linear span of
(non-associative) monomials in $x_{1},x_{2},\ldots,x_{n}$ of degree $k$. If

$\displaystyle f=f_{s}+f_{s+1}+\ldots+f_{t}\in M_{n},f_{j}\in T_{j},f_{s}\neq
0,f_{t}\neq 0,$

then $\deg(f)=t$ is the degree and $\mathrm{ldeg}(f)=s$ is the lower degree of
$f$ in the variables $x_{1},x_{2},\ldots,x_{n}$. Set also $\deg(0)=-\infty$
and $\mathrm{ldeg}(0)=\infty$.

Let

$\displaystyle\phi=(x_{1}+f_{1},x_{2}+f_{2},\ldots,x_{n}+f_{n}),$

where $f_{i}\in M_{n}^{2}$ for all $1\leq i\leq n$. Then we define
$\deg(\phi)$ as the maximum of all $\deg(f_{i}),1\leq i\leq n$, and
$\mathrm{ldeg}(\phi)$ as the minimum of all $\mathrm{ldeg}(f_{i}),1\leq i\leq
n$. If $\deg(\phi)=\mathrm{ldeg}(\phi)=k\geq 1$ then we say that $\phi$ is
homogeneous of degree $k$. In particular, if $k=2$ then $\phi$ is called
quadratic and if $k=3$ then it is called cubic.

The group of automorphisms $\mathrm{IAut}(M_{n})$ of $M_{n}$ that induces the
identity automorphism on $M_{n}/[M_{n},M_{n}]$ plays an important role in our
study. Obviously, $\mathrm{IAut}(M_{n})$ is the set of all automorphisms of
$M_{n}$ of lower degree $\geq 2$. The elements of $\mathrm{IAut}(M_{n})$ are
called $\mathrm{IA}$-automorphisms.

Here we describe a set of generators of the group of tame
$\mathrm{IA}$-automorphisms and almost tame $\mathrm{IA}$-automorphisms.

######  Lemma 1.

$(a)$ The group of tame $\mathrm{IA}$-automorphisms
$\mathrm{TAut}(M_{n})\cap\mathrm{IAut}(M_{n})$ is generated by all linear
conjugates of elementary automorphisms of the form (5) where
$f\in[M_{n},M_{n}]$ does not depend on $x_{1}$.

$(b)$ The group almost tame $\mathrm{IA}$-automorphisms
$\mathrm{TAut}(M_{n})\cap\mathrm{IAut}(M_{n})$ is generated by all linear
conjugates of Chein automorphisms of the form (5).

Proof. Every tame automorphism $\phi$ can be represented as

$\displaystyle\phi=\alpha_{1}\phi_{1}\alpha_{2}\phi_{2}\ldots\phi_{k}\alpha_{k+1},$

where $\phi_{i}$ is an elementary automorphism of the form (5) and
$\alpha_{i}$ is linear for all $i$. Since $\phi\in\mathrm{IAut}(M_{n})$ for
all $i$ it follows that

$\displaystyle\alpha_{1}\alpha_{2}\ldots a_{k+1}=\mathrm{id}.$

since $\phi_{i}\in\mathrm{IAut}(M_{n})$ for all $i$. Then
$\alpha_{k+1}=(\alpha_{1}\ldots\alpha_{k})^{-1}$ and, consequently,

$\displaystyle\phi=(\alpha_{1}\phi_{1}\alpha_{1}^{-1})(\alpha_{1}\alpha_{2}\phi_{2}(\alpha_{1}\alpha_{2})^{-1})\ldots(\alpha_{1}\ldots\alpha_{k}\phi_{k}(\alpha_{1}\ldots\alpha_{k})^{-1})$

The statement $(b)$ of the lemma can be proved similarly. $\Box$

##  3\. Jacobian matrices

In this section, we define and state some necessary information about Fox
derivatives and Jacobian matrices from [43, 56, 60, 62]. In fact, we define
Fox derivatives more directly for free metabelian Lie algebras.

First we give another representation of the free metabelian algebra $M_{n}$
over $K$ with free generators $x_{1},\ldots,x_{n}$. Let $Y_{n}$ be the abelian
Lie algebra with a linear basis $y_{1},y_{2},\ldots,y_{n}$. The universal
enveloping algebra $U(Y_{n})$ of $Y_{n}$ is the polynomial algebra
$U=U(Y_{n})=K[y_{1},\ldots,y_{n}]$ in the variables
$y_{1},y_{2},\ldots,y_{n}$. Let

$\displaystyle U=U_{0}\oplus U_{1}\oplus\ldots\oplus U_{k}\oplus\ldots$

be the degree grading of $U$, i.e., $U_{k}$ is the linear span of all
homogeneous of degree $k$ monomials in $y_{1},y_{2},\ldots,y_{n}$. If

$\displaystyle a=a_{s}+a_{s+1}+\ldots+a_{t}\in U,a_{j}\in U_{j},a_{s}\neq
0,a_{t}\neq 0,$

then $\deg(a)=t$ is the degree and $\mathrm{ldeg}(a)=s$ is the lower degree of
$a$ in the variables $y_{1},y_{2},\ldots,y_{n}$. Set $\deg(0)=-\infty$ and
$\mathrm{ldeg}(0)=\infty$ as usual.

Let $T_{n}$ be a free right $U$-module with basis $t_{1},\ldots,t_{n}$. Turn
the direct sum

$\displaystyle M=Y_{n}\oplus T_{n}$

into a Lie algebra by

$\displaystyle[a+t,b+s]=tb-sa,$

where $a,b\in Y_{n}$ and $t,s\in T_{n}$. Then the subalgebra of $M$ generated
by

$\displaystyle x_{i}=y_{i}+t_{i},\ \ 1\leq i\leq n,$

is the free metabelian Lie algebra $M_{n}$ with the free generators
$x_{1},\ldots,x_{n}$. In the case of groups the corresponding embedding is
called the Magnus embedding [25] and in the case of Lie algebras it follows
from Shmel’kin’s Theorem on wreath products [54].

Notice that $M_{n}^{\prime}=[M_{n},M_{n}]\subseteq T_{n}$ and for any $m\in
M_{n}^{\prime}$ and $y_{i}$ we get

$\displaystyle my_{i}=[m,x_{i}].$

Every element $m\in M_{n}^{\prime}$ can be written in the form

$\displaystyle m=\sum_{1\leq i<j\leq n}[x_{i},x_{j}]a_{ij},\ \ a_{ij}\in U.$

Let $f\in M_{n}$ be an arbitrary element $M_{n}$. Then $f$ is uniquely
represented as

$\displaystyle f=y+t,y\in Y_{n},t\in T_{n}.$

Moreover, $t$ is uniquely represented as

$\displaystyle t=t_{1}d_{1}+\ldots+t_{n}d_{n},\ \ \ \ d_{1},\ldots,d_{n}\in
K[y_{1},\ldots,y_{n}]=U.$

Set $d_{i}=\frac{\partial f}{\partial x_{i}}$ for all $1\leq i\leq n$ and

$\displaystyle\partial(f)=(d_{1},\ldots,d_{n})^{t}=\big{(}\frac{\partial
f}{\partial x_{1}},\ldots,\frac{\partial f}{\partial x_{n}}\big{)}^{t},$

where $t$ means the transpose. We call $\frac{\partial f}{\partial
x_{1}},\ldots,\frac{\partial f}{\partial x_{n}}$ the Fox derivatives of $f$
(see [56, 60]).

Let $e_{1},e_{2},\ldots,e_{n}$ be the columns of the identity matrix $I=I_{n}$
of order $n$.

The statement of the next lemma is well known (see [60, Lemma 5],[62]).

######  Lemma 2.

Let $a=(a_{1},a_{2},\ldots,a_{n})^{t}\in(U^{n})^{t}$ be an arbitrary column of
dimension $n$ over $U$. Then the following conditions are equivalent.

$(1)$ $a=\partial(f)$ for some $f\in[M_{n},M_{n}]$;

$(2)$ $Ya=y_{1}a_{1}+\ldots+y_{n}a_{n}=0$ where $Y=(y_{1},\ldots,y_{n})$;

$(3)$ $a=\sum_{i<j}(e_{i}y_{j}-e_{j}y_{i})a_{ij},\ a_{ij}\in U$.

Proof. If $i<j$ and $a\in U$, then

$\displaystyle\partial([x_{i},x_{j}]a)=e_{i}y_{j}a-e_{j}y_{i}a=(e_{i}y_{j}-e_{j}y_{i})a.$

Consequently, $(1)$ implies $(3)$. Obviously, $(3)$ implies $(2)$ since
$Y(e_{i}y_{j}-e_{j}y_{i})=0$ for all $i<j$.

Suppose that $(2)$ holds. Then $a_{n}=y_{1}b_{1}+\ldots+y_{n-1}b_{n-1}$ for
some $b_{j}\in U$. Set $g=[x_{n},x_{1}]b_{1}+\ldots+[x_{n},x_{n-1}]b_{n-1}$.
Then $\frac{\partial g}{\partial x_{n}}=a_{n}$ and the last component of
$a^{\prime}=a-\partial(g)$ is zero. Moreover, if
$a^{\prime}=(a_{1}^{\prime},\ldots,a_{n-1}^{\prime},0)$ then
$y_{1}a_{1}^{\prime}+\ldots+y_{n-1}a_{n-1}^{\prime}=0$. Leading an induction
on the number of nonzero components of $a$, we may assume that there exists
$h\in[M_{n},M_{n}]$ such that $a^{\prime}=\partial(h)$. Then
$a=\partial(g+h)$. $\Box$

Every endomorphism $\phi$ of $M_{n}$ induces an endomorphism of
$Y_{n}=M_{n}/[M_{n},M_{n}]$, which can be uniquely extended to an endomorphism
of $U=U(Y_{n})$. This endomorphism will be denoted again by $\phi$ for
simplicity of notation. If $f\in U$, then the image $\phi(f)$ will be denoted
by $f^{\phi}$.

We define the Jacobian matrix of $\phi=(f_{1},\ldots,f_{n})$ by

$\displaystyle J(\phi)=\left[\frac{\partial f_{j}}{\partial
x_{i}}\right]_{1\leq i,j\leq n}=\left[\partial(f_{1})\ \ldots\
\partial(f_{n})\right].$

If $\phi=(f_{1},\ldots,f_{n})$ and $\psi=(g_{1},\ldots,g_{n})$ then

$\displaystyle\phi\circ\psi(x_{i})=g_{i}(f_{1},\ldots,f_{n})$

for all $1\leq i\leq n$. The corresponding Chain Rule gives [56, 60, 62] that

(6) $\displaystyle J(\phi\circ\psi)=J(\phi)J(\psi)^{\phi}.$

The statement of the next lemma is also well known (see [56],[60, Theorem 2]).

######  Lemma 3.

An endomorphism $\phi$ of $M_{n}$ is an automorphism if and only if the
Jacobian matrix $J(\phi)$ is invertible.

Here is a description of Chein automorphisms.

######  Corollary 1.

An endomorphism of $M_{n}$ of the form (5) is an automorphism if and only if
$f$ belongs to the ideal generated by all $[x_{s},x_{t}]$, where $s,t\neq 1$.

Proof. Let $\phi=(x_{1}+f,x_{2},\ldots,x_{n})$ where $f\in M_{n}^{2}$. The
Jacobian matrix of $\phi$ is $J(\phi)=I+[\partial(f)\ 0\ldots 0]$. This matrix
is invertible if and only if $1+\frac{\partial f}{\partial x_{1}}$ is
invertible in $U$ or, equivalently, $\frac{\partial f}{\partial x_{1}}=0$. We
show that $\frac{\partial f}{\partial x_{1}}=0$ if and only if $f$ belongs to
the ideal generated by all $[x_{s},x_{t}]$, where $s,t\neq 1$.

The basis (4) of $M_{n}^{2}$ is convenient to work with $\frac{\partial
f}{\partial x_{n}}=0$. Any $f\in M_{n}^{2}$ can be written in the form

$\displaystyle
f=f^{\prime}+[x_{n},x_{1}]a_{1}+[x_{n},x_{2}]a_{2}+\ldots+[x_{n},x_{n-1}]a_{n-1},$

where $f^{\prime}$ belongs to the ideal generated by all $[x_{s},x_{t}]$ with
$s,t\neq n$ and $a_{i}\in K[y_{i},\ldots,y_{n}]$ for all $i$. Since
$\frac{\partial f^{\prime}}{\partial x_{n}}=0$ it follows that

$\displaystyle\frac{\partial f}{\partial
x_{n}}=y_{1}a_{1}+y_{2}a_{2}+\ldots+y_{n-1}a_{n-1}.$

If

$\displaystyle y_{1}a_{1}+y_{2}a_{2}+\ldots+y_{n-1}a_{n-1}=0$

then the substitution $y_{1}=\ldots=y_{n-2}=0$ gives that $a_{n-1}=0$.
Repeated substitution $y_{1}=\ldots=y_{n-3}=0$ gives $a_{n-2}=0$. Continuing
the same discussions, we get $a_{n-1}=a_{n-2}=\ldots=a_{1}=0$. Threfore
$\frac{\partial f}{\partial x_{n}}=0$ if and only if $f=f^{\prime}$. $\Box$

Consider the set of Chein automorphisms of the form

(7) $\displaystyle(x_{1}+[x_{2},x_{3}]a,x_{2},\ldots,x_{n}),\ a\in U.$

These automorphisms are elementary if $a$ does not depend on $y_{1}$.

In this paper we denote by $\mathrm{Mat}_{n}(U)$ the set of all square
matrices of order $n$ over $U$ (since the more standard notation $M_{n}(U)$
can be confused with the notation of the free metabelian Lie algebra $M_{n}$).
As usual, denote by $GL_{n}(U)$ and $SL_{n}(U)$ the general linear group and
the special linear group of matrices of order $n$ over $U$, respectively. The
identity matrix of order $n$ is already denoted by $I$.

In Section 2 we mentioned that the group $GL_{n}(K)$ will be identified with
the group of all linear automorphisms of $M_{n}$. We meant that each linear
automorphism $\alpha$ will be identified with its Jacobian matrix $J(\alpha)$.
This means that $\alpha$ is a matrix and instead of $J(\alpha)$ in our
calculations we will write $\alpha$.

######  Lemma 4.

$(a)$ If $\phi,\psi\in\mathrm{IAut}(M_{n})$ then

$\displaystyle J(\phi\psi)=J(\phi)J(\psi);$

$(b)$ If $\phi\in\mathrm{IAut}(M_{n})$ and $\alpha\in GL_{n}(K)$ then

$\displaystyle J(\alpha\phi\alpha^{-1})=\alpha J(\phi)^{\alpha}\alpha^{-1};$

$(c)$ The map

$\displaystyle J:\mathrm{IAut}(M_{n})\to SL_{n}(U)$

is an injective homomorphism of groups and $G=J(\mathrm{IAut}(M_{n}))$ is the
set of all matrics from $SL_{n}(U)$ of the form

$\displaystyle I+A,$

where $A\in\mathrm{Mat}(U)$ is a matrix such that $YA=0$.

Proof. Since $\phi$ induces an identity automorphism on $U$, $(a)$ is a direct
consequence of (6). Using (6) we also get

$\displaystyle
J(\alpha\phi\alpha^{-1})=J(\alpha(\phi\alpha^{-1}))=J(\alpha)J(\phi\alpha^{-1})^{\alpha}=J(\alpha)(J(\phi)J(\alpha^{-1})^{\phi})^{\alpha}=J(\alpha)(J(\phi)J(\alpha^{-1}))^{\alpha}$
$\displaystyle=J(\alpha)J(\phi)^{\alpha}J(\alpha^{-1})^{\alpha}=J(\alpha)J(\phi)^{\alpha}J(\alpha^{-1})=J(\alpha)J(\phi)^{\alpha}J(\alpha)^{-1}=\alpha
J(\phi)^{\alpha}\alpha^{-1}.$

In this calculation we again used that $\phi$ induces an identity automorphism
on $U$, $J(\alpha^{-1})\in GL_{n}(K)$ is a constant matrix, and that
$J(\alpha^{-1})=J(\alpha)^{-1}$ by (6).

If $\phi\in\mathrm{IAut}(M_{n})$, then

$\displaystyle\phi=(x_{1}+f_{1},x_{2}+f_{2},\ldots,x_{n}+f_{n}),$

where $f_{i}\in[M_{n},M_{n}]$ for all $i$, and

$\displaystyle J(\phi)=I+A,\ A=\left[\partial(f_{1})\ \ldots\
\partial(f_{n})\right].$

We get $YA=0$ by Lemma 2. If $J(\phi)=I$ then $\partial(f_{i})=0$ for all $i$.
Notice that for any $f\in[M_{n},M_{n}]$ we get
$f=(t_{1},\ldots,t_{n})\partial(f)$ by the definition of $\partial(f)$.
Consequently, $f_{i}=(t_{1},\ldots,t_{n})\partial(f_{i})=0$. This means
$\phi=\mathrm{id}$, i.e., $J$ is injective.

Conversely, let $A\in\mathrm{Mat}_{n}(U)$ such that $YA=0$. By Lemma 2 there
exist $g_{1},g_{2},\ldots,g_{n}\in[M_{n},M_{n}]$ such that
$A=\left[\partial(g_{1})\ \ldots\ \partial(g_{n})\right]$. Set
$\psi=(x_{1}+g_{1},x_{2}+g_{2},\ldots,x_{n}+g_{n})$. Then $J(\psi)=I+A$. If
$I+A\in SL_{n}(U)$ then $\psi$ is an automorphism by Lemma 3 and
$\psi\in\mathrm{IAut}(M_{n})$. $\Box$

##  4\. Matrix notations

Using Lemma 4, we will represent automorphisms from $\mathrm{IAut}(M_{n})$ by
their Jacobian matrices and make calculations in the language of Jacobian
matrices. Results formulated in the language of matrices will be also
reformulated in the language of automorphisms.

Thus $G$ represents the set of all Jacobian matrices of automorphisms from
$\mathrm{IAut}(M_{n})$. Set
$T=J(\mathrm{TAut}(M_{n})\cap\mathrm{IAut}(M_{n}))$ and
$S=J(\mathrm{ATAut}(M_{n})\cap\mathrm{IAut}(M_{n}))$. We have

$\displaystyle T\subseteq S\subseteq G.$

In order to prove that $\mathrm{ATAut}(M_{n})=\mathrm{Aut}(M_{n})$ it is
sufficient to prove that
$\mathrm{ATAut}(M_{n})\cap\mathrm{IAut}(M_{n})=\mathrm{IAut}(M_{n})$ since
both groups include all linear automorphisms. Thus, it is sufficient to prove
that $S=G$.

Let $E_{ij}$ a square matrix with $1$ in the $(i,j)$ position and zeros
elsewhere. The order of $E_{ij}$ is usually clear from the context. Recall
that

$\displaystyle E_{ij}E_{kl}=\delta_{jk}E_{il},$

where $\delta_{jk}$ is the Kronecker delta. A matrix of the form

$\displaystyle I+aE_{ij},\ i\neq j,\ a\in U,$

is called elementary.

If $i<j$, then for any $k$ denote by $F_{ijk}$ the $n\times n$ matrix with
$y_{j}$ in position $(i,k)$, $-y_{i}$ in position $(j,k)$, and zeros in other
places, that is,

$\displaystyle F_{ijk}=y_{j}E_{ik}-y_{i}E_{jk}.$

In other words, the $k$th column of $F_{ijk}$ is equal to
$\partial([x_{i},x_{j}])=e_{i}y_{j}-e_{j}y_{i}$, and the rest are zeros. It is
easy to check that

$\displaystyle F_{ijk}F_{stm}=\left\\{\begin{array}[]{ccc}y_{t}F_{ijm}&{\rm
if}&k=s,\\\ -y_{s}F_{ijm}&{\rm if}&k=t,\\\ 0&{\rm if}&k\neq s,t\\\
\end{array}\right.,$ $\displaystyle
E_{ij}F_{kst}=y_{s}E_{ij}E_{kt}-y_{k}E_{ij}E_{st}=\left\\{\begin{array}[]{ccc}y_{s}E_{it}&{\rm
if}&j=k,\\\ -y_{k}E_{it}&{\rm if}&j=s,\\\ 0&{\rm if}&j\neq k,s\\\
\end{array}\right.,$

and

$\displaystyle
F_{kst}E_{ij}=y_{s}E_{kt}E_{ij}-y_{k}E_{st}E_{ij}=\left\\{\begin{array}[]{ccc}F_{skj}&{\rm
if}&t=i,\\\ 0&{\rm if}&t\neq i\\\ \end{array}\right..$

We refer to these three formulas as the multiplication tables but often use
them without reference. Properties of the product of matrices of types

$\displaystyle I+A+B=(I+A)(I+B),$

which is true if $AB=0$, and

$\displaystyle I+(a+b)C=(I+aC)(I+bC),$

which is valid if $C^{2}=0$, to simplify terminology, will be called
additivity.

For example, if $f=\sum_{i<j}[x_{i},x_{j}]a_{ij}$, then the Jacobian matrix of
the endomorphism (5) is $I+\sum_{i<j}a_{ij}F_{ij1}$, and, by additivity,

$\displaystyle I+\sum_{i<j}a_{ij}F_{ij1}=\prod_{i<j}(I+a_{ij}F_{ij1}).$

Moreover,

$\displaystyle I+(a+b)F_{ij1}=(I+aF_{ij1})(I+bF_{ij1}).$

Consider the set of matrices of the form

(11) $\displaystyle I+aF_{231},\ a\in U.$

These matrices are Jacobian matrices of Chein automorphisms (7).

Here are two remarks:

$(a)$ If $A\in G$ and $\alpha\in GL_{n}(K)$ then $\alpha
A^{\alpha}\alpha^{-1}\in G$ by Lemma 4$(b)$. This matrix will be called the
linear conjugate of $A$ by $\alpha$.

$(b)$ If $J(\phi)=I+A\in S$ and $A$ is a homogeneous matrix, then, usually, it
is easy to show that $I+\mu A\in S$ for any $\mu\in K$. This can be done by
adding some coefficients in the proofs. For example, this can be done when
conjugating matrices (11) using $(x_{1},\mu x_{2},x_{3},\ldots,x_{n})$. For
this reason we ignore the difference between the coefficients of this type.

######  Corollary 2.

$(a)$ The group $T$ is generated by all linear conjugates of matrices of the
form (11) where $a$ does not depend on $y_{1}$.

$(b)$ The group $S$ is generated by all linear conjugates of matrices of the
form (11).

Proof. By Lemma 1, every automorphism
$\mathrm{TAut}(M_{n})\cap\mathrm{IAut}(M_{n})$ is a product of linear
conjugates of automorphisms of the form (5). Notice that every automorphism of
the form (5) is itself a product of linear conjugates of automorphisms of the
form (7) with $a\in K[y_{2},\ldots,y_{n}]$. Then $J(\tau)$ is a matrix of the
form (11). If $\alpha\in GL_{n}(K)$ then

$\displaystyle J(\alpha\tau\alpha^{-1})=\alpha J(\tau)^{\alpha}\alpha^{-1}$

by Lemma 4, that is, $J(\alpha\tau\alpha^{-1})$ is the linear conjugate of
$J(\tau)$ by $\alpha$. If $\phi,\psi\in\mathrm{IAut}(M_{n})$ then
$J(\phi\psi)=J(\phi)J(\psi)$ by the same lemma. This proves $(a)$.

The statement $(b)$ can be checked similarly. $\Box$

##  5\. Tame automorphisms

In this section we assume that either $n\geq 4$ and $K$ is a field of
characteristic $\neq 3$ or $n\geq 5$.

######  Lemma 5.

Every matrix of the form (11) belongs to $T$ if $\mathrm{ldeg}(a)\geq 2$.

Proof. The proof practically repeats the proof of Theorem 1 from [37].

Due to the additivity of (11) with respect to $a$, we can assume that
$a=y_{1}^{i_{1}}y_{2}^{i_{2}}\ldots y_{n}^{i_{n}}$. If $i_{1}=0$ then
$I+aF_{231}\in T$ by the definition of $T$. Assume that $i_{1}\geq 1$.

Case 1. Suppose that $i_{j}\geq 1$ for some $2\leq j\leq n-1$. Set

$\displaystyle A=I+y_{1}^{i_{1}}\ldots y_{j}^{i_{j}-1}F_{23n},\ \
B=I-y_{j+1}^{i_{j+1}}\ldots y_{n}^{i_{n}}F_{jn1}.$

A direct calculation using the multiplication tables gives that

$\displaystyle[A,B]=ABA^{-1}B^{-1}=I+aF_{231}\in T$

since $A,B\in T$.

Case 2. Let $i_{2}=i_{3}=\ldots=i_{n-1}=0$ and $i_{n}\neq 0$. Then
$a=y_{1}^{i_{1}}y_{n}^{i_{n}}$ and we get

$\displaystyle
aF_{231}=y_{1}^{i_{1}}y_{3}y_{n}^{i_{n}-1}F_{2n1}-y_{1}^{i_{1}}y_{2}y_{n}^{i_{n}-1}F_{3n1}$

by the Jacobi identity. If $i_{n}\geq 2$ then

$\displaystyle I+y_{1}^{i_{1}}y_{3}y_{n}^{i_{n}-1}F_{2n1},\
I-y_{1}^{i_{1}}y_{2}y_{n}^{i_{n}-1}F_{3n1}\in T$

by Case 1. Consequently,

$\displaystyle
I+aF_{231}=(I+y_{1}^{i_{1}}y_{3}y_{n}^{i_{n}-1}F_{2n1})(I-y_{1}^{i_{1}}y_{2}y_{n}^{i_{n}-1}F_{3n1})\in
T.$

If $i_{n}=1$, then set $\alpha=I+E_{42}$. Notice that
$I+y_{1}^{i_{1}}y_{2}F_{231}\in T$ by Case 1. Hence

$\displaystyle(I+E_{42})(I+y_{1}^{i_{1}}y_{2}F_{231})^{\alpha}(I-E_{42})=I+y_{1}^{i_{1}}y_{2}F_{231}+y_{1}^{i_{1}}(y_{4}F_{231}-y_{2}F_{341})-y_{1}^{i_{1}}y_{4}F_{341}$
$\displaystyle=(I+y_{1}^{i_{1}}y_{2}F_{231})(I+y_{1}^{i_{1}}(y_{4}F_{231}-y_{2}F_{341}))(I-y_{1}^{i_{1}}y_{4}F_{341})\in
T.$

Since the first and the last factors of this product belong to $T$ it follows
that

$\displaystyle C=I+y_{1}^{i_{1}}(y_{4}F_{231}-y_{2}F_{341})\in T.$

The Jacobi identity implies that

$\displaystyle y_{4}F_{231}=y_{3}F_{241}-y_{2}F_{341}.$

Consequently,

$\displaystyle
3aF_{231}=y_{1}^{i_{1}}[(y_{4}F_{231}-y_{2}F_{341})+(y_{4}F_{231}+y_{3}F_{241})]$

and, by additivity,

$\displaystyle
I+3aF_{231}=[I+y_{1}^{i_{1}}(y_{4}F_{231}-y_{2}F_{341})][I+y_{1}^{i_{1}}(y_{4}F_{231}+y_{3}F_{241})].$

The first factor is $C$ and belongs to $T$. The second factor also belongs to
$T$ since

$\displaystyle(23)(C^{-1})^{(23)}(23)=(23)((I+y_{1}^{i_{1}}(y_{4}F_{231}-y_{2}F_{341}))^{-1})^{(23)}(23)$
$\displaystyle=(23)(I-y_{1}^{i_{1}}(y_{4}F_{231}-y_{2}F_{341}))^{(23)}(23)=I+(y_{4}F_{231}+y_{3}F_{241}),$

where $(23)=(x_{1},x_{3},x_{2},x_{4},\ldots,x_{n})$. Consequently,
$I+3aF_{231}\in T$. If the characteristic of $K$ is not $3$, then its
conjugation by $(x_{1},3x_{2},x_{3},\ldots,x_{n})$ gives that $I+aF_{231}\in
T$.

If $n\geq 5$, then conjugating (11) by $(4n)$, we can make $i_{4}>0$. Case 1
gives that $I+aF_{231}\in T$.

Case 3. Let $i_{2}=i_{3}=\ldots=i_{n-1}=i_{n}=0$. In this case
$a=y_{1}^{i_{1}}$ and $s=i_{1}\geq 2$ since $\mathrm{ldeg}(a)\geq 2$. Case 2
gives that $I+y_{4}y_{1}^{s-1}F_{231}\in T$. Set $\beta=I+E_{14}\in
GL_{n}(K)$. Using the multiplication tables we get

$\displaystyle\beta(I+y_{4}y_{1}^{s-1}F_{231})^{\beta}\beta^{-1}=I+(y_{4}+y_{1})y_{1}^{s-1}F_{231}-(y_{4}+y_{1})y_{1}^{s-1}F_{234}$
$\displaystyle=(I+(y_{4}+y_{1})y_{1}^{s-1}F_{231})(I-(y_{4}+y_{1})y_{1}^{s-1}F_{234})$
$\displaystyle=(I+y_{4}y_{1}^{s-1}F_{231})(I+y_{1}^{s}F_{231})(I-y_{4}y_{1}^{s-1}F_{234})(I-y_{1}^{s}F_{234})\in
T.$

Since all factors of this product except $I+y_{1}^{s}F_{231}$ belong to $T$,
this implies that $I+y_{1}^{s}F_{231}$ also belongs to $T$. $\Box$

Here is a reformulation of this lemma in the language of automorphisms.

######  Corollary 3.

Every Chein automorphism of the form (5) with $\mathrm{ldeg}(f)\geq 4$ is
tame.

Proof. By Corollary 1, any such Chein automorphism is a product of linear
conjugates of Chein automorphisms of the form (7) with $\mathrm{ldeg}(a)\geq
2$. Then $J(\tau)$ is a matrix of the form (11) and belongs to $T$ by Lemma 5.
$\Box$

This corollary contradicts the results of [40].

######  Corollary 4.

Assume that either $n\geq 4$ and $K$ is a field of characteristic $\neq 3$ or
$n\geq 5$. Then the group of all almost tame automorphisms
$\mathrm{ATAut}(M_{n})$ is generated by all linear automorphisms, as well as a
quadratic automorphism (1) and a cubic automorphism (2).

Proof. The group of all tame automorphisms $\mathrm{TAut}(M_{n})$ is generated
by all linear automorphisms, as well as a quadratic automorphism (1) [37]. All
Chein automorphisms of the form (7) with $\mathrm{ldeg}(a)\geq 2$ belong to
$\mathrm{TAut}(M_{n})$. By the additivity of automorphisms of the form (7)
with respect to $a$, we may assume that $a$ is homogeneous of degree $\leq 1$.
If $a$ does not depend on $y_{1}$ then (7). Again, by additivity, we may
assume that $a=y_{1}$. In this case (7) is the cubic automorphism (2). $\Box$

######  Lemma 6.

Every matrix of the form

$\displaystyle A(a)=I+ay_{1}F_{121}+ay_{2}F_{122}$

belongs to $T$ if $\mathrm{ldeg}(a)\geq 2$.

Proof. By Lemma 5,

$\displaystyle
B=I-by_{1}F_{341}-by_{2}F_{342}=(I-by_{1}F_{341})(I-by_{2}F_{342})\in T$

if $\mathrm{ldeg}(b)\geq 1$. Set $C=I+F_{123}\in T$. Then

$\displaystyle[B,C]=BCB^{-1}C^{-1}=I+by_{1}y_{4}F_{121}+by_{2}y_{4}F_{122}=A(y_{4}b)\in
T.$

Consequently, the statement of the lemma is true if $y_{4}$ divides $a$.
Obviously, the same is true if $y_{i}$ divides $a$ for all $3\leq i\leq n$. By
the additivity of $A(a)$ with respect to $a$, we may assume that $a\in
K[y_{1},y_{2}]$.

Set $\alpha=I+E_{14}\in T$. If $b\in K[y_{1},y_{2}]$ then, using the
multiplication tables, we get

$\displaystyle\alpha
A(y_{4}b)^{\alpha}\alpha^{-1}=I+y_{1}(y_{4}+y_{1})bF_{121}+y_{2}(y_{4}+y_{1})bF_{122}-y_{2}(y_{4}+y_{1})bF_{124}$
$\displaystyle=A(y_{4}b)A(y_{1}b)(I-y_{2}(y_{4}+y_{1})bF_{124})\in T.$

Since $A(y_{4}b)\in T$ and $I-y_{2}(y_{4}+y_{1})bF_{124}\in T$ by Lemma 5,
this implies that $A(y_{1}b)\in T$. Therefore the statement of the lemma is
true if $a\in K[y_{1},y_{2}]$ and $y_{1}$ divides $a$. Obviously, the same is
true for $y_{2}$. $\Box$

######  Corollary 5.

Every automorphism of the form

$\displaystyle(x_{1}+[x_{1},x_{2}]y_{1}a,x_{2}+[x_{1},x_{2}]y_{2}a,x_{3},\ldots,x_{n})$

is tame if $\mathrm{ldeg}(a)\geq 2$.

Proof. By Lemma 6 the Jacobian matrices of these automorphisms belong to $T$.
$\Box$

######  Corollary 6.

Every exponential automorphism of lower degree $\geq 5$ is tame.

Proof. Let

$\displaystyle\exp(\mathrm{ad}(m))=(x_{1}+[m,x_{1}],x_{2}+[m,x_{2}],\ldots,x_{n}+[m,x_{n}])$

be an exponential automorphism with $\mathrm{ldeg}(m)\geq 4$. By the
additivity with respect to $m$ we may assume that $m=[x_{1},x_{2}]a$ with
$\mathrm{ldeg}(a)\geq 2$. Then the Jacobian matrix of $\exp(\mathrm{ad}(m))$
is

$\displaystyle
I+\sum_{i}y_{i}aF_{12i}=(I+y_{1}aF_{121}+y_{2}aF_{122})\prod_{i\geq
3}(I+y_{i}aF_{12i}).$

This matrix belongs to $T$ by Lemmas 5 and 6. $\Box$

This corollary contradicts the main result of [5].

##  6\. Almost tame automorphisms

For the remainder of this article, we always assume that $n\geq 4$ and $K$ is
an arbitrary field. By Corollary 2, every matrix of the form (11) and its
conjugates belong to $S$. For this reason we do not need the proof of Lemma 5,
where the condition that either $n\geq 4$ and $K$ is a field of characteristic
$\neq 3$ or $n\geq 5$ was used. The other proofs of Section 5 did not use this
condition.

######  Lemma 7.

Every matrix of the form

$\displaystyle C(a)=I+ay_{1}F_{121}+ay_{2}F_{122}$

with $\mathrm{ldeg}(a)\geq 1$ belongs to $S$.

Proof. For any $b\in U$ we have

$\displaystyle
B=I-by_{1}F_{341}-by_{2}F_{342}=(I-by_{1}F_{341})(I-by_{2}F_{342})\in S.$

Repeating the proof of Lemma 6 with this data, we get that $C(a)\in S$ for all
$a$ with $\mathrm{ldeg}(a)\geq 1$. $\Box$

Notice that Lemma 7 does not include only the matrix

(12) $\displaystyle I+y_{1}F_{121}+y_{2}F_{122}$

of the same type, which is the Jacobian matrix of (3).

######  Corollary 7.

Every exponential automorphism of lower degree $\geq 4$ is almost tame.

Proof. Using the decomposition of the Jacobian matrix of an exponential
automorphism from the proof of Corollary 6, we conclude that it belongs to $S$
by Lemma 7 if its lower degree $\geq 3$. Consequently, every exponential
automorphism of lower degree $\geq 4$ belongs to $\mathrm{ATAut}(M_{n})$.
$\Box$

##  7\. Special matrices

We use matrices instead of matrix notation in many places because it makes the
calculations easier to understand. To some extent this is due to the
application of Suslin’s theorem [59].

The main aim of this section is to prove that every matrix of the form

(13) $\displaystyle I+\begin{bmatrix}y_{n}C&\begin{matrix}0\\\ .\\\ .\\\ .\\\
0\end{matrix}\\\ \begin{matrix}*&*&\ldots&*\end{matrix}&0\end{bmatrix}\in G$

belongs to $S$. We will provide several lemmas to the proof of this result. A
matrix of the form (13) will be called special.

######  Lemma 8.

Let $B$ be any matrix from $SL_{n-1}(U,y_{n}U)$. Then $B=I_{n-1}+y_{n}C$ for
some matrix $C\in\mathrm{Mat}_{n-1}(U)$ and there exists a unique special
matrix $A$ such that

$\displaystyle A=I+\left[\begin{array}[]{cc}y_{n}C&0\\\ &0\\\
\end{array}\right].$

Proof. Obviously, $B=I_{n-1}+y_{n}C$ for some matrix
$C\in\mathrm{Mat}_{n-1}(U)$. If $C=[c_{ij}]$ then

$\displaystyle A=I+\begin{bmatrix}y_{n}D&\begin{matrix}0\\\ .\\\ .\\\ .\\\
0\end{matrix}\\\
\begin{matrix}-\sum_{i=1}^{n-1}y_{i}d_{i1}&-\sum_{i=1}^{n-1}y_{i}d_{i2}&\ldots&-\sum_{i=1}^{n-1}y_{i}d_{i(n-1)}\end{matrix}&0\end{bmatrix}$

belongs to $G$ by Lemma 2. If $C=0$ then $A=I$ by Lemma 2. This implies the
uniqueness of $A$. $\Box$

If $B\in SL_{n}(U)$ is a matrix of the form $B=\begin{bmatrix}C&0\\\
0&I_{n-k}\end{bmatrix}$, where $C\in SL_{k}(U)$ for some $1\leq k\leq n$, then
we say that $B$ is $k$-reducible and $B_{k}=C$ will be called its
$k$-reduction. Calculations with $k$-reducible matrices from $G$ can be done
by using their $k$-reductions.

Set $A_{n-1}=B$ for matrices $A,B$ satisfying the conditions of Lemma 8. We
use this notation for special matrices despite the fact that $A$ is not
$n-1$-reducible in general. But Lemma 16 allows us to use the matrix $B$
instead of $C$ in some calculations. Moreover, if $B$ is $k$-reducible for
some $1\leq k\leq n-1$, then $B_{k}$ will also be called the $k$-reduction of
the special matrix $A$ and the special matrix $A$ will be called special
$k$-reducible. Let us specify how can we use $k$-reductions of special
matrices. Using (6) it is easy to check the following statements:

$(1)$ If $A$ and $B$ are two special $k$-reducible matrices, then $AB$ is a
special $k$-reducible and $(AB)_{k}=A_{k}B_{k}$;

$(2)$ If $A$ is special $k$-reducible and $B\in SL_{n}(U)$ (not necessarily
$B\in G$) is $k$-reducible, then $A^{B}=BAB^{-1}$ is special $k$-reducible and
$(A^{B})_{k}=B_{k}A_{k}B_{k}^{-1}$;

$(3)$ If $A$ is special $k$-reducible and $\alpha\in SL_{n}(K)$ is
$k$-reducible, then the linear conjugation $\alpha A^{\alpha}\alpha^{-1}$ of
$A$ by $\alpha$ is special $k$-reducible and $(\alpha
A^{\alpha}\alpha^{-1})_{k}=\alpha_{k}A_{k}^{\alpha_{k}}\alpha_{k}^{-1}$.

######  Lemma 9.

Every special matrix of the form

$\displaystyle A(h,g)=I+hgF_{1n1}+hg^{2}F_{2n1}-hF_{1n2}-hgF_{2n2}$

belongs to $S$.

Proof. We have

$\displaystyle A(h,g)_{2}=I_{2}+y_{n}h\begin{bmatrix}g&-1\\\
g^{2}&-g\end{bmatrix}.$

$(a)$ Let $\alpha=I-\lambda E_{21},\lambda\in K$. Then

$\displaystyle
A(h,g)_{2}^{\alpha}=I_{2}+y_{n}h^{\alpha}\begin{bmatrix}g^{\alpha}&-1\\\
(g^{\alpha})^{2}&-g^{\alpha}\end{bmatrix}$

and

$\displaystyle\alpha_{2}A(h,g)_{2}^{\alpha}\alpha_{2}^{-1}=\begin{bmatrix}1&0\\\
-\lambda&1\end{bmatrix}\left(I_{2}+y_{n}h^{\alpha}\begin{bmatrix}g^{\alpha}&-1\\\
(g^{\alpha})^{2}&-g^{\alpha}\end{bmatrix}\right)\begin{bmatrix}1&0\\\
\lambda&1\end{bmatrix}$
$\displaystyle=I_{2}+y_{n}h^{\alpha}\begin{bmatrix}(g^{\alpha}-\lambda)&-1\\\
(g^{\alpha}-\lambda)^{2}&-(g^{\alpha}-\lambda)\end{bmatrix}=A(h^{\alpha},(g^{\alpha}-\lambda))_{2}.$

Consequently, $\alpha
A(h,g)^{\alpha}\alpha^{-1}=A(h^{\alpha},(g^{\alpha}-\lambda))$. This means
that $A(h,g)\in S$ if and only if $A(h^{\alpha},(g^{\alpha}-\lambda))\in S$.
Using this we may assume that $g(0,\ldots,0)=0$.

$(b)$ Set $B=I-fF_{231}\in S$. Then

$\displaystyle B_{3}=\begin{bmatrix}1&0&0\\\ -y_{3}f&1&0\\\
y_{2}f&0&1\end{bmatrix}.$

Direct calculations give that

$\displaystyle
B_{3}A(h,g)_{3}B_{3}^{-1}=I+y_{n}h\begin{bmatrix}(g-y_{3}f)&-1&0\\\
(g-y_{3}f)^{2}&-(g-y_{3}f)&0\\\ y_{2}(g-y_{3}f)f&-y_{2}f&0\end{bmatrix}$
$\displaystyle=\left(I+y_{n}h\begin{bmatrix}(g-y_{3}f)&-1&0\\\
(g-y_{3}f)^{2}&-(g-y_{3}f)&0\\\
0&0&0\end{bmatrix}\right)\left(I+y_{n}h\begin{bmatrix}0&0&0\\\ 0&0&0\\\
y_{2}(g-y_{3}f)f&-y_{2}f&0\end{bmatrix}\right)$
$\displaystyle=A(h,g-y_{3}f)_{3}C_{3},$

where

$\displaystyle
C=I+y_{2}h(g-y_{3}f)fF_{3n1}-y_{2}hfF_{3n2}=(I+y_{2}h(g-y_{3}f)fF_{3n1})(I-y_{2}hfF_{3n2}).$

Notice that $C\in S$. Consequently, $A(h,g)\in S$ if and only if
$A(h,g-y_{3}f)\in S$. Of course, the same trick can be done with any variable
$y_{i}$ if $3\leq i\leq n-1$. Therefore, replacing $g$ by $g-y_{i}f$ for all
$3\leq i\leq n-1$ we may assume that

$\displaystyle g=G(y_{1},y_{2})-y_{n}a.$

$(c)$ Now we show how to eliminate $y_{n}a$. First we replace $g$ to

$\displaystyle g+y_{3}a=G(y_{1},y_{2})+y_{3}a-y_{n}a$

using $(b)$. If $\gamma=I+E_{3n}$, then it is easy to check that

$\displaystyle\gamma
A(h,g+y_{3}a)^{\gamma}\gamma^{-1}=A(h^{\gamma},(g+y_{3}a)^{\gamma}).$

We have

$\displaystyle(g+y_{3}a)^{\gamma}=g^{\gamma}+(y_{3}+y_{n})a^{\gamma}$
$\displaystyle=G(y_{1},y_{2})-y_{n}a^{\gamma}+(y_{3}+y_{n})a^{\gamma}=G(y_{1},y_{2})+y_{3}a^{\gamma}.$

Using $(b)$ again, we may assume that

$\displaystyle g=G(y_{1},y_{2}).$

(d) Let $f$ be a polynomial in $y_{1},y_{2}$. Then $g$ can be replaced by
$g+y_{i}f$, where $i=1,2$. For this we first replace $g$ by $g+y_{3}f$ using
$(b)$, i.e., we can assume that $g=G(y_{1},y_{2})+y_{3}f$. If
$\delta=I+E_{i3}$, then

$\displaystyle\delta
A(h,g)^{\delta}\delta^{-1}=(I-h^{\delta}g^{\delta}F_{1n1}-h^{\delta}(g^{\delta})^{2}F_{2n1})A(h^{\delta},g^{\delta}).$

Since $I-h^{\delta}g^{\delta}F_{1n1}-h^{\delta}(g^{\delta})^{2}F_{2n1}\in S$
it follows that $A(h,g)\in S$ if and only if $A(h^{\delta},g^{\delta})\in S$.

Notice that

$\displaystyle g^{\gamma}=G(y_{1},y_{2})+y_{3}f^{\gamma}+y_{i}f^{\gamma}.$

Using $(b)$, we can replace $g^{\gamma}$ by

$\displaystyle G(y_{1},y_{2})+y_{i}f^{\gamma}.$

This means that we can replace $g=G(y_{1},y_{2})$ by $G(y_{1},y_{2})+y_{i}f$
for any $f$. Then we can make $g=0$. Notice that $A(h,0)=I-hF_{1n2}\in S$.
$\Box$

######  Corollary 8.

Every automorphism of the form

$\displaystyle(x_{1}+[x_{1},x_{n}]hg+[x_{2},x_{n}]hg^{2},x_{2}+[x_{1},x_{n}]h+[x_{2},x_{n}]hg,x_{3},\ldots,x_{n})$

belongs to $W$.

######  Lemma 10.

Every matrix of the form

$\displaystyle A(h,f,g)=I+hfgF_{1n1}+hg^{2}F_{2n1}-hf^{2}F_{1n2}-hfgF_{2n2}$

belongs to $S$.

Proof. The proof of the lemma repeats the proof given on pages 90–91 of [4].
Notice that $A(h,f,g)$ is an induced $2$-reducible matrix and

$\displaystyle A(h,f,g)_{2}=I_{2}+y_{n}h\begin{bmatrix}fg&-f^{2}\\\
g^{2}&-fg\end{bmatrix}$

We have

$\displaystyle A(h,f,g)_{3}=\begin{bmatrix}1&0&0\\\ 0&1&0\\\
g&-f&1\end{bmatrix}\begin{bmatrix}1&0&-y_{n}hf\\\ 0&1&-y_{n}hg\\\
0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\\ 0&1&0\\\
-g&f&1\end{bmatrix}\begin{bmatrix}1&0&y_{n}hf\\\ 0&1&y_{n}hg\\\
0&0&1\end{bmatrix}.$

In matrix notation, this can be written as

$\displaystyle A(h,f,g)=[(I+gE_{31}-fE_{32}),(I-hfF_{1n3}-hgF_{2n3})]$
$\displaystyle=(I-hfF_{1n3}-hgF_{2n3})^{(I+gE_{31}-fE_{32})}(I+hfF_{1n3}+hgF_{2n3})$

The last factor belongs to $S$. Therefore it is sufficient to prove that

$\displaystyle(I-hfF_{1n3}-hgF_{2n3})^{(I+gE_{31}-fE_{32})}=(I-hfF_{1n3})^{(I+gE_{31}-fE_{32})}(I-hgF_{2n3})^{(I+gE_{31}-fE_{32})}$

belongs to $S$. Moreover, it is sufficient to prove that the first factor of
the last product belongs to $S$ since its conjugation by the linear
automorphism $(12)$ gives a matrix that has the form of the second factor. We
have

$\displaystyle\begin{bmatrix}1&0&0\\\ 0&1&0\\\
g&-f&1\end{bmatrix}\begin{bmatrix}1&0&-y_{n}hf\\\ 0&1&0\\\
0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\\ 0&1&0\\\
-g&f&1\end{bmatrix}=I_{3}+y_{n}h\begin{bmatrix}fg&-f^{2}&-f\\\ 0&0&0\\\
fg^{2}&-f^{2}g&-fg\end{bmatrix}$
$\displaystyle=\left(I_{3}+y_{n}h\begin{bmatrix}0&-f^{2}&0\\\ 0&0&0\\\
0&-f^{2}g&0\end{bmatrix}\right)\left(I_{3}+y_{n}(hf)\begin{bmatrix}g&0&-1\\\
0&0&0\\\ g^{2}&0&-g\end{bmatrix}\right),$

which means that

$\displaystyle(I-hfF_{1n3})^{(I+gE_{31}-fE_{32})}=(I-hf^{2}F_{1n2}-hf^{2}gF_{3n2})$
$\displaystyle\cdot(I+(hf)gF_{1n1}+(hf)g^{2}F_{3n1}-(hf)F_{1n3}-(hf)gF_{3n3}).$

The first factor of this product belongs to $S$ and the second factor belongs
to $S$ by Lemma 9. $\Box$

######  Corollary 9.

Every automorphism of the form

$\displaystyle(x_{1}+[x_{1},x_{n}]hfg+[x_{2},x_{n}]hg^{2},x_{2}+[x_{1},x_{n}]hf^{2}+[x_{2},x_{n}]hfg,x_{3},\ldots,x_{n})$

belongs to $W$.

First, let us recall some results of Cohn [12] and Suslin [59] on generators
of special linear groups over polynomial algebras.

In 1966 P. Cohn [12] proved that the matrix

$\displaystyle\begin{bmatrix}1+x_{1}x_{2}&x_{2}^{2}\\\
-x_{1}^{2}&1-x_{1}x_{2}\\\ \end{bmatrix}$

is not a product of elementary matrices over the polynomial algebra
$K[x_{1},x_{2}]$.

Denote by $E_{k}(U)$ the subgroup of $SL_{k}(U)$ generated by all elementary
matrices, that is, by all matrices of the form

$\displaystyle I_{k}+aE_{ij},\ a\in U.$

The well-known Suslin’s Theorem [59] states that if $k\geq 3$ then
$SL_{k}(U)=E_{k}(U)$, that is, every matrix from $SL_{k}(U)$ is a product of
elementary matrices. If $I$ is an ideal of $U$ then the natural map $U\to U/I$
induces the natural maps $GL_{k}(U)\to GL_{k}(U/I)$ and $GL_{k}(U)\to
GL_{k}(U/I)$. Denote by $GL_{k}(U,I)$ the kernel of $GL_{k}(U)\to GL_{k}(U/I)$
and by $SL_{k}(U,I)$ the kernel of $SL_{k}(U)\to SL_{k}(U/I)$. Denote by
$E_{k}(U,I)$ the normal closure in $E_{k}(U)$ of the set of all elementary
matrices of the form $I+aE_{ij}$, where $i\neq j$ and $a\in I$.

The next corollary of Suslin’s results is formulated in [4].

######  Proposition 1.

Let $k\geq 3$ and $I=y_{n}U$. Then

(1) $SL_{k}(U,I)=E_{k}(U,I)$;

(2) $E_{k}(U,I)$ is generated by the set of all matrices of the form

$\displaystyle I+h(fe_{i}+ge_{j})(ge_{i}-fe_{j})^{t}$

where $i<j$, $h\in I$, and $f,g\in U$.

Proof. We repeat a short proof given in [4]. By [59], every matrix $\Phi\in
SL_{k}(U,I)$ is a product of elementary matrices $\Phi=E_{1}\ldots E_{s}$. Any
$a\in U=K[y_{1},\ldots,y_{n}]$ can be written in the form

$\displaystyle a=y_{n}b+c,\ \ b\in U,c\in K[y_{1},\ldots,y_{n-1}],$

and, consequently,

$\displaystyle I+aE_{ij}=(I+y_{n}bE_{ij})(I+cE_{ij}).$

Thus, $E_{i}=H_{i}G_{i}$ where $H_{i}\in E_{k}(U,I)$ and $G_{i}\in
E_{k}(K[y_{1},\ldots,y_{n-1}])$. Then

$\displaystyle\Phi=H_{1}(G_{1}H_{2}G_{1}^{-1})(G_{1}G_{2}H_{3}G_{2}^{-1}G_{1}^{-1})\ldots(G_{1}\ldots
G_{s-1}H_{s}G_{s-1}^{-1}\ldots G_{1}^{-1})(G_{1}\ldots G_{s}).$

Substituting $y_{n}=0$ we get $G_{1}\ldots G_{s}=1$. Since

$\displaystyle G_{1}\ldots G_{i-1}H_{i}G_{i-1}^{-1}\ldots G_{1}^{-1}\in
E_{k}(U,I)$

it follows that $\Phi\in E_{k}(U,I)$.

By [59, Corollary 1.4], the group $E_{k}(U,I)$ is generated by the set of all
matrices of the form

$\displaystyle I_{k}+hu(u_{j}e_{i}-u_{i}e_{j})^{t}$

where $h\in I$, $u=(u_{1},\ldots,u_{k})$ is a unimodular column over $U$, and
$1\leq i<j\leq k$. Recall that a column $u$ is called unimodular if the ideal
generated by its components contains the identity element. We have

$\displaystyle
I_{k}+hu(u_{j}e_{i}-u_{i}e_{j})^{t}=(I_{k}+h(u_{i}e_{i}+u_{j}e_{j})(u_{j}e_{i}-u_{i}e_{j})^{t})$
$\displaystyle\prod_{l\neq i,j}(I_{k}+hu_{l}e_{l}(u_{j}e_{i})^{t})\prod_{l\neq
i,j}(I_{k}-h(u_{l}e_{l})^{t}u_{i}e_{j}).$

This proves $(b)$. $\Box$

Notice that if $i=1,j=2$, then the matrix from Proposition 1$(2)$ is
$2$-reducible and its $2$-reduction is

$\displaystyle I_{2}+h\left[\begin{array}[]{cc}fg&-f^{2}\\\ g^{2}&-fg\\\
\end{array}\right],$

which is a generalization of the Cohn matrix mentioned above.

######  Proposition 2.

Every special matrix belongs to $S$.

Proof. Since $n\geq 4$ it follows by Proposition 1 that $SL_{n-1}(U,y_{n}U)$
is generated by the set of all matrices

$\displaystyle B=I_{n-1}+y_{n}h(fe_{i}+ge_{j})(ge_{i}-fe_{j})^{t},\ 1\leq
i<j\leq n-1.$

It is sufficient to prove that for any such matrix $B$ there exists a special
matrix $A$ from $S$ such that its special $(n-1)$-reduction is $B$. Without
loss of generality we may assume that $i=1$ and $j=2$, that is,

$\displaystyle B=I_{n-1}+y_{n}h(fe_{1}+ge_{2})(ge_{1}-fe_{2})^{t}.$

Then for

$\displaystyle A=I+hfgF_{1n1}+hg^{2}F_{2n1}-hf^{2}F_{1n2}-hfgF_{2n2}$

we get $A_{n-1}=B$ and $A$ belongs to $S$ by Lemma 10. $\Box$

######  Corollary 10.

Every automorphism of the form

$\displaystyle(x_{1}+f_{1},x_{2}+f_{2},\ldots,x_{n-1}+f_{n-1},x_{n}),$

where $f_{1},f_{2},\ldots,f_{n-1}$ belong to the ideal generated by
$[x_{1},x_{n}],\ldots,[x_{n-1},x_{n}]$, belongs to $W$.

##  8\. Inductive lemma

The following lemma is part of the excellent proof given in [4, p. 99–100],
and we will repeat this for completeness. Since we have no exponential
automorphisms of degree $3$, we first present this statement with a slight
restriction. Using this lemma, in the next section we will deal with
exponential automorphisms of degree $3$. Then we can prove this lemma in full
form. Fortunately, the proof given here works in the general case.

######  Lemma 11.

Let $A=I+[a_{ij}]\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)\cap G$,
$\mathrm{ldeg}(a_{ij})\geq 3$, and
$a_{1n},a_{2n},\ldots,a_{(n-1)n}\in\sum_{j=s}^{n-1}y_{j}U+y_{n}^{2}U$. Then
there exists $B\in S$ such that $B^{-1}A\in SL_{n}(U,\sum_{j=s}^{n-1}y_{j}U)$.

Proof. For any $a\in U$ and any matrix $R$ over $U$ let $\overline{a}$ and
$\overline{R}$ be the result of setting $y_{s}=y_{s+1}=\ldots=y_{n-1}=0$ in
$a$ and $R$ respectively.

Let

$\displaystyle M=\begin{bmatrix}I_{n-1}&\begin{matrix}0\\\ .\\\ .\\\ .\\\
0\end{matrix}\\\
\begin{matrix}y_{1}&y_{2}&\ldots&y_{n-1}\end{matrix}&y_{n}\end{bmatrix}.$

Then

$\displaystyle\overline{M}=\begin{bmatrix}I_{n-1}&\begin{matrix}0\\\ .\\\ .\\\
.\\\ 0\end{matrix}\\\
\begin{matrix}y_{1}&\ldots&y_{s-1}&0&\ldots&0\end{matrix}&y_{n}\end{bmatrix}.$

We have

$\displaystyle
MAM^{-1}=I_{n}+\begin{bmatrix}[a_{ij}-y_{n}^{-1}y_{j}a_{in}]&\begin{matrix}y_{n}^{-1}a_{1n}\\\
.\\\ .\\\ .\\\ y_{n}^{-1}a_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&0\end{bmatrix}$
$\displaystyle=\begin{bmatrix}I_{n-1}&\begin{matrix}y_{n}^{-1}a_{1n}\\\ .\\\
.\\\ .\\\ y_{n}^{-1}a_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}\left(I_{n}+\begin{bmatrix}[a_{ij}-y_{n}^{-1}y_{j}a_{in}]&\begin{matrix}0\\\
.\\\ .\\\ .\\\ 0\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&0\end{bmatrix}\right),$
$\displaystyle\overline{M}\ \overline{A}\
\overline{M}^{-1}=\begin{bmatrix}I_{n-1}&\begin{matrix}y_{n}^{-1}\overline{a}_{1n}\\\
.\\\ .\\\ .\\\ y_{n}^{-1}\overline{a}_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}\left(I_{n}+\begin{bmatrix}[\overline{a}_{ij}-y_{n}^{-1}y_{j}\overline{a}_{in}]&\begin{matrix}0\\\
.\\\ .\\\ .\\\ 0\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&0\end{bmatrix}\right),$

and

$\displaystyle\overline{M}\ \overline{A}\
\overline{M}^{-1}=\begin{bmatrix}I_{n-1}&\begin{matrix}y_{n}\overline{d}_{1n}\\\
.\\\ .\\\ .\\\ y_{n}\overline{d}_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}\left(I_{n}+\begin{bmatrix}[y_{n}\overline{d}_{ij}]&\begin{matrix}0\\\
.\\\ .\\\ .\\\ 0\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&0\end{bmatrix}\right),$

where $\overline{d}_{ij}\in K[y_{1},\ldots,y_{s-1},y_{n}]$. The last equality
is possible since $\overline{a}_{ij}\in y_{n}K[y_{1},\ldots,y_{s-1},y_{n}]$
for all $i,j$ and $\overline{a}_{in}\in
y_{n}^{2}K[y_{1},\ldots,y_{s-1},y_{n}]$ for all $1\leq i\leq n-1$. Moreover,
notice that $\mathrm{ldeg}(\overline{d}_{ij})\geq 2$ for all $1\leq i,j\leq
n-1$ and $\mathrm{ldeg}(\overline{d}_{in})\geq 1$ for all $1\leq i\leq n-1$
since $\mathrm{ldeg}(\overline{a}_{ij})\geq 3$ for all $i,j$.

For any $1\leq k\leq n-1$, let

$\displaystyle\delta_{k}=I_{n}+\begin{bmatrix}0&0&\ldots&0\\\ .&.&\ldots&.\\\
.&.&\ldots&.\\\ .&.&\ldots&.\\\ 0&0&\ldots&0\\\
y_{1}y_{n}\overline{d}_{kn}&y_{2}y_{n}\overline{d}_{kn}&\ldots&y_{n}^{2}\overline{d}_{kn}\\\
0&0&\ldots&0\\\ .&.&\ldots&.\\\ .&.&\ldots&.\\\ .&.&\ldots&.\\\
0&0&\ldots&0\\\
-y_{1}y_{k}\overline{d}_{kn}&-y_{2}y_{k}\overline{d}_{kn}&\ldots&-y_{n}y_{k}\overline{d}_{kn}\end{bmatrix}$
$\displaystyle=\prod_{l\neq
k,n}(I_{n}+y_{l}\overline{d}_{kn}F_{knl})\cdot(I_{n}+y_{k}\overline{d}_{kn}F_{knk}-y_{n}\overline{d}_{kn}F_{knn}).$

Notice that all factors of the product $\prod_{l\neq
k,n}(I_{n}+y_{l}\overline{d}_{kn}F_{knl})$ belong to $S$ and

$\displaystyle
I_{n}+y_{k}\overline{d}_{kn}F_{knk}-y_{n}\overline{d}_{kn}F_{knn}\in S$

by Lemma 7 since $\mathrm{ldeg}(\overline{d}_{kn})\geq 1$ (this is the only
place where the restriction $\mathrm{ldeg}(a_{ij})\geq 3$ was used).
Consequently, $\delta_{k}\in S$.

Notice that

$\displaystyle\overline{M}\overline{\delta}_{1}\ldots\overline{\delta}_{n-1}\overline{M}^{-1}=\begin{bmatrix}I_{n-1}&\begin{matrix}y_{n}\overline{d}_{1n}\\\
.\\\ .\\\ .\\\ y_{n}\overline{d}_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&0\end{bmatrix}$

and

$\displaystyle\overline{M}(\overline{\delta}_{1}\ldots\overline{\delta}_{n-1})^{-1}\Phi\overline{M}^{-1}=\begin{bmatrix}[y_{n}\overline{d}_{ij}]&\begin{matrix}0\\\
.\\\ .\\\ .\end{matrix}\\\ \begin{matrix}0&\ldots\end{matrix}&0\end{bmatrix}.$

Let

$\displaystyle\epsilon=I_{n}+\begin{bmatrix}y_{n}\overline{d}_{11}&\ldots&y_{n}\overline{d}_{1(n-1)}&0\\\
.&\ldots&.&.\\\ .&\ldots&.&.\\\ .&\ldots&.&.\\\
y_{n}\overline{d}_{(n-1)1}&\ldots&y_{n}\overline{d}_{(n-1)(n-1)}&0\\\
-\sum_{j=1}^{n-1}y_{j}\overline{d}_{j1}&\ldots&-\sum_{j=1}^{n-1}y_{j}\overline{d}_{j(n-1)}&0\end{bmatrix}.$

By Proposition 2, $\epsilon\in S\cap SL_{n}(U,\sum_{j=s}^{n}y_{j}U)$, and

$\displaystyle M\epsilon
M^{-1}=\overline{M}\overline{\epsilon}\overline{M}^{-1}=\begin{bmatrix}[y_{n}\overline{d}_{ij}]&\begin{matrix}0\\\
.\\\ .\\\ .\end{matrix}\\\ \begin{matrix}0&\ldots\end{matrix}&0\end{bmatrix}.$

Consequently,

$\displaystyle\overline{M}\overline{\epsilon}^{-1}(\overline{\delta}_{1}\ldots\overline{\delta}_{n-1})^{-1}\overline{A}\
\overline{M}^{-1}$ $\displaystyle\overline{M}\overline{B}^{-1}\overline{A}\
\overline{M}^{-1}=I_{n},$

where $B=\delta_{1}\ldots\delta_{n-1}\epsilon$. This means that $B^{-1}A\in
S\cap SL_{n}(U,\sum_{j=s}^{n-1}y_{j}U)$. $\Box$

##  9\. Exceptional automorphism

Now we prove that the automorphism (3) is almost tame using the ideas of
approximation [8] and the results of the preceding sections. Recall that the
Jacobian matrix of (3) is the matrix (12).

######  Lemma 12.

The matrix (12) belongs to $S$.

Proof. We have

$\displaystyle A=I+y_{2}F_{231}-\lambda y_{1}F_{231}\in S.$

Let $\alpha=I+\lambda E_{12}$. Then $B=\alpha A^{\alpha}\alpha^{-1}\in S$.
Since $A$ and $\alpha$ are $3$-reducible it follows that $B$ is $3$-reducible
and

$\displaystyle
B_{3}=\alpha_{3}A_{3}^{\alpha}\alpha_{3}^{-1}=\begin{bmatrix}1&\lambda&0\\\
0&1&0\\\ 0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\\ (y_{2}-\lambda
y_{1})y_{3}&1&0\\\ -(y_{2}-\lambda
y_{1})y_{2}&0&1\end{bmatrix}^{\alpha}\begin{bmatrix}1&-\lambda&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}$ $\displaystyle=\begin{bmatrix}1&\lambda&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\\ y_{2}y_{3}&1&0\\\
-y_{2}(y_{2}+\lambda y_{1})&0&1\end{bmatrix}\begin{bmatrix}1&-\lambda&0\\\
0&1&0\\\ 0&0&1\end{bmatrix}$ $\displaystyle=\begin{bmatrix}1+\lambda
y_{2}y_{3}&-\lambda^{2}y_{2}y_{3}&0\\\ y_{2}y_{3}&1-\lambda y_{2}y_{3}&0\\\
-y_{2}(y_{2}+\lambda y_{1})&\lambda y_{2}(y_{2}+\lambda y_{1})&1\end{bmatrix}$

Set $C=(I-y_{2}F_{231})(I+\lambda^{2}y_{2}F_{132})B$. Then

$\displaystyle C_{3}=\begin{bmatrix}1&0&0\\\ -y_{2}y_{3}&1&0\\\
y_{2}^{2}&0&1\end{bmatrix}\begin{bmatrix}1&\lambda^{2}y_{2}y_{3}&0\\\ 0&1&0\\\
0&-\lambda^{2}y_{2}y_{1}&1\end{bmatrix}\begin{bmatrix}1+\lambda
y_{2}y_{3}&-\lambda^{2}y_{2}y_{3}&0\\\ y_{2}y_{3}&1-\lambda y_{2}y_{3}&0\\\
-y_{2}(y_{2}+\lambda y_{1})&\lambda y_{2}(y_{2}+\lambda y_{1})&1\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1+\lambda
y_{2}y_{3}+\lambda^{2}y_{2}^{2}y_{3}^{2}&-\lambda^{3}y_{2}^{2}y_{3}^{2}&0\\\
-\lambda y_{2}^{2}y_{3}^{2}(1+\lambda y_{2}y_{3})&1-\lambda
y_{2}y_{3}+\lambda^{3}y_{2}^{3}y_{3}^{3}&0\\\
\lambda(-y_{1}y_{2}+y_{2}^{3}y_{3})(1+\lambda y_{2}y_{3})&\lambda
y_{2}^{2}+\lambda^{3}y_{1}y_{2}^{2}y_{3}-\lambda^{3}y_{2}^{4}y_{3}^{2}&1\end{bmatrix}$
$\displaystyle=I_{3}+\lambda\begin{bmatrix}y_{2}y_{3}&0&0\\\
0&-y_{2}y_{3}&0\\\
-y_{1}y_{2}&y_{2}^{2}&0\end{bmatrix}+y_{2}^{2}y_{3}C^{\prime},$

where

$\displaystyle
C^{\prime}=\begin{bmatrix}\lambda^{2}y_{3}&-\lambda^{3}y_{3}&0\\\ -\lambda
y_{3}(1+\lambda y_{2}y_{3})&\lambda^{3}y_{2}y_{3}^{2}&0\\\
\lambda(y_{2}-\lambda y_{1}+\lambda
y_{2}^{2}y_{3})&\lambda^{3}y_{1}-\lambda^{3}y_{2}^{2}y_{3}&0\end{bmatrix}.$

Set also $D=(I+y_{2}F_{231})(I-\lambda^{2}y_{2}F_{132})B^{-1}$. Then

$\displaystyle D_{3}=\begin{bmatrix}1-\lambda
y_{2}y_{3}+\lambda^{2}y_{2}^{2}y_{3}^{2}&-\lambda^{3}y_{2}^{2}y_{3}^{2}&0\\\
-\lambda y_{2}^{2}y_{3}^{2}(1-\lambda y_{2}y_{3})&1+\lambda
y_{2}y_{3}-\lambda^{3}y_{2}^{3}y_{3}^{3}&0\\\
\lambda(y_{1}y_{2}+y_{2}^{3}y_{3})(1-\lambda y_{2}y_{3})&-\lambda
y_{2}^{2}+\lambda^{3}y_{1}y_{2}^{2}y_{3}+\lambda^{3}y_{2}^{4}y_{3}^{2}&1\end{bmatrix}$

and

$\displaystyle(23)_{3}D_{3}^{(23)}(23)_{3}=\begin{bmatrix}1-\lambda
y_{2}y_{3}+\lambda^{2}y_{2}^{2}y_{3}^{2}&0&-\lambda^{3}y_{2}^{2}y_{3}^{2}\\\
\lambda(y_{1}y_{3}+y_{2}y_{3}^{3})(1-\lambda y_{2}y_{3})&1&-\lambda
y_{3}^{2}+\lambda^{3}y_{1}y_{2}y_{3}^{2}+\lambda^{3}y_{2}^{2}y_{3}^{4}\\\
-\lambda y_{2}^{2}y_{3}^{2}(1-\lambda y_{2}y_{3})&0&1+\lambda
y_{2}y_{3}-\lambda^{3}y_{2}^{3}y_{3}^{3}\end{bmatrix}$
$\displaystyle=I_{3}+\lambda\begin{bmatrix}-y_{2}y_{3}&0&0\\\
y_{1}y_{3}&0&-y_{3}^{2}\\\
0&0&y_{2}y_{3}\end{bmatrix}+y_{2}y_{3}^{2}D^{\prime},$ $\displaystyle
D^{\prime}=\begin{bmatrix}\lambda^{2}y_{2}&0&-\lambda^{3}y_{2}\\\
\lambda(y_{3}-\lambda y_{1}-\lambda
y_{2}y_{3}^{2})&0&\lambda^{3}y_{1}+\lambda^{3}y_{2}y_{3}^{2}\\\ -\lambda
y_{2}(1-\lambda y_{2}y_{3})&0&-\lambda^{3}y_{2}^{2}y_{3}\end{bmatrix}.$

Let $E=C(23)D^{(23)}(23)\in S$. A direct calculation gives that

$\displaystyle
E_{3}=C_{3}(23)_{3}D_{3}^{(23)}(23)_{3}=(I_{3}+\lambda\begin{bmatrix}y_{2}y_{3}&0&0\\\
0&-y_{2}y_{3}&0\\\
-y_{1}y_{2}&y_{2}^{2}&0\end{bmatrix}+y_{2}^{2}y_{3}C^{\prime})$
$\displaystyle\times(I_{3}+\lambda\begin{bmatrix}-y_{2}y_{3}&0&0\\\
y_{1}y_{3}&0&-y_{3}^{2}\\\
0&0&y_{2}y_{3}\end{bmatrix}+y_{2}y_{3}^{2}D^{\prime})=I_{3}+\lambda\begin{bmatrix}0&0&0\\\
y_{1}y_{3}&-y_{2}y_{3}&-y_{3}^{2}\\\
-y_{1}y_{2}&y_{2}^{2}&y_{2}y_{3}\end{bmatrix}$
$\displaystyle+\lambda^{2}y_{2}y_{3}\begin{bmatrix}-y_{2}y_{3}&0&0\\\
-y_{1}y_{3}&0&y_{3}^{2}\\\
2y_{1}y_{2}&0&-y_{2}y_{3}\end{bmatrix}+(I_{3}+\lambda\begin{bmatrix}y_{2}y_{3}&0&0\\\
0&-y_{2}y_{3}&0\\\
-y_{1}y_{2}&y_{2}^{2}&0\end{bmatrix}+y_{2}^{2}y_{3}C^{\prime})y_{2}y_{3}^{2}D^{\prime}$
$\displaystyle+y_{2}^{2}y_{3}C^{\prime}+\lambda
y_{2}^{2}y_{3}C^{\prime}\begin{bmatrix}-y_{2}y_{3}&0&0\\\
y_{1}y_{3}&0&-y_{3}^{2}\\\
0&0&y_{2}y_{3}\end{bmatrix}=I_{3}+\lambda\begin{bmatrix}0&0&0\\\
y_{1}y_{3}&-y_{2}y_{3}&-y_{3}^{2}\\\
-y_{1}y_{2}&y_{2}^{2}&y_{2}y_{3}\end{bmatrix}+E^{\prime},$

where all elements of $E^{\prime}$ are divisible by $y_{3}$ and all elements
of the last column of $E^{\prime}$ are divisible by $y_{3}^{2}$. Let

$\displaystyle P=(I+\lambda y_{1}F_{231})(I-\lambda y_{2}F_{232}-\lambda
y_{3}F_{233})=I+\lambda y_{1}F_{231}-\lambda y_{2}F_{232}-\lambda
y_{3}F_{233}.$

Then

$\displaystyle P_{3}=I_{3}+\lambda\begin{bmatrix}0&0&0\\\
y_{1}y_{3}&-y_{2}y_{3}&-y_{3}^{2}\\\
-y_{1}y_{2}&y_{2}^{2}&y_{2}y_{3}\end{bmatrix}.$

Then

$\displaystyle P_{3}^{-1}E_{3}=(I_{3}-\lambda\begin{bmatrix}0&0&0\\\
y_{1}y_{3}&-y_{2}y_{3}&-y_{3}^{2}\\\
-y_{1}y_{2}&y_{2}^{2}&y_{2}y_{3}\end{bmatrix})(I_{3}+\lambda\begin{bmatrix}0&0&0\\\
y_{1}y_{3}&-y_{2}y_{3}&-y_{3}^{2}\\\
-y_{1}y_{2}&y_{2}^{2}&y_{2}y_{3}\end{bmatrix}+E^{\prime})$
$\displaystyle=I_{3}+(I_{3}-\lambda\begin{bmatrix}0&0&0\\\
y_{1}y_{3}&-y_{2}y_{3}&-y_{3}^{2}\\\
-y_{1}y_{2}&y_{2}^{2}&y_{2}y_{3}\end{bmatrix})E^{\prime}=I_{3}+P^{\prime}.$

Notice that all elements of $P^{\prime}$ are of lower degree $\geq 4$,
divisible by $y_{3}$, and all elements of the last column are divisible by
$y_{3}^{2}$. Thus the matrix $(3n)(P^{-1}E)^{(3n)}(3n)\in SL_{n}(U,y_{n}U)$
and satisfies the conditions of Lemma 11 for $s=n$. By Lemma 11, there exists
a matrix $Q$ such that $Q^{-1}(3n)(P^{-1}E)^{(3n)}(3n)\in SL_{n}(U,0)$, that
is, $Q^{-1}(3n)(P^{-1}E)^{(3n)}(3n)=I$. Consequently,
$(3n)(P^{-1}E)^{(3n)}(3n)=Q\in S$ and $P^{-1}E\in S$. Since $E\in S$ it
follows that $P\in S$. Notice that the first factor $I+\lambda y_{1}F_{231}$
of $P$ belongs to $S$. Consequently,

$\displaystyle(I-\lambda y_{1}F_{231})P=I-\lambda y_{2}F_{232}-\lambda
y_{3}F_{233}\in S.$

Conjugating the inverse of this matrix we can get the matrix (12). $\Box$

######  Corollary 11.

The automorphism (3) is almost tame.

##  10\. Elimination of a variable

In this section, we prove the main proposition to reduce the number of
variables in matrices from $G$ modulo $S$, repeating the proofs from [4, pages
99–100], with some minimal necessary changes. We present two lemmas to avoid
yet another use of inner automorphisms in [4].

######  Lemma 13.

There exists a $3$-reducible matrix $C$ in $T$ such that $C_{3}$ after setting
$y_{2}=y_{3}=0$ becomes equal to

$\displaystyle I_{3}+\lambda y_{1}E_{13}=I_{3}+\begin{bmatrix}0&0&0\\\
0&0&0\\\ \lambda y_{1}&0&0\end{bmatrix},$

where $\lambda\in K$.

Proof. Set

$\displaystyle A=I+F_{231},\ \alpha=I+\lambda E_{12},\
B=I-\lambda^{2}F_{132}.$

Then

$\displaystyle
A_{3}\alpha_{3}(A_{3}^{-1})^{\alpha}\alpha_{3}^{-1}=\begin{bmatrix}1&0&0\\\
y_{3}&1&0\\\ -y_{2}&0&1\end{bmatrix}\begin{bmatrix}1&\lambda&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\\ -y_{3}&1&0\\\
y_{2}&0&1\end{bmatrix}^{\alpha}\begin{bmatrix}1&-\lambda&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}$ $\displaystyle=\begin{bmatrix}1&0&0\\\ y_{3}&1&0\\\
-y_{2}&0&1\end{bmatrix}\begin{bmatrix}1&\lambda&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\\ -y_{3}&1&0\\\ y_{2}+\lambda
y_{1}&0&1\end{bmatrix}\begin{bmatrix}1&-\lambda&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}$ $\displaystyle=\begin{bmatrix}1-\lambda
y_{3}&\lambda^{2}y_{3}&0\\\ -\lambda y_{3}^{2}&1+\lambda
y_{3}+\lambda^{2}y_{3}^{2}&0\\\ \lambda y_{1}+\lambda y_{2}y_{3}&-\lambda
y_{2}-\lambda^{2}y_{1}-\lambda^{2}y_{2}y_{3}&1\end{bmatrix}$

and

$\displaystyle
B_{3}A_{3}\alpha_{3}(A_{3}^{-1})^{\alpha}\alpha_{3}^{-1}=\begin{bmatrix}1&-\lambda^{2}y_{3}&0\\\
0&1&0\\\ 0&\lambda^{2}y_{1}&1\end{bmatrix}\begin{bmatrix}1-\lambda
y_{3}&\lambda^{2}y_{3}&0\\\ -\lambda y_{3}^{2}&1+\lambda
y_{3}+\lambda^{2}y_{3}^{2}&0\\\ \lambda y_{1}+\lambda y_{2}y_{3}&-\lambda
y_{2}-\lambda^{2}y_{1}-\lambda^{2}y_{2}y_{3}&1\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1-\lambda
y_{3}+\lambda^{3}y_{3}^{3}&-\lambda^{3}y_{3}^{2}-\lambda^{4}y_{3}^{3}&0\\\
-\lambda y_{3}^{2}&1+\lambda y_{3}+\lambda^{2}y_{3}^{2}&0\\\ \lambda
y_{1}+\lambda y_{2}y_{3}-\lambda^{3}y_{1}y_{3}^{2}&-\lambda
y_{2}-\lambda^{2}y_{2}y_{3}+\lambda^{3}y_{1}y_{3}+\lambda^{4}y_{1}y_{3}^{2}&1\end{bmatrix}.$

If $C=BA\alpha(A^{-1})^{\alpha}\alpha^{-1}$ then $C\in T$ and
$C_{3}=B_{3}A_{3}\alpha_{3}(A_{3}^{-1})^{\alpha}\alpha_{3}^{-1}$ after setting
$y_{2}=y_{3}=0$ gives $I_{3}+\lambda y_{1}E_{13}$. $\Box$

######  Lemma 14.

Let

$\displaystyle I_{n}+[a_{ij}]\in SL_{n}(U,y_{1}U+y_{2}U)\cap G.$

Then $\mathrm{ldeg}(a_{ij})\geq 2$ for all $1\leq i,j\leq 2$.

Proof. Let

$\displaystyle A=[a_{ij}]=A_{1}+\ldots+A_{k},$

where all elements of $A_{i}$ are homogeneous of degree $i$ for all $1\leq
i\leq k$ and let $A_{1}=[b_{ij}]$. Since $a_{ij}\in y_{1}U+y_{2}U$ then
$b_{ij}\in Ky_{1}+Ky_{2}$. Moreover, since $YA=0$ by Lemma 2, we get
$YA_{1}=0$. By Lemma 2 there exist quadratic elements $f_{1},\ldots,f_{n}\in
M_{n}$ such that

$\displaystyle A_{1}=[b_{ij}]=[\partial(f_{1})\ldots\partial(f_{n})].$

Therefore all components of $\partial(f_{i})$ belong to $Ky_{1}+Ky_{2}$.

Let

$\displaystyle f_{i}=\sum_{k=1}^{n-1}\alpha_{k}[x_{n},x_{k}]+f_{i}^{\prime},$

where $f_{i}^{\prime}$ does not depend on $x_{n}$ and
$\alpha_{1},\ldots,\alpha_{n-1}\in K$. If $\alpha_{j}\neq 0$ for some $1\leq
j\leq n-1$ then

$\displaystyle\frac{\partial f_{i}}{\partial
x_{j}}=-\alpha_{j}x_{n}+\frac{\partial f_{i}^{\prime}}{\partial x_{j}}\notin
Ky_{1}+Ky_{2}.$

Consequently, $\alpha_{1}=\ldots=\alpha_{n-1}=0$ and $f_{i}$ does not depend
on $x_{n}$. The same discussions give that $f_{i}$ does not depend on
$x_{3},\ldots,x_{n}$. Hence $f_{i}=\lambda_{i}[x_{1},x_{2}]$ for all $i$. Then

$\displaystyle
A_{1}=\lambda_{1}F_{121}+\lambda_{2}F_{122}+\ldots+\lambda_{n}F_{12n}$

and the trace $\mathrm{tr}(A_{1})$ of $A_{1}$ is $\lambda_{1}y_{2}-\lambda
y_{1}$.

Recall that the determinat of $I+A$ is $1$. The homogeneous part of degree $1$
of the determinant of $I+A_{1}+\ldots+A_{k}$ is $\mathrm{tr}(A_{1})$.
Consequently, $\mathrm{tr}(A_{1})=0$. The equation $\lambda_{1}y_{2}-\lambda
y_{1}=0$ gives that $\lambda_{1}=\lambda_{2}=0$. This means that $b_{ij}=0$
and $\mathrm{ldeg}(a_{ij})\geq 2$ for all $1\leq i,j\leq 2$. $\Box$

######  Lemma 15.

Let

$\displaystyle A=I+[a_{ij}]\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)\cap G.$

Then there exists $R\in S$ such that $R^{-1}A=I+[e_{ij}]\in
SL_{n}(U,\sum_{j=s}^{n}y_{j}U)\cap G$ and
$e_{1n},e_{2n},\ldots,e_{n-1n}\in\sum_{j=s}^{n-1}y_{j}U+y_{n}^{2}U$.

Proof. For any $a\in U$ and any matrix $R$ over $U$ let $\overline{a}$ and
$\overline{R}$ be the result of setting $y_{s}=y_{s+1}=\ldots=y_{n-1}=0$ in
$a$ and $R$, respectively.

By Lemma 2,

$\displaystyle y_{1}a_{1j}+y_{2}a_{2j}+\ldots y_{n}a_{nj}=0,\ 1\leq j\leq n.$

Consequently,

$\displaystyle y_{1}\overline{a}_{1j}+\ldots
y_{s-1}\overline{a}_{(s-1)j}+y_{n}\overline{a}_{nj}=0,\ 1\leq j\leq n.$

Since $A=I+[a_{ij}]\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)$ it follows that
$a_{ij}\in\sum_{j=s}^{n}y_{j}U$ and, consequently, $\overline{a}_{ij}\in
y_{n}K[y_{1},\ldots,y_{s-1},y_{n}]$, that is,
$\overline{a}_{ij}=y_{n}\overline{b}_{ij}$ for some $\overline{b}_{ij}\in
K[y_{1},\ldots,y_{s-1},y_{n}]$. Therefore
$\overline{A}=I+[\overline{a}_{ij}]=I+[y_{n}\overline{b}_{ij}]$ and the
preceding equality gives that

$\displaystyle y_{1}\overline{b}_{1j}+\ldots
y_{s-1}\overline{b}_{(s-1)j}+y_{n}\overline{b}_{nj}=0,\ 1\leq j\leq n.$

If $(\overline{c}_{1n},\ldots,\overline{c}_{(n-1)n})$ is the result of setting
$y_{n}=0$ in $(\overline{b}_{1n},\ldots,\overline{b}_{(n-1)n})$ then from this
we get

(16) $\displaystyle y_{1}\overline{c}_{1n}+\ldots
y_{s-1}\overline{c}_{(s-1)n}=0,\ 1\leq j\leq n.$

First we show that there exists a matrix $R\in S\cap
SL_{n}(U,\sum_{j=s}^{n}y_{j}U)$ such that

$\displaystyle\overline{R}=C=\begin{bmatrix}I_{n-1}&\begin{matrix}y_{n}\overline{c}_{1n}\\\
.\\\ .\\\ .\\\ y_{n}\overline{c}_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}.$

belongs to $S$. In fact,

$\displaystyle
C=\begin{bmatrix}I_{n-1}&\begin{matrix}y_{n}\overline{c}_{1n}\\\ .\\\ .\\\
.\\\ y_{n}\overline{c}_{(s-1)n}\\\ 0\\\ .\\\ .\\\ .\\\ 0\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}\begin{bmatrix}I_{n-1}&\begin{matrix}0\\\
.\\\ .\\\ .\\\ 0\\\ y_{n}\overline{c}_{sn}\\\ .\\\ .\\\ .\\\
y_{n}\overline{c}_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}.$

The first factor belongs to $G$ by (16) and Lemma 2, and, consequently,
belongs to $S$. The second factor is equal to the product

$\displaystyle D=\prod_{k=s}^{n-1}(I+y_{n}\overline{c}_{kn}E_{kn})$

Let

$\displaystyle\overline{c}_{kn}=y_{1}f_{1}+\ldots+y_{s-1}f_{s-1}+\lambda,$

where $f_{i}\in K[y_{1},\ldots,y_{s-1}]$ and $\lambda\in K$. Then

$\displaystyle
y_{n}\overline{c}_{kn}=y_{1}y_{n}f_{1}+\ldots+y_{s-1}y_{n}f_{s-1}+y_{n}\lambda.$

Let

$\displaystyle P_{k}=\begin{bmatrix}I_{n-1}&\begin{matrix}-y_{k}y_{n}f_{1}\\\
.\\\ .\\\ .\\\ -y_{k}y_{n}f_{s-1}\\\ 0\\\
y_{1}y_{n}f_{1}+\ldots+y_{s-1}y_{n}f_{s-1}\\\ 0\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}\in S.$

Let $s=n-1$. Then $k=n-1$ and $a_{ij}\in y_{n-1}U+y_{n}U$. By Lemma 14,
$\mathrm{ldeg}(a_{ij})\geq 2$ for all $n-1\leq i,j\leq n$. By construction the
linear part of $y_{n}\overline{c}_{kn}$ is obtained from the linear part of
$a_{(n-1)n}$. Consequently, $\lambda=0$ and
$\overline{P}_{k}=I+y_{n}\overline{c}_{kn}E_{kn}$.

Let $s<n-1$, i.e., the set $\\{s,\ldots,n\\}$ contains at least $3$ elements.
Then, by Lemma 13, there exists a matrix $Q\in T$ such that

$\displaystyle\overline{Q}=I_{n}+\lambda y_{n}E_{kn}.$

Consequently, $QP\in S$ and $\overline{QP}=I+y_{n}\overline{c}_{kn}E_{kn}$.

Therefore in both cases there exists a matrix $\gamma_{i}\in S$ such that
$\overline{\gamma_{i}}=I+y_{n}\overline{c}_{kn}E_{kn}$. Then
$\overline{\gamma_{s}\ldots\gamma_{n-1}}=D$. Consequently, there exists a
matrix $R\in S\cap SL_{n}(U,\sum_{j=s}^{n}y_{j}U)$ such that $\overline{R}=C$.
Then $R^{-1}A\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)\cap G$. Let
$R^{-1}A=I_{n}+[e_{ij}]$. Recall that, by construction,
$\overline{a}_{kn}=y_{n}\overline{c}_{kn}+y_{n}^{2}\overline{d}_{kn}$ for all
$1\leq k\leq n-1$, where $\overline{d}_{kn}\in K[y_{1},\ldots,y_{s-1},y_{n}]$.
Then

$\displaystyle\overline{R^{-1}A}=C^{-1}(I+[\overline{a}_{ij}])=\begin{bmatrix}I_{n-1}&\begin{matrix}-y_{n}\overline{c}_{1n}\\\
.\\\ .\\\ .\\\ -y_{n}\overline{c}_{(n-1)n}\end{matrix}\\\
\begin{matrix}0&\ldots&0\end{matrix}&1\end{bmatrix}\left(I+\begin{bmatrix}[\overline{a}_{ij}]_{i,j\leq
n-1}&\begin{matrix}\overline{a}_{1n}\\\ .\\\ .\\\ .\\\
\overline{a}_{nn}\end{matrix}\end{bmatrix}\right)$
$\displaystyle=I+\begin{bmatrix}[\overline{a}_{ij}]_{i,j\leq
n-1}&\begin{matrix}\overline{a}_{1n}-y_{n}\overline{c}_{1n}\\\ .\\\ .\\\ .\\\
\overline{a}_{(n-1)n}-y_{n}\overline{c}_{(n-1)n}\\\
\overline{a}_{nn}\end{matrix}\end{bmatrix}=I+\begin{bmatrix}[\overline{a}_{ij}]_{i,j\leq
n-1}&\begin{matrix}y_{n}^{2}\overline{d}_{1n}\\\ .\\\ .\\\ .\\\
y_{n}^{2}\overline{d}_{(n-1)n}\\\ \overline{a}_{nn}\end{matrix}\end{bmatrix}$

Then $\overline{e}_{kn}=y_{n}^{2}\overline{d}_{kn}$. This means that
$e_{kn}\in\sum_{j=s}^{n-1}y_{j}U+y_{n}^{2}U$ for all $1\leq k\leq n-1$. The
matrix $R^{-1}A$ satisfies the statement of the lemma. $\Box$

For convenience, we formulate the following complete version of the lemma 11.

######  Lemma 16.

Let $A=I+[a_{ij}]\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)\cap G$ and
$a_{1n},a_{2n},\ldots,a_{(n-1)n}\in\sum_{j=s}^{n-1}y_{j}U+y_{n}^{2}U$. Then
there exists $B\in S$ such that $B^{-1}A\in SL_{n}(U,\sum_{j=s}^{n-1}y_{j}U)$.

Proof. Notice that the restriction $\mathrm{ldeg}(a_{ij})\geq 3$ was used only
once to apply that the matrix

$\displaystyle
I_{n}+y_{k}\overline{d}_{kn}F_{knk}-y_{n}\overline{d}_{kn}F_{knn}$

belong to $S$ by Lemma 7 since $\mathrm{ldeg}(\overline{d}_{kn})\geq 1$. By
Lemma 13, it belongs to $S$ anyway. Repeating the proof of Lemma 11, we get
the statement of the lemma. $\Box$

######  Proposition 3.

Let

$\displaystyle A=I+[a_{ij}]\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)\cap G.$

Then there exists $B\in S$ such that $B^{-1}A\in
SL_{n}(U,\sum_{j=s}^{n-1}y_{j}U)$.

Proof. By Lemma 15, there exists a matrix $C\in S$ such that
$C^{-1}A=I+[c_{ij}]\in SL_{n}(U,\sum_{j=s}^{n}y_{j}U)$ and
$c_{1n},c_{2n},\ldots,c_{n-1n}\in\sum_{j=s}^{n-1}y_{j}U+y_{n}^{2}U$. By Lemma
16, there exists a matrix $D\in S$ such that $D^{-1}C^{-1}A\in
SL_{n}(U,\sum_{j=s}^{n-1}y_{j}U)$. $\Box$

##  11\. Main results

######  Theorem 1.

Every automorphism of a free metabelian Lie algebra $M_{n}$ of rank $n\geq 4$
over an arbitrary field $K$ is almost tame.

Proof. First we show that $S=G$. Let $A\in G$. Then

$\displaystyle A\in SL_{n}(U,\sum_{j=1}^{n}y_{j}U).$

By Proposition 3, there exist matrices $B_{1},\ldots,B_{n}\in S$ such that

$\displaystyle B_{s}^{-1}\ldots B_{1}^{-1}A\in
SL_{n}(U,\sum_{j=s+1}^{n}y_{j}U)$

for all $s<n$ and

$\displaystyle B_{n}^{-1}\ldots B_{1}^{-1}A\in SL_{n}(U,0)=\\{I\\}.$

Consequently,

$\displaystyle B_{n}^{-1}\ldots B_{1}^{-1}A=I$

and

$\displaystyle A=B_{1}\ldots B_{n}\in S.$

This proves that $S=G$. Therefore
$\mathrm{ATAut}(M_{n})\cap\mathrm{IAut}(M_{n})=\mathrm{IAut}(M_{n})$. It
follows that $\mathrm{ATAut}(M_{n})=\mathrm{Aut}(M_{n})$ since both groups
contain all linear automorphisms. $\Box$

######  Theorem 2.

The group of all automorphisms $\mathrm{Aut}(M_{n})$ of a free metabelian Lie
algebra $M_{n}$ of rank $n\geq 4$ over a field $K$ of characteristic $\neq 3$
is generated by all linear automorphisms, as well as a quadratic automorphism
(1) and a cubic automorphism (2).

######  Corollary 12.

The group of all automorphisms $\mathrm{Aut}(M_{n})$ of a free metabelian Lie
algebra $M_{n}$ of rank $n\geq 5$ over any field $K$ is generated by all
linear automorphisms, as well as a quadratic automorphism (1) and a cubic
automorphism (2).

Theorem 2 and Corollary 12 are direct corollaries of Corollary 4 and Theorem
1.

## Acknowledgments

The author would like to thank the Max Planck Institute für Mathematik for its
hospitality and excellent working conditions, where some part of this work has
been done. The author thanks Professors V. Dotsenko, M. Jibladze, L. Makar-
Limanov, and I. Shestakov for useful discussions.

The project is supported by the grant AP14872073 of the Ministry of Education
and Science of the Republic of Kazakhstan.

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