# Results on ${\textrm{K}_{1}}$ of general quadratic groups

Rabeya Basu and Kuntal Chakraborty  Rabeya Basu – Indian Institute of Science
Education and Research (IISER) Pune, India<EMAIL_ADDRESS><EMAIL_ADDRESS>Kuntal Chakraborty – Indian Institute of Science
Education and Research (IISER) Pune, India<EMAIL_ADDRESS>
###### Abstract

In the first part of this article, we discuss the relative cases of
Quillen–Suslin’s local-global principle for the general quadratic (Bak’s
unitary) groups, and its applications for the (relative) stable and unstable
${\textrm{K}_{1}}$-groups. The second part is dedicated to the graded version
of the local-global principle for the general quadratic groups and its
application to deduce a result for Bass’ nil groups.

###### keywords:

Graded rings, Form rings, Bak’s unitary groups, Bass’ nil groups

11E57, 13A02, 13C10, 15A63, 19B14, 19D35, 20F18 32 2024 1
https://doi.org/10.46298/cm.9855

##  1 Introduction

In [RB] and [basu2], the first author has discussed many results in classical
K-theory for the absolute cases, related to the Serre’s problem on projective
modules. In this article, following the tricks used in [BRK], we are going to
consider some problems for the relative cases; viz Quillen–Suslin’s local-
global principle of the transvection subgroups,
${\textrm{K}_{1}}$-stabilization and the structure of unstable
${\textrm{K}_{1}}$-groups of the general quadratic (Bak’s unitary) groups over
associative rings which are finite over the center. For previous results on
these problems we refer to the works of Bak–Basu–Rao–Khanna for the local-
global principle (L-G principle) in [brk], [BBR], [BRK], Bak–Petrov–Tang for
${\textrm{K}_{1}}$-stabilization in [BTP], and Bak–Harzat–Vavilov for
solvability of unstable ${\textrm{K}_{1}}$-groups in [Bak], [HV], [HR]. For
the linear case, the graded version of L-G principle was studied by Chouinard
in [C], and by Gubeladze in [gubel], [gubel2]. In [BS], the first author and
M.K. Singh deduced an analog for the traditional classical groups. In the
second part of the article, we have deduced an analog for the transvection
subgroups of the general quadratic groups over graded rings. As an
application, by using a recent result on Higman linearization due to V.
Kepoeiko (cf. [V], [V1]), we could revisit a problem on absence of torsion in
Bass’ nil group of the general quadratic groups, and its graded analog. We
refer to [STI], [BL1], [WEL], [WEL1], and [RB] for previous results in this
direction.

##  2 Preliminaries

Let us recall some necessary definitions and the key lemmas.

######  Definition 2.1 (cf. [Bak1]).

Let $R$ be an (not necessarily commutative) associative ring with identity,
and with involution $-:R\rightarrow R$, $a\mapsto\overline{a}$. Let
$\lambda\in C(R)$ = center of $R$ be an element with the property
$\lambda\overline{\lambda}=1$. We define additive subgroups of $R$:

$\Lambda_{max}=\\{a\in R\mid a=-\lambda\overline{a}\\}~{}\text{and
}~{}\Lambda_{min}=\\{a-\lambda\overline{a}\mid a\in R\\}.$

One checks that $\Lambda_{max}$ and $\Lambda_{min}$ are closed under the
conjugation operation $a\mapsto\overline{x}ax$ for any $x\in R$. A
$\lambda$-form parameter on $R$ is an additive subgroup $\Lambda$ of $R$ such
that $\Lambda_{min}\subseteq\Lambda\subseteq\Lambda_{max}$, and
$\overline{x}\Lambda x\subseteq\Lambda$ for all $x\in R$. A pair $(R,\Lambda)$
is called a form ring.

######  Remark 2.2.

For a from ring $(R,\Lambda)$, we can extend the involution $-:R\to R$ to an
involution $-:R[X]\to R[X]$ by setting $\overline{X}=X$.

######  Lemma 2.3.

Let $(R,\Lambda,\lambda)$ be a form ring. Then $(R[X],\Lambda[X],\lambda)$ is
also a form ring with respect to the involution obtained by extending, as in
the previous remark.

######  Proof 2.4.

Let us consider the subgroups $\Lambda_{min}(R[X])$ and $\Lambda_{max}(R[X])$.
It can be checked that $\Lambda_{min}(R[X])=\Lambda_{min}(R)[X]$, and
$\Lambda_{max}(R[X])=\Lambda_{max}(R)[X]$. Hence it follows

$\Lambda_{min}(R[X])\subseteq\Lambda[X]\subseteq\Lambda_{max}(R[X]).$

To prove that it is closed under conjugation, let us consider elements
$a(X)\in R[X]$ and $b(X)\in\Lambda[X]$. Using double induction on the degrees
of $a(X)$ and $b(X)$, we will show that

$\overline{a(X)}b(X)a(X)\in\Lambda[X].$

We first show that the statement is true for $\deg(b(X))=0$. Assume
$b(X)=b\in\Lambda$. Hence we need to prove $\overline{a(X)}ba(X)\in\Lambda[X]$
for all $a(X)\in R[X]$. Clearly, this is obvious when $\deg(a(X))=0$. For a
linear polynomial $a(X)=a_{0}+a_{1}X$,

$\displaystyle\overline{a(X)}ba(X)$
$\displaystyle=(\overline{a}_{0}+\overline{a}_{1}X)b(a_{0}+a_{1}X)$
$\displaystyle=\overline{a}_{0}ba_{0}+(\overline{a}_{0}ba_{1}+\overline{a}_{1}ba_{0})X+\overline{a}_{1}ba_{1}X^{2}$
$\displaystyle=\overline{a}_{0}ba_{0}+\overline{(a_{0}+a_{1})}b(a_{0}+a_{1})X-(\overline{a}_{0}ba_{0}+\overline{a}_{1}ba_{1})X+\overline{a}_{1}ba_{1}X^{2}.$

Hence the statement is true for $\deg(b(X))=0$ and $\deg(a(X))=1$. Assume the
statement for $\deg(b(X))=0$ is true for all $a(X)$ with $\deg(a(X))<m$.
Suppose $\deg(a(X))=m$. Then, $a(X)=a_{m-1}(X)+a_{m}X^{m}$, where $a_{m-1}(X)$
is a polynomial of degree at most $m-1$, and $a_{m}\in R$. Hence, we have

$\displaystyle\overline{a(X)}ba(X)=$
$\displaystyle(\overline{a_{m-1}(X)}+\overline{a}_{m}X^{m})b(a_{m-1}(X)+a_{m}X^{m})$
$\displaystyle=$
$\displaystyle\overline{a_{m-1}(X)}ba_{m-1}(X)+\overline{a_{m-1}(X)}ba_{m}X^{m}+\overline{a}_{m}ba_{m-1}(X)X^{m}+\overline{a}_{m}ba_{m}X^{2m}.$

Now

$\displaystyle\overline{a_{m-1}(X)}ba_{m}+\overline{a}_{m}ba_{m-1}(X)=$
$\displaystyle=\overline{(a_{m-1}(X)+a_{m})}b(a_{m-1}(X)+a_{m})-(\overline{a_{m-1}(X)}ba_{m-1}(X)+\overline{a}_{m}ba_{m}).$

Hence by induction it follows that $\overline{a(X)}ba(X)\in\Lambda[X]$. Assume
the statement is true for $\deg(b(X))<n$. Let $b^{\prime}(X)$ be a polynomial
in $\Lambda[X]$ of degree $n$. Then we can write
$b^{\prime}(X)=b^{\prime\prime}(X)+b_{n}X^{n}$, where $b^{\prime\prime}(X)$ is
a polynomial in $\Lambda[X]$ of degree at most $n-1$ and $b_{n}\in\Lambda$.
Hence by induction

$\displaystyle\overline{a(X)}b^{\prime}(X)a(X)=$
$\displaystyle\overline{a(X)}(b^{\prime\prime}(X)+b_{n}X^{n})a(X)$
$\displaystyle=$
$\displaystyle\overline{a(X)}b^{\prime\prime}(X)(a(X))+(\overline{a(X)}b_{n}a(X))X^{n}\in\Lambda[X].$

To define Bak’s unitary group or the general quadratic group, we fix a (non-
zero) central element $\lambda\in R$ with $\lambda\overline{\lambda}=1$, and
then consider the form

$\psi_{n}=\begin{pmatrix}0&{\operatorname{I}}_{n}\\\
\lambda{{\operatorname{I}}}_{n}&0\end{pmatrix}.$

Bak’s Unitary or General Quadratic Groups ${\operatorname{GQ}}$:

${\operatorname{GQ}}(2n,R,\lambda)~{}=~{}\\{\sigma\in{\operatorname{GL}}(2n,R)\,|\,\overline{\sigma}\psi_{n}\sigma=\psi_{n}\\}.$

### Elementary Quadratic Matrices:

Let $\rho:\\{1,2,\ldots,n\\}\to\\{n+1,n+2,\ldots,2n\\}$ be defined by
$\rho(i)=n+i$. Let $e_{ij}$ be the matrix with $1$ in the $ij$-th position and
$0$’s elsewhere. For $a\in R$, and $1\leq i,j\leq n$, we define

$q\varepsilon_{ij}(a)={\operatorname{I}}_{2n}+ae_{ij}-\overline{a}e_{\rho(j)\rho(i)}$
for $i\neq j$,

$qr_{ij}(a)=\left\\{\begin{array}[]{ll}{\operatorname{I}}_{2n}+ae_{i\rho(j)}-\lambda\overline{a}e_{j\rho(i)}&\text{for}~{}i\neq
j\\\
{\operatorname{I}}_{2n}+ae_{\rho(i)j}&\text{for}~{}i=j,\end{array}\right.$
$ql_{ij}(a)=\left\\{\begin{array}[]{ll}{\operatorname{I}}_{2n}+ae_{\rho(i)j}-\overline{\lambda}\overline{a}e_{\rho(j)i}&\text{for}~{}i\neq
j\\\
{\operatorname{I}}_{2n}+ae_{\rho(i)j}&\text{for}~{}i=j.\end{array}\right.$

(Note that for the second and third type of elementary matrices, if $i=j$,
then we get $a=-\lambda\overline{a}$, and hence it forces that
$a\in\Lambda_{max}(R)$. One checks that these above matrices belong to
${\operatorname{GQ}}(2n,R,\Lambda)$; cf. [Bak1].) $n$-th Elementary Quadratic
Group ${\operatorname{EQ}}(2n,R,\Lambda)$: The subgroup generated by
$q\varepsilon_{ij}(a),qr_{ij}(a)\text{and }ql_{ij}(a)$, for $a\in R$ and
$1\leq i,j\leq n$. For uniformity we denote the elementary generators of
${\operatorname{EQ}}(2n,R,\Lambda)$ by $\eta_{ij}(*)$. It is clear that the
stabilization map takes generators of ${\operatorname{EQ}}(2n,R,\Lambda)$ to
the generators of ${\operatorname{EQ}}(2(n+1),R,\Lambda)$.

######  Remark 2.5.

Throughout this article we shall assume that all ideals of $R$ are involution-
invariant, i.e., if $J$ is an ideal, then $\overline{J}=J$. We also assume
that $2n\geq 6$.

######  Definition 2.6.

The relative general quadratic subgroup of ${\operatorname{GQ}}(2n,R,\Lambda)$
with respect to the ideal $J$ is defined by the set
$\\{\alpha\in{\operatorname{GQ}}(2n,R,\Lambda)\mid\alpha\equiv{\rm
I}_{2n}\pmod{J}\\}$ and it is denoted by
${\operatorname{GQ}}(2n,R,\Lambda,J)$.

######  Definition 2.7.

Let $(R,\Lambda)$ be a form ring and $J\subset R$ be an ideal. The subgroup of
${\operatorname{GQ}}(2n,R,\Lambda)$ generated by the matrices of the form
$\eta_{ij}(x)\eta_{ji}(a)\eta_{ij}(x)^{-1}$, where $x\in R$ and $a\in J$, is
called the relative elementary subgroup and is denoted by
${\operatorname{EQ}}(2n,R,\Lambda,J)$.

######  Notation 2.7.

A row $(a_{1},a_{2},\dots,a_{n})\in R^{n}$ is said to be unimodular if there
exists a vector $(b_{1},b_{2},\dots,b_{n})\in\nolinebreak R^{n}$ such that
$\sum_{i=1}^{n}a_{i}b_{i}=1$. The set of all unimodular rows of length $n$ is
denoted by ${\rm Um}_{n}(R)$. We denote the set of all unimodular rows of
length $n$ which are congruent to $e_{1}=(1,0,\dots,0)$ modulo the ideal $J$
by ${\rm Um}_{n}(R,J)$. For an ideal $J\subset R$ the extended ideal
$J\otimes_{R}R[X]$ of $R[X]$ is denoted by $J[X]$. We will mostly use
localizations with respect to two types of multiplicatively closed subsets of
$R$, viz. $S=\\{1,s,s^{2},\dots\\}$, where $s\in R$ is a non-nilpotent non-
zero divisor; and $S=R\setminus\mathfrak{m}$ for some
$\mathfrak{m}\in\operatorname{Max}(R)$. By $J_{s}[X]$ and
$J_{\mathfrak{m}}[X]$ we shall mean the extension of $J[X]$ in $R_{s}[X]$ and
$R_{\mathfrak{m}}[X]$ respectively.

######  Definition 2.8.

Let $(R,\Lambda)$ be a form ring and $J\subset R$ be an ideal of $R$. The
excision ring of $R$ with respect to the ideal $J$ is denoted by $R\oplus J$
and is defined by the set $\\{(r,i)\mid r\in R,i\in J\\}$ with the addition
defined by $(r,i)+(s,j)=(r+s,i+j)$, and the multiplication defined by
$(r,i)(s,j)=(rs,rj+is+ij)$. We can extend the involution $-$ to the ring
$R\oplus J$ by setting $\overline{(r,i)}=(\overline{r},\overline{i})$. The
element $(\lambda,0)\in R\oplus J$ satisfies
$(\lambda,0)(\overline{\lambda},0)=(1,0)$, where $(1,0)$ is the identity
element of $R\oplus J$. We can observe that the additive subgroup
$(\Lambda\oplus J)\cap\Lambda_{max}(R\oplus J)$ satisfies the properties of
$(\lambda,0)$-parameter on $R\oplus J$. We fix the notation $\Gamma\oplus J$
for the subgroup $(\Lambda\oplus J)\cap\Lambda_{max}(R\oplus J)$. Hence we get
the form ring $(R\oplus J,\Gamma\oplus J)$.

Recall that there is a natural map $f:R\oplus J\rightarrow R$ given by
$f(r,i)=r+i$. This map induces a canonical homomorphism on
${\operatorname{GQ}}(2n,R\oplus J,\Gamma\oplus J)$. We shall use the same
notation $f$ to denote this map. The following two key lemmas are proved in
[BRK] for the traditional classical groups. Proofs are similar for the general
quadratic groups.

######  Lemma 2.9.

For $\alpha\in{\operatorname{EQ}}(2n,R,\Lambda,J)$, there exists a matrix
$\widetilde{\alpha}\in{\operatorname{EQ}}(2n,R\oplus J,\Gamma\oplus J)$ such
that $f(\widetilde{\alpha})=\alpha$.

######  Lemma 2.10.

For $\alpha\in{\operatorname{GQ}}(2n,R,\Lambda,J)$, there exists a matrix
$\widetilde{\alpha}\in{\operatorname{GQ}}(2n,R\oplus J,\Gamma\oplus J)$ such
that $f(\widetilde{\alpha})=\alpha$.

##  3 Relative L-G Principle for the Transvection Subgroups

It is known that any module finite ring (i.e., finite over its center) $R$ can
be written as a direct limit of its finitely generated subrings. Also,
${\operatorname{G}}(R,\lambda)=\underset{\longrightarrow}{\lim}\,{\operatorname{G}}(R_{i},\lambda_{i})$,
where the limit is taken over all finitely generated subring of $R$. Hence, we
can assume that $C(R)$ is Noetherian. For the rest of this section, we shall
consider $R$ to be a module finite ring with identity. The local-global
principle for the transvection subgroups (absolute cases) of the full
automorphism groups was established in [BBR] for the traditional classical
groups. Then, in [basu2] it was generalized for the general quadratic groups.
In this section we deduce the relative L-G principle for the transvection
subgroups. The following results are proved in [BRK] for traditional classical
groups of free modules, and the steps of the proof for the general quadratic
groups are identical. Therefore, we state these results without proof. For any
column vector $v\in(R^{2n})^{t}$ we consider the row vector
$\widetilde{v}=\overline{v}^{t}\psi_{n}$.

######  Definition 3.1.

We define a map $M:(R^{2n})^{t}\times(R^{2n})^{t}\rightarrow M(2n,R)$ and the
inner product $\langle\>,\>\rangle$ as follows:

$M(v,w)=v.\widetilde{w}-\overline{\lambda}\overline{w}\widetilde{v},~{}\langle
v,w\rangle=\widetilde{v}.w$

######  Lemma 3.2.

Let $(R,\Lambda)$ be a form ring and
$v\in{\operatorname{EQ}}(2n,R,\Lambda,J)e_{1}$. Let $w\in J^{2n}$ be a column
vector such that $\langle v,w\rangle=0$. Then ${\rm
I}_{2n}+M(v,w)\in{\operatorname{EQ}}(2n,R,\Lambda,J)$.

######  Theorem 3.3 (Relative L-G principle).

Let $R$ be a ring and $J\subset R$ be an ideal of $R$. Let
$\alpha(X)\in{\operatorname{GQ}}(2n,R[X],\Lambda[X],J[X])$, with
$\alpha(0)={\rm I}_{2n}$ be such that for every maximal ideal
$\mathfrak{m}\in{\rm Max}(C(R))$, we have
$\alpha_{\mathfrak{m}}(X)\in{\operatorname{EQ}}(2n,R_{\mathfrak{m}}[X],\Lambda_{\mathfrak{m}}[X],J_{\mathfrak{m}}[X])$.
In that case,
$\alpha(X)\in\nolinebreak{\operatorname{EQ}}(2n,R[X],\Lambda[X],J[X])$.

We recall some definitions and fix notations.

######  Definition 3.4.

Let $(R,\Lambda)$ be a form ring and $P$ be a right $R$-module. A map
$f:P\times P\to R$ is said to be a sesquilinear form if
$f(pa,qb)=\overline{a}f(p,q)b$ for all $p,q\in P$ and $a,b\in R$. A map
$q:P\to R/\Lambda$ is said to be a quadratic form if $q(p)=f(p,p)+\Lambda$,
where $f$ is a sesquilinear form on $P$. With respect to a sesquilinear form
on $P$, we can define an associated $\lambda$-Hermitian form $h:P\times P\to
R$ by $h(p,q)=f(p,q)+\lambda\overline{f(q,p)}$. The triplet $(P,h,q)$ is
called a quadratic module.

######  Definition 3.5.

Let $(P,q,h)$ be a quadratic module and ${\operatorname{GL}}(P)$ be the full
automorphism group of $P$. The quadratic module $P$ is said to be non-singular
if $P$ is a projective $R$-module and the associated $\lambda$-Hermitian form
is non-singular. For a non-singular $P$, the general quadratic group of $P$ is
defined as follows:

${\operatorname{GQ}}(P,\Lambda,q,h)=\\{\alpha\in{\operatorname{GL}}(P)\mid
h(\alpha u,\alpha v)=h(u,v),q(\alpha u)=q(u)\\}$

We will denote by ${\operatorname{GQ}}(P,\Lambda,J)$ the set
$\\{\alpha\in{\operatorname{GQ}}(P,\Lambda)\mid\alpha\equiv{\rm
Id}\pmod{JP}\\}$.

######  Definition 3.6.

Let $(P,h,q)$ be a quadratic module over $(R,\Lambda)$ and $J\subset R$ be an
ideal of $R$. Let $u,v\in P$ and $a\in R$ be such that $f(u,u)\in\Lambda$,
$h(u,v)=0$ and $f(v,v)=a\pmod{\Lambda}$. Then the transvection map
$\sigma=\sigma_{u,v,a}:P\rightarrow P$ is defined by

$\sigma(x)=x+uh(v,x)-v\overline{\lambda}h(u,x)-u\overline{\lambda}ah(u,x).$

The set of all transvections of $P$ will be denoted as
${\operatorname{T}}(P,\Lambda)$. A map
$\sigma\in{\operatorname{T}}(P,\Lambda)$ is said to be a transvection relative
to $J$ if either $u$ or $v$ belongs to the submodule $JP$. The set of all
transvections relative to the ideal $J$ will be denoted by
${\operatorname{T}}(P,\Lambda,J)$.

######  Definition 3.7.

Let $(P,h,q)$ be a quadratic module over a form ring $(R,\Lambda)$ and
$J\subset R$ be an ideal of $R$. Let $Q$ be the quadratic module
$P\perp\mathbb{H}(R)$, where $\mathbb{H}(R)$ denotes the hyperbolic form
$R\perp R^{*}$. Then the transvections which are of the form

$q=(p,a,b)\mapsto(p-aq,a,b+h(p,q)),$

or,

$q=(p,a,b)\mapsto(p-bq,a+h(p,q),b),$

where $a\in R$, $b\in R^{*},p,q\in P$, are called elementary transvections.
The set of all elementary transvections is denoted by
${\operatorname{ET}}(Q,\Lambda)$. An elementary transvection is said to be
elementary transvection relative to $J$ if $q\in JQ$. The subgroup of
${\operatorname{ET}}(Q,\Lambda)$ generated by elementary transvections
relative to $J$ is denoted by ${\operatorname{ET}}(JQ)$. And we denote
${\operatorname{ET}}(Q,\Lambda,JQ)$ by the normal closure of
${\operatorname{ET}}(JQ)$ in ${\operatorname{ET}}(Q,\Lambda)$. We also use the
notation ${\operatorname{ET}}(Q,\Lambda,J)$ for the group
${\operatorname{ET}}(Q,\Lambda,JQ)$.

######  Notation 3.7.

Let $P$ be a finitely generated quadratic $R$ module of rank $2n$ with a fixed
form $\langle\>,\>\rangle$. Denote by $Q$ the module $P\perp\mathbb{H}(R)$and
by $Q[X]$ the module $(P\perp\mathbb{H}(R))[X]$. We assume that the rank of
the quadratic module is $2n\geq 6$.

We shall also assume the following two hypotheses: $(H1)$ For every maximal
ideal $\mathfrak{m}$ of $R$, the quadratic module $Q_{\mathfrak{m}}$ is
isomorphic to $R_{\mathfrak{m}}^{2n+2}$ for the standard bilinear form
$\mathbb{H}(R_{\mathfrak{m}}^{n+1})$. $(H2)$ For every non-nilpotent $s\in R$,
if the projective module $M_{s}$ is a free $R_{s}$-module, then the quadratic
module $M_{s}$ is isomorphic to $R_{s}^{2n+2}$ for the standard bilinear form
$\mathbb{H}(R^{n+1})$.

######  Lemma 3.8 ([RB, Lemma 5.7], [BRK, Lemma 3.6]).

The group

${\operatorname{GQ}}(2n,R[X],\Lambda[X],(X))\cap{\operatorname{EQ}}(2n,R[X],\Lambda[X],J[X])$

is generated by the elements of the form

$\varepsilon\eta_{ij}(Xh(X))\varepsilon^{-1},$

where $\varepsilon\in{\operatorname{EQ}}(2n,R,\Lambda)$, $h(X)\in\nolinebreak
J[X]$ and $\eta_{ij}(*)$ are the elementary generators of the group
${\operatorname{EQ}}(2n,R[X],\Lambda[X])$.

######  Lemma 3.9 ([BRK, Corollary 3.8]).

If $\eta=\eta_{1}\eta_{2}\dots\eta_{r}$, where each $\eta_{j}$ is an
elementary generator, and $h(Y)\in J[Y]$, then there are elements
$h_{t}(X,Y)\in J[X,Y]$ such that

$\eta\eta_{pq}(X^{2^{r}m}h(Y))\eta^{-1}=\prod_{t=1}^{k}\eta_{p_{t}q_{t}}(X^{m}h(X,Y)).$

We now recall some standard results:

######  Lemma 3.10.

Let $R$ be a ring and $K$ a finitely presented left (right) $R$-module, and
let $L$ be any left (right) $R$-module. Then we have a natural isomorphism:

$f:{\rm Hom}_{R}(K,L)[X]\rightarrow{\rm Hom}_{R[X]}(K[X],L[X]).$

######  Lemma 3.11.

Let $S$ be a multiplicatively close subset of a ring $R$. Let $K$ be a
finitely presented $R$-module and $L$ be any $R$-module. Then we have a
natural isomorphism

$g:S^{-1}({\rm Hom}_{R}(K,L))\rightarrow{\rm Hom}_{S^{-1}R}(S^{-1}K,S^{-1}L).$

The following lemma is used frequently (sometimes in a subtle way) in the
proof of the main results.

######  Lemma 3.12 (cf. [HV, Lemma 5.1]).

Let $A$ be Noetherian ring and $0\neq s\in A$. Then there exists a natural
number $k$ such that the homomorphism

${\operatorname{G}}(A,s^{k}A,s^{k}\lambda)\rightarrow{\operatorname{G}}(A_{s},\lambda_{s})$

induced by the localization homomorphism $A\rightarrow A_{s}$ is injective.

Now we prove the relative L-G principle by using Lemma 2.9.

######  Proposition 3.13 (Relative Dilation Principle).

Let $R$ be an almost commutative ring (i.e., an associative ring which is
finite over its center $C(R)$) and $J\subset R$ be an ideal. Let $P$ and $Q$
be as in 3. Let $s$ be a non-nilpotent element of $R$ such that $P_{s}$ is a
free module. Let $\sigma(X)\in{\operatorname{GQ}}(Q[X],\Lambda[X],J[X])$ with
$\sigma(0)=Id$and suppose additionally that
$\sigma_{s}(X)\in{\operatorname{EQ}}(2n+2,R_{s}[X],\Lambda_{s}[X],J_{s}[X])$.
Then there exists
$\widehat{\sigma}(X)\in{\operatorname{ET}}(Q[X],\Lambda[X],J[X])$ and $l>0$
such that $\widehat{\sigma}(X)$ localizes to $\sigma(bX)$ for some
$b\in(s^{l})$ and $\widehat{\sigma}(0)=Id$.

######  Proof 3.14.

Since elementary transvections can always be lifted, then we may assume that
$R$ is reduced. We will show that there exists $l>0$ such that

$\sigma(bX)\in{\operatorname{ET}}(Q[X],\Lambda[X],J[X])$

for all $b\in(s^{k})$ and for all $k\geq l$. As $\sigma(0)={\rm Id}$, by Lemma
3.8, we can write
$\sigma_{s}(X)=\prod_{k}\gamma_{k}\eta_{i_{k}j_{k}}(X\lambda_{k}(X))\gamma_{k}^{-1}$,
where $\gamma_{k}\in{\operatorname{EQ}}(2n+2,R_{s},\Lambda_{s})$, and
$\lambda_{k}(X)\in J_{s}[X]$. Hence, by the proof of Lemma 2.9, there exists
$\widetilde{\sigma}_{(s,0)}(X)\in{\operatorname{EQ}}(2n+2,(R_{s}\oplus
J_{s})[X],(\Gamma_{s}\oplus J_{s})[X])$ such that

$\widetilde{\sigma}_{(s,0)}(X)=\prod_{k}\widetilde{\gamma_{k}}\widetilde{\eta_{i_{k}j_{k}}}(0,X\lambda_{k}(X))\widetilde{\gamma_{k}}^{-1},$

where $\phi_{s}(\widetilde{\gamma_{k}})=\gamma_{k}$,
$(0,X\lambda_{k}(X))\in(R\oplus J)_{(s,0)}[X]$, the
$\widetilde{\eta_{i_{k}j_{k}}}(*)$ are the elementary generators of
${\operatorname{EQ}}(2n+2,(R_{s}\oplus J_{s})[X],(\Gamma_{s}\oplus J_{s})[X])$
and $\phi:R\oplus J\rightarrow R$ is defined by $\phi((a,i))=a+i$ and
$\Gamma_{s}\oplus J_{s}=(\Gamma\oplus J)_{s}$ is the localization of
$\Gamma\oplus J$ with respect to the set $\\{1,s,s^{2},\dots\\}$. Hence, for
$d>0$, we have that

$\widetilde{\sigma}_{(s,0)}(XT^{2d})=\prod_{k}\widetilde{\gamma}_{k}\widetilde{\eta_{i_{k}j_{k}}}((0,XT^{2d}\lambda_{k}(XT^{2d}))\widetilde{\gamma}_{k}^{-1},$

for some $\gamma_{k}\in{\operatorname{EQ}}(2n+2,R_{s},\Lambda_{s})$. Using
Lemma 3.9 and standard commutator formulas (see [Bak1, Lemma 3.16, pg. 43]),
we get that
$\widetilde{\sigma}_{(s,0)}(XT^{2d})=\prod_{t}\widetilde{\eta_{p_{t}q_{t}}}(T\mu_{t}(X))$,
for some $\mu_{t}(X)\in(R\oplus J)_{(s,0)}[X]$ with $p_{t}=1$ or $q_{t}=1$.
Since $P_{s}$ is a free $R_{s}$-module, then we have

$(P\oplus J)_{(s,0)}[X,T]\cong(R\oplus J)_{(s,0)}^{2n}[X,T]\cong(P\oplus
J)_{(s,0)}[X,T]^{*}.$

Thus using the isomorphism, polynomials in $(P\oplus J)_{(s,0)}[X,T]$ can be
regarded as linear forms. First we consider the case: $p_{t}=1$. Let
$p_{1}^{*},p_{2}^{*},\dots,p_{n}^{*},p_{-1}^{*},\dots,p_{-n}^{*}$ be the
standard basis of $(P\oplus J)_{(s,0)}$. Let $s^{m}p_{i}^{*}\in P\oplus J$ for
some $m>0$ and $i=\pm 1,\pm 2,\dots,\pm n$. Let $e_{\pm i}^{*}$ be the
standard basis of $(R\oplus J)^{2n}$. Then for $q_{t}=\pm i$, consider the
element $T\mu_{t}(X)e_{\pm i}^{*}$ as an element in $(P\oplus
I)_{(s,0)}[X,T]^{*}$. As $(P\oplus J)_{(s,0)}$ is free, by Lemma 3.10, we can
say $T\mu_{t}(X)e_{\pm i}^{*}$ is a polynomial in $T$. Again, by Lemma 3.11,
there exists $k_{1}>0$ such that $k_{1}$ is the maximum power of $(s,0)$
occurring in the denominator of $\mu_{t}(X)e_{\pm i}^{*}$. Choose
$l_{1}\geq\max(k,m)$. Now consider the case $q_{t}=1$. Then, for $p_{t}=\pm
j$, $T\mu_{t}(X)e_{\pm j}^{*}\in(P\oplus J)_{(s,0)}[X,T]$. By a similar
argument we can consider $T\mu_{t}(X)e_{\pm j}^{*}$ as a polynomial in $T$ and
hence there exists $k_{2}>0$ such that $k_{2}$ is maximum power of $(s,0)$
occurred in $\mu_{t}e_{\pm j}^{*}$. Now choose
$l_{2}\geq\nolinebreak\max(k_{2},m)$. For $l\geq\max(l_{1},l_{2})$, under the
transformation $T\mapsto(s,0)^{l}T$,
$\widetilde{\sigma}_{(s,0)}((b,0)XT^{2d})$ is defined over $(Q\oplus J)[X,T]$,
i.e., there exists some element

$\widetilde{\widehat{\sigma}}(X,T)\in{\operatorname{ET}}((Q\oplus J)[X,T])$

such that
$\widetilde{\widehat{\sigma}}_{(s,0)}(X,T)=\widetilde{\sigma}_{(s,0)}((b,0)XT^{2d})$.
Replacing $T=(1,0)$ and using Lemma 3.12, one gets
$\widetilde{\sigma}((b,0)X)\in{\operatorname{ET}}(Q[X]\oplus
J[X],(\Gamma\oplus J)[X],0\oplus J[X])$. Hence, the result follows by applying
$\phi$.

Consequence: Relative L-G principle for the transvection subgroups:

######  Theorem 3.15.

Let $R$ be an almost commutative ring and $J\subset R$ be an ideal. Let $P$
and $Q$ be as in 3. Let
$\sigma(X)\in{\operatorname{GQ}}(Q[X],\Lambda[X],J[X])$ with $\sigma(0)=Id$.
If
$\sigma_{\mathfrak{m}}(X)\in{\operatorname{EQ}}(2n+2,R_{\mathfrak{m}}[X],\Lambda_{\mathfrak{m}}[X],J_{\mathfrak{m}}[X])$
for all $\mathfrak{m}\in{\rm Max}(C(R))$, then we have
$\sigma(X)\in{\operatorname{ET}}(Q[X],\Lambda[X],J[X])$.

######  Proof 3.16.

Follows by arguing as in the proof of [basu2, Lemma 3.10], and using
Proposition 3.13.

##  4 Relative Stability for Quadratic ${\textrm{K}_{1}}$

The aim of this section is to establish the ${\textrm{K}_{1}}$-stability of
the relative transvection groups as an application of Theorem 3.15. For the
absolute case we refer to [BTP] and [basu2].

######  Definition 4.1.

Let $R$ be an associative ring with identity and $J\subset R$ an ideal.
Consider the ring $D$ = $\\{(a,b)\in R\times R:a-b\in J\\}$ with addition and
multiplication defined component wise. We call it the double ring relative to
the ideal $J$. For a form ring $(R,\Lambda)$, one extends the involution of
$-:R\rightarrow R$ to the ring $D$, defining $-:D\rightarrow D$ by
$\overline{(a,b)}=(\overline{a},\overline{b})$. We fix the element
$(\lambda,\lambda)$ and define $\Lambda^{\prime}$ =
$\\{(a,b)\in\Lambda\times\Lambda\mid a-b\in J\\}$. Then one can show that
$(D,\Lambda^{\prime})$ is a form ring.

######  Definition 4.2.

Let $(R,\Lambda,\lambda)$ and $(S,\Lambda^{\prime},\lambda^{\prime})$ be two
form rings. A ring homomorphism $f:R\rightarrow S$ is said to be a morphism of
form rings if $f(\overline{r})=\overline{f(r)}$,
$f(\Lambda)\subseteq\Lambda^{\prime}$ and $f(\lambda)=\lambda^{\prime}$.

######  Lemma 4.3.

Let $(R,\Lambda)$ be a form ring and $J\subset R$ be a two sided ideal. Then
the form rings $(D,\Lambda^{\prime})$ and $(R\oplus J,\Lambda\oplus J)$ are
isomorphic.

######  Proof 4.4.

Consider the homomorphisms $f:D\rightarrow R\oplus J$ defined by
$f(a,b)=(a,b-a)$ and $g:R\oplus J\rightarrow D$ defined by $g(a,i)=(a,a+i)$.
It can be checked that both $f$ and $g$ are form homomorphisms. They are
inverses of each other.

######  Lemma 4.5 ([keshari, Proposition 3.1]).

Let $A$ be a commutative Noetherian ring of (Krull) dimension $d$ and
$J\subset A$ be an ideal. Then the ring $D$ is also a commutative ring of
dimension $d$.

######  Proof 4.6.

Clearly $D_{A}$ is a commutative Noetherian ring. We first prove that
dimension of $A\oplus J$ is $d$, and the rest follows from Lemma 4.3. Clearly
$A$ can be identified with the sub-ring $\\{(r,0)\mid r\in A\\}$ of $A\oplus
J$. The element $(0,i)$ is integral over $A$ since we have that
$(0,i)^{2}-(i,0)(0,i)=(0,0)$. Hence every element of $A\oplus J$ is integral
over $R$, and therefore $\dim(A)=\dim(A\oplus J)$.

Recall ${\textrm{K}_{1}}$-stability result of Bak–Petrov–Tang (cf. [BTP]) for
general quadratic groups in the absolute case.

######  Theorem 4.7 ([BTP]).

Let $R$ be an almost commutative with $\dim(C(R))=d$. Consider the form ring
$(R,\Lambda)$. Then the stabilization map

$\frac{{\operatorname{GQ}}(2n,R,\Lambda)}{{\operatorname{EQ}}(2n,R,\Lambda)}\rightarrow\frac{{\operatorname{GQ}}(2n+2,R,\Lambda)}{{\operatorname{EQ}}(2n+2,R,\Lambda)}$

is an isomorphism for $2n\geq\max(6,2d+4)$.

We prove the above result in the relative case.

######  Theorem 4.8.

Let $R$ be a form ring which is finitely generated over its center $C(R)$,
with $\dim(C(R))=d$ and let $J\subset R$ be a two-sided ideal of $R$. Then the
stabilization map

$\frac{{\operatorname{GQ}}(2n,R,\Lambda,J)}{{\operatorname{EQ}}(2n,R,\Lambda,J)}\rightarrow\frac{{\operatorname{GQ}}(2n+2,R,\Lambda,J)}{{\operatorname{EQ}}(2n+2,R,\Lambda,J)}$

is an isomorphism for $2n\geq\max(6,2d+4)$.

######  Proof 4.9.

Consider the stabilization map
$\phi:{\operatorname{KQ}}_{1,2n}\rightarrow{\operatorname{KQ}}_{1,2n+2}$,
where

${\operatorname{KQ}}_{1,2n}(R,\Lambda,J)=\frac{{\operatorname{GQ}}(2n,R,\Lambda,J)}{{\operatorname{EQ}}(2n,R,\Lambda,J)}.$

By [HO, 5.3.22], we have following sequences are exact

${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{GQ}}(2n,R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{GQ}}(2n,D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{GQ}}(2n,R,\Lambda)}$ |  | ${\ignorespaces 1,}$ |
---|---|---|---|---|---|---|---|---|---
${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,R,\Lambda)}$ |  | ${\ignorespaces 1,}$ |
$\scriptstyle{\ignorespaces i}$$\scriptstyle{\ignorespaces
p_{2}}$$\scriptstyle{\ignorespaces i}$$\scriptstyle{\ignorespaces p_{2}}$

where the map $i$ is induced from the map $M_{2n}(J)\rightarrow M_{2n}(D)$
given by $i(\alpha)=(\alpha,{\rm I}_{2n})$ and the map $p_{2}$ is induced from
the projection map $p_{2}:D\rightarrow R$ given by $p_{2}((a,b))=b$. Thus we
have the following commutative diagram:

|  | ${\ignorespaces 1}$ |  | ${\ignorespaces 1}$ |  | ${\ignorespaces 1}$ |  |  |
---|---|---|---|---|---|---|---|---|---
${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,R,\Lambda)}$ |  | ${\ignorespaces 1}$ |
${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{GQ}}(2n,R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{GQ}}(2n,D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{GQ}}(2n,R,\Lambda)}$ |  | ${\ignorespaces 1}$ |
${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1,2n}(R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1,2n}(D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1,2n}(R,\Lambda)}$ |  | ${\ignorespaces 1}$ |
|  | ${\ignorespaces 1}$ |  | ${\ignorespaces 1}$ |  | ${\ignorespaces 1}$ |

Taking limits over $n$ in the lower row, we get an exact sequence

${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1}(R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1}(D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1}(R,\Lambda)}$ |  | ${\ignorespaces 1,}$ |
---|---|---|---|---|---|---|---|---|---

and for any $n$, we get a homomorphism of exact sequences:

${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1,2n}(R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1,2n}(D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1,2n+2}(R,\Lambda)}$ |  | ${\ignorespaces 1}$ |
---|---|---|---|---|---|---|---|---|---
${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1}(R,\Lambda,J)}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1}(D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{KQ}}_{1}(R,\Lambda)}$ |  | ${\ignorespaces 1.}$ |

For $2n\geq 2d+4$ the right most map is an isomorphism by Theorem 4.7. Since
$R$ is finitely generated over $C(R)$, then it can be checked that $D$ is
finitely generated over $C(D)$. Since $C(D)$ is a double ring of $C(R)$
relative to the ideal $C(R)\cap J$, by Lemma 4.5, we get $\dim(C(D))=d$. Hence
for $2n\geq 2d+4$, the middle homomorphism is also an isomorphism by Theorem
4.7. Hence it follows that the left most map is an isomorphism.

Now we prove the ${\textrm{K}_{1}}$-stability of relative transvection groups
as an application of relative local-global principle of transvection groups.
For this we need to recall the following result of Vaserstein.

######  Lemma 4.10 ([basu2, Lemma 4.2]).

Let $(R,\Lambda)$ be a form ring finitely generated over $C(R)$ with Krull
dimension of $C(R)=d$ and $J\subset R$ be an two-sided ideal of $R$. Let $P,Q$
be as in Notation 3. Let the rank of $Q$ be $2n\geq{\rm Max}(6,2d+2)$. Then
the group of elementary transvections
${\operatorname{ET}}(Q\perp\mathbb{H}(R),J)$ acts transitively on the set
${\rm Um}(Q\perp\mathbb{H}(R),J)$ of unimodular elements which are congruent
to $(0,\dots,0,1,0)$ modulo $J$.

######  Theorem 4.11.

Let $(R,\Lambda)$ be a form ring finitely generated over $(C(R))$ with Krull
dimension $d$ and let $J\subset R$ be a two-sided ideal. Let $P,Q$ be as
defined in Notation 3 and let the rank of $Q$ be $2n\geq\max(6,2d+4)$. Then
the stabilization map

$i_{2n}:{\operatorname{KQ}}_{1,2n}(Q,J)\rightarrow{\operatorname{KQ}}_{1,2n+2}(Q\perp\mathbb{H}(R),J)$

is an isomorphism.

######  Proof 4.12.

The surjectivity part follows from Lemma 4.10. In order to prove the
injectivity part, let $\alpha\in{\operatorname{GQ}}(Q,J)$ be such that
$\widetilde{\alpha}=\alpha\perp\rm Id$ lies in
${\operatorname{ET}}(M\perp\mathbb{H}(R),J)$. Let $\varphi(X)$ be the isotropy
between $\widetilde{\alpha}$ and $\rm Id$. Now by similar argument as given in
[basu2, Theorem 4.4], we can get an isotropy
$\widetilde{\varphi}(X)\in{\operatorname{GQ}}(Q[X],J[X])$ between $\alpha$ and
$\rm Id$. Now, by localizing $\widetilde{\varphi}$ at a maximal ideal
$\mathfrak{m}\in\operatorname{Max}(C(R))$,
$\widetilde{\varphi}_{\mathfrak{m}}(X)$ becomes (relative) stably elementary.
By Theorem 4.8 it follows that $\widetilde{\varphi}_{\mathfrak{m}}(X)$ is
actually (relative) elementary. Therefore, by the relative L-G principle
(Theorem 3.15) one gets
$\widetilde{\varphi}(X)\in{\operatorname{ET}}(Q[X]),J[X])$. Hence
$\alpha=\nolinebreak\widetilde{\varphi}(1)\in\nolinebreak{\operatorname{ET}}(Q,J)$.

##  5 Structure of Unstable Quadratic ${\textrm{K}_{1}}$ Groups; Relative Case

We devote this section to discuss the study of nilpotent property of
(relative) unstable ${\rm K_{1}}$-groups. Throughout this section we assume
$R$ is a commutative ring with identity, i.e., we are considering the trivial
involution and $n\geq 3$. By ${\operatorname{SQ}}(2n,R,\lambda)$ we shall
denote the subgroup of ${\operatorname{GQ}}(2n,R,\lambda)$ with matrices of
determinant 1, and analogously for the relative cases. Let $G$ be a group.
Define $Z^{i}$ inductively by $Z^{0}=G$, $Z^{1}=[G,G]$ and
$Z^{i}=[G,G^{i-1}]$. The group $G$ is said to be nilpotent if $Z^{r}=\\{e\\}$
for some $r>0$.

######  Definition 5.1.

A group $G$ is called nilpotent-by-abelian if it has a normal subgroup $H$
such that $H$ is nilpotent and $G/H$ is abelian.

In [Bak], A.Bak proved that the unstable ${\textrm{K}_{1}}(R)$ group of
${\operatorname{GL}}_{n}(R)$ is nilpotent-by-abelian for $n\geq 3$, and hence
${\textrm{K}_{1}}(R)$ is solvable. Later it was generalized by R. Hazrat for
the general quadratic groups over module finite rings (cf. [HR]). In [HV]
Hazrat–Vavilov revisited this problem for ordinary classical Chevalley groups
(that is, types A, C, and D) and finally extends it further to the exceptional
Chevalley groups (that is, types E, F, and G). A simpler and shorter proof is
given in [BBR] for the linear, symplectic and orthogonal groups (absolute
cases). In [RB], the first author proved this result for the general Hermitian
groups and the same proof also works for the general quadratic groups. The
relative cases are proved in [BBR] for the traditional classical groups. In
this article we prove the result for the relative general quadratic groups,
and consequently we get the result for the module cases as an application of
(relative) L-G principle for the transvection subgroups.

######  Theorem 5.2.

The quotient group
$\operatorname{SQ}(2n,R,\Lambda,J)/\operatorname{EQ}(2n,R,\Lambda,J)$ is
nilpotent for $n\geq 3$. The class of nilpotency is at most $\max(1,d+2-n)$
where $d=\dim R.$

######  Proof 5.3.

Note that for $n\geq d+2$, the quotient group

$G={\operatorname{SQ}}(2n,R,\Lambda,J)/{\operatorname{EQ}}(2n,R,\Lambda,J)$

is abelian and hence nilpotent. So we consider the case $n<d+2$. Let us fix a
natural number $n$. We prove the theorem by induction on $d=\dim R$. Let
$m=d+2-n$ and $\alpha=[\beta,\gamma]$ for some $\beta\in G$ and $\gamma\in
Z^{m-1}$. Clearly the result is true for $d=0$. Let $\widetilde{\beta}$ be the
pre-image of $\beta$ and $\widetilde{\gamma}$ be the pre-image of $\gamma$
under the map

${\operatorname{SQ}}(2n,R,\Lambda,J)\rightarrow G.$

Consider the double ring $(D,\Lambda^{\prime})$ as defined in 4.1. By [HO,
5.3.22], we have the following exact sequences:

${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{SQ}}(2n,R,J,\Lambda)}$ |  | ${\ignorespaces{\operatorname{SQ}}(2n,D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{SQ}}(2n,R,\Lambda)}$ |  | ${\ignorespaces 1}$ |
---|---|---|---|---|---|---|---|---|---
${\ignorespaces 1}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,R,J,\Lambda)}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,D,\Lambda^{\prime})}$ |  | ${\ignorespaces{\operatorname{EQ}}(2n,R,\Lambda)}$ |  | ${\ignorespaces 1}$ |
$\scriptstyle{\ignorespaces i}$$\scriptstyle{\ignorespaces
p_{2}}$$\scriptstyle{\ignorespaces i}$$\scriptstyle{\ignorespaces p_{2}}$

where the map $i$ is induced from the map $M_{2n}(J)\rightarrow M_{2n}(D)$
given by $i(\alpha)=(\alpha,\rm I_{2n})$ and $p_{2}$ is induced from the
projection map $p_{2}:D\rightarrow R$ given by $p_{2}((a,b))=b$. Now, in the
group
$H={\operatorname{SQ}}(2n,D,\Lambda^{\prime})/{\operatorname{EQ}}(2n,D,\Lambda^{\prime})$
we have $i(\widetilde{\gamma})\in H^{m}$. Since ${\rm dim}(R)={\rm dim(D)}$
(by Lemma 4.5), by the result of the absolute cases, we get
$[i(\widetilde{\beta}),i(\widetilde{\gamma})]\in{\operatorname{EQ}}(2n,D,\Lambda^{\prime})$.
It can be checked that
$i([\widetilde{\beta},\widetilde{\gamma}])=[i(\widetilde{\beta}),i(\widetilde{\gamma})]$
and the image of this element under $p_{2}$ is identity. Hence by the above
diagram one gets
$[\widetilde{\beta},\widetilde{\gamma}]\in{\operatorname{EQ}}(2n,R,J)$,
consequently $[\beta,\gamma]$ is trivial in $G$. Next, consider the pre-images
$\beta_{1}$ of $\widetilde{\beta}$ and $\gamma_{1}$ of $\widetilde{\gamma}$
under the map ${\operatorname{SQ}}(2n,R\oplus J,\Gamma\oplus
J)\rightarrow\nolinebreak{\operatorname{SQ}}(2n,R,\Lambda,J)$. Now by the
result of absolute cases we get $[\beta_{1},\gamma_{1}]\in\nolinebreak
EQ(2n,R\oplus J,\Gamma\oplus J)$. It can also be checked that the same
commutator $[\beta_{1},\gamma_{1}]$ is in ${\operatorname{EQ}}(2n,R\oplus
J,\Gamma\oplus J,0\oplus J)$. By projecting $R\oplus J$ onto $R$, one gets
$[\widetilde{\beta},\widetilde{\gamma}]\in{\operatorname{EQ}}(2n,R,J,\Lambda)$,
and hence $\alpha=\\{1\\}$ in $G$.

######  Corollary 5.4.

Let $(R,\Lambda)$ be a commutative form ring and $J\subset R$ be an ideal of
$R$. Then the quotient group
${\operatorname{GQ}}(2n,R,\Lambda,J)/{\operatorname{ET}}(2n,R,\Lambda,J)$ is
nilpotent-by-abelian for $n\geq 3$.

######  Corollary 5.5.

Let $(R,\Lambda)$ be a commutative form ring and $J\subset R$ be an ideal of
$R$. Consider the notation as in 3. Let $d=\dim(R)$ and $t=$ the local rank of
$Q$. The quotient group ${\operatorname{T}}(Q,J)/{\operatorname{ET}}(Q,J)$ is
nilpotent of class at most $\max(1,d+3-t/2)$.

######  Proof 5.6.

The proof is same as [BBR, Theorem 4.1].

##  6 Bass’ Nil Group ${\rm NKQ_{1}(R)}$

The group
${\textrm{NK}_{1}}(R)=\mathsf{ker}({\textrm{K}_{1}}(R[X])\rightarrow{\textrm{K}_{1}}(R))$;
$X=0$ is called the Bass’ nil-group of R. This is the subgroup consisting of
elements $[\alpha(X)]\in{\textrm{K}_{1}}(R[X])$ such that
$[\alpha(0)]=[{\operatorname{I}}]$, and hence ${\textrm{K}_{1}}(R[X])\cong{\rm
NK}_{1}(R)\oplus{\textrm{K}_{1}}(R)$. The aim of the next sections is to study
some properties of Bass nil-groups ${\rm NK_{1}}$ for the general quadratic
groups or Bak’s unitary groups. In this section we recall some basic
definitions and properties of the representatives of ${\rm NKQ_{1}}(R)$. We
represent any element of ${\rm M}_{2n}(R)$ as a matrix

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix},\text{where }a,b,c,d\in{\rm
M}_{n}(R).$

For a matrix as above, we call $\begin{pmatrix}a&b\end{pmatrix}$ the upper
half of $\alpha$. Let $(R,\lambda,\Lambda)$ be a form ring. By setting
$\overline{\Lambda}=\\{\overline{a}\mid a\in\Lambda\\}$ we get another form
ring $(R,\overline{\lambda},\overline{\Lambda})$. We can extend the involution
of $R$ to ${\rm M}_{n}(R)$ by setting $(a_{ij})^{*}=(\overline{a}_{ji})$. For
details, see [V], [V1].

######  Definition 6.1.

Let $(R,\lambda,\Lambda)$ be a form ring. A matrix $\alpha=(a_{ij})\in{\rm
M}_{n}(R)$ is said to be $\Lambda$-Hermitian if $\alpha=-\lambda\alpha^{*}$
and all the diagonal entries of $\alpha$ are contained in $\Lambda$. A matrix
$\beta\in{\rm M}_{n}(R)$ is said to be $\overline{\Lambda}$-Hermitian if
$\beta=-\overline{\lambda}\beta^{*}$ and all the diagonal entries of $\beta$
are contained in $\overline{\Lambda}$.

######  Remark 6.2.

A matrix $\alpha\in{\rm M}_{n}(R)$ is $\Lambda$-Hermitian if and only if
$\alpha^{*}$ is $\overline{\Lambda}$-Hermitian.

######  Lemma 6.3 ([V, Example 2]).

Let $\beta\in{\operatorname{GL}}_{n}(R)$ be a $\Lambda$-Hermitian matrix. Then
the matrix $\alpha^{*}\beta\alpha$ is $\Lambda$-Hermitian for every
$\alpha\in{\operatorname{GL}}_{n}(R)$.

######  Definition 6.4.

Let $\alpha=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in{\rm M}_{2n}(R)$ be a
matrix. Then $\alpha$ is said to be a $\Lambda$-quadratic matrix if one of the
following equivalent conditions holds:

  1. 1.

$\alpha\in{\operatorname{GQ}}(2n,R,\Lambda)$ and the diagonal entries of the
matrices $a^{*}c$, $b^{*}d$ are in $\Lambda$,

  2. 2.

$a^{*}d+\lambda c^{*}b={\operatorname{I}}_{n}$ and the matrices $a^{*}c$,
$b^{*}d$ are $\Lambda$-Hermitian,

  3. 3.

$\alpha\in{\operatorname{GQ}}(2n,R,\Lambda)$ and the diagonal entries of the
matrices $ab^{*},cd^{*}$ are in $\Lambda$,

  4. 4.

$ad^{*}+\lambda bc^{*}={\operatorname{I}}_{n}$ and the matrices $ab^{*}$,
$cd^{*}$ are $\Lambda$-Hermitian.

######  Lemma 6.5.

Let $\alpha=\begin{pmatrix}a&0\\\ 0&d\end{pmatrix}\in{\rm M}_{2n}(R)$. Then
$\alpha\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$ if and only if we have
$a\in{\operatorname{GL}}_{n}(R)$ and $d=(a^{*})^{-1}$.

######  Proof 6.6.

Let $\alpha\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$. In view of
condition 2. of Definition 6.4, we have, $a^{*}d={\operatorname{I}}_{n}$.
Hence $a$ is invertible and $d=(a^{*})^{-1}$. The converse holds by condition
2. of Definition 6.4.

######  Definition 6.7.

Let $\alpha\in{\operatorname{GL}}_{n}(R)$ be a matrix. A matrix of the form
$\begin{pmatrix}\alpha&0\\\ 0&(\alpha^{*})^{-1}\end{pmatrix}$ is denoted by
$\mathbb{H}(\alpha)$ and is said to be hyperbolic.

######  Remark 6.8.

In a similar way we can show that a matrix of the form
$T_{12}(\beta):=\begin{pmatrix}{\operatorname{I}}_{n}&\beta\\\
0&{\operatorname{I}}_{n}\end{pmatrix}$ is a $\Lambda$-quadratic matrix if and
only if $\beta$ is $\overline{\Lambda}$-Hermitian. Similarly, a matrix of the
form $T_{21}(\gamma):=\begin{pmatrix}{\operatorname{I}}_{n}&0\\\
\gamma&{\operatorname{I}}_{n}\end{pmatrix}$ is a $\Lambda$-quadratic matrix if
and only if $\gamma$ is $\Lambda$-Hermitian.

Likewise, in the quadratic case we can define the notion of
$\Lambda$-elementary quadratic groups in the following way:

######  Definition 6.9.

The $\Lambda$-elementary quadratic group is denoted by
${\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)$ and defined to be the group
generated by the $2n\times 2n$ matrices of the following forms:

  * •

$\mathbb{H}(\alpha)$, here $\alpha\in{\operatorname{E}}_{n}(R)$,

  * •

$T_{12}(\beta)$, where $\beta$ is $\overline{\Lambda}$-Hermitian, and

  * •

$T_{21}(\gamma)$, where $\gamma$ is $\Lambda$-Hermitian.

######  Lemma 6.10.

Let $A=\begin{pmatrix}\alpha&\beta\\\ 0&\delta\end{pmatrix}\in{\rm
M}_{2n}(R)$. Then $A\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$ if and
only if we have $\alpha\in{\operatorname{GL}}_{n}(R)$,
$\delta=(\alpha^{*})^{-1}$ and $\alpha^{-1}\beta$ is
$\overline{\Lambda}$-Hermitian. In this case

$A\equiv\mathbb{H}(\alpha)\pmod{{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)}.$

######  Proof 6.11.

Let $A\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$. Then, by condition 4.
of Definition 6.4, we have that $\alpha\delta^{*}={\operatorname{I}}_{n}$ and
$\alpha\beta^{*}$ is $\Lambda$-Hermitian. Hence $\alpha$ is invertible and
$\delta=(\alpha^{*})^{-1}$. For $\alpha^{-1}\beta$, we get

$(\alpha^{-1}\beta)^{*}=\beta^{*}(\alpha^{-1})^{*}=\alpha^{-1}(\alpha\beta^{*})(\alpha^{-1})^{*},$

which is $\Lambda$-Hermitian by Lemma 6.3. Hence $\alpha^{-1}\beta$ is
$\overline{\Lambda}$-Hermitian. Conversely, condition on $A$ fulfills
condition 4. of Definition 6.4. Hence $A$ is $\Lambda$-quadratic. Now , since
$\alpha^{-1}\beta$ is $\overline{\Lambda}$-Hermitian,
$T_{12}(-\alpha^{-1}\beta)\in{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)$, and
$AT_{12}(\alpha^{-1}\beta)=\mathbb{H}(\alpha)$. Thus
$A\equiv\mathbb{H}(\alpha)\pmod{{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)}$.

Arguing similarly one gets the following:

######  Lemma 6.12.

Let $B=\begin{pmatrix}\alpha&0\\\ \gamma&\delta\end{pmatrix}\in{\rm
M}_{2n}(R)$. Then $B\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$ if and
only if $\alpha\in{\operatorname{GL}}_{n}(R)$, $\delta=(\alpha^{*})^{-1}$ and
$\gamma$ is $\Lambda$-Hermitian. In this case

$B\equiv\mathbb{H}(\alpha)\pmod{{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)}.$

######  Lemma 6.13.

Let $\alpha=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$. If
$a\in{\operatorname{GL}}_{n}(R)$ then

$\alpha\equiv\mathbb{H}(a)\pmod{{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)}.$

######  Proof 6.14.

By same argument as given in Lemma 6.10, we have $a^{-1}b$ is
$\overline{\Lambda}$-Hermitian. Hence
$T_{12}(-a^{-1}b)\in{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)$, and
consequently $\alpha T_{12}(-a^{-1}b)=\begin{pmatrix}a&0\\\
c&d^{\prime}\end{pmatrix}\in{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)$ for
some $d^{\prime}\in{\operatorname{GL}}_{n}(R)$. Hence by Lemma 6.12, we get

$\alpha T_{12}(-a^{-1}b)\equiv
H(a)\pmod{{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)}.$

Hence $\alpha\equiv H(a)\pmod{{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda)}$.

######  Definition 6.15.

Consider the embedding:

${\operatorname{GQ}}^{\lambda}(2n,R,\Lambda)\rightarrow{\operatorname{GQ}}^{\lambda}(2n+2,R,\Lambda),\,\,\,$

We use the following notation:

${\operatorname{GQ}}^{\lambda}(R,\Lambda)=\underset{n=1}{\overset{\infty}{\cup}}{\operatorname{GQ}}^{\lambda}(2n,R,\Lambda),$
${\operatorname{EQ}}^{\lambda}(R,\Lambda)=\underset{n=1}{\overset{\infty}{\cup}}{\operatorname{EQ}}^{\lambda}(2n,R,\Lambda).$

In view of quadratic analog of Whitehead Lemma, we have that the group
${\operatorname{EQ}}^{\lambda}(R,\Lambda)$ coincides with the commutator of
${\operatorname{GQ}}^{\lambda}(R,\Lambda)$. Therefore, the group

${\rm
KQ_{1}}^{\lambda}(R,\Lambda):=\frac{{\operatorname{GQ}}^{\lambda}(R,\Lambda)}{{\operatorname{EQ}}^{\lambda}(R,\Lambda)}$

is well-defined. The class of a matrix
$\alpha\in{\operatorname{GQ}}^{\lambda}(R,\Lambda)$ in the group ${\rm
KQ_{1}}^{\lambda}(R,\Lambda)$ is denoted by $[\alpha]$. In this way we obtain
a ${\textrm{K}_{1}}$-functor ${\rm KQ_{1}}^{\lambda}$ acting from the category
of form rings to the category of abelian groups.

######  Definition 6.16.

In the quadratic case, the kernel of the group homomorphism

${\rm KQ_{1}}^{\lambda}(R[X],\Lambda[X])\rightarrow{\rm
KQ_{1}}^{\lambda}(R,\Lambda)$

induced from the form ring homomorphism
$(R[X],\Lambda[X])\rightarrow(R,\Lambda);X\mapsto 0$ is denoted by ${\rm
NKQ_{1}}^{\lambda}(R,\Lambda)$.

######  Remark 6.17.

Since the class of $\Lambda$-quadratic groups is a subclass of the class of
quadratic groups, the local-global principle holds for $\Lambda$-quadratic
groups. We use this fact throughout the next section.

##  7 Absence of torsion in ${\rm NKQ_{1}}^{\lambda}(R,\lambda)$

In [STI], J. Stienstra showed that ${\textrm{NK}_{1}}(R)$ is a
${\operatorname{W}}(R)$-module, where ${\operatorname{W}}(R)$ is the ring of
big Witt vectors (cf. [BL1] and [WEL1]). Consequently, in ([WEL], §3), C.
Weibel observed that if $k$ is a unit in a commutative local ring $R$, then
${\textrm{SK}_{1}}(R[X])$ has no $k$-torsion. Note that if $R$ is a
commutative local ring, then ${\textrm{SK}_{1}}(R[X])$ coincides with
${\textrm{NK}_{1}}(R)$; indeed, if $R$ is a local ring, then
${\operatorname{SL}}_{n}(R)={\operatorname{E}}_{n}(R)$ for all $n>0$.
Therefore, we may replace $\alpha(X)$ by $\alpha(X)\alpha(0)^{-1}$ and assume
that $[\alpha(0)]=[{\operatorname{I}}]$. In [RB1], the first author extended
Weibel’s result for arbitrary associative rings. In this section we prove the
analog result for $\lambda$-unitary Bass nil-groups, viz. ${\rm
NK_{1}GQ}^{\lambda}(R,\lambda)$, where $(R,\lambda)$ is the form ring as
introduced by A. Bak in [Bak1]. The main ingredient for our proof is an analog
of Higman linearization (for a subclass of Bak’s unitary group) due to V.
Kopeiko; cf. [V]. For the general linear groups, Higman linearization (cf.
[RB1]) allows us to show that ${\textrm{NK}_{1}}(R)$ has a unipotent
representative. The same result is not true in general for the unitary nil-
groups. Kopeiko’s results in [V], [V1] give a complete description of the
elements of ${\rm NK_{1}GQ}^{\lambda}(R,\lambda)$ which have (unitary)
unipotent representatives.

######  Definition 7.1.

For an associative unital ring $R$ we consider the truncated polynomial ring

$R_{t}=\frac{R[X]}{(X^{t+1})}.$

######  Lemma 7.2 (cf. [RB1], Lemma 4.1).

Let $P(X)\in R[X]$ be any polynomial. Then the following identity holds in the
ring $R_{t}$:

$(1+X^{r}P(X))=(1+X^{r}P(0))(1+X^{r+1}Q(X)),$

where $r>0$ and $Q(X)\in R[X]$, with $\deg(Q(X))<t-r$.

######  Proof 7.3.

Let us write $P(X)=a_{0}+a_{1}X+\cdots+a_{t}X^{t}$. Then we can express the
polynomial $P(X)$ as $P(X)=P(0)+XP^{\prime}(X)$ for some $P^{\prime}(X)\in
R[X]$. Now, in $R_{t}$

$\displaystyle(1+X^{r}P(X))(1+X^{r}P(0))^{-1}$
$\displaystyle=(1+X^{r}P(0)+X^{r+1}P^{\prime}(X))(1+X^{r}P(0))^{-1}$
$\displaystyle=1+X^{r+1}P^{\prime}(X)(1-X^{r}P(0)+X^{2r}(P(0))^{2}-\cdots)$
$\displaystyle=1+X^{r+1}Q(X),$

where $Q(X)\in R[X]$ with $\deg(Q(X))<t-r$. Hence the lemma follows.

######  Remark 7.4.

Iterating the above process we can write, for any polynomial $P(X)\in R[X]$,

$(1+XP(X))=\Pi_{i=1}^{t}(1+a_{i}X^{i})$

in $R_{t}$, for some $a_{i}\in R$. By ascending induction, it follows that the
$a_{i}$’s are uniquely determined.

######  Lemma 7.5.

Let $R$ be an associative ring, and $k\in\mathbb{Z}$ such that $kR=R$. Let
$P(X)$ be a polynomial in $R[X]$. Assume $P(0)$ lies in the center of $R$.
Then, if

$(1+X^{r}P(X))^{k^{r}}=1$

for some $r\geq 0$, we have that

$(1+X^{r}P(X))=(1+X^{r+1}Q(X))$

in the ring $R_{t}$, for some $Q(X)\in R[X]$ with $\deg(Q(X))<t-r$.

The following result is due to V. Kopeiko, cf. [V]. This is an analog of
Higman linearization for this special case.

######  Proposition 7.6.

Let $(R,\lambda)$ be a form ring. Then, every element of the group ${\rm
NKQ_{1}}^{\lambda}(R,\Lambda)$ has a representative of the form

$[a;b,c]_{n}=\begin{pmatrix}{\operatorname{I}}_{r}-aX&bX\\\
-cX^{n}&{\operatorname{I}}_{r}+a^{*}X+\cdots+(a^{*})^{n}X^{n}\end{pmatrix}\in{\operatorname{GQ}}^{\lambda}(2r,R[X],\Lambda[X])$

for some positive integers $r$ and $n$, where $a,b,c\in{\rm M}_{r}(R)$ satisfy
the following conditions:

  1. 1.

the matrices $b$ and $ab$ are Hermitian and also $ab=ba^{*}$,

  2. 2.

the matrices $c$ and $ca$ are Hermitian and also $ca=a^{*}c$,

  3. 3.

$bc=a^{n+1}$and $cb=(a^{*})^{n+1}$.

######  Corollary 7.7.

Let $R$ be an associative ring. Let $[\alpha]\in{\rm
NKQ_{1}}^{\lambda}(R,\Lambda)$ have the representation $[a;b,c]_{n}$ for some
$a,b,c\in{\rm M}_{n}(R)$ according to Proposition 7.6. Then, if it is true
that $({\rm I}_{r}-aX)\in{\operatorname{GL}}_{r}(R)$, we have
$[\alpha]=[\mathbb{H}({\rm I}_{r}-aX)]$ in ${\rm
NKQ_{1}}^{\lambda}(R,\Lambda)$.

######  Proof 7.8.

By Lemma 6.13, we have
$[a;b,c]_{n}\equiv\mathbb{H}({\operatorname{I}}_{r}-aX)\pmod{{\operatorname{EQ}}^{\lambda}(2r,R[X],\Lambda[X])}$.
Hence $[\alpha]=[\mathbb{H}({\operatorname{I}}_{r}-aX)]$ in ${\rm
NKQ_{1}}^{\lambda}(R,\Lambda)$.

######  Theorem 7.9.

Let $(R,\Lambda)$ be a form ring, where $R$ is an associative ring with $1$
and $k$ is an invertible integer, i.e., $kR=R$. Let

$[\alpha(X)]=\left[\begin{pmatrix}A(X)&B(X)\\\
C(X)&D(X)\end{pmatrix}\right]\in{\rm NKQ_{1}}^{\lambda}(R,\Lambda)$

with $A(X)\in{\operatorname{GL}}_{r}(R[X])$ for some $r\in\mathbb{N}$. Then
$[\alpha(X)]$ has no $k$-torsion.

######  Proof 7.10.

By Theorem 7.6, $[\alpha(X)]=[[a;b,c]_{n}]$ for some $a,b,c\in{\rm M}_{s}(R)$
and for some natural numbers $n$ and $s$. Note that in Step $1$ of the
Proposition 7.6, the invertibility of the top left corner of the matrix
$\alpha$ is not changed during the linearization process. The same is true for
the remaining steps of the Proposition 7.6, and so, since the top left corner
matrix is $A(X)\in{\operatorname{GL}}_{r}(R[X])$, we get
$({\operatorname{I}}_{s}-aX)\in{\operatorname{GL}}_{s}(R[X])$. Using Corollary
7.7, one gets $[\alpha(X)]=[\mathbb{H}({\operatorname{I}}_{s}-aX)]$. Now let
$[\alpha]$ have $k$-torsion, which implies that also
$[\mathbb{H}({\operatorname{I}}_{r}-aX)]$ has $k$-torsion. Since
$({\operatorname{I}}_{r}-aX)$ is invertible, it follows that $a$ is nilpotent.
Let $t$ be such that $a^{t+1}=0$. Since
$[({\operatorname{I}}_{r}-aX)]^{k}=[{\operatorname{I}}]$ in ${\rm
NKQ}_{1}^{\lambda}(R[X],\Lambda[X])$, by arguing as in [RB], we get
$[{\operatorname{I}}_{r}-aX]=[I]$ in ${\rm
NKQ}_{1}^{\lambda}(R[X],\Lambda[X])$. This completes the proof.

##  8 Graded Analog

We recall the well-known “Swan–Weibel homotopy trick”, which is the main
ingredient to handle the graded case. Let $R=R_{0}\oplus R_{1}\oplus
R_{2}\oplus\cdots$ be a graded ring. An element $a\in R$ is denoted by
$a=a_{0}+a_{1}+a_{2}+\cdots$, where $a_{i}\in R_{i}$ for each $i$, and all but
finitely many $a_{i}$’s are zero. Let $R_{+}=R_{1}\oplus R_{2}\oplus\cdots$. A
graded structure of $R$ induces a graded structure on ${\rm M}_{n}(R)$ (the
ring of $n\times n$ matrices with entries in $R$).

######  Definition 8.1.

Let $a\in R_{0}$ be a fixed element. We fix an element $b=b_{0}+b_{1}+\cdots$
in $R$ and define a ring homomorphism $\epsilon:R\rightarrow R[X]$ as follows:

$\epsilon(b)=\epsilon(b_{0}+b_{1}+\cdots)\;=\;b_{0}+b_{1}X+b_{2}X^{2}+\cdots+b_{i}X^{i}+\cdots.$

Then we evaluate the polynomial $\epsilon(b)(X)$ at $X=a$ and denote the image
by $b^{+}(a)$ i.e., $b^{+}(a)=\epsilon(b)(a)$. Note that
$\big{(}b^{+}(x)\big{)}^{+}(y)=b^{+}(xy)$. Observe, $b_{0}=b^{+}(0)$. We shall
use this fact frequently.

The above ring homomorphism $\epsilon$ induces a group homomorphism from
${\operatorname{GL}}(n,R)$ to ${\operatorname{GL}}(n,R[X])$ of rank $n$ for
every $n\geq 1$, i.e., for $\alpha\in{\operatorname{GL}}(n,R)$ we get a map

$\epsilon:{\operatorname{GL}}(n,R)\rightarrow{\operatorname{GL}}(n,R[X])\text{,
defined by}$
$\alpha=\alpha_{0}\oplus\alpha_{1}\oplus\alpha_{2}\oplus\cdots\mapsto\alpha_{0}\oplus\alpha_{1}X\oplus\alpha_{2}X^{2}\cdots,$

where $\alpha_{i}\in{\rm M}(n,R_{i})$. As above, for $a\in R_{0}$, we define
$\alpha^{+}(a)$ as $\alpha^{+}(a)=\epsilon(\alpha)(a).$ The graded dilation
lemma and graded local-global principle are proved in [BS] for linear,
symplectic and orthogonal groups. Arguing in similar manner one gets:

######  Theorem 8.2.

(Graded Local-Global Principle) Let $R=R_{0}\oplus R_{1}\oplus
R_{2}\oplus\cdots$ be an almost commutative graded ring with identity $1$. Let
$\alpha\in{\operatorname{GQ}}(2n,R,\Lambda)$ be such that
$\alpha\equiv{\operatorname{I}}_{2n}\pmod{R_{+}}$. If
$\alpha_{\mathfrak{m}}\in{\operatorname{EQ}}(2n,R_{\mathfrak{m}},\Lambda_{\mathfrak{m}})$,
for every maximal ideal $\mathfrak{m}\in\operatorname{Max}(C(R_{0}))$, then
$\alpha\in{\operatorname{EQ}}(2n,R,\Lambda).$

Moreover, the L-G principle for the elementary subgroups and their normality
property are equivalent.

######  Theorem 8.3.

Let $R=R_{0}\oplus R_{1}\oplus R_{2}\oplus\cdots$ be an almost commutative
graded ring with identity $1$. Then the following assertions are equivalent
for $2n\geq 6$:

  1. 1.

${\operatorname{EQ}}(2n,R,\Lambda)$ is a normal subgroup of
${\operatorname{GQ}}(2n,R,\Lambda)$.

  2. 2.

If $\alpha\in{\operatorname{GQ}}(2n,R,\Lambda)$ with $\alpha^{+}(0)={\rm
I}_{2n}$, and
$\alpha_{\mathfrak{m}}\in{\operatorname{EQ}}(2n,R_{\mathfrak{m}},\Lambda_{\mathfrak{m}})$
for every maximal ideal $\mathfrak{m}\in{\rm Max}(C(R_{0}))$, then
$\alpha\in{\operatorname{EQ}}(2n,R,\Lambda)$.

As an application of Theorem 8.2 and the Theorem 7.9, we obtain the following:

######  Theorem 8.4.

Let $R=R_{0}\oplus R_{1}\oplus\dots$ be an almost commutative graded ring with
$1$. Let $N=N_{0}+N_{1}+\dots+N_{r}\in{\rm M}_{r}(R)$ be a nilpotent matrix,
and let ${\operatorname{I}}$ denote the identity matrix. Let $k\in\mathbb{Z}$
be a unit in $R_{0}$. If $[({\operatorname{I}}+N)]^{k}=[{\operatorname{I}}]$
in ${\rm NKQ}_{1}^{\lambda}(R,\Lambda)$, then
$[{\operatorname{I}}+N]=[{\operatorname{I}}+N_{0}]$.

######  Proof 8.5.

Consider the ring homomorphism $f:R\rightarrow R[X]$ defined by

$f(a_{0}+a_{1}+\dots)=a_{0}+a_{1}X+\dots.$

Then

$\displaystyle[({\operatorname{I}}+N)^{k}]=[{\operatorname{I}}]$
$\displaystyle\Rightarrow
f([{\operatorname{I}}+N]^{k})=[f({\operatorname{I}}+N)]^{k}=[{\operatorname{I}}]$
$\displaystyle\Rightarrow[({\operatorname{I}}+N_{0}+N_{1}X+\dots+N_{r}X^{r})]^{k}=[{\operatorname{I}}].$

Let $\mathfrak{m}$ be a maximal ideal in $C(R_{0})$. By Theorem 7.9, we have

$[({\operatorname{I}}+N_{0}+N_{1}X+\dots+N_{r}X^{r})]=[{\operatorname{I}}]$

in ${\rm NKQ}_{1}^{\lambda}(R_{\mathfrak{m}},\Lambda_{\mathfrak{m}})$. Hence
by using the local-global principle we conclude

$[({\operatorname{I}}+N)]=[{\operatorname{I}}+N_{0}]$

in ${\textrm{NK}_{1}}{\operatorname{GQ}}^{\lambda}(R,\Lambda)$, as required.

### Acknowledgment

We thank the unknown referee for useful suggestions to improve the manuscript.
We also thank V. Kopeiko and Sergey Sinchuk for useful discussions on Bass’s
nil group. Research by the first author was supported by SERB-MATRICS grant
for the financial year 2020–2021. Research by the second author was supported
by IISER (Pune) post-doctoral research grant, 2020–2021.

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July 28, 2022February 9, 2023Roozbeh Hazrat