#  ${\bf{\rm NK}_{1}}$ of Bak’s unitary group over Graded Rings

Rabeya Basu  Indian Institute of Science Education and Research (IISER) Pune,
India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Kuntal Chakraborty
Indian Institute of Science Education and Research (IISER) Pune, India
<EMAIL_ADDRESS>
Research by the first author was supported by SERB-MATRICS grant for the
financial year 2020–2021. And, research by the second author was supported by
IISER (Pune) post-doctoral research grant.

Corresponding Author<EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract: For an associative ring $R$ with identity, we study the absence of
$k$-torsion in ${\rm NK_{1}GQ}(R)$; Bass nil-groups for the general quadratic
or Bak’s unitary groups. By using a graded version of Quillen–Suslin theory we
deduce an analog for the graded rings.

2020 Mathematics Subject Classification: 19-XX, 15-XX, 16-XX, 20-XX

Key words: General linear groups, Elementary subgroups, Quadratic forms,
Higman linearisation, $k$-torsion, Whitehead group - $\rm K_{1}$.

##  1\. Introduction

Let $R$ be an associative ring with identity element $1$. When $R$ is
commutative, we define ${\rm SK_{1}}(R)$ as the kernel of the determinant map
from the Whitehead group ${\rm K_{1}}(R)$ to the group of units of $R$. The
Bass nil-group ${\rm NK_{1}}(R)=\textnormal{ker}({\rm
K_{1}}(R[X])\rightarrow{\rm K_{1}}(R))$; $X=0$. i.e., the subgroup consisting
of elements $[\alpha(X)]\in{\rm K_{1}}(R[X])$ such that $[\alpha(0)]=[{\rm
I}]$. Hence ${\rm K_{1}}(R[X])\cong{\rm NK}_{1}(R)\oplus{\rm K_{1}}(R)$. The
aim of this paper is to study some properties of Bass nil-groups ${\rm
NK_{1}}$ for the general quadratic groups or Bak’s unitary groups.

It is well-known that for many rings, e.g. if $R$ is regular Noetherian,
Dedekind domain, or any ring with finite global dimension, the group ${\rm
NK}_{1}(R)$ is trivial. On the other hand, if it is non-trivial, then it is
not finitely generated as a group. e.g. if $G$ is a non-trivial finite group,
the group ring $\mathbb{Z}G$ is not regular. In many such cases, it is
difficult to compute ${\rm NK}_{1}(\mathbb{Z}[G])$. In [13], D.R. Harmon
proved the triviality of this group when $G$ is finite group of square free
order. C. Weibel, in [20], has shown the non-triviality of this group for $G$
= $\mathbb{Z}/2\oplus\mathbb{Z}/2$, $\mathbb{Z}/4$, and $D_{4}$. Some more
results are known for finite abelian groups from the work of R.D. Martin;
cf.[16]. It is also known (cf.[12]) that for a general finite group $G$, ${\rm
NK}_{1}(R[G])$ is a torsion group for the group ring $R[G]$. In fact, for
trivial ${\rm NK}_{1}(R)$, every element of finite order of ${\rm
NK}_{1}(R[G])$ is some power of the cardinality of $G$. For $R=\mathbb{Z}$,
this is a result of Weibel. In particular, if $G$ is a finite $p$-group ($p$ a
prime), then every element of ${\rm NK}_{1}(\mathbb{Z}[G])$ has $p$-primary
order. In [17], J. Stienstra showed that ${\rm NK_{1}}(R)$ is a ${\rm
W}(R)$-module, where ${\rm W}(R)$ is the ring of big Witt vectors (cf.[11] and
[19]). Consequently, in ([18], §3), C. Weibel observed that if $k$ is a unit
in $R$, then ${\rm SK_{1}}(R[X])$ has no $k$-torsion, when $R$ is a
commutative local ring. Note that if $R$ is a commutative local ring then
${\rm SK_{1}}(R[X])$ coincides with ${\rm NK_{1}}(R)$; indeed, if $R$ is a
local ring then ${\rm SL}_{n}(R)={\rm E}_{n}(R)$ for all $n>0$. Therefore, we
may replace $\alpha(X)$ by $\alpha(X)\alpha(0)^{-1}$ and assume that
$[\alpha(0)]=[{\rm I}]$. In [7], the first author extended Weibel’s result for
arbitrary associative rings. In this paper 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 [1]. The main
ingredient for our proof is an analog of Higman linearisation (for a subclass
of Bak’s unitary group) due to V. Kopeiko; cf.[15]. For the general linear
groups, Higman linearisation (cf.[6]) allows us to show that ${\rm NK_{1}}(R)$
has a unipotent representative. The same result is not true in general for the
unitary nil-groups. Kopeiko’s results in [15] explain a complete description
of the elements of ${\rm NK_{1}GQ}^{\lambda}(R,\Lambda)$ that have (unitary)
unipotent representatives. Followings are the main results in this article.

######  Theorem 1.1.

Let $[\alpha(X)]=\big{[}\begin{pmatrix}A(X)&B(X)\\\
C(X)&D(X)\end{pmatrix}\big{]}\in{\rm NK_{1}GQ}^{\lambda}(R,\Lambda)$ with
$A(X)\in{\rm GL}_{r}(R[X])$ for some $r\in\mathbb{N}$. Then $[\alpha(X)]$ has
no $k$-torsion if $kR=R$.

And, an analog for the graded rings:

######  Theorem 1.2.

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

In the proof of 1.2, we have used a graded version of Quillen–Suslin’s local-
global principle for Bak’s unitary group over graded rings. This unify and
generalize the results proved in [5], [7], [9], and [10].

######  Theorem 1.3.

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

##  2\. Preliminaries

Let $R$ be an associative ring with identity element $1$. Let ${\rm M}(n,R)$
denote the additive group of $n\times n$ matrices, and ${\rm GL}(n,R)$ denote
the multiplicative group of $n\times n$ invertible matrices. Let $e_{ij}$ be
the matrix with $1$ in the $ij$-th position and $0$’s elsewhere. The
elementary subgroup of ${\rm GL}(n,R)$ plays a key role in classical algebraic
K-theory. We recall,

######  Definition 2.1.

Elementary Group ${\rm E}(n,R)$: The subgroup of all matrices in ${\rm
GL}(n,R)$ generated by $\\{{\rm E}_{ij}(\lambda):\lambda\in R,i\neq j\\}$,
where ${\rm E}_{ij}(\lambda)={\rm I}_{n}+\lambda e_{ij}$, and $e_{ij}$ is the
matrix with $1$ in the $ij$-position and $0$’s elsewhere.

######  Definition 2.2.

For $\alpha\in{\rm M}(r,R)$ and $\beta\in{\rm M}(s,R)$, the matrix
$\alpha\perp\beta$ denotes its embedding in ${\rm M}(r+s,R)$ (here $r$ and $s$
are even integers in the non-linear cases), given by

$\alpha\perp\beta=\left(\begin{array}[]{cc}\alpha&0\\\
0&\beta\end{array}\right).$

There is an infinite counterpart: Identifying each matrix $\alpha\in{\rm
GL}(n,R)$ with the large matrix $(\alpha\perp\\{1\\})$ gives an embedding of
${\rm GL}(n,R)$ into ${\rm GL}(n+1,R)$. Let ${\rm
GL}(R)=\underset{n=1}{\overset{\infty}{\cup}}{\rm GL}(n,R)$, and ${\rm
E}(R)=\underset{n=1}{\overset{\infty}{\cup}}{\rm E}(n,R)$ be the corresponding
infinite linear groups.

As a consequence of classical Whitehead Lemma (cf.[3]) due to A. Suslin, one
gets

$[{\rm GL}(R),{\rm GL}(R)]={\rm E}(R).$

######  Definition 2.3.

The quotient group

${\rm K_{1}}(R)=\frac{{\rm GL}(R)}{[{\rm GL}(R),{\rm GL}(R)]}=\frac{{\rm
GL}(R)}{{\rm E}(R)}$

is called the Whitehead group of the ring $R$. For  $\alpha\in{\rm GL}(n,R)$,
let $[\alpha]$ denote its equivalence class in ${\rm K_{1}}(R)$.

In the similar manner we define ${\rm K_{1}}$ group for the other types of
classical groups; viz., the symplectic Whitehead group ${\rm K_{1}}{\rm
Sp}(R)$ and the orthogonal Whitehead group ${\rm K_{1}}{\rm O}(R)$.

This paper explores a uniform framework for classical type groups over graded
structures. Let us begin by recalling the concept of form rings and form
parameter as introduced by A. Bak in [1]. This allows us to give a uniform
definition for classical type groups.

######  Definition 2.4.

(Form rings): Let $R$ be an associative ring with identity, and with an
involution $-:R\rightarrow R$, $a\mapsto\overline{a}$. Let $\lambda\in C(R)$ =
the center of $R$, with the property $\lambda\overline{\lambda}=1$. We define
two additive subgroups of $R$, viz.

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

One checks that for any $x\in R$, $\Lambda_{\rm max}$ and $\Lambda_{\rm min}$
are closed under the conjugation operation $a\mapsto\overline{x}ax$.

A $\lambda$-form parameter on $R$ is an additive subgroup $\Lambda$ of $R$
such that $\Lambda_{\rm min}\subseteq\Lambda\subseteq\Lambda_{\rm max}$, and
$\overline{x}\Lambda x\subseteq\Lambda$ for all $x\in R$. i.e., a subgroup
between two additive groups which is also closed under the conjugation
operation. A pair $(R,\Lambda)$ is called a form ring.

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

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

For more details, see [7], and [8].

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

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

### Elementary Quadratic Matrices :

Let $\rho$ be the permutation, defined by $\rho(i)=n+i$ for $i=1,\dots,n$. For
$a\in R$, and $1\leq i,j\leq n$, we define

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

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

(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
$a\in\Lambda_{\rm max}(R)$. One checks that these above matrices belong to
$\rm GQ(2n,R,\Lambda)$; cf.[1].
$n$-th Elementary Quadratic Group ${\rm EQ}(2n,R,\Lambda)$:
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 ${\rm EQ}(2n,R,\Lambda)$ by $\eta_{ij}(*)$.

Stabilization map: There are standard embeddings:

${\rm GQ}(2n,R,\Lambda)\longrightarrow{\rm GQ}(2n+2,R,\Lambda)$

given by

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\mapsto\begin{pmatrix}a&0&b&0\\\
0&1&0&0\\\ c&0&d&0\\\ 0&0&0&1\end{pmatrix}.$

Hence we have

${\rm GQ}(R,\Lambda)=\underset{\longrightarrow}{\lim}\,\,{\rm
GQ}(2n,R,\Lambda)$.

It is clear that the stabilization map takes generators of ${\rm
EQ}(2n,R,\Lambda)$ to the generators of ${\rm EQ}(2(n+1),R,\Lambda)$. Hence we
have

${\rm EQ}(R,\Lambda)=\underset{\longrightarrow}{\lim}\,\,{\rm
EQ}(2n,R,\Lambda)$

There are standard formulas for the commutators between quadratic elementary
matrices. For details, we refer [1] (Lemma 3.16). In later sections there are
repeated use of those relations. The analogue of the Whitehead Lemma for the
general quadratic groups (cf.[1]) due to Bak allows us to write:

$[{\rm GQ}(R,\Lambda),{\rm GQ}(R,\Lambda)]=[{\rm EQ}(R,\Lambda),{\rm
EQ}(R,\Lambda)]={\rm EQ}(R,\Lambda).$

Hence we define the Whitehead group of the general quadratic group

${\rm K_{1}}{\rm GQ}=\frac{{\rm GQ}(R,\Lambda)}{{\rm EQ}(R,\Lambda)}.$

And, the Whitehead group at the level $m$

${\rm K}_{1,m}{\rm GQ}=\frac{{\rm GQ}_{m}(R,\Lambda)}{{\rm
EQ}_{m}(R,\Lambda)},$

where $m=2n$ in the non-linear cases.

Let $(R,\Lambda)$ be a form ring. We extend the involution of $R$ to the ring
$R[X]$ of polynomials by setting $\overline{X}=X$. As a result we obtain a
form ring $(R[X],\Lambda[X])$.

######  Definition 2.5.

The kernel of the group homomorphism

${\rm K_{1}GQ}(R[X],\Lambda[X])\rightarrow{\rm K_{1}GQ}(R,\Lambda)$

induced from the form ring homomorphism
$(R[X],\Lambda[X])\rightarrow(R,\Lambda):X\mapsto 0$ is denoted by ${\rm
NK_{1}GQ}(R,\Lambda)$. We often say it as Bass nilpotent unitary ${\rm
K_{1}}$-group of $R$, or just unitary nil-group.

From the definition it follows that

${\rm K_{1}GQ}(R[X],\Lambda[X])={\rm K_{1}GQ}(R,\Lambda)\oplus{\rm
NK_{1}GQ}(R,\Lambda).$

In this context, we will use following two types of localizations, mainly over
graded ring $R=R_{0}\oplus R_{1}\oplus R_{2}\oplus\cdots$.

  1. (1)

Principal localization: for a non-nilpotent, non-zero divisor $s$ in $R_{0}$
with $\overline{s}=s$, we consider the multiplicative subgroup
$S=\\{1,s,s^{2},\dots\\}$, and denote localized form ring by
$(R_{s},\Lambda_{s})$.

  2. (2)

Maximal localization: for a maximal ideal $\mathfrak{m}\in\rm Max(R_{0})$, we
take the multiplicative subgroup $S=R_{0}-\mathfrak{m}$, and denote the
localized form ring by $(R_{\mathfrak{m}},\Lambda_{\mathfrak{m}})$.

Blanket assumption: We always assume that $2n\geq 6$.

Next, we recall the well-known “Swan–Weibel’s 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 $a\in R$ will be denoted by
$a=a_{0}+a_{1}+a_{2}+\cdots$, where $a_{i}\in R_{i}$ for each $i$, and all but
finitely many $a_{i}$ are zero. Let $R_{+}=R_{1}\oplus R_{2}\oplus\cdots$.
Graded structure of $R$ induces a graded structure on ${\rm M}_{n}(R)$ (ring
of $n\times n$ matrices).

######  Definition 2.6.

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 $\epsilon:R\rightarrow R[X]$ given by

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

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 at the
${\rm GL}(2n,R)$ level for every $n\geq 1$, i.e., for $\alpha\in{\rm
GL}(2n,R)$ we get a map

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

where $\alpha_{i}\in{\rm M}(2n,R_{i})$. As above for $a\in R_{0}$, we define
$\alpha^{+}(a)$ as

$\alpha^{+}(a)=\epsilon(\alpha)(a).$

######  Notation 2.7.

By ${\rm GQ}(2n,R[X],\Lambda[X],(X))$ we shall mean the group of all quadratic
matrices over $R[X]$, which are ${\rm I}_{2n}$ modulo $(X)$. Also if $R$ is a
graded ring, then by ${\rm GQ}(2n,R,\Lambda,(R_{+}))$ we shall mean the group
of all quadratic matrices over $R$ which are ${\rm I}_{2n}$ modulo $R_{+}$.

The following lemma highlights very crucial fact which we use (repeatedly) in
the proof of “Dilation Lemma”.

######  Lemma 2.8.

Let $R$ be a Noetherian ring and $s\in R$. Then there exists a natural number
$k$ such that the homomorphism ${\rm GQ}(2n,R,\Lambda,s^{k}R)\rightarrow{\rm
GQ}(2n,R_{s},\Lambda_{s})$ $($induced by localization homomorphism
$R\rightarrow R_{s})$ is injective. Moreover, it follows that the induced map
${\rm EQ}(2n,R,\Lambda,s^{k}R)\rightarrow{\rm EQ}(2n,R_{s},\Lambda_{s})$ is
injective.

For the proof of the above lemma we refer [14], Lemma 5.1. Recall that any
module finite ring $R$ is direct limit of its finitely generated subrings.
Also, ${\rm G}(R,\Lambda)=\underset{\longrightarrow}{\lim}\,{\rm
G}(R_{i},\Lambda_{i})$, where the limit is taken over all finitely generated
subring of $R$. Thus, one may assume that $C(R)$ is Noetherian. Hence one may
consider module finite (form) rings $(R,\Lambda)$ with identity.

Now we recall few technical definitions and useful lemmas.

######  Definition 2.9.

A row $(a_{1},a_{2},\dots,a_{n})\in R^{n}$ is said to be unimodular if there
exists $(b_{1},b_{2},\dots,b_{n})\in R^{n}$ such that
$\sum_{i=1}^{n}a_{i}b_{i}=1$. The set of all unimodular rows of length $n$ is
denoted by ${\rm Um}_{n}(R)$.

For any column vector $v\in(R^{2n})^{t}$ we define the row vector
$\widetilde{v}=\overline{v}^{t}\psi_{n}$.

######  Definition 2.10.

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

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

Note that the elementary generators of the groups ${\rm EQ}(2n,R,\Lambda)$ are
of the form ${\rm I}_{2n}+M(*_{1},*_{2})$ for suitably chosen standard basis
vectors.

######  Lemma 2.11.

$($cf.[1]$)$ The group ${\rm E}(2n,R,\Lambda)$ is perfect for $n\geq 3$, i.e.,

$[{\rm EQ}(2n,R,\Lambda),{\rm EQ}(2n,R,\Lambda)]={\rm EQ}(2n,R,\Lambda).$

######  Lemma 2.12.

For all elementary generators of ${\rm GQ}(2n,R,\Lambda)$ we have the
following splitting property: for all $x,y\in R$,

$\eta_{ij}(x+y)=\eta_{ij}(x)\eta_{ij}(y).$

${{\bf Proof:}}$ See pg. 43-44, Lemma 3.16, [1].

######  Lemma 2.13.

Let $G$ be a group, and $a_{i},b_{i}\in G$, for $i=1,2,\ldots,n$. Then for
$r_{i}=\Pi_{j=1}^{i}a_{j}$, we have
$\Pi_{i=1}^{n}r_{i}b_{i}r_{i}^{-1}\Pi_{i=1}^{n}a_{i}$.

######  Lemma 2.14.

The group ${\rm GQ}(2n,R,\Lambda,R_{+})\cap{\rm EQ}(2n,R,\Lambda)$ generated
by the elements of the type $\varepsilon\eta_{ij}(*)\varepsilon^{-1}$, where
$\varepsilon\in{\rm EQ}(2n,R,\Lambda)$ and $*\in R_{+}$.

${{\bf Proof:}}$ Let $\alpha\in{\rm GQ}(2n,R,\Lambda,R_{+})\cap{\rm
EQ}(2n,R,\Lambda)$. Then we can write

$\alpha=\Pi_{k=1}^{r}\eta_{i_{k}j_{k}}(a_{k})$

for some element $a_{k}\in R$, $k=1,\dots,r$. We can write $a_{k}$ as
$a_{k}=(a_{0})_{k}+(a_{+})_{k}$ for some $(a_{0})_{k}\in R_{0}$ and
$(a_{+})_{k}\in R_{+}$. Using Lemma 2.12, we can write $\alpha$ as,

$\alpha=\Pi_{k=1}^{r}(\eta_{i_{k}j_{k}}(a_{0})_{k})(\eta_{i_{k}j_{k}}(a_{+})_{k}).$

Let $\epsilon_{t}=\Pi_{k=1}^{t}\eta_{i_{k}j_{k}}((a_{0})_{k})$ for $1\leq
t\leq r$. By the Lemma 2.13, we have

$\alpha=\left(\Pi_{k=1}^{r}\epsilon_{k}\eta_{i_{k}j_{k}}((a_{+})_{k})\epsilon_{k}^{-1}\right)\left(\Pi_{k=1}^{r}\eta_{i_{k}j_{k}}((a_{0})_{k})\right).$

Let us write
$A=\Pi_{k=1}^{r}\epsilon_{k}\eta_{i_{k}j_{k}}((a_{+})_{k})\epsilon_{k}^{-1}$
and $B=\Pi_{k=1}^{r}\eta_{i_{k}j_{k}}((a_{0})_{k})$. Hence $\alpha=AB$. Let
‘over-line’ denotes the quotient ring modulo $R_{+}$. Now going modulo
$R_{+}$, we have $\overline{\alpha}=\overline{AB}=\bar{A}\bar{B}=\overline{\rm
I}_{2n}\bar{B}=\overline{\rm I}_{2n}$, the last equality holds as
$\alpha\in{\rm GQ}(2n,R,\Lambda,R_{+})$. Hence, $\overline{B}=\overline{{\rm
I}}_{2n}$. Since the entries of $B$ are in $R_{0}$, it follows that $B={\rm
I}_{2n}$. Therefore it follows that

$\alpha=\Pi_{k=1}^{r}\epsilon_{k}\eta_{i_{k}j_{k}}((a_{+})_{k})\epsilon_{k}^{-1}.$

$\Box$

##  3\. Quillen–Suslin Theory for Bak’s Group over Graded Rings

###  3.1. Local–Global Principle

######  Lemma 3.1.

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

${{\bf Proof:}}$ Let $v=\varepsilon e_{1}$. Then we have ${\rm
I}_{2n}+M(v,w)=\varepsilon({\rm I}_{2n}+M(e_{1},w_{1}))\varepsilon^{-1}$,
where $w_{1}=\varepsilon^{-1}w$. Since $\langle e_{1},w_{1}\rangle=\langle
v,w\rangle=0$, we have $w_{1}^{T}=(w_{11},\dots,w_{1n-1},0,\dots,w_{12n})$.
Therefore, since $\lambda\bar{\lambda}=1$, we have

${\rm I}_{2n}+M(v,w)=\prod_{\begin{subarray}{c}1\leq j\leq n\\\ 1\leq i\leq
n-1\end{subarray}}\varepsilon
ql_{in}(-\bar{\lambda}\overline{w}_{1n+i})q\varepsilon_{jn}(-\bar{\lambda}\overline{w}_{1j})ql_{nn}^{-1}(*)\varepsilon^{-1}$

######  Lemma 3.2.

Let $R$ be a graded ring. Let $\alpha\in{\rm EQ}(2n,R,\Lambda)$. Then for
every $a\in R_{0}$ one gets $\alpha^{+}(a)\in{\rm EQ}(2n,R,\Lambda)$.

${{\bf Proof:}}$ Let $\alpha=\Pi_{k=1}^{t}({\rm
I}_{2n}+aM(e_{i_{k}},e_{j_{k}})),$ where $a\in R$ and $t\geq 1$. Hence for
$b\in R_{0}$, we have $\alpha^{+}(b)=\Pi_{k=1}^{t}({\rm
I}_{2n}+a^{+}(b)M(e_{i_{k}},e_{j_{k}}))$. Now taking $v=e_{i}$ and
$v=a^{+}(b)e_{j}$ we have $\langle v,w\rangle=0$ and ${\rm I}_{2n}+M(v,w)={\rm
I}_{2n}+a^{+}(b)M(e_{i},e_{j}))$ which belongs to ${\rm EQ}(2n,R,\Lambda)$ by
Corollary 3.1. Hence we have $\alpha^{+}(b)\in{\rm EQ}(2n,R,\Lambda)$ for
$b\in R_{0}$. $\Box$

######  Lemma 3.3.

(Graded Dilation Lemma) Let $\alpha\in{\rm GQ}(2n,R,\Lambda)$ with
$\alpha^{+}(0)={\rm I}_{2n}$ and $\alpha_{s}\in{\rm EQ}(2n,R_{s},\Lambda_{s})$
for some non-zero-divisor $s\in R_{0}$. Then there exists $\beta\in{\rm
EQ}(2n,R,\Lambda)$ such that

$\beta_{s}^{+}(b)=\alpha_{s}^{+}(b)$

for some $b=s^{l}$ and $l\gg 0$.

${{\bf Proof:}}$ Since $\alpha_{s}\in{\rm EQ}(2n,R_{s},\Lambda_{s})$ with
$(\alpha_{0})_{s}={\rm I}_{2n}$, then $\alpha_{s}=\gamma$, where
$\gamma_{ii}=1+g_{ii}$ where $g_{ii}\in(R^{+})_{s}$ and $\gamma_{ij}=g_{ij}$
for $i\neq j$, where $g_{ij}\in(R^{+})_{s}$. Choose $l$ large enough such that
every denominator of $g_{ij}$ for all $i,j$ divides $s^{l}$. Then by Lemma
3.2, we have $\alpha_{s}^{+}(s^{l})\in{\rm EQ}(2n,R_{s},\Lambda_{s})$. As all
denominator is cleared then $\alpha_{s}^{+}(s^{l})$ permits a natural
pullback. Hence we have $\alpha^{+}(s^{l})\in{\rm EQ}(2n,R,\Lambda).$ Call
this pullback as $\beta$. $\Box$

######  Lemma 3.4.

Let $\alpha_{s}\in{\rm EQ}(2n,R_{s},\Lambda_{s})$ with $\alpha_{s}^{+}(0)={\rm
I}_{2n}$. Then one gets

$\alpha_{s}^{+}(b+d)\alpha_{s}^{+}(d)^{-1}\in{\rm EQ}(2n,R,\Lambda)$

for some $s,d\in R_{0}$ and $b=s^{l},l\gg 0$.

${{\bf Proof:}}$ Since $\alpha_{s}\in{\rm EQ}(2n,R_{s},\Lambda_{s})$, we have
$\alpha_{s}^{+}(X)\in{\rm EQ}(2n,R_{s}[X],\Lambda_{s}[X])$. Let

$\beta^{+}(X)=\alpha^{+}(X+d)\alpha^{+}(d)^{-1},$

where $d\in R_{0}$. Then we have

$\beta^{+}_{s}(X)\in{\rm EQ}(2n,R_{s}[X],\Lambda_{s}[X])$

and $\beta^{+}(0)={\rm I}_{2n}$. Hence by Lemma 3.3, we have, there exists
$b=s^{l}$, $l\gg 0$, such that $\beta^{+}(bX)\in{\rm EQ}(2n,R[X],\Lambda[X])$.
Putting $X=1$, we get our desired result. $\Box$

Proof of Theorem 1.3 – Graded Local-Global Principle:
Since $\alpha_{\mathfrak{m}}\in{\rm
EQ}(2n,R_{\mathfrak{m}},\Lambda_{\mathfrak{m}})$ for all $\mathfrak{m}\in{\rm
Max}(C(R_{0}))$, for each $\mathfrak{m}$ there exists $s\in
C(R_{0})\setminus\mathfrak{m}$ such that $\alpha_{s}\in{\rm
EQ}(2n,R_{s},\Lambda_{s})$. Using Noetherian property we can consider a finite
cover of $C(R_{0})$, say $s_{1}+\dots+s_{r}=1$. From Lemma 3.3, we have
$\alpha^{+}(b_{i})\in{\rm EQ}(2n,R,\Lambda)$ for some $b_{i}=s_{i}^{l_{i}}$
with $b_{1}+\dots+b_{r}=1$. Now consider $\alpha_{s_{1}s_{2}\dots s_{r}}$,
which is the image of $\alpha$ in $R_{s_{1}s_{2}\dots s_{r}}$. By Lemma 2.8,
$\alpha\mapsto\alpha_{s_{1}s_{2}\dots s_{r}}$ is injective. Hence we can
perform our calculation in $R_{s_{1}s_{2}\dots s_{r}}$ and then pull it back
to $R$.

$\begin{split}\alpha_{s_{1}s_{2}\dots s_{r}}=&\alpha_{s_{1}s_{2}\dots
s_{r}}^{+}(b_{1}+b_{2}+\dots+b_{r})\\\
=&((\alpha_{s_{1}})_{s_{2}s_{3}\dots})^{+}(b_{1}+\dots+b_{r})((\alpha_{s_{1}})_{s_{2}s_{3}\dots})^{+}(b_{2}+\dots+b_{r})^{-1}\dots\\\
&((\alpha_{s_{i}})_{s_{1}\dots\hat{s_{i}}\dots
s_{r}})^{+}(b_{i}+\dots+b_{r})((\alpha_{s_{i}})_{s_{1}\dots\hat{s_{i}}\dots
s_{r}})^{+}(b_{i+1}+\dots+b_{r})^{-1}\\\ &((\alpha_{s_{r}})_{s_{1}s_{2}\dots
s_{r-1}})^{+}(b_{r})((\alpha_{s_{r}})_{s_{1}s_{2}\dots
s_{r-1}})^{+}(0)^{-1}\end{split}$

Observe that $((\alpha_{s_{i}})_{s_{1}\dots\hat{s_{i}}\dots
s_{r}})^{+}(b_{i}+\dots+b_{r})((\alpha_{s_{i}})_{s_{1}\dots\hat{s_{i}}\dots
s_{r}})^{+}(b_{i+1}+\dots+b_{r})^{-1}\in{\rm EQ}(2n,R,\Lambda)$ due to Lemma
3.4 (here $\hat{s_{i}}$ means we omit $s_{i}$ in the product
$s_{1}\dots\hat{s_{i}}\dots s_{r}$), and hence $\alpha_{s_{1}s_{2}\dots
s_{r}}\in{\rm EQ}(2n,R_{s_{1}\dots s_{r}},\Lambda_{s_{1}\dots s_{r}})$. This
proves $\alpha\in{\rm EQ}(2n,R,\Lambda)$. $\Box$

###  3.2. Normality and Local–Global

Next we are going to show that if $K$ is a commutative ring with identity and
$R$ is an associative $K$-algebra such that $R$ is finite as a left
$K$-module, then the normality criterion of elementary subgroup is equivalent
to the Local-Global principle for quadratic group. (One can also consider $R$
as a right $K$-algebra.)

######  Lemma 3.5.

$($Bass; cf.[4]$)$ Let $A$ be an associative $B$-algebra such that $A$ is
finite as a left $B$-module and $B$ be a commutative local ring with identity.
Then $A$ is semilocal.

######  Theorem 3.6.

$($cf.[7]$)$ Let $A$ be a semilocal ring $($not necessarily commutative$)$
with involution. Let $v\in{\rm Um}_{2n}(A)$.Then $v\in e_{1}{\rm EQ}(2n,A)$.
In other words the group ${\rm EQ}(2n,A)$ acts transitively on ${\rm
Um}_{2n}(A)$.

Before proving the next theorem we need to recall a theorem from [7]:

######  Theorem 3.7.

$($Local-Global Principle$)$ Let $A$ be an associative $B$-algebra such that
$A$ is finite as a left $B$-module and $B$ be a commutative ring with
identity.. If $\alpha(X)\in{\rm GQ}(2n,A[X],\Lambda[X])$, $\alpha(0)=\rm{\rm
I}_{2n}$ and $\alpha_{\mathfrak{m}}(X)\in{\rm
EQ}(2n,A_{\mathfrak{m}}[X],\Lambda_{\mathfrak{m}}[X])$ for every maximal ideal
$\mathfrak{m}\in{\rm Max}(B)$, then $\alpha\in{\rm EQ}(2n,A[X],\Lambda[X])$.

######  Theorem 3.8.

Let $K$ be a commutative ring with unity and $R=\oplus_{i=0}^{\infty}R_{i}$ be
a graded $K$-algebra such that $R_{0}$ is finite as a left $K$-module. Then
for $n\geq 3$ the following are equivalent:

$(1)$ ${\rm EQ}(2n,R,\Lambda)$ is a normal subgroup of ${\rm
GQ}(2n,R,\Lambda)$.

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

${{\bf Proof:}}$ $(1)\Rightarrow(2)$ We have proved the Lemma 3.1 for any form
ring with identity and shown that the local-global principle is a consequence
of Lemma 3.1. So, the result is true in particular if we have ${\rm
EQ}(2n,R,\Lambda)$ is a normal subgroup of ${\rm GQ}(2n,R,\Lambda)$.

$(2)\Rightarrow(1)$ Since polynomial rings are special case of graded rings,
the result follows by using the Theorem 3.7. Let $\alpha\in{\rm
EQ}(2n,R,\Lambda)$ and $\beta\in{\rm GQ}(2n,R,\Lambda)$. Then we have $\alpha$
can be written as product of the matrices of the form $({\rm I}_{2n}+\beta
M(*_{1},*_{2})\beta^{-1})$, with $\langle*_{1},*_{2}\rangle=0$ where $*_{1}$
and $*_{2}$ are suitably chosen basis vectors. Let $v=\beta*_{1}$. Then we can
write $\beta\alpha\beta^{-1}$ as a product of the matrices of the form ${\rm
I}_{2n}+M(v,w)$ for some $w\in R^{2n}$. We must show that each ${\rm
I}_{2n}+M(v,w)\in{\rm EQ}(2n,R,\Lambda)$.

Consider $\gamma={\rm I}_{2n}+M(v,w)$. Then $\gamma^{+}(0)={\rm I}_{2n}$. By
Lemma 3.5 we have the ring $S^{-1}R$ is semilocal where
$S=K\setminus\mathfrak{m}$, and $\mathfrak{m}\in{\rm Max}(K)$. Since $v\in{\rm
Um}_{2n}(R)$, then by Theorem 3.6, we have $v\in{\rm
EQ}(2n,S^{-1}R,S^{-1}\Lambda)e_{1}$. Therefore by applying Lemma 3.1 to the
ring $(S^{-1}R,S^{-1}\Lambda)$, we have $\gamma_{\mathfrak{m}}\in{\rm
EQ}(2n,R_{\mathfrak{m}},\Lambda_{\mathfrak{m}})$ for every maximal ideal
$\mathfrak{m}\in{\rm Max}(K)$. Hence by hypothesis we have $\gamma\in{\rm
EQ}(2n,R,\Lambda)$. This completes the proof. $\Box$

######  Remark 3.9.

We conclude that the local-global principle for the elementary subgroups and
their normality properties are equivalent.

##  4\. Bass Nil Group ${\rm{\rm NK_{1}}{\rm GQ}(R)}$

In this section recall some basic definitions and properties of the
representatives of ${\rm{\rm NK_{1}}{\rm GQ}(R)}$. We represent any element of
${\rm M}_{2n}(R)$ as $\begin{pmatrix}a&b\\\ c&d\end{pmatrix},$ where
$a,b,c,d\in{\rm M}_{n}(R)$. For $\alpha=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}$ we call $\begin{pmatrix}a&b\end{pmatrix}$ the upper half of
$\alpha$. Let $(R,\lambda,\Lambda)$ be a form ring. By setting
$\bar{\Lambda}=\\{\bar{a}:a\in\Lambda\\}$ we get another form ring
$(R,\bar{\lambda},\bar{\Lambda})$. We can extend the involution of $R$ to
${\rm M}_{n}(R)$ by setting $(a_{ij})^{*}=(\overline{a}_{ji})$.

######  Definition 4.1.

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 $\bar{\Lambda}$-Hermitian if
$\beta=-\bar{\lambda}\beta^{*}$ and all the diagonal entries of $\beta$ are
contained in $\bar{\Lambda}$.

######  Remark 4.2.

A matrix $\alpha\in{\rm M}_{n}(R)$ is $\Lambda$-Hermitian if and only if
$\alpha^{*}$ is $\bar{\Lambda}$-Hermitian. The set of all $\Lambda$-Hermitian
matrices forms a group under matrix multiplication.

######  Lemma 4.3.

[15, Example 2] Let $\beta\in{\rm GL}_{n}(R)$ be a $\Lambda$-Hermitian matrix.
Then the matrix $\alpha^{*}\beta\alpha$ is $\Lambda$-Hermitian for every
$\alpha\in{\rm GL}_{n}(R)$.

######  Definition 4.4.

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)

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

  2. (2)

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

  3. (3)

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

  4. (4)

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

######  Remark 4.5.

The set of all $\Lambda$-quadratic matrices of order $2n$ forms a group called
$\Lambda$-quadratic group. We denote this group by ${\rm
GQ}^{\lambda}(2n,R,\Lambda)$. If $2\in R^{*}$, then we have $\Lambda_{\rm
min}=\Lambda_{\rm max}$. In this case notions of quadratic groups and notions
of $\Lambda$-quadratic groups coincides. Also this happens when
$\Lambda=\Lambda_{\rm max}$. Hence quadratic groups are special cases of
$\Lambda$-quadratic groups. Other classical groups appear as
$\Lambda$-quadratic groups in the following way. Let $R$ be a commutative ring
with trivial involution. Then

${\rm GQ}^{\lambda}(2n,R,\Lambda)=\begin{cases}{\rm Sp}_{2n}(R),&\text{if
}\lambda=-1\text{ and }\Lambda=\Lambda_{\rm max}=R\\\ {\rm
O}_{2n}(R),&\text{if }\lambda=1\text{ and }\Lambda=\Lambda_{\rm
min}=0\end{cases}$

And for general linear group ${\rm GL}_{n}(R)$, we have, ${\rm GL}_{n}(R)={\rm
GQ}^{1}(2n,H(R),\Lambda=\Lambda_{\rm max})$, where $\mathbb{H}(R)$ denotes the
ring $R\oplus R^{op}$ with $R^{op}$ is the opposite ring of $R$ and the
involution on $\mathbb{H}(R)$ is defined by $\overline{(x,y)}=(y,x)$. Thus the
study of $\Lambda$-quadratic matrices unifies the study of quadratic matrices.

We recall following results from [15].

######  Lemma 4.6.

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

${{\bf Proof:}}$ Let $\alpha\in{\rm GQ}^{\lambda}(2n,R,\Lambda)$. In view of
$(2)$ of Definition 4.4, we have, $a^{*}d={\rm I}_{n}$. Hence $a$ is
invertible and $d=(a^{*})^{-1}$. Converse holds by $(2)$ of Definition 4.4.
$\Box$

######  Definition 4.7.

Let $\alpha\in{\rm GL}_{n}(R)$ 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 4.8.

In a similar way we can show that matrices of the form
$T_{12}(\beta):=\begin{pmatrix}{\rm I}_{n}&\beta\\\ 0&{\rm
I}_{n}\end{pmatrix}$ is $\Lambda$-quadratic matrix if and only if $\beta$ is
$\bar{\Lambda}$-Hermitian. And the matrix of the form
$T_{21}(\gamma):=\begin{pmatrix}{\rm I}_{n}&0\\\ \gamma&{\rm
I}_{n}\end{pmatrix}$ is $\Lambda$-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 4.9.

The $\Lambda$-elementary quadratic group is denoted by ${\rm
EQ}^{\lambda}(2n,R,\Lambda)$ and defined by the group generated by $2n\times
2n$ matrices of the form $\mathbb{H}(\alpha)$ where $\alpha\in{\rm E}_{n}(R)$,
$T_{12}(\beta)$ and $\beta$ is $\bar{\Lambda}$-Hermitian and $T_{21}(\gamma)$
is $\gamma$ $\Lambda$-Hermitian.

######  Lemma 4.10.

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

${{\bf Proof:}}$ Let $A\in{\rm GQ}^{\lambda}(2n,R,\Lambda)$. Then by $(4)$ of
Definition 4.4, we have $\alpha\delta^{*}={\rm I}_{n}$ 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 4.3. Hence $\alpha^{-1}\beta$ is
$\bar{\Lambda}$-Hermitian. Conversely, the condition on $A$ will fulfill the
condition $(4)$ of Definition 4.4. Hence $A$ is $\Lambda$-quadratic. Since
$\alpha^{-1}\beta$ is $\bar{\Lambda}$-Hermitian,

$T_{12}(-\alpha^{-1}\beta)\in{\rm EQ}^{\lambda}(2n,R,\Lambda)$

and $AT_{12}(\alpha^{-1}\beta)=\mathbb{H}(\alpha)$. Thus
$A\equiv\mathbb{H}(\alpha)\pmod{{\rm EQ}^{\lambda}(2n,R,\Lambda)}$. $\Box$

A similar proof will prove the following:

######  Lemma 4.11.

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

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

######  Lemma 4.12.

Let $\alpha=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in{\rm
GQ}^{\lambda}(2n,R,\Lambda)$. Then

$\alpha\equiv\mathbb{H}(a)\pmod{{\rm EQ}^{\lambda}(4n,R,\Lambda)}$

if $a\in{\rm GL}_{n}(R).$ Moreover, if $a\in{\rm E}_{n}(R)$, then
$\alpha\equiv\mathbb{H}(a)\pmod{{\rm EQ}^{\lambda}(2n,R,\Lambda)}$.

${{\bf Proof:}}$ By same argument as given in Lemma 4.10, we have $a^{-1}b$ is
$\Lambda$-Hermitian. Hence $T_{12}(-a^{-1}b)\in{\rm
EQ}^{\lambda}(2n,R,\Lambda)$, and consequently $\alpha
T_{12}(-a^{-1}b)=\begin{pmatrix}a&0\\\ c&d^{\prime}\end{pmatrix}\in{\rm
GQ}^{\lambda}(2n,R,\Lambda)$ for some $d^{\prime}\in{\rm GL}_{n}(R)$. Hence by
Lemma 4.11, we get

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

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

######  Definition 4.13.

Let $\alpha=\begin{pmatrix}a_{1}&b_{1}\\\ c_{1}&d_{1}\end{pmatrix}\in{\rm
M}_{2r}(R)$, $\beta=\begin{pmatrix}a_{2}&b_{2}\\\
c_{2}&d_{2}\end{pmatrix}\in{\rm M}_{2s}(R)$. As before, we define
$\alpha\perp\beta$, and consider an embedding

$\rm GQ^{\lambda}(2n,R,\Lambda)\rightarrow\rm
GQ^{\lambda}(2n+2,R,\Lambda),\,\,\,\alpha\mapsto\alpha\perp{\rm I}_{2}.$

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

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

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

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

######  Remark 4.14.

Likewise in the quadratic case, the kernel of the group homomorphism

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

induced from the form ring homomorphism
$(R[X],\Lambda[X])\rightarrow(R,\Lambda);X\mapsto 0$ is denoted by ${\rm
NK_{1}GQ}^{\lambda}(R,\Lambda)$. Since the $\Lambda$-quadratic groups are
subclass of the quadratic groups, the Local-global principle holds for
$\Lambda$-quadratic groups. We use this throughout for the next section.

##  5\. Absence of torsion in ${\rm NK_{1}}{\rm GQ}^{\lambda}(R,\Lambda)$

In this section we give the proof of Theorem 1.1 and Theorem 1.2. In [6], the
proof of the theorem for the linear case is based on two key results, viz. the
Higman linearisation, and a lemma on polynomial identity in the truncated
polynomial rings. Here we recall the lemma with its proof to highlight its
connection with the big Witt vectors. Recently, in [15], V. Kopeiko deduced an
analog of Higman linearisation process for a subclass of the general quadratic
groups.

######  Definition 5.1.

For a associative ring $R$ with unity we consider the truncated polynomial
ring

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

######  Lemma 5.2.

$($cf.[6], 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 $\deg(Q(X))<t-r$.

${{\bf Proof:}}$ Let us write $P(X)=a_{0}+a_{1}X+\cdots+a_{t}X^{t}$. Then we
can write $P(X)=P(0)+XP^{\prime}(X)$ 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. $\Box$
Remark. 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 will follow that
the $a_{i}$’s are uniquely determined. In fact, if $R$ is commutative then
$a_{i}$’s are the $i$-th component of the ghost vector corresponding to the
big Witt vector of $(1+XP(X))\in{\rm W}(R)=(1+XR[[X]])^{\times}$. For details
see ([11], $\mathcal{x}$I).

######  Lemma 5.3.

Let $R$ be a ring with $1/k\in R$ and $P(X)\in R[X]$. Assume $P(0)$ lies in
the center of $R$. Then

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

in the ring $R_{t}$ for some $r>0$ and $Q(X)\in R[X]$ with $\deg(Q(X))<t-r$.

Following result is due to V. Kopeiko, cf. [15].

######  Proposition 5.4.

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

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

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 5.5.

Let $[\alpha]\in{\rm NK_{1}}{\rm GQ}^{\lambda}(R,\Lambda)$ has the
representation $[a;b,c]_{n}$ for some $a,b,c\in{\rm M}_{n}(R)$ according to
Proposition 5.4. Then

$[\alpha]=[\mathbb{H}({\rm I}_{r}-aX)]$

in ${\rm NK_{1}}{\rm GQ}^{\lambda}(R,\Lambda)$ if $({\rm I}_{r}-aX)\in{\rm
GL}_{r}(R)$.

${{\bf Proof:}}$ By Lemma 4.12 we have $[a;b,c]_{n}\equiv\mathbb{H}({\rm
I}_{r}-aX)\pmod{{\rm EQ}^{\lambda}(2r,R[X],\Lambda[X])}$. Hence we have
$[\alpha]=[\mathbb{H}({\rm I}_{r}-aX)]$ in ${\rm NK_{1}}{\rm
GQ}^{\lambda}(R,\Lambda)$. $\Box$

Proof of Theorem 1.1:
By the Theorem 5.4, we have $[\alpha]=[[a;b,c]_{n}]$ for some $a,b,c\in{\rm
M}_{s}(R)$ and for some natural numbers $n$ and $s$. Note that in the Step $1$
of the Proposition 5.4, the invertibility of the first corner of the matrix
$\alpha$ will not be changed during the linearisation process. Also the
invertibility of the first corner is preserved in the remaining steps of the
Proposition 5.4. Therefore since the first corner matrix $A(X)\in{\rm
GL}_{r}(R[X])$, then we have $({\rm I}_{s}-aX)\in{\rm GL}_{s}(R[X])$. By
Corollary 5.5, we have $[\alpha]=[\mathbb{H}({\rm I}_{s}-aX)]$. Now let
$[\alpha]$ be a $k$-torsion. Then we have $[\mathbb{H}({\rm I}_{r}-aX)]$ is a
$k$-torsion. Since $({\rm I}_{r}-aX)$ is invertible, it follows that $a$ is
nilpotent. Let $a^{t+1}=0$. Since $[({\rm I}_{r}-aX)]^{k}=[{\rm I}]$ in ${\rm
K_{1}}{\rm GQ}^{\lambda}(R[X],\Lambda[X])$, then by arguing as given in [7],
we have $[{\rm I}_{r}-aX]=[I]$ in ${\rm K_{1}}{\rm
GQ}^{\lambda}(R[X],\Lambda[X])$. This completes the proof. $\Box$

Proof of Theorem 1.2 – (Graded Version):
Consider the ring homomorphism $f:R\rightarrow R[X]$ defined by

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

Then

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

Let $\mathfrak{m}$ be a maximal ideal in $R_{0}$. By Theorem 1.1, we have

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

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

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

in ${\rm NK_{1}}{\rm GQ}^{\lambda}(R,\Lambda)$, as required. $\Box$
Acknowledgment: We thank Sergey Sinchuk and V. Kopeiko for many useful
discussions.

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