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# On group rings of virtually abelian groups

Joan Claramunt  Fakultät für Mathematik und Informatik, Universität Leipzig,
Augustusplatz 10, 04109 Leipzig, Germany<EMAIL_ADDRESS>and  Łukasz Grabowski  Fakultät für Mathematik und Informatik, Universität
Leipzig, Augustusplatz 10, 04109 Leipzig, Germany<EMAIL_ADDRESS>leipzig.de

###### Abstract.

Let $\Gamma$ be a finitely generated torsion-free group. We show that the
statement of $\Gamma$ being virtually abelian is equivalent to the statement
that the $*$-regular closure of the group ring ${\mathbb{C}}[\Gamma]$ in the
algebra of (unbounded) operators affiliated to the group von Neumann algebra
is a central division algebra. More generally, for any field $k$, it is shown
that $k[\Gamma]$ embeds into a central division algebra in case $\Gamma$ is
virtually abelian. We take advantage of this result in order to develop a
criterion for existence of units in the group ring $k[\Gamma]$. We develop
this criterion in the particular case of $\Gamma$ being the Promislow’s group.

###### 2010 Mathematics Subject Classification:

Primary 16S34, 20C07; Secondary 16E50

Both authors were supported by the ERC Starting Grant “Limits of Structures in
Algebra and Combinatorics” No. 805495.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Central division algebras
    2. 2.2 Virtually abelian groups
    3. 2.3 Von Neumann regular and $*$-regular rings
    4. 2.4 Crossed products and group algebras
  3. 3 Central division algebras and virtually abelian groups
    1. 3.1 From virtually abelian groups to central division algebras
    2. 3.2 From central division algebras to virtually abelian groups
  4. 4 On the existence of units in group rings of virtually abelian groups
  5. 5 The case of the Promislow’s group
    1. 5.1 The $\omega$-degree of an element
    2. 5.2 Criteria for existence of $\omega$-conjugates with $\omega$-degree $(2,2)$

##  1\. Introduction

Recently Gardam [5] presented a counterexample to one of the famous
conjectures on group rings attributed to Kaplansky: the unit conjecture. This
conjecture asserts that the only units in the group ring of a torsion-free
group $\Gamma$, with coefficients in a field $k$, are the trivial ones, namely
the ones of the form $\lambda g$ with $\lambda\in
k\backslash\\{0\\},g\in\Gamma$. Gardam disproved this conjecture by explicitly
exhibiting a group $G$, a field $k$ and the non-trivial unit. His field is the
field $k=\mathbb{F}_{2}$ of order $2$, and the group $G$ is the Promislow’s
group, the unique torsion-free $3$-dimensional crystallographic group with
finite abelianization. Later on, Murray [8] disproved the conjecture by using
the same grup but over fields of arbitrary positive characteristic
$\text{char}(k)>0$ following the same lines as in [5].

The unit conjecture is still open when the coefficient field is a field of
characteristic zero. While we develop some interesting new criteria for the
existence of units in this case, we have not been able to achieve our initial
goal of settling this conjecture, even in the case of the Promislow’s group.

The initial point of our investigation is the following observation. Let
${\pazocal R}(\Gamma)$ be the $*$-regular closure of the group ring
${\mathbb{C}}[\Gamma]$ inside the algebra ${\pazocal U}(\Gamma)$ of (possibly
unbounded) operators affiliated to the group von Neumann algebra ${\pazocal
N}(\Gamma)$. When $\Gamma$ is torsion-free we may equivalently define
${\pazocal R}(\Gamma)$ to be the Ore localization of ${\mathbb{C}}[\Gamma]$.
Then ${\pazocal R}(\Gamma)$ is a central division algebra, i.e. it is a
division algebra which is finite-dimensional over its center. With this in
mind, the problem of finding units which lie specifically in
${\mathbb{C}}[\Gamma]$ has a flavor which is very similar to the problem of
determining the units in the ring of integers of a finite field extension of
the rational numbers.

In the case of the Promislov’s group, the division algebra ${\pazocal
R}(\Gamma)$ is $16$-dimensional over its center, and the center is a
$2$-dimensional extension of the field of rational functions in $3$ variables.
First let us note that, irrespectively of the unit conjecture, it is
interesting to ask which groups $\Gamma$ have the property that ${\pazocal
R}(\Gamma)$ is a central division algebra. The following theorem answers this
question.

######  Theorem (Theorem 3.1).

If $\Gamma$ is a finitely generated torsion-free group, then $\Gamma$ is
virtually abelian if and only if the $*$-regular closure ${\pazocal
R}(\Gamma):={\pazocal R}({\mathbb{C}}[\Gamma],{\pazocal U}(\Gamma))$ of
${\mathbb{C}}[\Gamma]$ in ${\pazocal U}(\Gamma)$ is a central division
algebra.

Let us now briefly describe how we develop a criterion for the existence of
units in ${\mathbb{C}}[\Gamma]$ in the general case when $\Gamma$ is an
arbitrary virtually abelian group. The details are contained in Section 4.
Suppose we are given a polynomial $p(x)$ of degree $d\geq 2$ with coefficients
in $\text{Z}(k[\Gamma])$, the center of $k[\Gamma]$. Suppose further that
there exists a root of $p(x)$ in $\text{Z}(k[\Gamma])$, call it $\kappa$. We
can then write $p(x)$ as the product $(x-\kappa)q(x)$ for some polynomial
$q(x)$ with coefficients also in $\text{Z}(k[\Gamma])$. Then if $T\in
k[\Gamma]$ is any element such that $p(T)=1$, we obtain

$1=(T-\kappa)q(T),$

so $T-\kappa$ is a unit in $k[\Gamma]$, which may be trivial or not.

Let us focus on the case when $\Gamma$ is the Promislow’s group $G$. First,
let us make some preliminary observations about units in $k[G]$ when
$\text{char}(k)\neq 2$. As we show in Propositions 5.14 and 5.15, the non-
trivial units in $k[G]$ can be classified into two types. Let us write
${\pazocal Z}$ for the center of ${\pazocal R}(\Gamma)$ and let $U$ be a non-
trivial unit in $k[G]$. Write $\omega^{U}=a+bs+ct+du\in k[G]$, where
$a,b,c,d\in k[{\mathbb{Z}}^{3}]$, and $s,t,u$ are the standard generator of
$G$. Then both of the fields ${\pazocal Z}(a+bs)$ and ${\pazocal Z}(ct+du)$
have the same degree over ${\pazocal Z}$ and this common degree is equal to
either $2$ or $4$. We say that $U$ is of type $2$ in the first case and of
type $4$ in the second case. A direct check shows that the units found by
Murray in [8] when $\text{char}{(k)}>2$ are of type $2$ (Proposition 5.21),
and as such it makes sense to ask about criteria for existence of units of
type $2$ when $\text{char}{(k)}=0$.

With this in mind, let us state our criterion for the existence of units of
type $2$ in ${\mathbb{C}}[G]$.

######  Theorem (Theorem 5.24).

Let $G$ be the Promislow group. Then there exist non-trivial units of type 2
in ${\mathbb{C}}[G]$ if and only if the following system of quadratic equation
in $9$ variables with coefficients in ${\pazocal Z}$ has non-trivial solutions
in ${\pazocal Z}$:

$\displaystyle\diamond\text{ }$
$\displaystyle(A^{2}-1)\omega_{x}^{2}+C_{1}^{2}(\kappa_{x}+1)(\kappa_{y}-1)-C_{2}^{2}(\kappa_{x}-1)(\kappa_{y}+1)+C_{3}^{2}(\kappa_{x}+1)(\kappa_{y}+1)-C_{4}^{2}(\kappa_{x}-1)(\kappa_{y}-1)$
$\displaystyle\phantom{(A^{2}-1)\omega_{x}^{2}}+2D_{1}^{2}(\kappa_{z}-1)-2D_{2}^{2}\omega_{x}^{2}(\kappa_{z}+1)+2D_{3}^{2}(\kappa_{z}+1)-2D_{4}^{2}\omega_{x}^{2}(\kappa_{z}-1)=0,$
$\displaystyle\diamond\text{ }$
$\displaystyle(\kappa_{x}-1)C_{2}C_{4}-(\kappa_{x}+1)C_{1}C_{3}=0,$
$\displaystyle\diamond\text{ }$
$\displaystyle(\kappa_{x}-1)C_{2}D_{2}+C_{3}D_{3}=0,$
$\displaystyle\diamond\text{ }$
$\displaystyle(\kappa_{x}+1)C_{1}D_{2}+C_{4}D_{3}=0,$
$\displaystyle\diamond\text{ }$
$\displaystyle(\kappa_{x}-1)C_{4}D_{4}+C_{1}D_{1}=0,$
$\displaystyle\diamond\text{ }$
$\displaystyle(\kappa_{x}+1)C_{3}D_{4}+C_{2}D_{1}=0.$

Here $\kappa_{g}=\frac{1}{2}(g+g^{-1})\in{\mathbb{C}}[G]$ and
$\omega_{g}=\frac{1}{2}(g-g^{-1})\in{\mathbb{C}}[G]$ for $g\in G$.

This criterion should be compared with the determinant condition stated in
[2], which gives a single polynomial equation of degree $4$ with $16$
variables. In this sense, our criterion gives a ‘simpler’ polynomial equation
than the determinant condition.

######  Remark.

It is not known if the group ring ${\mathbb{C}}[G]$ of the Promislow’s group
contains any non-trivial unitaries. In Theorem 5.26 we state an if-and-only-if
criterion similar to the theorem above for the existence of unitaries.

The paper is organized as follows. In Section 2 we collect the definitions of
von Neumann regular and $*$-regular rings, together with the definition of the
$*$-regular closure of a subring in a $*$-regular ring. We also recall the
construction of a crossed product of a ring with a group, from which the group
ring of a ring is a particular case of it. In Section 3 we present and prove
the first of our main results, namely the characterization of virtually
abelian groups $\Gamma$ in terms of algebraic properties of the $*$-regular
closure of the group ring ${\mathbb{C}}[\Gamma]$ inside ${\pazocal
U}(\Gamma)$, the algebra of (unbounded) operators affiliated to the group von
Neumann algebra ${\pazocal N}(\Gamma)$. We use part of this characterization,
in Section 4, to briefly present a general procedure for finding units in
group algebras of virtually abelian groups. We specialize, and further
analyze, this procedure in the case $\Gamma$ is the Promislow’s group $G$,
culminating in the criterias presented in Theorems 5.24 and 5.26, which give
necessary and sufficient conditions for finding non-trivial units (resp.
unitaries) in the group ring of $G$. This is done in Section 5.

##  2\. Preliminaries

###  2.1. Central division algebras

Let $k$ be any field. Throughout this section, and for the rest of the paper,
a $k$-algebra will be a unital, associative algebra $A$ over the field $k$.
Its center will be denoted by $\mathrm{Z}(A)$. It always holds that
$k\subseteq\mathrm{Z}(A)$. A $k$-algebra $A$ is called central if this
inclusion is in fact an equality, that is $k=\mathrm{Z}(A)$.

A $k$-algebra $A$ is a division algebra if every non-zero element $a\in A$ is
two-sided invertible, meaning that there are elements $b,c\in A$ such that
$ab=1$ and $ca=1$. In this case necessarily $b=c$ and this common element is
the unique element satisfying the two equations $ax=1,ya=1$. We will denote it
by $a^{-1}$, called the inverse of $a$. So a division algebra over $k$ is a
division ring which is also an algebra over $k$.

In this paper we adopt the following convention.

######  Definition 2.1.

A $k$-algebra $A$ will be called a central division algebra if

  1. a)

$A$ is central;

  2. b)

$A$ is a division algebra;

  3. c)

$A$ is finite-dimensional over $\mathrm{Z}(A)=k$.

We will denote by $[A:k]$ the dimension of $A$ over $k$.

More generally, if ${\pazocal R}$ denotes any division ring and if $V$ is a
left ${\pazocal R}$-module, then $[V:{\pazocal R}]$ will denote the dimension
of $V$ over ${\pazocal R}$.

###  2.2. Virtually abelian groups

A group $\Gamma$ is said to be virtually abelian if there exists an abelian
subgroup of finite index. Equivalently, if there exists a normal abelian
subgroup of finite index.

In this paper we will work with finitely generated groups only. In this case,
a finitely generated virtually abelian group $\Gamma$ contains a finitely
generated normal abelian subgroup $N$ of finite index. If moreover $\Gamma$ is
assumed to be torsion-free, then necessarily $N$ is isomorphic to
${\mathbb{Z}}^{n}$ for some $n\geq 1$.

###  2.3. Von Neumann regular and $*$-regular rings

A unital ring $R$ is (von Neumann) regular if for every element $x\in R$ there
exists $y\in R$ such that $xyx=x$. Note that the element $e=xy$ is an
idempotent in $R$, that is $e^{2}=e$.

A $*$-regular ring is a regular ring $R$ endowed with a proper involution
$*\colon R\to R$, proper meaning that $x^{*}x=0$ implies $x=0$. It is not
difficult to show that in a $*$-regular ring, the following hold. For every
element $x\in R$ there exist unique projections $e,f\in R$ (that is elements
$p\in R$ satisfying $p^{2}=p^{*}=p$) such that $xR=eR$ and $Rx=Rf$, and
moreover there exists a unique element $y\in fRe$, called the relative inverse
of $x$ and denoted by $\overline{x}$, such that $xy=e$ and $yx=f$. We denote
by $\text{LP}(x)$ the projection $e$, termed the left projection of $x$, and
by $\text{RP}(x)$ the projection $f$, termed the right projection of $x$.

For any subset $S$ of a $*$-regular ring $R$, there exists a smallest
$*$-regular subring of $R$ containing $S$, denoted by ${\pazocal R}(S,R)$ and
termed the $*$-regular closure of $S$ in $R$ [1, Proposition 6.2]. It always
contains the division closure ${\pazocal D}(S,R)$, which we recall it is
defined as the smallest subring of $R$ containing $S$ and closed under taking
inverse of elements, when they exist in $R$ (so if $x\in{\pazocal D}(S,R)$ is
invertible in $R$ with inverse $x^{-1}$, then $x^{-1}\in{\pazocal D}(S,R)$).

###  2.4. Crossed products and group algebras

Let $R$ be a unital ring and $\Gamma$ be any countable group. A crossed
product, denoted by $R\ast\Gamma$, is a unital ring in which every element can
be uniquely written as a finite sum

$\sum_{g\in\Gamma}r_{g}g$

with $r_{g}\in R$. The sum is given componentwise, and the product is
described by the rule

$(rg)\cdot(sh)=r\text{ }\sigma_{g}(s)\text{ }\tau(g,h)\text{ }gh$

for $r,s\in R$ and $g,h\in\Gamma$, where $\tau:\Gamma\times\Gamma\rightarrow
R^{\times}$ and $\sigma:\Gamma\rightarrow\text{Aut}(R)$ are maps satisfying
the following properties:

  1. a)

$\sigma_{1}=\text{id}_{R}$, and $\tau(1,h)=\tau(g,1)=1$ for all
$g,h\in\Gamma$;

  2. b)

$\tau(g,h)\tau(gh,f)=\sigma_{g}(\tau(h,f))\tau(g,hf)$ for all $g,h,f\in\Gamma$
(so $\tau$ is a $2$-cocycle for $\sigma$);

  3. c)

$\tau(g,h)\sigma_{gh}(r)=\sigma_{g}(\sigma_{h}(r))\tau(g,h)$ for all
$g,h\in\Gamma$, $r\in R$.

Here $R^{\times}$ is the group of units of $R$.

In the particular case that $\tau$ and $\sigma$ are the trivial maps, we
obtain the group ring $R[\Gamma]$. We will we particularly interested in the
case that $R=k$ is a field. If moreover $R=k$ is a field endowed with an
involution $*$, it linearly extends to an involution on $k[\Gamma]$ by the
rule

$(\lambda g)^{*}=\lambda^{*}g^{-1}$

for $\lambda\in k,g\in\Gamma$.

######  Definition 2.2.

An element $u\in k[\Gamma]$ is called a unit if there exists $v\in k[\Gamma]$
such that $1=uv=vu$. It will be called a trivial unit if $u=\lambda g$ for
some $\lambda\in k\backslash\\{0\\}$ and $g\in\Gamma$, otherwise it is called
a non-trivial unit.

Take now $k={\mathbb{C}}$ with complex conjugation as involution. Let
$l^{2}(\Gamma)$ denote the Hilbert space with orthonormal basis the elements
of $\Gamma$. Then ${\mathbb{C}}[\Gamma]$ acts faithfully on $l^{2}(\Gamma)$ by
left multiplication, and so we may identify ${\mathbb{C}}[\Gamma]$ as a subset
of ${\pazocal B}(l^{2}(\Gamma))$, i.e. as bounded operators on
$l^{2}(\Gamma)$. The group von Neumann algebra of $\Gamma$, denoted by
${\pazocal N}(\Gamma)$, is defined to be the weak closure of
${\mathbb{C}}[\Gamma]$ in ${\pazocal B}(l^{2}(\Gamma))$. Algebraically, it is
characterized as follows: ${\pazocal N}(\Gamma)$ consists of all bounded
operators $T\in{\pazocal B}(l^{2}(\Gamma))$ which commute with the action of
${\mathbb{C}}[\Gamma]$ on $l^{2}(\Gamma)$ given by right multiplication.

Also, we denote by ${\pazocal U}(\Gamma)$ the algebra of (possibly unbounded)
operators affiliated to the von Neumann algebra ${\pazocal N}(\Gamma)$. Again,
we can give an algebraic characterization of it: ${\pazocal N}(\Gamma)$ is an
Ore domain, and ${\pazocal U}(\Gamma)$ coincides with the classical ring of
quotients of ${\pazocal N}(\Gamma)$ [9, Proposition 2.8].

We denote by ${\pazocal D}(\Gamma):={\pazocal
D}({\mathbb{C}}[\Gamma],{\pazocal U}(\Gamma))$ the division closure of
${\mathbb{C}}[\Gamma]$ in ${\pazocal U}(\Gamma)$, and by ${\pazocal
R}(\Gamma):={\pazocal R}({\mathbb{C}}[\Gamma],{\pazocal U}(\Gamma))$ the
$*$-regular closure of ${\mathbb{C}}[\Gamma]$ in ${\pazocal U}(\Gamma)$. The
latter exists since ${\pazocal U}(\Gamma)$ is $*$-regular [9, Note 2.11]. We
have the inclusions

${\mathbb{C}}[\Gamma]\subseteq{\pazocal D}(\Gamma)\subseteq{\pazocal
R}(\Gamma)\subseteq{\pazocal U}(\Gamma).$

Note that for $H\leq\Gamma$ a subgroup of $\Gamma$, the inclusion
${\mathbb{C}}[H]\hookrightarrow{\mathbb{C}}[\Gamma]$ induces natural
embeddings ${\pazocal N}(H)\hookrightarrow{\pazocal N}(\Gamma)$, ${\pazocal
U}(H)\hookrightarrow{\pazocal U}(\Gamma)$ and ${\pazocal
R}(H)\hookrightarrow{\pazocal R}(\Gamma)$.

##  3\. Central division algebras and virtually abelian groups

Let $\Gamma$ be a finitely generated torsion-free group. In this section we
give a necessary and sufficient condition on the $*$-regular closure
${\pazocal R}(\Gamma)$ in order to guarantee that $\Gamma$ be virtually
abelian. The main theorem is the following.

######  Theorem 3.1.

Let $\Gamma$ be a finitely generated torsion-free group. The following
statements are equivalent.

  1. (1)

$\Gamma$ is virtually abelian.

  2. (2)

${\pazocal R}(\Gamma)$ is a central division algebra.

The rest of the section is devoted to prove Theorem 3.1 above, although some
tangential results will also be obtained.

###  3.1. From virtually abelian groups to central division algebras

Let us start by taking $\Gamma$ to be a finitely generated, torsion-free,
virtually abelian group. Our convention for conjugation is that
$h^{g}:=ghg^{-1}$ for $g,h\in\Gamma$.

Let $N\leq\Gamma$ be a maximal normal abelian subgroup of finite index, so
$N\cong{\mathbb{Z}}^{n}$ for some $n\geq 1$. Let $F:=\Gamma/N$, which is
finite. We have a natural action of $F$ on $N$ given by conjugation $n\mapsto
n^{f}$, $f\in F$, which linearly extends to an action of $F$ on $k[N]$, also
denoted by $p\mapsto p^{f}$, $f\in F$. Take $\tilde{F}$ to be a fixed right
transversal for $N$ i $\Gamma$ and, for simplicity, we take the representative
of $N$ to be the unit element $1\in\Gamma$, that is $\tilde{1}=1$. If
$\pi:\Gamma\rightarrow F$ denotes the natural quotient map, the image of an
element $\tilde{f}\in\tilde{F}$ will be simply denoted by $f$, that is
$f=\pi(\tilde{f})$. Therefore, for $p\in k[N]$ and $\tilde{f}\in\tilde{F}$,

$p^{\tilde{f}}=p^{f}.$

Fix now any field $k$ endowed with an involution $\ast$. Note that every
element of the group algebra $k[\Gamma]$ can be uniquely written as

$\sum_{\tilde{f}\in\tilde{F}}p_{\tilde{f}}\tilde{f}$

for some Laurent polynomials $p_{\tilde{f}}\in k[N]\cong k[x_{1}^{\pm
1},...,x_{n}^{\pm 1}]$. Therefore, $k[\Gamma]$ is obtained as a crossed
product

$k[\Gamma]=k[N]\ast\tilde{F}$

whose multiplicative structure is given by

$(p\tilde{f})\cdot(q\tilde{g})=p\text{ }q^{f}\text{ }\tau(f,g)\text{
}\widetilde{fg},$

where $\tau:F\times F\rightarrow N$ is the $2$-cocycle given by
$\tau(f,g)=\tilde{f}\text{ }\tilde{g}\text{ }\widetilde{fg}^{-1}$. In
particular, we see that the map
$F\rightarrow\text{Aut}(k[N]),f\mapsto\cdot^{f}$ is a group homomorphism.

Since $\Gamma$ is virtually abelian, the group ring $k[\Gamma]$ satisfies
Kaplansky’s zero-divisor conjecture, i.e. it is a domain [6, Theorem 1.4]. In
fact it is an Ore domain. We denote by ${\pazocal Q}:={\pazocal Q}(k[\Gamma])$
its Ore field of fractions, which is a division algebra over $k$.

The following notation will be used throughout the section.

######  Notation 3.2.

  1. a)

We will denote by ${\pazocal Z}$ the center of ${\pazocal Q}$, which is a
field.

  2. b)

For a subset $A\subseteq{\pazocal Q}$, we denote by ${\pazocal Z}(A)$ the
minimal division algebra over $k$ containing ${\pazocal Z}$ and the elements
of $A$.

  3. c)

For any division ring ${\pazocal R}$ and any $\alpha,\beta\in{\pazocal R}$
with $\beta\neq 0$, conjugation of $\alpha$ by $\beta$ will be denoted by
$\alpha^{\beta}:=\beta\alpha\beta^{-1}$.

We start by computing, in the case $k={\mathbb{C}}$, the division closure
${\pazocal D}(\Gamma)$ and the $*$-regular closure ${\pazocal R}(\Gamma)$ in
terms of ${\pazocal Q}$. It turns out that these three rings coincide.

######  Proposition 3.3.

For $k={\mathbb{C}}$, we have

${\pazocal Q}={\pazocal D}(\Gamma)={\pazocal R}(\Gamma).$

###### Proof.

The inclusions ${\pazocal Q}\subseteq{\pazocal D}(\Gamma)\subseteq{\pazocal
R}(\Gamma)$ are obvious. Now, ${\pazocal Q}$ is a division algebra, so it is
von Neumann regular. It is also $*$-closed and contains $k[\Gamma]$, so
${\pazocal R}(\Gamma)\subseteq{\pazocal Q}$, finishing the proof of the
proposition. ∎

Let us return to the general setting where $k$ is an arbitrary field with
involution $\ast$. We define ${\pazocal L}$ to be the field of fractions of
$k[N]\cong k[x_{1}^{\pm 1},...,x_{n}^{\pm 1}]$, which can be seen as a
subfield of ${\pazocal Q}$. In fact, it can be identified with
$k(x_{1},...,x_{n})$, the field of rational functions in $n$ variables
$x_{1},...,x_{n}$, and we will do so throughout the paper. Each element $f\in
F$ acts on ${\pazocal L}$ by conjugation, so it naturally induces an element
of $\text{Autl}({\pazocal L}/k)$, which we will denote by $\alpha_{f}$. It is
easy to see, using maximality of the abelian subgroup $N$, that the map
$\alpha\colon F\rightarrow\text{Aut}({\pazocal L}/k),f\mapsto\alpha_{f}$ is an
injective group homomorphism, so we identify $F$ with the subgroup
$H:=\text{im}(\alpha)$.

######  Proposition 3.4.

The following statements hold true.

  1. (i)

The ring ${\pazocal Q}$ is Artinian, and moreover

${\pazocal Q}={\pazocal L}\ast\tilde{F}$

for a suitable crossed product. In particular, $\tilde{F}$ is a basis of
${\pazocal Q}$ over ${\pazocal L}$, where ${\pazocal L}$-multiplication is
given from the left. Therefore $[{\pazocal Q}:{\pazocal L}]=|F|$.

  2. (ii)

If we denote by ${\pazocal L}^{H}$ the fixed subfield of ${\pazocal L}$
associated with the subgroup $H\leq\emph{Aut}({\pazocal L}/k)$, then

${\pazocal Z}={\pazocal L}^{H}.$

In particular, ${\pazocal Q}$ is finite-dimensional over ${\pazocal Z}$, and
in fact $[{\pazocal Q}:{\pazocal Z}]=|F|^{2}$.

###### Proof.

  1. (i)

Denote by ${\pazocal A}$ the subring of ${\pazocal Q}$ generated by ${\pazocal
L}$ and $\Gamma$. Every element of ${\pazocal A}$ is a finite sum of monomials

$x_{1}x_{2}\cdots x_{r}$

where either $x_{i}\in{\pazocal L}$ or $x_{i}\in\tilde{F}$. Note that if
$x\in{\pazocal L}$ and $g=n\tilde{f}\in\Gamma$ for unique elements $n\in
N,\tilde{f}\in\tilde{F}$, then

$gx=n\tilde{f}x=n\tilde{f}x\tilde{f}^{-1}\tilde{f}=nx^{\tilde{f}}\tilde{f}\in\sum_{\tilde{f}\in\tilde{F}}{\pazocal
L}\tilde{f}.$

With this we deduce that ${\pazocal A}=\sum_{\tilde{f}\in\tilde{F}}{\pazocal
L}\tilde{f}$. To prove that this sum is direct, suppose that we have

(3.1)
$\sum_{\tilde{f}\in\tilde{F}}x_{\tilde{f}}\tilde{f}=\sum_{\tilde{f}\in\tilde{F}}y_{\tilde{f}}\tilde{f}$

for some $x_{\tilde{f}},y_{\tilde{f}}\in{\pazocal L}$. Write

$x_{\tilde{f}}=p_{\tilde{f}}q_{\tilde{f}}^{-1},\quad
y_{\tilde{f}}=r_{\tilde{f}}s_{\tilde{f}}^{-1}$

where each $p_{\tilde{f}},q_{\tilde{f}},r_{\tilde{f}},s_{\tilde{f}}\in k[N]$,
$q_{\tilde{f}},s_{\tilde{f}}\neq 0$. By multiplying (3.1) on the left with the
element $h=\prod_{\tilde{f}\in\tilde{F}}q_{\tilde{f}}s_{\tilde{f}}\neq 0$ we
get

$\sum_{\tilde{f}\in\tilde{F}}x_{\tilde{f}}h\tilde{f}=\sum_{\tilde{f}\in\tilde{F}}y_{\tilde{f}}h\tilde{f},$

which is an equality in $k[\Gamma]$. But $k[\Gamma]=k[N]\ast\tilde{F}$, so
necessarily $x_{\tilde{f}}h=y_{\tilde{f}}h$ for every $\tilde{f}\in\tilde{F}$,
which implies $x_{\tilde{f}}=y_{\tilde{f}}$ as $h\neq 0$. This proves that the
sum $\sum_{\tilde{f}\in\tilde{F}}{\pazocal L}\tilde{f}$ is direct, and so
${\pazocal A}={\pazocal L}\ast\tilde{F}$ for a suitable crossed product. In
particular, ${\pazocal A}$ is Artinian, and we deduce that ${\pazocal
A}={\pazocal Q}$. Indeed, given $ab^{-1}\in{\pazocal Q}$ with $a,b\in
k[\Gamma]$, $b\neq 0$, then the chain of ideals

$b{\pazocal A}\supseteq b^{2}{\pazocal Q}\supseteq\cdots\supseteq
b^{n}{\pazocal A}\supseteq\cdots$

stabilizes, so we can find a positive integer $k\geq 1$ and $c\in{\pazocal A}$
such that $b^{k}=b^{k+1}c$. This implies that $b^{-1}=c\in{\pazocal A}$, and
so $ab^{-1}\in{\pazocal A}$, proving the equality ${\pazocal Q}={\pazocal
A}={\pazocal L}\ast\tilde{F}$.

The second part now easily follows.

  2. (ii)

Let us first show the inclusion ${\pazocal L}^{H}\subseteq{\pazocal Z}$. Given
$x\in{\pazocal L}^{H}$, it is clear that $x$ commutes with any $n\in N$. Take
now $\tilde{f}\in\tilde{F}$. We compute

$\displaystyle\tilde{f}x\tilde{f}^{-1}$
$\displaystyle=x^{\tilde{f}}=x^{f}=\alpha_{f}(x)=x,$

so $x$ also commutes with $\tilde{F}$. Therefore $x$ is central in
$k[\Gamma]$, and so also central in ${\pazocal Q}$.

For the other inclusion ${\pazocal Z}\subseteq{\pazocal L}^{H}$, we first note
the following. Due to maximality of $N\trianglelefteq\Gamma$, we have that
$\ker(N\xrightarrow{\cdot^{f}}N)\neq N$ for any $f\in F\backslash\\{1\\}$.
This says that, for any $\tilde{f}\in\tilde{F}\backslash\\{1\\}$, there exists
$n\in N$ such that $n^{f}\neq n$.

Take $x\in{\pazocal Z}$. Using part (i) we can uniquely write it as

$x=\sum_{\tilde{f}\in\tilde{F}}l_{\tilde{f}}\tilde{f}$

for some $l_{\tilde{f}}\in{\pazocal L}$. Fix
$\tilde{g}\in\tilde{F}\backslash\\{1\\}$, and consider $n\in N$ such that
$n^{g}\neq n$. Then $x$ commutes with $n$, so we have the equality

$\sum_{\tilde{f}\in\tilde{F}}l_{\tilde{f}}n^{f}\tilde{f}=\sum_{\tilde{f}\in\tilde{F}}l_{\tilde{f}}n\tilde{f}.$

By uniqueness, necessarily

$l_{\tilde{f}}(n^{f}-n)=0$

for all $\tilde{f}\in\tilde{F}$. In particular, $l_{\tilde{g}}=0$. We thus
conclude that $x=l_{\tilde{1}}\tilde{1}=l_{1}\in{\pazocal L}$. Now, given any
$f\in F$, we compute

$\alpha_{f}(x)=\tilde{f}x\tilde{f}^{-1}=x$

as $x\in{\pazocal Z}$. This says that $x\in{\pazocal L}^{H}$, and hence
establishes the other inclusion. The last statement now follows from part (i)
and the fact that ${\pazocal L}/{\pazocal L}^{H}$ is a finite Galois extension
with Galois group $H$.

∎

As a consequence of part (ii) of the above proposition, we see that ${\pazocal
Q}$ is finite-dimensional over its center ${\pazocal Z}$. This turns
${\pazocal Q}$ into a central division algebra when considered as an algebra
over ${\pazocal Z}$. Now one of the implications of Theorem 3.1 is immediate.

###### Proof of $(1)\implies(2)$ of Theorem 3.1.

Just use Proposition 3.3 and the conclusions above. ∎

###  3.2. From central division algebras to virtually abelian groups

We start the section by stating the following theorem.

######  Theorem 3.5.

Let $\Gamma$ be a finitely generated torsion-free group with ${\pazocal
R}(\Gamma)$ being a semisimple ring. Suppose that ${\pazocal R}(\Gamma)$ is
finitely generated as a $\emph{Z}({\pazocal R}(\Gamma))$-module. Then $\Gamma$
is virtually abelian.

Before proving it, we need a couple of technical lemmas.

######  Lemma 3.6.

Let $G$ be a group with ${\pazocal R}(G)$ being a semisimple ring. Let $H\leq
G$ be a subgroup, and define ${\pazocal K}:={\pazocal R}(G)\cap{\pazocal
U}(H)$. Then ${\pazocal K}$ is a regular ring.

Moreover, ${\pazocal R}(G)$ becomes a ${\pazocal K}$-module, and if ${\pazocal
R}(G)$ is finitely generated as a ${\pazocal K}$-module, then $[G:H]<+\infty.$

###### Proof.

First, note that ${\pazocal K}$ can be considered as a subring of ${\pazocal
U}(G)$ because of the embedding ${\pazocal U}(H)\hookrightarrow{\pazocal
U}(G)$. Also, note that all three rings ${\pazocal R}(G),{\pazocal U}(H)$ and
${\pazocal U}(G)$ are $*$-regular.

Let us first prove that ${\pazocal K}$ is regular. Let $x\in{\pazocal K}$ be
given. Since $x\in{\pazocal R}(G)\subseteq{\pazocal U}(G)$ and $x\in{\pazocal
U}(H)$, there exist unique projections $e_{1},f_{1}\in{\pazocal R}(G)$,
$e_{2},f_{2}\in{\pazocal U}(G)$ and $e_{3},f_{3}\in{\pazocal U}(H)$ such that

$x{\pazocal R}(G)=e_{1}{\pazocal R}(G),\quad x{\pazocal U}(G)=e_{2}{\pazocal
U}(G),\quad x{\pazocal U}(H)=e_{3}{\pazocal U}(H)$

and

${\pazocal R}(G)x={\pazocal R}(G)f_{1},\quad{\pazocal U}(G)x={\pazocal
U}(G)f_{2},\quad{\pazocal U}(H)x={\pazocal U}(H)f_{3},$

and moreover there exist unique elements $\overline{x}_{1}\in{\pazocal
R}(G),\overline{x}_{2}\in{\pazocal U}(G)$ and $\overline{x}_{3}\in{\pazocal
U}(H)$ such that $x\overline{x}_{i}=e_{i}$ and $\overline{x}_{i}x=f_{i}$ for
all $i=1,2,3$. In particular, note that $e_{i}x=x=xf_{i}$ for all $i=1,2,3$.
But now we have

$e_{1}{\pazocal U}(G)=x\overline{x}_{1}{\pazocal U}(G)\subseteq x{\pazocal
U}(G)=e_{1}x{\pazocal U}(G)\subseteq e_{1}{\pazocal U}(G),$ $e_{3}{\pazocal
U}(G)=x\overline{x}_{3}{\pazocal U}(G)\subseteq x{\pazocal
U}(G)=e_{3}x{\pazocal U}(G)\subseteq e_{3}{\pazocal U}(G),$

and so $e_{1}{\pazocal U}(G)=e_{3}{\pazocal U}(G)=x{\pazocal
U}(G)=e_{2}{\pazocal U}(G)$.By uniqueness, $e:=e_{1}=e_{2}=e_{3}$. Similarly,
$f:=f_{1}=f_{2}=f_{3}$. But now by uniqueness of the elements
$\overline{x}_{i}$ we also conclude that
$\overline{x}:=\overline{x}_{1}=\overline{x}_{2}=\overline{x}_{3}\in{\pazocal
R}(G)\cap{\pazocal U}(H)={\pazocal K}$. This proves that ${\pazocal K}$ is a
regular ring, since

$x\overline{x}x=ex=x.$

It is clear that ${\pazocal R}(G)$ is a (right) ${\pazocal K}$-module. Suppose
now that ${\pazocal R}(G)$ is generated by $n$ elements as a ${\pazocal
K}$-module, say by $\\{r_{1},...,r_{n}\\}\subseteq{\pazocal R}(G)$. This gives
us a surjective ${\pazocal K}$-module map

$\alpha\colon{\pazocal K}^{n}\twoheadrightarrow{\pazocal
R}(G),\quad(u_{1},...,u_{n})\mapsto\sum_{i=1}^{n}r_{i}u_{i}.$

Suppose, by way of contradiction, that $[G:H]=+\infty$, and take
$S=\\{s_{1},...,s_{n+1}\\}$ a set of transversals of $H$ consisting of $n+1$
elements. This set gives us a ${\pazocal K}$-module map

$\beta\colon{\pazocal K}^{n+1}\rightarrow{\pazocal
R}(G),\quad(u_{1},...,u_{n+1})\mapsto\sum_{i=1}^{n+1}s_{i}u_{i}.$

But since free modules are projective, we can lift $\beta$ to a ${\pazocal
K}$-module map $\gamma\colon{\pazocal K}^{n+1}\to{\pazocal K}$ such that

${{\pazocal K}^{n+1}}$${{\pazocal K}^{n}}$${{\pazocal
R}(G)}$$\scriptstyle{\gamma}$$\scriptstyle{\beta}$$\scriptstyle{\alpha}$

If it were the case that $\beta$ is injective, then so would be $\gamma$. But
regular rings are absolutely flat, so tensoring $\gamma$ with
$\text{id}_{{\pazocal R}(G)}$ would yield an injective ${\pazocal
R}(G)$-module map

$\gamma\otimes\text{id}_{{\pazocal R}(G)}\colon{\pazocal
R}(G)^{n+1}\to{\pazocal R}(G)^{n}.$

This is impossible as ${\pazocal R}(G)$ is a semisimple ring, hence Artinian.

We conclude that $\beta$ is not injective, which means that there exist
elements $u_{1},...,u_{n+1}\in{\pazocal K}$ not all zero such that

$u:=\sum_{i=1}^{n+1}s_{i}u_{i}=0.$

We now think of each $u_{i}\in{\pazocal K}\subseteq{\pazocal U}(H)$ as a
densely defined closed operator $u_{i}\colon D_{i}\to l^{2}(H)$ commuting with
the right action of $H$, with $D_{i}\subseteq l^{2}(H)$ dense in $l^{2}(H)$
and invariant under the right action of $H$. The operator $u\colon D\to
l^{2}(H)$ is defined on $D:=\bigcap_{i=1}^{n+1}D_{i}$, which is dense in
$l^{2}(H)$ because each $D_{i}$ is essentially dense (see [7, Lemma 8.3]). Now
for each $l\in D$, we compute

$0=u(l)=\sum_{i=1}^{n+1}s_{i}u_{i}(l),$

with each $s_{i}u_{i}(l)\in s_{i}l^{2}(H)\subseteq l^{2}(G)$. But since
$S=\\{s_{1},...,s_{n+1}\\}$ is a set of transversals of $H$ in $G$, the
Hilbert subspaces $s_{1}l^{2}(H),...,s_{n+1}l^{2}(H)$ are pairwise orthogonal.
We thus deduce that each $s_{i}u_{i}(l)=0$, that is $u_{i}(l)=0$. We conclude
that each operator $u_{i}$ is zero on the dense subspace $D$. Since each
$u_{i}$ is closed, this implies that $u_{i}\equiv 0$. This contradicts the
fact that at least one of the elements $u_{1},...,u_{n+1}$ is non-zero.
Therefore necessarily $[G:H]<+\infty$, and in fact we have proved the more
stronger result $[G:H]\leq n$. ∎

######  Lemma 3.7.

Let $G$ be a group and consider $\Delta(G)$ to be the subgroup of $G$
consisting of all elements whose conjugacy class in $G$ is finite. Then

$\emph{Z}({\pazocal U}(G))\subseteq{\pazocal U}(\Delta(G)).$

###### Proof.

Take $x\in\text{Z}({\pazocal U}(G))$, so we can write it as $x=ab^{-1}$ for
some $a,b\in{\pazocal N}(G)$. In fact, by using the polar decomposition and
the spectral theorem for elements of ${\pazocal U}(G)$, we can even take $a,b$
to be in the center of ${\pazocal N}(G)$. Thus in order to prove the lemma it
is enough to show the inclusion $\text{Z}({\pazocal N}(G))\subseteq{\pazocal
N}(\Delta(G))$. For this, we will use the identification of ${\pazocal N}(G)$
as a subspace of $l^{2}(G)$ given by

${\pazocal N}(G)\hookrightarrow l^{2}(G),\quad T\mapsto T(1).$

So take $T\in\text{Z}({\pazocal N}(G))$ and write it as $T=\sum_{g\in
G}\lambda_{g}g\in l^{2}(G)$. In particular $T$ commutes with every $h\in G$,
and so we deduce that $\lambda_{hgh^{-1}}=\lambda_{g}$ for all $g,h\in G$. So
if $g$ has infinite conjugacy class in $G$, the $l^{2}$-summability of $T$
implies that necessarily $\lambda_{g}=0$. This shows that $T$ is supported on
elements in $\Delta(G)$, which is what we wanted to prove. ∎

We are now ready to prove Theorem 3.5.

###### Proof of Theorem 3.5.

Write ${\pazocal Z}:=\text{Z}({\pazocal R}(\Gamma))$ for the center of
${\pazocal R}(\Gamma)$. Let $\Delta(\Gamma)$ be the subgroup of $\Gamma$
consisting of all elements $g\in G$ whose conjugacy class in $\Gamma$ is
finite. Consider

${\pazocal K}:={\pazocal R}(\Gamma)\cap{\pazocal U}(\Delta(\Gamma)).$

We claim that ${\pazocal Z}\subseteq{\pazocal K}$. For this, let
$z\in{\pazocal Z}=\text{Z}({\pazocal R}(\Gamma))$. In particular, $z$ commutes
with every $g\in\Gamma$, and hence it commutes with every element of
${\pazocal N}(\Gamma)$. Since ${\pazocal U}(\Gamma)$ is the classical ring of
quotients of ${\pazocal N}(\Gamma)$, we conclude that $z$ also commutes with
every element of ${\pazocal U}(\Gamma)$. This shows that

${\pazocal Z}\subseteq{\pazocal R}(\Gamma)\cap\text{Z}({\pazocal U}(\Gamma)).$

The claim follows due to Lemma 3.7.

Now since ${\pazocal R}(\Gamma)$ is finitely generated as a ${\pazocal
Z}$-module, it is also finitely generated as a ${\pazocal K}$-module since
${\pazocal Z}\subseteq{\pazocal K}$. Thus Lemma 3.6 tells us that
$[\Gamma:\Delta(\Gamma)]<+\infty$. In particular, since $\Gamma$ is finitely
generated and $\Delta(\Gamma)$ is a subgroup of finite index, then so is
$\Delta(\Gamma)$. Pick $S=\\{g_{1},..,g_{r}\\}$ to be a finite generating
subset of $\Delta(\Gamma)$, and consider the centralizer of $S$ in
$\Delta(\Gamma)$, that is

$C_{\Delta(\Gamma)}(S)=\\{h\in\Delta(\Gamma)\mid hgh^{-1}=g\text{ for all
}g\in S\\}.$

Note that

$C_{\Delta(\Gamma)}(S)=\bigcap_{i=1}^{r}C_{\Delta(\Gamma)}(\\{g_{i}\\}).$

For each $g_{i}\in S$, since its conjugacy class is finite, say with
$n_{i}\in{\mathbb{N}}$ elements, we can pick $h_{i1},...,h_{in_{i}}\in\Gamma$
such that the conjugacy class of $g_{i}$ consists of the elements

$\\{h_{i1}g_{i}h_{i1}^{-1},...,h_{in_{i}}g_{i}h_{in_{i}}^{-1}\\}.$

In this case, the set $J_{i}:=\\{h_{i1},...,h_{in_{i}}\\}$ contains a
transversal of $C_{\Delta(\Gamma)}(\\{g_{i}\\})$ in $\Delta(\Gamma)$. In
particular, $[\Delta(\Gamma):C_{\Delta(\Gamma)}(\\{g_{i}\\})]<+\infty$. Since
the intersection of finite index subgroups is a finite index subgroup, we
conclude that

$[\Delta(\Gamma):C_{\Delta(\Gamma)}(S)]<+\infty.$

But being $S$ a generating set for $\Delta(\Gamma)$ implies that the subgroup
$C_{\Delta(\Gamma)}(S)$ is central, and therefore abelian. Finally,

$[\Gamma:C_{\Delta(\Gamma)}(S)]=[\Gamma:\Delta(\Gamma)]\cdot[\Delta(\Gamma):C_{\Delta(\Gamma)}(S)]<+\infty,$

and so $C_{\Delta(\Gamma)}(S)$ is an abelian subgroup of $\Gamma$ with finite
index. This proves precisely that $\Gamma$ is virtually abelian, as required.
∎

We can finally prove the other implication of Theorem 3.1.

###### Proof of $2)\implies 1)$ of Theorem 3.1.

This is a special case of Theorem 3.5. ∎

##  4\. On the existence of units in group rings of virtually abelian groups

In this section we present a general procedure which can help in the
understanding of units in group rings of virtually abelian groups. This
procedure will be applied to the case of the Promislow’s group in Section 5.
We will follow the same notation as in Section 3.1. Since we have already
showed that ${\pazocal Q}$ is a central division algebra, the following result
from the theory of central division algebras will prove useful for us in the
sequel.

######  Theorem 4.1.

Let ${\pazocal R}$ be a division algebra, finite-dimensional over its center
$\emph{Z}({\pazocal R})$. The following statements hold.

  1. (1)

$[{\pazocal R}:\emph{Z}({\pazocal R})]=n^{2}$ for some $n\in{\mathbb{N}}$.
Furthermore, if ${\pazocal K}\subseteq{\pazocal R}$ is a maximal subfield,
then $[{\pazocal K}:\emph{Z}({\pazocal R})]=n$.

  2. (2)

(Skolem-Noether) Let ${\pazocal A}_{1},{\pazocal A}_{2}\subseteq{\pazocal R}$
be two isomorphic $\emph{Z}({\pazocal R})$-subalgebras. Then for any
$\emph{Z}({\pazocal R})$-algebra homomorphism $\phi\colon{\pazocal
A}_{1}\to{\pazocal A}_{2}$ there is an element $b\in{\pazocal R}$ such that
$\phi$ is given by conjugation by $b$. In symbols,

$\phi(a)=a^{b}$

for all $a\in{\pazocal A}_{1}$.

###### Proof.

Part (1) follows from [4, Theorem 3.10 and Corollary 3.17]. Part (2) is a
special case of [4, Theorem 3.14]. ∎

An immediate observation, due to Theorem 4.1 (1) and Proposition 3.4 (ii), is
that ${\pazocal L}\subseteq{\pazocal Q}$ is a maximal subfield of ${\pazocal
Q}$ which contains its center ${\pazocal Z}={\pazocal L}^{H}$.

We now expose the key idea on constructing units in $k[\Gamma]$. Let
$p(x)\in\text{Z}(k[\Gamma])[x]$ be a polynomial with coefficients in the ring
$\text{Z}(k[\Gamma])\subseteq{\pazocal Z}$, and assume that $1-p(x)$ is not
irreducible over $\text{Z}(k[\Gamma])$, so we can write

$1-p(x)=q_{1}(x)q_{2}(x)$

for some non-trivial polynomials $q_{1}(x),q_{2}(x)\in\text{Z}(k[\Gamma])[x]$.
We then observe that, if $T\in k[\Gamma]$ is a root of the polynomial $p(x)$,
then

$1=q_{1}(T)q_{2}(T),$

so $q_{1}(T)$ is a unit in $k[\Gamma]$, which can be either trivial or non-
trivial.

We will concentrate in polynomials of the form

$P_{\kappa,d}(x)=x^{d}-\kappa^{d}+1$

for some $\kappa\in\text{Z}(k[\Gamma])$ and $d\in{\mathbb{N}}$, for which in
this case

$1-P_{\kappa,d}(x)=\kappa^{d}-x^{d}=(\kappa-x)(x^{d-1}+\kappa
x^{d-2}+\cdots+\kappa^{d-2}x+\kappa^{d-1}).$

######  Definition 4.2.

Let $\omega\in k[\Gamma]$ be a root of $P_{\kappa,d}(x)$. An element
$T\in{\pazocal Q}$ is said to be an _$\omega$ -conjugate of degree $d$_ if
$T^{d}=\omega^{d}$. We call such a $T$ _trivial_ if it is of the form
$\xi\omega$ where $\xi\in k$ is a $d^{\text{th}}$ root of unity. If $T$ is not
trivial, we say that it is _non-trivial_.

For $\omega\neq 0$, Theorem 4.1 (2) implies that every $\omega$-conjugate is
conjugate, in ${\pazocal Q}$, to $\omega$. Hence the above definition is
consistent.

The following proposition summarizes the previous idea on finding units in
$k[\Gamma]$.

######  Proposition 4.3.

Let $\omega\in k[\Gamma]$ be a non-zero root of $P_{\kappa,d}(x)$, and let
$T\in{\pazocal Q}$ be an $\omega$-conjugate of degree $d$.

  1. (i)

The element $U_{T}:=\kappa-T$ is a unit in ${\pazocal Q}$, with inverse
$U_{T}^{-1}=T^{d-1}+\kappa T^{d-2}+\cdots+\kappa^{d-2}T+\kappa^{d-1}$.

  2. (ii)

$T\in k[\Gamma]$ if and only if $U_{T}\in k[\Gamma]$, in which case
$U_{T}^{-1}$ also belongs in $k[\Gamma]$.

Conversely, let $U\in{\pazocal Q}\backslash\\{0\\}$, and define
$T_{U}:=\omega^{U}=U\omega U^{-1}$.

  1. (i)

The element $T_{U}$ is an $\omega$-conjugate of degree $d$.

  2. (ii)

If moreover $U,U^{-1}\in k[\Gamma]$, then $T_{U}\in k[\Gamma]$.

###### Proof.

The first part is a simple computation:

$U_{T}U_{T}^{-1}=(\kappa-T)(T^{d-1}+\kappa
T^{d-2}+\cdots+\kappa^{d-2}T+\kappa^{d-1})=\kappa^{d}-T^{d}=\kappa^{d}-\omega^{d}=1.$

Similarly $U_{T}^{-1}U_{T}=1$. Moreover, from the definition $U_{T}=\kappa-T$
it is clear that $T\in k[\Gamma]$ if and only if $U_{T}\in k[\Gamma]$, in
which case

$U_{T}^{-1}=T^{d-1}+\kappa T^{d-2}+\cdots+\kappa^{d-2}T+\kappa^{d-1}\in
k[\Gamma]$

as well.

Conversely, let $U\in{\pazocal Q}\backslash\\{0\\}$ and define
$T_{U}:=\omega^{U}$. We compute

$T_{U}^{d}=U\omega^{d}U^{-1}=\omega^{d}$

as $\omega^{d}=\kappa^{d}-1\in{\pazocal Z}$. Hence $T_{U}$ is an
$\omega$-conjugate of degree $d$. Since $\omega\in k[\Gamma]$, it is clear
that if $U,U^{-1}\in k[\Gamma]$ then $T_{U}\in k[\Gamma]$ as well. ∎

Proposition 4.3 does not give conditions about the non-triviality of the units
$U_{T},U_{T}^{-1}$ in case they belong to $k[\Gamma]$. Nevertheless, we will
show in the next section that, in case $\Gamma$ is the Promislow’s group, the
non-triviality of $U_{T},U_{T}^{-1}\in k[\Gamma]$ is equivalent to the non-
triviality of the $\omega$-conjugate $T$, and analogously for the non-
triviality of $T_{U}$ in terms of the non-triviality of $U,U^{-1}\in
k[\Gamma]$.

##  5\. The case of the Promislow’s group

For a group $\Gamma$, it will be convenient for us to denote by $\overline{g}$
the inverse of $g\in\Gamma$.

In this section we will apply the general procedure displayed in Section 4 to
the case of the Promislow’s group. This group, which will be denoted by $G$,
is given by the presentation

$G=\langle s,t\mid
ts^{2}\overline{t}=\overline{s}^{2},st^{2}\overline{s}=\overline{t}^{2}\rangle.$

We define $u:=st$ together with $x:=s^{2},y:=t^{2}$ and $z:=u^{2}$. We can
then write the above presentation of the group as

$G=\langle s,t,x,y\mid
x^{t}=\overline{x},y^{s}=\overline{y},x=s^{2},y=t^{2}\rangle.$

Using this, it is clear that $G$ is given by the amalgam $G_{1}\ast_{H}G_{2}$,
with $G_{1}=\langle t,x\mid x^{t}=\overline{x}\rangle$, $G_{2}=\langle s,y\mid
y^{s}=\overline{y}\rangle$ and $H=\langle x,t^{2}\rangle=\langle
s^{2},y\rangle$. We note that $H\cong{\mathbb{Z}}^{2}$ and $G_{1}\cong
G_{2}\cong{\mathbb{Z}}\rtimes{\mathbb{Z}}$, where the automorphism
implementing the crossed product is the inversion automorphism. This shows
that $G$ is a torsion-free group.

Another important property of $G$ is that it fits in a non-split short exact
sequence

$1\rightarrow{\mathbb{Z}}^{3}\rightarrow
G\rightarrow{\mathbb{Z}}/2{\mathbb{Z}}\times{\mathbb{Z}}/2{\mathbb{Z}}\rightarrow
1,$

where ${\mathbb{Z}}^{3}$ is identified with the maximal abelian normal
subgroup $N=\langle x,y,z\rangle$, and
$F=G/N=\\{N,Ns,Nt,Nu\\}\cong{\mathbb{Z}}/2{\mathbb{Z}}\times{\mathbb{Z}}/2{\mathbb{Z}}$.
In particular, $G$ is a virtually abelian group. The natural action of $F$ on
$N$ by conjugation is explicitly described in Table 1 below.

conj | $x$ | $y$ | $z$
---|---|---|---
$N$ | $x$ | $y$ | $z$
$Ns$ | $x$ | $\overline{y}$ | $\overline{z}$
$Nt$ | $\overline{x}$ | $y$ | $\overline{z}$
$Nu$ | $\overline{x}$ | $\overline{y}$ | $z$
Table 1. The first column elements act on the first row elements by
conjugation.

We will fix the right transversal of $N$ in $G$ to be
$\tilde{F}=\\{1,s,t,u\\}$. The associated $2$-cocycle is explicitly written
down in Table 2.

$\tau(a,b)$ | $1$ | $s$ | $t$ | $u$
---|---|---|---|---
$1$ | $1$ | $1$ | $1$ | $1$
$s$ | $1$ | $x$ | $1$ | $x$
$t$ | $1$ | $\bar{x}y\bar{z}$ | $y$ | $\bar{x}\bar{z}$
$u$ | $1$ | $\bar{y}z$ | $\bar{y}$ | $z$
Table 2. The first column corresponds to the different values of $a$, and the
first row corresponds to the different values of $b$.

For the following discussion, we take $k$ to be any field with involution $*$
with characteristic different from $2$. For $g\in\\{x,y,z\\}$, we let

$\kappa_{g}:=\frac{1}{2}(g+\overline{g}),\text{
}\omega_{g}:=\frac{1}{2}(g-\overline{g})\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}].$

We note that

(5.1) $\kappa_{g}^{2}-\omega_{g}^{2}=1.$

Following Sections 3.1 and 4, we denote by ${\pazocal Q}$ the Ore field of
fractions of $k[G]$, by ${\pazocal L}$ the Ore field of fractions of $k[N]$,
namely ${\pazocal L}=k(x,y,z)$, and the group $F$ is identified, as a subgroup
of $\text{Aut}({\pazocal L}/k)$, with
$H=\\{\text{id},\alpha_{s},\alpha_{t},\alpha_{u}\\}$. Apart from the
automorphisms in $H$, we have three natural automorphisms of ${\pazocal L}$,
namely

$\iota_{x}:{\pazocal L}\rightarrow{\pazocal
L},\quad\iota_{x}(x)=\overline{x},\iota_{x}(y)=y,\iota_{x}(z)=z;$
$\iota_{y}:{\pazocal L}\rightarrow{\pazocal
L},\quad\iota_{y}(x)=x,\iota_{y}(y)=\overline{y},\iota_{y}(z)=z;$
$\iota_{z}:{\pazocal L}\rightarrow{\pazocal
L},\quad\iota_{z}(x)=x,\iota_{z}(y)=y,\iota_{z}(z)=\overline{z}.$

That is, for $g\in\\{x,y,z\\}$, $\iota_{g}$ is the automorphism induced by the
inversion map with respect to $g$. The next proposition is an extension of
Proposition 3.4 in the particular case of the Promislow’s group.

######  Proposition 5.1.

The following statements hold true.

  1. (i)

The set $\\{1,s,t,u\\}$ is a basis of ${\pazocal Q}$ over ${\pazocal L}$,
where ${\pazocal L}$-multiplication is given from the left. Therefore
$[{\pazocal Q}:{\pazocal L}]=4$.

  2. (ii)

When writing an element $\alpha\in{\pazocal Q}$ as $\alpha=a+bs+ct+du$ with
$a,b,c,d\in{\pazocal L}$, we get the formulas

$\displaystyle
a=\frac{1}{4}(\alpha+\alpha^{\omega_{x}}+\alpha^{\omega_{y}}+\alpha^{\omega_{z}}),\qquad
bs=\frac{1}{4}(\alpha+\alpha^{\omega_{x}}-\alpha^{\omega_{y}}-\alpha^{\omega_{z}}),$
$\displaystyle
ct=\frac{1}{4}(\alpha-\alpha^{\omega_{x}}+\alpha^{\omega_{y}}-\alpha^{\omega_{z}}),\qquad
du=\frac{1}{4}(\alpha-\alpha^{\omega_{x}}-\alpha^{\omega_{y}}+\alpha^{\omega_{z}}).$

  3. (iii)

The fixed field of the subgroup $H$ is exactly
$k(\kappa_{x},\kappa_{y},\kappa_{z},\xi)$, where
$\xi:=\omega_{x}\omega_{y}\omega_{z}$, which is an extension of ${\pazocal
K}:=k(\kappa_{x},\kappa_{y},\kappa_{z})$ of order $2$. In particular,
${\pazocal Z}={\pazocal K}(\xi)$.

  4. (iv)

We have $[{\pazocal Q}:{\pazocal Z}]=16$, and ${\pazocal L}$ is a maximal
subfield of ${\pazocal Q}$ which contains ${\pazocal Z}$.

###### Proof.

  1. (i)

This is part (i) of Proposition 3.4.

  2. (ii)

We conjugate $\alpha=a+bs+ct+du$ with the elements
$\omega_{x},\omega_{y},\omega_{z}$ to get

$\alpha^{\omega_{x}}=a+bs-ct-du,$ $\alpha^{\omega_{y}}=a-bs+ct-du,$
$\alpha^{\omega_{z}}=a-bs-ct+du.$

From here we obtain the displayed formulas.

  3. (iii)

Clearly the elements $\kappa_{x},\kappa_{y},\kappa_{z}$ and
$\xi=\omega_{x}\omega_{y}\omega_{z}$ are fixed by $H$, so ${\pazocal
K}(\xi)=k(\kappa_{x},\kappa_{y},\kappa_{z},\xi)\subseteq{\pazocal L}^{H}$.

Consider now the field extensions

${\pazocal K}\subseteq{\pazocal K}(\omega_{x})\subseteq{\pazocal
K}(\omega_{x},\omega_{y})\subseteq{\pazocal
K}(\omega_{x},\omega_{y},\omega_{z})={\pazocal L}.$

Since $\omega_{g}^{2}=\kappa_{g}^{2}-1\in{\pazocal K}$ for any
$g\in\\{x,y,z\\}$, all three inclusions are of degree at most $2$. We prove
that the first one is exactly $2$ by showing that $\omega_{x}\notin{\pazocal
K}$. But this is immediate, as $\iota_{x}(\omega_{x})=-\omega_{x}$, and
$\iota_{x}$ restricts to the identity over ${\pazocal K}$. The other
inclusions are analogously proven to be strict by using the other
automorphisms $\\{\iota_{y},\iota_{z}\\}$. Thus $[{\pazocal L}:{\pazocal
K}]=8$. On the other hand, we observe that ${\pazocal K}(\xi)$ is an extension
of ${\pazocal K}$ of order $2$, since
$\xi^{2}=(\kappa_{x}^{2}-1)(\kappa_{y}^{2}-1)(\kappa_{z}^{2}-1)\in{\pazocal
K}$ but, for example, $\iota_{x}(\xi)=-\xi$, so $\xi\notin{\pazocal K}$.

We now deduce that $[{\pazocal L}:{\pazocal K}(\xi)]=4$. But $[{\pazocal
L}:{\pazocal L}^{H}]=|H|=4$, hence ${\pazocal L}^{H}={\pazocal K}(\xi)$. The
statement now follows from part (ii) of Proposition 3.4, which tells us that
${\pazocal Z}={\pazocal L}^{H}$.

  4. (iv)

This is part (ii) of Proposition 3.4, together with Theorem 4.1 (1).

∎

For $g,h,j\in\\{x,y,z\\}$ we will denote by $\iota_{gh}$ the composition
$\iota_{g}\iota_{h}$, and by $\iota_{ghj}$ the composition
$\iota_{g}\iota_{h}\iota_{j}$. We note that conjugation by $s$ is equal to
$\iota_{yz}$ on ${\pazocal L}$, i.e. $\alpha_{s}=\iota_{yz}$. In fact,
$\alpha_{t}=\iota_{xz}$ and $\alpha_{u}=\iota_{xy}$, so
$H=\\{\text{id},\iota_{xy},\iota_{xz},\iota_{yz}\\}$ (see Table 1).

######  Corollary 5.2.

The set $\\{1,\omega_{x},\omega_{y},\omega_{z}\\}$ is a ${\pazocal Z}$-basis
for ${\pazocal L}$.

###### Proof.

Since ${\pazocal Z}={\pazocal L}^{H}$ by Proposition 3.4, we have $[{\pazocal
L}:{\pazocal Z}]=[{\pazocal L}:{\pazocal L}^{H}]=|H|=4$ and
$\text{Gal}({\pazocal L}/{\pazocal
Z})=H=\\{\text{id},\iota_{xy},\iota_{xz},\iota_{yz}\\}$. Now, linear
independence of the set $\\{1,\omega_{x},\omega_{y},\omega_{z}\\}$ follows by
applying the Galois automorphisms to a ${\pazocal Z}$-linear combination
$a+b\omega_{x}+c\omega_{y}+d\omega_{z}=0$, hence deducing $a=b=c=d=0$. ∎

Thus given an element $\alpha\in{\pazocal L}$, it can be uniquely written as

(5.2) $\alpha=a+b\omega_{x}+c\omega_{y}+d\omega_{z}$

with the coefficients $a,b,c,d\in{\pazocal Z}$ explicitly given by

(5.3) $\displaystyle
a=\frac{1}{4}(\alpha+\iota_{xy}(\alpha)+\iota_{xz}(\alpha)+\iota_{yz}(\alpha)),\quad
b=\frac{1}{4\omega_{x}}(\alpha-\iota_{xy}(\alpha)-\iota_{xz}(\alpha)+\iota_{yz}(\alpha)),$
$\displaystyle
c=\frac{1}{4\omega_{y}}(\alpha-\iota_{xy}(\alpha)+\iota_{xz}(\alpha)-\iota_{yz}(\alpha)),\quad
d=\frac{1}{4\omega_{z}}(\alpha+\iota_{xy}(\alpha)-\iota_{xz}(\alpha)-\iota_{yz}(\alpha)).$

In the next lemma we compute the center of $k[G]$.

######  Lemma 5.3.

We have $\emph{Z}(k[G])=k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$.

###### Proof.

We remark that by Proposition 5.1 (iv) we have $\text{Z}(k[G])={\pazocal
K}(\xi)\cap k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$, which looks like “almost” what
we claim, but nevertheless we do not use Proposition 5.1 in this proof.

We note that, given $p\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$, the combination

$p+p^{s}+p^{t}+p^{u}=p+\iota_{yz}(p)+\iota_{xz}(p)+\iota_{xy}(p)=\sum_{\gamma\in\text{Gal}({\pazocal
L}/{\pazocal Z})}\gamma(p)$

belongs to $\text{Z}(k[G])$. We thus consider the surjective linear map

$P\colon k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]\to\text{Z}(k[G]),\quad
P(p)=\frac{1}{4}\sum_{\gamma\in\text{Gal}({\pazocal L}/{\pazocal
Z})}\gamma(p).$

It then suffices to show that the image of $P$ equals
$k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$. We start by checking that $P(m)\in
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ when $m\in k[x^{\pm 1},y^{\pm
1},z^{\pm 1}]$ is a monomial. Furthermore, since $P$ commutes with
$\iota_{x}$, $\iota_{y}$, $\iota_{z}$, we may assume that $m\in k[x,y,z]$.

Let $\Gamma_{+}$ be the subgroup of $\text{Aut}({\pazocal L}/k)$ generated by
$\iota_{x},\iota_{y},\iota_{z}$, and $\Gamma_{-}$ be the group generated by
the linear maps $-\iota_{x},-\iota_{y},-\iota_{z}$. Let

$Q(m):=\frac{1}{8}\sum_{\gamma\in\Gamma_{+}}\gamma(m),\quad
R(m):=\frac{1}{8}\sum_{\gamma\in\Gamma_{-}}\gamma(m),$

and note that $P(m)=Q(m)+R(m)$.

We first consider the case when $m$ depends on a single variable, for example
$x$, thus $m=x^{d}$ with $d\geq 1$. We proceed by strong induction on the
exponent. For $d=1$ we have $P(m)=\kappa_{x}$ and the claim is true. Now
assume that the result holds for all degrees up to $d\geq 1$, and we compute

$2^{d+1}\kappa_{x}^{d+1}=\sum_{k=0}^{d+1}\binom{d+1}{k}x^{2k-(d+1)}=\sum_{k=0}^{d+1}\binom{d+1}{k}x^{d+1-2k}.$

By adding these two expressions and dividing by $2$, we get

$\displaystyle 2^{d+1}\kappa_{x}^{d+1}$
$\displaystyle=\sum_{k=0}^{d+1}\binom{d+1}{k}P(x^{d+1-2k})$
$\displaystyle=2P(x^{d+1})+\sum_{k=1}^{d}\binom{d+1}{k}P(x^{d+1-2k})$
$\displaystyle=2P(x^{d+1})+\sum_{k=1}^{\lfloor\frac{d+1}{2}\rfloor}\binom{d+1}{k}P(x^{d+1-2k})+\sum_{k=\lfloor\frac{d+1}{2}\rfloor+1}^{d}\binom{d+1}{k}\iota_{x}P(x^{2k-d-1}).$

We now use strong induction to deduce that $P(x^{d+1})\in
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$. The same argument works mutatis
mutandis for the variables $y$, $z$.

Let us now consider the case when $m$ depends on two variables, for example
$x$ and $y$. Thus take $m=x^{d}y^{e}$ with $d,e\geq 1$. Then

$P(m)=\frac{1}{4}(x^{d}y^{e}+x^{d}\bar{y}^{e}+\bar{x}^{d}y^{e}+\bar{x}^{d}\bar{y}^{e})=\frac{1}{4}(x^{d}+\bar{x}^{d})(y^{e}+\bar{y}^{e})=P(x^{d})P(y^{e}),$

and thus we are done by the case of a single variable. The cases for the
variables $x,z$ and $y,z$ are treated analogously.

Finally let us consider the case of three variables $m=x^{d}y^{e}z^{f}$, with
$d,e,f\geq 1$. We note that

$Q(m)=\frac{1}{8}(x^{d}+\bar{x}^{d})(y^{e}+\bar{y}^{e})(z^{f}+\bar{z}^{f})=P(x^{d})P(y^{e})P(z^{f}),\quad
R(m)=\frac{1}{8}(x^{d}-\bar{x}^{d})(y^{e}-\bar{y}^{e})(z^{f}-\bar{z}^{f}).$

Since $P(m)=Q(m)+R(m)$, we only need to deal with $R(m)$. We note that the
quotient

$\alpha:=\frac{x^{d}-\bar{x}^{d}}{x-\bar{x}}\frac{y^{e}-\bar{y}^{e}}{y-\bar{y}}\frac{z^{f}-\bar{z}^{f}}{z-\bar{z}}$

is invariant under the whole group $\text{Gal}({\pazocal L}/{\pazocal
K})=\Gamma_{+}$, and so belongs to ${\pazocal
K}=k(\kappa_{x},\kappa_{y},\kappa_{z})$. In fact, this quotient can be
explicitly computed to be

$\alpha=\sum_{i=-d+1}^{d-1}x^{i}\sum_{j=-e+1}^{e-1}y^{j}\sum_{k=-f+1}^{f-1}z^{k}=\Big{(}1+2\sum_{i=1}^{d-1}P(x^{i})\Big{)}\Big{(}1+2\sum_{j=1}^{e-1}P(y^{j})\Big{)}\Big{(}1+2\sum_{k=1}^{f-1}P(z^{k})\Big{)}\in
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi].$

Therefore $R(m)=\xi\alpha\in k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$.

This analysis proves the inclusion $\text{im}(P)\subseteq
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$. The other inclusion is clear, since
$\kappa_{x},\kappa_{y},\kappa_{z},\xi\in{\pazocal Z}\cap k[G]$. ∎

######  Remark 5.4.

An alternative proof of Lemma 5.3 can be obtained by noting that
$\kappa_{x},\kappa_{y},\kappa_{z},\xi$ are invariant functions and then using
Molien’s formula to check that they generate the whole algebra of invariant
functions.

We note the following simple corollary of Lemma 5.3.

######  Lemma 5.5.

Let $f,g\in k[\kappa_{x},\kappa_{y},\kappa_{z}]$ be such that $\frac{f}{g}\in
k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$. Then $\frac{f}{g}\in
k[\kappa_{x},\kappa_{y},\kappa_{z}]$. Informally, we may say that if $g|f$ in
$k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ then $g|f$ in
$k[\kappa_{x},\kappa_{y},\kappa_{z}]$.

###### Proof.

Since $\frac{f}{g}\in{\pazocal K}\subseteq{\pazocal Z}$ we deduce that
$\frac{f}{g}\in\text{Z}(k[G])$, so by Lemma 5.3 we have $\frac{f}{g}=p+q\xi$
with $p,q\in k[\kappa_{x},\kappa_{y},\kappa_{z}]$. Note that
$\frac{f}{g}=\iota_{x}(\frac{f}{g})=p-q\xi$ which shows that $q=0$. Thus
$\frac{f}{g}=p\in k[\kappa_{x},\kappa_{y},\kappa_{z}]$ and the proof is
complete. ∎

Let us gather together some information from the previous lemmas.

######  Proposition 5.6.

We have the equalities

$\displaystyle k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ $\displaystyle=k(x,y,z)\cap
k[G];$ $\displaystyle k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$
$\displaystyle=\emph{Z}(k[G])=k(\kappa_{x},\kappa_{y},\kappa_{z})(\xi)\cap
k[G];$ $\displaystyle k[\kappa_{x},\kappa_{y},\kappa_{z}]$
$\displaystyle=k(\kappa_{x},\kappa_{y},\kappa_{z})\cap k[x^{\pm 1},y^{\pm
1},z^{\pm 1}].$

###### Proof.

The first displayed equality is clear, the second one is the content of Lemma
5.3 and the third one follows from Lemma 5.5. ∎

###  5.1. The $\omega$-degree of an element

Regarding Section 4, we will concentrate in the study of
$\omega_{x}$-conjugates. For brevity, we will denote by $\omega$ the element
$\omega_{x}=\frac{1}{2}(x-\overline{x})$, and by $\kappa$ the element
$\kappa_{x}=\frac{1}{2}(x+\overline{x})$. Note that the key relation here is
Formula (5.1), namely

$\kappa^{2}-\omega^{2}=1.$

We specialize Definition 4.2 in this context.

######  Definition 5.7.

An element $T\in{\pazocal Q}$ is called an _$\omega$ -conjugate_ if
$T^{2}=\omega^{2}$. We say that $T$ is _non-trivial_ if $T\neq\pm\omega$.

By Theorem 4.1 (2), we have that $T$ is an $\omega$-conjugate if and only if
$T=\omega^{U}$ for some $U\in{\pazocal Q}\backslash\\{0\\}$; thus the name
_$\omega$ -conjugate_ is appropriate.

The following proposition is an extension of Proposition 4.3 in the particular
case of the Promislow’s group.

######  Proposition 5.8.

Let $T\in{\pazocal Q}$ be an $\omega$-conjugate, and define $U_{T}:=\kappa-T$.

  1. (i)

The element $U_{T}$ is a unit in ${\pazocal Q}$, with inverse
$U_{-T}=\kappa+T$.

  2. (ii)

$T\in k[G]$ if and only if $U_{T}\in k[G]$, in which case $U_{-T}$ also
belongs in $k[G]$.

  3. (iii)

$T$ is a non-trivial $\omega$-conjugate if and only if $U_{T}$ is a non-
trivial unit.

  4. (iv)

If the involution of $k$ restricts to the identity on its prime subfield, then
$T^{\ast}=-T$ if and only if $U_{T}$ is a unitary.

Conversely, let $U\in{\pazocal Q}\backslash\\{0\\}$, and define
$T_{U}:=\omega^{U}=U\omega U^{-1}$.

  1. (i)

The element $T_{U}$ is an $\omega$-conjugate.

  2. (ii)

If both $U,U^{-1}\in k[G]$, then $T_{U}\in k[G]$.

  3. (iii)

If $T_{U}$ is a non-trivial $\omega$-conjugate, then $U$ is a non-trivial
unit. The converse is also true if we assume $U\in k[G]$.

  4. (iv)

If $U$ is a unitary, then $T_{U}^{\ast}=-T_{U}$.

This is mostly Proposition 4.3. Nevertheless, since Proposition 5.8 gives
necessary and sufficient conditions for the non-triviality of the units, we
prefer to give a full detailed proof of it.

###### Proof of Proposition 5.8.

Since $\kappa\in{\pazocal Z}$ we have

$U_{T}U_{-T}=(\kappa-T)(\kappa+T)=\kappa^{2}-T^{2}=\kappa^{2}-\omega^{2}=1,$

which shows that $U_{T}$ is a unit in ${\pazocal Q}$ with inverse $U_{-T}$.
Moreover, since $\kappa\in k[G]$, it is clear that $T\in k[G]$ if and only if
$U_{T}\in k[G]$, in which case $U_{-T}\in k[G]$ as well. Furthermore, if the
involution $\ast$ is the identity over the prime subfield of $k$, then a
computation shows that $U_{T}^{*}=\kappa-T^{\ast}$, and so
$U_{T}^{*}=U_{T}^{-1}$ if and only if $T^{\ast}=-T$.

Let us show now that $U_{T}$ is a non-trivial unit in case $T$ is a non-
trivial $\omega$-conjugate. Suppose, by way of contradiction, that
$U_{T}=\lambda g$ for some $\lambda\in k\backslash\\{0\\}$ and $g\in G$. After
isolating $T$ in the equation $U_{T}=\lambda g$ and squaring, we get the
equation

$\lambda^{2}g^{2}-\lambda xg-\lambda\overline{x}g+1=0.$

This equation is inconsistent unless $g$ equals either $x$ or $\overline{x}$.
But in such cases we get $\lambda=1$, so $U_{T}=x^{\pm 1}$. This leads to
$T=\pm\omega$, which are the trivial $\omega$-conjugates, a contradiction.
Conversely, if $T=\pm\omega$, then

$U_{T}=\kappa-T=\kappa\mp\omega=x^{\mp 1},$

so $U_{T}$ is a trivial unit.

For the second part, we define $T_{U}:=U\omega U^{-1}$ and compute

$T_{U}^{2}=U\omega^{2}U^{-1}=\omega^{2}$

as $\omega^{2}\in{\pazocal Z}$. Hence $T_{U}$ is an $\omega$-conjugate. It is
clear that if both $U,U^{-1}\in k[G]$ then $T_{U}\in k[G]$. Furthermore, if
$U$ is a unitary, then $T_{U}=U\omega U^{\ast}$, and since
$\omega^{\ast}=-\omega$ we get

$T_{U}^{\ast}=U\omega^{\ast}U^{\ast}=-U\omega U^{\ast}=-T_{U}.$

Let us show now that $U$ is a non-trivial unit in case $T_{U}$ is a non-
trivial $\omega$-conjugate. Suppose, by way of contradiction, that $U=\lambda
g$ for some $\lambda\in k\backslash\\{0\\}$ and $g\in G$. Write $g=n\tilde{f}$
for $n\in N$ and $\tilde{f}\in\tilde{F}=\\{1,s,t,u\\}$. Then

$T_{U}=U\omega U^{-1}=n\omega^{\tilde{f}}\overline{n}\in\\{\omega,-\omega\\}$

as $\omega^{\tilde{f}}\in\\{\omega,-\omega\\}$. Thus $T_{U}$ is a trivial
$\omega$-conjugate, a contradiction. Conversely, suppose that $U\in k[G]$ and
that $T_{U}=\pm\omega$. Write $U$ as

$U=a+bs+ct+du$

with $a,b,c,d\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$. The case $T_{U}=\omega$
implies $U^{\omega}=U$, and so $U=a+bs$. The same holds for $U^{-1}$, and so
$U$ turns out to be a unit in $k[x^{\pm 1},y^{\pm 1},z^{\pm 1}][s]$. This
latter ring is a subring of $k[{\mathbb{Z}}^{2}\rtimes{\mathbb{Z}}]$, and such
groups are known not to admit non-trivial units. Hence $U$ is a trivial unit,
a contradiction.

The case $T_{U}=-\omega$ implies $U^{\omega}=-U$, and so $U=ct+du=(c+ds)t$.
This implies that $c+ds$ is again a unit in $k[x^{\pm 1},y^{\pm 1},z^{\pm
1}][s]$, and by the same argument as above $U$ must be a trivial unit, again a
contradiction. This finishes the proof. ∎

As a corollary, we get:

######  Corollary 5.9.

The existence of non-trivial units in the group ring $k[G]$ is equivalent to
the existence of non-trivial $\omega$-conjugates in $k[G]$.

If moreover the involution of $k$ restricts to the identity on its prime
subfield, then the existence of non-trivial unitaries in $k[G]$ is equivalent
to the existence of non-trivial $\omega$-conjugates in $k[G]$ which are anti-
selfadjoint, i.e. an $\omega$-conjugate $T\in k[G]$ satisfying $T^{\ast}=-T$.

Corollary 5.9 gives a condition equivalent to the existence of units, which
might conceivably be considered simpler than the determinant condition proved
in [2], in the following sense: the determinant condition written as an
equation over ${\pazocal Z}$ is a single equation of degree $4$ with $16$
variables, whereas the condition for being an $\omega$-conjugate is an
equation of degree $2$ with $16$ variables.

######  Remark 5.10.

Suppose that $U$ is a non-trivial unit in $k[G]$. Then by the above
proposition we get that $T_{U}$ is a non-trivial $\omega$-conjugate in $k[G]$.
Therefore $V:=\kappa-T_{U}$ is again a non-trivial unit in $k[G]$. Moreover,
its inverse is given by $V^{-1}=\kappa+T_{U}=2\kappa-V$. Thus $V$ satisfies

$2\kappa V-V^{2}=1.$

From this we obtain that there are non-trivial units in $k[G]$ if and only if
there are non-trivial units in $k[G]$ satisfying the equation

$X^{2}-2\kappa X+1=0.$

We will now study $\omega$-conjugates more closely. Take $\phi:{\pazocal
Q}\rightarrow{\pazocal Q}$ to be any $k$-automorphism of ${\pazocal Q}$ of
order $2$ fixing $\omega^{2}$, that is $\phi(\omega^{2})=\omega^{2}$. Then any
element $\alpha\in{\pazocal Q}$ can be uniquely written as

$\alpha=\alpha_{1}+\alpha_{-1}$

where $\alpha_{1},\alpha_{-1}\in{\pazocal Q}$ are eigenvectors of $\phi$ with
eigenvalues $1$ and $-1$, respectively. In fact, they are explicitly given by

$\alpha_{1}=\frac{1}{2}(\alpha+\phi(\alpha)),\quad\alpha_{-1}=\frac{1}{2}(\alpha-\phi(\alpha)).$

Moreover $\phi$ preserves $\omega$-conjugacy, in the sense that it
$T\in{\pazocal Q}$ is an $\omega$-conjugate, then $\phi(T)$ is also an
$\omega$-conjugate, simply because

$\phi(T)^{2}=\phi(T^{2})=\phi(\omega^{2})=\omega^{2}.$

######  Definition 5.11.

The $\phi$-degree of an element $\alpha\in{\pazocal Q}$ is the pair of natural
numbers

$\text{deg}_{\phi}(\alpha):=([{\pazocal Z}(\alpha_{1}):{\pazocal
Z}],[{\pazocal Z}(\alpha_{-1}):{\pazocal
Z}])\in{\mathbb{N}}\times{\mathbb{N}},$

where $\alpha_{1},\alpha_{-1}$ are the elements determining the
$\phi$-decomposition of $\alpha$.

We note that, for any $D\in{\pazocal Q}$, the division algebra ${\pazocal
Z}(D)$ is in fact a subfield of ${\pazocal Q}$ containing ${\pazocal Z}$. Thus
we may use Theorem 4.1 (1) to conclude that $[{\pazocal Z}(D):{\pazocal
Z}]\leq 4$, where we recall that $[{\pazocal Q}:{\pazocal Z}]=16$. Moreover,
since

${\pazocal Z}\subseteq{\pazocal Z}(D)\subseteq{\pazocal Q}$

necessarily $[{\pazocal Z}(D):{\pazocal Z}]$ must be divisible by $[{\pazocal
Q}:{\pazocal Z}]=16$. We thus get a finite number of options for the
$\phi$-degree of an element $\alpha\in{\pazocal Q}$, namely

$\text{deg}_{\phi}(\alpha)\in\\{(1,1),(1,2),(1,4),(2,1),(2,2),(2,4),(4,1),(4,2),(4,4)\\}.$

In Proposition 5.14 we further restrict the $\phi$-degree of an
$\omega$-conjugate $T\in{\pazocal Q}$ in case $T_{1},T_{-1}\neq 0$. First, we
establish a technical lemma.

######  Lemma 5.12.

Let $T\in{\pazocal Q}$ be an $\omega$-conjugate, $\phi:{\pazocal
Q}\rightarrow{\pazocal Q}$ a $k$-automorphism fixing $\omega^{2}$, and let
$T=T_{1}+T_{-1}$ be its $\phi$-decomposition. Then $T_{1}T_{-1}=-T_{-1}T_{1}$,
and moreover if both $T_{1},T_{-1}\neq 0$ we also have the equalities

$\emph{Z}({\pazocal Z}[T,T_{-1}])={\pazocal Z}(T_{1}^{2})={\pazocal
Z}(T_{-1}^{2})$

and

$[{\pazocal Z}(T_{1}):{\pazocal Z}]=[{\pazocal Z}(T_{-1}):{\pazocal Z}].$

Here ${\pazocal Z}[T_{1},T_{-1}]$ is the subalgebra of ${\pazocal Q}$
generated by $T_{1}$ and $T_{-1}$ over ${\pazocal Z}$. As a consequence, if
both $T_{1},T_{-1}\neq 0$, then

$\emph{deg}_{\phi}(T)\in\\{(1,1),(2,2),(4,4)\\}.$

###### Proof.

Let us write ${\pazocal R}:={\pazocal Z}[T_{1},T_{-1}]$. We compute

$\displaystyle T_{1}T_{-1}+T_{-1}T_{1}$
$\displaystyle=\frac{1}{4}\big{[}(T+\phi(T))(T-\phi(T))+(T-\phi(T))(T+\phi(T))\big{]}$
$\displaystyle=\frac{1}{2}(T^{2}-\phi(T)^{2})$
$\displaystyle=\frac{1}{2}(\omega^{2}-\omega^{2})=0.$

In particular, $T^{2}=T_{1}^{2}+T_{-1}^{2}=\omega^{2}\in{\pazocal Z}$, hence
${\pazocal Z}(T_{1}^{2})={\pazocal Z}(T_{-1}^{2})$.

Now we assume that $T_{1},T_{-1}\neq 0$, so in particular ${\pazocal R}$ is
not commutative. A direct computation shows that $T_{1}^{2}$ commutes with
$T_{-1}$, and $T_{-1}^{2}$ commutes with $T_{1}$, hence
$T_{1}^{2},T_{-1}^{2}\in\text{Z}({\pazocal R})$. This shows the inclusion

(5.4) ${\pazocal Z}(T_{1}^{2})\subseteq\text{Z}({\pazocal R}).$

######  Claim 5.13.

If either $T_{1}^{2}\in{\pazocal Z}$ or $T_{-1}^{2}\in{\pazocal Z}$, then
$[{\pazocal R}:{\pazocal Z}]\leq 4$.

###### Proof of Claim.

Due to the relation $T_{1}^{2}+T_{-1}^{2}=\omega^{2}\in{\pazocal Z}$, we see
that both $T_{1}^{2},T_{-1}^{2}\in{\pazocal Z}$. It follows that every element
of ${\pazocal R}$ can be written as a combination

$z_{1}+z_{2}T_{1}+z_{3}T_{-1}+z_{4}T_{1}T_{-1},\quad
z_{1},z_{2},z_{3},z_{4}\in{\pazocal Z}.$

The claim follows. ∎

Now ${\pazocal Z}\subseteq{\pazocal R}\subseteq{\pazocal Q}$, so the only
possibilities for $[{\pazocal R}:{\pazocal Z}]$ are in fact $1,2,4,8$ and
$16$. Let us check the statement of the lemma case by case.

  1. 1)

If $[{\pazocal R}:{\pazocal Z}]=16$, then ${\pazocal R}={\pazocal Q}$, and so
$\text{Z}({\pazocal R})={\pazocal Z}\subseteq{\pazocal Z}(T_{1}^{2})$. We are
done by (5.4).

  2. 2)

If $[{\pazocal R}:{\pazocal Z}]=8$, then by Theorem 4.1 (1) the quantity
$[{\pazocal R}:\text{Z}({\pazocal R})]$ must be a square dividing $8$, thus it
is either $1$ or $4$. But ${\pazocal R}$ is not commutative, so necessarily
$[{\pazocal R}:\text{Z}({\pazocal R})]=4$ and so $[\text{Z}({\pazocal
R}):{\pazocal Z}]=2$. Thus by (5.4) it is enough to argue that
$T_{1}^{2}\notin{\pazocal Z}$. But this follows from Claim 5.13.

  3. 3)

If $[{\pazocal R}:{\pazocal Z}]=4$, then again Theorem 4.1 (1) implies that
$\text{Z}({\pazocal R})={\pazocal Z}\subseteq{\pazocal Z}(T_{1}^{2})$. We are
again done by (5.4).

  4. 4)

The cases $[{\pazocal R}:{\pazocal Z}]=1$ and $2$ lead to the conclusion that
${\pazocal R}$ is commutative, which is a contradiction.

This analysis finishes the first equality. For the second one, simply note
that

$[{\pazocal Z}(T_{1}):{\pazocal Z}]=[{\pazocal Z}(T_{1}):{\pazocal
Z}(T_{1}^{2})][{\pazocal Z}(T_{1}^{2}):{\pazocal Z}]=[{\pazocal
Z}(T_{1}):{\pazocal Z}(T_{1}^{2})][\text{Z}({\pazocal R}):{\pazocal Z}],$
$[{\pazocal Z}(T_{-1}):{\pazocal Z}]=[{\pazocal Z}(T_{-1}):{\pazocal
Z}(T_{-1}^{2})][{\pazocal Z}(T_{-1}^{2}):{\pazocal Z}]=[{\pazocal
Z}(T_{-1}):{\pazocal Z}(T_{-1}^{2})][\text{Z}({\pazocal R}):{\pazocal Z}],$

and moreover $[{\pazocal Z}(T_{1}):{\pazocal Z}(T_{1}^{2})]=[{\pazocal
Z}(T_{-1}):{\pazocal Z}(T_{-1}^{2})]=2$, since both ${\pazocal
Z}(T_{1}^{2})={\pazocal Z}(T_{-1}^{2})=\text{Z}({\pazocal R})$ and $T_{1}$
does not commute with $T_{-1}$. Therefore we conclude that

$[{\pazocal Z}(T_{1}):{\pazocal Z}]=[{\pazocal Z}(T_{-1}):{\pazocal Z}].$

The statement of the $\phi$-degrees now follows. ∎

The first part of the following proposition shows that non-trivial
$\omega$-conjugates fall naturally into two groups, assuming that both
$T_{1},T_{-1}\neq 0$.

######  Proposition 5.14.

Let $T\in{\pazocal Q}$ be an $\omega$-conjugate, $\phi:{\pazocal
Q}\rightarrow{\pazocal Q}$ a $k$-automorphism fixing $\omega^{2}$, and let
$T=T_{1}+T_{-1}$ be its $\phi$-decomposition. Assume that $T_{1},T_{-1}\neq
0$.

Then either $T$ is a trivial $\omega$-conjugate or
$\emph{deg}_{\phi}(T)\in\\{(2,2),(4,4)\\}$. Moreover, in the particular case
that $T$ is non-trivial, the following are equivalent.

  1. (i)

$\deg_{\phi}(T)=(2,2)$.

  2. (ii)

$T_{1}^{2},T_{-1}^{2}\in{\pazocal Z}$.

  3. (iii)

$[{\pazocal Z}[T_{1},T_{-1}]:{\pazocal Z}]=4$.

###### Proof.

By Lemma 5.12, $\text{deg}_{\phi}(T)\in\\{(1,1),(2,2),(4,4)\\}$. Therefore to
establish the first part of the proposition it is enough to argue that
$\deg_{\phi}(T)\neq(1,1)$ in case $T$ is a non-trivial $\omega$-conjugate.
However, if $\deg_{\phi}(T)=(1,1)$, then $T\in{\pazocal Z}$ and the equation
$X^{2}=\omega^{2}$ would have more than two solutions in the field ${\pazocal
Z}(\omega)$, namely $\\{\pm\omega,\pm T\\}$, which cannot happen. This
concludes the first part of the proposition.

For the second part of the proposition, suppose that $T$ is a non-trivial
$\omega$-conjugate. We first prove (i) $\Rightarrow$ (ii). We note that
$\deg_{\phi}(T)=(2,2)$ implies that we can find $\alpha,\beta\in{\pazocal Z}$
such that $T_{1}^{2}=\alpha T_{1}+\beta$. Since $T_{1}^{2}$ commutes with
$T_{-1}$, we deduce that

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

i.e. $T_{1}^{2}=\beta\in{\pazocal Z}$. Therefore
$T_{-1}^{2}=\omega^{2}-T_{1}^{2}\in{\pazocal Z}$ also, and this proves (ii).

To establish (ii) $\Rightarrow$ (iii), let us assume that both
$T_{1}^{2},T_{-1}^{2}\in{\pazocal Z}$. By Claim 5.13 we get $[{\pazocal
Z}[T_{1},T_{-1}]:{\pazocal Z}]\in\\{1,2,4\\}$. Since $T_{1},T_{-1}\neq 0$ by
assumption, ${\pazocal Z}[T_{1},T_{-1}]$ is not commutative. Hence necessarily
$[{\pazocal Z}[T_{1},T_{-1}]:{\pazocal Z}]=4$.

Finally we establish (iii) $\Rightarrow$ (i), so assume $[{\pazocal
Z}[T_{1},T_{-1}]:{\pazocal Z}]=4$. Then by Lemma 5.12 we have

$4=[{\pazocal Z}[T_{1},T_{-1}]:{\pazocal Z}]=[{\pazocal
Z}[T_{1},T_{-1}]:\text{Z}({\pazocal Z}[T_{1},T_{-1}])][{\pazocal
Z}(T_{1}^{2}):{\pazocal Z}].$

Since ${\pazocal Z}[T_{1},T_{-1}]$ is not commutative, using Theorem 4.1 (1)
we deduce that $[{\pazocal Z}(T_{1}^{2}):{\pazocal Z}]=1$, i.e.
$T_{1}^{2}\in{\pazocal Z}$. Since $T_{1}\notin{\pazocal Z}$, necessarily
$[{\pazocal Z}(T_{1}):{\pazocal Z}]=2$ and thus, by the first part of the
proposition, we must have $\deg_{\phi}(T)=(2,2)$. This finishes the proof. ∎

Let us note an important improvement of Proposition 5.14 in a particular case.
Recall that conjugation by $\omega$ is an automorphism of ${\pazocal Q}$ of
order $2$ which preserves $k[G]$. Explicitly, we have

$(a+bs+ct+du)^{\omega}=a+bs-ct-du$

for $a,b,c,d\in{\pazocal L}=k(x,y,z)$. We will denote by $\omega$ itself the
automorphism of ${\pazocal Q}$ given by conjugation by $\omega$. The
$\omega$-decomposition of $T=a+bs+ct+du$ is given explicitly by $T_{1}=a+bs$
and $T_{-1}=ct+du$.

######  Proposition 5.15.

Let $T\in{\pazocal Q}$ be an $\omega$-conjugate and $T=T_{1}+T_{-1}$ be its
$\omega$-decomposition. If $T$ is non-trivial, then necessarily
$T_{1},T_{-1}\neq 0$.

Combining Propositions 5.15 and 5.14, we conclude that an $\omega$-conjugate
$T\in{\pazocal Q}$ is either trivial or
$\text{deg}_{\omega}(T)\in\\{(2,2),(4,4)\\}$, in which case the following
equivalences apply:

  1. (1)

$\deg_{\omega}(T)=(2,2)$.

  2. (2)

$T_{1}^{2},T_{-1}^{2}\in{\pazocal Z}$.

  3. (3)

$[{\pazocal Z}[T_{1},T_{-1}]:{\pazocal Z}]=4$.

###### Proof of Proposition 5.15.

If we denote by $\overline{k}$ the algebraic closure of $k$, then the
inclusion $k\subseteq\overline{k}$ gives an inclusion
$k[G]\hookrightarrow\overline{k}[G]$, which in turn extends to an inclusion of
the corresponding Ore field of fractions. Therefore we may assume, without
loss of generality, that $k$ is algebraically closed.

We first show that we can never have $T_{1}=0$. By way of contradiction, let
us assume that $T_{1}=0$. We have $\omega T_{-1}=-T_{-1}\omega$ and
$T_{-1}^{2}=\omega^{2}$, thus $k(\omega^{2})[\omega,T_{-1}]$ is a division
algebra with $[k(\omega^{2})[\omega,T_{-1}]:k(\omega^{2})]=4$, and whose
center is $k(\omega^{2})$. However, Tsen’s Theorem (see [3]) shows that the
field of rational functions in one variable over an algebraically closed field
cannot be the center of a finite-dimensional division algebra. This
contradiction shows that $T_{1}\neq 0$.

Now, in case $T_{-1}=0$ and $T=T_{1}\neq\pm\omega$, then ${\pazocal
Z}[\omega,T_{1}]$ is a field in which the equation $X^{2}=\omega^{2}$ has at
least $4$ different solutions, namely $\\{\pm\omega,\pm T_{1}\\}$, which is a
contradiction. This concludes the proof of the proposition. ∎

######  Definition 5.16.

A non-trivial unit $U\in k[G]$ is said to be of type 2 if
$\text{deg}_{\omega}(\omega^{U})=(2,2)$. Otherwise it is said to be of type 4,
i.e. in case $\text{deg}_{\omega}(\omega^{U})=(4,4)$.

######  Remark 5.17.

  1. 1)

Let us see that there are $\omega$-conjugates in ${\pazocal Q}$ whose
$\omega$-degree is $(4,4)$. Indeed, let $k={\mathbb{Q}}$ and let $U:=(1+x)+u$.
Then a direct check shows that

$U^{-1}=\frac{1+\bar{x}}{2+x+\bar{x}-z}-\frac{1}{2+x+\bar{x}-z}u,$

and so

$T:=\omega^{U}=\frac{2+x+\bar{x}+z}{2+x+\bar{x}-z}\omega-\frac{2(1+x)\omega}{2+x+\bar{x}-z}u.$

Note that here, if $T=T_{1}+T_{-1}$ denotes the $\omega$-decomposition of $T$,
we obtain

$T_{1}=\frac{2+x+\bar{x}+z}{2+x+\bar{x}-z}\omega$

and another computation shows that $T_{1}^{2}$ is not invariant under
$\iota_{z}$, so that $T_{1}^{2}\notin{\pazocal Z}$.

  2. 2)

On the other hand, there are no non-trivial $\omega$-conjugates in $k[G]$ of
the form $T=a+bt$, where $a,b\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$, by the
familiar argument using the unique product property. Indeed, we can replace
every occurrence of $y$ with $t^{2}$ to express $T$ as a polynomial in $t$
with coefficients in $k[x^{\pm 1},z^{\pm}]$, and it is easy to argue that
$T^{2}\in k[x^{\pm 1}]$ implies that in fact $T\in k[x^{\pm 1}]$, which then
implies that $T=\pm\omega$.

  3. 3)

In Proposition 5.21 below we use the results from [8] to find non-trivial
$\omega$-conjugates of $\omega$-degree $(2,2)$ in $k[G]$, whenever
$\text{char}(k)>2$. However, we were not able to find $\omega$-conjugates in
$k[G]$ with $\omega$-degree $(4,4)$.

For the purpose of searching for $\omega$-conjugates it is convenient to
assume some extra hypotheses about the coefficients. The following proposition
shows that, if there are non-trivial $\omega$-conjugates in $k[G]$ with
$\omega$-degree $(2,2)$, then there are also non-trivial $\omega$-conjugates
with the same $\omega$-degree, but whose $\omega$-decomposition $T_{1}+T_{-1}$
satisfies $T_{1}\omega\in{\pazocal Z}$.

######  Proposition 5.18.

Suppose that $T\in{\pazocal Q}$ is an $\omega$-conjugate with
$\deg_{\omega}(T)=(2,2)$, and consider the unit $U_{T}=\kappa-T$ given in
Proposition 5.8. Then the element

$S:=\omega^{U_{T}}$

has the following properties:

  1. (i)

$S$ is an $\omega$-conjugate with $\deg_{\omega}(S)=(2,2)$.

  2. (ii)

If $S=S_{1}+S_{-1}$ is the $\omega$-decomposition of $S$, then
$S_{1}=\alpha\omega$ for some $\alpha\in{\pazocal Z}$. If moreover $T\in k[G]$
then we can take $\alpha\in k[G]$.

  3. (iii)

If the involution of $k$ restricts to the identity on its prime subfield and
$T^{\ast}=-T$, then also $S^{\ast}=-S$.

###### Proof.

Let $T=T_{1}+T_{-1}$ be the $\omega$-decomposition of $T$. We have

$\displaystyle S_{1}$ $\displaystyle=\frac{1}{2}(S+S^{\omega})$
$\displaystyle=\frac{1}{2}(\omega^{U_{T}}+\omega^{\omega U_{T}})$
$\displaystyle=\frac{1}{2}\Big{(}(\kappa-
T_{1}-T_{-1})(\kappa+T_{1}-T_{-1})+(\kappa-
T_{1}+T_{-1})(\kappa+T_{1}+T_{-1})\Big{)}\omega$
$\displaystyle=(\kappa^{2}-T_{1}^{2}+T_{-1}^{2})\omega$
$\displaystyle=(\kappa^{2}-\omega^{2}+2T_{-1}^{2})\omega=(1+2T_{-1}^{2})\omega$

and

$\displaystyle S_{-1}$ $\displaystyle=\frac{1}{2}(S-S^{\omega})$
$\displaystyle=\frac{1}{2}(\omega^{U_{T}}-\omega^{\omega U_{T}})$
$\displaystyle=\frac{1}{2}\Big{(}(\kappa-
T_{1}-T_{-1})(\kappa+T_{1}-T_{-1})-(\kappa-
T_{1}+T_{-1})(\kappa+T_{1}+T_{-1})\Big{)}\omega$
$\displaystyle=2(T_{1}T_{-1}-\kappa
T_{-1})\omega=2(T_{1}-\kappa)T_{-1}\omega.$

Since $\deg_{\omega}(T)=(2,2)$, we see that $T_{1}-\kappa\neq 0$, hence
$S_{-1}\neq 0$ and therefore $\text{deg}_{\omega}(S)$ is either $(2,2)$ or
$(4,4)$. By Propositions 5.14 and 5.15, $1+2T_{-1}^{2}\in{\pazocal Z}$, and
hence $[{\pazocal Z}(S_{1}):{\pazocal Z}]\leq 2$. This establishes both (i)
and (ii) with $\alpha=1+2T_{-1}^{2}$. Observe that if $T\in k[G]$ then also
$T^{\omega}\in k[G]$, so $T_{-1}\in k[G]$ and $\alpha\in k[G]$.

As for (iii), we recall (see the proof of Corollary 5.9) that if $T$ is anti-
selfadjoint then $U_{T}$ is a unitary. In this case, we compute

$S^{\ast}=(\omega^{U_{T}})^{\ast}=(U_{T}\omega
U_{T}^{\ast})^{\ast}=U_{T}\omega^{\ast}U_{T}^{\ast}.$

Since $\omega^{\ast}=-\omega$, this concludes the proof that $S^{\ast}=-S$,
and the proof of the proposition. ∎

As corollaries, we get the following results.

######  Corollary 5.19.

The following hold true:

  1. (a)

There exists a non-trivial unit $U\in k[G]$ of type $2$ if and only if there
exists a non-trivial $\omega$-conjugate $S\in k[G]$ with
$\emph{deg}_{\omega}(S)=(2,2)$ of the form

$S=\alpha\omega+ct+du,$

with $\alpha\in k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and $c,d\in k[x^{\pm
1},y^{\pm 1},z^{\pm 1}]$.

  2. (b)

Assuming that the involution of $k$ restricts to the identity on its prime
subfield, there exists a non-trivial unitary $U\in k[G]$ of type $2$ if and
only if there exists a non-trivial anti-selfadjoint $\omega$-conjugate $S\in
k[G]$ with $\emph{deg}_{\omega}(S)=(2,2)$ of the form

$S=\alpha\omega+ct+du,$

with $\alpha\in k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and $c,d\in k[x^{\pm
1},y^{\pm 1},z^{\pm 1}]$.

###### Proof.

Given such a unit $U$ of type $2$, we let $T:=\omega^{U}$, which has
$\omega$-degree $(2,2)$ by assumption. In particular $T\in k[G]$ is non-
trivial, and so the unit $U_{T}:=\kappa-T$ is a non-trivial unit in $k[G]$ by
Proposition 5.8. Thus $S:=\omega^{U_{T}}$ is also a non-trivial
$\omega$-conjugate in $k[G]$ by the same proposition, with $\omega$-degree
$(2,2)$ and of the required form by Propositions 5.18 and 5.6.

Conversely, let $S\in k[G]$ be a non-trivial $\omega$-conjugate of the stated
form, with $\omega$-degree $(2,2)$. Then the associated unit $U_{S}:=\kappa-S$
is a non-trivial unit in $k[G]$ of type $2$ by Propositions 5.8 and 5.18. This
proves part (a).

Part (b) is completely analogous, taking into account the corresponding
statements relating unitarity of $U_{T}$ and anti-selfadjointness of $T$. ∎

######  Corollary 5.20.

There exists a non-trivial unit $U\in k[G]$ of type $2$ if and only if there
exists a non-trivial unit $V\in k[G]$ of the form

$V=(\kappa-\alpha\omega)-ct-du$

where $\alpha\in k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and $c,d\in k[x^{\pm
1},y^{\pm 1},z^{\pm 1}]$, whose inverse is
$V^{-1}=(\kappa+\alpha\omega)+ct+du$.

###### Proof.

If $U\in k[G]$ is a non-trivial unit of type $2$, then by Corollary 5.19 there
is a non-trivial $\omega$-conjugate $S\in k[G]$ of the form
$S=\alpha\omega+ct+du$ with $\alpha\in
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and $c,d\in k[x^{\pm 1},y^{\pm
1},z^{\pm 1}]$. By Proposition 5.8, the unit $V:=\kappa-S$ is non-trivial in
$k[G]$ with inverse $V^{-1}=\kappa+S$, which is of the desired form.

Conversely, suppose that $V=(\kappa-\alpha\omega)-ct-du$ is a non-trivial unit
in $k[G]$ with inverse $V^{-1}=(\kappa+\alpha\omega)+ct+du$. We compute

$\displaystyle\omega^{V}$ $\displaystyle=V\omega V^{-1}$
$\displaystyle=((\kappa^{2}-\alpha^{2}\omega^{2})+cc^{t}y+dd^{u}z)\omega+(cd^{t}\overline{x}\overline{z}+c^{u}d\overline{y})\omega
s+2(\kappa-\alpha\omega)\omega ct+2(\kappa-\alpha\omega)\omega dt$
$\displaystyle=(1+2\omega^{2}(1-\alpha^{2}))\omega+2(\kappa-\alpha\omega)\omega
ct+2(\kappa-\alpha\omega)\omega dt,$

where we have used that $VV^{-1}=1$. From here we conclude that
$(\omega^{V})_{1}^{2}\in{\pazocal Z}$, and so
$\text{deg}_{\omega}(\omega^{V})=(2,2)$ by Proposition 5.15. ∎

The following proposition, which uses the existence of non-trivial units in
$k[G]$ for the case $\text{char}(k)\neq 0$ recently proved in [5, 8], proves
existence of non-trivial $\omega$-conjugates of the first class, namely having
$\omega$-degree $(2,2)$.

######  Proposition 5.21.

Let $\Gamma=G$ be the Promislow’s group. If $\emph{char}(k)>2$ then there
exists a non-trivial unit in $k[G]$ of type $2$.

###### Proof.

Write $p:=\text{char}(k)>2$. In [8, Theorem 3] the author constructs non-
trivial units in $k[G]$. Explicitly, these units are given as follows.

For any choice of parameters $m,n\in{\mathbb{Z}}$, define
$h:=(1-z^{1-2n})^{p-2}\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ and elements
$a,b,c,d\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ by

$\displaystyle a:=(1+x)(1+y)(z^{n}+z^{1-n})h,$ $\displaystyle
b:=z^{m}[(1+x)(\overline{x}+\overline{y})+(1+\overline{y})(1+z^{2n-1})]h,$
$\displaystyle c:=z^{m}[(1+\overline{y})(x+y)z^{n}+(1+x)(z^{n}+z^{1-n})]h,$
$\displaystyle d:=z^{2n-1}+(4+x+\overline{x}+y+\overline{y})h.$

Then $U:=a+bs+ct+du$ is a non-trivial unit in $k[G]$, with inverse given by

$U^{-1}=a^{\prime}+b^{\prime}s+c^{\prime}t+d^{\prime}u,$

where

$a^{\prime}=\overline{x}a^{s},\quad b^{\prime}=-\overline{x}b,\quad
c^{\prime}=-\overline{y}c,\quad d^{\prime}=\overline{z}d^{s}.$

By Proposition 5.8, the element $T:=\omega^{U}$ is a non-trivial
$\omega$-conjugate in $k[G]$. We will show that
$\text{deg}_{\omega}(T)=(2,2)$, which will prove the proposition. We first
obtain the $\omega$-decomposition of $T$:

$T=\omega^{U}=U\omega
U^{-1}=(a+bs+ct+du)(a^{\prime}+b^{\prime}s-c^{\prime}t-d^{\prime}u)\omega,$

and so

$\displaystyle T_{1}=\frac{1}{2}(T+T^{\omega})=$
$\displaystyle\Big{[}(aa^{\prime}+xb(b^{\prime})^{s}-yc(c^{\prime})^{t}-zd(d^{\prime})^{u})$
$\displaystyle+(ab^{\prime}+b(a^{\prime})^{s}-\overline{x}\text{
}\overline{z}c(d^{\prime})^{t}-\overline{y}d(c^{\prime})^{u})s\Big{]}\omega,$
$\displaystyle T_{-1}=\frac{1}{2}(T-T^{\omega})=$
$\displaystyle\Big{[}(-ac^{\prime}-xb(d^{\prime})^{s}+c(a^{\prime})^{t}+\overline{y}zd(b^{\prime})^{u})t$
$\displaystyle+(-ad^{\prime}-b(c^{\prime})^{s}+\overline{x}y\overline{z}c(b^{\prime})^{t}+d(a^{\prime})^{u})u\Big{]}\omega.$

Since $(a^{\prime})^{s}=\overline{x}a$, $(c^{\prime})^{u}=-\bar{x}yc$ and
$(d^{\prime})^{t}=zd$, we see that

$ab^{\prime}+b(a^{\prime})^{s}=-\overline{x}ab+\overline{x}ba=0$

and also

$\overline{x}\text{
}\overline{z}c(d^{\prime})^{t}+\overline{y}d(c^{\prime})^{u}=\overline{x}cd-\overline{x}dc=0.$

Hence the expression accompanying the element $s$ in the expression of $T_{1}$
is exactly $0$, which leads to

$T_{1}=(aa^{\prime}+xb(b^{\prime})^{s}-yc(c^{\prime})^{t}-zd(d^{\prime})^{u})\omega=:\alpha\omega.$

We compute the expression (5.2) of $\alpha\in{\pazocal L}$ as a ${\pazocal
Z}$-linear combination using the formulas (5.3). In fact, a direct computation
shows that

$\displaystyle\alpha-\iota_{xy}(\alpha)-\iota_{xz}(\alpha)+\iota_{yz}(\alpha)=0,$
$\displaystyle\alpha-\iota_{xy}(\alpha)+\iota_{xz}(\alpha)-\iota_{yz}(\alpha)=0,$
$\displaystyle\alpha+\iota_{xy}(\alpha)-\iota_{xz}(\alpha)-\iota_{yz}(\alpha)=0.$

We thus conclude that $\alpha$ is a central element in ${\pazocal Q}$.
Therefore $T_{1}^{2}=\alpha^{2}\omega^{2}\in{\pazocal Z}$, and this already
implies that $\text{deg}_{\omega}(T)=(2,2)$ by Proposition 5.15. This finishes
the proof. ∎

It is an interesting question whether we can find non-trivial units in $k[G]$
of type $2$ in the case $\text{char}(k)=0$. Also, when the involution of $k$
restricts to the identity on its prime subfield, it is also interesting to ask
about non-trivial unitaries in $k[G]$ of type $2$. These questions are handled
in the next section.

###  5.2. Criteria for existence of $\omega$-conjugates with $\omega$-degree
$(2,2)$

In this section we provide criteria for existence of non-trivial units in
$k[G]$ of type $2$. They will be stated for an arbitrary field $k$ with
$\text{char}(k)\neq 2$, although in view of Proposition 5.21 they are
interesting only when $\text{char}(k)=0$.

In view of Corollary 5.19, we will study non-trivial $\omega$-conjugates
$T\in{\pazocal Q}$ of the form

(5.5) $T=a\omega+ct+du,$

where $a\in k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]\subseteq{\pazocal Z}$ and
$c,t\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]\subseteq{\pazocal L}$. Non-triviality
of $T$ implies that both $c,d\neq 0$. Indeed, if $c=0$ then $T=a\omega+du$ and
$U_{T}:=\kappa-T=(\kappa-a\omega)-du$ would be a non-trivial unit in $k[x^{\pm
1},y^{\pm 1},z^{\pm 1}][u]$. A similar argument as the one given in the proof
of Proposition 5.8 implies that $U_{T}$ could not be non-trivial, giving a
contradiction. Similarly, we can prove that $d\neq 0$. The criteria are
obtained by considering the equation

$T^{2}=\omega^{2}$

and writing it using explicit bases of ${\pazocal Q}$ over ${\pazocal Z}$
which are convenient for computations. Let us start by introducing such bases.

The first column of the following table introduces a basis of ${\pazocal L}$
over ${\pazocal Z}$ convenient for studying the coefficient $c$.

| s | t | u | $\gamma_{i}^{2}\bar{x}y$ | $(\gamma_{i}t)^{*}$
---|---|---|---|---|---
$\gamma_{1}:=(1+x)(1-\bar{y})$ | $-y$ | $\phantom{-}\bar{x}$ | $-\bar{x}y$ | $4(\kappa_{x}+1)(\kappa_{y}-1)$ | $-\gamma_{1}t$
$\gamma_{2}:=(1-x)(1+\bar{y})$ | $\phantom{-}y$ | $-\bar{x}$ | $-\bar{x}y$ | $4(\kappa_{x}-1)(\kappa_{y}+1)$ | $\phantom{-}\gamma_{2}t$
$\gamma_{3}:=(1+x)(1+\bar{y})$ | $\phantom{-}y$ | $\phantom{-}\bar{x}$ | $\phantom{-}\bar{x}y$ | $4(\kappa_{x}+1)(\kappa_{y}+1)$ | $\phantom{-}\gamma_{3}t$
$\gamma_{4}:=(1-x)(1-\bar{y})$ | $-y$ | $-\bar{x}$ | $\phantom{-}\bar{x}y$ | $4(\kappa_{x}-1)(\kappa_{y}-1)$ | $-\gamma_{4}t$
Table 3. First basis: $\gamma$-elements.

The columns 2-4 of Table 3 show how conjugation acts on the elements of the
basis, e.g. $\gamma_{1}^{s}=-y\cdot\gamma_{1}$. Note that those conjugation
relations actually prove that $\gamma_{i}$ form a basis. The final two columns
are for a future reference. We briefly note that $y,\bar{x}$ and $\bar{x}y$
are _not_ central elements.

The second basis, for studying the coefficient $d$, is presented in the
following table.

| s | t | u | $\delta_{i}^{2}z$ | $(\delta_{i}u)^{*}$
---|---|---|---|---|---
$\delta_{1}:=(1-\bar{z})\phantom{(x-\bar{x})}$ | $-z$ | $-z$ | $\phantom{-}1$ | $2(\kappa_{z}-1)$ | $-\delta_{1}u$
$\delta_{2}:=(x-\bar{x})(1+\bar{z})$ | $\phantom{-}z$ | $-z$ | $-1$ | $8\omega_{x}^{2}(\kappa_{z}+1)$ | $\phantom{-}\delta_{2}u$
$\delta_{3}:=(1+\bar{z})\phantom{(x-\bar{x})}$ | $\phantom{-}z$ | $\phantom{-}z$ | $\phantom{-}1$ | $2(\kappa_{z}+1)$ | $\phantom{-}\delta_{3}u$
$\delta_{4}:=(x-\bar{x})(1-\bar{z})$ | $-z$ | $\phantom{-}z$ | $-1$ | $8\omega_{x}^{2}(\kappa_{z}-1)$ | $-\delta_{4}u$
Table 4. Second basis: $\delta$-elements.

Let us also note the following relations for future reference.

(5.6)
$\displaystyle\frac{\gamma_{2}\delta_{2}}{\gamma_{3}\delta_{3}}=\frac{\gamma_{4}\delta_{4}}{\gamma_{1}\delta_{1}}$
$\displaystyle=\frac{2\omega_{x}(1-x)}{(1+x)}=\frac{2\omega_{x}(1-x^{2})}{1+2x+x^{2}}=\frac{-4\omega_{x}^{2}}{2\kappa_{x}+2}=-2(\kappa_{x}-1),$
$\displaystyle\frac{\gamma_{1}\delta_{2}}{\gamma_{4}\delta_{3}}=\frac{\gamma_{3}\delta_{4}}{\gamma_{2}\delta_{1}}$
$\displaystyle=\frac{2\omega_{x}(1+x)}{1-x}=-2(\kappa_{x}+1).$

From now on, we write

$c=c_{1}+c_{2}+c_{3}+c_{4},\quad d=d_{1}+d_{2}+d_{3}+d_{4},$

where, for all $1\leq i\leq 4$, we have $c_{i}=C_{i}\gamma_{i}$,
$d_{i}=D_{i}\delta_{i}$ for some $C_{i},D_{i}\in{\pazocal Z}$. We start with a
technical lemma.

######  Lemma 5.22.

We have $c\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ if and only if each $c_{i}\in
k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$. Similarly, $d\in k[x^{\pm 1},y^{\pm
1},z^{\pm 1}]$ if and only if each $d_{i}\in k[x^{\pm 1},y^{\pm 1},z^{\pm
1}]$.

###### Proof.

We will deal with the coefficient $c$, being the other one $d$ completely
analogous. After conjugation by $s,t,u$, we get

$\begin{cases}c=c_{1}+c_{2}+c_{3}+c_{4},\\\
c^{s}=y(-c_{1}+c_{2}+c_{3}-c_{4}),\\\
c^{t}=\bar{x}(c_{1}-c_{2}+c_{3}-c_{4}),\\\
c^{u}=\bar{x}y(-c_{1}-c_{2}+c_{3}+c_{4}).\end{cases}$

Therefore we get the formulas

(5.7) $\displaystyle c_{1}=\frac{c-\bar{y}c^{s}+xc^{t}-x\bar{y}c^{u}}{4},\quad
c_{2}=\frac{c+\bar{y}c^{s}-xc^{t}-x\bar{y}c^{u}}{4},$ $\displaystyle
c_{3}=\frac{c+\bar{y}c^{s}+xc^{t}+x\bar{y}c^{u}}{4},\quad
c_{4}=\frac{c-\bar{y}c^{s}-xc^{t}+x\bar{y}c^{u}}{4}.$

From here the lemma easily follows, since $c\in k[x^{\pm 1},y^{\pm 1},z^{\pm
1}]$ if and only if $c^{s},c^{t},c^{u}\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$. ∎

The next proposition gives necessary and sufficient conditions for $T$ as in
(5.5) so that $T^{2}\in{\pazocal Z}$.

######  Proposition 5.23.

Let $T$ be as in (5.5). Then the following conditions are equivalent.

  1. (i)

$T^{2}\in{\pazocal Z}$.

  2. (ii)

We have

(5.8) $cd^{t}\bar{x}\bar{z}+c^{u}d\bar{y}=0.$

  3. (iii)

The equation

$(\kappa-1)C_{2}C_{4}-(\kappa+1)C_{1}C_{3}=0$

is satisfied and (at least) one of the following is true:

    1. ($I$)

In case either $C_{1}\neq 0$ or $C_{4}\neq 0$, there exist
$\alpha,\beta\in{\pazocal Z}$ such that

$\displaystyle D_{1}=-2\beta(\kappa-1)C_{4},$ $\displaystyle D_{4}=\beta
C_{1},$ $\displaystyle D_{2}=\alpha C_{4},$ $\displaystyle
D_{3}=-2\alpha(\kappa+1)C_{1};$

    2. ($II$)

In case either $C_{2}\neq 0$ or $C_{3}\neq 0$, there exist
$\alpha,\beta\in{\pazocal Z}$ such that

$\displaystyle D_{1}=-2\beta(\kappa+1)C_{3},$ $\displaystyle D_{4}=\beta
C_{2},$ $\displaystyle D_{2}=\alpha C_{3},$ $\displaystyle
D_{3}=-2\alpha(\kappa-1)C_{2}.$

If any of these conditions hold, then

$T^{2}=a^{2}\omega^{2}+(c_{1}^{2}-c_{2}^{2}+c_{3}^{2}-c_{4}^{2})\bar{x}y+(d_{1}^{2}-d_{2}^{2}+d_{3}^{2}-d_{4}^{2})z$

and furthermore all nine summands are in ${\pazocal Z}$.

###### Proof.

In this proof we will freely use the conjugation tables (3) and (4).

For the implication $(i)\Rightarrow(ii)$, we simply compute

$T^{2}=(a\omega+ct+du)^{2}=(a^{2}\omega^{2}+cc^{t}y+dd^{u}z)+(cd^{t}\bar{x}\bar{z}+c^{u}d\bar{y})s.$

If we require $T^{2}\in{\pazocal Z}$, we necessarily need

$cd^{t}\bar{x}\bar{z}+c^{u}d\bar{y}=0,$

which is the statement in $(ii)$.

Now we show the equivalence $(ii)\Leftrightarrow(iii)$. We have

$\displaystyle c^{u}=\bar{x}y(-c_{1}-c_{2}+c_{3}+c_{4}),$ $\displaystyle
d^{t}=z(-d_{1}-d_{2}+d_{3}+d_{4}),$

thus equation (5.8) becomes

$\displaystyle 0$ $\displaystyle=cd^{t}\bar{x}y\bar{z}+c^{u}d$
$\displaystyle=(c_{1}+c_{2}+c_{3}+c_{4})(-d_{1}-d_{2}+d_{3}+d_{4})\bar{x}y+(-c_{1}-c_{2}+c_{3}+c_{4})(d_{1}+d_{2}+d_{3}+d_{4})\bar{x}y$
$\displaystyle=2\bar{x}y[(c_{3}+c_{4})(d_{3}+d_{4})-(c_{1}+c_{2})(d_{1}+d_{2})].$

This is equivalent to the equation

$(c_{1}+c_{2})(d_{1}+d_{2})=(c_{3}+c_{4})(d_{3}+d_{4}).$

By conjugating this last equation with $s$, $t$ and $u$, we get the following
system of non-linear equations:

$\begin{cases}(c_{1}+c_{2})(d_{1}+d_{2})=(c_{3}+c_{4})(d_{3}+d_{4}),\\\
(c_{1}-c_{2})(d_{1}-d_{2})=(c_{3}-c_{4})(d_{3}-d_{4}),\\\
(-c_{1}+c_{2})(d_{1}+d_{2})=(c_{3}-c_{4})(d_{3}+d_{4}),\\\
(c_{1}+c_{2})(-d_{1}+d_{2})=(c_{3}+c_{4})(d_{3}-d_{4}).\end{cases}$

This system is equivalent to the following system of equations:

$\begin{cases}c_{1}d_{1}-c_{4}d_{4}=0,\\\ c_{1}d_{2}-c_{4}d_{3}=0,\\\
c_{2}d_{1}-c_{3}d_{4}=0,\\\ c_{2}d_{2}-c_{3}d_{3}=0.\end{cases}$

Using the definition of the $c_{i}$’s and the $d_{i}$’s, and the relations
(5.6), those equations are equivalent to the following equations:

$\displaystyle 2(\kappa-1)C_{2}D_{2}+C_{3}D_{3}$ $\displaystyle=0$
$\displaystyle C_{1}D_{1}+2(\kappa-1)C_{4}D_{4}$ $\displaystyle=0$
$\displaystyle 2(\kappa+1)C_{1}D_{2}+C_{4}D_{3}$ $\displaystyle=0$
$\displaystyle C_{2}D_{1}+2(\kappa+1)C_{3}D_{4}$ $\displaystyle=0$

Note that, given $C_{1},C_{2},C_{3},C_{4}$, these are in fact two systems of
equations, the first system to determine $D_{2},D_{3}$ and the other system to
determine $D_{1},D_{4}$. Thus the conclusion is that equation (5.8) is
equivalent to the condition that the determinants of the above systems vanish,
i.e.

$(\kappa-1)C_{2}C_{4}-(\kappa+1)C_{1}C_{3}=0,$

together with the statements (I) and (II) of the proposition, depending on the
non-vanishing of the $C_{i}$’s.

Finally we show $(iii)\Rightarrow(i)$. We have already shown that $(iii)$
implies the vanishing of the second term in

$T^{2}=(a\omega+ct+du)^{2}=(a^{2}\omega^{2}+cc^{t}y+dd^{u}z)+(cd^{t}\bar{x}\bar{z}+c^{u}d\bar{y})s.$

Thus it is enough to show that $(iii)$ also implies that the term
$cc^{t}y+dd^{u}z$ is central. We have

$\displaystyle c^{t}=\bar{x}(c_{1}-c_{2}+c_{3}-c_{4}),$ $\displaystyle
d^{u}=d_{1}-d_{2}+d_{3}-d_{4},$

therefore

$\displaystyle cc^{t}y+dd^{u}z$
$\displaystyle=(c_{1}+c_{2}+c_{3}+c_{4})(c_{1}-c_{2}+c_{3}-c_{4})\bar{x}y+(d_{1}+d_{2}+d_{3}+d_{4})(d_{1}-d_{2}+d_{3}-d_{4})z$
$\displaystyle=\bar{x}y(c_{1}^{2}-c_{2}^{2}+c_{3}^{2}-c_{4}^{2})+z(d_{1}^{2}-d_{2}^{2}+d_{3}^{2}-d_{4}^{2})+2\bar{x}y(c_{1}c_{3}-c_{2}c_{4})+2z(d_{1}d_{3}-d_{2}d_{4}).$

Let us now argue that $c_{1}c_{3}-c_{2}c_{4}=d_{1}d_{3}-d_{2}d_{4}=0$. We
have, on one hand,

$\displaystyle c_{1}c_{3}-c_{2}c_{4}$
$\displaystyle=C_{1}C_{3}\gamma_{1}\gamma_{3}-C_{2}C_{4}\gamma_{2}\gamma_{4}$
$\displaystyle=(1-\bar{y}^{2})[(1+x^{2}+2x)C_{1}C_{3}-(1+x^{2}-2x)C_{2}C_{4}]$
$\displaystyle=2x(1-\bar{y}^{2})[(\kappa+1)C_{1}C_{3}-(\kappa-1)C_{2}C_{4}]=0.$

On the other hand,

$\displaystyle d_{1}d_{3}-d_{2}d_{4}$
$\displaystyle=D_{1}D_{3}\delta_{1}\delta_{3}-D_{2}D_{4}\delta_{2}\delta_{4}$
$\displaystyle=(1-\bar{z}^{2})(D_{1}D_{3}-4\omega^{2}D_{2}D_{4}).$

In case (I), this equals

$4(1-\bar{z}^{2})\alpha\beta C_{1}C_{4}[(\kappa+1)(\kappa-1)-\omega^{2}]=0,$

and in case (II), we get

$4(1-\bar{z}^{2})\alpha\beta C_{2}C_{3}[(\kappa+1)(\kappa-1)-\omega^{2}]=0.$

We thus obtain the expression

$T^{2}=a^{2}\omega^{2}+(c_{1}^{2}-c_{2}^{2}+c_{3}^{2}-c_{4}^{2})\bar{x}y+(d_{1}^{2}-d_{2}^{2}+d_{3}^{2}-d_{4}^{2})z.$

The fact that all nine summands are in ${\pazocal Z}$, i.e. that each
$\gamma_{i}^{2}\bar{x}y,\delta_{i}^{2}z\in{\pazocal Z}$, follows from looking
at the second last columns of Tables (3) and (4). This proves the implication
$(iii)\Rightarrow(i)$, and the proposition. ∎

We are now ready to present our main theorems.

######  Theorem 5.24.

There exists a non-trivial unit $U\in k[G]$ of type $2$ if and only if the
system of equations presented below have a solution

$(A,C_{1},C_{2},C_{3},C_{4},D_{1},D_{2},D_{3},D_{4},\alpha,\beta)\in{\pazocal
Z}^{11}$

different from $(\pm 1,0,0,0,0,0,0,0,0,0,0)$, such that $A\in
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and
$C_{i}\gamma_{i},D_{i}\delta_{i}\in k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$.

$\displaystyle\diamond\text{ }$
$\displaystyle(A^{2}-1)\omega_{x}^{2}+C_{1}^{2}(\kappa_{x}+1)(\kappa_{y}-1)-C_{2}^{2}(\kappa_{x}-1)(\kappa_{y}+1)+C_{3}^{2}(\kappa_{x}+1)(\kappa_{y}+1)-C_{4}^{2}(\kappa_{x}-1)(\kappa_{y}-1)$
$\displaystyle\phantom{(A^{2}-1)\omega_{x}^{2}}+2D_{1}^{2}(\kappa_{z}-1)-2D_{2}^{2}\omega_{x}^{2}(\kappa_{z}+1)+2D_{3}^{2}(\kappa_{z}+1)-2D_{4}^{2}\omega_{x}^{2}(\kappa_{z}-1)=0,$
$\displaystyle\diamond\text{ }$
$\displaystyle(\kappa_{x}-1)C_{2}C_{4}-(\kappa_{x}+1)C_{1}C_{3}=0,$
$\displaystyle\diamond\text{ }$ $\displaystyle\text{ In case $C_{1}\neq 0$ or
$C_{4}\neq 0$},\begin{cases}D_{1}=\beta(1-\kappa_{x})C_{4},\\\ D_{2}=\alpha
C_{4},\\\ D_{3}=-\alpha(1+\kappa_{x})C_{1},\\\ D_{4}=\beta
C_{1}.\end{cases}\text{ In case $C_{2}\neq 0$ or $C_{3}\neq
0$},\begin{cases}D_{1}=-\beta(1+\kappa_{x})C_{3},\\\ D_{2}=\alpha C_{3},\\\
D_{3}=\alpha(1-\kappa_{x})C_{2},\\\ D_{4}=\beta C_{2}.\end{cases}$

######  Remarks 5.25.

  1. 1)

We have preferred to write the equations in terms of the old notation
$\omega_{x},\kappa_{x}$ as it seems to be the most convenient form to reflect
the explicit appearance of those terms in comparison with the other elements
$\kappa_{y},\kappa_{z}$. Nevertheless, we will continue to use the notation
$\omega,\kappa$ for the proof of the theorem.

  2. 2)

As stated, the criterion in Theorem 5.24 includes 6 equations and 11
variables. However, the substitutions given by the last $4$ equations reduce
it to a system of 2 equations with $7$ variables
$(A,C_{1},C_{2},C_{3},C_{4},\alpha,\beta)$. The second equation can be used to
reduce further to a single equation with $6$ variables. We chose to state it
in an “extended” form as it seems that it is the form most amenable to further
analysis.

  3. 3)

One can also use the conclusion of Corollary 5.20 to prove Theorem 5.24.

###### Proof of Theorem 5.24.

By Corollary 5.19, the existence of a non-trivial unit $U\in k[G]$ of type $2$
is equivalent to the existence of a non-trivial $\omega$-conjugate $T\in k[G]$
with $\omega$-degree $(2,2)$ of the form

(5.9) $T=A\omega+ct+du,$

where $A\in k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and $c,d\in k[x^{\pm
1},y^{\pm 1},z^{\pm 1}]$. We have already noted that non-triviality of $T$
implies that both $c,d\neq 0$.

So first suppose that such $T$ exists. In particular
$T^{2}=\omega^{2}\in{\pazocal Z}$, and so Proposition 5.23 tells us that all
of the equations of the statement of the theorem, excluding the first one,
follow. But the first one also follows since

$\displaystyle\omega^{2}=T^{2}$
$\displaystyle=A^{2}\omega^{2}+(c_{1}^{2}-c_{2}^{2}+c_{3}^{2}-c_{4}^{2})\bar{x}y+(d_{1}^{2}-d_{2}^{2}+d_{3}^{2}-d_{4}^{2})z$
$\displaystyle=A^{2}\omega^{2}+C_{1}^{2}\gamma_{1}^{2}\bar{x}y-C_{2}^{2}\gamma_{2}^{2}\bar{x}y+C_{3}^{2}\gamma_{3}^{2}\bar{x}y-C_{4}^{2}\gamma_{4}^{2}\bar{x}y$
$\displaystyle\qquad+D_{1}^{2}\delta_{1}^{2}z-D_{2}^{2}\delta_{2}^{2}z+D_{3}^{2}\delta_{3}^{2}z-D_{4}^{2}\delta_{4}^{2}z$
$\displaystyle=A^{2}\omega^{2}+4C_{1}^{2}(\kappa+1)(\kappa_{y}-1)-4C_{2}^{2}(\kappa-1)(\kappa_{y}+1)+4C_{3}^{2}(\kappa+1)(\kappa_{y}+1)-4C_{4}^{2}(\kappa-1)(\kappa_{y}-1)$
$\displaystyle\qquad+2D_{1}^{2}(\kappa_{z}-1)-8D_{2}^{2}\omega^{2}(\kappa_{z}+1)+2D_{3}^{2}(\kappa_{z}+1)-8D_{4}^{2}\omega^{2}(\kappa_{z}-1),$

which is the first equation after a change of variables $2C_{i}\rightarrow
C_{i}$ and $2D_{2}\rightarrow D_{2},2D_{4}\rightarrow D_{4}$. Note that the
remaining equations of the theorem have been also changed accordingly under
these changes of variables. Lastly, all of the variables
$A,C_{i},D_{i},\alpha,\beta$ belong to ${\pazocal Z}$, and moreover $A\in
k[\kappa_{x},\kappa_{y},\kappa_{z},\xi]$ and each
$c_{i}=C_{i}\gamma_{i},d_{i}=D_{i}\delta_{i}\in k[x^{\pm 1},y^{\pm 1},z^{\pm
1}]$ by Lemma 5.22. Finally, since both $c,d\neq 0$, at least some
$c_{i},d_{j}\neq 0$, and hence the solution obtained is different from the
solution $(\pm 1,0,0,0,0,0,0,0,0,0,0)$.

Conversely, if such a solution of the stated system of equations exists, then
the element

$T=A\omega+\Big{(}\frac{C_{1}}{2}\gamma_{1}+\frac{C_{2}}{2}\gamma_{2}+\frac{C_{3}}{2}\gamma_{3}+\frac{C_{4}}{2}\gamma_{4}\Big{)}t+\Big{(}D_{1}\delta_{1}+\frac{D_{2}}{2}\delta_{2}+D_{3}\delta_{3}+\frac{D_{4}}{2}\delta_{4}\Big{)}u$

belongs to $k[G]$, and a straightforward computation using the equations of
the statement of the theorem shows that $T^{2}=\omega^{2}$, hence central.
Clearly the $\omega$-degree of $T$ is $(2,2)$, and non-triviality of $T$
follows from the non-triviality of the solution obtained, i.e. different from
$(\pm 1,0,0,0,0,0,0,0,0,0,0)$. ∎

We now present a variant of the previous theorem for anti-selfadjoint
$\omega$-conjugates, in case the involution of $k$ restricts to the identity
on its prime subfield.

######  Theorem 5.26.

Supose that the involution of $k$ restricts to the identity on its prime
subfield. Then there exists a non-trivial unitary $U\in k[G]$ of type $2$ if
and only if the system of equations from Theorem 5.24 has a solution

$(A,C_{1},C_{2}\xi,C_{3}\xi,C_{4},D_{1},D_{2}\xi,D_{3}\xi,D_{4},\alpha,\beta)\in{\pazocal
K}^{9}\times{\pazocal Z}^{2}$

different from $(\pm 1,0,0,0,0,0,0,0,0,0,0)$, such that $A\in
k[\kappa_{x},\kappa_{y},\kappa_{z}]$ and $C_{i}\gamma_{i},D_{i}\delta_{i}\in
k[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$.

For its proof, we first need some observations, which we will gather in the
following lemma. Recall that ${\pazocal Z}={\pazocal K}(\xi)$, which is an
extension of degree $2$ over ${\pazocal
K}=k(\kappa_{x},\kappa_{y},\kappa_{z})$.

######  Lemma 5.27.

Suppose that the involution of $k$ is the identity. Then ${\pazocal
Z}^{\ast}={\pazocal Z}$, and moreover $z^{\ast}=z$ for any $z\in{\pazocal K}$.
Also, the following hold true.

  1. i)

The element $z$ belongs to ${\pazocal K}$ if and only if $z^{\ast}=z$.

  2. ii)

The element $z\xi$ belongs to ${\pazocal K}$ if and only if $z^{\ast}=-z$.

###### Proof.

The statement ${\pazocal Z}^{\ast}={\pazocal Z}$ is clear. Note that
$z^{\ast}=z$ for any $z\in{\pazocal K}=k(\kappa_{x},\kappa_{y},\kappa_{z})$
since $\kappa_{x}^{\ast}=\kappa_{x},\kappa_{y}^{\ast}=\kappa_{y}$ and
$\kappa_{z}^{\ast}=\kappa_{z}$.

Now given $z\in{\pazocal Z}$, write it as $z=z_{1}+\xi z_{2}$ for some
$z_{1},z_{2}\in{\pazocal K}$. Then $z^{*}=z_{1}-\xi z_{2}$, and so the rest of
the lemma follows straightforwardly. ∎

Finally we are in a position to prove Theorem 5.26.

###### Proof of Theorem 5.26.

This is a particular case of Theorem 5.24, and the only thing which requires
further study is the condition $T^{\ast}=-T$. We write

$T=A\omega+C_{1}\gamma_{1}t+C_{2}\gamma_{2}t+C_{3}\gamma_{3}t+C_{4}\gamma_{4}t+D_{1}\delta_{1}u+D_{2}\delta_{2}u+D_{3}\delta_{3}u+D_{4}\delta_{4}u,$

and we use the last columns of Tables (3) and (4):

$\displaystyle T^{\ast}$
$\displaystyle=A^{\ast}\omega^{\ast}+C_{1}^{\ast}(\gamma_{1}t)^{\ast}+C_{2}^{\ast}(\gamma_{2}t)^{\ast}+C_{3}^{\ast}(\gamma_{3}t)^{\ast}+C_{4}^{\ast}(\gamma_{4}t)^{\ast}+D_{1}^{\ast}(\delta_{1}u)^{\ast}+D_{2}^{\ast}(\delta_{2}u)^{\ast}+D_{3}^{\ast}(\delta_{3}u)^{\ast}+D_{4}^{\ast}(\delta_{4}u)^{\ast}$
$\displaystyle=-A^{\ast}\omega-
C_{1}^{\ast}\gamma_{1}t+C_{2}^{\ast}\gamma_{2}t+C_{3}^{\ast}\gamma_{3}t-C_{4}^{\ast}\gamma_{4}t-D_{1}^{\ast}\delta_{1}u+D_{2}^{\ast}\delta_{2}u+D_{3}^{\ast}\delta_{3}u-D_{4}^{\ast}\delta_{4}u$

Thus the condition $T^{\ast}=-T$ is equivalent, by Lemma 5.27, to the
conditions

$A,C_{1},C_{2}\xi,C_{3}\xi,C_{4},D_{1},D_{2}\xi,D_{3}\xi,D_{4}\in{\pazocal
K}.$

These are precisely the condition in the statement of Theorem 5.26 which
differs from the statement of Theorem 5.24. The proof is complete. ∎

## Acknowledgments

The authors would like to thank Dawid Kielak for his helpful comments on the
first part of the paper.

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