On Four-Dimensional Unital Division Algebras over Fields of Characteristic not 2

TL;DR

This paper extends the classification of four-dimensional unital division algebras from finite fields to arbitrary fields of characteristic not 2, providing an explicit construction, isomorphism criteria, and automorphism group descriptions.

Contribution

It generalizes previous finite field results to all fields of characteristic not 2, offering an exhaustive construction and classification of these division algebras.

Findings

Explicit construction depending on quadratic field extensions and parameters

Isomorphism criteria in terms of parameters

Classification of automorphism groups for all such division algebras

Abstract

In [2], an exhaustive construction is achieved for the class of all 4-dimensional unital division algebras over finite fields of odd order, whose left nucleus is not minimal and whose automorphism group contains Klein's four-group. We generalize the approach of [2] towards all division algebras of the above specified type, but now admitting arbitrary fields k of characteristic not 2 as ground fields. For these division algebras we present an exhaustive construction that depends on a quadratic field extension of k and three parameters in k, and we derive an isomorphism criterion in terms of these parameters. As an application we classify, for o an ordered field in which every positive element is a square, all division o-algebras of the mentioned type, and in the finite field case we refine the Main Theorem of [2] to a classification even of the division algebras studied there. The category formed by the division k-algebras investigated here is a groupoid, whose structure we describe in a supplementary section in terms of a covering by group actions. In particular, we exhibit the automorphism groups for all division algebras in this groupoid.

Equations

$$\underline{n}=\{i\in N|1\le i\le n\}.$$

$$\mathrm{Aut}(A)=\{\phi \in {\mathrm{GL}}_k(A)|\phi (xy)=\phi (x)\phi (y)\forall x,y\in A\}.$$

$$\begin{array}[]{ccl}\mathscr{N}&=&\{A\in\mathscr{C}\ |\ A\ \mbox{is\ not associative}\},\\[4.30554pt] \mathscr{S}&=&\{A\in\mathscr{C}\ |\ A\ \mbox{is\ a\ central\ skew\ field\ over}\ k\}\ \mbox{and}\\[4.30554pt] \mathscr{K}&=&\{A\in\mathscr{C}\ |\ A\ \mbox{is\ a\ field\ extension\ of}\ k\}.\end{array}$$

$$\mathrm{Im}(\ell )=\{x\in \ell |\overline{x}=-x\}=\{x\in \ell |x^2\in k\}\setminus k^{\ast },$$

$$\begin{array}[]{lclcl}{S}\hskip 5.69046pt=\ {S}(\ell/k)&=&\{x\in\ell^{\ast}\ |\ x\overline{x}=1\}&=&{\rm ker}(n_{\ell/k}^{\ast}),\hskip 2.84544pt\ \mbox{and}\\[5.69046pt] {A}^{\ast}={A}^{\ast}(\ell/k)&=&\{x\in\ell^{\ast}\ |\ x^{2}\in k^{\ast}\}&=&(s_{\ell}^{\ast})^{-1}(k^{\ast})=k^{\ast}\sqcup({\rm Im}(\ell)\setminus\{0\}).\end{array}$$

$$N=N_r(A)=\{z\in A|(xy)z=x(yz)\forall x,y\in A\}$$

$${\dim }_k(A)={\dim }_N(A){\dim }_k(N).$$

$$\mathcal{C}(k)=\mathcal{N}(k)⨿\mathcal{S}(k)⨿\mathcal{K}(k)$$

$${\dim }_k(A)={\dim }_Z(A){\dim }_k(Z)={\dim }_Z(E{)}^2{\dim }_k(Z).$$

$$F=A^H=\{x\in A|\sigma (x)=x\forall \sigma \in H\}$$

$$\underline{c}=(c_1,c_2,c_3)\in k^3$$

$$q_{\underline{c}}:{\ell }^2\to \ell ,q_{\underline{c}}(x,y)=(1-c_1)x^2+c_{1}x\overline{x}-c_2{y}^2-c_{3}y\overline{y}.$$

$$C_{\ell }=\left\{\underline{c}\in k^3|q_{\underline{c}}^{-1}(0)=\{(0,0)\}\right\}$$

$$\left(\begin{array}[]{c}x\\ y\end{array}\right)\cdot\left(\begin{array}[]{c}w\\ z\end{array}\right)=\left(\begin{array}[]{cc}x&c_{2}y+c_{3}\overline{y}\\ y&(1-c_{1})x+c_{1}\overline{x}\end{array}\right)\left(\begin{array}[]{c}w\\ z\end{array}\right),$$

$$1_{A}=\left(\begin{array}[]{c}1\\ 0\end{array}\right).$$

$$1_{A}\ell=\left(\begin{array}[]{c}\ell\\ 0\end{array}\right)\subset A$$

$$1_{A}k\subset 1_A\ell \subset N_r(A)\subset A$$

$$\alpha\left(\begin{array}[]{c}x\\ y\end{array}\right)=\left(\begin{array}[]{r}x\\ -y\end{array}\right)\hskip 4.30554pt\mbox{and}\hskip 4.30554pt\beta\left(\begin{array}[]{c}x\\ y\end{array}\right)=\left(\begin{array}[]{c}\overline{x}\\ \overline{y}\end{array}\right),$$

$$a\cdot (bl)=(a\cdot b)l$$

$$a\cdot 1_{A}l=al$$

$$\det(L_{a})=\det\left(\begin{array}[]{cc}x&c_{2}y+c_{3}\overline{y}\\ y&(1-c_{1})x+c_{1}\overline{x}\end{array}\right)=q_{\underline{c}}(x,y).$$

$${\mathcal{F}}_{\ell }:C_{\ell }\to \mathcal{C},{\mathcal{F}}_{\ell }(\underline{c})=A(\ell ,\underline{c}),$$

$$\left\{\begin{array}[]{ccc}(j\cdot ji)\cdot j&=&j\cdot(ji\cdot j)\\ (ji\cdot j)\cdot j&=&ji\cdot(j\cdot j)\end{array}\right.$$

$$\left\{\begin{array}[]{rcl}-c_{1}c_{2}+(1-c_{1})c_{3}&=&0\\ c_{1}c_{2}+(1-c_{1})c_{3}&=&0,\end{array}\right.$$

$$\left\{\begin{array}[]{rcl}c_{1}c_{2}&=&0\\ (1-c_{1})c_{3}&=&0.\end{array}\right.$$

$${\ell }^{\ast }(\underline{c},\underline{d})=\left\{a\in {\ell }^{\ast }|\left(c_1,\frac{c_2}{a^2},\frac{c_3}{a\overline{a}}\right)=(d_1,d_2,d_3)\right\}.$$

$$\varphi_{a}:A(\ell,\underline{c})\to A(\ell,\underline{d}),\ \varphi_{a}\left(\begin{array}[]{c}x\\ y\end{array}\right)=\left(\begin{array}[]{r}x\\ ay\end{array}\right)$$

$$\psi_{a}:A(\ell,\underline{c})\to A(\ell,\underline{d}),\ \psi_{a}\left(\begin{array}[]{c}x\\ y\end{array}\right)=\left(\begin{array}[]{r}\overline{x}\\ a\overline{y}\end{array}\right)$$

$$A=U_1\oplus U_2\oplus U_3\oplus U_4=\underset{ij\in \mathrm{V}}{⨁}{B}_{ij},$$

$${\tau }_A(x,y)=\mathrm{tr}(L_{xy})=\mathrm{tr}(L_{\phi (xy)})=\mathrm{tr}(L_{\phi (x)\phi (y)})={\tau }_B(\phi (x),\phi (y))$$

Full text

We generalize the approach of [2] towards all division algebras of the above specified type, but now admitting arbitrary fields $k$ of characteristic not 2 as ground fields. For these division algebras we present an exhaustive construction that depends on a quadratic field extension of $k$ and three parameters in $k$, and we derive an isomorphism criterion in terms of these parameters. As an application we classify, for $k_{o}$ an ordered field in which every positive element is a square, all division $k_{o}$-algebras of the mentioned type, and in the finite field case we refine the Main Theorem of [2] to a classification even of the division algebras studied there.

The category formed by the division $k$-algebras investigated here is a groupoid, whose structure we describe in a supplementary section in terms of a covering by group actions. In particular, we exhibit the automorphism groups for all division algebras in this groupoid.

2010 Mathematics Subject Classification 12F10 $\cdot$ 15A21 $\cdot$ 17A35 $\cdot$ 17A36 $\cdot$ 17A60 $\cdot$ 20L05

Keywords Unital division algebra $\cdot$ Right nucleus $\cdot$ Klein’s four-group $\cdot$ Groupoid $\cdot$ Classification $\cdot$ Covering by group actions

1 Introduction

For any field $k$, we denote by $\mathscr{D}_{4}(k)$ the category of all 4-dimensional division algebras over $k$, not assumed to be associative, whose morphisms are the non-zero algebra morphisms. This category is in fact a groupoid, i.e. all morphisms in $\mathscr{D}_{4}(k)$ are isomorphisms.

The present article is devoted to the investigation of the full subgroupoid $\mathscr{C}(k)$ of $\mathscr{D}_{4}(k)$, formed by all $A\in\mathscr{D}_{4}(k)$ which are unital, such that $k1$ is properly contained in the right nucleus $N_{r}(A)$, and whose automorphism group admits Klein’s four-group as a subgroup. Our interest in the groupoid $\mathscr{C}(k)$ is twofold. Firstly, we would like to classify $\mathscr{C}(k)$ up to isomorphism, and secondly, we want to understand its categorical structure.

To motivate our first interest we note that, although classification problems generally tend to provoke deepened insight, classifications of division algebras are in fact rare. Against this background, the recent paper [2] stimulated our approach to a classification of $\mathscr{C}(k)$, which generalizes the approach of [2] as follows. Assume that ${\rm char}(k)\not=2$. Setting out from any quadratic field extension $\ell$ of $k$ and any scalar triple $\underline{c}=(c_{1},c_{2},c_{3})\in k^{3}$ that satisfies an implicitly defined $\ell$-admissibility condition, we construct a division algebra $A(\ell,\underline{c})\in\mathscr{C}(k)$. Up to isomorphism, every $A\in\mathscr{C}(k)$ can be constructed in this way. We say that two $\ell$-admissible triples $\underline{c}$ and $\underline{d}$ are $\ell^{\ast}$-equivalent, if the algebras $A(\ell,\underline{c})$ and $A(\ell,\underline{d})$ are isomorphic. We find a criterion that expresses $\ell^{\ast}$-equivalence of $\underline{c}$ and $\underline{d}$ in terms of $\underline{c},\underline{d}$ and $\ell^{\ast}$. At this stage, the classification problem of $\mathscr{C}(k)$ is reduced to the subproblems of displaying explicitly, for all quadratic extensions $k\subset\ell$, the subsets of all $\ell$-admissible triples $C_{\ell}\subset k^{3}$, of finding a transversal $T_{\ell}\subset C_{\ell}$ for the $\ell^{\ast}$-equivalence classes of $C_{\ell}$, and of eliminating redundancy of type $A(\ell,\underline{c})\ \simeq\ A(\ell^{\prime},\underline{d})$ with $\ell\ \not\simeq\ \ell^{\prime}$. The complexity of these three subproblems depends heavily on the nature of the ground field $k$, and only the second subproblem appears easy in general. For two types of fields $k$, however, we do achieve complete solutions to all three subproblems and thereby a classification of $\mathscr{C}(k)$, namely for ordered fields $k_{o}$ in which every positive element is a square, and for finite fields ${F}_{q}$ of odd order $q$. In the arithmetic case $k={Q}$, an attempt to solve the above three subproblems readily evolves into intriguing number theoretic questions. Implementing theorems of Fermat, Gauss and Legendre in pursuit of this approach, substantial progress towards a classification of $\mathscr{C}({Q})$ has been achieved in [9].

The second interest stems from our observation that general groupoids often admit a non-trivial covering by groupoids that arise naturally from group actions. These so-called group action groupoids are very concrete mathematical objects, even when the groupoid they cover originally was given in highly abstract terms. Due to this feature, we hold that the initiation of a study of coverings of groupoids by group actions is well motivated. In this respect, we treat the groupoid $\mathscr{C}=\mathscr{C}(k)$ as a test case. In Section 6, a covering of $\mathscr{C}$ by group actions is indeed established, provided that ${\rm char}(k)\not=2$. Its intrinsic nature turns out to vary interestingly with the various types of algebra structures which $\mathscr{C}$ comprises, namely non-associative division algebras, non-commutative skew fields, and fields. A language, appropriate to formulate this covering result concisely, is introduced in Subsection 6.1.

On a technical level, the covering results of Section 6 (Corollaries 6.5 and 6.7) emerge from the results of Sections 2-4 as follows. We exhibit a partition $C_{\ell}=C_{\ell}^{0}\sqcup C_{\ell}^{1}$ such that each $C_{\ell}^{\nu}$ is a union of $\ell^{\ast}$-equivalence classes in $C_{\ell}$, which for their part coincide with the orbits of a suitable group action on $C_{\ell}^{\nu}$, which in turn endows $C_{\ell}^{\nu}$ with the structure of a group action groupoid. Now, the constructions $\mathscr{F}_{\ell}^{\nu}:C_{\ell}^{\nu}\to\mathscr{C},\ \mathscr{F}_{\ell}^{\nu}(\underline{c})=A(\ell,\underline{c}),$ give rise to faithful and dense functors $\mathscr{F}_{\ell}^{\nu}:C_{\ell}^{\nu}\to\mathscr{C}_{\ell}^{\nu},\ \nu\in\{0,1\},$ where $\mathscr{C}_{\ell}=\mathscr{C}_{\ell}^{0}\amalg\mathscr{C}_{\ell}^{1}$ is the full subgroupoid of $\mathscr{C}$ formed by all $A\in\mathscr{C}$ admitting a filtration of subalgebras $k\subset\ell\subset N_{r}(A)\subset A$, such that $\ell$ is invariant under two distinct commuting order 2 automorphisms of $A$, with blocks $\mathscr{C}_{\ell}^{0}=\{A\in\mathscr{C}_{\ell}\ |\ |{\rm Aut}(A)|>4\}$ and $\mathscr{C}_{\ell}^{1}=\{A\in\mathscr{C}_{\ell}\ |\ |{\rm Aut}(A)|=4\}$. As $\ell$ ranges through all isoclasses of quadratic extensions of $k$ and $\nu$ ranges through $\{0,1\}$, the subgroupoids $\mathscr{C}_{\ell}^{\nu}$ cover $\mathscr{C}$. If we ignore the non-commutative skew fields in $\mathscr{C}$, then the faithful and dense functors $\mathscr{F}_{\ell}^{\nu}:C_{\ell}^{\nu}\to\mathscr{C}_{\ell}^{\nu}$ are even full, i.e. equivalences of categories.

2 Preliminaries

2.1 Conventions, notation, terminology and basic notions

In this article, the least natural number is [math]. For all $n\in{N}$, we set

[TABLE]

If $F$ is a field and $(m,n)\in{N}^{2}$, then $F^{m\times n}$ denotes the set of all $m\times n$-matrices with entries in $F$, and $F^{\ast}=(F\setminus\{0\},\cdot)$ is the multiplicative group of $F$. The square map $s_{F}:F\to F,\ s_{F}(x)=x^{2}$, induces a group endomorphism $s_{F}^{\ast}:F^{\ast}\to F^{\ast}$. The notation $X_{sq}=s_{F}(X)=\{x^{2}\ |\ x\in X\}$ will be used for all subsets $X\subset F$.

The cardinality of a set $Y$ is denoted by $|Y|$. Maps are written on the left of their arguments, and are composed on the left. The composite map $gf$ will also be denoted by $g\circ f$, if it is beneficial to clarity.

If $(Y_{i})_{i\in I}$ is a family of subclasses of a class $Z$, then $Y=\bigsqcup_{i\in I}Y_{i}$ expresses that $Y=\bigcup_{i\in I}Y_{i}$ and $Y_{i}\cap Y_{j}=\emptyset$ for all $i\not=j$. The object class of a category $\mathscr{A}$ is also denoted by $\mathscr{A}$, for simplicity. A family $(\mathscr{A}_{i})_{i\in I}$ of full subcategories of a category $\mathscr{A}$ will be called a covering of $\mathscr{A}$ if the identity of object classes $\mathscr{A}=\bigcup_{i\in I}\mathscr{A}_{i}$ holds true, and will be called a decomposition of $\mathscr{A}$ with blocks $\mathscr{A}_{i}$ if the identity of object classes $\mathscr{A}=\bigsqcup_{i\in I}\mathscr{A}_{i}$ holds true and ${\rm Mor}_{\mathscr{A}}(A_{i},A_{j})=\emptyset$ for all $(A_{i},A_{j})\in\mathscr{A}_{i}\times\mathscr{A}_{j}$ and all $i\not=j$. The decomposition of $\mathscr{A}$ into blocks $\mathscr{A}_{i}$ is also expressed by saying that $\mathscr{A}$ is the coproduct of $(\mathscr{A}_{i})_{i\in I}$ with cofactors $\mathscr{A}_{i}$, or by the formula $\mathscr{A}=\coprod_{i\in I}\mathscr{A}_{i}$.

Following [7], we say that a category $\mathscr{A}$ is svelte, in case its isoclasses form a set, denoted by $\mathscr{A}/\simeq$. In this article, the term classification always means classification up to isomorphism. More precisely, by a classification of a svelte category $\mathscr{A}$ we mean an explicitly displayed transversal $\mathscr{T}$ for $\mathscr{A}/\simeq$. Synonymously, we say that $\mathscr{T}$ classifies $\mathscr{A}$.

For any group $G$ with identity element $e$, we call $G^{\circ}=G\setminus\{e\}$ a punctured group. In writing $H<G$, we mean that $H$ is a subgroup of $G$. If $k$ is a field and $h<k^{\ast}$, then $\left(k^{\ast}/h\right)^{\circ}=(k^{\ast}/h)\setminus\{h\}$ is the punctured quotient group of $k^{\ast}$ by $h$. By Klein’s four-group we mean any non-cyclic group of order 4. The notation ${\rm G}={\rm C}_{2}\times{\rm C}_{2}$ with ${\rm C}_{2}=\{\pm 1\}$ for a specific multiplicative instance of Klein’s four-group, and ${\rm V}={Z}_{2}\times{Z}_{2}$ with ${Z}_{2}=\{\overline{0},\overline{1}\}$ for a specific additive instance of Klein’s four-group, will be kept throughout this paper.

The symbol $k$ always denotes a field. By a $k$-algebra $A$ we mean a vector space $A$ over $k$, equipped with an algebra structure, i.e. a $k$-bilinear map $A\times A\to A,\ (x,y)\mapsto xy$. The product $xy$ will occasionally be denoted by $x\cdot y$, to enhance comprehensibility. With every $k$-algebra $A$ and every element $a\in A$, we associate the $k$-linear operators $L_{a}:A\to A,$ $L_{a}(x)=ax$ and $R_{a}:A\to A,\ R_{a}(x)=xa$. A division algebra over $k$ is a non-zero $k$-algebra $A$ such that both $L_{a}$ and $R_{a}$ are bijective for all $a\in A\setminus\{0\}$. A $k$-algebra $A$ is called unital if it has an unity $1=1_{A}$, such that $1x=x=x1$ for all $x\in A$. For unital $k$-algebras $A$, we usually treat the canonical field isomorphism $k\ \tilde{\to}\ k1$ as identity $k=k1$, to ease notation.

A Hurwitz algebra over $k$ is a non-zero unital $k$-algebra $A$, admitting a non-degenerate multiplicative quadratic form $q:A\to k$, traditionally called the norm of $A$.

A skew field over $k$ is a unital associative division algebra over $k$. With each skew field $A$ over $k$ we associate its multiplicative group $A^{\ast}=(A\setminus\{0\},\cdot)$. A skew field over $k$ is called central if its centre equals $k$. The commutative skew fields $\ell$ over $k$ are just the field extensions $k\subset\ell$. Their dimension over $k$ is traditionally called degree of the field extension, and is denoted by $\dim_{k}(\ell)$ or $[\ell:k]$. Field extensions of degree 2 are briefly referred to as quadratic extensions.

A morphism from a $k$-algebra $A$ to a $k$-algebra $B$ is a $k$-linear map $\varphi:A\to B$ such that $\varphi(xy)=\varphi(x)\varphi(y)$ for all $x,y\in A$. An isomorphism of $k$-algebras is a bijective morphism. Thus, the automorphism group of a $k$-algebra $A$ is

[TABLE]

Its identity element will be denoted by ${I}={I}_{A}$. If $A$ is unital, then every automorphism $\varphi$ of $A$ fixes $k=k1$ elementwise. In particular, if a $k$-algebra $\ell$ is a field extension $k\subset\ell$, then ${\rm Aut}(\ell)$ coincides with the group of all $k$-automorphisms of the field $\ell$, also known as the Galois group of $\ell$ over $k$, and denoted ${\rm Gal}(\ell/k)$. Furthermore, if $A$ is a skew field over $k$, then every $a\in A^{\ast}$ determines an inner automorphism $\kappa_{a}\in{\rm Aut}(A)$, defined by $\kappa_{a}(x)=axa^{-1}$. A famous Theorem of Skolem and Noether asserts that, if $A$ is a central skew field over $k$, then ${\rm Aut}(A)=\{\kappa_{a}\ |\ a\in A^{\ast}\}$. In that case, the surjective group morphism $\kappa:A^{\ast}\to{\rm Aut}(A),\ \kappa(a)=\kappa_{a}$ induces the isomorphism $\overline{\kappa}:A^{\ast}/k^{\ast}\ \tilde{\to}\ {\rm Aut}(A),\ \overline{\kappa}(\overline{a})=\kappa_{a}$.

A morphism from a division $k$-algebra $A$ to a division $k$-algebra $B$ is understood to be a non-zero algebra morphism $\varphi:A\to B$. With this convention, every morphism of division algebras is injective, and every morphism of division algebras of equal finite dimension is an isomorphism [5, Proposition 2.3].

By a groupoid we mean a category, in which all morphisms are isomorphisms. Thus, for every field $k$ and all $n\in{N}$, the category $\mathscr{D}_{n}(k)$ of all $n$-dimensional division algebras over $k$ is a groupoid. (It may or may not be empty.) Consequently, every full subcategory of $\mathscr{D}_{n}(k)$ is a groupoid. In particular, the full subcategory $\mathscr{C}=\mathscr{C}(k)$ of $\mathscr{D}_{4}(k)$, introduced in Section 1, is a groupoid. It contains the full subgroupoids $\mathscr{N}=\mathscr{N}(k),\ \mathscr{S}=\mathscr{S}(k)$ and $\mathscr{K}=\mathscr{K}(k)$ formed by

[TABLE]

In the same vein, $\mathscr{D}_{2}(k)$ contains the full subgroupoid $\mathscr{Q}=\mathscr{Q}(k)$ formed by all quadratic extensions of $k$.

The groupoids $\mathscr{N},\mathscr{S},\mathscr{K}$ and $\mathscr{Q}$ will play a prominent role in the investigation of $\mathscr{C}$, throughout this article.

2.2 Generalities on quadratic extensions

Let $k\subset\ell$ be a quadratic extension, and ${\rm char}(k)\not=2$. Then $k\subset\ell$ is Galois, with Galois group ${\rm Gal}(\ell/k)=\langle\sigma\rangle$ of order 2. We write $\overline{x}=\sigma(x)$ for all $x\in\ell$. The set of all purely imaginary elements in $\ell$,

[TABLE]

is a 1-dimensional $k$-linear subspace in $\ell$, such that $\ell=k\oplus{\rm Im}(\ell)$.

The norm map $n_{\ell/k}:\ell\to k,\ n_{\ell/k}(x)=x\overline{x}$, induces a group morphism $n_{\ell/k}^{\ast}:\ell^{\ast}\to k^{\ast},\ n_{\ell/k}^{\ast}(x)=x\overline{x}$. Besides, we have the group endomorphism $s_{\ell}^{\ast}:\ell^{\ast}\to\ell^{\ast},\ s_{\ell}^{\ast}(x)=x^{2}$, introduced in Subsection 2.1.

For any subgroup $H<\ell^{\ast}$ with $\sigma(H)=H$, the symbol $H>\lhd\ {\rm Gal}(\ell/k)$ denotes the semi-direct product of $H$ and ${\rm Gal}(\ell/k)$, i.e. the group with underlying set $H\times{\rm Gal}(\ell/k)$ and binary operation $(a,\sigma^{i})(b,\sigma^{j})=(a\sigma^{i}(b),\sigma^{i+j})$. The semi-direct products $H>\lhd\ {\rm Gal}(\ell/k)$ that will arise naturally in our context involve three subgroups $H_{i}<\ell^{\ast}$, namely $(H_{1},H_{2},H_{3})=(\ell^{\ast},{S},{A}^{\ast})$, where the “unit sphere” ${S}$ and the “punctured axes” ${A}^{\ast}$ are defined by

[TABLE]

2.3 Decomposition of $\mathscr{C}(k)$

Let $k$ be any field. We show that every division algebra $A\in\mathscr{C}(k)$ either is non-associative, or a central skew field over $k$, or a field extension of $k$, and we supplement this decomposition statement on $\mathscr{C}(k)$ by equivalent characterizations of each block. To begin with, we recall some basic notions needed in this context, and notation that goes along with them.

Let $A$ be any $k$-algebra. A subalgebra of $A$ is a $k$-linear subspace $U\subset A$, such that $xy\in U$ for all $x,y\in U$. Every subalgebra $U$ of $A$ is itself a $k$-algebra. A subfield of $A$ is a subalgebra $U\subset A$, such that $U$ is a field. The right nucleus of $A$ is, by definition, the subset

[TABLE]

of $A$. Every division algebra $A$ over $k$ has no zero divisors, i.e. $xy=0$ holds in $A$ only if $x=0$ or $y=0$. If, conversely, $A$ has no zero divisors and is finite dimensional, then $A$ is a division algebra over $k$. It follows, that every subalgebra of any finite-dimensional division algebra over $k$ again is a finite-dimensional division algebra over $k$.

The full subgroupoids $\mathscr{C}(k),\mathscr{N}(k),\mathscr{S}(k),\mathscr{K}(k)$ of $\mathscr{D}_{4}(k)$ are defined in Section 1 and Subsection 2.1. Besides, we introduce the full subgroupoid $\mathscr{HD}_{4}(k)$ of $\mathscr{D}_{4}(k)$, formed by all 4-dimensional Hurwitz division algebras over $k$.

Lemma 2.1**.**

*Let $A$ be an algebra over a field $k$.

(i) The right nucleus $N$ of $A$ is an associative subalgebra of $A$.

(ii) If $A$ is unital, then $k\subset N\subset A$ is a filtration of subalgebras, and $N$ is a unital associative subalgebra of $A$. Moreover, $A$ is a right $N$-module, whose module structure $A\times N\to A$ is the restricted algebra structure of $A$.

(iii) If $A$ is a finite-dimensional unital division algebra over $k$, then $N$ is a skew field over $k$, and $A$ is a right vector space over $N$, such that*

[TABLE]

If in addition $\dim_{k}(N)=2$, then $N$ is a field.

Proof.

$(i)$ and $(ii)$ admit straightforward verifications.

$(iii)$ By (ii), $N$ is a unital associative subalgebra of the finite-dimensional division algebra $A$ over $k$. Hence $N$ is a unital associative division algebra over $k$, i.e. a skew field over $k$. Consequently, again by (ii), $A$ is a right vector space over $N$. If $(x_{i})_{i\in I}$ is an $N$-basis in $A$ and $(y_{j})_{j\in J}$ is a $k$-basis in $N$, then $(x_{i}y_{j})_{ij\in I\times J}$ is a $k$-basis in $A$. This proves the dimension formula. If $\dim_{k}(N)=2$, then $N$ admits a $k$-basis $(1,b)$, whence we conclude that $N$ is a commutative skew field, i.e. a field. ∎

Proposition 2.2**.**

For every field $k$, the decomposition

[TABLE]

holds true. Moreover, $\mathscr{N}(k)=\{A\in\mathscr{C}(k)\ |\ \dim_{k}(N)=2\}$, and $\mathscr{K}(k)$ is formed by all Galois extensions of $k$ whose Galois group is Klein’s four-group. If ${\rm char}(k)\not=2$, then $\mathscr{S}(k)\supset\mathscr{HD}_{4}(k)$.

Proof.

The object classes $\mathscr{N}(k),\mathscr{S}(k)$ and $\mathscr{K}(k)$ are pairwise disjoint, and closed under isomorphisms within the groupoid $\mathscr{C}(k)$. Therefore, in order to prove the asserted decomposition of $\mathscr{C}(k)$, it suffices to show that $\mathscr{C}(k)\subset\mathscr{N}(k)\sqcup\mathscr{S}(k)\sqcup\mathscr{K}(k)$. So, let $A\in\mathscr{C}(k)$ be given. If $A$ is not associative, then $A\in\mathscr{N}(k)$. If $A$ is associative, then $A$ is a skew field over $k$. Let $Z$ be the centre of $A$, and let $E$ be a maximal subfield of $A$. Then [10, Theorem 7.15 (i)] implies that $k\subset Z\subset E\subset A$ and

[TABLE]

Thus $\dim_{Z}(A)$ divides 4 and is a square, whence $\dim_{Z}(A)\in\{1,4\}$. If $\dim_{Z}(A)=4$, then $\dim_{k}(Z)=1$ shows that $A$ is a central skew field over $k$, i.e. $A\in\mathscr{S}(k)$. If $\dim_{Z}(A)=1$, then $\dim_{k}(Z)=4$ shows that $A$ is a commutative skew field over $k$, i.e. $A\in\mathscr{K}(k)$.

For all $A\in\mathscr{C}(k)$, the dimension formula in Lemma 2.1 (iii) implies that $\dim_{k}(N)\in\{2,4\}$. Since $A$ is associative if and only if $A=N$, we conclude that $A$ is not associative if and only if $\dim_{k}(N)=2$.

Let $A\in\mathscr{K}(k)$. Then Klein’s four-group appears as a subgroup $H<{\rm Aut}(A)$. By [8, Propositions V.3.7 and V.3.6], the fixed field

[TABLE]

is an intermediate field $k\subset F\subset A$ such that $F\subset A$ is a Galois extension with Galois group $H$, whose degree is $[A:F]=|{\rm Gal}(A/F)|=|H|=4$. It follows that $[F:k]=1$, i.e. $F=k$. Conversely, if $k\subset A$ is a Galois extension whose Galois group is Klein’s four-group, then $[A:k]=|{\rm Gal}(A/k)|=4$ and ${\rm Gal}(A/k)={\rm Aut}(A)$, so $A\in\mathscr{K}(k)$.

Classical theory of Hurwitz algebras states that every $A\in\mathscr{HD}_{4}(k)$ is a central skew field over $k$ [11, Theorem 1.6.2 and Proposition 1.9.1]. Therefore, in order to show that $A\in\mathscr{S}(k)$, it suffices to exhibit Klein’s four-group as a subgroup of ${\rm Aut}(A)$. For that purpose, we make use of the Skolem-Noether isomorphism $\overline{\kappa}:A^{\ast}/k^{\ast}\ \tilde{\to}\ {\rm Aut}(A),\ \overline{\kappa}(\overline{a})=\kappa_{a}$, explained in Subsection 2.1. If ${\rm char}(k)\not=2$, then [11, Corollary 1.6.3] asserts the existence of an orthogonal $k$-basis $(1,a,b,ab)$ in $A$. It follows that $\overline{a}\not=\overline{b}$ and, invoking [11, Proposition 1.2.3], that $\overline{a}^{2}=\overline{b}^{2}=\overline{1}$ and $\overline{a}\overline{b}=\overline{b}\overline{a}$. Accordingly, the subgroup $\langle\kappa_{a},\kappa_{b}\rangle<{\rm Aut}(A)$ is Klein’s four-group. ∎

The inclusion of object classes $\mathscr{S}(k)\supset\mathscr{HD}_{4}(k)$, which we just proved, is in fact an identity (Corollary 4.6). Our proof of the converse inclusion $\mathscr{S}(k)\subset\mathscr{HD}_{4}(k)$ must be postponed, as it is based on a closer analysis of the groupoid $\mathscr{C}(k)$, to be developed in Sections 3 and 4.

3 Constructions

Throughout this section, $k$ is a field of characteristic not 2, and $k\subset\ell$ is a quadratic extension. We refer to Subsection 2.2 for notation that goes along with these data. In Subsection 3.1 we define the subset $C_{\ell}\subset k^{3}$, formed by all $\ell$-admissible triples in $k^{3}$. In Subsection 3.2 we construct from any $\ell$-admissible triple $\underline{c}\in C_{\ell}$ a division algebra $A(\ell,\underline{c})\in\mathscr{C}(k)$, and we study how this construction behaves with respect to the decomposition of $\mathscr{C}(k)$. In Subsection 3.3 we associate with any pair $(\underline{c},\underline{d})\in C_{\ell}\times C_{\ell}$ a subset $\ell^{\ast}(\underline{c},\underline{d})\subset\ell^{\ast}$, and we construct from each $a\in\ell^{\ast}(\underline{c},\underline{d})$ two morphisms $\varphi_{a}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ and $\psi_{a}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ in $\mathscr{C}(k)$.

3.1 The notion of an $\ell$-admissible triple

A general function $f:\ell^{2}\to\ell$ with $f(0,0)=0$ is said to be anisotropic if $f^{-1}(0)=\{(0,0)\}$, and isotropic otherwise. With any triple of scalars

[TABLE]

we associate a function

[TABLE]

We say that a triple $\underline{c}\in k^{3}$ is $\ell$-admissible if the function $q_{\underline{c}}:\ell^{2}\to\ell$ is anisotropic. The $\ell$-admissible triples in $k^{3}$ form a subset

[TABLE]

of $k^{3}$. Depending on the quadratic extension $k\subset\ell$, the subset $C_{\ell}\subset k^{3}$ may or may not be empty.

3.2 Construction of objects in $\mathscr{C}(k)$

From any triple $\underline{c}\in k^{3}$ we construct a 4-dimensional $k$-algebra $A(\ell,\underline{c})$, defining it as the vector space $\ell^{2\times 1}$ over $k$, equipped with the algebra structure

[TABLE]

where the right hand side is a product of $\ell$-matrices.

Proposition 3.1**.**

*Let $k\subset\ell$ be a quadratic extension in characteristic not 2, and let $\underline{c}\in k^{3}$. Then the 4-dimensional $k$-algebra $A=A(\ell,\underline{c})$ has the following properties.

(i) $A$ is unital, with unity*

[TABLE]

(ii) The subspace

[TABLE]

is a $k$-subalgebra, canonically isomorphic to $\ell$, and such that

[TABLE]

*is a filtration of $k$-subalgebras.

(iii) The operators $\alpha$ and $\beta$ on $A$, defined by*

[TABLE]

*are algebra automorphisms of $A$, which generate Klein’s four-group.

(iv) $A(\ell,\underline{c})$ is a division algebra if and only if $\underline{c}$ is $\ell$-admissible.

Proof.

(i)-(ii) One verifies directly that 1_{A}=\left(\begin{array}[]{c}1\\ 0\end{array}\right) is a unity in $A$, and that $1_{A}\ell$ is a subalgebra of $A$, which is canonically isomorphic to $\ell$. By construction, $A$ is also a 2-dimensional right vector space over $\ell$, such that the identity

[TABLE]

holds for all $(a,b,l)\in A\times A\times\ell$. Setting $b=1_{A}$, we see that

[TABLE]

holds for all $(a,l)\in A\times\ell$. Now (1) and (2) imply $1_{A}\ell\subset N_{r}(A)$, which establishes the claimed filtration of $k$-subalgebras.

(iii) One verifies directly, that $\alpha$ and $\beta$ are distinct commuting automorphisms of $A$, both of order 2.

(iv) For all $a\in A$, the $k$-linear operator $L_{a}:A\to A$ is even $\ell$-linear, by (1). If a=\left(\begin{array}[]{c}x\\ y\end{array}\right), then

[TABLE]

Now, $A$ is a division algebra if and only if $\det(L_{a})\not=0$ for all $a\in A\setminus\{0\}$. Equivalently, the function $q_{\underline{c}}:\ell^{2}\to\ell$ is anisotropic, i.e. $\underline{c}$ is $\ell$-admissible. ∎

As a direct consequence of Proposition 3.1, we obtain the following construction of objects in $\mathscr{C}(k)$.

Corollary 3.2**.**

If $\underline{c}\in C_{\ell}$, then $A(\ell,\underline{c})\in\mathscr{C}(k)$.

Let us summarize and introduce new notation. For every quadratic extension $k\subset\ell$ in characteristic not 2, we constructed a map

[TABLE]

where $\mathscr{C}=\mathscr{C}(k)$. The decomposition $\mathscr{C}=\mathscr{N}\amalg\mathscr{S}\amalg\mathscr{K}$ (Proposition 2.2) induces via $\mathscr{F}_{\ell}$ a partition $C_{\ell}=C_{\ell}^{\mathscr{N}}\sqcup C_{\ell}^{\mathscr{S}}\sqcup C_{\ell}^{\mathscr{K}}$, whose subsets are implicitly defined by $C_{\ell}^{\mathscr{N}}=\mathscr{F}_{\ell}^{-1}(\mathscr{N}),\ C_{\ell}^{\mathscr{S}}=\mathscr{F}_{\ell}^{-1}(\mathscr{S})$, and $C_{\ell}^{\mathscr{K}}=\mathscr{F}_{\ell}^{-1}(\mathscr{K})$. For later applications, we proceed to display these subsets explicitly.

Proposition 3.3**.**

*For all $\underline{c}\in C_{\ell}$, the following holds true.

(i) $A(\ell,\underline{c})\in\mathscr{N}$ if and only if $(c_{1},c_{2})\not=(1,0)$ and $(c_{1},c_{3})\not=(0,0)$.

(ii) $A(\ell,\underline{c})\in\mathscr{S}$ if and only if $(c_{1},c_{2})=(1,0)$.

(iii) $A(\ell,\underline{c})\in\mathscr{K}$ if and only if $(c_{1},c_{3})=(0,0)$.*

Proof.

Given $\underline{c}\in C_{\ell}$, set A=A(\ell,\underline{c}),\ N=N_{r}(A),\ j=\left(\begin{array}[]{c}0\\ 1\end{array}\right)\in A, and choose $i\in{\rm Im}(\ell)\setminus\{0\}$. We will repeatedly use Proposition 2.2.

(i) Assume that $(c_{1},c_{2})=(1,0)$ or $(c_{1},c_{3})=(0,0)$. Then one verifies directly that $j\in N$. Since $A=1_{A}\ell\oplus j\ell=1_{A}\ell\oplus j\cdot(1_{A}\ell)$ by (2), and $1_{A}\ell\subset N$ by Proposition 3.1 (ii), we conclude that $A=N$. Thus $A$ is associative, i.e. $A\not\in\mathscr{N}$. Assume conversely that $A\not\in\mathscr{N}$, i.e. $A$ is associative. Then direct calculation shows that the system

[TABLE]

is equivalent to

[TABLE]

which in turn is equivalent to

[TABLE]

Since $1\not=0$ excludes $c_{1}=1-c_{1}=0$, and $q_{\underline{c}}(0,1)\not=0$ excludes $c_{2}=c_{3}=0$, we are left with the two alternatives $(c_{1},c_{2})=(1,0)$ or $(c_{1},c_{3})=(0,0)$.

(ii)-(iii) If $(c_{1},c_{2})=(1,0)$, then $A\in\mathscr{S}\amalg\mathscr{K}$ holds by (i). Moreover, $q_{\underline{c}}(0,1)\not=0$ implies $c_{3}\not=0$, which in turn implies $j\cdot ji\not=ji\cdot i$. So $A$ is not commutative, and hence $A\in\mathscr{S}$.

If $(c_{1},c_{3})=(0,0)$, then one verifies directly that $A$ is commutative. Together with (i), it follows that $A\in\mathscr{K}$.

In view of (i) and $\mathscr{S}\cap\mathscr{K}=\emptyset$, the “only if” statements in (ii) and (iii) are now direct consequences. ∎

3.3 Construction of morphisms in $\mathscr{C}(k)$

Every pair $(\underline{c},\underline{d})\in C_{\ell}\times C_{\ell}$ determines a subset $\ell^{\ast}(\underline{c},\underline{d})\subset\ell^{\ast}$, defined by

[TABLE]

Direct verification establishes the following lemma.

Lemma 3.4**.**

*For all $\underline{c},\underline{d}\in C_{\ell}$ and $a\in\ell^{\ast}$, the following holds true.

(i) $a\in\ell^{\ast}(\underline{c},\underline{d})$ if and only if $\overline{a}\in\ell^{\ast}(\underline{c},\underline{d})$.

(ii) If $a\in\ell^{\ast}(\underline{c},\underline{d})$, then the maps*

[TABLE]

and

[TABLE]

*are morphisms in $\mathscr{C}(k)$.

(iii) $\{\pm 1\}\subset\ell^{\ast}(\underline{c},\underline{c})$, and the automorphisms $\varphi_{-1}$ and $\psi_{1}$ of $A(\ell,\underline{c})$ coincide with the automorphisms $\alpha$ and $\beta$ of $A(\ell,\underline{c})$, displayed in Proposition 3.1 (iii).*

4 Reductions

Even in this section, $k$ is a field of characteristic not 2. Tracing the constructions of the previous section in the reverse direction, we achieve reductions of objects and morphisms in $\mathscr{C}(k)$, that can be phrased as follows. For every division algebra $A\in\mathscr{C}(k)$ there is a quadratic extension $k\subset\ell$ and an $\ell$-admissible triple $\underline{c}\in C_{\ell}$ such that $A(\ell,\underline{c})\ \tilde{\to}\ A$. For every quadratic extension $k\subset\ell$ and all $(\underline{c},\underline{d})\in C_{\ell}\times C_{\ell}$, the division algebras $A(\ell,\underline{c})$ and $A(\ell,\underline{d})$ are isomorphic if and only if the subset $\ell^{\ast}(\underline{c},\underline{d})\subset\ell^{\ast}$ is not empty.

These reduction statements can be proved in the way of a generalization of [2] from finite fields of odd order to general fields $k$ of characteristic not 2. This proof is elementary, but yet laborious and somewhat obscure. Instead, we present here a conceptual and more lucid approach, suggested by an anonymous referee, to whom credit must be given for this improvement. It is based on the insight, that every $A\in\mathscr{C}(k)$ is V-graded, and on the usefulness of the trace bilinear form of $A$.

4.1 V-grading and trace bilinear form of objects in $\hat{\mathscr{C}}(k)$

If $\varphi$ is an automorphism of a $k$-algebra $A$ and $\varepsilon\in k$ is an eigenvalue of $\varphi$, then we denote by $E_{\varphi}(\varepsilon)=\{x\in A\ |\ \varphi(x)=\varepsilon x\}$ the eigenspace of $\varphi$ associated with the eigenvalue $\varepsilon$.

Let $\alpha$ and $\beta$ be automorphisms of a $k$-algebra $A$. We say that $(\alpha,\beta)$ is a Kleinian pair for $A$ if $\langle\alpha,\beta\rangle$ is Klein’s four-group. Equivalently, $\alpha$ and $\beta$ are distinct commuting automorphisms of order 2.

Because the right nucleus of a division algebra $A\in\mathscr{C}(k)$ is irrelevant for the material to be expounded in this subsection, we introduce the full subgroupoid $\hat{\mathscr{C}}(k)$ of $\mathscr{D}_{4}(k)$, formed by all $A\in\mathscr{D}_{4}(k)$ which are unital and admit a Kleinian pair. With this definition, $\mathscr{C}(k)\subset\hat{\mathscr{C}}(k)\subset\mathscr{D}_{4}(k)$ is a filtration of full subgroupoids.

Recall our notation ${\rm G}={\rm C}_{2}\times{\rm C}_{2}$ and ${\rm V}={Z}_{2}\times{Z}_{2}$, for two specific instances of Klein’s four-group. The identity element in ${\rm G}$ is $e=(1,1)$, and general elements in ${\rm V}$ will be denoted by $ij=(\overline{i},\overline{j})$, for brevity.

Proposition 4.1**.**

For every division algebra $A\in\hat{\mathscr{C}}(k)$ with Kleinian pair $(\alpha,\beta)$, the subspaces $A_{ij}=A_{ij}^{\alpha\beta}=E_{\alpha}\left((-1)^{i}\right)\cap E_{\beta}\left((-1)^{j}\right)$ of $A$, $ij\in{\rm V}$, form a V-grading $A=\bigoplus_{ij\in{\rm V}}A_{ij}$ of $A$, such that $\dim_{k}(A_{ij})=1$ for all $ij\in{\rm V}$, and $A_{00}=k1_{A}$.

Proof.

The group algebra $k{\rm G}$ is semisimple, because ${\rm char}(k)\not|\ |{\rm G}|$. It admits four isoclasses of simple modules, represented by the 1-dimensional $k{\rm G}$-modules $S_{ij},\ ij\in{\rm V}$, given by $S_{ij}=k$ and $a1=(-1)^{i},\ b1=(-1)^{j}$.

Given $A$ and $(\alpha,\beta)$ as in the statement, we choose an isomorphism ${\rm G}\ \tilde{\to}\ \langle\alpha,\beta\rangle$ and compose it with the inclusion morphism $\langle\alpha,\beta\rangle\hookrightarrow{\rm Aut}(A)$, to obtain a group monomorphism $\rho:{\rm G}\hookrightarrow{\rm Aut}(A)$. This endows $A$ with the structure of a $k{\rm G}$-module, which decomposes into simple submodules

[TABLE]

where $B_{ij}=\bigoplus_{U_{h}\simeq S_{ij}}U_{h}$. For all $ij\in{\rm V}$, the inclusion $B_{ij}\subset A_{ij}$ holds by definition of $B_{ij}$ and $A_{ij}$, whence $B_{ij}=A_{ij}$ follows with Krull-Schmidt’s Theorem. Thus, we proved the direct sum decomposition $A=\bigoplus_{ij\in{\rm V}}A_{ij}$. Straightforward arguments show that this is a V-grading of $A$, such that $1\in A_{00}$. It remains to show that $m_{ij}=\dim_{k}(A_{ij})=1$ for all $ij\in{\rm V}$.

Let $\chi_{ij}:{\rm G}\to k$ be the character of $S_{ij}$. Then the characters of the $k{\rm G}$-modules $A$ and $k{\rm G}$ are $\chi_{A}=\sum_{ij\in{\rm V}}m_{ij}\chi_{ij}$ and $\chi_{k{\rm G}}=\sum_{ij\in{\rm V}}\chi_{ij}$, respectively. Since the irreducible characters $\chi_{ij}$ are linearly independent over $k$ [3, Theorem 30.12 (i)], it suffices to show that $\chi_{A}=\chi_{k{\rm G}}$.

So, let $g\in{\rm G}\setminus\{e\}$ and set $E_{g}(\varepsilon)=E_{\rho(g)}(\varepsilon)$ for all $\varepsilon\in\{\pm 1\}$. Then $A=E_{g}(1)\oplus E_{g}(-1)$, because $x=\frac{1}{2}(x+gx)+\frac{1}{2}(x-gx)$ holds for all $x\in A$. Now $g\not=e$ guarantees the existence of an element $u\in E_{g}(-1)\setminus\{0\}$. Since $A$ is a division algebra and $\rho(g)\in{\rm Aut}(A)$, the $k$-linear operator $L_{u}:A\to A,\ L_{u}(x)=ux$, is bijective and satisfies $L_{u}(E_{g}(1))\subset E_{g}(-1)$ and $L_{u}(E_{g}(-1))\subset E_{g}(1)$. It follows that $\dim_{k}(E_{g}(1))=\dim_{k}(E_{g}(-1))=2$, and hence that $\chi_{A}(g)=0=\chi_{k{\rm G}}(g)$. Moreover, $\chi_{A}(e)=4=\chi_{k{\rm G}}(e)$. Altogether, we proved that $\chi_{A}=\chi_{k{\rm G}}$. ∎

Proposition 4.2**.**

*(i) For every division algebra $A\in\hat{\mathscr{C}}(k)$, the trace bilinear form $\tau_{A}:A\times A\to k,\ \tau_{A}(x,y)={\rm tr}(L_{xy})$ is symmetric and non-degenerate.

(ii) Every morphism $\varphi:A\to B$ in $\hat{\mathscr{C}}(k)$ is orthogonal with regard to $\tau_{A}$ and $\tau_{B}$, i.e. $\tau_{A}(x,y)=\tau_{B}(\varphi(x),\varphi(y))$ for all $x,y\in A$.*

Proof.

(i) Every $A\in\hat{\mathscr{C}}(k)$ admits a V-grading $A=\bigoplus_{ij\in{\rm V}}A_{ij}$ (Proposition 4.1). Set $U=A_{01}\cup A_{10}\cup A_{11}$. Then ${\rm tr}(L_{u})=0$ for all $u\in U$. Choose a $k$-basis $(a_{ij})_{ij\in{\rm V}}$ in $A$ such that $a_{ij}\in A_{ij}$ for all $ij\in{\rm V}$. If $ij$ and $mn$ are distinct elements in ${\rm V}$, then $a_{ij}a_{mn}\in U$ and $a_{mn}a_{ij}\in U$ implies that $\tau_{A}(a_{ij},a_{mn})=0=\tau_{A}(a_{mn},a_{ij})$, whence symmetry of $\tau_{A}$ follows.

For every $x\in A\setminus\{0\}$ there exists an element $y\in A$ such that $xy=1$. Then $\tau_{A}(x,y)={\rm tr}(L_{1})=4\not=0$ shows that $\tau_{A}$ is non-degenerate.

(ii) Every morphism $\varphi:A\to B$ in $\hat{\mathscr{C}}(k)$ is a $k$-linear bijection. For all $z\in A$ and for every $k$-basis $\underline{a}$ in $A$, the matrix of $L_{z}$ in $\underline{a}$ equals the matrix of $L_{\varphi(z)}$ in $\varphi(\underline{a})$, whence ${\rm tr}(L_{z})={\rm tr}(L_{\varphi(z)})$ follows. Thus

[TABLE]

holds for all $x,y\in A$. ∎

Lemma 4.3**.**

*Let $A$ be a division algebra in $\hat{\mathscr{C}}(k)$, with Kleinian pair $(\alpha,\beta)$ and associated V-grading $A=\bigoplus_{ij\in{\rm V}}A_{ij}$. Then, for all $(ij,mn)\in{\rm V}\times{\rm V}$ and $(x_{ij},y_{mn})\in A_{ij}\times A_{mn}$, the following holds true.

(i) $\tau_{A}(x_{ij},y_{mn})=0$  if  $ij\not=mn$.

(ii) $\tau_{A}(x_{ij},y_{ij})=0$  if and only if  $x_{ij}y_{ij}=0$.*

Proof.

(i) The above proof of symmetry of $\tau_{A}$ contains even a proof of this statement.

(ii) The “if” part holds trivially true. If $x_{ij}y_{ij}\not=0$, then $x_{ij}y_{ij}\in k1\setminus\{0\}$ implies that $\tau_{A}(x_{ij},y_{ij})\not=0$. ∎

Proposition 4.4**.**

For every division algebra $A\in\hat{\mathscr{C}}(k)$ with Kleinian pair $(\alpha,\beta)$, the orthogonal supplement of $E_{\beta}(1)$ with regard to $\tau_{A}$ is

[TABLE]

Proof.

Let $A=\bigoplus_{ij\in{\rm V}}A_{ij}$ be the V-grading associated with $(\alpha,\beta)$. Then $E_{\beta}(1)=A_{00}\oplus A_{10}$ and $E_{\beta}(-1)=A_{01}\oplus A_{11}$. The claimed identity is now a straightforward consequence of Lemma 4.3. ∎

4.2 Reduction of objects in $\mathscr{C}(k)$

Let $A$ be a division algebra in $\mathscr{C}(k)$. We say that a subfield $\ell\subset A$ is Kleinian if there exists a Kleinian pair $(\alpha,\beta)$ for $A$, such that $\ell=E_{\beta}(1)\subset N_{r}(A)$. Since $E_{\beta}(1)=A_{00}\oplus A_{10}=k1_{A}\oplus A_{10}$ (Proposition 4.1), every Kleinian subfield of $\ell\subset A$ is a quadratic extension of $k$, such that $k\subset\ell\subset N_{r}(A)$.

Theorem 4.5**.**

*For every division algebra $A\in\mathscr{C}(k)$, the following holds true.

(i) There exists a Kleinian subfield $\ell\subset A$. If $A$ is not associative, then $\ell=N_{r}(A)$ is the unique Kleinian subfield of $A$.

(ii) For every Kleinian subfield $\ell\subset A$, there exists an $\ell$-admissible triple $\underline{c}\in C_{\ell}$, such that $A(\ell,\underline{c})\ \tilde{\to}\ A$.*

Proof.

(i) If $A$ is associative, choose any Kleinian pair $(\alpha,\beta)$ for $A$, and set $\ell=E_{\beta}(1)$. Then $\ell=A_{00}\oplus A_{10}=k1_{A}\oplus A_{10}$ (Proposition 4.1) shows that $\ell$ is a 2-dimensional unital associative subalgebra of the division algebra $A$, contained in $A=N_{r}(A)$, and hence is a Kleinian subfield of $A$.

If $A$ is not associative, then we choose an injective group homomorphism $\rho:{\rm G}\hookrightarrow{\rm Aut}(A)$. Every automorphism $\varphi$ of $A$ induces an automorphism $\varphi_{N}$ of $N_{r}(A)$. This assignment is a group homomorphism

[TABLE]

Due to Lemma 2.1 (iii) and Proposition 2.2, $k\subset N_{r}(A)$ is a quadratic extension, and hence ${\rm Aut}(N_{r}(A))={\rm Gal}(N_{r}(A)/k)$ has order 2. The composed group homomorphism $\nu\rho:{\rm G}\to{\rm Aut}(N_{r}(A))$ is therefore not injective. Choose $b\in{\rm ker}(\nu\rho)\setminus\{e\}$ and $a\in{\rm G}\setminus\{e,b\}$. Then $(\alpha,\beta)=(\rho(a),\rho(b))$ is a Kleinian pair for $A$, and $\beta_{N}=\nu\rho(b)={I}_{N_{r}(A)}$ shows, together with $\dim_{k}(E_{\beta}(1))=2=\dim_{k}(N_{r}(A))$, that $E_{\beta}(1)=N_{r}(A)$. From this information we conclude with Proposition 4.1, as in the associative case above, that $\ell=E_{\beta}(1)$ is a Kleinian subfield of $A$.

If $A$ is not associative and $\ell\subset A$ is any Kleinian subfield, then the filtration $k\subset\ell\subset N_{r}(A)$, combined with $\dim_{k}(\ell)=2=\dim_{k}(N_{r}(A))$, yields $\ell=N_{r}(A)$.

(ii) Given any Kleinian subfield $\ell\subset A$, there exists a Kleinian pair $(\alpha,\beta)$ for $A$, such that $\ell=E_{\beta}(1)\subset N_{r}(A)$. Let $A=\bigoplus_{ij\in{\rm V}}A_{ij}$ be the V-grading associated with $(\alpha,\beta)$. Then $\ell=A_{00}\oplus A_{10}$, $A_{00}=k1_{A}$ and $A_{10}={\rm Im}(\ell)$. Choose $u\in A_{10}\setminus\{0\}$ and $v\in A_{01}\setminus\{0\}$. Then $\ell=k1_{A}\oplus ku,\ \overline{u}=-u$, and $A=\ell\oplus v\ell$.

The $k$-linear bijection $\varphi=L_{v}^{-1}R_{v}\in{\rm GL}(A)$ satisfies

[TABLE]

Since $uv\in A_{10}A_{01}=A_{11}=vA_{10}$, it also satisfies

[TABLE]

whence $\varphi(u)=\zeta_{1}u$ for some $\zeta_{1}\in k^{\ast}$. Thus, $\varphi\in{\rm GL}(A)$ induces a $k$-linear bijection $\varphi_{\ell}\in{\rm GL}(\ell)$. With $c_{1}=\frac{1-\zeta_{1}}{2}$, the identity

[TABLE]

holds for all $x\in\{1_{A},u\}$, and hence for all $x\in\ell$. Accordingly,

[TABLE]

yields

[TABLE]

for all $x,z\in\ell$.

In the same vein, the $k$-linear bijection $\psi=R_{v}L_{v}\in{\rm GL}(A)$ satisfies $\psi(1_{A})=(v1_{A})v=vv=\zeta_{2}1_{A}$ for some $\zeta_{2}\in k^{\ast}$, and $\psi(u)=(vu)v=\zeta_{3}u$ for some $\zeta_{3}\in k^{\ast}$, since $(vu)v\in(A_{01}A_{10})A_{01}=A_{11}A_{01}=A_{10}=ku$. Thus, $\psi\in{\rm GL}(A)$ induces a $k$-linear bijection $\psi_{\ell}\in{\rm GL}(\ell)$. With $c_{2}=\frac{\zeta_{2}+\zeta_{3}}{2}$ and $c_{3}=\frac{\zeta_{2}-\zeta_{3}}{2}$, the identity

[TABLE]

holds for all $y\in\{1_{A},u\}$, and hence for all $y\in\ell$. Accordingly,

[TABLE]

holds for all $y,z\in\ell$.

Summarizing, we have found a triple $\underline{c}=(c_{1},c_{2},c_{3})\in k^{3}$, that satisfies (3) and (4) for all $x,y,z\in\ell$. We claim that the map

[TABLE]

is an isomorphism of $k$-algebras. Indeed, $\theta$ is $k$-linear by definition, and bijective since $\ell\oplus v\ell=A$. Using (3) and (4), we see that

[TABLE]

holds for all $x,y,w,z\in\ell$. The algebra isomorphism $\theta:A(\ell,\underline{c})\ \tilde{\to}\ A$ is thus established. It follows that $A(\ell,\underline{c})$ is a division algebra, and hence that $\underline{c}$ is $\ell$-admissible (Proposition 3.1 (iv)). ∎

Corollary 4.6**.**

For every field $k$ of characteristic not 2, the class $\mathscr{S}(k)$ of all 4-dimensional central skew fields over $k$ that admit a Kleinian pair coincides with the the class $\mathscr{HD}_{4}(k)$ of all 4-dimensional Hurwitz division algebras over $k$.

Proof.

We already know that $\mathscr{S}(k)\supset\mathscr{HD}_{4}(k)$ (Proposition 2.2). Conversely, for every $A\in\mathscr{S}(k)$ there exists a quadratic extension $k\subset\ell$ and an $\ell$-admissible triple $\underline{c}=(1,0,c_{3})\in C_{\ell}^{\mathscr{S}}$, such that $A(\ell,\underline{c})\ \tilde{\to}\ A$ (Theorem 4.5 and Proposition 3.3). Identifying $(x,y)\in\ell^{2}$ with \left(\begin{array}[]{c}x\\ y\end{array}\right)\in\ell^{2\times 1}, the function $q_{\underline{c}}:\ell^{2}\to\ell$ turns into a quadratic form

[TABLE]

which is non-degenerate because it is anisotropic, and multiplicative because

[TABLE]

holds for all $a,b\in A(\ell,\underline{c})$. Accordingly, $A(\ell,\underline{c})\in\mathscr{HD}_{4}(k)$. As $A\ \tilde{\to}\ A(\ell,\underline{c})$, it follows that $A\in\mathscr{HD}_{4}(k)$. ∎

4.3 Reduction of morphisms in $\mathscr{C}(k)$

Recall that all morphisms in $\mathscr{C}(k)$ are isomorphisms. We use the notation introduced in Subsection 3.3.

Theorem 4.7**.**

For every quadratic extension $k\subset\ell$ in characteristic not 2 and for all $(\underline{c},\underline{d})\in C_{\ell}\times C_{\ell}$, the division algebras $A(\ell,\underline{c})$ and $A(\ell,\underline{d})$ are isomorphic if and only if the subset $\ell^{\ast}(\underline{c},\underline{d})\subset\ell^{\ast}$ is not empty.

Proof.

Let $k\subset\ell$ and $(\underline{c},\underline{d})\in C_{\ell}\times C_{\ell}$ be given. If $\ell^{\ast}(\underline{c},\underline{d})\not=\emptyset$, then we can choose $a\in\ell^{\ast}(\underline{c},\underline{d})$ and construct the isomorphism $\varphi_{a}:A(\ell,\underline{c})\to A(\ell,\underline{d})$, defined in Lemma 3.4 (ii).

Conversely, suppose that an isomorphism $\mu:A(\ell,\underline{c})\to A(\ell,\underline{d})$ exists. Towards a proof of non-emptiness of $\ell^{\ast}(\underline{c},\underline{d})$ we proceed in four steps, wherein we denote the common unity 1_{A(\ell,\underline{c})}=1_{A(\ell,\underline{d})}=\left(\begin{array}[]{c}1\\ 0\end{array}\right) by $1_{A}$, and set j=\left(\begin{array}[]{c}0\\ 1\end{array}\right).

Step 1. There exists an isomorphism $\mu_{1}:A(\ell,\underline{c})\to A(\ell,\underline{d})$, such that $\mu_{1}(1_{A}\ell)=1_{A}\ell$.

Proof of Step 1. The existence of $\mu$ implies with Proposition 2.2, that

[TABLE]

Suppose $(\underline{c},\underline{d})\in C_{\ell}^{\mathscr{N}}\times C_{\ell}^{\mathscr{N}}$. The isomorphism $\mu$ induces an isomorphism of right nuclei, and $N_{r}(A(\ell,\underline{c}))=1_{A}\ell=N_{r}(A(\ell,\underline{d}))$ by Proposition 3.1 (ii) and Proposition 2.2, so $\mu(1_{A}\ell)=1_{A}\ell$ follows.

Suppose $(\underline{c},\underline{d})\in C_{\ell}^{\mathscr{S}}\times C_{\ell}^{\mathscr{S}}$. The isomorphism $\mu$ induces an isomorphism $\mu_{\iota}:1_{A}\ell\to\mu(1_{A}\ell)$ of subfields of $A(\ell,\underline{d})$. Since $A(\ell,\underline{d})$ is a central skew field over $k$, [10, Theorem 7.21] asserts that $\mu_{\iota}$ extends to an automorphism $\vartheta$ of $A(\ell,\underline{d})$. Now $\mu_{1}=\vartheta^{-1}\mu$ will do.

Suppose $(\underline{c},\underline{d})\in C_{\ell}^{\mathscr{K}}\times C_{\ell}^{\mathscr{K}}$. Choose $u\in{\rm Im}(1_{A}\ell)\setminus\{0\}$. Then $u^{2}=1_{A}\lambda$ for some $\lambda\in k\setminus k_{sq}$. Now $(\mu(u))^{2}=\mu(u^{2})=\mu(1_{A}\lambda)=1_{A}\lambda=u^{2}$ implies, since $A(\ell,\underline{d})$ is a field, that $\mu(u)=\pm u$. Because $1_{A}\ell=1_{A}k\oplus uk$, it follows that $\mu(1_{A}\ell)=1_{A}\ell$.

Step 2. There exists an isomorphism $\mu_{2}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ that fixes $1_{A}\ell$ elementwise.

Proof of Step 2. Let $\mu_{1}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ be an isomorphism as in Step 1. Then $\mu_{1}$ induces a Galois automorphism $\mu_{\ell}\in{\rm Gal}(\ell/k)$, and ${\rm Gal}(\ell/k)=\langle\sigma\rangle$ has order 2. If $\mu_{\ell}={I}_{\ell}$, then $\mu_{2}=\mu_{1}$ will do. If $\mu_{\ell}=\sigma$, then $\mu_{2}=\mu_{1}\psi_{1}$ will do, where $\psi_{1}\in{\rm Aut}(A(\ell,\underline{c}))$ is the automorphism defined in Lemma 3.4 (ii).

Step 3. If $\mu_{2}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ is an isomorphism that fixes $1_{A}\ell$ elementwise, then $\mu_{2}(j)=ja$ for some $a\in\ell^{\ast}$.

Proof of Step 3. Let $\mu_{2}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ be an isomorphism that fixes $1_{A}\ell$ elementwise, and let $(\beta,\alpha)$ be the Kleinian pair for $A(\ell,\underline{c})$ and $A(\ell,\underline{d})$, defined in Proposition 3.1 (iii). Then, by Proposition 4.4, the orthogonal supplement of $1_{A}\ell$ with regard to the trace bilinear form is

[TABLE]

in both $A(\ell,\underline{c})$ and $A(\ell,\underline{d})$. We conclude with Proposition 4.2 (ii) that

[TABLE]

holds for all $l\in\ell$. So $\mu_{2}(j)\in 1_{A}\ell^{\perp}=j\ell$, i.e. $\mu_{2}(j)=ja$ for some $a\in\ell^{\ast}$.

Step 4. If $\mu_{2}:A(\ell,\underline{c})\to A(\ell,\underline{d})$ and $a\in\ell^{\ast}$ are as in Step 3, then $a\in\ell^{\ast}(\underline{c},\underline{d})$.

Proof of Step 4. As $\mu_{2}$ fixes $1_{A}\ell$ elementwise, identity (2) in the proof of Proposition 3.1 shows that $\mu_{2}$ is right $\ell$-linear. Thus, the system of equations

[TABLE]

valid for all $(x,y)\in\ell^{2}$, gives rise to the system of equations

[TABLE]

which, evaluated in $x=i\in{\rm Im}(\ell)\setminus\{0\}$ and $y\in\{1,i\}$, yields $c_{1}=d_{1}$ and $(c_{2},c_{3})=(a^{2}d_{2},a\overline{a}d_{3})$, i.e. $a\in\ell^{\ast}(\underline{c},\underline{d})$. ∎

4.4 $\mathscr{L}$-coverings of $\mathscr{C}(k)$ and $\mathscr{N}(k),\mathscr{S}(k),\mathscr{K}(k)$

For every division algebra $A\in\mathscr{C}(k)$, we denote by ${\cal N}(A)$ the set of all Kleinian subfields of $A$. Then ${\cal N}(A)\subset\mathscr{Q}(k)$, where $\mathscr{Q}(k)$ is the groupoid formed by all quadratic extensions of $k$ (Subsection 2.1). Let $\mathscr{L}\subset\mathscr{Q}(k)$ be a transversal for the set $\mathscr{Q}(k)/\simeq$ of all isoclasses in $\mathscr{Q}(k)$. For each $\ell\in\mathscr{L}$ we define the full subgroupoid $\mathscr{C}_{\ell}(k)\subset\mathscr{C}(k)$ by its object class

[TABLE]

and likewise we define the full subgroupoids $\mathscr{N}_{\ell}(k),\mathscr{S}_{\ell}(k)$ and $\mathscr{K}_{\ell}(k)$ respectively of $\mathscr{N}(k),\mathscr{S}(k)$ and $\mathscr{K}(k)$ by their object classes

[TABLE]

Proposition 4.8**.**

The groupoid $\mathscr{C}(k)$ and its blocks $\mathscr{N}(k),\mathscr{S}(k),\mathscr{K}(k)$ admit coverings of the form

[TABLE]

Proof.

Every $A\in\mathscr{C}(k)$ has a Kleinian subfield $n\subset A$ (Theorem 4.5 (i)), and $n\ \tilde{\to}\ \ell$ for some $\ell\in\mathscr{L}$, so $A\in\mathscr{C}_{\ell}(k)$. This proves the covering statement for $\mathscr{C}(k)$. Taking $A$ in $\mathscr{N}(k),\mathscr{S}(k)$ and $\mathscr{K}(k)$ respectively, we obtain the covering statements for $\mathscr{N}(k),\mathscr{S}(k)$ and $\mathscr{K}(k)$. For all $\ell\in\mathscr{L}$, the decomposition $\mathscr{C}_{\ell}(k)=\mathscr{N}_{\ell}(k)\amalg\mathscr{S}_{\ell}(k)\amalg\mathscr{K}_{\ell}(k)$ holds by Proposition 2.2.

Let $\ell,m\in\mathscr{L}$. If $A\in\mathscr{N}_{\ell}(k)$ and $B\in\mathscr{N}_{m}(k)$ are isomorphic objects, then Theorem 4.5 (i), second statement, yields $\ell\ \tilde{\to}\ N_{r}(A)\ \tilde{\to}\ N_{r}(B)\ \tilde{\to}\ m$. This implies $\ell=m$, whence the decomposition statement for $\mathscr{N}(k)$ follows. ∎

The question of how to find a transversal $\mathscr{L}$ for $\mathscr{Q}(k)/\simeq$ arises naturally in the context of Proposition 4.8. It is answered on the level of the punctured quotient group $\left(k^{\ast}/k_{sq}^{\ast}\right)^{\circ}$, introduced in Subsection 2.1, by the following proposition, whose proof makes an exercise in basic algebra.

Proposition 4.9**.**

*For every field $k$ of characteristic not 2, the following statements hold true.

(i) If $a\in k^{\ast}\setminus k_{sq}^{\ast}$, then $k(\sqrt{a}~{})\in\mathscr{Q}(k)$.

(ii) If $\ell\in\mathscr{Q}(k)$, then $\ell\ =k(\sqrt{a}~{})$, for some $a\in k^{\ast}\setminus k_{sq}^{\ast}$.

(iii) For all $a,b\in k^{\ast}\setminus k_{sq}^{\ast}$, the coset identity $ak_{sq}^{\ast}=bk_{sq}^{\ast}$ holds if and only if $k(\sqrt{a}~{})\ \tilde{\to}\ k(\sqrt{b}~{})$.

(iv) A subset $L\subset k^{\ast}\setminus k_{sq}^{\ast}$ is a transversal for $\left(k^{\ast}/k_{sq}^{\ast}\right)^{\circ}$ if and only if the subset $\mathscr{L}=\left\{\left.k(\sqrt{a}~{})\right|a\in L\right\}\subset\mathscr{Q}(k)$ is a transversal for $\mathscr{Q}(k)/\simeq$.

(v) The order of the quotient group $k^{\ast}/k_{sq}^{\ast}$ and the cardinality of the set of isoclasses $\mathscr{Q}(k)/\simeq$ satisfy the identity $\left|k^{\ast}/k_{sq}^{\ast}\right|-1=\left|\mathscr{Q}(k)/\simeq\right|$.*

5 Classifications

We retain our assumption, that $k$ is a field of characteristic not 2. The present section is devoted to the problem of classifying the groupoid $\mathscr{C}(k)$. This problem is reduced in the first instance, by Proposition 4.8, to the problem of classifying $\mathscr{C}_{\ell}(k)$ for all $\ell$ in a transversal $\mathscr{L}$ for $\mathscr{Q}(k)/\simeq$, and such a transversal can be found in accordance with Proposition 4.9 (iv). For each $\ell\in\mathscr{L}$, the problem of classifying $\mathscr{C}_{\ell}(k)$ is in turn reduced, by Theorems 4.5 and 4.7, to the problem of classifying all $\ell$-admissible triples up to $\ell^{\ast}$-equivalence. The latter problem, when restricted to $C_{\ell}^{\mathscr{S}}\sqcup C_{\ell}^{\mathscr{K}}$, can be solved on the level of two punctured quotient groups of $k^{\ast}$.

These reductions, which we here only outlined, are made precise in Subsection 5.1 below, and explicit classifications will be derived from them in Subsections 5.3-5.4.

5.1 Reduction of the classification problem of $\mathscr{C}(k)$

Let $k\subset\ell$ be a quadratic extension in characteristic not 2. On the set $C_{\ell}$ of all $\ell$-admissible triples we introduce a binary relation $\sim$, defined by $\underline{c}\sim\underline{d}$ if and only if $\ell^{\ast}(\underline{c},\underline{d})\not=\emptyset$. It is in fact an equivalence relation, called $\ell^{\ast}$-equivalence on $C_{\ell}$. Its equivalence classes refine the partition $C_{\ell}=C_{\ell}^{\mathscr{N}}\sqcup C_{\ell}^{\mathscr{S}}\sqcup C_{\ell}^{\mathscr{K}}$, introduced in Subsection 3.2. The set of all $\ell^{\ast}$-equivalence classes of $C_{\ell},C_{\ell}^{\mathscr{N}},C_{\ell}^{\mathscr{S}}$ and $C_{\ell}^{\mathscr{K}}$ respectively will be denoted by $C_{\ell}/\ell^{\ast},C_{\ell}^{\mathscr{N}}/\ell^{\ast},C_{\ell}^{\mathscr{S}}/\ell^{\ast}$ and $C_{\ell}^{\mathscr{K}}/\ell^{\ast}$.

We proceed with the following improvement of Corollary 3.2.

Corollary 5.1**.**

If $\underline{c}\in C_{\ell}$, then $A(\ell,\underline{c})\in\mathscr{C}_{\ell}(k)$.

Proof.

If $\underline{c}\in C_{\ell}$, then $A(\ell,\underline{c})\in\mathscr{C}(k)$, by Corollary 3.2. Proposition 3.1 (ii)-(iii) shows that the subfield $1_{A}\ell\subset A(\ell,\underline{c})$ is Kleinian and is isomorphic to $\ell$. So $A(\ell,\underline{c})\in\mathscr{C}_{\ell}(k)$. ∎

It is thus justified to view the map $\mathscr{F}_{\ell}:C_{\ell}\to\mathscr{C}$, introduced subsequent to Corollary 3.2, as a map

[TABLE]

where $\mathscr{C}_{\ell}=\mathscr{C}_{\ell}(k)$. It induces the maps $\mathscr{F}_{\ell}:C_{\ell}^{\mathscr{N}}\to\mathscr{N}_{\ell},\mathscr{F}_{\ell}:C_{\ell}^{\mathscr{S}}\to\mathscr{S}_{\ell}$ and $\mathscr{F}_{\ell}:C_{\ell}^{\mathscr{K}}\to\mathscr{K}_{\ell}$, where $\mathscr{N}_{\ell}=\mathscr{N}_{\ell}(k),\mathscr{S}_{\ell}=\mathscr{S}_{\ell}(k)$ and $\mathscr{K}_{\ell}=\mathscr{K}_{\ell}(k)$.

Theorem 5.2**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, and for all subsets $T_{\ell},T_{\ell}^{\mathscr{N}},T_{\ell}^{\mathscr{S}}$ and $T_{\ell}^{\mathscr{K}}$ of $C_{\ell},C_{\ell}^{\mathscr{N}},C_{\ell}^{\mathscr{S}}$ and $C_{\ell}^{\mathscr{K}}$ respectively, the following statements hold true.

(i) $T_{\ell}$ is a transversal for $C_{\ell}/\ell^{\ast}$ if and only if $\mathscr{F}_{\ell}(T_{\ell})$ classifies $\mathscr{C}_{\ell}$.

(ii) $T_{\ell}^{\mathscr{N}}$ is a transversal for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$ if and only if $\mathscr{F}_{\ell}(T_{\ell}^{\mathscr{N}})$ classifies $\mathscr{N}_{\ell}$.

(iii) $T_{\ell}^{\mathscr{S}}$ is a transversal for $C_{\ell}^{\mathscr{S}}/\ell^{\ast}$ if and only if $\mathscr{F}_{\ell}(T_{\ell}^{\mathscr{S}})$ classifies $\mathscr{S}_{\ell}$.

(iv) $T_{\ell}^{\mathscr{K}}$ is a transversal for $C_{\ell}^{\mathscr{K}}/\ell^{\ast}$ if and only if $\mathscr{F}_{\ell}(T_{\ell}^{\mathscr{K}})$ classifies $\mathscr{K}_{\ell}$.*

Proof.

(i) Let $k\subset\ell$ and $T_{\ell}\subset C_{\ell}$ be given. Because the equivalences

[TABLE]

hold for all $\underline{c},\underline{d}\in T_{\ell}$ (Theorem 4.7), the subset $T_{\ell}\subset C_{\ell}$ is irredundant regarding $\ell^{\ast}$-equivalence if and only if the subset $\mathscr{F}_{\ell}(T_{\ell})\subset\mathscr{C}_{\ell}$ is irredundant regarding isomorphism.

Suppose that $T_{\ell}$ exhausts $C_{\ell}/\ell^{\ast}$, and let $A\in\mathscr{C}_{\ell}$. Then $A$ has a Kleinian subfield $n$, such that $\ell\ \tilde{\to}\ n$. Theorem 4.5 (ii) guarantees the existence of an $n$-admissible triple $\underline{c}\in C_{n}$, such that $A(n,\underline{c})\ \tilde{\to}\ A$. Moreover, $\ell\ \tilde{\to}\ n$ implies that $\underline{c}\in C_{\ell}$ and $A(\ell,\underline{c})\ \tilde{\to}\ A(n,\underline{c})$. By hypothesis on $T_{\ell}$, the triple $\underline{c}\in C_{\ell}$ is $\ell^{\ast}$-equivalent to some $\underline{d}\in T_{\ell}$. Applying Theorem 4.7, we arrive at the chain of isomorphisms $\mathscr{F}_{\ell}(\underline{d})=A(\ell,\underline{d})\ \tilde{\to}\ A(\ell,\underline{c})\ \tilde{\to}\ A(n,\underline{c})\ \tilde{\to}\ A$, which proves that $\mathscr{F}_{\ell}(T_{\ell})$ exhausts $\mathscr{C}_{\ell}/\simeq$.

Conversely, suppose that $\mathscr{F}_{\ell}(T_{\ell})$ exhausts $\mathscr{C}_{\ell}/\simeq$, and let $\underline{c}\in C_{\ell}$. Then $\mathscr{F}_{\ell}(\underline{c})\in\mathscr{C}_{\ell}$ (Corollary 5.1). By hypothesis on $T_{\ell}$, the division algebra $\mathscr{F}_{\ell}(\underline{c})$ is isomorphic to $\mathscr{F}_{\ell}(\underline{d})$, for some $\underline{d}\in T_{\ell}$. With Theorem 4.7 we deduce the $\ell^{\ast}$-equivalence $\underline{c}\sim\underline{d}$, which proves that $T_{\ell}$ exhausts $C_{\ell}/\ell^{\ast}$.

(ii)-(iv) These statements are proved analogously. ∎

Transversals for $C_{\ell}^{\mathscr{S}}/\ell^{\ast}$ and $C_{\ell}^{\mathscr{K}}/\ell^{\ast}$ can be found as follows.

Proposition 5.3**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, the following holds true.

(i) $C_{\ell}^{\mathscr{S}}=\{(1,0,c_{3})\ |\ c_{3}\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})\}$.

(ii) A subset $S_{\ell}\subset k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$ is a transversal for $(k^{\ast}/{\rm im}(n_{\ell/k}^{\ast}))^{\circ}$ if and only if $\hat{S}_{\ell}=\{(1,0,c_{3})\ |\ c_{3}\in S_{\ell}\}$ is a transversal for $C_{\ell}^{\mathscr{S}}/\ell^{\ast}$.

(iii) $C_{\ell}^{\mathscr{K}}=\{(0,c_{2},0)\ |\ c_{2}\in k^{\ast}\setminus\ell_{sq}^{\ast}\}$.

(iv) A subset $K_{\ell}\subset k^{\ast}\setminus\ell_{sq}^{\ast}$ is a transversal for $(k^{\ast}/(\ell_{sq}^{\ast}\cap k^{\ast}))^{\circ}$ if and only if $\hat{K}_{\ell}=\{(0,c_{2},0)\ |\ c_{2}\in K_{\ell}\}$ is a transversal for $C_{\ell}^{\mathscr{K}}/\ell^{\ast}$.*

Proof.

(i) By Proposition 3.3 (ii), $C_{\ell}^{\mathscr{S}}$ consists of all triples $\underline{c}=(1,0,c_{3})\in k^{3}$ such that the associated function $q_{\underline{c}}:\ell^{2}\to\ell,\ q_{\underline{c}}(x,y)=x\overline{x}-c_{3}y\overline{y}$ is anisotropic. This occurs if and only if $c_{3}\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$.

(ii) Let $S_{\ell}\subset k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$ be given. Then, for all $c,d\in S_{\ell}$, the coset identity $c({\rm im}(n_{\ell/k}^{\ast}))=d({\rm im}(n_{\ell/k}^{\ast}))$ holds if and only if $(1,0,c)$ and $(1,0,d)$ are $\ell^{\ast}$-equivalent. Together with (i), this establishes the claimed equivalence.

(iii) By Proposition 3.3 (iii), $C_{\ell}^{\mathscr{K}}$ consists of all triples $\underline{c}=(0,c_{2},0)\in k^{3}$ such that the associated function $q_{\underline{c}}:\ell^{2}\to\ell,\ q_{\underline{c}}(x,y)=x^{2}-c_{2}y^{2}$ is anisotropic. This occurs if and only if $c_{2}\in k^{\ast}\setminus\ell_{sq}^{\ast}$.

(iv) Let $K_{\ell}\subset k^{\ast}\setminus\ell_{sq}^{\ast}$ be given. Then, for all $c,d\in K_{\ell}$, the coset identity$c(\ell_{sq}^{\ast}\cap k^{\ast})=d(\ell_{sq}^{\ast}\cap k^{\ast})$ holds if and only if $(0,c,0)$ and $(0,d,0)$ are $\ell^{\ast}$-equivalent. Together with (iii), this establishes the claimed equivalence. ∎

Theorem 5.2 (iii)-(iv) and Proposition 5.3 have the following immediate consequence.

Corollary 5.4**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, the following holds true.

(i) A subset $S_{\ell}\subset k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$ is a transversal for $(k^{\ast}/{\rm im}(n_{\ell/k}^{\ast}))^{\circ}$ if and only if the associated set of central skew fields $\{A(\ell,(1,0,c_{3}))\ |\ c_{3}\in S_{\ell}\}$ classifies $\mathscr{S}_{\ell}$.

(ii) A subset $K_{\ell}\subset k^{\ast}\setminus\ell_{sq}^{\ast}$ is a transversal for $(k^{\ast}/(\ell_{sq}^{\ast}\cap k^{\ast}))^{\circ}$ if and only if the associated set of fields $\{A(\ell,(0,c_{2},0))\ |\ c_{2}\in K_{\ell}\}$ classifies $\mathscr{K}_{\ell}$.*

Let us summarize our present insight into the classification problem of $\mathscr{C}(k)$, where $k$ is any field of characteristic not 2. With support of Propositions 4.9 and 5.3 we may assume that a transversal $\mathscr{L}$ for $\mathscr{Q}(k)/\simeq$ is known and that, for each $\ell\in\mathscr{L}$, transversals $\hat{S}_{\ell}$ for $C_{\ell}^{\mathscr{S}}/\ell^{\ast}$ and $\hat{K}_{\ell}$ for $C_{\ell}^{\mathscr{K}}/\ell^{\ast}$ are known. Then $\bigcup_{\ell\in\mathscr{L}}\mathscr{F}_{\ell}(\hat{S}_{\ell})$ exhausts $\mathscr{S}(k)/\simeq$ and $\bigcup_{\ell\in\mathscr{L}}\mathscr{F}_{\ell}(\hat{K}_{\ell})$ exhausts $\mathscr{K}(k)/\simeq$, by Proposition 4.8 and Theorem 5.2 (iii)-(iv). In order to accomplish a classification of $\mathscr{C}(k)$ on this basis, it suffices in view of Propositions 2.2 and 4.8 and Theorem 5.2 (ii), to solve the following four remaining problems.

Problem 1. For each $\ell\in\mathscr{L}$, display the subset $C_{\ell}^{\mathscr{N}}\subset k^{3}$ explicitly.

Problem 2. For each $\ell\in\mathscr{L}$, find a transversal $T_{\ell}^{\mathscr{N}}$ for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$.

Problem 3. Find a subset $T^{\mathscr{S}}\subset\bigsqcup_{\ell\in\mathscr{L}}\left(\{\ell\}\times\hat{S}_{\ell}\right)$ such that the associated subset $\left\{A(\ell,\underline{c})\ |\ (\ell,\underline{c})\in T^{\mathscr{S}}\right\}\subset\bigcup_{\ell\in\mathscr{L}}\mathscr{F}_{\ell}(\hat{S}_{\ell})$ classifies $\mathscr{S}(k)$.

Problem 4. Find a subset $T^{\mathscr{K}}\subset\bigsqcup_{\ell\in\mathscr{L}}\left(\{\ell\}\times\hat{K}_{\ell}\right)$ such that the associated subset $\left\{A(\ell,\underline{c})\ |\ (\ell,\underline{c})\in T^{\mathscr{K}}\right\}\subset\bigcup_{\ell\in\mathscr{L}}\mathscr{F}_{\ell}(\hat{K}_{\ell})$ classifies $\mathscr{K}(k)$.

The complexity of Problems 1-4 depends heavily on the nature of the ground field $k$. It depends, in particular, on the cardinality $|\mathscr{Q}(k)/\simeq|=|k^{\ast}/k_{sq}^{\ast}|-1$ (Proposition 4.9 (v)).

The trivial case occurs when $\mathscr{Q}(k)=\emptyset$, or equivalently, when $k^{\ast}=k_{sq}^{\ast}$. Then Problems 1-4 are empty, and so is $\mathscr{C}(k)$.

The simplest non-trivial case occurs when $|\mathscr{Q}(k)/\simeq|=1$, or equivalently, when $|k^{\ast}/k_{sq}^{\ast}|=2$ . We express this situation by saying that $k$ has unique quadratic extension class. For such fields $k$, Problems 1-4 admit a further reduction, expounded in Subsection 5.2 below. On that basis, we achieve in Subsections 5.3 and 5.4 explicit classifications of $\mathscr{C}(k)$ for two types of ground fields $k$ having unique quadratic extension class, namely for all ordered fields $k$ in which every positive element is a square, and for all finite fields $k$ of odd order, respectively.

In the arithmetic case $k={Q}$ we have that $|\mathscr{Q}({Q})/\simeq|=\aleph_{0}$, and complete solutions to Problems 1-4 seem not to be known at present. Interesting partial solutions, with a distinct flavour of classic number theory, have recently been achieved by G. Hammarhjelm [9].

5.2 Fields with unique quadratic extension class

In this subsection, we study the decomposition $\mathscr{C}(k)=\mathscr{N}(k)\amalg\mathscr{S}(k)\amalg\mathscr{K}(k)$ (Proposition 2.2) in case $k$ has characteristic not 2 and unique quadratic extension class. It turns out that, for all such fields $k$, the block $\mathscr{K}(k)$ is empty, and the block $\mathscr{S}(k)$ consists of at most one isoclass (Proposition 5.7). As a consequence, the problem of classifying $\mathscr{C}(k)$ is reduced to a simplified version of the above Problems 1-4, formulated below as Problems $1^{\prime}$-$3^{\prime}$. We begin with two preparatory lemmas.

Lemma 5.5**.**

*Let $k\subset\ell$ be a quadratic extension in characteristic not 2. Choose $i\in{\rm Im}(\ell)\setminus\{0\}$, and set $i^{2}=-t$. Then the following holds true.

(i) ${\rm im}(n_{\ell/k}^{\ast})=\{x^{2}+ty^{2}\ |\ (x,y)\in k^{2}\setminus\{(0,0)\}\}$.

(ii) $\ell_{sq}^{\ast}\cap k^{\ast}=k_{sq}^{\ast}\sqcup-tk_{sq}^{\ast}$.

(iii) $k_{sq}^{\ast}<{\rm im}(n_{\ell/k}^{\ast})<k^{\ast}$.

(iv) $k_{sq}^{\ast}<\ell_{sq}^{\ast}\cap k^{\ast}<k^{\ast}$ and $\left|(\ell_{sq}^{\ast}\cap k^{\ast})/k_{sq}^{\ast}\right|=2$.*

Proof.

For every $a\in\ell^{\ast}$ there is a unique pair $(x,y)\in k^{2}\setminus\{(0,0)\}$, such that $a=x+iy$. Now $\overline{a}=x-iy$ implies $a\overline{a}=x^{2}+ty^{2}$, which proves (i). Also, $a^{2}\in k^{\ast}$ if and only if $xy=0$, and in that case $a^{2}=x^{2}$ or $a^{2}=-ty^{2}$. Since $-t\not\in k_{sq}^{\ast}$, this proves (ii). Setting $y=0$ in (i) one obtains (iii), while (ii) implies (iv). ∎

Lemma 5.6**.**

*Let $k$ be a field of characteristic not 2 that has unique quadratic extension class, represented by $\ell$. Then the following holds true.

(i) If $n_{\ell/k}^{\ast}:\ell^{\ast}\to k^{\ast}$ is not surjective, then ${\rm im}(n_{\ell/k}^{\ast})=k_{sq}^{\ast}$.

(ii) $\ell_{sq}^{\ast}\cap k^{\ast}=k^{\ast}$.*

Proof.

As $|k^{\ast}/k_{sq}^{\ast}|=2$ holds by hypothesis, (i) and (ii) follow directly from Lemma 5.5 (iii) and (iv). ∎

Proposition 5.7**.**

*Let $k$ be a field of characteristic not 2 that has unique quadratic extension class, represented by $\ell$. Then the following holds true.

(i)  $\mathscr{K}(k)=\emptyset$.

(ii) If $n_{\ell/k}^{\ast}$ is not surjective, then $\mathscr{S}(k)$ consists of precisely one isoclass, represented by $A(\ell,(1,0,c_{3}))$ for any $c_{3}\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$.

(iii) If $n_{\ell/k}^{\ast}$ is surjective, then $\mathscr{S}(k)=\emptyset$. Moreover,*

[TABLE]

Proof.

(i) Lemma 5.6 (ii) asserts that $(k^{\ast}/(\ell_{sq}^{\ast}\cap k^{\ast}))^{\circ}=\emptyset$. We conclude with Proposition 4.8 and Corollary 5.4 (ii) that $\mathscr{K}(k)=\mathscr{K}_{\ell}(k)=\emptyset$.

(ii) Lemma 5.6 (i) implies that $|(k^{\ast}/{\rm im}(n_{\ell/k}^{\ast}))^{\circ}|=|(k^{\ast}/k_{sq}^{\ast})^{\circ}|=1$. Hence every $c_{3}\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$ provides a transversal $S_{\ell}=\{c_{3}\}$ for $(k^{\ast}/{\rm im}(n_{\ell/k}^{\ast}))^{\circ}$. We conclude with Corollary 5.4 (i) and Proposition 4.8 that $\{A(\ell,(1,0,c_{3}))\}$ classifies $\mathscr{S}_{\ell}(k)=\mathscr{S}(k)$.

(iii) By hypothesis, $(k^{\ast}/{\rm im}(n_{\ell/k}^{\ast}))^{\circ}=\emptyset$. We conclude with Proposition 4.8 and Corollary 5.4 (i) that $\mathscr{S}(k)=\mathscr{S}_{\ell}(k)=\emptyset$. With (i) and Proposition 2.2, the identities $\mathscr{C}(k)=\mathscr{N}(k)$ and $C_{\ell}=C_{\ell}^{\mathscr{N}}$ follow. Assume that $c_{2}=0$ for some $\underline{c}\in C_{\ell}$. Then $\underline{c}=(c_{1},0,c_{3})\in k^{3}$, such that the associated function

[TABLE]

is anisotropic. So $c_{3}=-q_{\underline{c}}(0,1)\in k^{\ast}$. As $n_{\ell/k}^{\ast}$ is surjective, there exists an element $\eta\in\ell^{\ast}$ such that $\eta\overline{\eta}=c_{3}^{-1}$. Then $q_{\underline{c}}(1,\eta)=0$, which contradicts the anisotropy of $q_{\underline{c}}$. Thus $c_{2}\not=0$ for all $\underline{c}\in C_{\ell}$. ∎

Proposition 5.7, combined with Proposition 4.8 and Theorem 5.2 (ii), yields the following corollary.

Corollary 5.8**.**

*Let $k$ be a field of characteristic not 2 that has unique quadratic extension class, represented by $\ell$, and let $T_{\ell}^{\mathscr{N}}$ be a transversal for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$.

(i) If $n_{\ell/k}^{\ast}$ is not surjective and $c_{3}\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$, then*

[TABLE]

(ii) If $n_{\ell/k}^{\ast}$ is surjective, then

[TABLE]

For every field $k$ of characteristic not 2 that has unique quadratic extension class, represented by $\ell$, Corollary 5.8 reduces the problem of classifying $\mathscr{C}(k)$ to the following three problems.

Problem 1′.** Display the subset $C_{\ell}^{\mathscr{N}}\subset k^{3}$ explicitly.

Problem 2′.** Find a transversal $T_{\ell}^{\mathscr{N}}$ for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$.

Problem 3′.** Determine whether the group morphism $n_{\ell/k}^{\ast}:\ell^{\ast}\to k^{\ast}$ is surjective or not. If $n_{\ell/k}^{\ast}$ is not surjective, choose an element $c_{3}\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$.

The preceding results reveal a dichotomy of types, regarding fields $k$ that have characteristic not 2 and unique quadratic extension class. The first type is characterized by non-surjectivity of the norm morphism $n_{\ell/k}^{\ast}$, or equivalently, by non-emptiness of the groupoid $\mathscr{S}(k)$, whereas the second type is characterized by surjectivity of $n_{\ell/k}^{\ast}$, or equivalently, by emptiness of $\mathscr{S}(k)$.

In the remainder of the present section we classify the groupoid $\mathscr{C}(k)$ for two classes of fields $k$, having characteristic not 2 and unique quadratic extension class. The first class is formed by all square-ordered fields (Subsection 5.3), and all of these are of the first type. The second class is formed by all finite fields of odd order (Subsection 5.4), and all of these are of the second type.

5.3 Classification of $\mathscr{C}(k)$ for square-ordered fields $k$

Recall our notation $k_{sq}=\{x^{2}\ |\ x\in k\}$, valid for every field $k$. If $k$ is an ordered field, then $k$ has characteristic [math], the set $k_{\geq 0}=\{x\in k\ |\ x\geq 0\}$ of all non-negative elements in $k$ is defined, and the inclusion $k_{sq}\subset k_{\geq 0}$ holds. By a square-ordered field we mean an ordered field $k$ such that $k_{sq}=k_{\geq 0}$. The real number field ${R}$ and, more generally, all real closed fields $R$ are examples of square-ordered fields [8, VI, Theorem 2.3].

In this subsection, we classify $\mathscr{C}(k)$ for all square-ordered fields $k$ (Corollary 5.10), by way of establishing that $k$ has unique quadratic extension class, solving Problems $1^{\prime}$-$3^{\prime}$, and applying Corollary 5.8.

For technical reasons we introduce the partition $C_{\ell}^{\mathscr{N}}=C_{\ell}^{\mathscr{N}_{0}}\sqcup C_{\ell}^{\mathscr{N}_{1}}$, where $C_{\ell}^{\mathscr{N}_{0}}=\{\underline{c}\in C_{\ell}^{\mathscr{N}}\ |\ c_{2}=0\}$ and $C_{\ell}^{\mathscr{N}_{1}}=\{\underline{c}\in C_{\ell}^{\mathscr{N}}\ |\ c_{2}\not=0\}$. The set of all $\ell^{\ast}$-equivalence classes of $C_{\ell}^{\mathscr{N}_{0}}$ and $C_{\ell}^{\mathscr{N}_{1}}$ is denoted by $C_{\ell}^{\mathscr{N}_{0}}/\ell^{\ast}$ and $C_{\ell}^{\mathscr{N}_{1}}/\ell^{\ast}$, respectively.

Proposition 5.9**.**

*Let $k$ be a square-ordered field.

(i) Then $k$ has unique quadratic extension class, represented by $\ell=k\left(\sqrt{-1}~{}\right)$.

(ii) The map $n_{\ell/k}^{\ast}:\ell^{\ast}\to k^{\ast}$ is not surjective, and $-1\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$.

(iii) The subset $C_{\ell}\subset k^{3}$ admits the explicit display*

[TABLE]

(iv) The subsets $C_{\ell}^{\mathscr{N}_{0}}\subset C_{\ell}^{\mathscr{N}}$ and $C_{\ell}^{\mathscr{N}_{1}}\subset C_{\ell}^{\mathscr{N}}$ admit the explicit displays

[TABLE]

(v) The subsets

[TABLE]

*are transversals for $C_{\ell}^{\mathscr{N}_{0}}/\ell^{\ast}$ and $C_{\ell}^{\mathscr{N}_{1}}/\ell^{\ast}$, respectively.

(vi) The set $T_{\ell}^{\mathscr{N}}=T_{\ell}^{\mathscr{N}_{0}}\sqcup T_{\ell}^{\mathscr{N}_{1}}$ is a transversal for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$.*

Proof.

(i) Being a square-ordered field, $k$ satisfies the identities $k_{sq}^{\ast}=k_{>0}$ and $k^{\ast}\setminus k_{sq}^{\ast}=k_{<0}$, whence $-1\in k^{\ast}\setminus k_{sq}^{\ast}$ and $|k^{\ast}/k_{sq}^{\ast}|=2$. Statement (i) now follows from Proposition 4.9, (i) and (v).

(ii) General elements $a\in\ell^{\ast}$ are of the form $a=x+iy$, where $(x,y)\in k^{2}\setminus\{(0,0)\}$ and $i=\sqrt{-1}$. Thus $n_{\ell/k}^{\ast}(a)=a\overline{a}=x^{2}+y^{2}\in k_{>0}$ shows that $n_{\ell/k}^{\ast}$ is not surjective, and $-1\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$.

(iii) Set $C=\left\{\underline{c}\in k^{3}\ \left|\ c_{1}>\frac{1}{2}\ \wedge\ c_{3}<c_{2}<-c_{3}\right.\right\}$. We aim to show that $C_{\ell}=C$. We refer to Subsections 2.2 and 3.1.

Let $\underline{c}\in C_{\ell}$. For all $(x,y)\in k^{2}$ we have $(\overline{x},\overline{y})=(x,y)$, and hence

[TABLE]

If $c_{2}+c_{3}\geq 0$, then $c_{2}+c_{3}=\xi^{2}$ for some $\xi\in k$, whence $q_{\underline{c}}(\xi,1)=0$, contradicting $\underline{c}\in C_{\ell}$. So $c_{2}+c_{3}<0$.

For all $(x,y)\in k\times{\rm Im}(\ell)$ we have $(\overline{x},\overline{y})=(x,-y)$, and hence

[TABLE]

If $-c_{2}+c_{3}\geq 0$, then $-c_{2}+c_{3}=\xi^{2}$ for some $\xi\in k$, whence $q_{\underline{c}}(\xi,i)=0$, contradicting $\underline{c}\in C_{\ell}$. So $-c_{2}+c_{3}<0$.

For all $(x,y)\in{\rm Im}(\ell)\times k$ we have $(\overline{x},\overline{y})=(-x,y)$, and hence

[TABLE]

If $1-2c_{1}\geq 0$, then $\frac{1-2c_{1}}{-(c_{2}+c_{3})}\geq 0$ and hence $\frac{1-2c_{1}}{-(c_{2}+c_{3})}=\xi^{2}$ for some $\xi\in k$, whence $q_{\underline{c}}(i,\xi)=0$, contradicting $\underline{c}\in C_{\ell}$. So $1-2c_{1}<0$. Summarizing, we have shown that $\underline{c}\in C$.

Conversely, each $\underline{c}\in C$ determines a pair of quadratic forms $q_{\underline{c}}^{\mu}:\ell^{2}\to k,$ $\mu\in\underline{2}$, defined by

[TABLE]

Writing $(x,y)=(x_{1}+ix_{2},y_{1}+iy_{2})$ with $(x_{1},x_{2},y_{1},y_{2})\in k^{4}$, we obtain the identity

[TABLE]

where all of the coefficients $1,-(1-2c_{1}),-(c_{2}+c_{3})$ and $c_{2}-c_{3}$ are positive elements in $k$, by hypothesis. Thus, if $q_{\underline{c}}(x,y)=0$, then $q_{\underline{c}}^{1}(x,y)=0$ implies $(x,y)=(0,0)$. Accordingly, $\underline{c}\in C_{\ell}$.

(iv) This statement is a direct consequence of (iii), Proposition 3.3 (i), and the definition of the partition $C_{\ell}^{\mathscr{N}}=C_{\ell}^{\mathscr{N}_{0}}\sqcup C_{\ell}^{\mathscr{N}_{1}}$.

(v) In view of the definition of $\ell^{\ast}$-equivalence (Subsections 5.1 and 3.3), and because every $c_{3}<0$ can be written $c_{3}=-a\overline{a}$ for some $a\in\ell^{\ast}$, and every $c_{2}\not=0$ can be written $c_{2}=a^{2}$ for some $a\in\ell^{\ast}$, statement (v) is an easy consequence of (iv).

(vi) Since $C_{\ell}^{\mathscr{N}}/\ell^{\ast}=C_{\ell}^{\mathscr{N}_{0}}/\ell^{\ast}\sqcup C_{\ell}^{\mathscr{N}_{1}}/\ell^{\ast}$, (vi) follows directly from (v). ∎

Corollary 5.10**.**

Let $k$ be a square-ordered field, let $\ell=k\left(\sqrt{-1}~{}\right)$, and let $T_{\ell}^{\mathscr{N}}\subset C_{\ell}^{\mathscr{N}}$ be the subset displayed in Proposition 5.9 (v)-(vi). Then

[TABLE]

Proof.

Every ordered field $k$ has characteristic [math]. By Proposition 5.9, every square-ordered field $k$ has unique quadratic extension class, represented by $\ell$, the subset $T_{\ell}^{\mathscr{N}}\subset C_{\ell}^{\mathscr{N}}$ is a transversal for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$, and $-1\in k^{\ast}\setminus{\rm im}(n_{\ell/k}^{\ast})$. The classification statement is now a special case of Corollary 5.8 (i). ∎

5.4 Classification of $\mathscr{C}(k)$ for finite fields $k$ of odd order

Throughout this subsection, $k$ is a field of the second type, i.e. $k$ has characteristic not 2 and unique quadratic extension class, represented by $\ell$, such that the norm morphism $n_{\ell/k}^{\ast}:\ell^{\ast}\to k^{\ast}$ is surjective. All finite fields ${F}_{q}$ of odd order $q$ are fields of the second type.

We study the problem of classifying $\mathscr{C}(k)$, which, by Corollary 5.8 (ii), is reduced to Problems $1^{\prime}$ and $2^{\prime}$. The approach to Problems $1^{\prime}$ and $2^{\prime}$ which we present in the following discussion is a generalized and streamlined version of [2, p. 218-219], where the ground field is assumed to be ${F}_{q}$, with $q$ odd.

Let $k$ be a field of the second type, and let $k\subset\ell$ be a quadratic extension. We introduce the map

[TABLE]

and we note that $|x|=\sqrt{x}\ \overline{\sqrt{x}}=n_{\ell/k}(\sqrt{x})$ for all $x\in k$. Also, $|\cdot|$ induces a group endomorphism $|\cdot|^{\ast}:k^{\ast}\to k^{\ast}$, which is an automorphism if and only if $-1\in k_{sq}^{\ast}$. With any pair of scalars $(a,b)\in k^{2}$ we associate a function

[TABLE]

We say that a pair $(a,b)\in k^{2}$ is $\ell$-admissible if the function $h_{a,b}:\ell^{2}\to\ell$ is anisotropic. The $\ell$-admissible pairs in $k^{2}$ form a subset

[TABLE]

of $k^{2}$. Since $k$ has unique quadratic extension class, the subsets $C_{\ell}^{{\mathscr{N}}}\subset k^{3}$ and $B_{\ell}\subset k^{2}$ are in fact uniquely determined by $k$.

The following lemma reduces Problem $1^{\prime}$ to the problem of displaying the subset $B_{\ell}\subset k^{2}$ explicitly.

Lemma 5.11**.**

Let $k$ be a field of the second type, and let $\underline{c}\in k^{3}$. Then $\underline{c}\in C_{\ell}^{\mathscr{N}}$ if and only if $(1-c_{1})c_{2}\not=0$ and

[TABLE]

Proof.

Let $\underline{c}\in C_{\ell}^{\mathscr{N}}$. Then $c_{2}\not=0$ holds by Proposition 5.7 (iii). Since $n_{\ell/k}^{\ast}$ is surjective, there exists a $\xi\in\ell$ such that $\xi\overline{\xi}=c_{2}+c_{3}$. If $1-c_{1}=0$, then $q_{\underline{c}}(\xi,1)=0$ contradicts the anisotropy of $q_{\underline{c}}$. Hence $1-c_{1}\not=0$. By Lemma 5.6 (ii) there exist $\alpha,\beta\in\ell^{\ast}$ such that $(\alpha^{2},\beta^{2})=(1-c_{1},c_{2})$. Then $q_{\underline{c}}(x,y)=h_{a,b}(\alpha x,\beta y)$ holds for all $(x,y)\in\ell^{2}$. Since $q_{\underline{c}}$ is anisotropic, it follows that so is $h_{a,b}$. Thus $(a,b)\in B_{\ell}$.

Conversely, let $\underline{c}\in k^{3}$ be such that $(1-c_{1})c_{2}\not=0$ and $(a,b)\in B_{\ell}$. Reversing the last argument in the above reasoning we find that $q_{\underline{c}}$ is anisotropic, i.e. $\underline{c}\in C_{\ell}$, and $C_{\ell}=C_{\ell}^{\mathscr{N}}$ holds by Proposition 5.7 (iii). ∎

The next proposition aims at an explicit display of the subset $B_{\ell}\subset k^{2}$. The proof presented here generalizes a line of arguments, contained in [2, Proof of Proposition 1], from finite fields ${F}_{q}$ to fields $k$ of the second type.

Proposition 5.12**.**

*Let $k$ be a field of the second type, with associated subsets $M_{1}=\{m\in k\ |\ m^{2}-1\in k_{sq}\}$ and $M_{2}=\{m\in k\ |\ m^{2}-1\not\in k_{sq}^{\ast}\}$ of $k$. Then, for all $(a,b)\in k^{2}$, the following holds true.

(i) If $b=-a$, then $(a,b)\not\in B_{\ell}$.

(ii) If $b=a$, then $(a,b)\in B_{\ell}$ if and only if $1-a^{2}\not\in k_{sq}$.

(iii) If $b^{2}\not=a^{2}$, then $(a,b)\in B_{\ell}$ if and only if*

[TABLE]

Proof.

(i) The identity $h_{a,-a}(1,1)=0$ shows that $h_{a,-a}$ is isotropic.

For the remainder of the proof, let $(a,b)\in k^{2}$ such that $b\not=-a$. For all $(x,y)\in\ell^{2}$ such that $x^{2}-y^{2}\in k^{\ast}$, we set $t=x^{2}-y^{2}$ and $z=x+y$. Note that $(t,z)\in k^{\ast}\times\ell^{\ast}$. With this notation, the following statement (*) is proved in [2, Proof of Proposition 1].

(*) Let $(a,b)\in k^{2}$ with $b\not=-a$. Then $(a,b)\in B_{\ell}$ if and only if

[TABLE]

for all $(x,y)\in\ell^{2}$ with $x^{2}-y^{2}\in k^{\ast}$.

For every pair $(t,z)\in k^{\ast}\times\ell^{\ast}$ there is a pair $(x,y)\in\ell^{2}$ such that $x^{2}-y^{2}=t$ and $x+y=z$. Indeed, $(x,y)=\left(\frac{z^{2}+t}{2z},\frac{z^{2}-t}{2z}\right)$ will do. For each $z\in\ell^{\ast}$, the quadratic polynomial in $t$ appearing as left hand side in (*) has all of its coefficients in $k$, and its discriminant is

[TABLE]

Accordingly, (*) is equivalent to the following statement (**).

(**) Let $(a,b)\in k^{2}$ with $b\not=-a$. Then $(a,b)\in B_{\ell}$ if and only if $\Delta(z)\not\in k_{sq}$ for all $z\in\ell^{\ast}$.

We proceed to prove (ii) and (iii) by use of (**).

(ii) Let $(a,b)=(a,a)$ with $a\in k^{\ast}$. Then $\Delta(z)=1-a^{2}$ for all $z\in\ell^{\ast}$. Thus, (**) asserts that $(a,a)\in B_{\ell}$ if and only if $1-a^{2}\not\in k_{sq}$.

(iii) Let $(a,b)\in k^{2}$ with $a^{2}\not=b^{2}$. Following [2], we set $s=\frac{a+b}{2}$,

[TABLE]

With this notation, $\{\Delta(z)\ |\ z\in\ell^{\ast}\}=\{n^{2}-s^{2}\ |\ n\in N\}$. Hence (**) states that $(a,b)\in B_{\ell}$ if and only if $M_{s}\cap N=\emptyset$. Moreover, $M_{s}=sM_{1}$ and $N=1+\frac{a-b}{2}M_{2}$. To justify the latter identity, we observe with Hilbert’s Theorem 90 [8, V, Lemma 7.7] that

[TABLE]

where ${S}={\rm ker}(n_{\ell/k}^{\ast})$. Choose $i\in{\rm Im}(\ell)\setminus\{0\}$. Then $i^{2}=-t\in k^{\ast}\setminus k_{sq}^{\ast}$ and ${S}=\left\{w_{1}+iw_{2}\ \left|\ w_{1},w_{2}\in k\ \wedge\ w_{1}^{2}+tw_{2}^{2}=1\right.\right\}$ and $|k^{\ast}/k_{sq}^{\ast}|=2$ implies

[TABLE]

whence the claim follows. The above identities for $M_{s}$ and $N$ show that $M_{s}\cap N=\emptyset$ if and only if $\frac{a+b}{2}m_{1}\not=1+\frac{a-b}{2}m_{2}$ for all $(m_{1},m_{2})\in M_{1}\times M_{2}$. ∎

If $k$ is a finite field of odd order, then the above results can be supplemented by a counting argument, that readily yields solutions to Problems $1^{\prime}$ and $2^{\prime}$, and thereby a classification of $\mathscr{C}(k)$. In statements (i) and (ii) of Proposition 5.13 below we recover [2, Proposition 1] and [2, Main Theorem]. The latter statement solves Problem $1^{\prime}$. The counting argument used in the proof of Proposition 5.13 (i) is reproduced from [2, Proof of Proposition 1], to keep our presentation self-contained. In contrast, the transversal $T_{\ell}^{\mathscr{N}}$ that solves Problem $2^{\prime}$ (Proposition 5.13 (iii)) and the corresponding classification of $\mathscr{C}(k)$ (Corollary 5.14) are beyond the scope of [2].

Proposition 5.13**.**

*Let $k$ be a finite field of odd order. Then

(i) $B_{\ell}=\left\{(a,b)\in k^{2}\ \left|\ a=b\ \wedge\ 1-a^{2}\not\in k_{sq}\right.\right\}$,

(ii) $C_{\ell}^{\mathscr{N}}=\left\{\left.\left(c_{1},c_{2},\frac{-c_{1}|c_{2}|}{|1-c_{1}|}\right)\in k^{3}\ \right|\ (1-c_{1})c_{2}\not=0\ \wedge\ 1-2c_{1}\not\in k_{sq}\ \right\}$, and

(iii) $T_{\ell}^{\mathscr{N}}=\left\{\left.\left(c_{1},1,\frac{-c_{1}}{|1-c_{1}|}\right)\in k^{3}\ \right|\ c_{1}\not=1\ \wedge\ 1-2c_{1}\not\in k_{sq}\ \right\}$ is a transversal for $C_{\ell}^{\mathscr{N}}/\ell^{\ast}$.*

Proof.

(i) As every finite field $k$ of odd order is of the second type, Proposition 5.12 applies. Therefore it suffices to show that, for all $(a,b)\in k^{2}$ with $a^{2}\not=b^{2}$, the identity $(a+b)m_{1}=2+(a-b)m_{2}$ has a solution $(m_{1},m_{2})\in M_{1}\times M_{2}$.

Indeed, every $(a,b)\in k^{2}$ with $a^{2}\not=b^{2}$ determines a bijective map

[TABLE]

Setting $q=|k|$, we have that $|M_{1}|=\frac{q+1}{2}$ and $|M_{2}|=\frac{q+3}{2}$, and hence

[TABLE]

Thus, there exists a pair $(m_{1},m_{2})\in M_{1}\times M_{2}$ such that $m_{1}=\varphi(m_{2})$.

(ii) Applying Lemma 5.11 and (i), we obtain for all $\underline{c}\in k^{3}$ the chain of equivalences

[TABLE]

(iii) By definition of $\ell^{\ast}$-equivalence, and because every $c_{2}\in k^{\ast}$ can be written $c_{2}=a^{2}$ for some $a\in\ell^{\ast}$ (Lemma 5.6 (ii)), statement (iii) is a straightforward consequence of (ii). ∎

Corollary 5.14**.**

Let $k$ be a finite field of odd order, let $k\subset\ell$ be a quadratic extension, and let $T_{\ell}^{\mathscr{N}}$ be the transversal displayed in Proposition 5.13 (iii). Then $\mathscr{F}_{\ell}(T_{\ell}^{\mathscr{N}})\ \mbox{classifies}\ \mathscr{C}(k)=\mathscr{N}(k)$.

Proof.

Corollary 5.8 (ii) applies, and proves the statement. ∎

6 Kleinian coverings by group actions

In this last section, we investigate the structure of the groupoid $\mathscr{C}(k)$, still assuming that ${\rm char}(k)\not=2$. In order to reach this goal, we first present the concept of a covering of a groupoid by group actions, in great generality (Subsection 6.1). Next, we deepen our knowledge of the particular groupoid $\mathscr{C}(k)$ by gaining more insight into its morphism sets (Subsection 6.2). In the final Subsection 6.3, we mould the accumulated information on $\mathscr{C}(k)$ into the concept provided in Subsection 6.1. As an application, we determine the automorphism groups of all division algebras in $\mathscr{C}(k)$.

6.1 The concept of a covering of a groupoid by group actions

The concept to be presented here refines and extends the concept of a description of a groupoid by a group action, introduced in [5, Section 5].

Every left action $G\times X\to X,\ (g,x)\mapsto gx$ of a group $G$ on a set $X$ gives rise to a small groupoid ${}_{G}X$, with object set $X$ and morphism sets

[TABLE]

We say that a groupoid $\mathscr{X}$ is a group action groupoid if $\mathscr{X}=\ _{G}X$ for some left group action $G\times X\to X$.

For any groupoid $\mathscr{Y}$, a description of $\mathscr{Y}$ by a group action (or briefly, a description of $\mathscr{Y}$) is, by definition, a quadruple $(X,G,\gamma,\mathscr{F})$, composed of a set $X$, a group $G$, a group action $\gamma:G\times X\to X$, and a functor $\mathscr{F}:\ _{G}X\to\mathscr{Y}$ which is dense, faithful, and quasi-full in the sense that

[TABLE]

holds for all $(x,y)\in X\times X$. A description $(X,G,\gamma,\mathscr{F})$ of a groupoid $\mathscr{Y}$ is said to be full if the functor $\mathscr{F}$ is full, i.e. if $\mathscr{F}$ is an equivalence of categories.111Full descriptions in this sense were simply called descriptions in [5, Section 5], while descriptions in our present sense were not considered in [5, Section 5].

If $(X,G,\gamma,\mathscr{F})$ is a description of a groupoid $\mathscr{Y}$ and $T\subset X$ is a transversal for the orbit set $X/\gamma$ of the group action $\gamma$, then the subset $\mathscr{F}(T)\subset\mathscr{Y}$ is a transversal for the set $\mathscr{Y}/\simeq$ of all isoclasses of $\mathscr{Y}$. Accordingly, $\mathscr{Y}$ is svelte (Subsection 2.1), and $\mathscr{F}(T)$ classifies $\mathscr{Y}$.

Now, let $(\mathscr{Y}_{i})_{i\in I}$ be a family of full subgroupoids of a groupoid $\mathscr{Y}$, and suppose that a description $(X_{i},G_{i},\gamma_{i},\mathscr{F}_{i})$ of $\mathscr{Y}_{i}$ is given for each $i\in I$. We say that the family of descriptions $(X_{i},G_{i},\gamma_{i},\mathscr{F}_{i})_{i\in I}$ is a covering of $\mathscr{Y}$ by group actions if the identity of object classes $\mathscr{Y}=\bigcup_{i\in I}\mathscr{Y}_{i}$ holds true.

6.2 Morphism sets in $\mathscr{C}(k)$

Throughout this subsection, $k\subset\ell$ is a quadratic extension in characteristic not 2, and $\underline{c},\underline{d}\in C_{\ell}$ are $\ell$-admissible triples. Our study of the morphism sets in $\mathscr{C}(k)$ is based on a characterization of the special morphisms $\varphi_{a}$ and $\psi_{a}$ from $A(\ell,\underline{c})$ to $A(\ell,\underline{d})$, constructed in Subsection 3.3, among all morphisms from $A(\ell,\underline{c})$ to $A(\ell,\underline{d})$. Regarding the notation $1_{A}\ell$, see Proposition 3.1 (ii).

Proposition 6.1**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, and for all $\underline{c},\underline{d}\in C_{\ell}$, the following holds true.

(i) $\left\{\varphi_{a},\psi_{a}\left|\ a\in\ell^{\ast}(\underline{c},\underline{d})\right.\right\}=\left\{\mu\in{\rm Mor}_{\mathscr{C}(k)}(A(\ell,\underline{c}),A(\ell,\underline{d}))\left|\ \mu(1_{A}\ell)=1_{A}\ell\right.\right\}$.

(ii) If $(\underline{c},\underline{d})\in\left(C_{\ell}^{\mathscr{N}}\times C_{\ell}^{\mathscr{N}}\right)\sqcup\left(C_{\ell}^{\mathscr{K}}\times C_{\ell}^{\mathscr{K}}\right)$, then $\mu(1_{A}\ell)=1_{A}\ell$ holds for all morphisms $\mu\in{\rm Mor}_{\mathscr{C}(k)}(A(\ell,\underline{c}),A(\ell,\underline{d}))$.*

Proof.

(i) The inclusion “$\subset$” holds by definition of $\varphi_{a}$ and $\psi_{a}$. Conversely, let $\mu:A(\ell,\underline{c})\to A(\ell,\underline{d})$ be a morphism in $\mathscr{C}(k)$, such that $\mu(1_{A}\ell)=1_{A}\ell$. Then $\mu$ induces a Galois automorphism $\mu_{\ell}\in{\rm Gal}(\ell/k)$, and ${\rm Gal}(\ell/k)=\langle\sigma\rangle$ has order 2. If $\mu_{\ell}={I}_{\ell}$, then the arguments contained in the proof of Theorem 4.7, Step 3 and Step 4, show that $\mu=\varphi_{a}$ for some $a\in\ell^{\ast}(\underline{c},\underline{d})$. If $\mu_{\ell}=\sigma$, then the same reasoning applies to $\mu\psi_{1}$, where $\psi_{1}\in{\rm Aut}(A(\ell,\underline{c}))$, because $(\mu\psi_{1})_{\ell}={I}_{\ell}$. Thus $\mu\psi_{1}=\varphi_{a}$ and hence $\mu=\varphi_{a}\psi_{1}=\psi_{a}$, for some $a\in\ell^{\ast}(\underline{c},\underline{d})$.

(ii) See proof of Theorem 4.7, Step 1. ∎

Corollary 6.2**.**

If $(\underline{c},\underline{d})\in\left(C_{\ell}^{\mathscr{N}}\times C_{\ell}^{\mathscr{N}}\right)\sqcup\left(C_{\ell}^{\mathscr{K}}\times C_{\ell}^{\mathscr{K}}\right)$, then

[TABLE]

Recall from Proposition 3.1 (iii)-(iv) and Lemma 3.4 (iii) that, for all $\underline{c}\in C_{\ell}$, the division algebra $A(\ell,\underline{c})$ admits the Kleinian pair $(\alpha,\beta)=(\varphi_{-1},\psi_{1})$.

Proposition 6.3**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, and for all $\underline{c}\in C_{\ell}$, the following holds true.

(i) If $c_{2}=0$, then the subgroup $\langle\varphi_{-1},\psi_{1}\rangle<{\rm Aut}(A(\ell,\underline{c}))$ is proper.

(ii) If $c_{2}\not=0$, then $\langle\varphi_{-1},\psi_{1}\rangle={\rm Aut}(A(\ell,\underline{c}))$.*

Proof.

For definition of the subgroup ${S}<\ell^{\ast}$, see Subsection 2.2.

(i) If $c_{2}=0$, then $c_{3}\not=0$, whence $\ell^{\ast}(\underline{c},\underline{c})={S}$. Because $\frac{1+i}{1-i}\in{S}\setminus\{\pm 1\}$ for all $i\in{\rm Im}(\ell)\setminus\{0\}$, the first inclusion in

[TABLE]

is proper.

(ii) If $c_{2}\not=0$, then $\ell^{\ast}(\underline{c},\underline{c})=\{\pm 1\}$ and $\underline{c}\in C_{\ell}^{\mathscr{N}}\sqcup C_{\ell}^{\mathscr{K}}$ (Proposition 3.3). With this information, Corollary 6.2 implies that

[TABLE]

which completes the proof. ∎

6.3 The Kleinian coverings of $\mathscr{C}(k)$ by group actions

Let $k\subset\ell$ be a quadratic extension in characteristic not 2. In Subsection 4.4 we introduced the full subgroupoid $\mathscr{C}_{\ell}=\mathscr{C}_{\ell}(k)$ of $\mathscr{C}=\mathscr{C}(k)$. Now we decompose it into two blocks $\mathscr{C}_{\ell}^{0}$ and $\mathscr{C}_{\ell}^{1}$, defined by their object classes $\mathscr{C}_{\ell}^{0}=\{A\in\mathscr{C}_{\ell}\ |\ |{\rm Aut}(A)|>4\}$ and $\mathscr{C}_{\ell}^{1}=\{A\in\mathscr{C}_{\ell}\ |\ |{\rm Aut}(A)|=4\}$. Aiming for descriptions $Q_{\ell}^{\nu}=(C_{\ell}^{\nu},G_{\ell}^{\nu},\gamma_{\ell}^{\nu},\mathscr{F}_{\ell}^{\nu})$ of $\mathscr{C}_{\ell}^{\nu},\ \nu\in\{0,1\}$, we proceed to define the data constituting the quadruples $Q_{\ell}^{0}$ and $Q_{\ell}^{1}$.

The sets $C_{\ell}^{0}=\{\underline{c}\in C_{\ell}\ |\ c_{2}=0\}$ and $C_{\ell}^{1}=\{\underline{c}\in C_{\ell}\ |\ c_{2}\not=0\}$ are sets of $\ell$-admissible triples, partitioning $C_{\ell}$.

The groups $G_{\ell}^{0}=\ell^{\ast}>\lhd\ {\rm Gal}(\ell/k)$ and $G_{\ell}^{1}={A}^{\ast}>\lhd\ {\rm Gal}(\ell/k)$ are semi-direct products of the type mentioned in Subsection 2.2.

The group actions $\gamma_{\ell}^{\nu}:G_{\ell}^{\nu}\times C_{\ell}^{\nu}\to C_{\ell}^{\nu}$ are defined uniformly, for both $\nu\in\{0,1\}$, by

[TABLE]

The functors $\mathscr{F}_{\ell}^{\nu}:\ _{G_{\ell}^{\nu}}C_{\ell}^{\nu}\ \to\ \mathscr{C}_{\ell}^{\nu}$ are defined uniformly, for both $\nu\in\{0,1\}$, by $\mathscr{F}_{\ell}^{\nu}(\underline{c})\ =\ A(\ell,\underline{c})$ on objects $\underline{c}$, and by

[TABLE]

on morphisms $((a,\sigma^{i}),\underline{c},\underline{d})$.

It is easily verified that both maps $\gamma_{\ell}^{0}$ and $\gamma_{\ell}^{1}$ indeed are group actions, and, taking Propositions 3.1 (iv) and 6.3 as well as Lemma 3.4 (ii) into account, that both families of maps $\mathscr{F}_{\ell}^{0}$ and $\mathscr{F}_{\ell}^{1}$ indeed are functors.

We also note that, for both $\nu\in\{0,1\}$, the set $C_{\ell}^{\nu}$ is a union of $\ell^{\ast}$-equivalence classes in $C_{\ell}$, and the $\ell^{\ast}$-equivalence classes of $C_{\ell}^{\nu}$ coincide with the $G_{\ell}^{\nu}$-orbits of $C_{\ell}^{\nu}$. Thus, we are about to enhance our earlier approach with categorical and functorial structure.

Proposition 6.4**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, the following holds true.

(i) $\mathscr{C}_{\ell}=\mathscr{C}_{\ell}^{0}\amalg\mathscr{C}_{\ell}^{1}$.

(ii) $Q_{\ell}^{0}=(C_{\ell}^{0},G_{\ell}^{0},\gamma_{\ell}^{0},\mathscr{F}_{\ell}^{0})$ is a description of $\mathscr{C}_{\ell}^{0}$ by a group action.

(iii) $Q_{\ell}^{1}=(C_{\ell}^{1},G_{\ell}^{1},\gamma_{\ell}^{1},\mathscr{F}_{\ell}^{1})$ is a full description of $\mathscr{C}_{\ell}^{1}$ by a group action.*

Proof.

(i) The claimed decomposition holds by definition of $\mathscr{C}_{\ell}^{0}$ and $\mathscr{C}_{\ell}^{1}$.

(ii)-(iii) Both functors $\mathscr{F}_{\ell}^{0}$ and $\mathscr{F}_{\ell}^{1}$ are dense by Theorem 4.5 (ii) and Proposition 6.3, faithful by definition, and quasi-full by Theorem 4.7. Thus$Q_{\ell}^{0}$ is a description of $\mathscr{C}_{\ell}^{0}$ by a group action, and $Q_{\ell}^{1}$ is a description of $\mathscr{C}_{\ell}^{1}$ by a group action. In order to show that the functor $\mathscr{F}_{\ell}^{1}$ even is full, let $(\underline{c},\underline{d})\in C_{\ell}^{1}\times C_{\ell}^{1}$ and a morphism $\mu:A(\ell,\underline{c})\to A(\ell,\underline{d})$ in $\mathscr{C}_{\ell}^{1}$ be given. Then $(\underline{c},\underline{d})\in\left(C_{\ell}^{\mathscr{N}}\times C_{\ell}^{\mathscr{N}}\right)\sqcup\left(C_{\ell}^{\mathscr{K}}\times C_{\ell}^{\mathscr{K}}\right)$ by Proposition 3.3, whence Corollary 6.2 yields a morphism $((a,\sigma^{i}),\underline{c},\underline{d}):\underline{c}\to\underline{d}$ in $\ {}_{G_{\ell}^{1}}C_{\ell}^{1}$, such that $\mathscr{F}_{\ell}^{1}((a,\sigma^{i}),\underline{c},\underline{d})=\mu$. ∎

At last, the Kleinian covering of $\mathscr{C}(k)$ by group actions enters the stage. Again, we write $\mathscr{C}=\mathscr{C}(k),\ \mathscr{Q}=\mathscr{Q}(k)$ etc., for brevity.

Corollary 6.5**.**

Let $k$ be a field of characteristic not 2, and let $\mathscr{L}$ be a transversal for $\mathscr{Q}/\simeq$. Then the family

[TABLE]

of descriptions $Q_{\ell}^{\nu}$ of $\mathscr{C}_{\ell}^{\nu}$ is a covering of $\mathscr{C}$ by group actions.

Proof.

For each $(\ell,\nu)\in\mathscr{L}\times\{0,1\}$, the quadruple $Q_{\ell}^{\nu}$ is a description of $\mathscr{C}_{\ell}^{\nu}$ by a group action (Proposition 6.4 (ii)-(iii)). Furthermore, the full subgroupoids $\mathscr{C}_{\ell}^{\nu}\subset\mathscr{C}$ cover $\mathscr{C}$, as the identities of object classes

[TABLE]

hold by Proposition 4.8 and Proposition 6.4 (i)). ∎

We call the family $Q$, displayed in Corollary 6.5, the Kleinian covering of $\mathscr{C}$ by group actions. It admits a noteworthy refinement, which evolves from the decomposition of $\mathscr{C}_{\ell}^{0},\ \ell\in\mathscr{L}$, into the blocks $\mathscr{C}_{\ell}^{2}=\mathscr{N}_{\ell}\cap\mathscr{C}_{\ell}^{0}$ and $\mathscr{C}_{\ell}^{3}=\mathscr{S}_{\ell}$, via the following descriptions $Q_{\ell}^{\nu}=(C_{\ell}^{\nu},G_{\ell}^{\nu},\gamma_{\ell}^{\nu},\mathscr{F}_{\ell}^{\nu})$ of $\mathscr{C}_{\ell}^{\nu},\ \nu\in\{2,3\}$. The sets $C_{\ell}^{2}=C_{\ell}^{\mathscr{N}}\cap C_{\ell}^{0}$ and $C_{\ell}^{3}=C_{\ell}^{\mathscr{S}}$ are subsets of $C_{\ell}^{0}$, partitioning $C_{\ell}^{0}$ by Proposition 3.3. The groups $G_{\ell}^{2}$ and $G_{\ell}^{3}$, acting on these sets, are $G_{\ell}^{2}=G_{\ell}^{3}=G_{\ell}^{0}=\ell^{\ast}>\lhd\ {\rm Gal}(\ell/k)$. The group actions $\gamma_{\ell}^{\nu}:G_{\ell}^{\nu}\times C_{\ell}^{\nu}\to C_{\ell}^{\nu},$ $\nu\in\{2,3\}$, are both induced from $\gamma_{\ell}^{0}$, i.e. they satisfy formula (5). The functors $\mathscr{F}_{\ell}^{\nu}:\ _{G_{\ell}^{\nu}}C_{\ell}^{\nu}\ \to\ \mathscr{C}_{\ell}^{\nu},\ \nu\in\{2,3\}$, are both induced from $\mathscr{F}_{\ell}^{0}$, i.e. they satisfy $\mathscr{F}_{\ell}^{\nu}(\underline{c})\ =\ A(\ell,\underline{c})$ on objects, and formula (6) on morphisms.

Proposition 6.6**.**

*For every quadratic extension $k\subset\ell$ in characteristic not 2, the following holds true.

(i) $\mathscr{C}_{\ell}^{0}=\mathscr{C}_{\ell}^{2}\amalg\mathscr{C}_{\ell}^{3}$.

(ii) $Q_{\ell}^{2}=(C_{\ell}^{2},G_{\ell}^{2},\gamma_{\ell}^{2},\mathscr{F}_{\ell}^{2})$ is a full description of $\mathscr{C}_{\ell}^{2}$ by a group action.

(iii) $Q_{\ell}^{3}=(C_{\ell}^{3},G_{\ell}^{3},\gamma_{\ell}^{3},\mathscr{F}_{\ell}^{3})$ is a description of $\mathscr{C}_{\ell}^{3}$ by a group action.*

Proof.

(i) The decomposition $\mathscr{C}_{\ell}=\mathscr{N}_{\ell}\amalg\mathscr{S}_{\ell}\amalg\mathscr{K}_{\ell}$, valid by Proposition 4.8, induces the decomposition

[TABLE]

where the second identity stems from $\mathscr{S}_{\ell}\subset\mathscr{C}_{\ell}^{0}$ and $\mathscr{K}_{\ell}\cap\mathscr{C}_{\ell}^{0}=\emptyset$, valid by Theorem 4.5 (ii), Proposition 3.3 (ii)-(iii), and Proposition 6.3.

(ii)-(iii) Arguing as in the proof of Proposition 6.4 (ii)-(iii), one finds that $Q_{\ell}^{2}$ is a description of $\mathscr{C}_{\ell}^{2}$ by a group action, and $Q_{\ell}^{3}$ is a description $\mathscr{C}_{\ell}^{3}$ by a group action. Observing that $C_{\ell}^{2}\subset C_{\ell}^{\mathscr{N}}$, even fullness of the functor $\mathscr{F}_{\ell}^{2}$ is proved by the same arguments which proved fullness of $\mathscr{F}_{\ell}^{1}$, above. ∎

Corollary 6.7**.**

Let $k$ be a field of characteristic not 2, and let $\mathscr{L}$ be a transversal for $\mathscr{Q}/\simeq$. Then the family

[TABLE]

of descriptions $Q_{\ell}^{\nu}$ of $\mathscr{C}_{\ell}^{\nu}$ is a covering of $\mathscr{C}$ by group actions.

Proof.

For each $(\ell,\nu)\in\mathscr{L}\times\underline{3}$, the quadruple $Q_{\ell}^{\nu}$ is a description of $\mathscr{C}_{\ell}^{\nu}$ by a group action (Proposition 6.4 (iii) and Proposition 6.6 (ii)-(iii)). Moreover, the full subgroupoids $\mathscr{C}_{\ell}^{\nu}\subset\mathscr{C}$ cover $\mathscr{C}$, as the identities of object classes

[TABLE]

hold by Proposition 4.8, Proposition 6.4 (i) and Proposition 6.6 (i). ∎

We call the family $Q_{\rm ref}$, appearing in Corollary 6.7, the refined Kleinian covering of $\mathscr{C}$ by group actions. It enables us to display the automorphism groups, up to isomorphism, of all objects in $\mathscr{C}=\mathscr{C}(k)$. For that purpose, we decompose $\mathscr{N}=\mathscr{N}(k)$ into two blocks $\mathscr{N}^{0}$ and $\mathscr{N}^{1}$, defined by their object classes $\mathscr{N}^{0}=\{A\in\mathscr{N}\ |\ |{\rm Aut}(A)|>4\}$ and $\mathscr{N}^{1}=\{A\in\mathscr{N}\ |\ |{\rm Aut}(A)|=4\}$.

Corollary 6.8**.**

*Let $k$ be a field of characteristic not 2, and let $\mathscr{L}$ be a transversal for $\mathscr{Q}/\simeq$. Then the following holds true.

(i)   $\mathscr{C}=\mathscr{N}^{0}\amalg\mathscr{N}^{1}\amalg\mathscr{S}\amalg\mathscr{K}$.

(ii) If $A\in\mathscr{N}^{0}$, then $N_{r}(A)\ \tilde{\to}\ \ell$ for a unique $\ell\in\mathscr{L}$, and*

[TABLE]

*(iii) If $A\in\mathscr{N}^{1}$, then ${\rm Aut}(A)$ is Klein’s four-group.

(iv) If $A\in\mathscr{S}$, then ${\rm Aut}(A)\ \tilde{\to}\ A^{\ast}/k^{\ast}$.

(v) If $A\in\mathscr{K}$, then ${\rm Aut}(A)$ is Klein’s four-group.*

Proof.

(i) The decomposition $\mathscr{N}=\mathscr{N}^{0}\amalg\mathscr{N}^{1}$ holds by definition of $\mathscr{N}^{0}$ and $\mathscr{N}^{1}$. Together with Proposition 2.2, it yields the claimed decomposition of $\mathscr{C}$.

(ii) If $A\in\mathscr{N}^{0}$, then $N_{r}(A)$ is the unique Kleinian subfield of $A$, by Theorem 4.5 (i). In particular, $k\subset N_{r}(A)$ is a quadratic extension, so $N_{r}(A)\ \tilde{\to}\ \ell$ holds for a unique $\ell\in\mathscr{L}$. Consequently, $A\in\mathscr{N}_{\ell}\cap\mathscr{N}^{0}=\mathscr{N}_{\ell}\cap\mathscr{C}_{\ell}^{0}=\mathscr{C}_{\ell}^{2}$. The functor $\mathscr{F}_{\ell}^{2}:\ _{G_{\ell}^{2}}C_{\ell}^{2}\ \to\ \mathscr{C}_{\ell}^{2}$ is an equivalence of categories, by Proposition 6.6 (ii). Therefore $A\ \tilde{\to}\ A(\ell,\underline{c})$ for some $\underline{c}\in C_{\ell}^{2}=C_{\ell}^{\mathscr{N}}\cap C_{\ell}^{0}$, and ${\rm Aut}(A)\ \tilde{\to}\ {\rm Aut}(A(\ell,\underline{c}))\ \tilde{\to}\ {\rm Mor}_{{}_{G_{\ell}^{2}}C_{\ell}^{2}}(\underline{c},\underline{c})$. By definition of the group action groupoid ${}_{G_{\ell}^{2}}C_{\ell}^{2}$, and because $c_{2}=0$ and $c_{3}\not=0$, the latter automorphism group admits the following sequence of canonical isomorphisms and identities:

[TABLE]

(iii) If $A\in\mathscr{N}^{1}$, then ${\rm Aut}(A)$ contains Klein’s four-group as a subgroup and $|{\rm Aut}(A)|=4$, whence the statement follows.

(iv) If $A\in\mathscr{S}$, then $A$ is a central skew field over $k$. Therefore Skolem-Noether Theorem [10, Theorem 7.21] applies to $A$, whence the claimed isomorphism follows.

(v) If $A\in\mathscr{K}$, then $A$ is a Galois extension of $k$ whose Galois group is Klein’s four-group (Proposition 2.2), and ${\rm Aut}(A)={\rm Gal}(A/k)$. ∎

6.4 Epilogue

In retrospect, we see that the “construction and reduction” approach, presented in Sections 3 and 4, provides a sufficiently solid basis for a systematic treatment of the classification problem of $\mathscr{C}$. The information it conveys about morphisms in $\mathscr{C}$ is however minimal: it just gives a criterion for the existence of an isomorphism $A(\ell,\underline{c})\ \tilde{\to}\ A(\ell,\underline{d})$ in terms of $\underline{c}$ and $\underline{d}$ (Theorem 4.7). As soon as one changes the perspective and tries to understand the structure of the groupoid $\mathscr{C}$, more information about morphisms is required. The refined Kleinian covering of $\mathscr{C}$ meets this need. It supplements the construction and reduction approach by “local descriptions” of $\mathscr{C}$, i.e. descriptions of the full subgroupoids $\mathscr{C}_{\ell}^{\nu}$ by group actions, such that the object class $\mathscr{C}$ is covered by the object classes $\mathscr{C}_{\ell}^{\nu}$, where $(\ell,\nu)\in\mathscr{L}\times\underline{3}$. The proof of Corollary 6.8 (ii) illustrates the usefulness of this deepened insight.

The Kleinian coverings of $\mathscr{C}$ also exemplify a phenomenon which appears to be ubiquitous although far from evident, namely that a given groupoid $\mathscr{Y}$ admits an explicitly and constructively presented covering by group actions at all. Apart from the groupoid $\mathscr{C}$, the occurrence of this phenomenon has in fact been observed in various other contexts. As a sample, let us mention the groupoid of all 2-dimensional real division algebras [4, Theorem 3.3], the groupoid of all 4-dimensional absolute valued algebras [5, Proposition 5.3], and the groupoid of all 8-dimensional composition algebras over a field of characteristic not 2 [1, Corollary 4.6].

A general theory however, that would be able to outline and explain the scope of validity of this phenomenon, seems not to exist at present. Yet whenever it does occur for a given groupoid $\mathscr{Y}$, it provides a holistic picture of $\mathscr{Y}$ in the sense that the classification problem of $\mathscr{Y}$ locally can be viewed as the normal form problem of a group action, and, whenever a local description of $\mathscr{Y}$ is full, then the described full subgroupoid is equivalent to an explicitly presented group action groupoid.

The author of the present article enjoyed the privilege to visit the Mathematics Department of Zürich University during the years 1987-1992 as a member of the Algebra group, which then was led by Peter Gabriel. In numerous private discussions, fine and memorable ones, Gabriel repeatedly stressed his viewpoint of the primacy of the concept of a group action. He even based his last lectures in Linear Algebra, documented in [6], on that concept. The present article is a late outgrowth of the influental ideas that shaped the author during his stay at Zürich University. It confirms the conjecture that classification problems, in great generality, can be understood as normal form problems of group actions.