title section subsection [] [] [] [] [] [] [© by the authors]© by the authors
[© by the authors]© by the authors [] []

# Subalgebras of Octonion Algebras

Norbert Knarr, Markus J. Stroppel

###### Abstract

For an arbitrary octonion algebra, we determine all subalgebras. It turns out
that every subalgebra of dimension less than four is associative, while every
subalgebra of dimension greater than four is not associative. In any split
octonion algebra, there are both associative and non-associative subalgebras
of dimension four. Except for one-dimensional subalgebras spanned by
idempotents, any two isomorphic subalgebras are in the same orbit under
automorphisms.

###### Key words and phrases:

octonion, quaternion, composition algebra, subalgebra, automorphism

###### 1991 Mathematics Subject Classification:

17A75, 17A36, 17A45.

## Introduction

Octonion algebras are the most complicated examples of composition algebras.
Composition algebras are (not necessarily associative) algebras endowed with a
multiplicative quadratic form such that the associated polar form is not
degenerate. As a tool for the investigation of quadratic forms, they have
their roots in number theory. Four-dimensional composition algebras (also
called quaternion algebras) are ubiquitous in the theory of quadratic forms.
In the present paper, we are mainly concerned with composition algebras of
dimension $8$, also called _octonion algebras_ (or octaves).

It is known (see [22, 1.6.2]) that composition algebras only occur in
dimension $d\in\\{1,2,4,8\\}$. A composition algebra is called _split_ if the
quadratic form is isotropic. Over any field $F$, there are split composition
algebras of dimension $2$ (isomorphic to $F\times F$ with the quadratic form
mapping $(x,y)$ to $xy$), of dimension $4$ (isomorphic to $F^{2\times 2}$, the
quadratic form is the determinant) and of dimension $8$ (obtained by doubling
the split quaternion algebra $F^{2\times 2}$, see 1.4(i) below). These split
composition algebras are determined uniquely by their dimension and the
underlying field $F$ (see [22, 1.8.1]).

The existence of non-split composition algebras depends in an essential way on
properties of the ground field $F$. In particular, a non-split composition
algebra of dimension $2$ over $F$ is a separable quadratic extension field. A
non-split quaternion algebra exists if, and only if, there is a separable
quadratic extension and an anisotropic hermitian form in two variables over
that extension (cp. [15]). In particular, every composition algebra of
dimension greater than two over a finite field is split. A quaternion field
(i.e., a non-split quaternion algebra) over $F$ exists if, and only if, there
is a non-singular quadratic form on $F^{3}$ such that the polar form is not
zero (see [13, Th. 3.1.1]). A necessary (but not sufficient, see [13, Th.
3.1.6]) condition for the existence of a non-split octonion algebra over $F$
is of course the existence of a non-singular quadratic form on $F^{8}$.

Octonion algebras play a role in group theory: For instance, the automorphism
group of an octonion algebra is a group of type $\mathrm{G}_{2}$, the action
of the hyperplane orthogonal to $1$ gives the representation of that group on
a vector space of dimension $7$. Several important subgroups of that group can
conveniently be seen as stabilizers of certain elements or subspaces in the
octonion algebra (see [3]), and various exceptional isomorphisms between
classical groups can be understood using representations as such stabilizers.

We also remark that non-split octonion algebras of dimension $8$ coordinatize
Moufang planes that are not desarguesian, and that every non-desarguesian
Moufang plane is obtained in this way ([16], [18, Satz 50, p. 106 and
Abschnitt 6, pp. 157–186], [23, Ch. 19]). Split octonion algebras are used to
construct generalized hexagons (namely, the split Cayley hexagons, see [21, p.
207]).

In the present paper, we study the subalgebras of octonion algebras. The split
case is of special interest; we find many types of subalgebras in split
octonion algebras that do not have counterparts in non-split algebras. We also
prove that subalgebras that are isomorphic as abstract algebras are in the
same orbit under automorphisms of the octonion algebra (except if they are
one-dimensional, and spanned by $1$ and an idempotent $p\neq 1$,
respectively). See Figure 1 in Section 8 below for a graph showing the
relative position of (orbits of) subalgebras. As a byproduct, we obtain that
every subalgebra of dimension less than $4$ in a composition algebra is
associative (see 7.1). In split octonion algebras, there do occur subalgebras
of dimension $4$ that are not associative (see 2.3), and subalgebras of
dimensions $5$ and $6$, respectively (see 2.2); no subalgebra of dimension
greater than $4$ is associative (see 7.1).

Using our list of subalgebras, we identify the commutative subalgebras, see
7.3. It turns out that every commutative subalgebra is associative.

Maximal subalgebras of octonion algebras have been studied in [19], a gap in
that paper was noted in [6] and closed in [17]. Unary subalgebras (i.e.
subalgebras containing the neutral element of multiplication) of finite
octonion algebras have been determined in [8]. Our present treatment is
independent of those results.

##  1\. Composition algebras

######  1.1 Definition.

Let $F$ be a field, and let $(A,+,\cdot)$ be any (not necessarily unary, not
necessarily associative) algebra over $F$. A quadratic form $q\colon A\to F$
is called _multiplicative_ if $q(x\cdot y)=q(x)q(y)$ holds for all $x,y\in A$.

######  1.2 Lemma: multiplicative forms.

Let $(A,+,\cdot)$ be any algebra over $F$. Assume that $a,b\in A$ are such
that the maps $\rho_{a}\colon A\to A\colon x\mapsto xa$ and $\lambda_{b}\colon
A\to A\colon x\mapsto bx$ are bijections, and define the multiplication $*$ on
$A$ by $x*y:=\rho_{a}^{-1}(x)\cdot\lambda_{b}^{-1}(y)$.

  1. ((a))

The element $b\cdot a$ is a neutral element for the multiplication $*$.

  2. ((b))

The triplet $(\lambda_{b}^{-1}|\mathrm{id}|\rho_{a}^{-1})$ is an isotopism
from $(A,+,*)$ onto $(A,+,\cdot)$.

  3. ((c))

Assume that there is a non-zero multiplicative quadratic form $q$ on
$(A,+,\cdot)$.
Then $s:=q(b\cdot a)$ is not zero, and $sq(x*y)=q(x)q(y)$ holds for all
$x,y\in A$.
The form $s^{-1}q$ is thus a non-zero multiplicative form on $(A,+,*)$.

###### Proof.

We compute $(b\cdot a)*y=\rho_{a}^{-1}(b\cdot
a)\cdot\lambda_{b}^{-1}(y)=b\cdot\lambda_{b}^{-1}(y)=\lambda_{b}(\lambda_{b}^{-1}(y))=y$
and $x*(b\cdot a)=\rho_{a}^{-1}(x)\cdot\lambda_{b}^{-1}(b\cdot
a)=\rho_{a}^{-1}(x)\cdot a=\rho_{a}(\rho_{a}^{-1}(x))=x$. This shows that
$b\cdot a$ is a neutral element for the multiplication $*$.

For $x,y\in A$, we note
$\mathrm{id}(x*y)=x*y=\rho_{a}^{-1}(x)\cdot\lambda_{b}^{-1}(y)$. This is the
defining property of an isotopism from $(A,+,*)$ onto $(A,+,\cdot)$, cp. [12,
Section 1].

Pick $u\in A$ such that $q(u)\neq 0$, and let
$v:=\lambda_{b}^{-1}(\rho_{a}^{-1}(u))$. Then $0\neq q(u)=q((b\cdot v)\cdot
a)=q(b\cdot v)q(a)=q(b)q(v)q(a)=q(b)q(a)q(v)=q(b\cdot a)q(v)$ yields $q(b\cdot
a)\neq 0$. We compute
$sq({x*y})=sq\left(\rho_{a}^{-1}(x)\cdot\lambda_{b}^{-1}(y)\right)=\left(q(a)q(b)\right)q\left(\rho_{a}^{-1}(x)\right)q\left(\lambda_{b}^{-1}(y)\right)=q\left(\rho_{a}^{-1}(x)\right)q(a)q(b)q\left(\lambda_{b}^{-1}(y)\right)=q\left(a\cdot\rho_{a}^{-1}(x)\right)q\left(b\cdot\lambda_{b}^{-1}(y)\right)=q(x)q(y)$.
This implies the last assertion. ∎

######  1.3 Definition.

A _composition algebra_ over $F$ is a (not necessarily associative) algebra
$C$ over $F$ with a multiplicative quadratic form
${\mathrm{N}}:={\mathrm{N}}_{C}\colon C\to F$ (called the _norm form_) such
that the polar form of ${\mathrm{N}}$ is not degenerate. This polar form will
be written as $(x|y):={\mathrm{N}}(x+y)-{\mathrm{N}}(x)-{\mathrm{N}}(y)$. We
also assume that the algebra contains a neutral element for its
multiplication, denoted by $1$. (However, see 1.2 for the case of isotopes
without $1$.)

As usual, the ground field $F$ is embedded as $F1$ in $C$. The composition
algebra is called _split_ if it contains divisors of zero (i.e., non-zero
elements of norm zero, see 1.4(f) below). If no confusion is expected we write
multiplication as juxtaposition.

The first chapter of [22] gives a comprehensive introduction into composition
algebras over arbitrary fields, including the characteristic two case. We
collect the basic facts that we need in the present paper (for proofs, consult
[22]):

######  1.4 Properties of composition algebras.

Let $C$ be a composition algebra over $F$.

  1. ((a))

The map $\kappa\colon C\to C\colon
x\mapsto{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}:=(x|1)1-x$
is an involutory anti-automorphism, called the _standard involution_ of $C$.

  2. ((b))

The norm and its polar form are recovered from the standard involution as
${\mathrm{N}}(x)=x{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}={\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}x$ and
$(x|y)=x{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}+y{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}$. In
particular, we have the hyperplane $1^{\perp}=\left\\{{x\in
C}\left|\vphantom{x\in C{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}=-x}\right.\,{{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}=-x}\right\\}$
of _pure elements_.

  3. ((c))

For all $c,x,y\in C$, we have $(cx|y)=(x|{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{c}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{c}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{c}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{c}\hfil$\crcr}}}}y)$ and
$(xc|y)=(x|y{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{c}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{c}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{c}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{c}\hfil$\crcr}}}})$, see [22,
1.3.2].

  4. ((d))

In general, the multiplication is not associative, but weak versions of the
associative law are still valid; among them _Moufang’s identities_ [22, 1.4.1,
1.4.2]

$(ax)(ya)=a((xy)a),\quad a(x(ay))=((ax)a)y,\quad x(a(ya))=((xa)y)a.$

  5. ((e))

Artin’s Theorem (see [22, Prop. 1.5.2], [20, Thm. 3.1, p. 29]): For any two
elements $x,y\in C$, the subalgebra generated by $x$ and $y$ in $C$ is
associative.

  6. ((f))

An element $a\in C$ is invertible if, and only if, its norm is not zero; we
have $a^{-1}={\mathrm{N}}(a)^{-1}\,{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}$ in that
case. Thus a non-split composition algebra is a division algebra. Note that
Artin’s Theorem then implies $a^{-1}(ax)=x=a(a^{-1}x)=(xa)a^{-1}=(xa^{-1})a$,
for each $x\in C$.

  7. ((g))

Each element $a\in C$ is a root of a polynomial of degree $2$ over $F$,
namely, the polynomial $X^{2}-(a+{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}})X+{\mathrm{N}}(a)\in
F[X]$. We call
${\mathrm{T}}(a):={\mathrm{T}}_{C}(a):=a+{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}$ the
_trace_ of $a$ in $C$.

  8. ((h))

[12, 1.3] Every $F$-semilinear automorphism of $C$ commutes with the standard
involution. If $\dim_{F}C\geq 4$ then every $\mathbb{Z}$-linear automorphism
of $C$ is $F$-semilinear. Consequently, every $\mathbb{Z}$-linear automorphism
of such a $C$ is a semi-similitude of the norm form, and every $F$-linear
automorphism is an orthogonal map.

We write $\operatorname{Aut}(C)$ for the group of all $\mathbb{Z}$-linear
automorphisms (i.e., all bijections that are additive and multiplicative), and
$\operatorname{Aut}_{F}(C)$ for the group of all $F$-linear automorphisms.

  9. ((i))

[22, 1.5.3] If $D$ is a subalgebra of $C$ with $\dim_{F}C=2\dim_{F}D$ and
$D^{\perp}\cap D=\\{0\\}$ then $D^{\perp}=Dw$ holds for each $w\in D^{\perp}$
with ${\mathrm{N}}(w)\neq 0$, and the multiplication in $C=D\oplus D^{\perp}$
is given by
$(x+yw)(u+vw)=(xu-{\mathrm{N}}(w){\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}y)+(vx+y{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{u}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{u}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{u}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{u}\hfil$\crcr}}}})w$.

  10. ((j))

[22, 1.6.4] Every element of an octonion algebra is contained in a quaternion
subalgebra. In fact, every element of a split octonion algebra is contained in
a _split_ quaternion algebra, see [7, 5.5].

We use the most comprehensive notion of “subalgebra” here:

######  1.5 Definition.

A _subalgebra_ in a given (not necessarily unary, not necessarily associative)
algebra $B$ over $F$ is a vector subspace $A$ (over $F$) which is closed under
multiplication.

In particular, even if $B$ contains a neutral element of multiplication, we do
_not_ require it to lie in the subalgebra. A subalgebra $A$ may well contain a
neutral element for its multiplication; this will then be an idempotent
centralized by $A$, but need not be a unit element in $B$ (in general, it will
not even be centralized by $B$, and need not be invertible).

We remark that in [22, 1.2.1] much stronger properties are required for a
_composition subalgebra_ ; that is a subalgebra (in the sense defined above)
that contains $1$, and such that the restriction of the norm form has non-
degenerate polar form.

Note also that only subalgebras containing $1$ are considered in [8].

######  1.6 Lemma.

A vector subspace $V$ of a composition algebra $C$ is closed under the
standard involution $\kappa$ if, and only if, it either contains $1$ or is
contained in $1^{\perp}$.

###### Proof.

The assertion is obvious if $V\subseteq 1^{\perp}$. If $1\in V$ then
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}={\mathrm{T}}(v)-v$
lies in $V$, for each $v\in V$. Conversely, if $v\in V\smallsetminus
1^{\perp}$ then $0\neq{\mathrm{T}}(v)\in V\cap F$ yields $1\in V$. ∎

Note that a subalgebra will, in general, not be a composition algebra since
the polar form of the restriction of the norm form ${\mathrm{N}}$ to the
subalgebra may be degenerate.

######  1.7 Quaternion algebras.

In the context of abstract algebras over $F$, a _split quaternion algebra_ is
an algebra $H$ over $F$ which is isomorphic to the matrix algebra $F^{2\times
2}$. A _quaternion field_ can be defined as a skewfield (i.e., an associative
division algebra) with center $F$ and dimension $4$ over $F$, we call this a
_non-split quaternion algebra_ over $F$.

Both classes of quaternion algebras are completely contained in the class of
composition algebras; their members are exactly the composition algebras of
dimension $4$. The required (multiplicative) norm is obtained as the
determinant in the split case. For the non-split case, one uses a two-
dimensional subalgebra, and the fact that the tensor product with that
subalgebra yields a split quaternion algebra. See also [15].

A _quaternion subalgebra_ of an octonion algebra over $F$ is a subalgebra $H$
that is isomorphic (as an abstract algebra) to some quaternion algebra over
$F$.

For a quaternion subalgebra $H$ of an octonion algebra $\mathbb{O}$, we do not
require that the restriction of the norm ${\mathrm{N}}_{\mathbb{O}}$ to $H$
satisfies any non-degeneracy conditions, and we do not require _a priori_ that
$H$ contains the neutral element $1$ of $\mathbb{O}$. However, it turns out
that the quaternion algebra is embedded as nice as can be, see 4.1 below.

######  1.8 Lemma.

Let $\mathbb{O}$ be a _non-split_ octonion algebra over a field $F$ with
$\operatorname{char}{F}\neq 2$. Then any two elements of $\mathbb{O}$ are
contained in a quaternion subalgebra of $\mathbb{O}$.

###### Proof.

Consider $a,b\in\mathbb{O}$. If $a\in F+Fb$ (in particular, if $a\in F$) then
the assertion is obvious from the result that every single element $b$ of
$\mathbb{O}$ is contained in a quaternion algebra (this even holds if
$\mathbb{O}$ is split, or if $\operatorname{char}{F}=2$, see [22, 1.6.4]). So
assume that $a\notin F+Fb$. Then $A:=F+Fa$ is a quadratic extension field of
$F$, and $\mathbb{O}$ is a vector space over $A$. So $Ab$ has trivial
intersection with $A$, and $H:=A+Ab=F+Fa+Fb+Fab$ is a vector space of
dimension $4$ over $F$. Clearly, that vector space coincides with the
(associative) subalgebra generated by $\\{a,b\\}$ in $\mathbb{O}$. The
restriction ${\mathrm{N}}|_{H}$ of the norm form ${\mathrm{N}}_{\mathbb{O}}$
is non-singular, and then non-degenerate because $\operatorname{char}{F}\neq
2$. So $H$ is a (non-split) composition algebra (of dimension $4$), and thus a
quaternion field. ∎

######  1.9 Remark.

If $\mathbb{O}$ is a non-split octonion algebra of characteristic $2$ then
$\mathbb{O}$ contains commutative (and associative) subalgebras of dimension
$4$ (obtained as fixed point sets of involutory automorphisms, see [11, 4.5]).
In fact, such a subalgebra forms a totally inseparable field extension of
degree $4$ and exponent $1$ over $F$. Two elements generating such a
subalgebra do not lie in any common quaternion field, obviously, so 1.8 does
not extend to the characteristic two case.

######  1.10 Remark.

In a split octonion algebra, it is no longer true that any two elements lie in
a quaternion subalgebra. E.g., if $a,b$ are two nilpotent elements such that
$ab\neq 0$ then the algebra $A_{a,b}$ generated by them is a Heisenberg
algebra (see 2.7 below). Such a Heisenberg algebra contains two-dimensional
subalgebras with trivial multiplication (such as $Fa+Fab$), and is therefore
not contained in any quaternion algebra. So 1.8 does not extend to the split
case.

##  2\. Examples of subalgebras

The present section collects various examples that will serve as
representatives for orbits of subalgebras under
$\operatorname{Aut}_{F}(\mathbb{O})$. See 5.11, 6.2 and 6.4 below.

######  2.1 Lemma.

Let $\mathbb{H}$ be a split quaternion algebra, and construct the (split)
octonion algebra $\mathbb{O}:=\mathbb{H}+\mathbb{H}w$ with multiplication
$(a+xw)(b+yw)=ab+{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x+(ya+x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w\,.$ Let
$A$ be a subalgebra of $\mathbb{H}$, and let $R$ be a vector subspace of
$\mathbb{H}$ such that $RA\subseteq R$ and
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{R}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{R}\hfil$\crcr}}}}R:=\\{{{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x}\left|\vphantom{}\right.\,{x,y\in
R}\\}\subseteq A$. Then $A+Rw$ is a subalgebra of $\mathbb{O}$.

These conditions are satisfied if $A$ is a subalgebra with
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{A}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{A}\hfil$\crcr}}}}=A$ and $R$
is a right ideal in $A$.

###### Proof.

For $(a,x),(b,y)\in A\times R$, we use ${\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{R}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{R}\hfil$\crcr}}}}R\subseteq
A$ and $RA\subseteq R$ to see that the product
$({a+xw})({b+yw})=ab+{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x+(ya+x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w$ is of
the form $c+zw$ with $(c,z)\in A\times R$. If $R\subseteq
A={\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{A}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{A}\hfil$\crcr}}}}$ then
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{R}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{R}\hfil$\crcr}}}}R\subseteq
A$. ∎

A vector subspace $A$ of $\mathbb{O}$ satisfies
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{A}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{A}\hfil$\crcr}}}}=A$
precisely if either $1\in A$ or $A\subseteq 1^{\perp}$, see 1.6.

######  2.2 Examples.

Recall that each split quaternion algebra over $F$ is isomorphic to
$F^{2\times 2}$. We abbreviate $n_{0}=\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}\right)$ and $p_{0}=\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)$.

  1. ((a))

Apart from $F^{2\times 2}$ and $\\{0\\}$, each right ideal of $F^{2\times 2}$
has dimension $2$, and is of the form $R_{v}:=\left\\{{X\in F^{2\times
2}}\left|\vphantom{X\in F^{2\times 2}vX=0}\right.\,{vX=0}\right\\}$ for a row
vector $v=(v_{1},v_{2})\neq(0,0)$.

For any such $R_{v}$, the subalgebra $F^{2\times 2}+R_{v}w$ has dimension $6$.
We will prove that every subalgebra of dimension $6$ is in the orbit of this
subalgebra under $\operatorname{Aut}_{F}(\mathbb{O})$, see 6.2. Note that
$F^{2\times 2}+R_{(0,1)}w=\mathbb{O}(n_{0}w)+\mathbb{O}(p_{0}w)$.

  2. ((b))

Every $3$-dimensional subalgebra of $F^{2\times 2}$ is a conjugate of the
subalgebra $U:=\left\\{{\left(\begin{smallmatrix}r&t\\\
0&s\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}r&t\\\
0&s\end{smallmatrix}\right)r,s,t\in F}\right.\,{r,s,t\in F}\right\\}$ of upper
triangular matrices. The right ideals (apart from $\\{0\\}$ and $U$) in $U$
are $L:=\left\\{{\left(\begin{smallmatrix}r&t\\\
0&0\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}r&t\\\
0&0\end{smallmatrix}\right)r,t\in F}\right.\,{r,t\in F}\right\\}$,
$\kappa(L)=\left\\{{\left(\begin{smallmatrix}0&t\\\
0&s\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}0&t\\\
0&s\end{smallmatrix}\right)t,s\in F}\right.\,{t,s\in F}\right\\}$, and
$M_{y}:=\left\\{{\left(\begin{smallmatrix}0&y_{1}s\\\
0&y_{2}s\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}0&y_{1}s\\\
0&y_{2}s\end{smallmatrix}\right)s\in F}\right.\,{s\in F}\right\\}$ for
$y=\binom{y_{1}}{y_{2}}\in F^{2\times 1}$. By 2.1, this gives examples of
subalgebras of dimension $5$. We will prove that every subalgebra of dimension
$5$ is in the orbit of $U+Lw$ under $\operatorname{Aut}_{F}(\mathbb{O})$, see
6.2. Note that $U+Lw=n_{0}\mathbb{O}+\mathbb{O}n_{0}$.

  3. ((c))

Let $A$ be any subalgebra of $\mathbb{H}$, and let
$R:=\left\\{{\left(\begin{smallmatrix}r&t\\\
0&0\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}r&t\\\
0&0\end{smallmatrix}\right)r,t\in F}\right.\,{r,t\in F}\right\\}$. Then
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{R}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{R}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{R}\hfil$\crcr}}}}R=\\{0\\}$,
and $A+Rw$ is a subalgebra by 2.1. In this way, one finds examples of
subalgebras of dimension $\dim{A}+2\in\\{3,4,5\\}$. See 6.4(d)(ii).

  4. ((d))

The set $A:=\left\\{{\left(\begin{smallmatrix}r&t\\\
0&0\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}r&t\\\
0&0\end{smallmatrix}\right)r,t\in F}\right.\,{r,t\in F}\right\\}$ is a
subalgebra of $\mathbb{H}$ with ${\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{A}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{A}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{A}\hfil$\crcr}}}}\neq A$.
Using $R:=F\left(\begin{smallmatrix}0&1\\\ 0&0\end{smallmatrix}\right)$ and
2.1 we obtain a three-dimensional subalgebra $A+Rw$ in $\mathbb{O}$.
See 5.11(c)(ii). Note that $A+Rw=n_{0}\mathbb{O}\cap\mathbb{O}(n_{0}w)$.

######  2.3 Example.

In $\mathbb{O}=F^{2\times 2}+F^{2\times 2}w$ with $w^{2}=1$, consider
$n_{0}:=\left(\begin{smallmatrix}0&1\\\ 0&0\end{smallmatrix}\right)$ and
$p_{0}:=\left(\begin{smallmatrix}1&0\\\ 0&0\end{smallmatrix}\right)$. The
$4$-dimensional subalgebra
$n_{0}\mathbb{O}=\left\\{{\left(\begin{smallmatrix}r&s\\\
0&0\end{smallmatrix}\right)+\left(\begin{smallmatrix}0&t\\\
0&u\end{smallmatrix}\right)w}\left|\vphantom{\left(\begin{smallmatrix}r&s\\\
0&0\end{smallmatrix}\right)+\left(\begin{smallmatrix}0&t\\\
0&u\end{smallmatrix}\right)wr,s,t,u\in F}\right.\,{r,s,t,u\in F}\right\\}$
(obtained as in 2.1 with $A=Fp_{0}+Fn_{0}$ and
$R=(Fn_{0}+F{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}})w$) is
not associative. In fact, we have
$p_{0}\bigl{(}({\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}w)(n_{0}w)\bigr{)}=p_{0}({\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{n_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{n_{0}}\hfil$\crcr}}}}\,{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}})=-n_{0}$
while $\bigl{(}p_{0}({\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}w)\bigr{)}(n_{0}w)=\bigl{(}({\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}p_{0})w\bigr{)}(n_{0}w)=0$.

In the present paper, we will prove that every $3$-dimensional subalgebra of
any octonion algebra is associative (see 5.11 for algebras that do not contain
$1$; in the case where $1\in A$ and $\dim{A}=3$ choose a basis $1,a,b$ for
$A$, note that $a,b$ generate an associative subalgebra $B$ by 1.4(e), and
that $A=F+B$ is associative), and that every subalgebra of dimension greater
than $4$ is non-associative (see 5.11 and 6.2).

######  2.4 Lemma.

A subalgebra of $\mathbb{O}$ contains $1$ if, and only if, it contains an
invertible element.

###### Proof.

Assume that $A$ contains an invertible element $a$. Then
${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}a=(a+{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}})a-a^{2}$ is
a non-trivial element of $F\cap A$, and $1\in A$ follows. ∎

######  2.5 Corollary.

Every subalgebra of dimension greater than four contains $1$. ∎

For any $x,y\in\mathbb{O}$, the subalgebra generated by $\\{1,x,y\\}$ in
$\mathbb{O}$ is associative (by Artin’s Theorem, see 1.4(e)). We investigate
the borderline between associative and non-associative subalgebras in more
detail:

######  2.6 Lemma.

Let $A$ be a subalgebra of an octonion algebra $\mathbb{O}$ over $F$. If $A$
is generated by two elements then $\dim{A}\leq 4$; in fact, we have
$A\subseteq F+Fx+Fy+Fxy$ if $A$ is generated by $\\{x,y\\}$.

###### Proof.

We abbreviate $\xi:=x+{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}$ and
$\eta:=y+{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}$. Then
$\xi$ and $\eta$ lie in $F$, and $yx=(yx+{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{yx}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{yx}\hfil$\crcr}}}})-{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{yx}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{yx}\hfil$\crcr}}}}$ lies in
$F+F({\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{yx}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{yx}\hfil$\crcr}}}})$. Thus it
suffices to note that ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{yx}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{yx}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{yx}\hfil$\crcr}}}}={\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}\,{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}=(\xi-x)(\eta-y)=\xi\eta-\xi
y-\eta x+xy$ lies in $F+Fx+Fy+Fxy$. ∎

######  2.7 Heisenberg subalgebras.

Take a nilpotent element $a\neq 0$ (i.e., any non-zero element $a$ of norm and
trace $0$), and pick a nilpotent element $b\in
a^{\perp}\smallsetminus({F+Fa})$. Then ${\mathrm{N}}(ab)=0$, and
$(ab|1)=(b|{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}})=(b|-a)=0$
(see 1.4(c)) gives $ba=-ab$. Moreover, computation in the (associative)
subalgebra generated by $a$ and $b$ yields $a(ab)=-{\mathrm{N}}(a)b=0$,
$(ab)a=-(ba)a{{}=0}$. Analogously, we find $b(ab)=0=(ab)b$ and $(ab)(ab)=0$.
So the linear span of $\\{a,b,ab\\}$ is closed under multiplication, and forms
a subalgebra $A_{a,b}$.

There are two cases:

  1. ((a))

If $ab=0$ then $ba={\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{ab}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{ab}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{ab}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{ab}\hfil$\crcr}}}}=0$, and
$A_{a,b}=Fa+Fb$ is a two-dimensional subalgebra such that the restriction
${\mathrm{N}}|_{A_{a,b}}$ is trivial. We then conclude that the multiplication
on $A_{a,b}$ is trivial (i.e., every product of two elements in $A_{a,b}$
equals $0$).

  2. ((b))

If $ab\neq 0$ then $ab\notin Fa+Fb$ because $ab=ra+sb$ with $r,s\in F$ implies
$0=a(ab)=ra^{2}+s(ab)=s(ab)$ and $0=(ab)b=r(ab)+sb^{2}=r(ab)$, leading to the
contradiction $s=0=r$. In this case, the algebra $A_{a,b}$ has dimension $3$,
and is isomorphic to the Heisenberg algebra over $F$.

See 5.6 below for a characterization of Heisenberg algebras in $\mathbb{O}$,
and a proof that the Heisenberg subalgebras form a single orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$.

######  2.8 Examples.

In $\mathbb{O}=F^{2\times 2}+F^{2\times 2}w$ with $w^{2}=1$, the element
$p_{0}=\left(\begin{smallmatrix}1&0\\\ 0&0\end{smallmatrix}\right)$ is an
idempotent (i.e., $p_{0}^{2}=p_{0}$), the elements
$n_{0}=\left(\begin{smallmatrix}0&1\\\ 0&0\end{smallmatrix}\right)$ and
$n_{0}w$ are nilpotent, and so is $p_{0}w$.

Computing $n_{0}(p_{0}w)=n_{0}w$ and $n_{0}(n_{0}w)=0$, we obtain that
$Fn_{0}+Fn_{0}w=A_{n_{0},n_{0}w}$ is a two-dimensional algebra as in 2.7(a),
and that the $3$-dimensional algebra
$A_{n_{0},p_{0}w}={Fn_{0}+Fp_{0}w+Fn_{0}w}$ is an example of a Heisenberg
algebra, as in 2.7(b). Those algebras are obtained by the method given in 2.1,
using $A=Fn_{0}$ and $R\in\\{Fn_{0},Fn_{0}+Fp_{0}\\}$.

##  3\. Centralizers

######  3.1 Lemma.

Let $\mathbb{O}$ be an octonion algebra over $F$, and consider
$v\in\mathbb{O}\smallsetminus F$.

  1. ((a))

If ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}=v$ then
$\operatorname{char}{F}=2$, and the centralizer
$\operatorname{C}_{\mathbb{O}}(v)$ of $v$ in $\mathbb{O}$ has dimension $6$.
In this case, the centralizer is not closed under multiplication. That
centralizer consists of pure elements; in fact we have
$\operatorname{C}_{\mathbb{O}}(v)=\\{1,v\\}^{\perp}$.

  2. ((b))

If ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}\neq v$ (in
particular, if $\operatorname{char}{F}\neq 2$) then
$\dim{\operatorname{C}_{\mathbb{O}}(v)}=2$ if
${\mathrm{N}}(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})\neq 0$,
and $\dim{\operatorname{C}_{\mathbb{O}}(v)}=4$ if
${\mathrm{N}}(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0$.

  3. ((c))

If $\dim{\operatorname{C}_{\mathbb{O}}(v)}=4$ then $\operatorname{char}{F}\neq
2$, there exists $n\in F+Fv$ with $n^{2}=0\neq n$ (so $\mathbb{O}$ is split),
and $\operatorname{C}_{\mathbb{O}}(v)$ is a subalgebra.

###### Proof.

First of all, we choose a quaternion subalgebra $H$ of $\mathbb{O}$ containing
$v$ (see 1.4(j)), and write $\mathbb{O}=H+Hw$ as a double of $H$, with
multiplication

$(a+xw)(b+yw)=ab-{\mathrm{N}}(w){\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x+(ya+x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w\,.$

Now $a+xw$ centralizes $v$ precisely if $av=va$ and
$x(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0$. In any
case, this means that $a$ belongs to the two-dimensional centralizer $F+Fv$ of
$v$ in $H$.

If ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}=v$ then
$v\notin F$ implies $\operatorname{char}{F}=2$. In this case, $a\in F+Fv$ but
we have no restriction on $x$, and the centralizer has dimension $6$. However,
the product $ab-{\mathrm{N}}(w){\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x+(ya+x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w$ of
elements $(a+xw)$ and $(b+yw)$ of the centralizer belongs to that centralizer
only if ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x$ lies in
$F+Fv$.

Now assume ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}\neq v$. If
${\mathrm{N}}(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})\neq 0$
then $a+xw\in\operatorname{C}_{\mathbb{O}}(v)$ implies $x=0$, and
$\operatorname{C}_{\mathbb{O}}(v)=F+Fv$ has dimension $2$. If
${\mathrm{N}}(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0$ then
$H$ is a split quaternion algebra (isomorphic to $F^{2\times 2}$), and the
annihilator $\left\\{{x\in H}\left|\vphantom{x\in
Hx(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0}\right.\,{x(v-{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0}\right\\}$
has dimension $2$. Then
$\operatorname{C}_{\mathbb{O}}(v)=\left\\{{a+xw}\left|\vphantom{a+xwa\in
F+Fv,x(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0}\right.\,{a\in
F+Fv,x(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=0}\right\\}$
has dimension $4$. That centralizer is a subalgebra by 2.1. ∎

######  3.2 Example.

Consider $n_{0}=\left(\begin{smallmatrix}0&1\\\ 0&0\end{smallmatrix}\right)$
and $p_{0}=\left(\begin{smallmatrix}1&0\\\ 0&0\end{smallmatrix}\right)$ in
$\mathbb{O}=F^{2\times 2}+F^{2\times 2}w$, with $w^{2}=1$. If
$\operatorname{char}{F}\neq 2$ then
$\operatorname{C}_{\mathbb{O}}(n_{0})=F+Fn_{0}+Fn_{0}w+F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}w$. This
centralizer forms an associative subalgebra of $\mathbb{O}$, isomorphic to

$\left\\{{\left(\begin{smallmatrix}\alpha&\beta&\gamma&\delta\\\
0&\alpha&0&0\\\ 0&-\delta&\alpha&0\\\
0&\gamma&0&\alpha\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}\alpha&\beta&\gamma&\delta\\\
0&\alpha&0&0\\\ 0&-\delta&\alpha&0\\\
0&\gamma&0&\alpha\end{smallmatrix}\right)\alpha,\beta,\gamma,\delta\in
F}\right.\,{\alpha,\beta,\gamma,\delta\in F}\right\\}.$

Note that
$\operatorname{C}_{\mathbb{O}}(n_{0})=F+(n_{0}\mathbb{O}\cap\mathbb{O}n_{0})$
holds in this case.

If $\operatorname{char}{F}=2$ then $n_{0}-{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{n_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{n_{0}}\hfil$\crcr}}}}=0$, and
$\operatorname{C}_{\mathbb{O}}(n_{0})=F+Fn_{0}+F^{2\times 2}w$ has dimension
$6$.

######  3.3 Corollary.

  1. ((a))

If $\operatorname{char}{F}=2$ then the centralizer
$\operatorname{C}_{\mathbb{O}}(v)$ has dimension $6$ if $v\in
1^{\perp}\smallsetminus F$, and it has dimension $2$ if
$v\in\mathbb{O}\smallsetminus 1^{\perp}$.

  2. ((b))

In any case, the centralizer $\operatorname{C}_{\mathbb{O}}(v)$ of
$v\in\mathbb{O}\smallsetminus F$ has dimension $2$ if, and only if, the
subalgebra $F+Fv$ is a composition algebra. Then
$F+Fv=\operatorname{C}_{\mathbb{O}}(v)$.

  3. ((c))

If $\operatorname{char}{F}\neq 2$ then
$\dim\operatorname{C}_{\mathbb{O}}(v)=4$ holds if
$v\in\mathbb{O}\smallsetminus F$ and
$0={\mathrm{N}}(v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})=-(v-{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}})^{2}$. In
that case, we note that the centralizers of $v$ and of
$v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}=2v-{\mathrm{T}}(v)$
then coincide, and that $v-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{v}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{v}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{v}\hfil$\crcr}}}}$ is
nilpotent. As every non-trivial nilpotent element of $\mathbb{O}$ lies in the
same orbit under $\operatorname{Aut}_{F}(\mathbb{O})$ (see 4.4 below), all
these centralizers are in the orbit of the one described in 3.2. ∎

######  3.4 Remark.

If $\operatorname{C}_{\mathbb{O}}(v)$ has dimension $4$ then there exists
$n\in({1^{\perp}\cap({F+Fv})})\smallsetminus F$ such that $v\in F+n$, and
$\operatorname{C}_{\mathbb{O}}(v)=F+(n\mathbb{O}\cap\mathbb{O}n)=\operatorname{C}_{\mathbb{O}}(n)$.
See 3.2 for an explicit example, and also 3.7 below. This is not a composition
algebra: the restriction of the norm form has Witt index $3$, and is
degenerate. The non-invertible elements form a hyperplane, which is an ideal.
So $\operatorname{C}_{\mathbb{O}}(v)$ is a local algebra, in this case.

######  3.5 Corollary.

If $v\notin F$ then no quaternion subalgebra is contained in
$\operatorname{C}_{\mathbb{O}}(v)$. ∎

A vector subspace $V$ in a composition algebra $C$ is called _totally
singular_ if ${\mathrm{N}}_{C}(V)=\\{0\\}$. The maximal totally singular
subspaces in split octonion algebras are well known:

######  3.6 Theorem.

Let $\mathbb{O}$ be a split octonion algebra.

  1. ((a))

For any $a\in\mathbb{O}$, we have
$a\mathbb{O}=\ker\lambda_{{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}}$ (see [24,
Lemma 1]).

  2. ((b))

If ${\mathrm{N}}(a)=0$ but $a\neq 0$ then $a\mathbb{O}$ and $\mathbb{O}a$ are
maximal totally singular subspaces, and every maximal totally singular
subspace is of this form (see [24, Theorem 3]).

  3. ((c))

For $a,b\in\mathbb{O}\smallsetminus\\{0\\}$ with
${\mathrm{N}}(a)=0={\mathrm{N}}(b)$ we have (see [24, Theorem 4, Theorem 5])

    1. ((i))

$a\mathbb{O}=b\mathbb{O}\iff Fa=Fb$,

    2. ((ii))

$\dim(a\mathbb{O}\cap b\mathbb{O})=2\iff(a|b)=0\land\dim(Fa+Fb)=2$;
then $a\mathbb{O}\cap b\mathbb{O}=a({\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}}\mathbb{O})=b({\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}\mathbb{O})$,

    3. ((iii))

$a\mathbb{O}\cap b\mathbb{O}=\\{0\\}\iff(a|b)\neq 0$,

    4. ((iv))

$a\mathbb{O}\cap\mathbb{O}b=Fab\iff ab\neq 0$,

    5. ((v))

$\dim(a\mathbb{O}\cap\mathbb{O}b)=3\iff ab=0$;
then $a\mathbb{O}\cap\mathbb{O}b=\left\\{{ay}\left|\vphantom{ayy\in
b^{\perp}}\right.\,{y\in
b^{\perp}}\right\\}=\left\\{{xb}\left|\vphantom{xbx\in
a^{\perp}}\right.\,{x\in a^{\perp}}\right\\}$.

As the standard involution interchanges
$\left\\{{a\mathbb{O}}\left|\vphantom{a\mathbb{O}a\in\mathbb{O}}\right.\,{a\in\mathbb{O}}\right\\}$
with
$\left\\{{\mathbb{O}a}\left|\vphantom{\mathbb{O}aa\in\mathbb{O}}\right.\,{a\in\mathbb{O}}\right\\}$,
the list above covers all possibilities for the relative position (i.e., the
size of the intersection) of maximal totally singular subspaces in
$\mathbb{O}$. ∎

Consider a $3$-dimensional subspace $T$ such that ${\mathrm{N}}|_{T}=0$. Since
the intersection $a\mathbb{O}\cap a^{\prime}\mathbb{O}$ has dimension at most
$2$ if $a\mathbb{O}\neq a^{\prime}\mathbb{O}$, there is precisely one subspace
$a\mathbb{O}$ and precisely one subspace $\mathbb{O}b$ such that $T\subseteq
a\mathbb{O}$ and $T\subseteq\mathbb{O}b$ (i.e., such that
$T=a\mathbb{O}\cap\mathbb{O}b$).

######  3.7 Examples.

Consider $p_{0}=\left(\begin{smallmatrix}1&0\\\ 0&0\end{smallmatrix}\right)$,
and $n_{0}=\left(\begin{smallmatrix}0&1\\\ 0&0\end{smallmatrix}\right)$ in the
split octonion algebra $\mathbb{O}:=F^{2\times 2}+F^{2\times 2}w$, where
$w^{2}=1$.

  1. ((a))

The set $(p_{0}w)\mathbb{O}\cap\mathbb{O}(n_{0}w)=Fp_{0}+Fp_{0}w+Fn_{0}w$ is a
subalgebra of dimension $3$, and every element is obtained as a product in the
subalgebra. This subalgebra is isomorphic to the associative algebra

$\left\\{{\left(\begin{matrix}\alpha&0&0&\beta\\\ 0&\alpha&0&0\\\
0&\gamma&0&0\\\ 0&0&0&0\\\
\end{matrix}\right)}\left|\vphantom{\left(\begin{matrix}\alpha&0&0&\beta\\\
0&\alpha&0&0\\\ 0&\gamma&0&0\\\ 0&0&0&0\\\
\end{matrix}\right)\alpha,\beta,\gamma\in F}\right.\,{\alpha,\beta,\gamma\in
F}\right\\}\,;$

it is also obtained by the construction in 2.1 with $A=Fp_{0}$ and
$R=Fp_{0}+Fn_{0}$.

  2. ((b))

The algebra $n_{0}\mathbb{O}\cap\mathbb{O}n_{0}=\left\\{{\alpha n_{0}+(\beta
n_{0}+\gamma{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}})w}\left|\vphantom{\alpha
n_{0}+(\beta n_{0}+\gamma{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}})w\alpha,\beta,\gamma\in
F}\right.\,{\alpha,\beta,\gamma\in F}\right\\}$ has multiplication

$(\alpha p_{0}+(\beta p_{0}+\gamma n_{0})w)(\xi p_{0}+(\eta p_{0}+\zeta
n_{0})w)=(\beta\zeta-\gamma\eta)n_{0};$

this is a Heisenberg algebra, as in 2.7(b)). It is neither isomorphic nor
anti-isomorphic to $(p_{0}w)\mathbb{O}\cap\mathbb{O}(n_{0}w)$. This Heisenberg
algebra is also obtained by the construction in 2.1, with $A=Fn_{0}$ and
$R=Fn_{0}+F{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}$.

##  4\. Automorphisms of (split) composition algebras

The following is a variant of the well known Skolem-Noether Theorem (e.g., see
[10, 4.6, p. 222] or [4, 5.1.7, p. 183]). Note, however, that we deal with
subalgebras of a non-associative algebra, and do not require a priori that the
subalgebras contain $1$.

######  4.1 Lemma.

Let $\mathbb{O}$ and $\mathbb{O}^{\prime}$ be octonion algebras over $F$, and
let $H<\mathbb{O}$ and $H^{\prime}<\mathbb{O}^{\prime}$ be quaternion
subalgebras. Then both $H$ and $H^{\prime}$ contain the neutral element $1$ of
$\mathbb{O}$ and of $\mathbb{O}^{\prime}$, respectively. Moreover, we have:

  1. ((a))

If $\mathbb{O}$ and $\mathbb{O}^{\prime}$ are isomorphic then every non-
constant $F$-linear multiplicative map from $H$ to $H^{\prime}$ extends to an
isomorphism from $\mathbb{O}$ onto $\mathbb{O}^{\prime}$.

  2. ((b))

If $H$ or $H^{\prime}$ is split then every non-constant $F$-linear
multiplicative map from $H$ to $H^{\prime}$ extends to an isomorphism from
$\mathbb{O}$ onto $\mathbb{O}^{\prime}$.

###### Proof.

Let $e$ and $e^{\prime}$ be the neutral elements of multiplication in $H$ and
$H^{\prime}$, respectively. Then $H$ is a quaternion subalgebra of
$\mathbb{O}$ centralizing the idempotent $e$. From 3.5 we infer $e\in F$ and
then $e=1$.

Now consider a non-constant $F$-linear multiplicative map $\beta\colon H\to
H^{\prime}$. As $H$ is a simple algebra, the ideal $\ker\beta$ is trivial, and
$\beta$ is an isomorphism by dimension reasons. The restrictions
${\mathrm{N}}_{\mathbb{O}}|_{H}$ and
${\mathrm{N}}_{\smash{\mathbb{O}^{\prime}}}|_{\smash{H^{\prime}}}$ of the
norms are isometric because the quaternion subalgebras are isomorphic. By
Witt’s Cancellation Theorem ([25, Satz 4, p. 34], [1, Folgerung, p. 160], see
[5, p. 21, p. 35] or [9, 5.1, 12.12]) the restrictions
${\mathrm{N}}_{\mathbb{O}}|_{H^{\perp}}$ and
${\mathrm{N}}_{\smash{\mathbb{O}^{\prime}}}|_{\smash{H^{\prime}}^{\perp}}$ of
the norms are isometric, as well. We pick any element $w\in H^{\perp}$ with
${\mathrm{N}}(w)\neq 0$, and find $w^{\prime}\in\smash{H^{\prime}}^{\perp}$
with
${\mathrm{N}}_{\mathbb{O}}(w^{\prime})={\mathrm{N}}_{\smash{\mathbb{O}^{\prime}}}(w)$.
The formula (see 1.4(i)) for the multiplication in $\mathbb{O}=H+Hw$ and in
$\mathbb{O}^{\prime}=H^{\prime}+H^{\prime}w^{\prime}$, respectively,
immediately yields that mapping $a+xw\in H+Hw$ to
$\beta(a)+\beta(x)w^{\prime}\in H^{\prime}+H^{\prime}w^{\prime}$ is an
automorphism of $\mathbb{O}$.

If $H$ or $H^{\prime}$ is split then the existence of an isomorphism between
the quaternion algebras implies that both are split, and so are the octonion
algebras $\mathbb{O}$ and $\mathbb{O}^{\prime}$. So the latter are isomorphic,
and our first assertion applies. ∎

######  4.2 Remark.

Let $\mathbb{O}=H+Hw$ be obtained by doubling a quaternion subalgebra $H$,
using $w\in H^{\perp}$. The $F$-linear automorphisms of $\mathbb{O}$ that
leave $H$ invariant can be given explicitly: they are the maps
$\alpha_{s,t}\colon H+Hw\to H+Hw\colon a+xw\mapsto sas^{-1}+(txs^{-1})w$,
where $s,t$ are invertible elements in $H$ with
${\mathrm{N}}(s)={\mathrm{N}}(t)$.

In fact, the restriction of a given automorphism
$\alpha\in\operatorname{Aut}_{F}(\mathbb{O})$ with $\alpha(H)=H$ to $H$ gives
an $F$-linear automorphism of $H$, which is inner by the Skolem-Noether
Theorem (e.g., see [10, 4.6, p. 222] or [4, 5.1.7, p. 183]). So there exists
$s\in H$ with ${\mathrm{N}}(s)\neq 0$ and $\alpha(x)=sxs^{-1}$ for each $x\in
H$. Now $\alpha(w)$ lies in $H^{\perp}$, and is of the form $\alpha(w)=uw$
with $u\in H$. As $\alpha$ preserves the norm, we have ${\mathrm{N}}(u)=1$,
and obtain $\alpha(xw)=\alpha(y)\,\alpha(w)=(sxs^{-1})(uw)=(usxs^{-1})w$, for
each $x\in H$. Putting $t:=us$ gives the description as claimed.

######  4.3 Corollary.

In any octonion algebra $\mathbb{O}$, two quaternion subalgebras are in the
same orbit under $\operatorname{Aut}_{F}(\mathbb{O})$ if, and only if, they
are isomorphic (as $F$-algebras). ∎

######  4.4 Lemma.

Let $\mathbb{O}$ be an octonion algebra over a field $F$. Then non-central
elements of $\mathbb{O}$ are in the same orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$ if, and only if, they have the same norm
and the same trace.

###### Proof.

Clearly elements in the same orbit have the same norm and trace.

Now consider $a_{1},a_{2}\in\mathbb{O}\smallsetminus F$ with
${\mathrm{N}}(a_{1})={\mathrm{N}}(a_{2})$ and
${\mathrm{T}}(a_{1})={\mathrm{T}}(a_{2})$. Applying Witt’s theorem, we find
$b_{j}\in a_{j}^{\perp}$ with ${\mathrm{T}}(b_{1})={\mathrm{T}}(b_{2})=1$ and
${\mathrm{N}}(b_{1})={\mathrm{N}}(b_{2})\neq 0$.

If ${\mathrm{N}}(a_{1})\neq 0$ then $H_{j}:=F+Fb_{j}+(F+Fb_{j})a_{j}$ is a
quaternion algebra, and the $F$-algebras $H_{1}$ and $H_{2}$ are isomorphic
because their norm forms are isometric (see [22, 1.7.1]). By 4.1(a), any
$F$-linear isomorphism from $H_{1}$ onto $H_{2}$ extends to an automorphism of
$\mathbb{O}$. If $\alpha\in\operatorname{Aut}_{F}(\mathbb{O})$ is such an
extension then the image $\alpha(a_{1})\in H_{2}$ has the same norm and trace
as $a_{2}$, and is thus a conjugate of $a_{2}$ in $H_{2}$. Adapting $\alpha$
by the extension of a suitable (inner) automorphism of $H_{2}$, we thus obtain
an $F$-linear automorphism of $\mathbb{O}$ mapping $a_{1}$ to $a_{2}$.

If ${\mathrm{N}}(a_{1})=0$ then $a_{j}$ is contained in a split quaternion
subalgebra $H_{j}$ (see 1.4(j)). Since any two split quaternion algebras over
$F$ are isomorphic, we can proceed as above. ∎

##  5\. Totally singular and totally isotropic subalgebras

In this section, we study subalgebras consisting of elements of norm $0$,
i.e., such that the underlying vector space is _totally singular with respect
to the quadratic form ${\mathrm{N}}$_. These are just those subalgebras that
do not contain $1$; see 2.4. Throughout this section, we write
$\mathbb{O}=\mathbb{H}+\mathbb{H}w$ as the double of the split quaternion
algebra $\mathbb{H}=F^{2\times 2}$ with an element $w$ of norm $-1$ (i.e.,
with $w^{2}=1$); then

$(a+xw)(b+yw)=ab+{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x+(ya+x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}})w$

holds for all $a,b,x,y\in F^{2\times 2}$; see 1.4.(i).

######  5.1 Lemma.

If $A$ is a subalgebra of $\mathbb{O}$ with $A\subseteq 1^{\perp}$ then $A$ is
totally isotropic (with respect to the polar form of the norm), and
$\dim{A}\leq 4$.

###### Proof.

For $a,b\in A$, we compute $(a|b)=(1|{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}b)=(1|ab)=0$,
using $ab\in A\subseteq 1^{\perp}$. ∎

Note that a subalgebra contained in $1^{\perp}$ need not be totally _singular_
if $\operatorname{char}{F}=2$; for instance, consider $F+Fn_{0}$, or a totally
inseparable field of degree $4$ and exponent $1$, over a field $F$ of
characteristic two.

######  5.2 Lemma.

Let $p$ be any element with ${\mathrm{N}}(p)=0$ but ${\mathrm{T}}(p)\neq 0$.
Then $\dim{A}\leq 2$ holds for every subalgebra contained in $p\mathbb{O}$ or
in $\mathbb{O}p$, and every two-dimensional algebra in $p\mathbb{O}$ (or in
$\mathbb{O}p$) contains $p$.

###### Proof.

Replacing $p$ by ${\mathrm{T}}(p)^{-1}p$ we do not change $p\mathbb{O}$. Using
4.4, we may therefore assume $p=\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)$. Then
$p\mathbb{O}=\left\\{{\left(\begin{smallmatrix}a_{1}&a_{2}\\\
0&0\end{smallmatrix}\right)+\left(\begin{smallmatrix}x_{1}&0\\\
x_{2}&0\end{smallmatrix}\right)w}\left|\vphantom{\left(\begin{smallmatrix}a_{1}&a_{2}\\\
0&0\end{smallmatrix}\right)+\left(\begin{smallmatrix}x_{1}&0\\\
x_{2}&0\end{smallmatrix}\right)wa_{1},a_{2},x_{1},x_{2}\in
F}\right.\,{a_{1},a_{2},x_{1},x_{2}\in F}\right\\}$.

Let $A$ be a subalgebra in $p\mathbb{O}$, and consider the product of two
arbitrary elements of $A$. Then

$\begin{array}[]{rl}\left(\left(\begin{matrix}a_{1}&a_{2}\\\
0&0\end{matrix}\right)+\left(\begin{matrix}x_{1}&0\\\
x_{2}&0\end{matrix}\right)w\right)\left(\left(\begin{matrix}b_{1}&b_{2}\\\
0&0\end{matrix}\right)+\left(\begin{matrix}y_{1}&0\\\
y_{2}&0\end{matrix}\right)w\right)\\\\[4.30554pt]
=\left(\begin{matrix}a_{1}b_{1}&a_{1}b_{2}\\\
y_{1}x_{2}-y_{2}x_{1}&0\end{matrix}\right)+\left(\begin{matrix}y_{1}a_{1}&y_{1}a_{2}-x_{1}b_{2}\\\
y_{2}a_{1}&y_{2}a_{2}-x_{2}b_{2}\end{matrix}\right)w&\in
p\mathbb{O}\end{array}$

yields the conditions $y_{1}x_{2}-y_{2}x_{1}=0$, $y_{1}a_{2}-x_{1}b_{2}=0$,
and $y_{2}a_{2}-x_{2}b_{2}=0$. This means that the matrix

$\left(\begin{matrix}a_{2}&x_{1}&x_{2}\\\ b_{2}&y_{1}&y_{2}\\\
\end{matrix}\right)$

has rank at most one. So the quotient space $(A+Fp)/Fp$ has dimension at most
one.

The assertions about $\mathbb{O}p$ are obtained by applying the anti-
automorphism $\kappa$. ∎

The maximal totally singular subspaces in $\mathbb{O}$ (with respect to the
quadratic form ${\mathrm{N}}={\mathrm{N}}_{\mathbb{O}}$) are just the sets
$a\mathbb{O}$ and $\mathbb{O}a$, with $a\in\mathbb{O}\smallsetminus\\{0\\}$
such that ${\mathrm{N}}(a)=0$, and one knows
$a\mathbb{O}=\ker(\lambda_{{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}})$ and
$\mathbb{O}a=\ker(\rho_{{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}})$ (see
3.6(a), (b)).

We characterize the subalgebras among these subspaces:

######  5.3 Lemma.

Let $a\in\mathbb{O}\smallsetminus\\{0\\}$ with ${\mathrm{N}}(a)=0$. Then the
following are equivalent.

  1.   2. ((a))

$a\mathbb{O}$ is a subalgebra.

  3. ((b))

$\mathbb{O}a$ is a subalgebra.

  4. ((c))

${\mathrm{T}}(a)=0$.

  5. ((d))

$a^{2}=0\neq a$.

  6. ((e))

After identification of $\mathbb{O}$ with a double of $F^{2\times 2}$, we have
$a$ in the $\operatorname{Aut}_{F}(\mathbb{O})$-orbit of $n_{0}$.

If one (and then any) of these conditions is satisfied then the algebras
$a\mathbb{O}$ and $\mathbb{O}a$ are anti-isomorphic (via $\kappa$) but not
isomorphic.

These subalgebras have dimension $4$ and contain no invertible elements at
all. None of these subalgebras is associative (see 2.3), and none of them is
contained in $1^{\perp}$.

###### Proof.

Since the restriction ${\mathrm{N}}|_{1^{\perp}}$ is a non-degenerate
quadratic form, each totally singular subspace of $1^{\perp}$ has dimension at
most $3$. Therefore, we have $a\mathbb{O}\not\subseteq 1^{\perp}$ for each
$a\neq 0$. If ${\mathrm{T}}(a)\neq 0$ we apply 5.2 to see that neither
$a\mathbb{O}$ nor $\mathbb{O}a$ is a subalgebra.

So assume ${\mathrm{T}}(a)=0$ now. Then
$\ker(\lambda_{a})=\ker(\lambda_{{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}})=a\mathbb{O}$.
Using $a^{2}=-{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}a=0$ and one
of Moufang’s identities (see 1.4(d)) we verify that
$(a\mathbb{O})(a\mathbb{O})$ is contained in $\ker(\lambda_{a})$ by computing
$a\left((ax)(ay)\right)=\left((a(ax))a\right)y=\left((a^{2}x)a\right)y=0$. So
$a\mathbb{O}$ is a subalgebra. Applying the anti-automorphism $\kappa$ we
translate this observation to $\mathbb{O}a$. Moreover, we see that
$a\mathbb{O}$ and $\mathbb{O}a$ are anti-isomorphic if
${\mathrm{T}}(a)=0={\mathrm{N}}(a)$.

It remains to show that the algebras $a\mathbb{O}$ and $\mathbb{O}a$ are not
isomorphic. To this end, we note that $a$ is a non-trivial element of
$a\mathbb{O}$ such that $a(a\mathbb{O})=\\{0\\}$ while no element $b$ of
$a\mathbb{O}\smallsetminus\\{0\\}$ satisfies $(a\mathbb{O})b=\\{0\\}$; in
fact, the kernel $\mathbb{O}{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{b}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{b}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{b}\hfil$\crcr}}}}$ of
$\rho_{b}$ never coincides with $a\mathbb{O}$ (see 3.6(c)). ∎

######  5.4 Corollary.

Not every maximal totally singular subspace is a subalgebra of $\mathbb{O}$.
The group $\operatorname{Aut}(\mathbb{O})$ is not transitive on the set of all
maximal totally singular subspaces. ∎

######  5.5 Lemma.

If $A$ is a subalgebra of $\mathbb{O}$ containing an element $e$ that is a
neutral element for the multiplication in $A$ then either $e\in\\{0,1\\}$, or
$\mathbb{O}$ is split and $e$ belongs to the unique
$\operatorname{Aut}_{F}(\mathbb{O})$-orbit of elements of trace $1$ and norm
$0$. In the latter case, we have $A=Fe$.

###### Proof.

In any case, the subspace $F+Fe$ is an associative and commutative subalgebra
of $\mathbb{O}$. Assume $e\notin\\{0,1\\}$.

The element $e$ is an idempotent, and centralizes $A$. From $e^{2}=e$ and
$e\notin\\{0,1\\}$ we infer that the minimal polynomial of $e$ in $F+Fe$ is
$X^{2}-X$. Comparing this with the general form of the minimal polynomial
$X^{2}-{\mathrm{T}}(a)X+{\mathrm{N}}(a)$ for any element $a\in\mathbb{O}$, we
find ${\mathrm{T}}(e)=1$ and ${\mathrm{N}}(e)=0$. This implies $e\notin F$,
and $e$ is a non-trivial element of norm $0$ in $\mathbb{O}$. In particular,
the algebra $\mathbb{O}$ is split. From 4.4 we know that the non-central
elements of given norm and trace form a single orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$.

Now $A$ is contained in the centralizer of the element $e$ of norm $0$ and
trace $1$. We compute ${\mathrm{N}}(e-{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{e}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{e}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{e}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{e}\hfil$\crcr}}}})=-1$ and
use 3.1 to obtain $\dim\operatorname{C}_{\mathbb{O}}(e)=2$. This yields
$A\leq\operatorname{C}_{\mathbb{O}}(e)=F+Fe$, and $A=Fe$ follows because
$A=Ae\leq(F+Fe)e=Fe$. ∎

######  5.6 Proposition.

Let $A$ be a $3$-dimensional subalgebra of $\mathbb{O}$, and assume
$A\subseteq 1^{\perp}$. Then $\mathbb{O}$ is split; we write
$\mathbb{O}=F^{2\times 2}+F^{2\times 2}w$ as double of the split quaternion
algebra $F^{2\times 2}$, with $w^{2}=1$.

  1. ((a))

If $\operatorname{char}{F}\neq 2$ then $A$ is a Heisenberg algebra.

  2. ((b))

If $\operatorname{char}{F}=2$ then either $A$ is a Heisenberg algebra, or
$A=F+V$ with a two-dimensional subalgebra $V$ such that $v^{\prime}v=0$ holds
for all $v^{\prime},v\in V$.

  3. ((c))

The Heisenberg algebras form a single orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$, represented by
${n_{0}\mathbb{O}\cap\mathbb{O}n_{0}}$.

  4. ((d))

The subalgebras of the form $F+V$ with a two-dimensional subalgebra $V$ with
trivial multiplication form a single orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$, represented by ${F+Fn_{0}+Fn_{0}w}$.

###### Proof.

In each of the following cases, we will see that $A\smallsetminus\\{0\\}$
contains elements of norm $0$. So $\mathbb{O}$ is split.

Assume first that $\operatorname{char}{F}\neq 2$. From 5.1 we then know that
$A$ is totally isotropic, and infer that $A$ is totally singular. If there are
$a,b\in A$ such that $ab\neq 0$ then $A={Fa+Fb+Fab}$ is a Heisenberg algebra,
see 2.7. So assume that $ab=0$ for all $a,b\in A$. Up to an automorphism of
$\mathbb{O}$ (see 4.4), we may assume that
$n_{0}=\left(\begin{smallmatrix}0&1\\\ 0&0\end{smallmatrix}\right)$ lies in
$A$. Then $A$ is contained in the intersection of the left annihilator
$\left\\{{c+xw}\left|\vphantom{c+xwcn_{0}=0,x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{n_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{n_{0}}\hfil$\crcr}}}}=0}\right.\,{cn_{0}=0,x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{n_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{n_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{n_{0}}\hfil$\crcr}}}}=0}\right\\}=F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0}+(F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0})w=\mathbb{O}n_{0}$
with the right annihilator
$\left\\{{c+xw}\left|\vphantom{c+xwn_{0}c=0,x{n_{0}}=0}\right.\,{n_{0}c=0,x{n_{0}}=0}\right\\}=F{p_{0}}+Fn_{0}+(F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0})w=n_{0}\mathbb{O}$,
see 3.6(a). That intersection is the $3$-dimensional subspace
$n_{0}\mathbb{O}\cap\mathbb{O}n_{0}=Fn_{0}+(F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0})w$,
and must coincide with $A$. However, the product
$(n_{0}w)({\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}w)=p_{0}n_{0}=n_{0}$
is not zero. This contradiction shows that $A$ is a Heisenberg algebra.

If $\operatorname{char}{F}=2$ then $1$ lies in $1^{\perp}$, and may thus be a
member of $A$. In that case, the restriction of ${\mathrm{N}}$ to $A$ cannot
be non-singular because otherwise $A$ would contain an extension field of
degree two, and $\dim_{F}A$ would be even. If $1\notin A$ then
${\mathrm{N}}(A)=\\{0\\}$ anyway, see 2.4. So we may assume $n_{0}\in A$ (see
4.4). We also note that $A\subseteq 1^{\perp}$ is commutative if
$\operatorname{char}{F}=2$ because $xy={\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{xy}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{xy}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{xy}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{xy}\hfil$\crcr}}}}={\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}=yx$ holds
for all $x,y\in A$. If $1\notin A$ then ${\mathrm{N}}(A)=\\{0\\}$, and we
invoke 2.7 as above to obtain that $A$ is a Heisenberg algebra. If $1\in A$
then $A\subseteq\operatorname{C}_{\mathbb{O}}(n_{0})=F+Fn_{0}+F^{2\times 2}w$,
and there exists $y\in F^{2\times 2}\smallsetminus\\{0\\}$ with $yn_{0}\in Fy$
such that $A=F+Fn_{0}+Fyw$. Then $y\in
Fn_{0}+F{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}$, and
we obtain $0=n_{0}^{2}=(yw)^{2}=n_{0}(yw)=(yw)n_{0}$. In particular, the
subspace $Fn_{0}+Fyw$ forms a subalgebra with trivial multiplication.

It remains to show that the isomorphism type of the subalgebras in question
determines their orbit under $\operatorname{Aut}_{F}(\mathbb{O})$. If $A$ is a
Heisenberg algebra, we may (up to an automorphism of $\mathbb{O}$) assume that
$n_{0}$ lies in $A$, and that $n_{0}A=\\{0\\}=An_{0}$. We then find that $A$
coincides with the two-sided annihilator
$n_{0}\mathbb{O}\cap\mathbb{O}n_{0}=Fn_{0}+(F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0})w$,
see 3.6(a). Therefore, these Heisenberg algebras are in a single orbit.

Finally, consider a two-dimensional subalgebra $V$ with trivial
multiplication. Every element of $V\smallsetminus\\{0\\}$ is in the
$\operatorname{Aut}_{F}(\mathbb{O})$-orbit of $n_{0}$, we may (and will)
assume $n_{0}\in V$. Then $V=Fn_{0}+Fy$, where $y$ belongs to the two-sided
annihilator
$n_{0}\mathbb{O}\cap\mathbb{O}n_{0}={Fn_{0}+(F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0}})w$
of $n_{0}$. If $n_{0}\in V\not\subseteq Fn_{0}+Fn_{0}w$ then $V$ contains an
element of the form $({\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+rn_{0})w$,
with $r\in F$. Using $s=1$ and $t:=\left(\begin{smallmatrix}0&1\\\
-1&r\end{smallmatrix}\right)\in\mathbb{H}$, we obtain an automorphism
$\alpha_{1,t}\colon a+xw\mapsto a+(tx)w$ of $\mathbb{O}$ with
$\alpha_{1,t}(n_{0})=n_{0}$ and
$\alpha_{1,t}(({\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+rn_{0})w)=n_{0}w$,
as required. ∎

######  5.7 Lemma.

Let $p,n$ be two non-trivial elements (necessarily of norm $0$) with
$p^{2}=p\neq 1$, $pn=n$, $n^{2}=0=np$. Then $Fp+Fn$ is a subalgebra. These
subalgebras form a single orbit under $\operatorname{Aut}_{F}(\mathbb{O})$.

###### Proof.

The vector subspace $E:=F+Fp$ is a (split) composition algebra. The subspace
$\\{1,p,n\\}^{\perp}$ has dimension $5$, and contains an element $u$ with
${\mathrm{N}}(u)\neq 0$. So $H:=E+Eu$ is a (split) quaternion algebra. We note
that $(pu|n)=(u|{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}n)=(u|(1-p)n)=(u|0)=0$,
and infer that $n$ lies in $H^{\perp}$.

From 4.1 we know that there is $\alpha\in\operatorname{Aut}_{F}(\mathbb{O})$
with $\alpha(H)=\mathbb{H}$ (${}=F^{2\times 2}$). Then $\alpha(p)$ is a
conjugate of $p_{0}:=\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)$ in $\mathbb{H}$, and the corresponding inner
automorphism of $\mathbb{H}$ extends to an automorphism $\beta$ of
$\mathbb{O}$. So we may assume $p=p_{0}$ and $H=\mathbb{H}$, while
$n\in\mathbb{H}^{\perp}$. Then there exists $y\in\mathbb{H}$ such that $n=yw$,
and $0=np=(yw)p=(y{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}})w$ yields
$y{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}=0$. This
means that $y=\left(\begin{smallmatrix}y_{1}&0\\\
y_{2}&0\end{smallmatrix}\right)$. There exists
$s\in\mathrm{SL}_{2}{(F)}\subseteq\mathbb{H}$ such that
$s\binom{y_{1}}{y_{2}}=\binom{1}{0}$, and $a+xw\mapsto a+(sx)w$ is an
automorphism of $\mathbb{O}$ mapping $n=yw$ to $p_{0}w$. ∎

######  5.8 Example.

The elements $p_{0}:=\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)$ and $n_{0}:=\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}\right)$ in $F^{2\times 2}$ satisfy the assumptions in
5.7; they generate the subalgebra
$L:=Fp_{0}+Fn_{0}=\left\\{{\left(\begin{smallmatrix}r&s\\\
0&0\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}r&s\\\
0&0\end{smallmatrix}\right)r,s\in F}\right.\,{r,s\in F}\right\\}$ of
$F^{2\times 2}$.

This subalgebra $L$ is not invariant under the standard involution; its image
under $\kappa$ is ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{L}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{L}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{L}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{L}\hfil$\crcr}}}}={\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{Fp_{0}+Fn_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{Fp_{0}+Fn_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{Fp_{0}+Fn_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{Fp_{0}+Fn_{0}}\hfil$\crcr}}}}=\left\\{{\left(\begin{smallmatrix}0&s\\\
0&r\end{smallmatrix}\right)}\left|\vphantom{\left(\begin{smallmatrix}0&s\\\
0&r\end{smallmatrix}\right)r,s\in F}\right.\,{r,s\in F}\right\\}$. The
algebras $L$ and  $\hfil\textstyle{L}$ are not isomorphic.

######  5.9 Lemma.

Let $m,n$ be two non-trivial elements of $1^{\perp}$ with
${\mathrm{N}}(m)=0={\mathrm{N}}(n)$ and $mn=0$. Then $nm=0$, and
$Fm+Fn=m\mathbb{O}\cap n\mathbb{O}$ is a subalgebra of dimension $2$. These
subalgebras form a single orbit under $\operatorname{Aut}_{F}(\mathbb{O})$,
and for any such algebra $A$, the stabilizer of $A$ in
$\operatorname{Aut}_{F}(\mathbb{O})$ acts two-transitively on the set of one-
dimensional subspaces of $A$.

###### Proof.

We know from 5.3 that $m\mathbb{O}$ and $n\mathbb{O}$ are subalgebras, so
$m\mathbb{O}\cap n\mathbb{O}$ is a subalgebra, as well. As $m,n$ lie in
$1^{\perp}$, we have $nm={\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{mn}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{mn}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{mn}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{mn}\hfil$\crcr}}}}=0$.

Using 4.4, we may assume $n=n_{0}:=\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}\right)\in\mathbb{H}$. There are $b,y\in\mathbb{H}$ such
that $m=b+yw$, and $0=mn_{0}=bn_{0}-(y{n_{0}})w$ yields $bn_{0}=0$ and
$yn_{0}=0$. On the other hand, we have $0=n_{0}m=n_{0}b+(yn_{0})w$, and obtain
$n_{0}b=0$, as well. This yields $b\in Fn_{0}$ and
$y=\left(\begin{smallmatrix}0&y_{1}\\\ 0&y_{2}\end{smallmatrix}\right)$ with
$y_{1},y_{2}\in F$. Now $Fm+Fn_{0}=F(m-b)+Fn_{0}$, there exists
$s\in\mathrm{SL}_{2}{(F)}\subseteq\mathbb{H}$ such that $sy=n_{0}$, and
$a+xw\mapsto a+(sx)w$ is an automorphism of $\mathbb{O}$ mapping $m-b=yw$ to
$n_{0}w$.

For the last assertion, it now suffices to consider
$A=\left\\{{\left(\begin{smallmatrix}r&s\\\
0&0\end{smallmatrix}\right)w}\left|\vphantom{\left(\begin{smallmatrix}r&s\\\
0&0\end{smallmatrix}\right)wr,s\in F}\right.\,{r,s\in F}\right\\}$. The
stabilizer of $A$ in $\operatorname{Aut}_{F}(\mathbb{O})$ contains the maps
$a+xw\mapsto s^{-1}as+(xs)w$, where $s\in\mathrm{SL}_{2}{(F)}<\mathbb{H}$.
Clearly, these maps form a group that acts two-transitively on the set of one-
dimensional subspaces of $A$, as claimed. ∎

######  5.10 Corollary.

The group $\operatorname{Aut}_{F}(\mathbb{O})$ acts transitively on the set of
all pairs $(Fm,Fn)$ of different subspaces spanned by non-trivial elements
$m$, $n$ in $\mathbb{O}$ with
${\mathrm{N}}(m)={\mathrm{N}}(n)={\mathrm{T}}(m)={\mathrm{T}}(n)=0$ and
$mn=0$. ∎

######  5.11 Totally singular subalgebras.

Let $A$ be a subalgebra of $\mathbb{O}$, and assume $1\notin A$. Then $A$ is
totally singular, and $\dim{A}\leq 4$. In order to give representatives for
the orbits of subalgebras, we write $\mathbb{O}=\mathbb{H}+\mathbb{H}w$ as the
double of the split quaternion algebra $\mathbb{H}:=F^{2\times 2}$ with $w$
such that $w^{2}=1$, and use $p_{0}:=\left(\begin{smallmatrix}1&0\\\
0&0\end{smallmatrix}\right)$ and $n_{0}:=\left(\begin{smallmatrix}0&1\\\
0&0\end{smallmatrix}\right)$.

  1. ((a))

If $\dim{A}=1$ then $A=Fa$ with $a\in\mathbb{O}$ such that
${\mathrm{N}}(a)=0$. These subalgebras form two orbits under
$\operatorname{Aut}_{F}(\mathbb{O})$, represented, respectively, by $Fp_{0}$,
and by $Fn_{0}$. Note that $F$ and $Fp_{0}$ are isomorphic as algebras, under
an isomorphism that maps $1\in F$ to
$p_{0}\in\mathbb{O}\smallsetminus\\{1\\}$.

  2. ((b))

Totally singular subalgebras of dimension $2$ form three orbits under
$\operatorname{Aut}_{F}(\mathbb{O})$:

    1. ((i))

Algebras with a left identity for the multiplication, such as $Fp_{0}+Fn_{0}$
(see 5.7).

    2. ((ii))

Algebras with a right identity for the multiplication, obtained by applying
some anti-automorphism (such as $\kappa$) to algebras with left identity (see
5.7).

    3. ((iii))

Algebras with trivial multiplication, such as $(Fp_{0}+Fn_{0})w$ (see 5.9).

Each one of the algebras in the latter orbit is contained in $1^{\perp}$,
while none of those in the first two of these orbits is contained in
$1^{\perp}$.

  3. ((c))

Totally singular subalgebras of dimension $3$ form two orbits under
$\operatorname{Aut}_{F}(\mathbb{O})$, namely:

    1. ((i))

Heisenberg algebras $Fx+Fy+Fn$ with $xy=n=-yx$, obtained as
$n\mathbb{O}\cap\mathbb{O}n$ for $n\in\mathbb{O}$ such that $n^{2}=0\neq n$.
These subalgebras are contained in $1^{\perp}$.

    2. ((ii))

Algebras of the form $m\mathbb{O}\cap\mathbb{O}n$ where $m,n\in\mathbb{O}$ are
linearly independent, with $m^{2}=0=n^{2}=0=mn=nm$. None of these subalgebras
is contained in $1^{\perp}$.

  4. ((d))

Totally singular subalgebras of dimension $4$ form two orbits under
$\operatorname{Aut}_{F}(\mathbb{O})$, namely:

    1. ((i))

$n\mathbb{O}$ for $n\in\mathbb{O}$ such that $n^{2}=0\neq n$.

    2. ((ii))

$\mathbb{O}n$ for $n\in\mathbb{O}$ such that $n^{2}=0\neq n$.

In particular, two totally singular subalgebras of $\mathbb{O}$ are isomorphic
if, and only if, they are in the same orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$.

The totally singular subalgebras of dimension at most $3$ are associative. The
totally singular subalgebras of dimension $4$ are not associative.

###### Proof.

By 2.4, our assumption $1\notin A$ implies ${\mathrm{N}}(A)=\\{0\\}$, and $A$
is totally singular. As the quadratic form ${\mathrm{N}}$ is not degenerate,
we have $\dim{A}\leq\frac{1}{2}\dim\mathbb{O}=4$.

If $\dim{A}=1$ then $A=Fa$ with $a\in\mathbb{O}$ with ${\mathrm{N}}(a)=0$. If
${\mathrm{T}}(a)\neq 0$ then ${\mathrm{T}}(a)^{-1}a$ has trace $1$, and lies
in the orbit of $p_{0}$ under $\operatorname{Aut}_{F}(\mathbb{O})$ by 4.4. If
${\mathrm{T}}(a)=0$ then $a$ lies in the orbit of $n_{0}$ under
$\operatorname{Aut}_{F}(\mathbb{O})$.

Now assume that $A=Fx+Fy$ has dimension $2$. Then $A$ is associative, and
there are $r,s\in F$ such that $xy=rx+sy$. Without loss, we may assume $y\in
1^{\perp}$. Then $y^{2}=0$, and $0=xy^{2}=(xy)y=rxy+sy^{2}=rxy$ yields $xy\in
Fy$.

If $A\subseteq 1^{\perp}$, we also have $x^{2}=0$, and $xy=0$ follows. In this
case, the algebra $A$ is in the orbit of $(Fp+Fn)w=Fpw+Fnw$ under
$\operatorname{Aut}_{F}(\mathbb{O})$, see 5.9.

If $A\not\subseteq 1^{\perp}$, we may assume ${\mathrm{T}}(x)=1$. Then
$x^{2}=x$ yields $sy=xy=x(xy)=s^{2}y$, and $s\in\\{0,1\\}$ follows. From
$0=(x|y)={\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}y+{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x=(1-x)y-yx$
we then infer $yx=y-xy=y-sy$. Thus $A$ is in the orbit of $Fp+Fn$ if $s=1$,
and in the orbit of $F{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p}\hfil$\crcr}}}}+Fn$ if
$s=0$, see 5.9.

Now assume $\dim{A}=3$. Then there are $x,y\in\mathbb{O}$ with
${\mathrm{N}}(x)=0={\mathrm{N}}(y)$ and $xy=0$ such that
$A=x\mathbb{O}\cap\mathbb{O}y$. From 5.2 we infer $x,y\in 1^{\perp}$.

If $Fx=Fy$ then $A$ is in the orbit of $n_{0}\mathbb{O}\cap\mathbb{O}n_{0}$ by
4.4, and
$n_{0}\mathbb{O}\cap\mathbb{O}n_{0}=Fn_{0}+({F{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{p_{0}}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{p_{0}}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{p_{0}}\hfil$\crcr}}}}+Fn_{0}})w\subseteq
1^{\perp}$. If $Fx\neq Fy$ then 5.10 yields that $A$ is in the orbit of
$(p_{0}w)\mathbb{O}\cap\mathbb{O}(n_{0}w)=Fp_{0}+(Fp_{0}+Fn_{0})w$, cp.
3.7(a). That subalgebra contains $p_{0}\notin 1^{\perp}$. We know from 3.7
that these subalgebras are associative.

Finally, assume $\dim{A}=4$. Then either $A=x\mathbb{O}$ or $A=\mathbb{O}x$
holds for some $x\in\mathbb{O}$ with ${\mathrm{N}}(x)=0$, and 5.2 yields
${\mathrm{T}}(x)=0$. Thus either $A$ is in the orbit of $n_{0}\mathbb{O}$, or
$A$ is in the orbit of $\mathbb{O}n_{0}$, see 4.4. We know from 2.3 that these
subalgebras are not associative. ∎

##  6\. Subalgebras that are not totally singular

######  6.1 Lemma.

Assume that $A$ is a subalgebra of $\mathbb{O}$. Consider the radicals
$R:=A^{\perp}\cap A$ of the restriction of the polar form, and
$Q:=\left\\{{x\in R}\left|\vphantom{x\in
R{\mathrm{N}}(x)=0}\right.\,{{\mathrm{N}}(x)=0}\right\\}$ of the restriction
${\mathrm{N}}|_{A}$ of the quadratic form, respectively.

  1. ((a))

We have $AR\subseteq R$, and $RA\subseteq R$.

  2. ((b))

In any case, both $Q$ and $R$ are subalgebras of $\mathbb{O}$.

  3. ((c))

If the restriction of the polar form to $A$ is not trivial then $R$ is totally
singular. So $R=Q$, and $A/R$ is a composition algebra.

  4. ((d))

If the restriction of the polar form to $A$ is trivial then $A=R$. So either
$A=Q$ is totally singular, or $R\neq Q$. In the latter case, we have
$\operatorname{char}{F}=2$, the algebra $A$ is commutative, and contains $F$.

###### Proof.

If $A$ contains no invertible element, then the restriction of the norm form
to $A$ is zero, and so is the restriction of the polar form. Then $R=A$ and
assertion (a) is trivial.

So assume that $A$ contains invertible elements; then $1\in A$ and $A$ is
closed under $\kappa$ by 1.6. For $x\in R$ and $a,b\in A$, we compute
$(ax|b)=(x|{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{a}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{a}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{a}\hfil$\crcr}}}}b)=0$. So
$ax\in R$, and $AR\subseteq R$ follows. The argument for $RA\subseteq R$ is
completely symmetric to the one just given. So assertion (a) is established,
and we know that $R$ is as subalgebra.

In any case, the set $Q$ is closed under multiplication. If
$\operatorname{char}{F}\neq 2$ then $Q=R$. If $\operatorname{char}{F}=2$ then
the restriction ${\mathrm{N}}|_{R}\colon R\to F$ is a semi-linear map (with
respect to the Frobenius endomorphism), and $Q$ is the kernel of that map. So
$Q$ is a vector subspace, and then a subalgebra.

Now assume that the restriction of the polar form to $A$ is not trivial. We
show first that $R$ consists of elements of norm $0$; i.e., $R=Q$. By our
assumption, there exist $a,b\in A$ such that $(a|b)\neq 0$. For $x\in R$ we
use ${\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}},b{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}},b{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}x\in R$ and
compute $0=(a|b{\mathchoice{\vbox{\halign{#\cr\kern 1.0pt\cr\kern
1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}x)=(a|b)x{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}$. This
gives ${\mathrm{N}}(x)=0$, as claimed.

Using assertion (a), it is straightforward to check (by the usual arguments
known for associative algebras) that addition and multiplication on $A/R$ are
well-defined by $(a+R)(b+R)=(a+b)+R$, and $(a+R)(b+R)=(ab)+R$. The induced
norm $M(a+R):={\mathrm{N}}(a)$ is well-defined, and multiplicative on $A/R$.
The polar form of $M$ is non-degenerate, and $1\in A$ follows from the fact
that the subalgebra $A$ is not totally singular (see 2.4). So indeed $A/R$ is
a composition algebra.

Finally, assume that the restriction of the polar form to $A$ is trivial,
i.e., $R=A$. If $Q\neq R$ then $R$ contains an invertible element $b$. Then
$0=(b|b)=2{\mathrm{N}}(b)$ yields $\operatorname{char}{F}=2$, and
$b^{2}={\mathrm{N}}(b)\in(F\cap R)\smallsetminus\\{0\\}$ shows $F\subseteq R$.
For $x,y\in R$, we now have $0=(x|y)={\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{x}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{x}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{x}\hfil$\crcr}}}}y+{\mathchoice{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\displaystyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\textstyle{y}$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptstyle{y}\hfil$\crcr}}}{\vbox{\halign{#\cr\kern
1.0pt\cr\kern 1.0pt\hrulefill\crcr\kern
1.0pt\nointerlineskip\cr$\hfil\scriptscriptstyle{y}\hfil$\crcr}}}}x=-xy-yx$,
and use $\operatorname{char}{F}=2$ to infer that $R$ is commutative. ∎

######  6.2 Subalgebras of dimension greater than four.

Let $A$ be a subalgebra of $\mathbb{O}$. If $\dim{A}>4$ then either
$A=\mathbb{O}$, or $\mathbb{O}$ is split and one of the following cases
occurs:

  1. ((a))

$\dim{A}=5$, and there exists a nilpotent element $n$ such that
$A=n\mathbb{O}+\mathbb{O}n$. These subalgebras form a single orbit under
$\operatorname{Aut}_{F}(\mathbb{O})$.

  2. ((b))

$\dim{A}=6$, and there exist two linearly independent nilpotent elements $m,n$
with $mn=0$ such that $A=\\{m,n\\}^{\perp}$. Then
$A=m\mathbb{O}+n\mathbb{O}=\mathbb{O}m+\mathbb{O}n$. These subalgebras form a
single orbit under $\operatorname{Aut}_{F}(\mathbb{O})$.

###### Proof.

We know that $1\in A$ because $\dim{A}>4$ implies that $A$ is not totally
singular. From 5.1 we know $A\not\subseteq 1^{\perp}$. We show first that
there is a two-dimensional subalgebra $B\subseteq A$ such that the restriction
of the polar form to $B$ is not degenerate; then $B$ is a composition algebra.

If $\operatorname{char}{F}=2$ we just pick $b\in A$ such that $b\notin
1^{\perp}$. Then $b\notin F\subseteq 1^{\perp}$, and $B:=F+Fb\subseteq A$ is a
two-dimensional composition algebra, as required.

If $\operatorname{char}{F}\neq 2$ we first note that $1^{\perp}\cap A$ is not
totally singular. In fact, that intersection has at least dimension $4$. The
restriction of the polar form to $1^{\perp}$ is not degenerate (here we use
$\operatorname{char}{F}\neq 2$), and does not contain any totally isotropic
subspace of dimension greater than $3$. So there exists $a\in 1^{\perp}\cap A$
such that ${\mathrm{N}}(a)\neq 0$, and $B:=F+Fa\subseteq A$ is a composition
algebra, as required.

Now consider $B^{\perp}\cap A$. If there exists an invertible element $u\in
B^{\perp}\cap A$ then $H:=B+Bu$ is a quaternion subalgebra in $A$. If there is
still an invertible element $v\in H^{\perp}\cap A$ then $\mathbb{O}=H+Hv$
coincides with $A$. So we may assume that $H^{\perp}\cap A$ consists of
nilpotent elements. Then the polar form vanishes on that intersection, and we
find that $H^{\perp}\cap A$ equals the radical $R:=A^{\perp}\cap A$ of $A$.
Choose $w\in H^{\perp}$ with ${\mathrm{N}}(w)=-1$. Then $\mathbb{O}=H+Hw$ is
obtained by doubling, and $R\subseteq Hw$. For $a,x\in H$, we find
$a(xw)=(xa)w$, and $R$ is a non-trivial module over the opposite algebra of
$H\cong F^{2\times 2}$. This yields $\dim{R}\geq 2$, and any two-dimensional
subspace $L\subseteq R$ gives $A\leq R^{\perp}\leq L^{\perp}$. So
$A=L^{\perp}$ and $L=R$ hold for dimension reasons. The multiplication formula
1.4.(i) shows $H^{\perp}H^{\perp}\subseteq H$, and 6.1(a) yields $RR\subseteq
R\cap H\subseteq H^{\perp}\cap H=\\{0\\}$. So the multiplication in $R$ is
trivial, and $R$ is in the orbit of $(Fp_{0}+Fn_{0})w$, see 5.9 and 5.11. We
obtain that $A$ is in the orbit of
$\left((Fp_{0}+Fn_{0})w\right)^{\perp}=H+(Fp_{0}+Fn_{0})w$, which is a
subalgebra by 2.1 (see 2.2).

It remains to consider the case where $B^{\perp}\cap A$ consists of nilpotent
elements. Then the norm vanishes on that subspace, and $Q:=B^{\perp}\cap A$ is
the radical of the quadratic form ${\mathrm{N}}|_{A}$. As $Q$ is a totally
singular subspace of $1^{\perp}$, we have $\dim{Q}\leq 3$ and then $\dim{Q}=3$
and $\dim{A}=5$. The subspace $Q$ forms a totally singular subalgebra of
dimension $3$, and is contained in $1^{\perp}$. From 5.11 we know that $Q$ is
in the orbit of $n_{0}\mathbb{O}\cap\mathbb{O}n_{0}$, and $A=Q^{\perp}$ is in
the orbit of
$(n_{0}\mathbb{O}\cap\mathbb{O}n_{0})^{\perp}=n_{0}\mathbb{O}+\mathbb{O}n_{0}$.
∎

######  6.3 Lemma.

Let $A$ be a $3$-dimensional subalgebra with $1\in A$ and $\dim{R}\geq 2$.
Then $\dim{Q}\geq 2$.

###### Proof.

We note first of all that $A$ does not contain any $2$-dimensional vector
subspace $V$ such that the restriction ${\mathrm{N}}|_{V}$ of the norm form is
non-singular: for any $v\in V\smallsetminus\\{0\\}$, the subspace $v^{-1}V$
would be a subalgebra without zero divisors (i.e., a quadratic extension field
of $F$), and the dimension of $A$ would be even.

If $\dim{Q}<2$ then $Q\neq R$, and $\operatorname{char}{F}=2$ follows.
Consider $y\in R\smallsetminus Q$. If $y\in R$ then $A\subseteq 1^{\perp}$,
and $A=R$ holds by 5.1. If $y\notin F$ then $\operatorname{char}{F}=2$ yields
$1\perp 1$, and we obtain $A=R$, again. Now each singular element of $A$ lies
in $Q$. We pick a vector space complement $V$ for $Q$ in $A$ and reach the
contradiction that $N|_{V}$ is non-singular. ∎

After 5.11 and 6.2, it remains to treat subalgebras that are not totally
singular (so they contain $1$) and have dimension at most $4$.

######  6.4 Unary subalgebras of dimension at most four.

Let $A$ be a subalgebra of an octonion algebra $\mathbb{O}$, with $1\in A$. If
$\mathbb{O}$ is split, we write $\mathbb{O}=\mathbb{H}+\mathbb{H}w$, with