Albert's twisted field construction using division algebras with a multiplicative norm
S. Pumpl\"un
Albert's classical construction of twisted fields generates unital division algebras out of cyclic field extensions. We apply a generalized version to unital division algebras with a multiplicative norm and give conditions for the resulting twisted algebras to be division. Four- and eight-dimensional real unital division algebras with large derivation algebras are constructed out of Hamilton's quaternion and Cayley's octonion algebra.
1504.00188.tex
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\keywords{Real division algebras, Albert's twisted fields, composition algebra.}


\date{19.9.2014}
\author{S. Pumpl\"un}
\email{susanne.pumpluen@nottingham.ac.uk}
\address{School of Mathematical Sciences\\
University of Nottingham\\
University Park\\
Nottingham NG7 2RD\\
United Kingdom
}

\begin{document}

\title[Twisted algebras]
{Albert's twisted field construction using division algebras with a multiplicative norm}
\maketitle


\begin{abstract}
 Albert's classical construction of twisted fields generates unital division algebras out of cyclic field extensions.
 We apply a generalized version to unital division algebras with a multiplicative norm and
 give conditions for the resulting twisted algebras to be division.
  Four- and eight-dimensional real unital division
algebras with large derivation algebras are constructed out of Hamilton's quaternion and Cayley's octonion algebra.
\end{abstract}


\section*{Introduction}


Albert's classical construction resulting in twisted finite semifields \cite{A1}, \cite{A2}
 is a well-known tool to build $n$-dimensional unital nonassociative division algebras out of an $n$-dimensional cyclic
 field extension $K/F$  with Galois group
  ${\rm Gal}(K/F)=\langle \sigma\rangle$. For any choice of $c\in K$ such that $N_{K/F}(c)\not=1$ and $1\leq i,j< n$,
   $K$ equipped with  the new multiplication
  $$x\circ y=xy-c\sigma^i(x)\sigma^j(y),$$
 is a division algebra over $F$. For finite base fields $F$, Kaplanski's trick is then used to associate to any such
 presemifield $(K,\circ)$ an isotopic semifield. Any semifield isotopic to $(K,\circ)$ is called a {\it twisted field}.
In 1996, Menichetti  proved that if $n$ is prime and $q$ large enough, any division algebra of dimension $n$ over
$\mathbb{F}_q$ is either a field or a twisted field \cite{M}. Twisted fields also play a role in the classification of
three-dimensional nonassociative algebras \cite{DG}.

  Moreover, in the theory of semifields,
 the commutative twisted fields play a prominent role: For $c=-1$, the division algebra obtained from a
 finite field extension $K/F$ of odd degree, $F$ of odd characteristic, which is
 given by the multiplication $x\circ y=xy+\alpha(x)\alpha^{-1}(y)$ for a non-trivial $\alpha\in {\rm Gal}(K/F)$,
 $\alpha^{-1}\not=\alpha$, is a presemifield.
 Although this $(K,\circ)$ itself is not commutative,
its isotope $x\star y=x\circ \alpha(y)=x\alpha(y)+\sigma(x)y$ is, thus the associated unital isotope
 obtained by using Kaplanski's trick
 a commutative twisted field. All commutative twisted semifields are obtained this way.
This construction works for every base field of odd characteristic  which
admits a cyclic field extension of degree
$n$ and yields a central commutative division algebra for all odd $n>2$.



In the following, we consider the following set-up:
 Let $A$ be a not necessarily unital,
finite-dimensional algebra over $F$, which possesses a multiplicative norm $N:A\to F$ of degree $n$, i.e.
 $N(xy)=N(x)N(y)$ for all $x,y\in A$, where juxtaposition stands for the algebra multiplication of $A$.
 We then take an element $c\in A$ and two  similarities $f,g$ of $N$ to define a new multiplication on
 $A$.
  Since $A$ need not be commutative or associative, there are several options for a possible generalization
  of Albert's approach:
The possibilities include defining a new multiplication on $A$ as
 $$x\circ y=xy-c(f(x)g(y)) \text{ or } x\circ y=xy-(cf(x))g(y),$$
but we can also define a multiplicative
 structure by putting $c$ in the middle, i.e.
 $$x\circ y=xy-(f(x)c)g(y) \text{ or } x\circ y=xy-f(x)(cg(y)),$$
 or on the right-hand side:
$$x\circ y=xy-(f(x)g(y))c \text{ or } x\circ y=xy-f(x)(g(y)c).$$
We can moreover choose to swap around the factors in the second part of the equation and define
 $$x\circ y=xy-c(f(y)g(x))$$
 and so on, exhausting all possible combinations of this type. Starting from a division algebra $A$, for the right choice
 of $c\in A^\times$, the algebras $(A,\circ)$ will
  be division algebras. We then apply Kaplanski's trick to $(A,\circ)$.
   Our choice of taking the unit element of $A$ when applying Kaplanski's trick to unital algebras helps to compute the inverses
   of the left and right multiplication in order to write down the multiplication $*$ explicitly.
Different choices of elements in $A$ used for Kaplanski's trick yield isotopic unital division algebras $(A,*)$.
  In the
  theory of semifields this is not relevant as algebras are usually classified up to isotopy.

After the preliminaries in Section 1, Section 2 explains the construction process generalizing Albert's approach and when
 the new twisted algebras
are division.
 We give examples of unital division algebras of dimension
 $n^2>4$ over $F$ which contain a commutative twisted field of odd degree $n$,
 provided the field $F$ has odd characteristic and permits a cyclic division algebra of odd degree $n$.
Examples of  twisted algebras, where we can explicitly compute the twisted multiplication, are given in Section 3.
We obtain some results on their automorphism groups and derivation algebras.
Although our main interest is to understand the twisted algebras, we will  discuss
 the algebras $(A,\circ)$ used in the construction process in Section 4.
Examples of real division algebras $(A,\circ)$ and $(A,*)$, most with large derivation algebras are given in
Corollary \ref{cor:real}, Examples \ref{ex:1} and \ref{ex:2}.

The problem with the construction probably becomes most apparent in the explicitly computed examples of Section 3:
Although it is rather straightforward to check whether a given twisted algebra is division or not, it is not easy to
check whether it may or may not be isomorphic to already known algebras due to its usually rather complicated multiplicative
structure.


\section{Preliminaries}

Let $F$ be a field.

\subsection{Nonassociative algebras}

 By an ``$F$-algebra'' we mean a finite-dimensional nonassociative algebra over $F$.
 A nonassociative algebra $A\not=0$ is called a {\it division algebra} if for any $a\in A$, $a\not=0$,
the left multiplication  with $a$, $L_a(x)=ax$,  and the right multiplication with $a$, $R_a(x)=xa$, are bijective.
$A$ is a division algebra if and only if $A$ has no zero divisors \cite[pp. 15, 16]{Sch}.

For an $F$-algebra $A$, commutativity is measured by the {\it commutator} $[x, y] = xy - yx$ and
associativity is measured by the {\it associator} $[x, y, z] = (xy) z - x (yz)$.
The {\it left nucleus} of $A$ is defined as ${\rm Nuc}_l(A) = \{ x \in A \, \vert \, [x, A, A]  = 0 \}$, the
{\it middle nucleus} of $A$ is
defined as ${\rm Nuc}_m(A) = \{ x \in A \, \vert \, [A, x, A]  = 0 \}$ and  the {\it right nucleus} of $A$ is
defined as ${\rm Nuc}_r(A) = \{ x \in A \, \vert \, [A,A, x]  = 0 \}$.
Their intersection ${\rm Nuc}(A) = \{ x \in A \, \vert \, [x, A, A] = [A, x, A] = [A,A, x] = 0 \}$ is
 the {\it nucleus} of $A$. The nucleus is an associative
subalgebra of $A$ and $x(yz) = (xy) z$ whenever one of the elements $x, y, z$ is in
${\rm Nuc}(A)$.
 An anti-automorphism $\sigma:A\to A$ of period 2 is called an {\it involution} on $A$.
 Since $A$ is a finite-dimensional algebra over $F$, the Lie algebra ${\rm Aut}(A)$
 of automorphisms of $A$, viewed as algebraic group,
 is a subalgebra of the derivation algebra ${\rm Der}(A)$ and for $F=\mathbb{R}$, we have
${\rm dim}{\rm Aut}(A)={\rm dim}{\rm Der}(A)$. For $c\in A$, define
$${\rm Aut}_c(A)=\{f\in {\rm Aut}(A)\,|\, f(c)=c\}
\text{ and } {\rm Der}_c(A)=\{D\in {\rm Der}(A)\,|\, D(c)=0\}.$$
If $c\in F$ then ${\rm Aut}_c(A)={\rm Aut}(A)$ and ${\rm Der}_c(A)={\rm Der}(A)$.


\subsection{Isotopes}


 Following the notation introduced in \cite[Section 1]{P}, denote the set of possibly non-unital algebra structures on an
 $F$-vector space $V$
by ${\rm Alg}(V)$. Given $A\in {\rm Alg}(V)$, we write $xAy$ for the product of $x,y\in V$ in the algebra, if
it is not clear  from the context which multiplication is used.


For $f,g,h\in {\rm Gl}(V)$ define the algebra $A^{(f,g,h)}$, called an
{\it isotope} of $A$, as $V$ together with the new multiplication
$$xA^{(f,g,h)}y=h(f(x)A g(y)) \quad\quad x,y\in V.$$
$A^{(f,g,h)}$ is a division algebra if $A$ is division. Two algebras $A, A'\in {\rm Alg}(V)$ are called {\it isotopic} if
$xAy=h(f(x) A' g(y))$ for all $ x,y\in V.$ If $f=g=h^{-1}$ then $A\cong A'$.
The group $({\rm Gl}(V))^3$ yields a partition of the set ${\rm Alg}(V)$ into isotopy classes. In particular, since every algebra
isotopic to a division algebra is division itself, it yields a partition of the subset of division algebras in ${\rm Alg}(V)$.


For $h=id$, we call $A^{(f,g)}=A^{(f,g,h)}$ a {\it principal Albert isotope} of $A$.
Let $G={\rm Gl}(V)\times{\rm Gl}(V)$ be the direct product of two copies of the full linear group of $V$. It acts on
${\rm Alg}(V)$ by means of principal Albert isotopes:
The above defines a right action of $G$ on ${\rm Alg}(V)$ which is compatible with passing to the opposite algebra, i.e.,
 $(A^{(f,g)})^{op}=(A^{op})^{(f,g)}$.
Division algebras are principal Albert isotopes of unital division algebras \cite[1.5]{P}.

\subsection{Kaplanski's trick}
Let $L_a:A\to A, x\mapsto ax$ denote the left multiplication with $a\in A^\times$, and $R_a:A\to A, x\mapsto xa$
 the right
 multiplication.
 Fix  $a,b\in A^\times$. The unital division algebras isotopic to a division algebra $A$ then are, up to isomorphism, the algebras $A'$ given by
the new multiplication
$$(xAa) A' (bAy)=x Ay,$$
i.e.
$$x A' y=(R_a^{-1}x) A (L_b^{-1}y)$$
for all $x,y\in A$. $A'$ is a division algebra with identity element $aAb$ and is an isotope of
$A$ \cite[Proposition 9]{M}.
If we choose $a=b$, this construction is commonly known as {\it Kaplanski's trick}.
Kaplanski's trick applied to an algebra $A$ yields a unital algebra which is an Albert isotope of $A$.
Two unital algebras obtained from $A$ by choosing two elements $a,a'\in A^\times$ and using Kaplanski's trick
with $a$ and $a'$, respectively, are isotopic to each other.





\subsection{Multiplicative forms of degree $n$}

A map $\varphi:V\to F$  on a finite-dimensional $F$-vector space $V$
is called a {\it form of degree $d$} over $F$, if
  $\varphi(a v)=a^d\varphi(v)$ for all $a\in F$, $v\in V$ and such that the map $\theta : V \times
\dots \times V \to F$ defined by
 $$\theta(v_1,\dots,v_d)=
  \sum_{1\leq i_1< \dots<i_l\leq d}(-1)^{d-l}\varphi(v_{i_1}+ \dots +v_{i_l})$$
($1\leq l\leq d$)  is a $d$-linear form over $F$, i.e., $\theta : V \times\dots \times V \to F$ ($d$-copies) is an
$F$-multilinear map where $\theta (v_1,\dots, v_d)$ is invariant under all permutations of its variables.


 A form of degree $d$ is called {\it nondegenerate} if $\theta (v, v_{2}, \dots, v_d) = 0$ for all $v_i \in V$ implies that $v = 0$.
We will only look at nondegenerate forms. A form $\varphi$ of degree $n$ is called {\it anisotropic} if
$\varphi(x)=0$ implies that $x=0$, else it is called {\it isotropic}.

A nondegenerate form $\varphi$ of degree $d$ on an  $n$-dimensional vector space is
\emph{is multiplicative}, if $\varphi(x)\varphi(y)=\varphi(z)$  where $x$, $y$ are systems of $n$
indeterminates and where each $z_l$ is a bilinear form in $x,y$ with coefficients in $F$. In this case the vector space
$V=F^n$ admits a bilinear map $V\times V\rightarrow V$ which can be viewed as the multiplicative structure of a
nonassociative $F$-algebra $A$ defined on $V$ and so $\varphi(vAw)=\varphi(v)\varphi(w)$ for all $v, w\in V$.



\subsection{Composition algebras} A quadratic form $N_A \colon A \to F$ on an algebra $A$ is {\it multiplicative}  if
$N_A(uv)=N_A(u)N_A(v)$ for all $u,v\in A$.
An algebra $A$ is called a {\it composition algebra} over $F$
if it admits a multiplicative quadratic form $N_A \colon A \to F$. The form $N_A$ is unique  \cite[p.~454 ff.]{KMRT}.
 It is called the {\it norm} of $A$ and we will often just write $N=N_A$.
A unital composition algebra is called a {\it Hurwitz algebra}. Hurwitz algebras are quadratic alternative and $N(1_A)=1$;
the norm of a Hurwitz algebra $C$ is the unique nondegenerate quadratic form
 on $A$ that is multiplicative. A quadratic alternative algebra is a Hurwitz algebra if
and only if its norm is nondegenerate \cite[4.6]{McC}. Hurwitz algebras
exist only in dimensions 1, 2, 4 or 8. Those of dimension~2 are exactly the
quadratic \'etale $F$-algebras, those of dimension 4 exactly the well-known
 quaternion algebras. The ones of dimension 8 are called {\it octonion algebras}. The conjugation
 $\overline{x} = T_{C}(x)1_A - x$ of a Hurwitz algebra
 $C$ is a scalar involution, called the {\it standard involution} of $C$, where $T_A \colon A \to F$, $T_{A}(x) = N_{A}
 (1_A, x)$, is the {\it trace} of $A$.


Let $D$ be a Hurwitz algebra over $F$ with canonical involution $\can:D\to D$. Let  $c\in F^\times$.
 Then the $F$-vector space
 $A=D\oplus D$ can be made into a unital algebra over $F$ via the multiplications
 $$ (u,v)(u',v')=(uu'+c \bar v'v,v'u+v\bar u')$$
for $u,u',v,v'\in D$. The unit element of  $A$ is given by $1_A=(1,0)$.
$A$ is called the {\it Cayley-Dickson} doubling of $D$.



Every composition algebra is a principal Albert isotope of a Hurwitz algebra: There are isometries
$\varphi_1,\varphi_2$ of
the norm $N_C$ for a suitable Hurwitz algebra $C$ over $F$ such that its multiplication can be written as
$$x\star y=\varphi_1(x)C\varphi_2(y)$$
\cite{KMRT}.






\section{How to obtain division algebras out of a given division algebra with multiplicative norm using Albert's approach}

 Unless stated otherwise, let $A$ be a
finite-dimensional nonassociative division algebra over $F$,  not necessarily unital, which possesses a multiplicative norm
 $N:A\to F$ of degree $n$.
 Let $O(N)$ denote the group of
 isometries and $S(N)$ the group of similarities of $N$.
 We choose  $c\in A^\times$ and some $h_i,h,f,g\in S(N)$ with similarity factors
 $d_i,d,\alpha,\beta\in F^\times$, $1\leq i\leq 3$, respectively.

\subsection{The general construction}

Take the isotope $(A,\cdot)=A^{(h_1,h_2,h_3)}$, then $N(x \cdot y)=d_1d_2d_3N(x)
N(y)$ for all $x,y\in A$. If $N$ is anisotropic then $(A,\cdot)$ is a division algebra.

 Generalizing Albert's approach, we have many options to define  new multiplications $\circ$ on $(A,\cdot)$, for instance via
 $$x\circ y=x\cdot y-c\cdot h(f(x)\cdot g(y)),$$
 $$  x\circ y=x\cdot y-h((c\cdot f(x))\cdot g(y)),$$
 $$x\circ y=x\cdot y-h(f(x)\cdot c)\cdot g(y),$$
 $$   x\circ y=x\cdot y-h(f(x)\cdot (c\cdot g(y))),$$
  $$x\circ y=x\cdot y-h((f(x)\cdot g(y))\cdot c),$$
  $$  x\circ y=x\cdot y-h(f(x)\cdot (g(y)\cdot c))\text{ etc. }$$
for all $x,y\in A$.

For noncommutative algebras, we can also swap around the factors $f(x)$ and $g(y)$ in the second part of the equation.

  We next choose an
$a\in A^\times$ to apply Kaplanski's trick
and thus obtain a unital algebra from $(A, \circ)$  we denote by $(A,*_{a})$.
Then the same method of proof as used by Albert  yields for any of these multiplications $\circ$:

\begin{theorem} \label{thm:main}
Let $A$ be an algebra over $F$ with an anisotropic multiplicative norm $N$ of degree $n$
and  $(A,\cdot)=A^{(h_1,h_2,h_3)}$. If
$$N(c)\not= \frac{1}{\alpha\beta d d_1 d_2 d_3},$$
then $(A,\circ)$ is a division algebra and Kaplanski's trick using $a\in A^\times$
yields a unital division algebra $(A,*_{a})$ over $F$.
\end{theorem}

\begin{proof}
 Since $N$ is anisotropic, the isotope $(A,\cdot)$ is a division algebra.
Let $x,y\in A$ be non-zero. Suppose that $x\circ y=0$ and rearrange the resulting equation so that one side
consists of the term $xy$.
Apply the norm on both sides. For instance, if $x\circ y=x \cdot y-c \cdot h(f(x)\cdot g(y))$ then this
 yields $d_1 d_2 d_3 N(x)N(y)= d_1 d_2 d_3 N(c) d d_1 d_2 d_3 N(f(x))N(g(y))$,
 thus $N(x)N(y)= d d_1 d_2 d_3\alpha \beta N(c) N(x)N(y)$. Cancelling the
non-zero term $N(x)N(y)$, this contradicts our assumption. This argument applies to all multiplications $\circ$ of the type
indicated above.
\end{proof}

\begin{example}
 Let $A$ be a unital algebra over $F$ with an anisotropic multiplicative norm $N$ of degree $n$
and  $(A,\cdot)=A^{(h_1,h_2,h_3)}$.
For instance, consider $x\circ y=x \cdot y-c \cdot(f(x)\cdot g(y))$ then
$$x\circ y=x \cdot y-c \cdot(f(x)\cdot g(y))=h_3(h_1(x)h_2(y))-c \cdot    (h_3(      h_1(f(x))  h_2(g(y))  ))$$
$$=h_3(h_1(x)h_2(y))-h_3(h_1(c)h_2( (h_3(  h_1(f(x))  h_2(g(y))  ))).$$
Now
$$h_3^{-1}(h_1^{-1}(x)\circ h_2^{-1}(y))=x y-h_1(c) h_2(  h_3( h_1(f(h_1^{-1}(x))))  h_2(g( h_2^{-1}(y) ))  ))$$
$$= xy-d h_2((h_3(h_1(f(h_1^{-1}(x)))h_2(g( h_2^{-1}(y) )))), $$
 therefore here $(A,\circ)$ is isotopic to an algebra $(A,\diamond)$ with multiplication of type
$$x\diamond y= xy-d H( F(x)G(y) ),$$
where $d\in A^\times$ and $H,F,G\in S(N)$ are suitably chosen.
\end{example}

Thus if we are interested in understanding these algebras (even only up to isotopy),
 it would make sense to start the construction with unital algebras $A$ possessing a multiplicative norm
and then  define multiplications $\circ$  on $A$ as suggested above.



\subsection{}
Let $A$ be a unital algebra with multiplicative norm $N$. We will look at the possible cases which occur for $h=id$ to keep the
investigation within reasonable length. We  define  new multiplications $\circ_i$ on $A$ via
\begin{enumerate}
\item $x\circ_{(1)} y=xy-c(f(x)g(y)),$
\item $  x\circ_{(2)} y=xy-(cf(x))g(y),$
\item $x\circ_{(3)} y=xy-(f(x)c)g(y),$
\item $   x\circ_{(4)} y=xy-f(x)(cg(y)),$
 \item $x\circ_{(5)} y=xy-(f(x)g(y))c,$
 \item $  x\circ_{(6)} y=xy-f(x)(g(y)c)$
\end{enumerate}
for all $x,y\in A$.

For noncommutative algebras $A$, we can swap around the factors in the second part of the equation and  define
\begin{enumerate}
\setcounter{enumi}{6}
\item $x\circ_{(7)} y=xy-c(f(y)g(x)),$
\item $x\circ_{(8)} y=xy-(cf(y))g(x),$
\item $x\circ_{(9)} y=xy-(f(y)c)g(x),$
\item $  x\circ_{(10)} y=xy-f(y)(cg(x)),$
\item $x\circ_{(11)} y=xy-(f(y)g(x))c,$
\item $ x\circ_{(12)} y=xy-f(y)(g(x)c)$
\end{enumerate}
for all $x,y\in A$.

If we simply write $(A,\circ)$, this can be any of the twelve different multiplications just introduced.
We also simply write $(A,\circ)=(A,\circ_{(i)})$, $i=1,\dots,12$ from now on, if it is clear from the context which multiplication
$\circ_{(i)}$ we use.


Denote by $L_x:A\to A, x\mapsto x\circ y$ and $R_y:A\to A, y\mapsto x\circ y$ the left and right multiplication
on $(A,\circ)$. Denote by $e=1_A=1$ the unit element of $A$ and define a new multiplication using Kaplanski's trick via
$$x*_{(i)}y=R_e^{-1}(x)\circ_{(i)} L_e^{-1}(y)$$
for all $x,y\in A$. We call the unital algebra $(A,*_{(i)})$  a {\it twisted algebra}. By construction,
$(A,*_{(i)})$ is isotopic to $(A,\circ_{(i)})$. $(A,*_{(i)})$ is a division algebra if and only so is $(A,\circ_{(i)})$.

If we simply write $(A,*)$, this can be any of the twelve different multiplications just introduced.
We also  write $(A,*)=(A,*_{(i)})$, $i=1,\dots,12$, if it is clear from the context which multiplication
$*_{(i)}$ we use.
To determine a twisted multiplication $(A,*)$ explicitly, we have to compute
  the inverses of the maps $L_e$ and $R_e$.

\begin{lemma}
Let  $B$ be a subalgebra of $A$ with $c\in {B}^\times$.
Assume $f|_{B},g|_{B}\in S(N_{B})$. Then $(B,\circ)$ is a subalgebra of $(A,\circ)$
and $(B,*)$ is a subalgebra of $(A,*)$.
\end{lemma}

\begin{proof}
The first part of the assertion is trivial. Now let $x,y\in B$. Since $e=1_A\in B$ and $(B,\circ)$ is a subalgebra of $(A,\circ)$,
we know that $x\circ e,$ $e\circ y\in B$, and thus the restricted maps
$R_e:B\to B$, $L_e:B\to B$ are isomorphisms onto $B$. We conclude that $x*y=R_e^{-1}(x)\circ L_e^{-1}(y)\in B$ as well.
\end{proof}

 \begin{theorem} \label{thm:main3} (cf. \cite[p.~85]{M})
 Let $K$ be a cyclic field extension of $F$ of degree $n$ with norm $N_K$ and ${\rm Gal}(K/F)=\langle \sigma\rangle.$
  For $c\in K^\times$, define
  $$x\circ y= xy-c\sigma^s(x)\sigma^t(y), \quad 0\leq s,t\leq n-1.$$
  If $s$ or $t$ is prime to $n$ and $(K,\circ)$ is a division algebra then $N_K(c)\not=1$.
\end{theorem}

From Theorem \ref{thm:main} and \ref{thm:main3} we obtain:

\begin{corollary}  \label{cor:important}
Let $A$ be a unital algebra over $F$ with anisotropic multiplicative norm $N$.
\\
(i)  If
 $$N(c)\not= \frac{1}{\alpha \beta}$$
  then $(A,\circ)$ is a division algebra. The associated twisted algebra $(A,*)$ is a division algebra.
  \\ (ii)
Let $K/F$ be a cyclic field extension  of degree $n$  with ${\rm Gal}(K/F)=\langle \sigma\rangle$,
which is a subalgebra of $A$.  Suppose that
$f|_{K},g|_{K}\in S(N_{K})$  and
 $f|_{K}(x)= a\sigma^s(x)$, $g|_{K}(x)= b\sigma^t(x)$, where $a,b,\in K$
 and $s$ or $t$ is prime to $n$.
If $c\in K^\times$ then $$(A,\circ) \text{ is a division algebra if and only if }N(c)\not=1/\alpha\beta.$$
 In particular, this holds if $A$ is a quaternion or octonion division algebra and
 $c\in A^\times$ such that $K=F(c)$ is a separable
field extension and
 $s$ or $t$ is 1.
  \end{corollary}



\begin{proof}  (i) is clear.
\\ (ii) Obviously, $N(a)=\alpha,$ $N(b)=\beta$.
If $c\in K^\times$ then $(K,\circ)$ is a subalgebra of $(A,\circ)$ with multiplication
$$x\circ y=xy-abc \sigma^s(x) \sigma^t(y).$$
 By Theorem \ref{thm:main}, if $N_K(c)=N(c)\not=1/\alpha\beta$ then $(A,\circ)$ is division. Conversely,
if $(A,\circ)$ is a division algebra then so is $(K,\circ)$. By assumption, $s$ or $t$ is prime to $n$,
 so by Theorem  \ref{thm:main3},
$N_K(c)\not=1/\alpha\beta$.
\\ This applies to the set-up that  $K=F(c)$ is a quadratic separable field extension of $F$
with non-trivial automorphism $\sigma$ contained in
a quaternion or octonion division algebra $A$ , if
 $s$ or $t$ is 1.
\end{proof}



\begin{remark}
(i)  Any element $a\in A^\times$, $1_A\not=a$,
 yields a unital algebra $(A,*_a)$
 isotopic to $(A,*)$.  That means different choices of elements to use Kaplanski's trick produce algebras which also
would deserve investigation. (For division algebras over finite fields, it is considered sufficient to classify algebras up to
 isotopy.)
We will focus on the above defined isotopes $(A,*)$ unless specified otherwise, i.e.
choose the unit element $A$ when applying Kaplansky's trick.
 The inverses $L_e^{-1}$ and $ R_e^{-1}$ then seem easier to find.
\\ (ii)  Let $F$ be a field of characteristic 0 or $> d$. The
 unital division algebras with multiplicative norms are exactly the Hurwitz division
algebras and the central simple associative division algebras over  separable field extensions $K$ of $F$ \cite{Sch}.
They all have anisotropic norms.
\\ (iii) The algebras $(A,\circ_{(i)})$ can be isotopic to the algebra $A$:
 Suppose $A$ is a unital algebra over $F$ with an anisotropic multiplicative norm $N$ of degree $n$
and $f\in {\rm Aut}(A)$. Then
$$  x\circ_{(1)} y=
(id -cf)(xy) \text{ and } x\circ_{(11)} y=xy-f(xy)c.$$
For all $c\in A$ such that $N(c)\not=1$, $(A,\circ_{(1)})$ and $(A,\circ_{(11)})$ are  division by Theorem \ref{thm:main}
 but also
 isotopic to $A$, since
 the maps $id- cf$ and $F(x)=x-f(x) c$ are bijective.  If $A$ is associative, thus
$(A,*_{(1)})\cong (A,*_{(11)})\cong A$.
\end{remark}

\begin{example} For $n>2$, unital division algebras of dimension $n^2$ containing commutative
twisted fields of dimension $n$ as subalgebras can be constructed out of cyclic division algebras
as follows:

Let $A=(K/F,\sigma,d)$ be a cyclic division algebra over $F$ of degree $n$,
${\rm Gal}(K/F)=\langle \sigma\rangle$.
 For all $c\in A$ such that $N(c)\not=1/\alpha\beta$,
the associated twisted algebra $(A,*_{(i)})$, $i=1,3,5,7,9,11$ is a division algebra (Theorem \ref{thm:main}).
 If   $c\in K$ and $f|_K,g|_K\in S(N_K)$,
   then
   $(A,*_{(i)})$ contains the $n$-dimensional unital subalgebra $(K,*)$.

  Moreover, if
 $f|_{K}(x)= a\sigma^s(x)$, $g|_{K}(x)= b\sigma^t(x)$, where $a,b,\in K$, and $s$ or $t$ is prime to $n$, by Corollary \ref{cor:important}
 (ii), $(A,*_{(i)})$ is a division algebra if and only if $N(c)\not=1/\alpha\beta$.
If additionally $s\not=t$, $s\not=0$, $t\not=0$, then
$(K,*)$ is a twisted field.

In particular, suppose $n$ is odd, $s+t=n$, $s\not=0$, $t\not=0$ and $abc=-1$.
Then $f|_{K}=a\sigma^s,g|_{K}=b\sigma^t=(f|_{K})^{-1}\in S(N_{K})$. The isotope
$x\diamond_{(i)} y=x\circ_{(i)} f(y)=xy+f(x) f^{-1}(y)$ of $(K,\circ_{(i)})$ is a commutative subalgebra of the isotope
$(A,\diamond_{(i)})$ of $(A,\circ_{(i)})$, defined as
$x\diamond_{(i)} y=x\circ_{(i)} h(y)$. Every unital division algebra $(A,*'_{(i)})$
 associated to $(A,\diamond_{(i)})$ by applying Kaplanski's trick hence
contains the $n$-dimensional unital commutative twisted field  $(K,*')$ as subalgebra.
\end{example}

It was pointed out by Kantor \cite{K} that although it at first appears as if Albert's construction of twisted fields yields a
large number of different (i.e., non-isotopic) semifields, this is actually not the case, as it only produces fewer than
$o$ non-isotopic semifields of order $o$.

Menichetti gives a geometric condition for when a division algebra of dimension $n$ is isomorphic to a twisted field,
involving its
left and right zero divisor hypersurfaces
(\cite[Corollary 32]{M}).

Even the question when the division algebras we construct are isotopic seems a difficult one, because of the numerous
ways of how to build a `twisted' multiplication $*_{(i)}$, starting with a nonassociative
algebra instead of a field extension, and will be left open here.



\section{Examples where we can explicitly compute the twisted multiplication}

Let $A$ be a unital algebra with anisotropic multiplicative norm $N$, $c\in A^\times$ and $f\in S(N)$.
In order to explicitly write down the multiplication of a twisted algebra $(A,*)$ we compute the inverse of
$$F(x)=x-cf(x) \text{ and } G(x)=x-f(x)c.$$
We can do so in special cases.

\begin{example}
For $c\not=1$ and $f=id$,
$F(x)=x-cx=(1-c)x \text{ and } G(x)=x-xc=x(1-c),$
therefore
$$F^{-1}(x)=(1-c)^{-1}x \text{ and } G^{-1}(x)=x (1-c)^{-1}.$$

\end{example}


\subsection{Central simple algebras}

Let $A$ be a central associative division algebra over $F$ with norm $N$.

\begin{example} \label{ex:isom}
For $f,g\in S(N)$, the multiplications
$$ x\circ_{(1)} y=xy-cf(x)y=(id-cf)(x)y,$$
$$ x\circ_{(1)} y=xy-cxg(y)=x(id-cg)(y)
\text{ for } c\in F^\times,$$
$$ x\circ_{(1)} y=xy-cf(x)f(y)=(id-cf)(xy) \text{ if } f\in {\rm Aut}(A),$$
$$  x\circ_{(5)} y=xy-xg(y)c=x(y-g(y)c),$$
  $$  x\circ_{(5)} y=xy-f(x)f(y)c=T(xy)  \text{ if }  f\in {\rm Aut}(A),  \text{ with }  T(z)=z-f(z)c,$$
$$  x\circ_{(7)} y=xy-c f(y)f(x)=(id-cf)(xy)  \text{ if }  f(xy)=f(y)f(x), $$
$$    x\circ_{(11)} y=xy-f(y)f(x)c=xy-f(xy)c=S(xy)  \text{ if } f(xy)=f(y)f(x) \text{ with } S(z)=z-f(z)c,$$
all yield isotopes of $A$, provided the  maps $id-cf$, $T$ etc. are bijective, which they are when we choose $c\in A$ suitable
for $(A,\circ)$ to be division.
Therefore in these  cases, the twisted algebras $(A,*)$ are
 isomorphic to $A$.
\end{example}

Suppose $f,g\in {\rm Aut}(A)$, $f\not=id$, $g\not=id$.


\begin{lemma}\label{le:explicit}
If  $f^n=id$ for some $n\geq2$ and
$1\not=cf(c)f^2(c)\cdots f^{n-1}(c)$, then
$$F^{-1}(x)=(1-cf(c)f^2(c)\cdots f^{n-1}(c))^{-1}(x+c f(x)+cf(c)f^2(x)+cf(c)f^2(c)f^3(x)+$$
$$\dots+ cf(c)\cdots
f^{n-2}(c)f^{n-1}(x) ),$$
$$G^{-1}(x)=(1-f^{n-1}(c) \cdots f^2(c) f(c) c )^{-1}(x+ f(x)c+f^2(x)f(c) c+f^3(x)f^2(c)f(c)c+$$
$$\dots+ f^{n-1}(x)f^{n-2}(c) \cdots f^2(c) f(c) c ).$$
\end{lemma}

The proof is straightforward.

\begin{remark}
Let  $f^n=id$ for some $n\geq2$.
\\
(i) Assume $N(c)\not=1$, so that $(A,\circ)$ is a division algebra. If $1=cf(c)f^2(c)\cdots f^{n-1}(c)$
then $N(c)^n=1$, hence $N(c)$ is a primitive $n$th root of unity contained in $F$.
\\ (ii)  We have $f(x)=dxd^{-1}$ for some $d\in A^\times$ with $d^n\in F^\times$.
Easy examples for this setup can be found looking at cyclic algebras $A=(K/F,\sigma,a)$ of degree $n$.
\end{remark}

\begin{example}
Suppose $f^n=id$, $g^n=id$ for some $n\geq2$  and
$1\not=cf(c)f^2(c)\cdots f^{n-1}(c)$, $1\not=cg(c)g^2(c)\cdots g^{n-1}(c)$, then  Lemma \ref{le:explicit} yields:

$$ x*_{(1)}y=R_e^{-1}(x)\circ L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-cf(R_e^{-1}(x))g(L_e^{-1}(y))$$
$$=(1-cf(c)f^2(c)\cdots f^{n-1}(c))^{-1} (x+c f(x)+cf(c)f^2(x)+cf(c)f^2(c)f^3(x)+\dots$$
$$+ cf(c)\cdots f^{n-2}(c)f^{n-1}(x) )$$
$$ (1-cg(c)g^2(c)\cdots g^{n-1}(c))^{-1}(y+c g(y)+cg(c)h^2(y)+cg(c)g^2(c)g^3(y)+\dots+ cg(c)\cdots g^{n-2}(c)g^{n-1}(y) )$$
$$-cf(   (1-cf(c)f^2(c)\cdots f^{n-1}(c))^{-1}  (x+c f(x)+cf(c)f^2(x)+cf(c)f^2(c)f^3(x)+$$
$$\dots+ cf(c)\cdots f^{n-2}(c)f^{n-1}(x) )     )  $$
$$  g(  (1-cg(c)g^2(c)\cdots g^{n-1}(c))^{-1}(y+c g(y)+cg(c)h^2(y)+cg(c)g^2(c)g^3(y)+$$
$$\dots+ cg(c)\cdots
g^{n-2}(c)g^{n-1}(y) )  )$$
and
$$ x*_{(3)}y=R_e^{-1}(x)\circ L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-f(R_e^{-1}(x))cg(L_e^{-1}(y)) $$
$$ = (1-f^{n-1}(c) \cdots f^2(c) f(c) c )^{-1}(x+ f(x)c+f^2(x)f(c) c+f^3(x)f^2(c)f(c)c+\dots$$
$$+ f^{n-1}(x)
f^{n-2}(c) \cdots f^2(c) f(c) c )$$
$$(1-cg(c)g^2(c)\cdots g^{n-1}(c))^{-1}(y+c g(y)+cg(c)g^2(y)+cg(c)g^2(c)g^3(y)+\dots+ cg(c)\cdots g^{n-2}(c)g^{n-1}(y) )$$
$$
-f(   (1-f^{n-1}(c) \cdots f^2(c) f(c) c)   )^{-1}$$
$$f(x+ f(x)c+f^2(x)f(c) c+f^3(x)f^2(c)f(c)c+\dots+ f^{n-1}(x)f^{n-2}(c) \cdots f^2(c) f(c) c     )$$
$$c g(  (1-cg(c)g^2(c)\cdots g^{n-1}(c))   )^{-1}$$
$$g(  y+c g(y)+cg(c)g^2(y)+cg(c)g^2(c)g^3(y)+\dots+ cg(c)\cdots g^{n-2}(c)g^{n-1}(y) )    ) $$
$$ = (1-f^{n-1}(c) \cdots f^2(c) f(c) c )^{-1}(x+ f(x)c+f^2(x)f(c) +f^3(x)f^2(c)f(c)c+\dots$$
$$+ f^{n-1}(x)f^{n-2}(c) \cdots f^2(c) f(c) c )$$
$$(1-cg(c)g^2(c)\cdots g^{n-1}(c))^{-1}(y+c g(y)+cg(c)g^2(y)+cg(c)g^2(c)g^3(y)+\dots+ cg(c)\cdots g^{n-2}(c)g^{n-1}(y) )$$
$$
- (1-cf^{n-1}(c) \cdots f^2(c) f(c))   )^{-1}$$
$$(f(x)+ f^2(x)f(c) c+f^3(x)f^2(c)f(c)+\dots+ x f^{n-1}(c)f^{n-2}(c) \cdots f^2(c) f(c)    )$$
$$c (1-g(c)g^2(c)\cdots g^{n-1}(c)c)   )^{-1}$$
$$(  g(y)+g(c) g^2(y)+g(c)g^2(c)g^3(y)+g(c)g^2(c)g^3(c)g^4(y)+\dots+ g(c)g^2(c)\cdots g^{n-1}(c)(y) )    ). $$
The remaining multiplications
$$  x*_{(5)}y=
R_e^{-1}(x)L_e^{-1}(y)-f(R_e^{-1}(x))g(L_e^{-1}(y))c,$$
$$  x*_{(7)}y=
R_e^{-1}(x)L_e^{-1}(y)-cg(L_e^{-1}(y))f(R_e^{-1}(x)), $$
$$   x*_{(9)}y=
R_e^{-1}(x)L_e^{-1}(y)-g(L_e^{-1}(y))cf(R_e^{-1}(x)), $$
$$    x*_{(11)}y=
R_e^{-1}(x)L_e^{-1}(y)-g(L_e^{-1}(y))f(R_e^{-1}(x))c$$
can be explicitly expanded analogously. Note that for $f=g$, $(A,*_{(1)})=(A,*_{(5)})=A$ (Remark \ref{ex:isom}).
\end{example}



\begin{proposition}\label{prop:aut}
Suppose $f^n=id$, $g^n=id$ for some $n\geq2$  and
$1\not=cf(c)f^2(c)\cdots f^{n-1}(c)$, $1\not=cg(c)g^2(c)\cdots g^{n-1}(c)$.
\\ (i) If $F\in {\rm Aut}(A)$ such that $F\circ f=f\circ F$, $F\circ g=g\circ F$ and $F(c)=c$,  then $F\in {\rm Aut}(A,*)$.
\\ (ii) If $D\in {\rm Der}(A)$ such that $D(c)=0$, $D(f(x))=f(D(x))$ and $D(g(x))=g(D(x))$ for all $x\in A$, then
$D\in {\rm Der}(A,*)$.

\end{proposition}


The proof is a simple computation.

Let $\tau$ be an involution on $A$.

\begin{lemma}
If  $c\in F^\times$ such that
$1\not=c \tau(c)$, then the inverse of $F(x)=x-c\tau(x)$ is given by
$$F^{-1}(x)=(1-c \tau(c))^{-1}(x+c \tau(x) ).$$
\end{lemma}

The proof is straightforward.

\begin{remark}
Note that if $c\in A\setminus F$, $1\not=c \tau(c)$, then we only know that
$F^{-1}(x)=(1-c \tau(c))^{-1}(x+c \tau(x) )$
for all $x\in A$ with $x\tau(c)=\tau(c)x$, e.g. for all $x\in F(c)$.
\end{remark}


We can, for instance, explicitly compute the following multiplications for all $c\in F^\times$, $c\not=\pm 1$:
$$(7.1)\quad x\circ_{(7)} y=xy-c\tau(y)x,$$
$$(7.2)\quad x\circ_{(7)} y=xy-cy\tau(x),$$
$$(7.3)\quad x\circ_{(7)} y=xy-c\tau(x)\tau(y).$$
We obtain
$$(7.1)\quad x*_{(7)}y
=\frac{1}{(1-c)(1-c\tau(c))}(x\circ (y+c\tau(y)))$$
$$= \frac{1}{(1-c)(1-c\tau(c))}( xy-c\tau(y)x+cx\tau(y)-c^2yx) $$
$$(7.2)\quad x*_{(7)}y
=\frac{1}{(1-c)(1-c\tau(c))}((x+c\tau(x))\circ y)$$
$$=\frac{1}{(1-c)(1-c\tau(c))}(xy-cy\tau(x)+c\tau(x)y-c^2 yx )$$
$$(7.3)\quad x*_{(7)}y
=\frac{1}{(1-c\tau(c))^2}((x+c\tau(x))\circ (y+c\tau(y)))$$
$$= (1-c\tau(c))^{-2} ( (1-c\tau(c)^{2})xy-c(1-c)\tau(x)\tau(y)+c(1-\tau(c))(x\tau(y)+\tau(x)y)  ). $$


An easy calculation shows:

\begin{proposition}\label{prop:invol}
Let $c\in F^\times$, $c\not=\pm 1$.
For the multiplications $*_{(7)}$ in $(7.1), (7.2)$ and $(7.3)$, we obtain:
\\ (i) If $F\in {\it Aut}(A)$ such that $F\circ\tau=\tau\circ F$, then $F\in {\rm Aut}(A,*_{(7)})$.
\\ (ii) If $D\in {\rm Der}(A)$ such that $D(\tau(x))=\tau(D(x))$ for all $x\in A$, then
$D\in {\rm Der}(A,*_{(7)})$.
\end{proposition}

\begin{corollary}
 Suppose $A$ contains a quaternion subalgebra $C$ with canonical involution $\sigma$ and assume that
$\tau|_C=\sigma$. In cases $(7.1), (7.2)$ and $(7.3)$ with $c\in F^\times$, $c\not=\pm 1$:
\\ (i) $\{F\in {\rm Aut}(A)\,|\, F(x)=dxd^{-1}, \, d\in C\}\subset {\rm Aut}(A,*_{(7)})$ and
${\rm Aut}(A,*_{(7)})$
contains a subgroup isomorphic to $SU(2)$.
\\ (ii) $\{F\in {\rm Der}(A)\,|\, D(x)=ax-xa, \, a\in C\}\subset {\rm Der}(A,*_{(7)}).$
and $su(2)$ is isomorphic to a subalgebra of ${\rm Der}(A,*_{(7)})$.
\end{corollary}

\begin{proof}
(i) We know that
$F(x)=dxd^{-1}$ for some $d\in A^\times$, therefore $F(\tau(x))=d\tau(x)d^{-1}$,
$\tau(F(x))=\tau(dxd^{-1})=\tau(d^{-1})\tau(x)\tau(d) $, so
$F(\tau(x))=\tau(F(x))$ iff $d\tau(x)d^{-1}=\tau(d^{-1})\tau(x)\tau(d)$ if and only if
$\tau(x)=d^{-1}\tau(d^{-1})\tau(x)\tau(d)d$  if and only if
$x\tau(d)d=\tau(d)dx$ for all $x\in A$ iff $\tau(d)d\in {\rm Comm}(A)=F$.
 Now  $\tau(d)d=\sigma(d)d=N_C(d)\in F^\times$ for all $d\in C^\times$.
 The canonical embedding
$$\{F\in {\rm Aut}(C)\,|\, F(x)=dxd^{-1}, \, d\in C\}\hookrightarrow \{F\in {\rm Aut}(A,*_{(7)})\,|\, F(x)=dxd^{-1}\}$$
$$(x\mapsto dxd^{-1})\mapsto (x\mapsto dxd^{-1})$$
 implies that $SU(2)\cong {\rm Aut}(C)$ is isomorphic to a subgroup of ${\rm Aut}(A,*_{(7)})$.
 \\ (ii) We have $D(x)=ax-xa$ for some $a\in A^\times$, therefore $D(\tau(x))=\tau(D(x))$ if and only if
 $(a+\tau(a))\tau(x)=\tau(x)(a+\tau(a))$ for all $x\in A$, which is equivalent to $a+\tau(a)\in {\rm Comm}(A)=F$.
 Now $a+\tau(a)=a+\sigma(a)$ for all $d\in D^\times$.
The canonical embedding
$$\{D\in {\rm Der}(C)\,|\, D(x)=ax-xa, \, a\in C\}\hookrightarrow \{D\in {\rm Der}(A,*_{(7)})\,|\, D(x)=ax-xa\}$$
$$(x\mapsto ax-xa)\mapsto (x\mapsto ax-xa)$$
implies that $su(2)$ is isomorphic to a subalgebra of ${\rm Der}(A,*_{(7)})$.
\end{proof}

If the multiplication $\circ$ is defined using one involution and one automorphism of the type used in Lemma
\ref{le:explicit} or using
two different involutions, $(A,*)$ can be explicitly computed analogously and
 results corresponding to Propositions \ref{prop:aut} and \ref{prop:invol} hold.



\subsection{Hurwitz algebras}

Let $A$ be a quaternion or octonion division algebra over $F$.

\begin{lemma} \label{le:18}
If
 $h\in {\rm Aut}(A)$ is a reflection of $A$, i.e. $h^2=id$,  and $c\in F^\times$ such that
$1\not=c^2$, then
$$F^{-1}(x)=(1-c^2)^{-1}(x+c h(x)).$$
\end{lemma}

Note that if $c\in A\setminus F$ in Lemma \ref{le:18}, then we only know that
$F^{-1}(x)=(1-c^2)^{-1}(x+c h(x) )$
for all $x\in {\rm Nuc}_r(A)$.




\begin{proposition}\label{prop:19}
 Let $f,g\in {\rm Aut}(A)$, $f\not=id$, $g\not=id$
be two reflections.
 \\ (i) Let  $c\in F^\times$ then
$$ x*_{(1)}y=xy-cf(x)g(y)=(1-c^2)^{-2}  (1-c^3)xy+(1+c)^{-2}c(1-c)^{-1}(  xg(y)+f(x)y+f(x)g(y)  )$$
if $c^2\not=1$.
\\
(ii) Let $A$ be a quaternion division algebra over $F$ and   $c\in A^\times$.
 For
$cf(c)\not=1$, $cg(c)\not=1$ we  compute the multiplication $*$ explicitly:
\\
For $  x\circ_{(1)} y=xy-cf(x)g(y)$,
$$  x*_{(1)}y=(1-cf(c))^{-1} (x+c f(x) ) (1-cg(c))^{-1}(y+c g(y) )$$
$$-c(   (1-f(c)c))^{-1}  (f(x)+f(c)x )     )  (1-g(c)c)^{-1}(g(y)+g(c)y )  ).$$
For $ x\circ_{(3)} y=xy-f(x)cg(y), $
$$  x*_{(3)}y= (1- f(c) c )^{-1}(x+ f(x)c )(1-cg(c))^{-1}(y+c g(y))$$
$$- (1-c f(c))   )^{-1}(f(x)+ xf(c)  )c (1-g(c)c  )^{-1}(  g(y)+g(c)y )    ).$$
For $ x\circ_{(5)} y=xy-f(x)g(y)c,$
$$  x*_{(5)}y=
(1-f(c)c)^{-1}(x+f(x)c) (1-g(c)c)^{-1}(x+g(x)c)$$
$$-( 1-cf(c))^{-1}(f(x)+xf(c))    )( 1-cg(c))^{-1}(g(x)+xg(c)) )c,$$
For $  x\circ_{(7)} y=xy-cg(y)f(x), $
$$  x*_{(7)}y=(1-cf(c))^{-1}(x+cf(x)) (1-cg(c))^{-1}(x+cg(x))$$
$$-c(  (1-g(c)c))^{-1}(g(x)+g(c)x)  ) (1-f(c)c))^{-1}(f(x)+f(c)x),  $$
For $  x\circ_{(9)} y=xy-g(y)cf(x), $
$$  x*_{(9)}y=(1-cf(c))^{-1}(x+cf(x))  (1-g(c)c)^{-1}(x+g(x)c)$$
$$ -(1-cg(c))^{-1}(g(x)+xg(c)) c ( 1-f(c)c)^{-1} ( f(x)+f(c)x)   $$
For $    x\circ_{(11)} y=xy-g(y)f(x)c$,
$$  x*_{(11)}y=
(1-cf(c))^{-1}(x+cf(x)) (1-g(c)c)^{-1}(x+g(x)c) $$
$$ -(  1-cg(c))^{-1}(g(x)+xf(c))(  1-f(c)c)^{-1}(f(x)+f(c)x)  c.$$
(iii)
 Let $A$ be a quaternion division algebra over $F$.
Suppose $c\in A\setminus F$  such that $c\not=1$ and $cg(c)\not=1$.
Then, choosing $f=id$, for $x\circ_{(1)} y=xy-c x g(y)$,
$$  x*_{(1)}y
= (1-c)^{-1}x (1-cg(c))^{-1}  (y+c g(y) )-c (1-c)^{-1}x (1-g(c)c)^{-1}  (g(y)+g(c) y ).$$
If $N(c)\not=1$, $(A,*)$ is a division algebra in (i), (ii) and (iii).
\end{proposition}


\begin{proof} (i) We have
$$ x*_{(1)}y=R_e^{-1}(x)\circ L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-cf(R_e^{-1}(x))g(L_e^{-1}(y))$$
$$=(1-c^2)^{-1} (x+c f(x) ) (1-c^2)^{-1}(y+c g(y) )$$
$$-cf(   (1-c^2)^{-1}  (x+c f(x) )     )  g(  (1-c^2)^{-1}(y+c g(y) )  )$$
$$=(1-c^2)^{-2} [(x+c f(x) )(y+c g(y) )-c (f(x)+c x ) (g(y)+c y)  ]$$
$$=(1-c^2)^{-2}  [xy+cxg(y)+cf(x)y+c^2f(x)g(y)-cf(x)g(y)-c^2f(x)y-c^2xg(y)-c^3xy]$$
$$=(1-c^2)^{-2}  [(1-c^3)xy+(c-c^2)xg(y)+(c-c^2)f(x)y+(c^2-c)f(x)g(y)]$$
$$=(1-c^2)^{-2}  (1-c^3)xy+(1+c)^{-2}c(1-c)^{-1}(  xg(y)+f(x)y+f(x)g(y)  ).$$
(ii) The computation in (i) yields the first two cases.
$$  x*_{(5)}y=R_e^{-1}(x)\circ_{(5)} L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-f(R_e^{-1}(x))g(L_e^{-1}(y))c$$
$$=(1-f(c)c)^{-1}(x+f(x)c) (1-g(c)c)^{-1}(x+g(x)c)-f( (1-f(c)c)^{-1}(x+f(x)c)    )g( (1-g(c)c)^{-1}$$
$(x+g(x)c) )c$
$$=(1-f(c)c)^{-1}(x+f(x)c) (1-g(c)c)^{-1}(x+g(x)c)-f( 1-f(c)c)^{-1}f(x+f(x)c)    )g( 1-g(c)c)^{-1}$$
$g(x+g(x)c) )c$
$$=(1-f(c)c)^{-1}(x+f(x)c) (1-g(c)c)^{-1}(x+g(x)c)-( 1-cf(c))^{-1}(f(x)+xf(c))    )( 1-cg(c))^{-1}$$
$(g(x)+xg(c)) )c$


$$  x*_{(7)}y=R_e^{-1}(x)\circ_{(7)} L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-cg(L_e^{-1}(y))f(R_e^{-1}(x)) $$
$$=(1-cf(c))^{-1}(x+cf(x)) (1-cg(c))^{-1}(x+cg(x))-cg(  (1-cg(c))^{-1}(x+cg(x))  )f( (1-cf(c))^{-1}$$
$(x+cf(x)) ) $
$$=(1-cf(c))^{-1}(x+cf(x)) (1-cg(c))^{-1}(x+cg(x))-cg(  (1-cg(c)))^{-1}g((x+cg(x))  )f( (1-cf(c)))^{-1}$$
$f(x+cf(x))  $
$$=(1-cf(c))^{-1}(x+cf(x)) (1-cg(c))^{-1}(x+cg(x))-c(  (1-g(c)c))^{-1}(g(x)+g(c)x)  ) (1-f(c)c))^{-1}$$
$(f(x)+f(c)x)  $


$$   x*_{(9)}y=R_e^{-1}(x)\circ_{(9)} L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-g(L_e^{-1}(y))cf(R_e^{-1}(x)) $$
$$=(1-cf(c))^{-1}(x+cf(x))  (1-g(c)c)^{-1}(x+g(x)c) -g((1-g(c)c)^{-1}(x+g(x)c))cf(  (1-cf(c))^{-1}$$
$(x+cf(x))  ) $
$$=(1-cf(c))^{-1}(x+cf(x))  (1-g(c)c)^{-1}(x+g(x)c) -g(1-g(c)c)^{-1}g(x+g(x)c)) c f( 1-cf(c))^{-1}$$
$  f(x+cf(x))   $
$$=(1-cf(c))^{-1}(x+cf(x))  (1-g(c)c)^{-1}(x+g(x)c) -(1-cg(c))^{-1}(g(x)+xg(c)) c ( 1-f(c)c)^{-1} $$
$( f(x)+f(c)x)   $


$$  x*_{(11)}y=R_e^{-1}(x)\circ_{(11)} L_e^{-1}(y)=R_e^{-1}(x)L_e^{-1}(y)-g(L_e^{-1}(y))f(R_e^{-1}(x))c$$
$$=(1-cf(c))^{-1}(x+cf(x)) (1-g(c)c)^{-1}(x+g(x)c)  -g(  (1-g(c)c)^{-1}(x+g(x)c) )f(  (1-cf(c))^{-1}$$
$(x+cf(x))  )c$
$$=(1-cf(c))^{-1}(x+cf(x)) (1-g(c)c)^{-1}(x+g(x)c)  -g(  1-g(c)c)^{-1}g(x+g(x)c) )f(  1-cf(c))^{-1}$$
$f(x+cf(x))  c$
$$=(1-cf(c))^{-1}(x+cf(x)) (1-g(c)c)^{-1}(x+g(x)c)  -(  1-cg(c))^{-1}(g(x)+xf(c))(  1-f(c)c)^{-1}$$
$(f(x)+f(c)x)  c$
\\ (iii) is analogous.
\end{proof}

\begin{remark}
Note that if $f=g$ in cases (i) and (ii) of Proposition \ref{prop:19}, then $(A,*)$ is clearly an isotope of $A$.

In (i), the choice of $f=id$ or $g=id$  gives isotopic algebras to $A$. Again, the multiplications
can be explicitly computed and looks similar.

We also do not explicitly compute the remaining possible cases in (iii), where we choose the identity map for one map
and a reflection for the other in multiplications $\circ_{(i)}$ for $i\in\{1,3,5,7,9,11\}$, since results
and proofs are analogous to the case given in (iii).
Some of these will yield quaternion algebras again, see Example \ref{ex:isom}.
\end{remark}



The proofs of the following results are straightforward but tedious calculations:

\begin{lemma}
Let $f,g\in {\rm Aut}(A)$, $f\not=id$, $g\not=id$   be two reflections, i.e. $f^2=id$, $g^2=id$.
Suppose that either $c\in F^\times$  such that
$c^2\not=1$ or that $A$ is a quaternion algebra and $c\in A^\times$ such that $N(c)\not=1$ and
$cf(c)\not=1$, respectively  $cg(c)\not=1$, depending on the multiplication below.
We look at some possible cases for $*_{(7)}$ which all yield division algebras:
\\
 $(7.1)\quad x\circ_{(7)} y=xy-cf(y)x$, so
$$  x*_{(7.1)}y=(1-c)^{-1}x (1-cf(c))^{-1}  (y+c f(y) )-c ( 1-f(c)c)^{-1}(f( y)+f(c) y  )(1-c)^{-1} y.$$
 $(7.2)\quad x\circ_{(7)} y=xy-c y g(x)$, so
$$  x*_{(7.2)}y=(1-cg(c))^{-1} (x+c g(x) ) (1-c)^{-1}y-c (1-c)^{-1}y   (1-g(c)c)^{-1}(g( x)+g(c) x  ).$$
 $(7.3)\quad x\circ_{(7)} y=xy-c f(y) g(x)$, so
$$  x*_{(7.3)}y=(1-cg(c))^{-1} (x+c g(x) ) (1-cf(c))^{-1}  (y+c f(y) )-c  (1-f(c)c) )^{-1} (f( y)+f(c) y  )$$
 $(1-g(c)c )^{-1}(g( x)+g(c) x  ).$
\end{lemma}



\begin{lemma}
Let $f,g\in {\rm Aut}(A)$, $f\not=id$, $g\not=id$
be two reflections.
\\ (a)
Suppose $c\in F^\times$, then
any $F\in {\rm Aut}(A)$ such that $f\circ F=f\circ F$, $g\circ F=g\circ F$ lies in ${\rm Aut}(A,*_{(1)})$.
\\ (b) Let $F\in {\rm Aut}_c (A)$ and suppose that $c\in F^\times$ if $A$ is an octonion algebra.
 Then
 \\ $F\in {\rm Aut} (A,*_{(7.1)})$ if $F\circ f=f\circ F$,
 \\ $F\in {\rm Aut} (A, ,*_{(7.2)})$ if $F\circ g=g\circ F$ and
\\ $F\in {\rm Aut} (A, ,*_{(7.3)})$ if $F\circ f=f\circ F$ and $F\circ g=g\circ F$.
\end{lemma}




\begin{corollary}
 Let $A$ be a quaternion algebra and $f,g\in {\rm Aut}(A)$, $f\not=id$, $g\not=id$
be two reflections,  $f(x)=sxs^{-1}$ and $g(x)=txt^{-1}$.
Then
$$\{F\in {\rm Aut}_c(A) \,|\, F(x)=dxd^{-1},\, d\in F(s) \}\subset \{F\in {\rm Aut}_c(A) \,|\, F\circ f=f\circ F \}\subset {\rm Aut} (A,*_{(7.1)}),$$
$$\{F\in {\rm Aut}_c(A) \,|\, F(x)=dxd^{-1},\, d\in F(t) \}\subset \{F\in {\rm Aut}_c(A) \,|\, g\circ f=g\circ F \}\subset {\rm Aut} (A,*_{(7.2)}),$$
$$\{F\in {\rm Aut}_c(A) \,|\, F(x)=dxd^{-1},\, d\in F(t)\cap F(s) \}\subset
\{F\in {\rm Aut}_c(A) \,|\, F\circ f=f\circ F, \,\,F\circ g=g\circ F \}$$
$\subset {\rm Aut} (A,*_{(7.3)}).$
\end{corollary}

\begin{lemma}
If $\tau$ is an involution on $A$  and $c\in F^\times$ such that
$1\not=c \tau(c)$, then
$$F^{-1}(x)=(1-c \tau(c))^{-1}(x+c \tau(x) ).$$
\end{lemma}


Then for $c\in F^\times$:
$$(1)\quad x*y= (1-c\tau(c))^{-2} ( (1-c\tau(c)^{2})xy-c(1-c)\tau(x)\tau(y)+c(1-\tau(c))(x\tau(y)+\tau(x)y)  ), $$
$$(7.1)\quad x*_{(7.1)}y= \frac{1}{(1-c)(1-c\tau(c))}( xy-c\tau(y)x+cx\tau(y)-c^2yx), $$
$$(7.2)\quad x*_{(7.2)}y=\frac{1}{(1-c)(1-c\tau(c))}(xy-cy\tau(x)+c\tau(x)y-c^2 yx ).$$


\begin{proposition} \label{prop:25}
Let $A$ be a quaternion or octonion division algebra and $c\in F^\times$ such that $c\not=\pm 1$
and $c\tau(c)\not=1$. Let $(A,*)$ with $*$ from the cases $(1)$, $(7.1)$ and $(7.2)$.
\\ (i)
If $f\in {\rm Aut}(A)$ such that $f\circ\tau=\tau\circ f$ then $f\in {\rm Aut}(A,*)$.
\\ (ii) If $\tau$ is the canonical involution of $A$ then $(A,*)$ is division, and
$$ SU(2)\cong {\rm Aut}(A)\subset  {\rm Aut}(A,*)$$
if $A$ is a quaternion algebra and
$$ G_2\cong {\rm Aut}(A)\subset  {\rm Aut}(A,*)$$
if $A$ is an octonion algebra.
\\ (iii) If $\tau$ is the canonical involution of $A$ and $A$ a quaternion algebra then
$$ su(2)\cong {\rm Der}(A)\subset  {\rm Der}(A,*).$$
\end{proposition}

\begin{proof}
(i) and (iii) are straightforward calculations, (ii)  follows from (i) and Theorem \ref{thm:main3}.
\end{proof}


More generally, similar considerations yield the following:

\begin{proposition}\label{prop:26}
Let $A$ be a quaternion or octonion division algebra over $F$ with canonical involution $\sigma$, $c\in A^\times$ such that
$N(c)\not=1$ and  $f,g\in \{ id,\sigma\}$. Take any possible definition of $\circ$ using these $f$ and $g$.
If $A$ has dimension 8, we additionally assume that $c\in F^\times$. Then
$${\rm Aut}_c(A)\subset {\rm Aut}(A,*).$$
 In particular, for $c\in F^\times$,
$$ SU(2)\cong {\rm Aut}(A)\subset  {\rm Aut}(A,*)$$
if $A$ is a quaternion algebra and
$$ G_2\cong {\rm Aut}(A)\subset  {\rm Aut}(A,*)$$
if $A$ is an octonion algebra.

\end{proposition}

\begin{proof} This follows from the fact that the canonical involution commutes with any automorphism of $A$.
\end{proof}

We point out that in the above setting,  $(A,*)$ is division if and only if $N(c)\not= 1$.
Using the classification of real division algebras \cite{B-O1} we obtain:


\begin{corollary}\label{cor:real}
Let $F=\mathbb{R}$ and $A$ a quaternion or octonion division algebra over $\mathbb{R}$  with canonical involution $\sigma$.
Let  $c\in \mathbb{R}^\times$ such that $c\not=\pm 1$. Take any possible definition of $\circ$ using $f,g\in \{ id,\sigma\}$. Then
$$ su(2)\cong {\rm Der}(A,*)$$
if $A$ is a quaternion division algebra and
$$ G_2\cong  {\rm Der}(A,*)$$
if $A$ is an octonion division algebra. In particular, this is true for the multiplications  $(1), (7.1)$, $(7.2)$ with $\tau=\sigma$
from Proposition \ref{prop:25}.
\end{corollary}


\section{The algebras $(A,\circ)$}



In this section, let $A$ be an algebra over $F$ with a multiplicative norm $N$ of degree $n$, $c\in A^\times$.



\begin{lemma} Let $f,g\in O(N)$.
\\ (i) Let $H\in {\rm Aut}_c(A)$
 such that $H(f(x))=f(H(x))$ and $H(g(x))=g(H(x))$ for all
$x\in A$. Then $H\in {\rm Aut}(A,\circ)$.
\\ (ii)  If $D\in {\rm Der}(A)$ such that $D(c)=0$, $D(f(x))=f(D(x))$ and $D(g(x))=g(D(x))$ for all $x\in A$, then
$D\in {\rm Der}(A,\circ)$.
\end{lemma}

\begin{proof}
(i) Consider $x\circ_{(1)} y=xy-c(f(x)g(y))$.
For $H\in {\rm Aut}(A)$, $H(x\circ_{(1)} y)=H(x)H(y)-H(c)(H(f(x))H(g(y)))$ and $H(x)H(y)=H(x)H(y)-c(f(H(x))g(H(y)))$.
So if $H(c)=c$ and $H(f(x))=f(H(x))$ as well as $H(g(x))=g(H(x))$ for all $x\in A$, $H\in {\rm Aut}(A,\circ)$.
A similar argument applies for all $\circ_{(i)}$.
\\ (ii) The proof is a simple computation.
\end{proof}

\begin{proposition} \label{cor:35}
Let $A$ be a central simple associative algebra over $F$ and $f,g\in {\rm Aut}(A)$, $f(x)=sxs^{-1}$ and
$g(x)=txt^{-1}$ for suitable $s,t\in A^\times$.
\\ (i) If $c\in A^\times\setminus F$ then
$$\{ d_a\in {\rm Der}(A)\,|\, a\in F(s)\cap F(t) \cap  F(c) \}\subset\{ d_a\in {\rm Der}_c(A)\,|\, a\in F(s)\cap F(t)\}$$
$$\subset \{ d_a\in {\rm Der}_c(A)\,|\, f,g\in {\rm Aut}_a(A)  \}\subset {\rm Der}(A,\circ).$$
If $c\in F^\times$ then
$$\{ d_a\in {\rm Der}(A)\,|\, a\in F(s) \cap F(t)  \}\subset
\{ d_a\in {\rm Der}(A)\,|\, f,g\in {\rm Aut}_a(A)  \}\subset {\rm Der}(A,\circ).$$
\\ (ii)  Suppose $c\in A^\times\setminus F$.
If  $s,t\in F(c)$ or $c\in F(s)\cap F(t)$ then
$$d_c\in {\rm Der}(A,\circ).$$
\end{proposition}

\begin{proof}
(i)
Every derivation of $A$ is of the kind
$d_a(x)=ax-xa$ for all $x\in A$. Let $f,g\in {\rm Aut}(A)$, then $f(x)=sxs^{-1}$ and
$g(x)=txt^{-1}$ for suitable $s,t\in A^\times$ and  $D(f(x))=f(D(x))$ and $D(g(x))=g(D(x))$ implies that if
$f(a)=a$, $g(a)=a$, then $d_a\in {\rm Der}(A,\circ)$. Now  $f(a)=a$ is the same as $sa=as$, so that this holds for all
$a\in F(s)$, and analogously, $g(a)=a$ is the same as $ta=at$, so that this holds for all
$a\in F(t)$.  If $c\in F^\times$ then $D(c)=0$. The assertion follows.
\\ (ii) We know that $0\not=d_c\in  {\rm Der}_c(A)$ by (i). Now if $s,t\in F(c)$ then $f(c)=c$ and $g(c)=c$ and so
$d_c\in {\rm Der}(A,\circ)$. Alternatively, if $c\in F(s)\cap F(t)$ then also $f(c)=c$ and $g(c)=c$.
\\ (iii) is now clear.
\end{proof}



\begin{example} \label{ex:1}

Let Let $F=\mathbb{R}$, $A=\mathbb{H}$   and $f,g\in S(N_\mathbb{H})$ with similarity
factors $\alpha$ and $\beta$.  For all $c\in \mathbb{H}$ such that $N_\mathbb{H}(c)\not=1/\alpha\beta$,
 $(A,\circ)$ is a division algebra.



By Corollary \ref{cor:35} and \cite{B-O1}, if $f,g\in {\rm Aut}(\mathbb{H})$, $f(x)=sxs^{-1}$,
$g(x)=txt^{-1}$ with $s,t\in \mathbb{H}^\times$ and
$c\in \mathbb{H}^\times$ then
 ${\rm dim}\, {\rm Der}(\mathbb{H},\circ)=1$ or ${\rm Der}(\mathbb{H},\circ)\cong su(2)$.

 Since $\mathbb{C}$ is contained in $A$,
 if $f|_\mathbb{C},g|_\mathbb{C}\in S(N_\mathbb{C})$ and $c\in \mathbb{C}$, $(A,\circ)$ contains the two-dimensional subalgebra $(\mathbb{C},\circ)$.
 Choose any $d\in A^\times$ and apply Kaplanski's trick. This yields a unital
 division algebra $(A,*_d)$.  If $d\in \mathbb{C}^\times$,
$(\mathbb{C},*_d)$ is a unital subalgebra, hence isomorphic to $\mathbb{C}$.
\end{example}


 \begin{example} \label{ex:2}
 Let $F=\mathbb{R}$ and $\mathbb{O}$ Cayley's octonion division algebra with norm $N_\mathbb{O}$.
\\ (i)
Let $c\in \mathbb{R}^\times$, then  for all possible choices of $\circ$ with
$f,g\in \{ id,\sigma\}$, $(\mathbb{O},\circ)$
 is division if and only if $N(c)\not=1$,
$G_2\cong {\rm Aut}(\mathbb{O})\subset {\rm Aut}(\mathbb{O},\circ)$ by Proposition \ref{prop:26}
and thus
$$G_2\cong {\rm Aut}(\mathbb{O},\circ) \text{ and } G_2\cong {\rm Der}(\mathbb{O},\circ)$$
by \cite{B-O1}.
\\ (ii) Let $f,g\in S(N_\mathbb{O})$ with similarity
factors $\alpha$ and $\beta$.
For all $c\in \mathbb{O}^\times$ such that $N_\mathbb{O}(c)\not=1/\alpha\beta$, $(\mathbb{O},\circ)$ is a division algebra.


If $f|_\mathbb{H},g|_\mathbb{H}\in S(N_\mathbb{H})$
and $c\in \mathbb{H}^\times$, $(\mathbb{O},\circ)$ contains the four-dimensional
 subalgebra $(\mathbb{H},\circ)$.
Then choose any $d\in \mathbb{O}^\times$ to
  apply Kaplanski's trick to $(\mathbb{O},\circ)$. This yields a unital division algebra $(\mathbb{O},*_d)$.
$(\mathbb{H},*_d)$ is a unital  four-dimensional subalgebra  and
contains $\mathbb{C}$ as subalgebra
if $f|_\mathbb{C},g|_\mathbb{C}\in S(N_\mathbb{C})$ and $c\in \mathbb{C}$.
 \end{example}


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\end{document}

1504.00188.extracted_bib
{A1 Albert, A. A., {Generalized twisted fields. Pacific J. Math. 11 (1961), 1-8.
1504.00188.extracted_bib
{A2 Albert, A. A., {Isotopy for generalized twisted fields. An. Acad. Brasil. Ci. 33 (1961), 265-275.
1504.00188.extracted_bib
{As-Pu Astier, A., Pumpl\"{un, S., {Nonassociative quaternion algebras over rings. Israel J. Math. 155 (2006), 125-147.
1504.00188.extracted_bib
{B-O1 Benkart, G. M., Osborn, J. M., {The derivation algebra of a real division algebra. Amer. J. of Math. 103 (6) (1981), 1135-1150.
1504.00188.extracted_bib
{DG Deajim, A., Grant, D., {Space time codes and nonassociative division algebras arising from elliptic curves. Contemp. Math. 463 (2008), 29-44.
1504.00188.extracted_bib
{E-My Elduque, A., Myung, H. C., {On mutations of Cayley-Dickson algebras. Contemp. Math. 184 (1995), 97-110.
1504.00188.extracted_bib
{J Jacobson, N., Some applications of Jordan norms to involutorial simple associative algebras. Adv. in Math. 48 (2) (1983), 149-165.
1504.00188.extracted_bib
{K Kantor, W. M., {Finite semifields. Finite geometries, groups and compositions, 103-114, Walter DeGruyter GmbH and Co. KG, Berlin 2006.
1504.00188.extracted_bib
{KMRT Knus, M. A., Merkurjev, A., Rost, M., Tignol, J.-P., ``The Book of Involutions'', AMS Coll. Publications, vol. 44 (1998).
1504.00188.extracted_bib
{McC McCrimmon, K., {Nonassociative algebras with scalar involution, Pacific J. of Math. 116 (1) (1985), 85-108.
1504.00188.extracted_bib
{M Menichetti, G., {$n$-dimensional division algebras over a field with a cyclic extension of degree $n$. Geom. Dedicata 63 (1996), 69-94.
1504.00188.extracted_bib
{P-I P\'erez-Izquierdo, J.M., {Division composition algebras through their derivation algebras. J. Algebra 303 (2006), 1-29.
1504.00188.extracted_bib
{P Petersson, H. P., {The classification of two-dimensional nonassociative algebras, Result. Math. 37 (2000), 120-154.
1504.00188.extracted_bib
{Sch Schafer, R. D., {Forms permitting composition. Adv. Math. 4 (1970), 127-148.
1504.00188.extracted_bib
{W Waterhouse, W.C., {Linear maps preserving reduced norms. Linear Alg. Appl. 43 (1982), 197-200.