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\begin{document}

\noindent
{\large \bf Additional notes \#5.}

\bigskip

\noindent
The Cayley-Dickson construction of the $2^n$-onions: 
${\BR}, {\BC}, {\BH}, {\BO}, \ldots$.

\medskip

In these notes we describe a recursive construction
of some famous (not necessarily associative)
${\BR}$-algebras: the real numbers, the complex numbers,
the quaternions, the octonions, the sedenions
(or hexadecanions), etc. The $n$-th constructed 
${\BR}$-algebra $\m a_n$ has the following properties:
\begin{enumerate}
\item[(1)] $\m a_n$ is a $2^n$-dimensional real vector space.
(Hence the nickname {\it $2^n$-onions} for the entire class.)
\item[(2)] The first five operations of 
$\m a_n = \lb A_n; \cdot, +, -, 0, 1, {}^*\rb$ 
are the same type as those of a ring, and $\m a_n$
satisfies all the usual defining equations for rings
except possibly the associative law of multiplication.
\item[(3)] The operation $x^*$ is a unary operation
called {\it conjugation}. The laws of conjugation are
$(x+y)^* = x^*+y^*$, $(xy)^* = y^*x^*$ and $(x^*)^* = x$.
\item[(4)] Each element of each $\m a_n$ has an inverse.
But because $\m a_n$ is nonassociative for $n\geq 3$
we are not guaranteed that division is possible.
In other words, for each $x$ there is an element
$x^{-1}$ such that $xx^{-1} = x^{-1}x = 1$, 
but for a given product $xy$ there may be no
element $\bar{x}$ such that $\bar{x}(xy) = y$.
(In fact, for $n = 3$ division is possible, but 
not for $n\geq 4$.)
\end{enumerate}

We begin the construction with 
$\m a_0 = {\BR}$ where conjugation is defined to
be the identity function: $x^* = x$.

If we have already constructed $\m a_n$, then this
is how we construct $\m a_{n+1}$.
Choose a symbol, say $i$, that does not appear
in $\m a_n$. $\m a_{n+1}$ will consist of elements
$a+bi$ where $a, b\in \m a_n$. The new element $i$
will satisfy the following conditions: $i^2 = -1, i^* = -i$, and
\begin{enumerate}
\item[(1)] $(bi)c = (bc^*)i$ if $b, c\in A_n$.
\item[(2)] $a(di) = (da)i$ if $a, d\in A_n$
\item[(3)] $(bi)(di) = -d^*b$ if $b, d\in A_n$.
\end{enumerate}
The operations of $\m a_{n+1}$ are defined by
\begin{enumerate}
\item[(1)] $(a+bi)(c+di) = (ac-d^*b) + (da+bc^*)i$.
(This definition follows from distributivity
and the laws above.)
\item[(2)] $(a+bi) + (c+di) = (a+c) + (c+d)i$.
\item[(3)] $-(a+bi) = (-a) + (-b)i$,
\item[(4)] $0 = 0+0i$.
\item[(5)] $1 = 1+0i$.
\item[(6)] $(a+bi)^* = a^* + i^*b^* = a^* - bi$.
\end{enumerate}
\bigskip

It is not hard to see that $\m a_1 = {\BC}$, and that
$(a+bi)^* = a-bi$ is usual complex conjugation.
To construct $\m a_2 = {\BH}$ from $\m a_1 = {\BC}$
we should add a new symbol $j$ that does not appear in ${\BC}$.
Then a typical element of ${\BH}$ has the form
$\alpha + \beta j$ with $\alpha, \beta \in {\BC}$.
If $\alpha = a+bi$ and $\beta = c+di$, then 
\[\alpha + \beta j = (a+bi) + (c+di)j = a+bi + cj+dij, \;\;\;
a,b,c,d\in {\BR}.\]
One usually write $k$ for $ij$, and usually expresses
a typical element of ${\BH}$ as simply
$a+bi+cj+dk, \;a,b,c,d\in {\BR}$. There is symmetry in
this form, since $i^2 = j^2 = k^2 = -1$, and
$ij = -ji = k, jk = -kj = i, ki = -ik = j$.
Note that $(a+bi+cj+dk)^* = a-bi-cj-dk$.

The quaternion algebra ${\BH}$ is the last of the $2^n$-onions
that is associative, hence the last one that is a ring
in the sense that we have defined rings. It is the simplest
division ring. Since $\BH$ is associative, 
to prove that it is a division ring
it is enough to show that each element has an inverse:
if $\alpha = a+bi+cj+dk$, then $\alpha^{-1} = 
\alpha^*/(\alpha\alpha^*)$. This makes sense
because $\alpha\alpha^* = a^2 + b^2 + c^2 + d^2$
is a real number, and any real number in ${\BH}$
has an inverse in ${\BH}$ since ${\BR}\subseteq {\BH}$.

The set of elements $\{\pm 1, \pm i, \pm j, \pm k\}$
is closed under multiplication. They form the quaternion group.

The {\it square norm}
of a quaternion $\alpha = a+bi+cj+dk$
is $|\alpha|^2 = (\alpha\alpha^*) = 
a^2+b^2+c^2+d^2$, which is a positive real number,
so it makes sense to define the {\it norm}
to be $|\alpha| = (\alpha\alpha^*)^{\frac{1}{2}} = 
\sqrt{a^2+b^2+c^2+d^2}$. Since the real numbers in 
${\BH}$ commute with all members of ${\BH}$ we have
\[|\alpha\beta|^2 = (\alpha\beta)(\alpha\beta)^*
= (\alpha\beta)(\beta^*\alpha^*)
= \alpha(\beta\beta^*)\alpha^* = \alpha\alpha^*\beta\beta^*
= |\alpha|^2|\beta|^2.\]
Hence $|\alpha\beta| = |\alpha| |\beta|$. 
This can be used to prove that the set 
$\{\alpha\in {\BH}\;|\;|\alpha| = 1\}$ of unit quaternions
is a group under multiplication. 

The properties of quaternion multiplication
help to prove a result in number
theory. Euler tried for decades to prove that 
every positive integer is a sum of four squares.
His plan was to show 
(1) every prime is a sum of four squares, and 
(2) if $m$ and $n$ are both sums of four squares,
then so is $mn$.
Euler proved (2), but could not prove (1).
(Lagrange eventually proved (1).) Euler's proof of 
(2) is remarkable, since the quaternions were not
available to him. But now it is easy to prove (2):

Assume that $m = a^2 + b^2 + c^2 + d^2$ and $n = 
e^2 + f^2 + g^2 + h^2$. Then for the
{\it integer quaternions} $\alpha = a+bi+cj+dk$
and $\beta = e+fi+gj+hk$ we have $m = |\alpha|^2$
and $n = |\beta|^2$. Thus $mn = |\alpha\beta|^2$.
Since the product of integer quaternions
$\alpha$ and $\beta$ is again an integer quaternion
$\alpha\beta = p+qi+rj+sk$
we get that $mn = |\alpha\beta|^2 = 
p^2+q^2+r^2+s^2$ is a sum of four squares!
(Using quaternion multiplication one can easily
write it all out:
\[\begin{array}{rl}
(a^2 + b^2 + c^2 + d^2)(e^2 + f^2 + g^2 + h^2) &=
(ae-bf-cg-dh)^2\\
&+(af+be+ch-dg)^2\\
&+(ag-bh+ce+df)^2\\
&+(ah+bg-cf+de)^2.
\end{array}\]
It's hard to imagine how Euler discovered this
before the quaternions were invented!)
\end{document}

