Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed

Theorem

Let $A = \left({A_F, \oplus}\right)$ be a $*$-algebra.

Let $A' = \left({A_F, \oplus'}\right)$ be constructed from $A$ using the Cayley-Dickson construction.

Let the conjugation operator on $A$ be $*$.

Let $\left({a, b}\right), \left({c, d}\right) \in A'$.

In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:

$a \oplus b =: a b$

$x \oplus' y =: x y$

The context will make it clear which is meant.

Then:

$\ds \left({a, b}\right) + \left({a, b}\right)^*$

$=$

$\ds \left({a, b}\right) + \left({a^*, -b}\right)$

$\ds $

$=$

$\ds \left({a + a^*, b - b}\right)$

$\ds $

$=$

$\ds \left({a + a^*, 0}\right)$

As $A$ is itself nicely normed, we have that $a + a^*$ is real.

Hence it follows that $\left({a + a^*, 0}\right)$ is real.

$=$

$\ds \left({a, b}\right) \left({a^*, -b}\right)$

$\ds $

$=$

$\ds \left({a a^* - \left({-b b^*}\right), a^* b + \left({-a^* b}\right)}\right)$

$\ds $

$=$

$\ds \left({\left \Vert {a} \right \Vert^2 + \left \Vert {b} \right \Vert^2, 0}\right)$

We have by definition of norm that $\left \Vert {a} \right \Vert^2 + \left \Vert {b} \right \Vert^2$ is real.

Hence it follows that $\left({\left \Vert {a} \right \Vert^2 + \left \Vert {b} \right \Vert^2, 0}\right)$ is real.

It follows from reversing the argument that if $A'$ is not nicely normed then nor will $A$ be.

$\blacksquare$