Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed

Theorem

Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.

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

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

Let $\tuple {a, b}, \tuple {c, d} \in A'$.

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:

$\ds \tuple {a, b} + \tuple {a, b}^*$

$=$

$\ds \tuple {a, b} + \tuple {a^*, -b}$

$\ds $

$=$

$\ds \tuple {a + a^*, b - b}$

$\ds $

$=$

$\ds \tuple {a + a^*, 0}$

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

Hence it follows that $\tuple {a + a^*, 0}$ is real.

$=$

$\ds \tuple {a, b} \tuple {a^*, -b}$

$\ds $

$=$

$\ds \tuple {a a^* - \tuple {-b b^*}, a^* b + \tuple {-a^* b} }$

$\ds $

$=$

$\ds \tuple {\norm a^2 + \norm b^2, 0}$

We have by definition of norm that $\norm a^2 + \norm b^2$ is real.

Hence it follows that $\tuple {\norm a^2 + \norm b^2, 0}$ is real.

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

$\blacksquare$