Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed

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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.


Then $A'$ is a nicely normed algebra if and only if $A$ is also a nicely normed algebra.


Proof

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.


Let $A$ be a nicely normed algebra.

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.


Next:

\(\ds \tuple {a, b} \tuple {a, b}^*\) \(=\) \(\ds \tuple {a, b} \tuple {a^*, -b}\)
\(\ds \) \(=\) \(\ds \tuple {a a^* - \tuple {-b b^*}, a^* b + \tuple {-a^* b} }\)
\(\ds \) \(=\) \(\ds \tuple {\norm a^2 + \norm b^2, 0}\) as $A$ is nicely normed

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.

$\blacksquare$


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