# Cayley-Dickson Construction forms Star-Algebra

## Theorem

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

Let $A' = \struct {A'_F, \oplus'} = \struct {A, \oplus}^2$ be the algebra formed from $A$ by the Cayley-Dickson construction.

Then $A'$ is also a $*$-algebra.

## Proof

### Bilinearity of $\oplus'$

First we need to show that $\oplus'$ is bilinear.

$(1): \quad$ Let $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {c, d} \in A'$.

Then:

 $\ds$  $\ds \paren {\tuple {a_1, b_1} + \tuple {a_2, b_2} } \oplus' \tuple {c, d}$ $\ds$ $=$ $\ds \tuple {a_1 + a_2, b_1 + b_2} \oplus' \tuple {c, d}$ $\ds$ $=$ $\ds \tuple {\paren {a_1 + a_2} \oplus c - d \oplus \paren {b_1 + b_2}^*, \paren {a_1 + a_2}^* \oplus d + c \oplus \paren {b_1 + b_2} }$ $\ds$ $=$ $\ds \tuple {a_1 \oplus c + a_2 \oplus c - d \oplus b_1^* - d \oplus b_2^*, a_1^* \oplus d + a_2^* \oplus d + c \oplus b_1 + c \oplus b_2}$ as $\oplus$ is bilinear $\ds$ $=$ $\ds \tuple {a_1 \oplus c - d \oplus b_1^*, a_1^* \oplus d + c \oplus b_1} + \tuple {a_2 \oplus c - d \oplus b_2^*, a_2^* \oplus d + c \oplus b_2}$ $\ds$ $=$ $\ds \paren {\tuple {a_1, b_1} \oplus' \tuple {c, d} } + \paren {\tuple {a_2, b_2} \oplus' \tuple {c, d} }$

Similarly (and equally tediously) we can show that:

$\tuple {c, d} \oplus' \paren {\tuple {a_1, b_1} + \tuple {a_2, b_2}} = \paren {\tuple {c, d} \oplus' \tuple {a_1, b_1}} + \paren {\tuple {c, d} \oplus' \tuple {a_2, b_2}}$

$(2): \quad$ Let $\tuple {a, b}, \tuple {c, d} \in A'$ and $\alpha, \beta \in \R$.

Then:

 $\ds$  $\ds \paren {\alpha \tuple {a, b} } \oplus' \tuple {c, d}$ $\ds$ $=$ $\ds \tuple {\alpha a, \alpha b} \oplus' \tuple {c, d}$ as $\oplus$ is bilinear $\ds$ $=$ $\ds \tuple {\alpha a \oplus c - d \oplus \alpha b^*, \alpha a^* \oplus d + c \oplus \alpha b}$ $\ds$ $=$ $\ds \alpha \tuple {a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}$ as $\oplus$ is bilinear $\ds$ $=$ $\ds \alpha \tuple {a, b} \oplus' \tuple {c, d}$

Similarly:

$\tuple {a, b} \oplus' \paren {\tuple {c, d} \beta} = \tuple {a, b} \oplus' \tuple {c, d} \beta$

So $\oplus'$ has been shown to be a bilinear mapping.

### Conjugate Nature of $*'$

We have that:

$\forall \tuple {a, b} \in A': {\tuple {a, b}^*}' = \tuple {a^*, -b}$

So:

 $\ds$  $\ds {\paren { {\tuple {a, b}^*}'}^*}'$ $\ds$ $=$ $\ds {\tuple {a^*, -b}^*}'$ $\ds$ $=$ $\ds \tuple {\paren {a^*}^*, -\paren {-b} }$ $\ds$ $=$ $\ds \tuple {a, b}$ as $*$ is a conjugation on $A$

Finally:

 $\ds$  $\ds {\paren {\tuple {a, b} \oplus' \tuple {c, d} }^*}'$ $\ds$ $=$ $\ds {\tuple {a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}^*}'$ $\ds$ $=$ $\ds \tuple {\paren {a \oplus c - d \oplus b^*}^*, -\paren {a^* \oplus d + c \oplus b} }$ $\ds$ $=$ $\ds \tuple {c^* \oplus a^* - b \oplus d^*, -\paren {c \oplus b + a^* \oplus d} }$ $\ds$ $=$ $\ds \tuple {c^*, -d} \oplus' \tuple {a^*, -b}$ $\ds$ $=$ $\ds {\tuple {c, d}^*}' \oplus' {\tuple {a, b}^*}'$

thus proving that $*'$ is a conjugation on $A'$.

Hence the result.

$\blacksquare$