Cayley-Dickson Construction forms Star-Algebra
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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.
$\blacksquare$