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Octonion conjugation is an anti-involution

Statement

For $O$ of Real octonions by Cayley–Dickson doubling, octonion conjugation satisfies $\overline{\overset{ˉ}{x}} = x$ and $\overline{x y} = \overset{ˉ}{y} \overset{ˉ}{x}$ . If $x = (a, b)$ with $a, b \in H$ , then

$x + \overset{ˉ}{x} = (\hat{2 a_{0}}, 0), x \overset{ˉ}{x} = \overset{ˉ}{x} x = (\hat{N (a) + N (b)}, 0),$

so its trace and norm are real scalars.

Facts & Assumptions

Given: $x = (a, b)$ and $y = (c, d)$ in the pair model for $O$ .

[F1]

$O = H \times H$ has product $(a, b) (c, d) = (a c - \overset{ˉ}{d} b, d a + b \overset{ˉ}{c})$ and conjugation $(a, b) \mapsto (\overset{ˉ}{a}, - b)$ (Real octonions by Cayley–Dickson doubling).

[F2]

Quaternion conjugation is an involution, reverses products, and satisfies $a + \overset{ˉ}{a} = \hat{2 a_{0}}$ (Quaternion conjugation is an anti-involution).

[F3]

$a \overset{ˉ}{a} = \overset{ˉ}{a} a = \hat{N (a)}$ for every quaternion $a$ ( $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ).

Proof

1.1givenF1F2algebra

Applying quaternion conjugation twice gives $\overline{\overline{(a, b)}} = (\overset{ˉ}{\overset{ˉ}{a}}, b) = (a, b)$ . Also $(a, b) + \overline{(a, b)} = (a + \overset{ˉ}{a}, 0) = (\hat{2 a_{0}}, 0)$ by [F2].

2.1givenF1F2step 1.1algebra

Using [F1] and reversing quaternion products by [F2], $\overline{(a, b) (c, d)} = (\overline{a c - \overset{ˉ}{d} b}, - (d a + b \overset{ˉ}{c})) = (\overset{ˉ}{c} \overset{ˉ}{a} - \overset{ˉ}{b} d, - d a - b \overset{ˉ}{c}) .$ The same pair-product formula gives $\overline{(c, d)} \overline{(a, b)} = (\overset{ˉ}{c}, - d) (\overset{ˉ}{a}, - b) = (\overset{ˉ}{c} \overset{ˉ}{a} - \overset{ˉ}{b} d, - b \overset{ˉ}{c} - d a),$ which is equal because quaternion addition is commutative. Thus $\overline{x y} = \overset{ˉ}{y} \overset{ˉ}{x}$ .

3.1givenF1F3step 1.1algebra∎

Substitution in the pair formula [F1] and the quaternion norm identities [F3] gives $(a, b) (\overset{ˉ}{a}, - b) = (a \overset{ˉ}{a} + \overset{ˉ}{b} b, 0) = (\hat{N (a) + N (b)}, 0), (\overset{ˉ}{a}, - b) (a, b) = (\overset{ˉ}{a} a + \overset{ˉ}{b} b, 0) = (\hat{N (a) + N (b)}, 0) .$ Hence $x \overset{ˉ}{x} = \overset{ˉ}{x} x = (\hat{N (a) + N (b)}, 0)$ , completing all claims.

Depends on

Used by

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Sources

N. Jacobson, Lectures on Quadratic Jordan Algebras, §8 (standard reference, not scraped)
Holger P. Petersson, Albert Algebras, §§3.1–3.2 (standard reference, not scraped)