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Some authors, e.g. Baez, Ward, defined multiplication of octonions by formula $ (a,b) \cdot^B (c,d)=(ac-db^*, cb+a^*d) \textrm{ for } a,b,c,d\in \mathbb H, $ some others, e.g. Springer & Veldkamp, N. Jacobson, by $ (a,b)\cdot^S(c,d)=(ac-d^*b, da+bc^*) \textrm{ for } a,b,c,d\in \mathbb H $ ($a^*$ denotes the conjugate quaternion to $a$).

Which of these multiplications is better?

Are algebras $\mathbb O=\mathbb H \times \mathbb H$ with multiplications $\cdot^B$ and $\cdot^S$ (and the standard addition and multiplications by real numbers) isomorphic?

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2 Answers 2

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Yes: the isomorphism is given by $(a,b) \mapsto (a,b^*)$.

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  • $\begingroup$ Exactly. The $\cdot ^B$ formula assume we express octonions in shape $a+\iota b$ while for $\cdot ^S$ formula we express them in shape $a+b\iota$ where $a,b \in \mathbb H$. Exchanging quaternion with $\iota$ conjugate quaternion, namely $b\iota=\iota\bar b$. This proves the isomorphism. Now I observed, that this what Gro-Tsen explained below, sorry. Still I prefer to use Greek "jota" or Hebrew letter "jod" for extra element. $\endgroup$
    – user21230
    Commented May 23, 2017 at 11:12
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The "S" formula you give corresponds to writing octonions in the form $q+r\ell$ where $q,r\in\mathbb{H}$ and $\ell$ is a unit octonion orthogonal to $\mathbb{H}$: we have $$(q+r\ell)(q'+r'\ell) = (qq'-r'^*r)+(r'q+rq'^*)\ell$$ which is your "S" formula, written with a more readable notation. The "B" formula you give corresponds to writing octonions in the form $q+\ell r$: we have $$(q+\ell r)(q'+\ell r') = (qq'-r'r^*)+\ell(q^*r'+q'r)$$ They can be deduced from one another by using the fact that $q + \ell r = q + r^*\ell$ (a consequence of either formula) which gives the isomorphism you ask for: it's an easy exercise to check that either formula gives the other by applying this isomorphism (e.g., $q(\ell r') = q(r'^*\ell) = (r'^* q)\ell = \ell(q^* r')$ applying the "S" formula for the middle equality).

As to which is best, it really depends whether you prefer to write $\mathbb{O}$ as $\mathbb{H} \oplus \mathbb{H}\ell$ or $\mathbb{H} \oplus \ell\mathbb{H}$. Personally prefer the former (so, "S"), but that's maybe just because I'm right-handed.

Incidentally, as for how to remember the formula, I think it's better to keep it split as (1) $q(r'\ell)=(r'q)\ell$, (2) $(r\ell)q' = (rq'^*)\ell$ and (3) $(r\ell)(r'\ell) = -r'^*r$ (obviously the purely quaternionic part doesn't need any remembering): the main difficulty is to remember the order of the factors. It's possible to recover them by remembering that $\ell$ anti-commutes with purely imaginary quaternions and anti-associates with a pair of orthogonal imaginary quaternions, but I think a better mnemonic is that for a unit quaternion $w$, the map fixing $\mathbb{H}$ and taking $r\ell$ to $(wr)\ell$ should be an automorphism of the octonions: this tells us, for example, that $r'$ should be on the left in the formula (1), because we need $q((wr')\ell)$ to be $(wr'q)\ell$, and it similarly constrains (2) and (3) so as to make mistakes impossible.

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  • $\begingroup$ I prefer the $\cdot ^S$ formula as well :) I will not reveal whether I am right or left-handed here :) Note that if we assume that quaternion commutes with extra element $\iota$ then we can obtain $\mathbb C\otimes \mathbb H$ = $M_2\mathbb C$, namely formula is:$ (a,b)\cdot (c,d)=(ac-bd, ad+bc) \textrm{ for } a,b,c,d\in \mathbb H $ $\endgroup$
    – user21230
    Commented May 23, 2017 at 11:23

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