Once octonions product is studied, together with the relations with $Spin(8)$ and $SO(8)$ geometry (see for instance Robert Bryant's notes), one realises that the key fact bringing all the phenomena of triality etc. is the following: that there exist some octonions $u_1,\ldots,u_k$ such that

$$R_{u_k}\ldots R_{u_1}=-Id\qquad\qquad L_{u_k}\ldots L_{u_1}=Id,$$

where $L_u$ and $R_u$ denote left and right product with $u$ respectively.

This may be found by the Lie group properties of $Spin(8)$ and $SO(8)$. But

May some set $u_1,\ldots,u_k$ be described explicitely?

Any idea is welcome.


Although it is not directly related to the question, it is interesting to note the vital importance of the claim. Following Briant's notes, we consider the group of maps $\mathbb{O}\oplus\mathbb{O}\longrightarrow\mathbb{O}\oplus\mathbb{O}$ generated by the elements $L_u\oplus R_u:(a,b)\longmapsto(ua,bu)$ with $\|u\|=1$. Of course, we are dealing with some group $G\subset SO(8)\times SO(8)$. Then: if we read the previous paragraphs, it happens that $G$ is the group $Spin(8)$ defined in terms of Clifford algebras, and the key property of this $Spin(8)$ being really a double cover of $SO(8)$ is supported partially by the fact that the inclussion $G\subset SO(8)\times SO(8)$ is not a vacuous one, as could be the diagonal $SO(8)\subset SO(8)\times SO(8)$, because $(Id,-Id)\in G$. That is why the sign is crucial.

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    $\begingroup$ Doesn't the standard octonion basis do the job? It looks like it to me, unless I'm out by a sign. $\endgroup$ Feb 27, 2015 at 13:39
  • $\begingroup$ Can you elaborate a bit on how this is the key fact bringing everything together? $\endgroup$
    – Vincent
    Feb 27, 2015 at 13:56
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    $\begingroup$ @Jjm I think the standard basis does work, I checked. $\endgroup$ Feb 27, 2015 at 19:20
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    $\begingroup$ I would be happy to, but it can be seen just by using the/a multiplication table, such as the one shown here: en.wikipedia.org/wiki/Octonion. You just need to check that $(((((((\mathfrak{i} \ \mathfrak{1}) \ \mathfrak{2}) \ \mathfrak{3}) \ \mathfrak{4}) \ \mathfrak{5}) \ \mathfrak{6}) \ \mathfrak{7})$ gives $\pm \mathfrak{i}$ (where $1, \mathfrak{1}, \ldots, \mathfrak{7}$ is the basis) for each basis member $\mathfrak{i}$, and that when you do it the other way around you get the opposite. It doesn't take too long to check directly. $\endgroup$ Feb 27, 2015 at 21:47
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    $\begingroup$ Note that there's nothing special about this basis (and I prefer not to think the octonions have a 'standard' basis at all), and any Cayley frame (i.e. $G_2$-related to this one) will work as well. $\endgroup$ Feb 27, 2015 at 21:50

1 Answer 1


Let $u_1,...,u_8$ be any eight perpendicular unit octonions. Then product $L_{u_1}L_{\bar{u_2}}...L_{u_7}L_{\bar{u_8}}=I$ and $R_{u_1}R_{\bar{u_2}}...R_{u_7}R_{\bar{u_8}}=-I$ (each even element is conjugated).

Alternatively you can take seven perpendicular imaginary unit octonions, then you can skip conjugation and reverse sign at $I$.

I have looked briefly Robert Bryant's notes. I am using the same definition for Spin(9) and Spin(10). This is interesting for me, although he didn't obtain $G_{2,8}^+$ = $Spin(10)/(Spin(8)\times Spin(2))$. This is dimension 1 of projective space over $\mathbb C\otimes \mathbb O$.


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