Once octonions product is studied, together with the relations with $Spin(8)$ and $SO(8)$ geometry (see for instance [Robert Bryant's notes](http://www.math.duke.edu/~bryant/Spinors.pdf)), 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.


**EDIT:**


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.



Octonions product: inversion in the right and identity in the left
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 $\mathbb C\otimes \mathbb O$.


