3
$\begingroup$

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.

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.

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ Doesn't the standard octonion basis do the job? It looks like it to me, unless I'm out by a sign. $\endgroup$
    – Paul Reynolds
    Commented Feb 27, 2015 at 13:39
  • $\begingroup$ Can you elaborate a bit on how this is the key fact bringing everything together? $\endgroup$
    – Vincent
    Commented Feb 27, 2015 at 13:56
  • 1
    $\begingroup$ @Jjm I think the standard basis does work, I checked. $\endgroup$
    – Paul Reynolds
    Commented Feb 27, 2015 at 19:20
  • 2
    $\begingroup$ 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. $\endgroup$
    – Paul Reynolds
    Commented Feb 27, 2015 at 21:47
  • 2
    $\begingroup$ 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. $\endgroup$
    – Paul Reynolds
    Commented Feb 27, 2015 at 21:50

1 Answer 1

Reset to default
1
$\begingroup$

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 $\mathbb C\otimes \mathbb O$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .