6
$\begingroup$

Let $f$ be an automorphism of the algebra of octonions. Is it true that $f$ preserves some quaternionic subalgebra? Has the statement an elementary proof?

$\endgroup$
6
$\begingroup$

Seen as a map of $8$-dimensional Euclidean vector spaces, $f$ is obviously (special) orthogonal, so we can find an orthonormal basis on which is has a block diagonal form of $2\times 2$ rotation matrices, and certainly at least $u,v$ unit octonions, orthogonal to each other and both orthogonal to $1$ (=pure imaginary) such that $f(u) = \cos\theta\cdot u + \sin\theta\cdot v$ and $f(v) = -\sin\theta\cdot u + \cos\theta\cdot v$. Then $1,u,v,uv$ span (and in fact, are an orthonormal basis of) a quaternion algebra which is preserved by $f$ (incidentally, the unit imaginary octonion $uv$ is also preserved by $f$).

$\endgroup$
2
$\begingroup$

I would start by finding imaginary octonion $i$ fixed by $f$. This we can conclude from knowing that $f$ is orthogonal map of $R^7$ perpendicular to $1$. Now we apply argument used in Gro-Tsen answer. Each element in $SO_7$ has fixed vector.

Next I would observe that multiplication by $i$ defines complex structure on perpendicular $R^6$. This complex structure is preserved by $f$, because $f(\color {red}ix)=f(\color{red}i)f(x)=\color{red}if(x)$.

We have proved that $f$ belongs to $U_3$. Now, there exists basis $\color{blue}u,\color{green}v,\color{brown}w$ in $\mathbb C^3$ such that $f$ is diagonal in it. This means $f(\color{blue}u)=e^{\color{red}i\alpha}\color{blue}u$. The subalgebra $\langle1,\color{red}i,\color{blue}u,\color{red}i\color{blue}u\rangle$ is quaternion subalgebra and it is fixed by $f$.

For example:

1) $\color{red}i\color{blue}u=-\color{blue}u\color{red}i$ because $\color{red}i,\color{blue}u$ are perpendicular and imaginary;

2) $(\color{red}i\color{blue}u)\color{blue}u=\color{red}i\color{blue}u^2=-\color{red}i$ (Moufang identity).

These are geometrical intuitions which I use in octonions. To have strict proof we need to have definition of octonion multiplication.

Side comment - interestingly for two perpendicular imaginary octonions $x,y$ element $(L_xR_y)^2$ is octonion automorphism. It is identity on quaternion subalgebra generated by $x, y$ and minus identity on perpendicular 4-space. Such elements form submanifold in $G_2$ being 8-dimensional exceptional symmetric space $G$ of all subalgebras $\{\mathbb H \subset \mathbb O\}$.

Second comment is that this formula $(L_xR_y)^2$ gives automorphism for octonions over finite fields. We need carefully define what "imaginary octonion" means in this case.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.