How do you find the 3rd roots of hypercomplex imaginary units?

Question

In my instance I want to find the 3rd root of the octonion imaginary unit $e_4$. I am working on simplifying the octonion $$ o=5+2e_1 \sqrt[3]{e_4}+3e_2 \sqrt[3]{e_4}+2e_3 \sqrt[3]{e_4} $$ $$ =5+2e_1e_4^{\frac{1}{3}}+3e_2e_4^{\frac{1}{3}}+2e_3e_4^{\frac{1}{3}} $$ to a form with only one imaginary unit per $Im(o)$, but I am stuck at trying to simplify $\sqrt[3]{e_4}$ so I can multiply the imaginary units with each other. How would I do that? Or is there a way to skip the step and multiply the units without solving the 3rd root?

Answer 1

For some purposes it's useful to write $\mathbb{O}=\mathbb{R}\oplus\mathbb{R}^7$, i.e. to treat octonions like a formal sum of a scalar and a 7D vector. Generalizing from the quaternions, the square roots of $-1$ are precisely the unit vectors, which form a $S^6\subset\mathbb{R}^7$, and every quaternion has a polar form $r\exp(\theta\mathbf{u})$ with $r\ge0$ positive, $0\le\theta\le\pi$ convex, and $\mathbf{u}$ a unit vector (which is unique for nonreal octonions, and for nonzero real octonions $\theta=0$ or $\pi$ according to sign and $\mathbf{u}\in S^6$ is arbitrary).

It satisfies Euler's $\exp(\theta\mathbf{u})=\cos(\theta)+\sin(\theta)\mathbf{u}$. As a result, any unit vector $\mathbf{u}$ has three cube roots, each $\exp(\theta\mathbf{u})$ with $\displaystyle\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}$ corresponding to $\displaystyle\pm\frac{\sqrt{3}}{\,2}+\frac{\mathbf{u}}{2},-\mathbf{u}$.