# $\mathbb{S}^1$-action and octonionic multiplication can be associated

Let $\mathbb{S}^7$ be the unit sphere of $\mathbb{R}^8$, which can be identified with the unit octonions. The circle $\mathbb{S}^1$ naturally acts on $\mathbb{S}^7$ by complex multiplication:

$$z \cdot x = (z x_1, z x_2, z x_3, z x_4),$$

where $z \in \mathbb{S}^1$, $x = (x_1, x_2, x_3, x_4) \in \mathbb{S}^7$ and $x_i \in \mathbb{C}$.

Denote octionic multiplication by $\ast$.

Is it true that

$$z \cdot (x \ast y) = (z \cdot x) \ast y, \quad \forall z \in \mathbb{S^1}, \, \forall x, y \in \mathbb{S}^7?$$

• This is very easy to check explicitly. What did you have difficulty with, when you attempted that? – xyzzyz Oct 13 '17 at 21:09

Hint: when $x=1$ the equality says $z\cdot y=z\ast y$. Do you think that's true?
• @EduardoLonga Don't guess, check. Consider the Cayley-Dickson multiplication laws, viewing $\Bbb O$ as $\Bbb H\oplus\Bbb H\ell$. Does octonionic multiplication by $i$ in the second factor match multiplication by $i$ on $\Bbb H$? In other words, check $z=i$ and $y=j\ell$. – anon Oct 30 '17 at 2:02