Division and multiplication that preserve Euclidean norms

Question

I am looking for ways to define

$$\frac{1}{x}\in \mathbb{R}^n\quad \quad and\quad \quad x\cdot y\in \mathbb{R}^n ,$$ where $x,y\in \mathbb{R}^n$ such that $$\left\|\frac{1}{x}\right\|=\frac{1}{\|x\|}\quad \quad and \quad \quad \|x \cdot y\|=\|x\|\|y\|,$$ where $\|\cdot\|$ is the $n$-dimensional Euclidean norm. Moreover, I need for all $x,y,z\in \mathbb{R}^n,$

1. $$(x\cdot y) \cdot z=x \cdot (y \cdot z)$$

2. $$x\cdot y=y \cdot x$$

3. $$x\cdot \frac{1}{x}=\frac{1}{x}\cdot x=1,$$ where $1$ is the identity of the operation $\cdot$ and $\|1\|=1.$

4. $$ (\alpha x)\cdot y=\alpha (x \cdot y)=x \cdot (\alpha y),$$ where $\alpha \in \mathbb{R}.$

These are all I need.

We have this when $n=2.$ Using the devision and multiplication in complex numbers, we can define $$\frac{1}{(a,b)}=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2} \right)$$ and $$(a,b)\cdot (c,d)=(ac-bd,ad+bc).$$ Could somebody help me? Thank you so much.

Masih.

Answer 1

If we have such a multiplication, then $x\cdot y = |x|\, |y| (e_x\cdot e_y)$ and $x^{-1}=(1/|x|)e_x^{-1}$, so we can focus on the unit sphere. We are then asked to make $S^{n-1}$ an abelian group, with the extra property that $$ (-u)\cdot v = u\cdot (-v) = - (u\cdot v) $$ (this comes from requirement 4, with $\alpha=-1$). Conversely, such a group structure gives a multiplication with the desired properties.

The only (small) obstacle to this is the requirement on the antipodal points, but we can just decompose $S^{n-1}= A\dot{\cup}-A$, then map $A$ bijectively to $\{e^{i\varphi} : 0\le\varphi <\pi\}\subseteq S^1$ and map $-A$ to the corresponding antipodal points on $S^1$ and then use multiplication of complex numbers.

If a continuous multiplication is desired, then we're in trouble and Todd's comment applies.