How to prove that Rigid rotation is a linear map? A rigid rotation of $\mathbb{R}^n$ is a map $f: \mathbb{R}^n \to \mathbb{R}^n$ such that $d(f(x), f(y)) = d(x ,y)$ for all vectors $x$ and $y$ in $\mathbb{R}^n$. Show that a rigid rotation of $\mathbb{R}^n$ in which the origin is fixed is a linear map.
I believe we can have only $48$ such different rigid rotations of the form $(x, y, z)$, $(y, x, z)$, $(-x, y, z)$ etc. but I'm not sure how to reason this. How can I prove this more rigorously?   
 A: Suppose $f:\mathbb R^n\to\mathbb R^n$ and


*

*$f(0)=0$ and

*for all $x,y\in\mathbb R^n$ we have $d(x,y) = d(f(x),f(y))$.


We would like to prove that $f(x+y)= f(x)+f(y)$.
We know that $f(x)$ and $f(y)$ and $f(x+y)$ are at the same distance from $0$ and from each other as $x$ and $y$ and $x+y$ from $0$ and from each other.
For any $x,y\in\mathbb R^n$ there is a dot-product $x\cdot y\in \mathbb R$, and there is a norm $\|x\|=\sqrt{x\cdot x}$, and we have $d(x,y)=\|x-y\|$.
If we can show $\Big(d(f(x+y),f(x)+f(y))\Big)^2=0$, then we've shown $f(x+y)=f(x)+f(y)$.
The polarization identity says
$$
x\cdot y = \frac{\|x+y\|^2 - \|x-y\|^2} 4.
$$
I'll leave the proof of that as an exercise.

\begin{align}
& \Big(d(f(x+y),f(x)+f(y))\Big)^2 = \|f(x+y) - f(x) - f(y)\|^2 \\[10pt]
= {} & \|f(x+y)\|^2 + \|f(x)\|^2 + \|f(y)\|^2 - 2f(x+y)\cdot f(x) - 2f(x+y)\cdot f(y) + 2f(x)\cdot f(y).
\end{align}
Next I would see if the polarization identity can finish this off.
You also need $f(cx)=cf(x)$ if $c$ is a scalar.  You know that $\|cx\| = |c|\|x\|$.
