If $G:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ is distance-preserving, $\exists p_{0}$ and a linear isometry $F$ such that $G(p)=F(p)+p_{0}.$

Question

I need to show that, if $G:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ is distance-preserving, $\exists p_{0}\in\mathbb{R}^{3}$ and a linear isometry $F$ such that $G(p)=F(p)+p_{0}.$

My definition of linear isometry is a linear map $F:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ such that $\left<F(v_{1}),F(v_{2})\right>=\left<v_1,v_2\right>.$

My idea is to find some isometry $F$ such that $G'(p)=F'(p)$. From the hypothesis, I get

$$\left<G(p)-G(q),G(p)-G(q)\right>=\left<p-q,p-q\right>. $$ Now, fix $q\in\mathbb{R}^{3}$ as a constant, and differentiate both sides.

$$2\left<G'(p),G(p)-G(q)\right>=2\left<1,p-q\right>$$ Dividing by $2$ and taking the square both sides,

$$|G'(p)(G(p)-G(q))|^{2}=|p-q|^{2}\implies |G'(p)|=\frac{|p-q|^{2}}{|G(p)-G(q)|^{2}}=1 $$ if $G(p)\neq G(q)$. But, that way, I found an expression for $|G'|$, not for $G'$. And I still have a problem with $G(p)=G(q)\iff p=q$. What can I do?

Answer 1

It is a standard result that all rigid motions in $\mathbb R^n$ are given by orthogonal linear transformation composted with a translation.

I found a complete proof here. Basically there are 3 steps:

1. Let $G(0) = p_0$. Then $G_0(x) = G(x) - p_0$ is also distance preserving and $G_0(0) = 0$ (Theorem 2.1 in the note).

2. One proves that $G_0$ satisfies $ G_0(x) \cdot G_0(y) = x\cdot y$ (Thereom 2.2 in the note).

3. One then proves that $G_0$ is linear (Theorem 2.4 in the note).