Existence of rigid transformations $T\colon \mathbb{R}^2 \to \mathbb{R}$

Question

I've been trying to prove no transformation that maps $\mathbb{R}^2$ to $\mathbb{R}$ is rigid, but instead I found a possible candidate for it is $T(v)$ = $\vert\vert v \vert\vert$. However, I remain unconvinced. The process I followed is presented here:

Let's assume there is a transformation such that $T\colon \mathbb{R}^2 \to \mathbb{R}$ and $\vert\vert T(v)-T(u) \vert\vert = \vert\vert v-u \vert\vert $ $\forall v,u\in\mathbb{R^2}$. Let v=(x,y), u=(a,b). Since rigid transformations preserve the dot product as well, we will have that $T(v)\cdot T(u) = v\cdot u$ (this will come in handy later on). Once we detect $T(v)-T(u)\in\mathbb{R}$, we can define this subtraction to be $z_1-z_2,\ z_1,z_2\in\mathbb{R}$. Same thing can be done with the dot product, such that $T(v)\cdot T(u) = z_1z_2$. Now, when we use the fact that $\vert\vert T(v)-T(u) \vert\vert = \sqrt{(z_1-z_2)^2} = z_1-z_2$, and $\vert\vert v-u \vert\vert = \sqrt{(x-a)^2+(y-b)^2}$, we sill obtain that our transformation must obey the following equation in order to be rigid:

$$z_1^2-2z_1z_2+z_2^2=x^2+y^2+a^2+b^2-2(xa+yb) = x^2+y^2+a^2+b^2-2(v\cdot u)$$

Since the transformation preserves the dot product, we will have that $2z_1z_2 = 2(v\cdot u)$, so we can substitute and simplify in order to get:

$$z_1^2 + z_2^2 = x^2+y^2 + a^2+b^2$$

Which can give us the values $z_1 = T(v) = \vert\vert v\vert\vert$ and $z_2 = T(u) = \vert\vert u\vert\vert$.

Is this okay? Did I omit something that presents a contradiction? How can I prove its existence or non-existence?

Thanks a lot.

Answer 1

If by rigid transformation you mean a map $T : \Bbb{R}^2 \to \Bbb{R}$ such that $$|T(x)-T(y)| = \|x-y\|_2,\quad \text{ for all }x,y \in \Bbb{R}^2$$ then it doesn't exist.

First assume that $T(0)=0$. Then for every $x \in \Bbb{R}^2$ we have $$|T(x)| = |T(x)-T(0)| = \|x-0\| = \|x\|.$$ Now for any $x,y \in \Bbb{R}^2$ we have \begin{align} |T(x)|^2 -2T(x)T(y) + |T(y)|^2 &= |T(x) - T(y)|^2 \\ &= \|x-y\|_2^2\\ &= \|x\|_2^2 - 2\langle x,y\rangle + \|y\|_2^2\\ &= |T(x)|^2 - 2\langle x,y\rangle + |T(y)|^2\\ \end{align} and hence $$T(x)T(y) = \langle x,y\rangle.$$

Now let $e_1 = (1,0)$ and $e_2 = (0,1)$ so we have $$|T(e_1)| = \|e_1\| = 1,$$ $$|T(e_2)| = \|e_2\| = 1,$$ $$T(e_1)T(e_2) = \langle e_1,e_2\rangle = 0$$ which is a contradiction. Therefore such a map doesn't exist.

Now for a general rigid transformation $T : \Bbb{R}^2 \to \Bbb{R}$ notice that $$T' : \Bbb{R}^2 \to \Bbb{R}, \quad T'(x) := T(x)-T(0)$$ is again a rigid transformation which now satisfies $T'(0)= 0$. But we know that this cannot exist so $T$ also cannot exist.

The same approach can be used to show that there are no rigid transformations $T : \Bbb{R}^m \to \Bbb{R}^n$ when $m \ne n$.

Answer 2

The transformation $T:\mathbb{R}^2 \to \mathbb{R}$ given by $T(v)=\|v\|$ is not rigid. For example it doesn't preserve the distance between $v_1=(1,0)$ and $v_2=(0,1)$: $$ d_{\mathbb{R}^2}(v_1,v_2) = \|v_1-v_2\| = \|(1,0)-(0,1)\|=\|(1,-1)\|=\sqrt{2} $$ while $$ d_{\mathbb{R}}(T(v_1),T(v_2)) = |\|v_1\|-\|v_2\|| = |1-1|=0. $$