# Distance-preserving maps are compositions of rotations and translations

1. Dec 21, 2012

### Fredrik

Staff Emeritus
How do you prove that a bijection $f:\mathbb R^n\to\mathbb R^n$ such that $d(f(x),f(y))=d(x,y)$ for all $x,y\in\mathbb R^n$ is a composition of a rotation and a translation? (d=Euclidean metric).

My first thought is to define $g=f-f(0)$, prove that g is a bijection that preserves distances too, and then try to prove that g is a rotation. I recently learned how to prove that maps that take straight lines to straight lines are linear, so I'm guessing that the strategy involves proving that g takes straight lines to straight lines.

Maybe we can use that g preserves circles around 0 in the sense that if C is such a circle and $x\in C$, then $g(x)\in C$. I imagine that we would take an arbitrary line L, and then look at points where it intersects some circle around 0, but I don't see how to turn this idea into a proof. I'm not sure if I should try to prove that g takes straight lines to straight lines first, or if there's a more direct approach.

2. Dec 21, 2012

### micromass

Staff Emeritus
You may want to start by actually defining what a rotation is. If you define rotations as elements of $O_n$, the orthogonal group, then what you say is known as the Mazur-Ulam theorem. See http://en.wikipedia.org/wiki/Mazur-Ulam_theorem The proof of the Mazur-Ulam theorem is not very difficult and can be found on a pdf in the wiki page.

3. Dec 22, 2012

### Fredrik

Staff Emeritus
Excellent. That's exactly what I wanted. I thought it would be much harder. Thanks.

4. Dec 22, 2012

### mathwonk

the usual method is to first determine all motions preserving a given point, show these are all in your group, and then show that also your motions are transitive. I.e. since translations are transitive, it suffices to show that all isometries preserving the origin are rotations, but of course they are not, since you have included only orientation preserving motions as rotations and translations, so you need also reflections. Of course you can define rotations as all isometries preserving the origin as micromass seems to suggest, but that is not the usual definition. Rather I think they are usually defined as the component SO (special orthogonal group) of the identity, in the full orthogonal group.

perhaps more revealing than simply defining the difficulty away, is to use a structure theorem such as that in herstein, that shows every orthogonal matrix has in some orthogonal basis the structure of a block matrix made up of sines and cosines, i.e. actual plane rotations, and a bunch of -1's, i.e. some 180 rotations and possibly one reflection. the theorem micromass refers to, is something i have been assuming as obvious, i.e. distance preserving maps preserving the origin are linear, essentially by the parallelogram law. maybe i should be more rigorous on that score. I.e. there are two steps 1) distance preserving maps are affine, 2) orientation preserving linear maps are composed of plane rotations in orthogonal planes.

Last edited: Dec 22, 2012
5. Dec 23, 2012

### Bacle2

I think you can use the fact that the metric is generated by an inner-product; then isometries must preserve the inner-product. you can then show that the only maps that preserve the real inner-product are rotations and translations. Of course this would not work on spaces where the inner-product does not generate the metric.

6. Dec 23, 2012

### pasmith

It is true that a map that preserves an inner product must preserve the norm induced by that product, and so the metric induced by the norm. But the converse is not the case. For example, consider $f : \mathbb{R}^n \to\mathbb{R}^n : x \mapsto x + a$.

Then $f(x) \cdot f(y) = x \cdot y + x \cdot a + y \cdot a + a \cdot a$ which is not identically equal to $x \cdot y$ unless $a = 0$. But $\|f(x) - f(y)\| = \|x + a - y - a\| = \|x - y\|$.

Also: Let $e_i$, $1 \leq i \leq n$, be the standard basis on $\mathbb{R}^n$. For $\sigma \in S_n$, define a linear map $A_\sigma : \mathbb{R}^n \to \mathbb{R}^n$ by $e_i \mapsto e_{\sigma(i)}$. Then $A_\sigma$ preserves inner products since
$$e_i \cdot e_j = \delta_{ij} = \delta_{\sigma(i)\sigma(j)} = e_{\sigma(i)} \cdot e_{\sigma(j)} = A_\sigma e_i \cdot A_\sigma e_j$$
because $i = j$ if and only if $\sigma(i) = \sigma(j)$.

But there are maps $A_\sigma$ which do not correspond to rotations: swapping $e_1$ and $e_2$, for example, which is a reflection in the surface $x_1 = x_2$.