Welcome to mathimatikoi.org forum; Enjoy your visit here.

## Isometry

### Isometry

We consider an isometry $\varphi:\mathbb{R}^n \rightarrow \mathbb{R}^n$. Prove that the isometry $\varphi$ can be writen as a unique sum ( decomposition ) of a vertical geometric transformation as well as a constant. Use the above to determine all isometries of $\mathbb{R}^2$.

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

- Tolaso J Kos
- Administrator
**Posts:**865**Joined:**Sat Nov 07, 2015 6:12 pm**Location:**Larisa-
**Contact:**

### Re: Isometry

The uniqueness part is easy. Since $\varphi$ is clearly continuous (in fact Lipschitz with constant $1$) it suffices to prove that $\varphi$ is affine i.e.

$$\varphi\left(\frac{x+y}{2}\right)=\frac{\varphi(x)+\varphi(y)}{2}$$

So fix $x,y$ and let $z=\frac{x+y}{2}$. Then $x-z=z-y=\frac{x-y}{2}$ and hence by the triangle inequality

\[\|x-y\|=\|\varphi(x)-\varphi(y)\| \le \|\varphi(x)-\varphi(z)\|+\|\varphi(z)-\varphi(y)\|=\|x-z\|+\|z-y\|=\|x-y\|\]

Hence equality must hold in the triangle inequality whence $\varphi(z)$ must be collinear with $\varphi(x),\varphi(y)$ and hence their midpoint (since it's also at the same distance from both the points).

$$\varphi\left(\frac{x+y}{2}\right)=\frac{\varphi(x)+\varphi(y)}{2}$$

So fix $x,y$ and let $z=\frac{x+y}{2}$. Then $x-z=z-y=\frac{x-y}{2}$ and hence by the triangle inequality

\[\|x-y\|=\|\varphi(x)-\varphi(y)\| \le \|\varphi(x)-\varphi(z)\|+\|\varphi(z)-\varphi(y)\|=\|x-z\|+\|z-y\|=\|x-y\|\]

Hence equality must hold in the triangle inequality whence $\varphi(z)$ must be collinear with $\varphi(x),\varphi(y)$ and hence their midpoint (since it's also at the same distance from both the points).

**Imagination is much more important than knowledge.**

## Create an account or sign in to join the discussion

You need to be a member in order to post a reply

## Create an account

Not a member? register to join our community

Members can start their own topics & subscribe to topics

It’s free and only takes a minute

## Sign in

### Who is online

Users browsing this forum: No registered users and 1 guest