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## 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
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### 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).

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