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


Analytic Geometry, Differential Geometry, Manifolds, Riemannian Geometry, Euclidean Geometry
Post Reply
User avatar
Articles: 0
Posts: 174
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia



Post by Riemann » Sun Oct 30, 2016 6:22 pm

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 $\mathbb{R}^2$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
User avatar
Tolaso J Kos
Administration team
Administration team
Articles: 2
Posts: 861
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa

Re: Isometry


Post by Tolaso J Kos » Sun Sep 22, 2019 8:48 am

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.


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.
Post Reply