# Isometric?

• Oct 10th 2010, 10:23 AM
Showcase_22
Isometric?
Quote:

Let $\displaystyle X=\{A,\ B,\ C, \ D\}$ with $\displaystyle d(A,D)=2$, but all other distances equal to 1.

d is a metric.

Prove that the metric space X is not isometric to any subset of $\displaystyle \mathbb{E}^n$ for any n.

Hint: can you realise $\displaystyle X$ as a subset of a sphere $\displaystyle S^2$ of appropriate radius, with the spherical "great circle" metric?
I think I have to take two points in $\displaystyle X$ and show that any mapping from $\displaystyle X$ to $\displaystyle \mathbb{E}^n$ would result in a different distance between them.

Here's my attempt:

Take a sphere of radius 1 and the points A and D.

Since $\displaystyle d(A,D)=2$, A and D are on opposite sides of the sphere.

Suppose there exists an isometry $\displaystyle \phi:X \rightarrow \sigma$ where $\displaystyle \sigma$ is is the plane used in stereographic projection (also a subset of $\displaystyle \mathbb{E}^n$).

Then $\displaystyle \phi(A)=\phi(D)$. Hence $\displaystyle d(\phi(A),\phi(D))=0$ which is not equal to $\displaystyle d(A,D)=2$.

Hence $\displaystyle \phi$ cannot be an isometry.
• Oct 10th 2010, 11:26 AM
Opalg
Quote:

Originally Posted by Showcase_22
I think I have to take two points in $\displaystyle X$ and show that any mapping from $\displaystyle X$ to $\displaystyle \mathbb{E}^n$ would result in a different distance between them.

Here's my attempt:

Take a sphere of radius 1 and the points A and D.

Since $\displaystyle d(A,D)=2$, A and D are on opposite sides of the sphere.

Suppose there exists an isometry $\displaystyle \phi:X \rightarrow \sigma$ where $\displaystyle \sigma$ is is the plane used in stereographic projection (also a subset of $\displaystyle \mathbb{E}^n$).

Then $\displaystyle \phi(A)=\phi(D)$. Hence $\displaystyle d(\phi(A),\phi(D))=0$ which is not equal to $\displaystyle d(A,D)=2$.

Hence $\displaystyle \phi$ cannot be an isometry.

I'm not sure that the hint is very helpful, and I don't think that stereographic projection is needed.

If such an isometry $\displaystyle \phi$ exists then the points $\displaystyle \phi (A),\ \phi (B),\ \phi (C)$ form an equilateral triangle with side 1, in some 2-dimensional subspace of $\displaystyle \mathbb{E}^n$. So do the points $\displaystyle \phi (B),\ \phi (C),\ \phi (D)$. Thus the images of the four points lie in some 3-dimensional subspace of $\displaystyle \mathbb{E}^n$, so we may as well assume that n=3.

The points $\displaystyle \phi (A)$ and $\displaystyle \phi (D)$ must both lie at a distance $\displaystyle \sqrt3/2$ from the midpoint of the line segment joining $\displaystyle \phi (B)$ and $\displaystyle \phi (C)$. So the distance from $\displaystyle \phi (A)$ to $\displaystyle \phi (D)$ is at most $\displaystyle \sqrt3<2$, contradicting the condition that $\displaystyle \phi$ is an isometry.