# Isometric?

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

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

d is a metric.

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

Hint: can you realise $X$ as a subset of a sphere $S^2$ of appropriate radius, with the spherical "great circle" metric?
I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ 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 $d(A,D)=2$, A and D are on opposite sides of the sphere.

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

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

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

Originally Posted by Showcase_22
I think I have to take two points in $X$ and show that any mapping from $X$ to $\mathbb{E}^n$ 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 $d(A,D)=2$, A and D are on opposite sides of the sphere.

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

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

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

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