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We call two metric spaces $(X,d), (Y,d')$ isometric if there are inverse functions $f: X \to Y$, $g: Y \to X$ with $d(x,x') = d'(f(x), f(x'))$ and $d'(y,y') = d(g(y), g(y'))$.

An example of topologically equivalent but non-isometric metric spaces is the unit circle and the circle of radius $2$ with the standard Euclidean metric in $\mathbb R^2$. Another example would be $\mathbb R^2$ with the Euclidean metric versus $\mathbb R^2$ with the $\max$-metric. (Right?)

It's easy to see that the homeomorphisms in both cases are not isometries. But how can I show that no such isometry can exist at all?

blue
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2 Answers2

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For your first example pair using Euclidean metrics:

In the circle of radius 2, there are two points which are distance $3$ apart. If there were even a single isometry with the unit circle, the unit circle would also have to have a pair of points distance $3$ apart. What do you think?

(Added later)

I had hoped the idea for the first example would inspire you for the second, but I can be less subtle. Consider the unit circles $\Bbb R^2$ under the regular metric and max metric. An isometry of $\Bbb R^2$ between these two would have to map one unit circle to the other, and the same applies to the inverse of the isometry. So, the isometry (after restriction) is actually an isometry of one unit circle onto the other.

Consider a corner $C$ of the unit sphere in the max-metric, and let $f(C)$ be its image on the Euclidean unit circle under the isometry. Now a circle of radius $2$ around $f(C)$ in the Euclidean metric intersects the Euclidean unit circle in one spot (at the diametric opposite of $f(C)$). Is the same true for the circle of radius $2$ around $C$ in the max-metric?

Another argument I thought of that I'm not confident about is this: I think isometries should preserve length. If that's the case, then you can notice that the perimeter of the unit circle in the max-metric is $8$, but in the Euclidean metric it's $2\pi$, and that seems like a giveaway that no isometry is possible. However, I worry about holes in analytical arguments like this that I haven't dealt with in years... The first argument is more elementary.

rschwieb
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Another way to show that no such isometry exists is to show that the equality set, defined for any metric space $(X,d)$ and $x,z \in X$ as $$E_d(x,z)=\{y \in X :d(x,z)=d(x,y)+d(yz,z)\},$$ is not preserved since it is the case that if $$f:(X,d) \rightarrow (Y,d')$$ is a surjective isometry, then for any $x,z \in X,$ we have $$f(E_d(x,z))=E_d{'}(f(x),f(z)).$$

Zeta10
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