Anti-isometry that permits existence of metric space that inverts relations of another metric space

Question

Suppose we have two spaces with respective metrics $d_1$ and $d_2$ and an (anti-)isometry $f$ that maps objects from the first space to the second. I'm interested in the setting where the following conditions are satisfied for all arbitrary points $x,y,z$ in the first space:

1. $d_1(x,y) < d_1(x,z) \iff d_2(f(x),f(y))>d_2(f(x),f(z))$
2. $d_1(x,y) > d_1(x,z) \iff d_2(f(x),f(y))<d_2(f(x),f(z))$

What are the necessary and sufficient conditions on the two spaces and metrics for such an (anti-)isometry to exist?

Constructing examples of such objects isn't difficult (e.g. if the number of objects in the first space is finite, then we can compute the pairwise distances in ascending order and then reverse the order to get the pairwise distances for the second space), so I've tried some simulation experiments involving sampling points in one space, reversing the rank order of pairwise distances, and then using embedding methods like multi-dimensional scaling to try and recover embeddings (that respect the new pairwise distances) in the second space.

So far, I've convinced myself that this is generally not possible in Euclidean space even if one space is much higher-dimensional than other. I don't have a proof of this but I'm guessing it's a consequence of triangle inequality.

I'd also appreciate any references to papers that discuss properties of anti-isometries, I've only been able to find a couple that mention it in passing.

Answer 1

Your gut feeling is right, this is not possible, provided that both spaces are rich enough not to be discrete.

For the duration of this post, I'm going to write $f(x)$ as $fx$.

1. Choose an arbitrary distance. Now choose two points $x,y$ in the first space $(X_1)$ located at that distance or closer. Say, $d_1(x,y)=a$.
2. Say, $d_2(fx,fy)=b$. Now, there are points in $X_1$ located closer to $x$ than $a$ (non-discreteness guarantees that). Then their images in $X_2$ are located farther from $fx$ than $b$. Choose one such point: $z$. Say, $d_2(fx,fz)=b+\varepsilon>b$ and $d_1(x,z)<a$.
3. Look at any point $fw$ in the $\varepsilon$-neighborhood of $fz$. Thanks to the triangle inequality, $d_2(fx,fw)>b$, hence $d_1(x,w)<a$.
4. Now look at the triangle $(x,z,w)\in X_1$. Two sides are bounded from above, then so is the third side: $d_1(z,w)<2a$.
5. But wait, this was true for any point $fw\in B_\varepsilon(fz)\subset X_2$. What about the other points of $X_2$? They are farther away frow $fz$, hence their preimages are closer to $z$, hence the same upper bound is true for all points $w\in X_1$.
6. So the whole space $X_1$ fits in a ball with radius $2a$. But wait again; what is $a$? That's an arbitrarily small distance. The space is smaller than anything. There is no space at all.

So it goes.