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I have been away from math for a while so be gentle if this is not very rigorous or if I am redefining already defined objects. The essential question that started me on this was the following. In the plane every circle has one center. On the surface of the sphere every circle has two centers. Okay, so let's formalize this notion.

Let $(X, d)$ be a metric space and let $C_r(x) = \{y \in X \ | \ d(x-y) = r \}$. Let $D \subseteq \mathbb{R}$ be the set of all distances between any two points of $X$. Then consider the following function:

$$f: X \times D \rightarrow P(X)$$

$$(x, r) \mapsto \{ z \in X \ | \ \exists s \in D \text{ with } C_s(z) = C_r(x) \}$$

We know that $x \in f(x,r), \forall (x,r) \in X \times D$. So this set is never empty and contains the original point. It is also reflexive in the sense that if $y \in f(x,r)$ for some $r \in D$ then $x \in f(y,s)$ for some $s \in D$. It would be nice if we could make this an equivalence relation. We can say that a metric space $(X,d)$ is regular if $f(x,r) = f(x,s) \ \forall r,s \in \mathbb{D}$. For a regular space $(X,d)$ we can define the following equivalence relation:

$$x \sim y \iff \exists r,s \in D \text{ with } C_r(x) = C_s(y)$$

Returning to our examples:

a) $\mathbb{R}^n$ with the metric induced by any $\mathbb{L}_p$ norm. Every equivalence class contains only the point itself.

b) The surface of a sphere with the "shortest path on the surface" distance. The equivalence classes are the sets of antipodal points.

c) A cone. I believe the center of the base and the top of the cone are in the same equivalence class, while every other point is in its own class.

We can say that a regular metric space is uniform if every equivalence class is of equal size. For a regular uniform metric space we can call the size of the equivalence class the degree of it. So that the first two examples would be regular uniform metric spaces but the 3rd would be regular but not uniform. The degree of $\mathbb{R}^n$ would be 1, while the degree of the surface of the unit sphere would be 2.

So here are some obvious questions I haven't been able to find answers to.

  1. Are there any other non-trivial regular uniform metric spaces?

  2. Can we find a regular metric space for every degree $n$? @user44191 Provided an affirmative answer to this question. A discrete set of $n$ points such that all distances between them are unique is a regular uniform metric space with degree $n$.

  3. What can we say about the p-adic numbers $\mathbb{Q}_p$? They wouldn't be regular in the above sense. Consider for example the set of points that are distance 1 from 0. Then every other number will be a center of these numbers. We can make the same argument for the set of points that are distance $\frac{1}{p}$ from 0. If this was an equivalence relation then from the above to sets we should get that all points are in the same equivalence class. But looking back at the set of points with distance 1 from 0, we see that those points are not the center of it's own set as for example $|5|_3 = 1$ and $|2|_3 = 1$ but $|5-2|_3 \neq 1$. But I guess they are uniform.

  4. What can we say about the surface of $S^n$? What can we say about a torus? A torus of genus 2? etc

  5. Are there any equivalent definitions of uniformity or regularity?

  6. Is there another condition other than regularity that would make this notion an equivalence class?

  7. What can we say about the set of spaces on which this notion is an equivalence class?

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  • $\begingroup$ For question 2: consider the $n$-point discrete space where the distances are all distinct. If I understand correctly, this should be a regular metric space with degree $n$. $\endgroup$ – user44191 Nov 10 '19 at 4:46
  • $\begingroup$ Definition of $C_r(x)$ features $d(x-y)$ where $x$ belongs to an abstract space $X$ while $y\in\Bbb R^n$. Something is not exactly right. $\endgroup$ – Wlod AA Nov 10 '19 at 4:49
  • $\begingroup$ If $X$ is regular then it seems to me (am I wrong?) that arbitrary $x\in X$ can be equivalent only to itself and not to any other point. $\endgroup$ – Wlod AA Nov 10 '19 at 4:58
  • $\begingroup$ @WlodAA I believe $y \in X$ is the intended definition; the sphere seems to be regular, with antipodes equivalent, if I'm not mistaken. $\endgroup$ – user44191 Nov 10 '19 at 5:00
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    $\begingroup$ Considering $\mathbb{T}^2$ as a product of the circle with itself, we can define a "taxicab metric" on it by adding the distances on both projections; I think this gives us a regular metric space, with $(a, b) \sim (a + \frac{1}{2}, b + \frac{1}{2})$. Not sure though - you may want to check. $\endgroup$ – user44191 Nov 12 '19 at 23:01
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This is a partial answer:

Assume $X$ is a connected, regular space. I claim there are no triples $x \sim y \sim z$.

Proof:

Assume otherwise. First, note that if $x \sim y$, then $C_{d(x, y)}(x) = \{y\}$. This is because $C_{d(x, y)}(x) = C_r(y)$ for some $r$, but as $y \in C_{d(x, y)}(x)$, we have that $y \in C_r(y)$, so $r = 0$, and $C_0(y) = \{y\}$.

Assume $x \sim y$. Then let $f: X \rightarrow \mathbb{R}_{\geq 0}^2, f(z) = (d(x, z), d(y, z))$. The image includes $(d(x, y), 0)$ and $(0, d(x, y))$; the above note implies that the image cannot otherwise cross the lines $(d(x, y), a)$ or $(b, d(x, y))$. Then $\{(a, b): a, b > d(x, y)\}$ is open, as is $\{(a, b): a < d(x, y) \, \text{OR} \, b < d(x, y)\}$; their union contains all of $\mathbb{R}_{\geq 0}^2$ except points on those two lines (and the latter includes the two points above), so their preimages are open and contain all of $X$. Therefore, as $X$ is connected, one of the preimages must be empty - and $(d(x, y), 0)$ and $(0, d(x, y))$ are both in the latter, so the former must be empty. Therefore, for any $z \in X$, we must have that $d(x, z) < d(x, y)$ or $d(y, z) < d(x, y)$ (or both).

Now assume $x \sim y \sim z$. Then by the first note, we have that $d(x, y) \neq d(x, z) \neq d(y, z)$; one of the three distances must therefore be smallest, say (without loss of generality) $d(x, y)$. But then $d(x, z) > d(x, y)$ and $d(y, z) > d(x, y)$, contradicting the above paragraph.


Some additional reasoning:

Using the above, I'm going to rephrase the question slightly. Given two points $x, y \in X$, define $f_{x, y}: X \rightarrow \mathbb{R}_{\geq 0}^2, z \mapsto (d(x, z), d(y, z))$. Then $x \sim y$ iff $f(X)$ meets each horizontal and vertical line at most once, that is, if $d(x, z) = d(x, w)$, then $d(y, z) = d(y, w)$, and vice versa.

Using this definition, I claim that if $x \sim y$, then for any $z \in X$, we have that $d(x, z) \leq d(x, y)$. In other words, $y$ is the most distant point from $x$.

Assume otherwise. Note that by the above proof, we have that $d(y, z) < d(x, y)$. Then consider the disjoint open sets $A, B \subseteq \mathbb{R}_{\geq 0}^2$:

$A := \{(a, b): a < d(x, y), b > d(y, z)\}$

$B := \{(a, b): a > d(x, y) \, OR \, b < d(y, z)\}$

Clearly, $f(x) = (0, d(x, y)) \in A, f(y) = (d(x, y), 0) \in B$, and both are in $f(X)$. But $f(X)$ is connected, so there must be some point $f(w)$ in neither of these open sets. Therefore either $d(x, w) = d(x, y)$ and $d(y, w) > d(y, z)$, or $d(x, w) < d(x, y), d(y, w) = d(y, z)$. In the former case, $d(y, w) = d(y, y) = 0$, but we assumed $w \neq y$. In the latter case, $d(x, w) = d(x, z)$, but we assumed $d(x, z) > d(x, y)$, and $d(x, y) > d(x, w)$. So this is a contradiction, meaning our assumption was wrong, and if $x \sim y$, then $y$ is the most distant point from $x$.

If $X$ is a uniform, connected regular space of degree $2$, then this gives us a nice "antipode" map. I strongly suspect this map is continuous.

We can then define a function $g: [0, d(x, y)] \rightarrow [0, d(x, y)]$ such that if $d(x, z) = c$, then $d(y, z) = g(c)$. The proof above should be relatively easy to adapt to prove that $g(c)$ is increasing and continuous.


It's not hard to see that if $X, Y$ are regular with degree $2$ and the circle distances are additive (i.e. $C_r(x) = C_{c - r}(y)$ for some $c$), then $X \times Y$ is also regular with degree $2$ if the metric used is $d((x_1, y_1), (x_2, y_2)) = d(x_1, x_2) + d(y_1, y_2)$. This applies to the torus - in fact, it gives an uncountable number of metrics on the torus, as the metric on either factor circle can be "stretched" by any constant.

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  • $\begingroup$ I am really glad you are still working on this. It seems this was another case of Maschler spaces. I followed your argument "the above note implies that the image cannot otherwise cross the lines". I will try to understand more of it tomorrow. $\endgroup$ – Heraiwa Nov 11 '19 at 5:28
  • $\begingroup$ @Heraiwa If you draw the lines in the plane, the argument should be pretty clear: the image can't cross a certain horizontal line or a certain vertical line, except at the axis - so as the image is connected, it can't have anything on the other side of both of those lines. The way I worded it was somewhat formal because I thought it would be clearer. $\endgroup$ – user44191 Nov 11 '19 at 6:07

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