# distance regular metric spaces

A metric space (V,d) will be called distance regular if for every distances a>0, b, c a nonnegative integer p(a,b,c) is defined, so that whenever d(B,C)=a, there are precisely p(a,b,c) points A such that d(A,B)=c, d(A,C)=b.

The Euclidean plane is an example: p(a,b,c)=0,1, or 2 when the triangle inequality for a,b,c, correspondingly, fails, turns into equality, or is strict.

If we also require that p(a,b,c)>0 whenever the triangle inequality does not fail, then I conjecture that this is the only possibility for the parameters p(a,b,c). That is, there may be many non-isomorphic examples, but the parameters will be the same for all of them. (Thanks to Heather for this clarification.)

Has anybody formulated/proved/refuted this conjecture before? It looks very natural.

UPD. I should have mentioned this: "for every distances a>0, b, c" means all nonnegative reals, and the same for "whenever the triangle inequality does not fail". In particular this means that all positive real distances actually occur.

UPD2. After a week trial, the question seems to be new, open, and interesting. Anton suugested a line of attack, and I believe I can write down the proof of the first step: that p(a,b,c)=1 when the triangle inequality turns into equality. Fedja produced examples showing that this first step is indeed essential. I'm adding the "open problem" tag to the question.

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If I understand correctly, there exist other examples. E.g., as a subset of Euclidean 3-space, take the union of the planes where z=0 and z=1. Maybe you could add the hypothesis that the space is connected? – Martin M. W. Dec 28 '09 at 16:13
Does not work. The number p(1, 1/2,1/2) is not defined: if we take points B,C at distance 1 in the same plane, there is a midpoint, if in different planes, there is none – Dima Fon-Der-Flaass Dec 28 '09 at 16:16
Oh, OK. Good point. – Martin M. W. Dec 28 '09 at 16:18
What about the hyperbolic plane? It's symmetrical enough for distance-regularity. As for your condition that any triangle-inequality-satisfying triple (a, b, c) be realized as a triangle: I think it holds, since we can achieve c = b-a and c = b+a, and thus, since distances vary continously with pairs of points, everything in between. – macbeth Dec 28 '09 at 17:57
Heather: yes, this is another example. But the parameters p(a,b,c) are the same. – Dima Fon-Der-Flaass Dec 28 '09 at 18:16

Note that any metric with unique infinite geodesics on $\mathbb R^2$ has this property. In particular, hyperbolic plane as noted by Heather Macbeth. It also includes Minkowski plane with smooth ball and all complete negatively curved Riemannian metrics on $\mathbb R^2$.

So it is better to ask:

• Is it true that the function $p$ is the only possible? ($p(a,b,c)=0,1,$ or $2$ when the triangle inequality for a,b,c, correspondingly, fails, turns into equality, or is strict.)

• Is it true that space has to be homeomorphic to $\mathbb R^2$?

Sketch for locally compact case You condition implies existence of line (i.e. infinite minimizing geodesic) through any pair of points. Such line must be unique; otherwise for some values $p(a,b,a+b)=\infty$ for some values $a,b$.

If space is locally compact then line depends continuousely on the points. For some points $a_1, a_2, a_3, a_4$ take geodesic $a_1a_2$ connect each point to $a_3$ and connect each point of this trianle to $a_4$. This way we get a cone-map of 3-simplex in your space. If it is nondegenerate (i.e. the image does not coinsides with image of its boundary) then it would be easy to get a triple of small numbers $a,b,c$ such that $p(a,b,c)=\infty$ thus we arrive to a contradiction.

Thus the space must be 2-dimensional and with unique infinite geodesics --- I bet it should be homeomorphic to $\mathbb R^2$. Then it follows that $p(a,b,c)=0,1$ or $2$ as defined above.

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It all would be nice if not for the following. Using the transfinite induction, you can easily construct a set in $\mathbb R^3$ that intersects each circle by any number of points greater than 2 you want depending on the radius of the circle only. Indeed, the cardinality of the set of all circles is continuum. Now take any complete ordering of this set so that each beginning interval has cardinality less than continuum and start adding points to the least circle that still has too few points on it. Clearly, the points that can overload other circles should lie on some circles with at least 3 points already, but those are fewer than continuum at each step of induction, so they intersect our circle by less than continuum points and there are continuum points on our circle to choose from. So, at each step we can make a good choice.

Needless to say, this can be generalized quite a lot, so I'm not sure if there are any restrictions on $p$ whatsoever as long as we talk about arbitrary metric spaces. Now, if we impose additional restrictions, the situation changes a lot. The minimal restriction to impose is that the space be complete. Can we make nontrivial examples now?

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This is a beautiful idea. I think a small modification is needed, since the set X you choose at present doesn't necessarily have complete regularity for triples where the triangle inequality is equality. But we can fix this, say, by asking also that X intersects each line in exactly 2 points -- i.e., that no three points in X are collinear. Then p(a, b, a-b) = p(a, a-b, b) = p(a-b, a, b) = 0. I'm not sure if in such a construction p can take nonzero values for the "equality" triples, though. – macbeth Dec 29 '09 at 19:12
You can not construct a geodesic space this way (i.e. all points can be joined by minimizing geodesic). AND our space is geodesic. But it should work if $p>0$ for non-degenerate triangles only... – Anton Petrunin Dec 30 '09 at 4:12

Consider the unit sphere. Then p(π, π/2, π/2) = ∞. The north and south poles are distance π apart, and every point on the equator is distance π/2 from them.

Forcing every positive real distance to occur does not change things. Cut a hole somewhere in the sphere and replace it with a cylinder stretching off to infinity. That is, the new surface is an infinite lollipop. Still, p(π, π/2, π/2) = ∞.

The problem here is that the sphere is very far from having unique geodesics.

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The problem is that these considerations are irrelevant: in the original question all values p(a,b,c) are explicitly required to be finite. – Dima Fon-Der-Flaass Dec 29 '09 at 8:29
I disagree. Once you have that p(a,b,c) can be infinite, by perturbing the sphere you can make it take any finite value you'd like. – Tom LaGatta Dec 29 '09 at 23:05
Consider a ball with n ridges. For example, this object ( beads1.co.uk/Biagi-Beads/Biagi-BS-silver-symbols/thumbs/… ), along with a spherical cap at the top and the bottom. Let x be the point at the top of the cap. Let d be the distance to its antipodal point along one of the ridges. Then p(d,d/2,d/2) = n. – Tom LaGatta Dec 29 '09 at 23:12