# Is there a 3d equivalent of this picture?

This question arises apropos of an earlier question I asked that was (VERY!!!) helpfully answered by Anton Petrunin:

Fitting a mesh to a density function

The picture below is the image of a regular equilateral triangular lattice in the complex plane under the map $z \mapsto z^2$:

(for whatever reason, I seem to be missing a few links in the picture above, but I hope this is sufficient). This picture has two nice visual properties, namely that the points are distributed in a "regular" or "predictable" way, and that the density of points becomes more concentrated as we move towards the origin. My question is, are there configurations of points in 3 dimensions that also have these two visual properties? Owing to Liouville's theorem, I realize that I won't be nearly as fortunate in constructing a nice and easy conformal mapping, so I'm just looking for any possible way to place points in $\mathbb{R}^3$ where I could get some kind of behavior similar to that pictured above. The best I can think of would be some sort of concentric geodesic buckyball-type structures, or possibly the cartesian image of a regular lattice in spherical coordinates, but I'm wondering if there's anything that's more interesting.

I should also mention that I don't necessary need the configuration to be "completely regular". That is, I'm happy to tolerate occasional points that are inconsistent with the rest of the configuration; there should just be some kind of visual consistency that one would readily observe by inspection.

• 1. It should be pointed out that the density does not increase indefinitely inside the "black spot" but there is some minimum distance (if the points were really generated as indicated in the text) 2. Would you tolerate, for example (in the planar example) occasional vertices with 5 or 7 incident triangles, or should it be "completely regular"? – Günter Rote Sep 18 '13 at 21:45
• Günter: Thanks for the comments! As for 1., yes, indeed it is true that the density pictured does have a finite limit since this was generated from a finite hexagonal mesh. As for 2., yes, I am certainly happy to tolerate occasional irregular vertices, I'm just trying to get away from a configuration that's completely random. Will update the question accordingly. – John Gunnar Carlsson Sep 18 '13 at 21:59
• If you are happy with infinite density, you can use a conformal transformation in 3D - just take a regular lattice and replace all distances to the origin by their inverses. – user25199 Sep 18 '13 at 22:19
• Carl, good call! This never would have occurred to me but suits my purposes quite well. – John Gunnar Carlsson Sep 18 '13 at 23:29
• Doesn't this conformal transform, while indeed giving infinite density close in, give zero density outside a certain circle? – Aaron Meyerowitz Sep 19 '13 at 9:19

The restriction to conformal maps is a natural one, as it means that there is no affine distortion in the neighbourhood of a point. Specifically, the Voronoi cells of the points will not be oblated or prolated, which is not the case for non-conformal maps. Once we insist on that restriction, Liouville's theorem insists that the only possibilities are Möbius transformations (compositions of Euclidean transformations and geometric inversions).

It seems like a good idea to invert a lattice, and the dense face-centred cubic lattice is more aesthetically pleasing than the ordinary cubic lattice. Geometric inversion of the face-centred cubic lattice ($A_3 \cong D_3$ in Sphere Packings, Lattices and Groups) yields an arrangement of points, all of which are contained within the unit ball, and the density diverges to infinity as you approach the origin. Specifically, we take the set of points:

$$\{ \dfrac{(x,y,z)}{x^2+y^2+z^2} : x,y,z \in \mathbb{Z}, \enspace x+y+z \equiv 1 \mod 2\}$$

Here's a three-dimensional rotating view of the configuration, where I've made the points semitransparent:

Is this what you're looking for? Unlike your two-dimensional configuration, this is bounded and infinitely dense in any open ball containing the origin.

As we can see in the picture, here are the rules.

1. The pattern has the symmetries of a triangle. I made with blue three rays starting at the origin, and making equal angles.
2. For each blue ray, there is an array of parabolas which are symmetric with respect to that ray.
3. The parabolas symmetric with respect to the same line are distributed so that the distance between the apexes of two consecutive parabolas is larger, when the distance to the center is larger. This follows from the complex map used.
4. Also, the parabolas are larger as the distance to the center increases.
5. The dots are at the intersections between pairs of parabolas.

Now, that we know the rules, it is pretty straightforward to generalize to $\mathbb R^3$.

1. Take a regular tetrahedron, with the center in the origin.
2. Draw four blue rays, starting from origin, and going through the four vertices of the tetrahedron.
3. Build sequences of paraboloids which have as revolution axes the blue rays.
4. Make sure the paraboloids are at the same distances, and have the same ratios, as in the planar case.
5. Draw the dots at the intersections made by triples of paraboloids.
• Can we see the result? – Jérémy Blanc Sep 18 '13 at 22:29
• Unlike in the planar case, the triple intersections of paraboloids will typically not be 4-tuple intersections. Call the 4 types of paraboloids A, B, C, and D. Then it's probably good to divide 3-space into 4 regions. In the first region, only look at intersections of paraboloids from the A,B,C families. In the 2nd region, only look at intersections of paraboloids from the A,B,D families. etc. – André Henriques Sep 18 '13 at 22:46
• @Jérémy Blanc: I didn't draw the result. I just gave instructions how to be done. We can visualize. – Cristi Stoica Sep 19 '13 at 4:45
• @ Cristi Stoica. I just wanted to say that, since the 2d picture was nice, would be good to see the 3d picture. – Jérémy Blanc Sep 19 '13 at 6:25
• @Jérémy Blanc. I know, I would like to see it too, so if anyone finds time to do this, I hope will share it with us. If anyone would be interested in doing this, I would suggest that the paraboloids are not displayed, because otherwise we would not be able to see the forest. Just display the dots, and maybe the parabolas which are the intersections of pairs of paraboloids. And also use four colors for the dots, corresponding to the ways to choose three paraboloids out of four. If at some point I will decide to invest time in making this, I will ping you. – Cristi Stoica Sep 19 '13 at 7:51