# What determines if a hyperbolic tiling with ideal vertices is regular?

I'd like to say I need this community's help in clearing my mind of the clutter that leads me to this contradiction:

As an example, the Wikipedia article for the Infinite-order triangular tiling provides this image as a coloring of this tiling, but also provides this image, but with the comment "nonregular".

All ideal triangles are congruent to each other.

The interior angles of an ideal triangle are all zero.

As such, I see no reason for one tiling to claim to be regular, while the other one is supposed not to be regular.

It seems to me to answer this question would require a more sophisticated definition of "regular" than I hold. Additionally, I think I might also lack knowledge of how distinct ideal point are, if there is a notion of distance between them and how to represent them notationally.

Can someone help?

If it was just a matter of congruent ideal triangles, then there could be a lot of "regular" ideal triangle tesselations. The key idea, I think, is that the tiling should have symmetries such as hyperbolic reflections, rotations or translations which preserve distances and angles. The ideal triangle points all have angles of zero and distance between them is meaningless, but it is important for non-ideal points. Although the "alternated color tiling" and recursive non-regular tiling look somewhat similar, there is a key difference. Notice how the pattern along the boundary of the disk is different between them. In the non-regular tiling the pattern is very homogeneous, whereas in the other tiling there are different spacing of the ideal vertices along the boundary. The result of a hyperbolic reflection, rotation or translation of the non-regular tiling makes the boundary pattern non-homogeneous while the other tiling will just be similar to what it was before because it is symmetric under hyperbolic reflections, rotations and ranslations. That is what makes it "regular". The process that created the recursive non-regular tiling did not use hyperbolic reflections, rotations and translations while the other tiling did.

Here's another answer which identifies the extra feature that must be preserved in a "regular" tiling of the hyperbolic plane $\mathbb H$ by ideal triangles.

Consider an ideal triangle $\Delta(\xi \eta \zeta) \subset \mathbb H$ with ideal vertices $\xi,\eta,\zeta \in \partial\mathbb H$. There exists a unique point $x \in \overline{\eta\zeta}$ such that the ray $\overline{x \xi}$ is perpendicular to the line $\overline{\eta\zeta}$; let's refer to this point $x$ as the "footprint" of the triangle $\Delta(\xi\eta\zeta)$ on its side $\overline{\eta\zeta}$. The footprints $y$ and $z$ of $\Delta(\xi\eta\zeta)$ on its sides $\overline{\xi\zeta}$ and $\overline{\xi\eta}$ are similarly defined.

One can characterize footprints by an intrinsic property of the ideal triangle: the footprint of an ideal triangle on one of its sides is the unique point $x$ on that side which is the finite endpoint of an infinite geodesic ray $R$ entirely contained in the interior of that triangle. By uniqueness of $R$, the footprint $x$ and the ray $R$ must be fixed by any reflection isometry of the triangle which fixes the opposite vertex, and it follows that the ray $R$ must meet the side at right angles at the point $x$.

If an ideal triangular tiling of $\mathbb H$ is to be regarded as "regular", then there should be many isometric symmetries of the tiling.

For example, consider an edge $\overline{\xi\eta}$ of the tiling. This edge is a side of two different triangles of the tiling, which I'll denote $\triangle(\xi\eta\zeta_1)$ and $\triangle(\xi\eta\zeta_2)$. Let $z_1,z_2$ be the footprints on $\overline{\xi\eta}$ of the two triangles $\triangle(\xi\eta\zeta_1)$ and $\triangle(\xi\eta\zeta_2)$, respectively.

If this tiling is regular, then there should be an isometric symmetry of the tiling which reflects across the line $\zeta_1,\zeta_2$, interchanging $\xi$ and $\eta$. But a little thought suggests that this is impossible in general, unless $z_1 = z_2$ in which case it is possible.

In fact, this gives a necessary and sufficient condition for an ideal hyperbolic tiling of $\mathbb H$ to be regular:

An ideal triangular hyperbolic tiling of $\mathbb H$ is regular if and only if for each edge of the tiling, the footprints on that edge of the two adjacent triangles are the same point on that edge.

If you look at the tiling in the second link you provided, you can find an edge whose two adjacent triangles have unequal footprints: not one of the sides of the central triangle, but instead a side of one of the triangles adjacent to the central triangle.