I am interested in octagonal tilings of the hyperbolic plane. My understanding is that one can construct such a tiling by choosing the fundamental domain to be a regular octagon such that the sum of the interior angles is $2\pi$ (which can also be stated in terms of a condition on the area of the octagon, and thus this octagon can be found by scaling).
I am curious whether there exist other octagonal tilings with non-regular octagons. It seems as though there are many degrees of freedom in the construction. From what I can tell, you can choose your fundamental domain for the octagon to be a polygon constructed from 8 points on the hyperbolic plane such that:
- The pairs of edges that map onto each other have the same length. This is 4 constraints.
- The sum of the interior angles is $2\pi$. This is just one more constraint.
My understanding is that the Poincare polygon theorem shows that any polygon satisfying the above constraints is the fundamental domain for some discrete group acting on the hyperbolic plane.
Naively therefore, it seems that there are 16 variables (2 for each vertex) and 5 constraints, so I think there should be a large 11-dimensional
family of non-regular octagonal tilings.
My question is, therefore:
Are there non-regular octagonal tilings of the hyperbolic plane, and how can they be constructed?
If possible, I would really appreciate some "algorithm" that I could use to produce such a tiling.
Best Answer
For the octagonal tiling you describe, in which each angle is $2\pi/8$ and all sides have equal length, 8 octagons meet around each vertex.
Any single one of the octagon tiles is a fundamental domain for the group generated by reflections in the 8 sides of the octagon (this might not be exactly the same as the group you alluded to in your post... perhaps you were thinking of gluing opposite sides of the octagon? ... but you were not clear on this point, so I'll stick with the reflection group).
According to the Poincare polygon theorem, that reflection group has Coxeter group presentation \begin{align*} \langle a,b,c,d,e,f,g,h &\mid a^2 = b^2 = ... = h^2 = 1, \\ (ab)^4 &= (bc)^4 = (cd)^4 = (de)^4 = (ef)^4 = (fg)^4 = (gh)^4 = (ha)^4 = 1\rangle \end{align*} But the Poincare polygon theorem can be applied to any other octagon having $2\pi/8$ angles at each vertex, and every such octagon determines a tiling with reflection group having the same presentation (and, hence, isomorphic).
And your intuition is correct: there are enough degrees of freedom to allow for the existence of other such octagons.
In fact one can compute the dimension of the deformation space, i.e. the space of $2\pi/8$ angled octagons. If I have calculated correctly, I believe there is a 5-dimensional space of such octagons. If the vertices are listed in cyclic order as $P_1,...,P_8$, then the side $[P_8,P_1]$ is connected to each of the other five non adjacent sides $[P_1,P_2]$, ... , $[P_6,P_7]$ by a common perpendicular, and the lengths of those 5 perpendiculars are 5 independent parameters which, I think, uniquely determine the octagon (subject to the constraint that all angles are $\pi/8$).