Imaginary values in Schlafli symbol?

I was browsing Wikipedia and came across this hyperbolic tiling: The provided Schlafli symbol is {12i,3}. The {12i}gons seem to have centers beyond the ideal points. What is this?

Page I found it on: https://en.wikipedia.org/wiki/Order-3_apeirogonal_tiling in *n32 symmetry mutation of regular tilings: {n,3}

• please post the link of the article where you found this Aug 11, 2017 at 6:50
• It seem to have to do something with a pseudogon see en.wikipedia.org/wiki/Apeirogon#Hyperbolic_geometry it is an apeirogon along a hypercycle Aug 11, 2017 at 21:53
• It seems to be called a pseudogon Chinese Wikipedia There are also many other pseudogon tilings and their duals. Aug 12, 2017 at 0:25

This is new to me, I'd love to see a link to the page you found it on.

The only way I can think of to make sense of it is this:

In the hyperbolic plane, if you do a translation then a rotation, you might find that you have a finite order transformation (ie, you eventually get back where you started from), or you might not. If your transformation is of order 6, you'll be tracing out hexagons. If you make the edges longer, you'll trace out heptagons. If you made the edges only slightly longer, you would trace out something in between a hexagon and a heptagon, giving fractional entries in the Schlafli symbol.

You could find a formula linking the edge length and the number of sides of your polygon. You'll find that for certain edge lengths, the number of sides of your polygons goes to infinity, giving you your classic tilings with apeirogons, $\{\infty,n\}$.

But what happens if you make the edge length longer?

I'm guessing (I haven't done the calculations) that the formula gives complex entries in the Schlafli symbol, and the transformation traces out the pattern in the picture you posted. As a discrete group, the symmetry group of $\{12i,3\}$ will be isomorphic to that of $\{\infty,3\}$, but clearly they are not the same pattern drawn on $H^2$.

I imagine $\{4,4i\}$ will have squares whose corners appear to disappear off the projection of $H^2$ onto the disk. I really, really want to see what $\{4i,4i\}$ looks like!

Now I want to see what $\{4i, 4i\}$ looks like :)