Message #2334

From: Don Hatch <>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Sat, 14 Jul 2012 15:51:39 -0400

On Fri, Jul 13, 2012 at 06:25:47PM -0500, Roice Nelson wrote:
> I’d love to see a picture of this thing too. Consider the {7,3,3} such
> that a vertex is at the origin, so 4 cells meet there. If we could
> calculate the size of the circle associated with one of these cells (I
> don’t know how to do this), we could start with that one. We’d generate
> a {3,7} tiling inside that circle. I suspect the triangles in it are
> precisely the same as those in the Poincare disk (?). Then we use
> Mobius transformations to copy this template {3,7} tiling all over the
> plane.
> I think we could leverage the Apollonian gasket to generate the list of
> needed Mobius transforms, because even though the {3,7} boundary circles
> aren’t kissing, the (non-Euclidean) centers of all the circles are still
> the same as that of the gasket. So the list of transforms will be the
> same list used to generate an Apollonian from a starting circle.
> I don’t think the construction I suggested works. I think it was
> incorrect of me to assume the centers of the {7,3,3} circles would
> coincide with the centers of the gasket (this is perhaps only true for the
> first 4 circles). Using the Mobius transforms of the Apollonian gasket as
> I suggested would leave empty space.
> So I’m not sure how one would go about constructing the {3,3,7} picture.
> This stuff can be hard to think about!
> Roice

If you can just figure out the coordinates
where three incident edges of one {3,3} of the {3,3,7} meet the sphere,
that will give you one of the little spherical triangles…
Then just transform that one spherical triangle
by symmetries of the {3,3,7}
(3 generators suffice, in any of several ways);
that should give the whole picture.


Don Hatch