Message #2296

From: Don Hatch <>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Mon, 02 Jul 2012 03:13:22 -0400

Hi Nan,

Heh, I try not to make judgements…
I think {3,3,7} is as legit as {7,3,3}
(in fact I’d go so far as to say they are the same object,
with different names given to the components).
But pragmatically, {7,3,3} seems easier to get a grip on,
in a viewer program such as yours which focuses naturally
on the vertices and edges.

Perhaps the best way to get a feeling for {3,3,7}
would be to view it together with the {7,3,3}?
Maybe one color for the {7,3,3} edges,
another color for the {3,3,7} edges,
and a third color for the edges formed where
the faces of one intersect the faces of the other.
And then perhaps, optionally,
the full outlines of the characteristic tetrahedra?
There are 6 types of edges in all (6 edges of a characteristic tet);
I wonder if there’s a natural coloring scheme
using the 6 primary and secondary colors.


On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> Hi Don,
> Nice to see you here. Here are my thoughts about {3,3,7} and the things
> similar to it.
> Just like when we constructed {7,3,3} we were not able locate the cell
> centers, when we consider {3,3,7} we have to sacrifice the vertices. Let’s
> start by considering something simpler in lower dimensions.
> For example, in 2D, we could consider a hyperbolic "triangle" for which
> the sides don’t meet even at the circle of infinity. The sides are
> ultraparallel. Since there’s no "angle", the name "triangle" is not
> appropriate any more. I’ll call it a "trilateral", because it does have
> three sides (the common triangle is also a trilateral in my notation).
> Here’s a tessellation of H2 using trilaterals, in which different colors
> indicate different trilaterals.
> I constructed it as follows:
> In R2, a hexagon can be regarded as a truncated triangle, that is, when
> you extend the first, third, and fifth side of a hexagon, you get a
> triangle. In H3, when you extend the sides of a hexagon, you don’t always
> get a triangle in the common sense: sometimes the extensions don’t meet.
> But I claim you always get a trilateral. So I started with a regular {6,4}
> tiling, and applied the extensions to get the tiling of trilaterals.
> And I believe we can do similar things in H3: extend a properly scaled
> truncated tetrahedron to construct a "tetrahedron" with no vertices.
> Fortunately the name "tetrahedron" remains valid because hedron means face
> rather than vertices. But I have never done an illustration of it yet.
> Then, maybe we can go ahead and put seven of them around an edge and make
> a {3,3,7}.
> I agree that these objects are not conventional at all. We lost something
> like the vertices. But just like the above image, they do have nice
> patterns and are something worth considering.
> So, what do you think about them? Do they sound more legit now?
> Nan
> — In, Don Hatch <hatch@…> wrote:
> > {3,3,7} less so… its vertices are not simply at infinity (as in
> {3,3,6}),
> > they are "beyond infinity"…
> > If you try to draw this one, none of the edges will meet at all (not
> even at
> > infinity)… they all diverge! You’ll see each edge
> > emerging from somewhere on the horizon (although there’s no vertex
> > there) and leaving somewhere else on the horizon…
> > so nothing meets up, which kind of makes the picture less satisfying.
> > If you run the formula for edge length or cell circumradius, you’ll get,
> not infinity,
> > but an imaginary or complex number (although the cell in-radius is
> finite, of
> > course, being equal to the half-edge-length of the dual {7,3,3}).

Don Hatch