# Message #2296

From: Don Hatch <hatch@plunk.org>

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.

Don

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.

>

> http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif

>

> 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 4D_Cubing@yahoogroups.com, 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

hatch@plunk.org

http://www.plunk.org/~hatch/