# Message #2295

From: schuma <mananself@gmail.com>

Subject: Re: Hyperbolic Honeycomb {7,3,3}

Date: Sat, 30 Jun 2012 23:29:32 -0000

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}).