# Message #2335

From: Don Hatch <hatch@plunk.org>

Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}

Date: Sun, 15 Jul 2012 05:31:52 -0400

Here’s a more detailed outline of how I’d go about drawing

the intersection of {3,3,7} with the sphere at infinity.

I’ll probably do this in a few days if no one else does it first.

To get started, consider the center cell

of a cell-centered {3,3,7}.

The intersection of that cell with the sphere at infinity

consists of three little regular spherical triangles;

we need to know the euclidean coords of one of these

little spherical triangles.

We can find that in 4 steps:

Step 1: compute the cell mid-radius r31{3,3,7}.

```
The reference paper "Regular Honeycombs In Hyperbolic Space" by HSM Coxeter<br>
doesn't give a direct formula for cell mid-radius r31,<br>
but it gives a formula for the cell in-radius r32.<br>
From that, we can use the identity<br>
sin(A)/sinh(a) = sin(B)/sinh(b) = sin(C)/sinh(c)<br>
on the right hyperbolic triangle<br>
formed by the cell center, face center, and edge center (draw it):<br>
sin(pi/2)/sinh(r31) = sin(pi/r)/sinh(r32)<br>
i.e. 1/sinh(r31) = sin(pi/r)/sinh(r32)<br>
i.e. r31 = asinh(sinh(r32)/sin(pi/r))<br>
So, plug in the formula for r32 from the paper:<br>
r32{p,q,r} = acosh(sin(pi/p)*cos(pi/r)/sqrt(1-cos(pi/p)^2-cos(pi/q)^2))<br>
and now we have a formula for r31{p,q,r}, the cell mid-radius:<br>
r31{p,q,r} = asinh(sinh(acosh(sin(pi/p)*cos(pi/r)/sqrt(1-cos(pi/p)^2-cos(pi/q)^2)))/sin(pi/r))<br>
(This can certainly be simplified, using the identity<br>
sinh(acosh(x)) = sqrt(x^2-1), and probably further simplifications,<br>
but I'm not bothering here.)<br>
(And note, I'm not positive I got all of the above exactly right,<br>
but the method should be sound.)<br>
Substitute p=3,q=3,r=7 to get the desired cell-mid-radius of {3,3,7}.
```

Step 2: from r31, compute the euclidean distance

from the origin to an edge of the center cell in the poincare ball model.

```
If I recall correctly, that will be tanh(r31/2).
```

Step 3: from that, compute the actual coords of the two endpoints-at-infinity

of one edge of the center cell.

```
For definiteness, align the center cell<br>
with the regular euclidean tetrahedron<br>
with verts:<br>
(-1,-1,1)<br>
(-1,1,-1)<br>
(1,-1,-1)<br>
(1,1,1)<br>
The center of the cell's edge closest to joining -1,-1,1 to 1,1,1<br>
lies on the +z axis, so by Step 2 this edge center is:<br>
(0,0,tanh(r31/2))<br>
The two endpoints-at-infinity of that edge<br>
will be (-sqrt(1/2),-sqrt(1/2),0) and (sqrt(1/2),sqrt(1/2),0)<br>
"translated by (0,0,tanh(r31/2))",<br>
i.e. transformed by the translation<br>
that takes the origin to that edge center (0,0,tanh(r31/2)).
Recall that for any points p,t in the poincare ball<br>
(of any number of dimensions), p translated by t is<br>
given by the magic formula:<br>
((1 + 2*p.t + p.p)*t + (1-t.t)*p) / (1 + 2*p.t + p.p * t.t)<br>
where "." denotes dot product. (Hope I wrote that down right.)<br>
So plug in:<br>
t = (0,0,tanh(r31/2))<br>
p = (sqrt(1/2),sqrt(1/2),0)<br>
(a bunch of terms simplify and drop out since p.p=1 and p.t=0, but whatever);<br>
The resulting endpoint coords are (a,a,b) for some a,b<br>
(then the other endpoint is (-a,-a,b), but we don't need that at this point).
```

Step 4: rotate one of those endpoints-at-infinity

around the appropriate axis

to get the other two vertices of the little spherical triangle.

The three spherical triangle vertices are:

(a,a,b)

(b,a,a)

(a,b,a)

(where a,b are the result of step 3).

=========================================================================

So now we have one little spherical triangle.

Now, choose a set of 3 generators for the rotational symmetry group

of {3,3,7}, and use them repeatedly to send the triangle everywhere.

There are lots of choices of 3 generators; here’s one:

A: plain old euclidean rotation by 120 degrees about the vector (1,1,1)

B: plain old euclidean rotation by 120 degrees about the vector (1,1,-1)

C: rotation by 2*pi/7 about an edge of {3,3,7}.

for this, we can use the composition of:

translate the edge center to the origin

(i.e. translate the origin to minus the edge center)

followed by plain old euclidean rotation of 2*pi/7 about this edge-through-the-origin

followed by translating the origin back to the original edge center

A specific edge center, and the translation formula,

can be found in Step 3 above.

Don

On Sat, Jul 14, 2012 at 03:51:39PM -0400, Don Hatch wrote:

>

>

> 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

>

–

Don Hatch

hatch@plunk.org

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