Message #2335

From: Don Hatch <>
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 &quot;Regular Honeycombs In Hyperbolic Space&quot; 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)&#58;<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&#58;<br>
    r32&#123;p,q,r&#125; = acosh(sin(pi/p)&#42;cos(pi/r)/sqrt(1-cos(pi/p)&#94;2-cos(pi/q)&#94;2))<br>
and now we have a formula for r31&#123;p,q,r&#125;, the cell mid-radius&#58;<br>
    r31&#123;p,q,r&#125; = asinh(sinh(acosh(sin(pi/p)&#42;cos(pi/r)/sqrt(1-cos(pi/p)&#94;2-cos(pi/q)&#94;2)))/sin(pi/r))<br>
(This can certainly be simplified, using the identity<br>
sinh(acosh(x)) = sqrt(x&#94;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 &#123;3,3,7&#125;.

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&#58;<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&#58;<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>
&quot;translated by (0,0,tanh(r31/2))&quot;,<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&#58;<br>
    ((1 + 2&#42;p.t + p.p)&#42;t + (1-t.t)&#42;p) / (1 + 2&#42;p.t + p.p &#42; t.t)<br>
where &quot;.&quot; denotes dot product.  (Hope I wrote that down right.)<br>
So plug in&#58;<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:
(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.


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