# Message #2336

From: Andrey <andreyastrelin@yahoo.com>

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

Date: Mon, 16 Jul 2012 14:35:26 -0000

Don,

I think it’s much more convenient to work not with sphere, but with boundary plane in half-space Poincare model. Every plane in H3 inersects with this plane by circle, and transformations of H3 are equivalent to Moebius transformations of boundary plane. And probably {3,3,8} will give better formulae for coordinates and movements.

Andrey

— In 4D_Cubing@yahoogroups.com, Don Hatch <hatch@…> wrote:

>

> 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

> doesn’t give a direct formula for cell mid-radius r31,

> but it gives a formula for the cell in-radius r32.

> From that, we can use the identity

> sin(A)/sinh(a) = sin(B)/sinh(b) = sin(C)/sinh(c)

> on the right hyperbolic triangle

> formed by the cell center, face center, and edge center (draw it):

> sin(pi/2)/sinh(r31) = sin(pi/r)/sinh(r32)

> i.e. 1/sinh(r31) = sin(pi/r)/sinh(r32)

> i.e. r31 = asinh(sinh(r32)/sin(pi/r))

> So, plug in the formula for r32 from the paper:

> r32{p,q,r} = acosh(sin(pi/p)*cos(pi/r)/sqrt(1-cos(pi/p)^2-cos(pi/q)^2))

> and now we have a formula for r31{p,q,r}, the cell mid-radius:

> 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))

> (This can certainly be simplified, using the identity

> sinh(acosh(x)) = sqrt(x^2-1), and probably further simplifications,

> but I’m not bothering here.)

> (And note, I’m not positive I got all of the above exactly right,

> but the method should be sound.)

> 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

> with the regular euclidean tetrahedron

> with verts:

> (-1,-1,1)

> (-1,1,-1)

> (1,-1,-1)

> (1,1,1)

> The center of the cell’s edge closest to joining -1,-1,1 to 1,1,1

> lies on the +z axis, so by Step 2 this edge center is:

> (0,0,tanh(r31/2))

> The two endpoints-at-infinity of that edge

> will be (-sqrt(1/2),-sqrt(1/2),0) and (sqrt(1/2),sqrt(1/2),0)

> "translated by (0,0,tanh(r31/2))",

> i.e. transformed by the translation

> that takes the origin to that edge center (0,0,tanh(r31/2)).

>

> Recall that for any points p,t in the poincare ball

> (of any number of dimensions), p translated by t is

> given by the magic formula:

> ((1 + 2*p.t + p.p)*t + (1-t.t)*p) / (1 + 2*p.t + p.p * t.t)

> where "." denotes dot product. (Hope I wrote that down right.)

> So plug in:

> t = (0,0,tanh(r31/2))

> p = (sqrt(1/2),sqrt(1/2),0)

> (a bunch of terms simplify and drop out since p.p=1 and p.t=0, but whatever);

> The resulting endpoint coords are (a,a,b) for some a,b

> (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