# Message #2339

From: Don Hatch <hatch@plunk.org>

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

Date: Wed, 18 Jul 2012 05:26:10 -0400

This looks great! The equator and tetrahedral tiling

are essential for getting context… I’d be totally lost

without them.

About the possibility that half of the surface

doesn’t get filled in…

I guess that’s possible, but it sure would be a surprise!

Let me sleep on that one.

I have a different question though…

When I was describing how I’d draw this thing,

I thought the little triangles would be

spherical triangles, that is, bounded by geodesics

(i.e. arcs of great circles).

But that’s not true… they are actually

bounded by non-geodesic circular arcs on the sphere.

But, I thought, it would be reasonable to

draw the first picture of it with them approximated

by geodesics… or even by straight line segments.

I see you didn’t draw straight line segments,

but I can’t tell– are you drawing geodesics?

Or are you drawing the real things?

Don

On Tue, Jul 17, 2012 at 05:43:05PM -0500, Roice Nelson wrote:

>

>

> I made a little progress with this, using info from Don’s email, but

> working in the plane rather than with the sphere. My best image so far,

> with 20k tetrahedra (80k triangles), is here:

> http://www.gravitation3d.com/roice/math/%7B3,3,7%7D_sphere_at_inf.png

> For reference, the blue circle is the sphere equator, and the green arcs

> are a tetrahedral tiling on the sphere. (The incircles of that tiling

> were the 4 starting circles for the previous {3,3,inf} image.)

> Clearly, much improvement could be made as far as filling things in.

> However, I still wanted to post because there are some areas which

> genuinely look like they will NEVER get filled in, no matter how much one

> recursed. These "holes" in the picture seem to be the dual areas of what

> is getting filled in (notice the 3 holes at the vertices of the

> tetrahedral tiling). Does this make sense to anyone? I’m wondering if it

> is possible that reflections of the infinite fundamental tetrahedron never

> reach certain portions of H3. The more likely scenario is that I’m doing

> something incorrectly :)

> Roice

> On Mon, Jul 16, 2012 at 9:35 AM, Andrey <andreyastrelin@yahoo.com> wrote:

>

> 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

>

> ————————————

>

> Yahoo! Groups Links

>

>

>

>

–

Don Hatch

hatch@plunk.org

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