Message #2326

From: Don Hatch <hatch@plunk.org>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Thu, 12 Jul 2012 01:18:27 -0400

Hi Roice,

Sorry I forgot to follow up on this message earlier.

On Wed, Jul 04, 2012 at 02:48:31PM -0500, Roice Nelson wrote:
>
>
> Hi Don,
> I’m loving this whole thread. Lots of cool ideas being thrown out by you
> and Andrey!
> I really like the fractal image you’re envisioning on the sphere at
> infinity. This past weekend, I was playing with recursive circle
> inversions, and I had no idea it would apply to a discussion like this at
> the time. Here is an image close to the the picture you describe. The
> difference is that each circle in the gasket is filled with a {3,inf}
> rather than a {3,7}.

I’m having trouble getting my email client to see or detach the image you’re
referring to, so I can only imagine it at the moment…

> I’m actually wondering, is the Apollonian gasket the result for {inf,3,3}?
> For {7,3,3}, I’m thinking the initial 4 circles in the circle packing
> would be smaller (not tangent), and would approach the Apollonian as p ->
> inf. In the {inf,3,3} case, the 4 cells that meet at the origin in Nan’s
> applet could represent the 4 initial circles in the gasket. So the cells
> that meet at the origin meet again at the sphere at infinity! (although
> not all at the one location this time) For the {7,3,3}, my intuition says
> the initial 4 cells don’t meet again at the sphere at infinity. I’m
> curious what you think about these speculations.

Hmm. I’m not *positive* that the circles for {7,3,3} meet,
but it seems to be implied by the line of thinking I was following,
starting with the {3,3,7}.
The cells of the {3,3,7} tile all of H3, right?
So they don’t leave any regions of H3 unfilled…
so that makes me think that there’s no open neighborhood on the sphere-at-infinity
that’s left untouched by the cells…
and the intersection of each cell with the sphere-at-infinity
is a spherical triangle…
so these spherical triangles must tile the entire sphere,
7 at each vertex (of this spherical tiling).
So we have 7 spherical triangles at each vertex, covering the whole sphere…
in other words, this tiling on the sphere is (combinatorially) a disjoint union of {3,7}’s,
each one projected onto the sphere… presumably each {3,7} onto
a spherical-disk region of the sphere (I can’t imagine it would be
a different shape).
Since the union of all these spherical-disks cover the sphere,
that means they do kiss…
and if I’m imagining this right, the spherical circles
bounding these spherical disks
would form the intersection of sphere with the dual {7,3,3}.
Does that seem right to you?

And then, it seems to me that the exact same construction
goes through for any {3,3,n}/{n,3,3},
leading to exactly the same Appolonian gasket for each of them…
and so presumably {3,3,inf}/{inf,3,3} would yield the same Appolonian gasket as well
(though I’m having a hard time visualizing
the {3,3,inf} directly).

Don

> Aside: as traditionally shown (e.g. on wikipedia), the Apollonian gasket
> is a stereographic projection of the pattern on the sphere at infinity we
> are discussing here, which I think is neat. 3 of the 4 cells jump out
> visually, and the 4th is inverted - the outside of the whole pattern.
> Best,
> Roice
>
> On Tue, Jul 3, 2012 at 2:51 PM, Don Hatch <hatch@plunk.org> wrote:
>
> Oh wait!
> I realized I got last part wrong, just after I hit the "send" button :-)
>
> The picture would start with an Apollonian gasket (see wikipedia
> article)
> of circles on the sphere;
> this is the intersection of the {7,3,3} with the sphere at infinity.
> Then each circle in the gasket is filled with a {3,7},
> the final result being the intersection of the {3,3,7} with the sphere
> at infinity.
> So, it isn’t true that there are isolated cluster points
> in the *center* of each circle; the clustering is
> towards the *boundary* of each circle. I think I have
> a clear picture in my head of what this looks like now.
>
> A cell of the {3,3,7} would touch the sphere
> in 4 spherical triangles (its "feet"),
> each foot in a different one of four mutually kissing circles
> of the gasket, I think.
> Don
>
> On Tue, Jul 03, 2012 at 03:29:38PM -0400, Don Hatch wrote:
> >
> >
> > Okay I think maybe I follow you now…
> > But each face formed by truncation…
> > it’s a triangle, not a hexagon, right?
> > In fact it’s a spherical triangle, on the sphere at infinity,
> right?
> > All of these spherical triangles, of different apparent sizes,
> > would tile the sphere, 7 at each vertex…
> > but with some "cluster points" which are the limit points
> > of infinitely many of these triangles of decreasing size.
> > I’d like to see a picture of this– it shouldn’t be too hard to
> generate
> > (together with the spherical circles
> > that are the intersection of the dual {7,3,3} with the sphere
> > at infinity, in a different color… I think each cluster point
> > would be the center of one of these circles).
> >
> > Don
> >
> > On Tue, Jul 03, 2012 at 02:43:52PM -0000, Andrey wrote:
> > >
> > >
> > > Yes, dihedral… I mean angle between hexagonal and triangular
> faces of
> > > truncated tetrahedron. By the selection of truncating planes it
> will be
> > > pi/2. And angles between hexagonal faces are 2*pi/7.
> > >
> > > Andrey
> > >
> > > — In 4D_Cubing@yahoogroups.com, Don Hatch <hatch@…> wrote:
> > > >
> > > > Hi Andrey,
> > > >
> > > > I’m not sure if I’m understanding correctly…
> > > > is "behedral angle" the same as "dihedral angle"?
> > > > If so, isn’t the dihedral angle going to be 2*pi/7,
> > > > since, by definition, 7 tetrahedra surround each edge?
> > > >
> > > > Don
> > > >
> > > >
> > > > On Tue, Jul 03, 2012 at 05:14:47AM -0000, Andrey wrote:
> > > > >
> > > > >
> > > > > Hi all,
> > > > > About {3,3,7} I had some idea (but it was long ago…). We
> know that
> > > its
> > > > > cell is a tetragedron that expands infinitely beyond
> "vertices". For
> > > each
> > > > > 3 its faces we have a plane perpendicular to them (that cuts
> them is
> > > the
> > > > > narrowest place). It we cut {3,3,7} cell by these planes, we
> get
> > > truncated
> > > > > tetrahedron with behedral angles = 180 deg. What happens if
> we
> > reflect
> > > it
> > > > > about triangle faces, and continue this process to infinity?
> It will
> > > be
> > > > > some "fractal-like" network inscribed in the cell of {3,3,7}
> -
> > regular
> > > > > polyhedron with infinite numer of infinite faces but with no
> > vertices.
> > > I’m
> > > > > sure that it has enough regular patterns of face coloring,
> and it
> > may
> > > be a
> > > > > good base for 3D puzzle.
> > > > >
> > > > > Andrey
> > > > >
> > > > > — In 4D_Cubing@yahoogroups.com, Don Hatch <hatch@> wrote:
> > > > > >
> > > > > > Hi Nan,
> > > > > >
> > > > > > Heh, I try not to make judgements…
> > > > > > I think {3,3,7} is as legit as {7,3,3}
> > > > > > (in fact I’d go so far as to say they are the same object,
> > > > > > with different names given to the components).
> > > > > > But pragmatically, {7,3,3} seems easier to get a grip on,
> > > > > > in a viewer program such as yours which focuses naturally
> > > > > > on the vertices and edges.
> > > > > >
> > > > > > Perhaps the best way to get a feeling for {3,3,7}
> > > > > > would be to view it together with the {7,3,3}?
> > > > > > Maybe one color for the {7,3,3} edges,
> > > > > > another color for the {3,3,7} edges,
> > > > > > and a third color for the edges formed where
> > > > > > the faces of one intersect the faces of the other.
> > > > > > And then perhaps, optionally,
> > > > > > the full outlines of the characteristic tetrahedra?
> > > > > > There are 6 types of edges in all (6 edges of a
> characteristic
> > tet);
> > > > > > I wonder if there’s a natural coloring scheme
> > > > > > using the 6 primary and secondary colors.
> > > > > >
> > > > > > Don
> > > > > >
> > > > > >
> > > > > > On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> > > > > > >
> > > > > > >
> > > > > > > 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}).
> > > > > > >
> > > > > > >
> > > > > >
> > > > > > –
> > > > > > Don Hatch
> > > > > > hatch@
> > > > > > http://www.plunk.org/~hatch/
> > > > > >
> > > > >
> > > > >
> > > >
> > > > –
> > > > Don Hatch
> > > > hatch@…
> > > > http://www.plunk.org/~hatch/
> > > >
> > >
> > >
> >
> > –
> > Don Hatch
> > hatch@plunk.org
> > http://www.plunk.org/~hatch/
> >
> >
>
> –
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
> ————————————
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–
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/