Message #2329

From: Don Hatch <hatch@plunk.org>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Thu, 12 Jul 2012 20:12:31 -0400

Hey Roice,

I take it all back.
Now I think you’re totally right…
it’s not an Apollonian gasket for n < infinity–
the circles don’t kiss.
I think my reasoning was good up through where I said
the sphere is covered by a disjoint union of {3,7}’s…
that much is true, but it doesn’t imply the circles kiss…
and in fact I now think you’re right, they don’t.
(It’s more obvious if we consider n=6 too…
maybe you were thinking of that that all along, but I missed it.)
I think the circles are dense on the surface of the sphere, but no two
of them kiss each other.

Now I really want to see a picture of this thing!

Don

On Thu, Jul 12, 2012 at 01:18:27AM -0400, Don Hatch wrote:
>
>
> 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/
> >
> > ————————————
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
>
> –
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
>

–
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/