# Message #2313

From: Don Hatch <hatch@plunk.org>

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

Date: Thu, 05 Jul 2012 16:39:07 -0400

Hi Andrey,

Okay, I finally get it (yes, I’m aware this is the 4th time I’ve said

that– this time for sure!). The picture really helped; thanks.

So this convex fractal object you’re talking about– it’s actually

contained in just one {3,3} of the {3,3,7}.

It’s surprising to me (although maybe it shouldn’t be)

that "most" of what’s going on in one of the {3,3}’s

is in each of the small-looking corners of it,

which are in reality infinitely bigger than the big-looking middle part.

This line of thinking really brings that fact to light.

It’s also really strange that this thing is convex.

That seems, intuitively, to contradict the fact that it’s got fat parts

and skinny parts separating the fat parts…

in euclidean space this would be a contradiction,

but in hyperbolic space, apparently it isn’t.

It’s an interesting thing to meditate on.

I wonder if there’s some stronger notion of convexity

formalizing this intuitive notion. Something like the following

additional condition:

whenever you tie a string around the object,

it should be possible to continuously slide

the string off the object without stretching it.

(Your convex object would fail to meet this criterion.)

Questions remaining in my mind about this object:

(1) You refer to the hexagon edges formed by truncation as the "short sides"–

maybe it’s obvious, but it’s not clear to me at this point

whether they are shorter or longer than the other sides (the remainders

of the edges that got truncated)… or maybe even equal,

making it a regular hexagon. Is it obvious to you? If so, how?

(2) Can the object be formed by erasing some parts

of some other more familiar uniform honeycomb?

(like Nan obtained the picture of the {3,ultrainfinity}

by erasing some of the edges of the {6,4})

(3) If you reflect the object about each of its faces

to form a kaleidescope, do the copies of the object tile

all of H3? Or does it still leave some regions empty?

Don

On Wed, Jul 04, 2012 at 12:01:57PM -0000, Andrey wrote:

>

>

> Hi, Don

> Let’s start with triangle with vertices "beyond infinity". Two its edges

> are "ultraparallel" and they have common perpendicular. If you cut triange

> by three such perpendiculars, you’ll get irregular hexagon with 6 right

> angles. Now you can reflect it about its short sides and continue this

> process to infinity. In result you’ll get very strange object - it is

> convex, it’s infinite, all its boundaries are straight lines and it’s

> regular - you for every two edges there is a movement of H2 that moves one

> edge to another and moves the whole object to itself.

> Something like this:

> http://groups.yahoo.com/group/4D_Cubing/photos/album/772706687/pic/856386394/view

> Red lines are sides of hexagons.

> You see that each hexagon is connected to three another and they make an

> acyclic graph (tree) together.

>

> The same is for tetrahedra in H3. You have a convex objects, bounded by

> planes (and faces of this object are exactly like H2 pattern), each

> truncated tetrahedron is connected to 4 others and together they make

> regular polyhedron with tree structure (all nodes of graph are

> equivalent).

>

> Points on edges of tetrahedron with the minimal distance are not infinite

> - they are inside H3, so whole object is "real".

>

> Andrey

>

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

> >

> > Hi Andrey,

> >

> > Maybe I finally get what you mean now…

> > there is a unique plane that contains

> > the three points at infinity where three edges of the tet come closest

> to meeting;

> > I didn’t see that before.

> > But are you sure the tet faces meet that plane

> > at a right angle as you claimed?

> > I believe the tet faces meet the sphere-at-infinity at right angles;

> > I don’t think both can be true.

> >

> > And I don’t understand what you mean by "a structure of regular infinite

> > 4-graph without loops" at all.

> > By "4-graph", do you mean every node has degree 4,

> > and by "without loops", do you mean a tree?

> > But I don’t see any such tree in what we’re talking about, so I’m lost

> :-(

> >

> > Don

> >

> > On Wed, Jul 04, 2012 at 04:03:38AM -0000, Andrey wrote:

> > >

> > >

> > > And cutting triangles are planar (i.e. H2), not sperical.

> > > Their position is selected so that triangles have minimal possible

> size

> > > (in the narrowest point of the "vertex"). I’ll try to draw one face of

> the

> > > object, but it will be not easy.

> > > Looks like this combination of truncated tetrahedra will be convex in

> H3

> > > (and have a structure of regular infinite 4-graph without loops).

> > >

> > > Andrey

> > >

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

> > > >

> > > > On Tue, Jul 03, 2012 at 03:29:38PM -0400, I wrote:

> > > > > But each face formed by truncation…

> > > > > it’s a triangle, not a hexagon, right?

> > > >

> > > > Sorry, mental lapse on my part!

> > > > The hexagons you’re talking about

> > > > are what’s left of the *original* faces when you truncate

> > > > a tetrahedron.

> > > > Don’t know where my mind went.

> > > >

> > > > Don

> > > >

> > >

> > >

> >

> > –

> > Don Hatch

> > hatch@…

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

> >

>

>

–

Don Hatch

hatch@plunk.org

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