Message #2316

From: Don Hatch <>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Fri, 06 Jul 2012 03:47:22 -0400

On Fri, Jul 06, 2012 at 06:37:50AM -0000, Andrey wrote:
> Hi, Don
> 1) you are right - cutting segments in {3*} may be shorter, or longer, or
> equal to remaining sides of trilateral. If you build it from {6,4} they
> will be exactly equal.
> BTW, my calculations show that dihedral angle of {3*,3} based on this face
> is acos(2/3)=48.19 deg - somewhere between 2*pi/8 and 2*pi/7.

Ah okay, it took me a few minutes to realize the significance of that.
That tells me that if we started with a {3,3} cell of the {3,3,7}
(which is what I was assuming)
we will NOT get a regular hexagon
(since the dihedral angle of that cell is exactly 2*pi/7),
in fact we will not get a regular hexagon
when starting with any {3,3,n}
(since the dihedral angle of the cell would be 2*pi/n).
That’s disappointing.

So when starting with {3,3,7}, the hexagon isn’t regular…
but I’m still not sure which edges are longer and which are shorter.
(And I guess whichever it is for {3,3,7},
it will be the opposite for all {3,3,n>=8}, since the switchover
is somewhere between 2*pi/8 and 2*pi/7…
assuming some kind of monotonicity, which seems likely.)

> 2) Yes, take {6,4}, then paint one hexagon, then paint every second its
> neighbour, and repeat process. You’ll get an example of {3*} this way.

Ah right, good. That’s the 2d face of it…
Okay… and I was wondering if the same process could be used in 3d, to find
your 3d convex object (I think you’re calling it {3*,3})
directly, as a sub-structure of some more "ordinary" uniform honeycomb.
But given the answer to 1), I think probably not.

> And funny thing - if you do it with 3D model of {6,4} based on truncated
> octahedra honeycomb, you’ll get finite object - eight faces of one cell.
> 3) In H2 answer is probably yes. And somehow graph of copies of {3*} will
> be the same regular 3-tree as a graph of hexagons of {3*}. I think that
> you can fill H3 with copies of {3*,3} (cut form cell of {3,3,n}) without
> overlapping, but not sure if the way from one copy to each other will take
> finite number of steps.
> And I called this object "fractal" but in reality it’s not - it is regular
> object that has exactly one scale. We see it as fractal when try to draw
> it on the Poincare disk, but it’s just a result of our projection.

Yes, good point.


> Andrey
> — In, Don Hatch <hatch@…> wrote:
> >
> > 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
> >

Don Hatch