Message #2315

From: Andrey <andreyastrelin@yahoo.com>
Subject: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Fri, 06 Jul 2012 06:37:50 -0000

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.

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. 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.

Andrey

— In 4D_Cubing@yahoogroups.com, 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
>