# Message #2316

From: Don Hatch <hatch@plunk.org>

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.

Don

>

> 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

> >

>

>

–

Don Hatch

hatch@plunk.org

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