Polytopes of the numbers B2Z4

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Polytopes of the numbers B2Z4

Postby wendy » Thu Jan 22, 2015 9:12 am

The polytopes of the number system B2Z4 are the sort of things that play in my mind at night.

B2 is base 2, that is, it's the set Z, with closure of z/2. That means if it comes out exactly in binary, it is in B2.

Z4 is the span of chords of the {4}, it corresponds to A+Bq, where q=sqrt(2).

Polyhedra of B2Z4, as with all fractional systems, are infinitely dense. None the same, they are coordinate-discrete: it is possible to prove that points are never reached by the vertices.

The easiest way to make polyhedra in this set, is to replace squares in rCO or C with octagons, and then continue the cover throughout. This is one of the reasons the 'W' notation was invented for: some of the polygons simply don't close. Wx means that the short-chord square of the polygon is x. You make a 1:1:sqrt(x) triangle, and continue marking vertices of the polygon until you give up.

A simple example, is to suppose the squares of the rCO become octagons, and that the resulting {8, A, 8/3} is a euclidean tiling. A evaluates to W(1/2). A second way you can place octagons in rCO is to assume that the octagons in the middle faces are sample faces, giving a cell of {8,B,8}. You then find B as W((2-q)/2). And so on.

The symmetry of this group obviously includes an octahedral group in a non-obvious way. If you take a polyhedron o3xUo, and then suppose that the three verticies of one of the triangles represent the rectangular group, the six vertices of the three connected triangles, not already used are the other six mirrors of this group.

Last night, i was playing around with {8,Wc,3}, where c= 1/(2+q). It is interesting because this implies a branch, in RK's notation of *a 8/4 *c - that is, the edges cross by fours into an asterisk, and that x8oWCx, actually has {8/2} inverse squares, as the middle face.

It's probably a little more complex for most people, but this is what i exercise my mind on.
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Re: Polytopes of the numbers B2Z4

Postby Klitzing » Thu Jan 22, 2015 1:35 pm

Sorry Wendy, I cannot tell whether I'm able to follow your ideas or not. I just cannot follow your description... :(

A simple example, is to suppose the squares of the rCO become octagons, ...

- all 3 squares per vertex become octagons (then you'd get 60 + 3*135 = 465 degrees, i.e. not euclidean)?
- only the triangle-adjacent ones (then you'd get 60 + 2*135 + 90 = 420 degrees, i.e. still beyond euclidean)?
- only the cubical squares (then 60 + 2*90 + 135 = 375 degrees, i.e. still beyond euclidean)?

... and that the resulting {8, A, 8/3} is a euclidean tiling.

what does this "{ ... }" notation mean here?
- is it a Schläfli symbol?
- is it a vertex configuration?

A second way you can place octagons in rCO is to assume that the octagons in the middle faces ...

what are the middle faces here?

... are sample faces, ...

what you mean by "sample faces" here?

... giving a cell of {8,B,8}.

(same problem as above).

RK's notation of *a 8/4 *c - that is, the edges cross by fours into an asterisk

where do you refer to me?
- wrt. "*a", "*c"?
- or wrt. "8/4 ... that is, the edges cross by fours"?
For, in my reading "8/4" means an octagon, which winds 4 times around, that is it just describes a single edge, covered 8 times in total.

actually has {8/2} inverse squares

{8/2} in my notation means an octagon that winds twice around. So it would look like a prograde square.
Perhaps you are rather refering to {8/6}, i.e. a octagon with such a high winding number that it looks like a retrograde square?

--- rk
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Re: Polytopes of the numbers B2Z4

Postby Klitzing » Thu Jan 22, 2015 3:18 pm

Do you mean by {A, B, C} in context to rCO that this codes somhow the vertex configuration?
That is, the vertex configuration of rCO is [3, 4a, 4b, 4a] (as a cycle), and your code then might read like {3, 4a, 4b}, i.e. accordingly {A, B, C} then codes a vertex configuration [A, B, C, B]?

Under that premise your {8, A, 8/3} would evaluate into [8, A, 8/3, A]. Thence, in order to get euclidean, providing just A=4.

And your {8, B, 8} would evaluate into [8, B, 8, B]. Thence, in order to get euclidean, providing just B=8/3.

But your {8, C, 3}, then representing [8, C, 3, C], would not provide such a simple solution for C...
Edit: well, C = 48/13 would do.

--- rk
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Re: Polytopes of the numbers B2Z4

Postby wendy » Fri Jan 23, 2015 12:56 am

These are more like {5,5/2}, but infinitely dense things. Some thing like {8,B,8} is a euclidean tiling, of cells with octagon faces and a dihedral angle of 45 deg. The use of a euclidean tiling helps sets B. {} is a schläfli symbol, we're dealing only with regular figures here.

When I write {p/d}, it means a thing with p points, where p(x) is connected to p(x+d). I don't typically use Grunbaum's wound polygons.
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Re: Polytopes of the numbers B2Z4

Postby Klitzing » Sun Jan 25, 2015 10:22 am

Ah, this makes a lot more sense, then.
{ ... } is a "true" Schläfli symbol, and "tiling" then means just euclidean polytope, rather than flat 2D fabric.

I for one use here the general term "tesselation" instead, while restricting "tiling" for 2D.
But yours on the other hand is 3D, thus it rather describes a "honeycomb".

In fact you were deeling with {8, oo, 8/3} = x-8-o-oo-o-8/3-o, with {8, oo, 8} = x-8-o-oo-o-8-o, resp. with x-8-o-oo-o-3-o. And you were pointing out, that even so {8, oo} clearly is a hyperbolic (2D) tiling, when taken without multiwrappings, it well can be given a representation within euclidean 3D space, i.e. a polyhedron with regular octagons only, infinitely many at all the vertices. And being an infinite structure, that one would never close again, that is at least not within finite reach. Furthermore you pointed out, that right this non-closing fact leaves a degree of freedom, e.g. the dihedral angle between the octagons. Accordingly you are free to impose an edge figure being a {8/3}, a {8}, resp. a {3} there. Or speaking in dihedral angles again, these would be then 135°, 45°, resp. 120°.

If labeling the octagon vertices alphabetically, and using A and B for the ones at the connecting edge of 2 adjoining octagons, then one might consider the lacing distance between the C Vertices, which provides a different measure for that degree of freedom. Speaking of octagons, one calculates easily that, when the dihedral angle would be given as alpha, that C-C-distance then would be sqrt(1-cos(alpha)).

Esp. we'd get here
  • sqrt(1-cos(135°)) = sqrt((2+sqrt(2))/2)
  • sqrt(1-cos(45°)) = sqrt((2-sqrt(2))/2)
  • sqrt(1-cos(120°)) = sqrt(3/2)

Thus starting with the incidence matrix for the abstract polytope {8, oo} (i.e. independent of its hyperbolic or eucliedean realisation)
Code: Select all
x-8-o-oo-o (N,M -> oo)

.   .    . | 8N |   M |  M
-----------+----+-----+---
x   .    . |  2 | 4NM |  2
-----------+----+-----+---
x-8-o    . |  8 |   8 | NM
one derives accordingly the incidence matrix for that first of your honeycombs:
Code: Select all
x-8-o-oo-o-8/3-o  (N,M,K,L,P -> oo)

.   .    .     . | 2NLP |    MK |   4MK |   8M
-----------------+------+-------+-------+-----
x   .    .     . |    2 | NMKLP |     8 |    8
-----------------+------+-------+-------+-----
x-8-o    .     . |    8 |     8 | NMKLP |    2
-----------------+------+-------+-------+-----
x-8-o-oo-o     . |   8L |   4LP |    LP | 2NMK

Sure, describing just abstract polytopes, the same matrix likewise describes the second of your honeycombs too, just omit that "/3" part in the leftmost descriptive column.

For the last one you'd then get similarily:
Code: Select all
x-8-o-oo-o-3-o  (N,M,K,L,P -> oo)

.   .    .   . | 8NLP |    2MK |    3MK |   6M
---------------+------+--------+--------+-----
x   .    .   . |    2 | 8NMKLP |      3 |    3
---------------+------+--------+--------+-----
x-8-o    .   . |    8 |      8 | 3NMKLP |    2
---------------+------+--------+--------+-----
x-8-o-oo-o   . |   8L |    4LP |     LP | 6NMK


Here note, that this "oo" (infinity symbol) only servs in the sense of a numerator, resp. in the sense of an abstract polytope. But considering the corresponding vertex figure of those "cells" x-8-o-oo-o each, it becomes clear that those surely ask for multiwrappings in the sense of some {n/d} polygram.

To spot onto that we first consider a general flat polygon or polygram. If that one has an vertex angle phi, then the corresponding shortchord would be 2 sin(phi/2). Reverse engineering we derive from our above calculated shortchords CC, that this angle phi can be calculated according to phi = 2 arcsin(CC/2). Accordingly we derive in these three cases
  • phi_{oo, 8/3} = 2 arcsin( sqrt((2+sqrt(2))/2) /2) = 81,578941881850578219234172802659°
  • phi_{oo, 8} = 2 arcsin( sqrt((2-sqrt(2))/2) /2) = 31,39971480991904220976984000837°
  • phi_{oo, 3} = 2 arcsin( sqrt(3/2) /2) = 75,522487814070076121228965200873°

On the other hand we have for some "normal" polygram {n/d} that this vertex angle phi is being calculated as phi = 180°.(1-2d/n). Sure, here we always have q=n/d to be a rational number. But above we supposedly get real numbers r instead. Those would be calculated in these 3 cases according to r = 360°/(180°-phi)
  • r_{oo, 8/3} = 3,6577538067903974646494807498297
  • r_{oo, 8} = 2,4226063869225328714883422611807
  • r_{oo, 3} = 3,4457175756572188802726257456335
I.e. those "oo" rather should be some numbers lim_(n,d -> oo) n/d with limiting values r. And it is that these different real numbers r were refered to by Wendy's A, B, and C.

--- rk
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Re: Polytopes of the numbers B2Z4

Postby wendy » Sun Jan 25, 2015 10:30 am

For infinite polygons, i use W, which is a kind of infinity, followed by the shortchord square. Should W become rational, it is resolved. This allows me to see if one infinitagon is different or same as another. It's pretty easy to see the angles are not rational, but you know how to pick a chord of a rational angle.

I've been thinking of using your notation to mark the tiling separate, ie x8oAo, holo: *c8/3o. But it needs work.
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Re: Polytopes of the numbers B2Z4

Postby Klitzing » Sun Jan 25, 2015 11:52 am

Okay, you seem to denote these realvalued quotients of infinitudes then according to r = "W" & (CC)², where & denotes string concatenation, and CC is that shortchord value.

Accordingly you'll have
  • A = 3,6577538067903974646494807498297 = W ((2+sqrt(2))/2) = W 1,7071067811865475244008443621048
  • B = 2,4226063869225328714883422611807 = W ((2-sqrt(2))/2) = W 0,29289321881345247559915563789515
  • C = 3,4457175756572188802726257456335 = W 3/2 = W 1,5

--- rk
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Re: Polytopes of the numbers B2Z4

Postby Klitzing » Sun Jan 25, 2015 12:52 pm

Just to add for convenience some more such rather frequently used values:
Code: Select all
phi =  36°  =>  r = 5/2 = W((3-sqrt(5))/2) = W 0,3819660112501051517954131656344
phi =  45°  =>  r = 8/3 = W(2-sqrt(2))     = W 0,5857864376269049511983112757903
phi =  60°  =>  r = 3   = W 1
phi =  90°  =>  r = 4   = W 2
phi = 108°  =>  r = 5   = W((3+sqrt(5))/2) = W 2,618033988749894848204586834366
phi = 120°  =>  r = 6   = W 3
phi = 135°  =>  r = 8   = W(2+sqrt(2))     = W 3,4142135623730950488016887242097
phi = 180°  =>  r = oo (here not being wrapped!) = W 4

--- rk

PS: and even more general, for any rational value r = n/d you'd get r = Wz with z = (2.sin(90°(1 - 2d/n)))².
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