infinte repetitions of some finite layer sequence

Higher-dimensional geometry (previously "Polyshapes").

Re: infinte repetitions of some finite layer sequence

Postby wendy » Thu Oct 13, 2016 11:35 am

If you take a replacement sequence L -> LS, S -> L, then you generate a fibonacci ruler. That is, you get LSL LS LSL LSL LS LSL LS LSL LSL LS LSL ...

If instead of making L = 1.618033&c, you put it as -0.618033... then the ruler is _bounded_. That is, it varies from -0.618033 to +1, and gives an L segment whenever the number is positive, and an S segment when it's negative. This binding restricts what kinds of tiles might fit together in the quasicrystal, makes the quasicrystal as 'iso-bound'.

When you plot the stuff out on a plane of 1, Ø, the effect is that the infinite part goes from edge to edge, but the bound part is restricted by the two parallel lines iv=-0.618 and iv=1.

By moving the origin along the band, one can effectively change the sequence as given above, so instead of starting at 0, one is starting somewhere else.

The entire line then changes according to where you are standing on it, in that it will oscillate between two values separated by Ø, but at different points.

If one were now to consider instead of penrose tiles, a tiling of pentagons, then this would be wound infinitely, but it's much like the reduction of the 1,Ø plane onto a single line. The effect of iso-bound is caused by the fact that molecules can not occupy the same position, and one is left with a zigzag that is roughly twice the line width, caused by binding rules. If you were to stand on a pentagon in a {5,10/3}, you would only see those pentagons, and parts of pentagons, that a straight ray without turning would lead. This means that when you come to a vertex, you don't see the full 10/3, but only 3 1/3 of the pentagons around it.

Moving around the pentagon you are standing in, would mean that different parts of the 10/3 come into view, and different segments of remote pentagons are seen. It is rather similar to drawing a 2-space through a 4d lattice, and mapping only the nearest points.

Because the construction is iso-bounded, one can not have an infinite repeating pattern, but the amount that can repeat depends on how small the iso-image is to the bounding circle. That is, if the isomorph is only 1/10 of the circle, you can not repeat the image more than ten times before a spacer is called for.

The heptagon example you give in your thesis is governed by the same rules in two separate images, and it is sufficient to show that _any_one_ of the images is bounded. This is because an infinite repeating pattern is infinite in _every_ image. So the heptagonal system with L, M, S, gives rise to two images one in L', M', S', and the second in L", M", S". Since the first is necessarily infinite, then either the second or third might be finite.

The same arguments are used when one divvies up the hyperbolic tilings to see what goes into what. For example, {5,3,4} is hyperbolic, but {5/2,3,4} is discrete (real, infinitely dense). On the other hand, {5,3,6} and {5/2,3,6} are both hyperbolic. In the first pair, {5,3,4} and {5/2,3,4} are the projections of a thin section of six-space, rather as the pentagonal zigzag thrice over. So if we find on isomorphism, any other real:hyperbolic pairs, there is a chance of joining.

With the heptagon series, the cycle is 7, 7/2, 7/3 and 14, 14/5, 14/3. So we use the usual curvature rule to show that {7,3}, and {7,4} can have the same symmetry, but {7,6} can't. This is because the first pair give H,S,S as their coordinate, but {7,6} goes H,H,S (only 7/3,6 can be started in spherical space). If one gets, eg S,H,S, it is fair to rotate these so that the first isomorph becomes prime. Note that the 7's and 14's must both be rotated in step. So {7,14} gives isomorphs {7/2, 14,5} and {7/3, 14/3}, which is the same HSS as the {7,3} set, and this can exist as a subgroup.
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Re: infinte repetitions of some finite layer sequence

Postby Klitzing » Thu Oct 13, 2016 2:55 pm

Okay, much better, Wendy!

Your "iso-bound" thus is nothing but a different term for the existance of an acceptance domain within perp space. And the lifting into embedding space here generally goes by algebraically conjugate values for the respective components (para space and the various perp spaces). :]

Your transfer onto hyperbolic tilings so looks new to me. There I should get more used to, I think, in order to fully grasp the details you're mentioning... :cry:


For "true" cut and project tilings with a Pisot-Vijayaraghavan scaling number (just one algebraic conjugate is > 1, all others are < 1) I should further mention some Inflation symmetry. Then you could use a concentric scaled down acceptance domain within perp space(s), providing a correspondingly scaled up tiling in para space, which looks essentially the same, but is using only a corresponding subset of the former tiling Vertices, i.e. uses larger tiles.

Would something like that also apply to the hyperbolic tilings, kind of a superposition of those, with some vertex coincidences and thus resulting (up to potential edge scalings) within the same curvature?

Sometimes those quasiperiodic inflation symmetries would allow for a directly deduced localized set of substition rules. - Supposedly this would be the same for such sets of hyperbolic tilings (of every dimension).

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Re: infinte repetitions of some finite layer sequence

Postby wendy » Fri Oct 14, 2016 11:42 am

Hyperbolic space is essentially exponential, in that you can do something like L -> LS, S -> L, so each new layer is at the same scale. This kind of expansion is in physics called a 'cascade'.

If the cascade is 'critical', it would expand at the same rate as space. If it were 'sub-critical', it would expand at less than space, and while getting ever bigger, space is expanding faster too. If it is 'super-critical', then it would expand faster than space, and become infinitely dense. All three, as well as 'iso-binding', occur in hyperbolic space.

If you take for example, oPx3xPo, the hexagons unfold without overlap onto x6o4o, in such a way that the same hexagon is identified with its copy, moved P spaces. That is if you count P polygons on the same side of a line through x6o4o, you end up with the same copy of it. In the cases where P=5 or P=6, this means there is a one-to-one mapping of these hexagons directly from infinite H3 onto infinite H2, and in such a way that the mapping is vanishingly small (that is, there are more copies of a given cell, then there are different cells).

The simple explanation comes from writing say, decimal digits into binary (super-), decimal (critical), or hexadecimal (sub-critical) columns. Decimal - on - decimal, evidently fills up all the space, and so fully covers whatever space it is given to it. Sub-critical gives rise to something decimal on hexadecimal: there are numbers you can't reach with decimal digits, and they tend to be the more common form. So for example A0 (dec 160) can not be written with simple decimal digits. The figures described in this way form 'yickles', or areas bounded by unbounded lines. The super-critical examples, come by writing decimal numbers in binary columns, so 1000 = 200 = 40 = 8 = 120 = dec 8 etc. The more steps involved, the more one has to go.

Isomorphism might be described as a projection of something higher into different spaces, at precise angles. The cyclotomic numbers, which lie at the base of all of the numbers we use here, are projected onto the same space, so it is always possible to find the transforms or conjucates of a figure. By considering the isomorphic forms and what kind of spaces their isomorphs are, one can consider the viability of does this go into that.

Regarding the actual substitution rules, there is one known for Marek's tiling of (3,5,5,5), it has a radial constant of 2.618033&c, and one might readily suppose the real reflex of this is (3,5/2,5/2,5/2). I shudder to think of all of the variations here. There are an infinitude of these things, but i don't think anyone really has pushed these hyperbolic quasi-lattices. According to Norman Johnson, they would form some sort of quasi-periodic substitution as in your thesis, but the notion that one can start tiling the plane with penrose tilings might mean that this might not be entirely correct.

Since we know that every {p,q} exists in some space, for integer p,q we would then note that these all suggest isomorphs, which are necessarily super-critical, (hyperbolic) or isobound. (spherical) But two different figures can not be related to each other unless their isomorphs are in the same kind of space either.

Just as we don't work with the spherical arcs of polyhedra as spherical tilings, we don't work with the hyperbolic arcs of hyperbolic tilings. Instead, we draw circles on the space, and span the circle with a euclidean plane. The distances are measured on this plane, which supprisingly has euclidean geometry! What is interisting is when you get something like o3x5o, regardless of size or space, the euclidean chords are in the same ratio. The function corresponds to the chord of the arc on the sphere, and the choh on the hyperbolic space. The matrix spreadsheet that i wrote for you already deals in these chords and hcords.

The lengths of the chords, and the radius, measured against an edge, are bounded by the number system one might derive from the symbol. Different number systems means that it's incompatible, except when all the isomorphs are bounded. Thus {5,3,3} Z5 can contain {3,3,3} Z1, but not the vertices of o3x4x3o Z4, because Z5 contains Z1 but not Z4.

Because {5,3,4} for example, is hyperbolic, it's essentially exponential, and thus its isomorph must have exponential branching too. This is different to the linear growth you get in euclidean lattices, or even quasi-lattices. {5/2,3,4} is piecewise finite, that means that if you strike an incidence matrix, there are no infinities except on the main diagonal. Piecewise finite tilings are constructed from 'algebraic integers'. particularly cyclotomic numbers. It is possible to figure out from the dynkin graph, the integer system involved, and the various directional incomeasurables involved.

The through-system is the simplest product that contains all of the branches. The general pattern is Z5Z3Z2, but the last two are identical to Z, a subset of Z5. There is no loop, so we don't have to find any loop constant. But the last node here is across a 4 branch, and the 'bridge constant' here is sqrt(2). So if the first part belongs to 1+sqrt(5), the second part belongs to sqrt(2)+sqrt(10). CZn is the cyclotomic numbers given by the span of the equation x^n=-1. The real part is Zn, which is the span of chords of an {n}. So CZ2 = gaussian numbers, CZ3 = eisenstein integers, and CZ1 is the real integers, all have a real component Z = Z1 = Z2 = Z3.

Where there is a loop, such as o3o3o3o4z, the 'through system' is Z2Z3 = Z. But the bridge constant is alternated as you go across branches, for odd branches, the bridge constant is 1. For even numbers, it goes for n=2, mod 8, one gets G(n/2) [something nasty], for g=4 mod 8, it's sqrt(2), for g=6 mod 8, it's sqrt(n/2).

So here we see that if you go around a loop, the values are 1, 1, 1, 1, sqrt(2) and so each node is marked [1, sqrt(2)] This means you have to add this into the system if it is not already part of the numbers. o3o3o3o4z is thus unrelated to any of the other symmetries made of 3 and 4, except prehaps o3o4o4o4z.

We then note that with hyperbolic tilings, something like {18,6} can not contain {6,6}, for the simple reason that one of the isomorphs here is the bounded {18/7,6}, and a spherical tiling can not contain a hyperbolic sub-polytope.

Regarding PV numbers. Every cyclotomic system must contain PV numbers, and there is a method of finding these. It's even possible to find the direction of greatest convergence. But i don't think that PV numbers are the only ones that can be powers of expansion.
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Re: infinte repetitions of some finite layer sequence

Postby Klitzing » Fri Oct 14, 2016 5:10 pm

what shall I say?
that all sounds very interesting ...
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Re: infinte repetitions of some finite layer sequence

Postby quickfur » Fri Oct 14, 2016 7:10 pm

An interesting question occurred to me about hyperbolic space. We know that {5,3} (i.e., a tiling of pentagons at 3 to a vertex) is finite; it is none other than the regular dodecahedron, which can be regarded as a tesselation of spherical space. Now, {6,3} cannot be embedded in a spherical space; it requires a space of curvature 0 (or less?), and is in fact a tesselation of the flat Euclidean plane. And {7,3} requires a space of negative curvature, i.e., a hyperbolic space. In general, it seems that {n,3} can be embedded in a spherical space (i.e., positive curvature) if n<6; a flat space if n=6, and a hyperbolic space if n>6. My question is, can {6,3} be embedded in hyperbolic space too?

More generally, can {n,3} be embedded in a space with lower curvature than it "needs"? I can think of how to embed {5,3} in a flat plane, for example: the stereographic projection comes to mind, where the outer pentagon wraps around the other 11, and so the embedding is finite. One intuitive way to think about it is that the curvature of the plane produces "more space" than {5,3} "needs", so it only occupies a finite amount of the plane as opposed to being a tesselation that covers the entire space.

Now wrt to {6,3}, since the tesselation needs "less space" than the hyperbolic plane "has", does that mean that an embedding of {6,3} into the hyperbolic plane should have a finite extent? It cannot cover the entire hyperbolic plane, can it? If so, we have the interesting situation where an infinite polytope can be embedded into a hyperbolic space without occupying the entire space!
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Re: infinte repetitions of some finite layer sequence

Postby Klitzing » Fri Oct 14, 2016 7:41 pm

quickfur wrote:My question is, can {6,3} be embedded in hyperbolic space too?

Yes it can. Consider e.g. {6,3,3}.
... does that mean that an embedding of {6,3} into the hyperbolic plane should have a finite extent?

No. As {6,3} is euclidean, it happens to be tangential to the circle of infinity, whereas hyperbolics would intersect it.
Actually, within {6,3,3} the {6,3} becomes an infinite polyhedron.

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Re: infinte repetitions of some finite layer sequence

Postby wendy » Fri Oct 14, 2016 11:49 pm

The art of circle-drawing renders polytopes into a net of euclidean chords. So the same chords that apply to o3x5o, apply to it not only when it is a spherical tesellation, such as the midsection of x3o3o5o. The same net of chords apply to it as a face of o3o3x5o and even as o4o3x5o and the verf of x3o5oAo.

A euclidean tiling is also a net of chords. It expands parabolically, but x^2 is no match for 2^x, and so while the length of horizon is getting bigger, the horizon is expanding faster still. In projections centred on a point, this means that x^2 / 2^x disappears to a point.

You can have hyperbolic tilings as cells. In essence, {8,3,4} has cells {8,3}, in the same way that {6,3,2} has cells {6,3}. Where the second the two cells are half-spaces, in the first, they are proper tilings of cells, with right-angles cells. The truncates o8o3x4x and o6o3x2x consist of flat cells o8o3x and o6o3x, which sandwiches layers of tC and tips resp. The o8o3x cell in o8o3x4o is larger than the flat o8o3x, but they are each a net of euclidean chords, the ratio of corresponding chords is constant (as in the sphere), and here, larger rather than smaller. So you don't need special cases for flat vs curved figures. The circle art works on all isocurves, the definition of straight becomes 'having the same curvature as space'. This is like all circles on the sphere work the same, but great circles are concentric with the sphere.

This of course causes some issue with 'what is a polytope'. You can have something like an {8,3} that has right-angled margins, and from the local prespective, looks like a polyhedron. A {5,3,3,} cell in o3x5o3o3o is sufficiently large that you can't tell if it's infinite or not. Even the rings of {3,3,3,5} give a million vertices in the tenth ring.
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