Underlying space for non-convex polytopes

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

Underlying space for non-convex polytopes

Postby quickfur » Thu Jul 14, 2011 8:04 pm

If convex n-polytopes tile the n-sphere, then what kind of space do non-convex n-polytopes tile, if any? In particular, what kind of space do the star polytopes tile? Does it depend on their Euler characteristic?

(I'm quite ashamed to admit that I'm ignorant of such basic matters in spite of being able to visualize many 4D objects. :P)
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Re: Underlying space for non-convex polytopes

Postby wendy » Sat Jul 16, 2011 5:36 am

You can get non-convex polytopes to tile space. You can even have space tiled by convex polytopes in a non-convex way.

For example, consider the tiling of hexagons {6,3}. The hexagons can be stellated to make a hexagram {6/2} to give a tiling of hexagrams {6/2,6}. The central hexagons are undisturbed (just like 5,3 -> 5/2,5), the points raise flat pyramids over the adjacent hexagons.

The next stellation is to replace the hexagons of (6/2,6) with regular hexagons. The vertex figure gets stellated, and if one preserves depth by the notion that the top is the one furtherest from some centre, then the thing looks like a {5,5/2}

A different kind of margin is a 'pleat-margin'. Where most cell-walls have cells on the opposite side, giving a margin-angle of 180°, a pleat-margin has both cells on the same side, giving a margin-angle of 0°. One counts one cell as facing 'in' or negative density. It often occurs in the isomorphic polytopes.

For example, if one considers the square-triangle stripped tiling, one can pleat this, so the triangles are folded under the squares. It still does a single cover, if you subtract the number of triangles from the number of squares at each point. It's a simple example of replacing sqrt(3) by -sqrt(3).

Similarly, one gets pleats, and the replacement of the vertex-pentagon by a pentagram, if one makes the triangle-square angle a pleat in the snub(4,4). The vertex figure is (3,3,4,3,4), but this is made to form a pentagram bu reversing directions when changing number.
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Re: Underlying space for non-convex polytopes

Postby quickfur » Sat Jul 16, 2011 2:13 pm

That is all very nice and good, but what I wanted to know was, what sort of space is the non-convex polytope itself tiling? For example, the regular (convex) polygons tile a circle, in the sense that their edges are a tiling of the circle, but what do the edges of a {5/2} tile? Or, for that matter, since the faces of convex polyhedra are a tiling of the sphere, then the faces of, say, {5/2, 3} form a tiling of what?
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Re: Underlying space for non-convex polytopes

Postby Secret » Mon Jul 18, 2011 6:59 am

If non convex polygons tile a circle, and non convex polygloss tile a sphere
Can we say the non convex polygloss tile a n sphere?

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Re: Underlying space for non-convex polytopes

Postby wendy » Mon Jul 18, 2011 7:25 am

Star-polytopes live in a variation of ordinary space, where in/out is replaced by an integer density. One can have instead of in (d1) and out (d0), any sort of integer.

Consider the pentagram. The surface is formed by the five lines that run from point to point. There are only five vertices at the points, the crossings on the pentagons are not vertices, as we shall soon demonstrate. The core of a pentagon is d2, the points are d1. When one uses pentagrams as faces of the {5/2,5}, the points have a density of 1, the core has a density of three, the difference is 2, that of the pentagon core.

The gradient of density is the outvector, which defines surface. Some parts of the surface can be interior (eg in the pentagram, the line between the point and the core), but at all points, the surface is a change of density. The full edge from point to point has an out-vector of 1, since everything on one side is d1 less than everything on the other side.

Volume is still moment of out-vector, but because now parts can belong to different surface elements, the spaces can have density weightings, a d3 region is counted three times in the volumne. The usual manner for evaluating volume is to find the volumes of the various flags, (simplex vertex edge-centre hedron-centre etc), by the number of flags. This is essentially face X depth, but applied recursively. One sees that this counts points in several flags (like a d3 region) that many times.

The surface of a polytope, whether regular or starry, has to be "sealed", that is the total radiant out-vector from any point not on the surface has to be constant in every direction. Without this, the volume ceases to be a numeric value, and becomes a vector moment.

This is why in the PG, surface and perimeter are different. A perimeter encloses all of the referenced points, the periform is the enclosed area. The surface is the gradient of density, can have elements internal to the periform. The periform of the pentagram is the zigzag decagon that encloses it, while the surface and the space occupied by the five lines. When the surface is taken as one, then it must have unbroken density, that is at each point, the total density gives an integer. (At the vertex, there are two edges of d1, so these make a continuity: for any full line containing an edge, there is a change from d1 to d0 at the vertex, so we know that the surface of the line exists there.
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Re: Underlying space for non-convex polytopes

Postby quickfur » Thu Sep 08, 2011 1:40 am

Thanks for the explanation, Wendy.

Now I'm wondering, if convex polytopes are tilings of the n-sphere, and tessellations are tilings of n-space, and hyperbolic tessellations are tilings of hyperbolic n-space, then what's the non-convex equivalent of tilings of n-space, hyperbolic or otherwise? Are there such things as starry tesselations?
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Re: Underlying space for non-convex polytopes

Postby wendy » Thu Sep 08, 2011 7:12 am

There are indeed star-tesselations. Densities are shown in square brackets after the polytope, as [d5].

In hyperbolic space, there is always {p,p/2} and {p/2,p}. When P is even, this becomes a compound of three parts. When p is odd, it is a star-tesselation on the lines of {5/2,5} and {5,5/2}. In higher dimensions, there are star-tiles like {5/2,5,3,3} [d4], {5,5/2,5,3,} [d6] and their duals. There are star tilings in H3, but none are regular. {5,5/2,5,3:} is a subgroup of order 4 of {5,3,4}, yields stars.

When one takes {6/2} and the stella octangula as stars on the line of {5/2}, one gets

{6/2,6} and {6,6/2} of order 3. To this, there is the 'asterix-star', formed by edges crossing in an asterix. Thence one has {6 * 3}, of d4, and {6/2*6} {6*6/2} of d12. This completes the regular stars of {6,3}

The simplest asterix-star is {4/2}, or +. These occur in the stellation of every {p,4} to give {p+p} [d2]. The stella octangula is an example of this, with p=3. But it occurs with all p, p=4 in euclidean space, and p>4 in hyperbolic. One finds also an {8/4} type edge in {8++8} [d6].

The stella octangula occurs as faces of stella tegmata, {3+3,4}, and as vertex figures of {4,3+3}. These both tile 4-space, as {3+3,4,3} [d4], {4,3+3,4} [d6], and {3,4,3+3} [d4], and again in H5, as {3+3,4,3,3} [d5], {4,3+3,4,3},[d10] {3,4,3+3,4} [d10], and {3,3,4,3+3}[d5].

In H3, one finds also {6/2,6,3} [d4], {6,6/2,6} [d6], {3,6,6/2} [d4], {4+4,4} and {4,4+4} both [d3]. The tilings {3+3,4,3+3}, {6/2,6,6/2} and {4+4+4} are all infinitely dense: the first two imply {3,3+3} and {3,6/2} which are also infinitely dense.

Over about 5 dimensions, stellated forms tend not to occur greatly.
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Re: Underlying space for non-convex polytopes

Postby Klitzing » Wed Jul 31, 2013 3:56 pm

Hmm, quickfur's questions sounds like aiming to understand a global versus a local application of rules.

Convex D-polytopes resemble (i.e. tile) globally a D-sphere, euclidean D-space, hyperbolic D-space globally. Okay. But beyond?

Just as Norman Johnson once was applying snubbing rules not as a global concept but rather as a local one, he came up with what he's calling holosnubs: i.e. applying the concept of vertex alternations locally well allows for using e.g. polyhedra for bases, which do not have all even numbered faces only. In case one just has to circle around some face-polygon twice instead in order to come back with the right parity. This is why he calls them "holo": they would use not half of the vertices of the base, but they will use all instead.

The same now applies here too. A pentagon is nothing but a piecewise straightened circle. Now wind up that very procedure twice around the center, and you'll get the pentagram. That is: locally it is quite the same, just with respect to the global view it differs.

Thus, after all, you just attach facets dyadically as usual.

Sure, this opens kind a box of Pandorra: In order to get a non-convex polytope, you might use non-convex building blocks (facets) like the small stellated dodecahedron {5/2, 5}, having 5 pentagrams per vertex. Or you could use still convex facets in a non-convex way, resulting in a non-convex vertex figure, like the great dodecahedron {5, 5/2}, having 5 convex pentagons around each vertex, but those would circle twice around (in the sense of a pentagram). - In fact, in principle any link of the dynkin symbol could carry some fractional number. (It just is a different question, which of those would result in a still finite covering.)

Esp. that small stellated dodecahedron {5/2, 5} shows some further principle: you'll attach the pentagrams edge-wise dyadically (= 2 per edge, or in general: 2 per (D-2)-face). You shall not worry about the tips of the pentagrams peaking into the others realm. This is just a global effect, not a local issue. Only in the global view you thereafter would have to deal with "densities", i.e. with multiply covered regions, when going form "outside" towards the body center (if existent). Even negative densities might occur occasionally.

With respect to euclidean space tilings and hyperbolical ones you usually stick with the tiling space itself. Kind of sticking, when considering a dodecahedron, just with the corresponding tiling of the sphere (2-space). But usually you'd consider a dodecahedron rather as some globally more or less spherical structure, embedded within euclidean 3-space! This is what a polytope is (in contrast to a mere tiling). In the same way, you should get used to understand euclidean (or hyperbolic) D-space tilings as being embedded into euclidean (D+1)-space (for hyperbolics this clearly would provide some kilt-mode - but again, we are interested in local considerations only!). Therefore, when considering {7/2, 7} you similarily to the small stellated dodecahedron would have edge-wise connections of the heptagrams, providing some "pyramids" from their tips atop the central regions of the neighbouring heptagram.

Consider the 2 pictures, being displayed at my webpage, showing the tilings {7/2, 7} and {7, 7/2} (where only the outer parts of the faces have been highlighted in order to come around the multiple coverage problem for display reasons). - Btw. "sissid" (small stellated dodecahedron = {5/2, 5}) and "gad" (great dodecahedron = {5, 5/2}) are linked there for reference too.

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Re: Underlying space for non-convex polytopes

Postby wendy » Sat Aug 03, 2013 11:59 am

If you have ever looked into a kaleidoscope, you would see lots of images. But when you lay out the reflective regions on space, light travels a straight path. It is that the usual symmetry of the kaleidoscope (* 3 3 3), has no overlap.

When two mirrors join, you can not see more than 180 degrees of the possible images at any point. Suppose the mirrors are set at 54 degrees, the symmetry angle of the decagram 10/3. You might be standing at a point like 300 degrees. You can see of the image, everything from 120 to 480 degrees. Edges run between multiples of 108, so you can see bits of the edge 108 to 216, all of 216 to 324, and 324 to 432, and bits of 432 to 540. To see beyond this, you have to walk around the decagram.

If you were standing at 'C', at 300 degrees, you could see an observer B at 180 degrees, and D at 420 degrees, but no others (supposing a difference of 120 degrees). An observer standing at F (660 degrees), is standing where you are, but he is on a different 'plane' so to speak. No one can see C and F together, the best is you can see half of C and half of F.

Now, the kaleidoscope produces triangles. If we were to tile the plane with trianlges like 36-54-90 degrees (ie centre and two edges of the pentagon), we could see right around the 36 and 90 point completely, but the two 54s need to be walked around. When you are laying out these tiles that you can see, you can't 'bend light' around a 54-degree pole. So if you draw a line from yourself to a pole, and you are laying tiles around it, and going behind the left, you can't see anything that is further around than the line between you and the pole.

So even though this is the symmetry of the tiling {5,10/3}, what you see is a broken penrose-style tiling of single pentagons and bits of pentagons, to a single density. Move a little bit, and the outer bits will change.
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