Powertopes

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

Powertopes

Postby ubersketch » Fri Feb 02, 2018 9:43 pm

I need an explanation for powertopes.
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Re: Powertopes

Postby Mercurial, the Spectre » Sun Feb 04, 2018 3:23 pm

Bowers has a nice definition on these shapes as well as on the hi.gher.space wiki.

Powertopes are basically objects that exist only in perfect-power dimensions such as 4, 8, 9, etc. because it is made out of two shapes, one being the base shape A, and an exponent shape B.

The powertope is constructed by defining A as any geometric shape, and B as a specific shape with brick symmetry (for every coordinate (a,b,...,z), there are coordinates which represent all sign combinations, so for a rectangle with one coordinate (3,4), you have (3,-4), (-3,4), and (-3,-4) as coordinates.). Having brick symmetry entails that if B is n-dimensional, then it has a topological order of 2^n. It is denoted as A^B and written as the B of A.

Then you consider B 's vertices as a compound of congruent n-cuboids (ex. an octagon is represented as a compound of two congruent rectangles and a cuboctahedron is represented as a compound of 3 squares lying perpendicular to each other), then out of it, you take several cartesian products of shapes that are similar but not necessarily congruent to A representing the n-cuboids. (hence, for a triangle^cuboid, you can have a cartesian product of three similar triangles of different sizes that depend on the edge lengths of the cuboid). If you are taking a pentagon^octagon, you have two pentagon-pentagon duoprisms of sizes NxM and MxN lying perpendicular to each other; the convex hull is the powertope. The number of cartesian products is dependent on the number of n-cuboids present within B (usually number of vertices / 2^n).

In fact, B can be curved as long as it has brick symmetry (such as a sphere but not a cone), though it has to be noted that the powertope is also curved.

Focusing on isogonal 4D shapes, the n-n duoprisms are equal to n^square, where n is a regular n-gon. Cells are 2n n-gonal prisms with a symmetry of 8n^2. Vertex figures are generally tetragonal disphenoids; the 4-4 duoprism (tesseract) is instead a regular tetrahedron.

For the octagon as an exponent, consider it as having D4 symmetry (basically a truncated square). Then the powertope is non-trivial and is formed from what I described above on the pentagon^octagon powertope, now for n^octagon, you have two n-n duoprisms of sizes NxM and MxN lying perpendicular to each other. Cells are 2n n-gonal prisms connected by n^2 rectangular trapezoprisms (D2d symmetry like tetragonal disphenoids) with the same symmetry as an n-n duoprism. Vertex figures are triangular bipyramids (C2v symmetry) that may instead be realized as compounds of two tetrahedra (since they are generally not corealmic).

Due to that, the grand antiprism is closely related to the decagon^octagon as an alternation, except that two new tetrahedra fill each deleted vertex.

That is all I can say about powertopes and I would love to see curved ones.

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Re: Powertopes

Postby username5243 » Sun Feb 04, 2018 4:51 pm

Would I be correct in saying that these could be represented in Klitzing's "&#zx" notation? An octagon in 2D can be given by xw wx&#zx (as the vertices of two rectangles). So an n-gon^octagon would appear to be xw-n-oo wx-n-oo&#zy (for some y).

As for rounded things, I don't know. I'm fairly sure square^circle is the shape thw eiki calls "duocrind" - I'm fairly sure it can be given by the equation max(x^2,y^2)+max(z^2,w^2) = 1, though I don't know if anyone's done a render of it. As for circular bases, circle^square is the duocylinder, circle^circle is glome. I'm not sure what circle^octagon would look like - it seems it would have to be the hull of two differently oriented duocylinders.

Now in 6D things get interesting. In addition to anything with cubic symmetry, I'm fairly sure anything with dodecahedral symmetry can be a power. Both the icosahedron and dodecahedron can be represented as a sum of rectangles and cubes, so I'm fairly sure the rest of the dodecahedral family of uniforms can be represented here too, meaning some really complex shapes like hexagon^great rhombicosidodecahedron should be possible.

Bowers also seems to suggest there are starry version of these (x^octagram etc), but I'm not sure what these look like either.
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Re: Powertopes

Postby wendy » Mon Feb 05, 2018 1:56 pm

The &#xt notation is not really good for powertopes, although the most ancient form comes from one of the power-topes.

Powertopes exist in all dimensions: the four regular solids, and the cubic tiling are the powers of points (for the triangle-tetrahedron-pentachoron series), and lines (for the rest). So something like "pentagon ^3 " can be read in five different ways, resulting in polytopes of 8, 6, 6, 6, and 4 dimensions.

A triangle is the third pyramid power of its vertices, the tetrahedron is the fourth. In a triangle, you can select a vertex, and draw rays to the opposite side. This kind of ray is a 'draught' or drawing, and in some way resembles the way chewing-gum works. Every point of one base is connected to every point of the second base, by a line. We can make a tetrahedron by drawing a triangle to a vertex, or we can draw it from one edge to the opposite edge, in the process, making rectangles of sides ab, where a+b is the original edge-length.

A cube can be made by way of repeating the points of one base, for each point of the second base. So you get for a square, an edge, and then a parallel line for the next point, and so forth to the opposite edge. A cylinder, for example, can be made of repeated sticks like matches, or by a stack of coins. In each case, you can see the repetition of the base across the extent of a second base. This is the prism product. Prism comes from a greek word meaning 'off-cut', as one might cut off a long stick of that section.

An octahedron is a drawing of surface. This is the tegum product. Tegum means to cover. Here the bases cross at right angles, and we cover these bases by drawing the surface: that is, the surface of the octahedron is lines etc drawn between the surface of the square (ie the perimeter) and the surface of the line (eg the ends of it). This particular figure is from whence the pyramid product was found.

The crind product yields spheres. It has no easy explanation, but if a solid's surface is regarded as a function at 1, and the remainder of points at proportional lengths, then the crind product gives a root-sum-square of the radial functions of the various bases.

The comb product requires at least polygons, but is a 'repetition of the surfaces'. So if you take say, a large circle, and for each point, repeat a small circle, you would turn the thing into a torus. A torus made by turning a page into a circle, and then ideally connecting the ends of this tube into a tube, would give the same effect. The page is the prism-product of the surfaces of two circles, and the new solid has a surface which is a product of the surfaces of a previous example.
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Re: Powertopes

Postby Mercurial, the Spectre » Mon Feb 05, 2018 2:48 pm

Mercurial, the Spectre wrote:Bowers has a nice definition on these shapes as well as on the hi.gher.space wiki.

Powertopes are basically objects that exist only in perfect-power dimensions such as 4, 8, 9, etc. because it is made out of two shapes, one being the base shape A, and an exponent shape B.

The powertope is constructed by defining A as any geometric shape, and B as a specific shape with brick symmetry (for every coordinate (a,b,...,z), there are coordinates which represent all sign combinations, so for a rectangle with one coordinate (3,4), you have (3,-4), (-3,4), and (-3,-4) as coordinates.). Having brick symmetry entails that if B is n-dimensional, then it has a topological order of 2^n. It is denoted as A^B and written as the B of A.

Then you consider B 's vertices as a compound of congruent n-cuboids (ex. an octagon is represented as a compound of two congruent rectangles and a cuboctahedron is represented as a compound of 3 squares lying perpendicular to each other), then out of it, you take several cartesian products of shapes that are similar but not necessarily congruent to A representing the n-cuboids. (hence, for a triangle^cuboid, you can have a cartesian product of three similar triangles of different sizes that depend on the edge lengths of the cuboid). If you are taking a pentagon^octagon, you have two pentagon-pentagon duoprisms of sizes NxM and MxN lying perpendicular to each other; the convex hull is the powertope. The number of cartesian products is dependent on the number of n-cuboids present within B (usually number of vertices / 2^n).

In fact, B can be curved as long as it has brick symmetry (such as a sphere but not a cone), though it has to be noted that the powertope is also curved.

Focusing on isogonal 4D shapes, the n-n duoprisms are equal to n^square, where n is a regular n-gon. Cells are 2n n-gonal prisms with a symmetry of 8n^2. Vertex figures are generally tetragonal disphenoids; the 4-4 duoprism (tesseract) is instead a regular tetrahedron.

For the octagon as an exponent, consider it as having D4 symmetry (basically a truncated square). Then the powertope is non-trivial and is formed from what I described above on the pentagon^octagon powertope, now for n^octagon, you have two n-n duoprisms of sizes NxM and MxN lying perpendicular to each other. Cells are 2n n-gonal prisms connected by n^2 rectangular trapezoprisms (D2d symmetry like tetragonal disphenoids) with the same symmetry as an n-n duoprism. Vertex figures are triangular bipyramids (C2v symmetry) that may instead be realized as compounds of two tetrahedra (since they are generally not corealmic).

Due to that, the grand antiprism is closely related to the decagon^octagon as an alternation, except that two new tetrahedra fill each deleted vertex.

That is all I can say about powertopes and I would love to see curved ones.

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I figured out that the dimensions need to be merely composite rather than being perfect powers for non-trivial powertopes... oh well...
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Re: Powertopes

Postby Klitzing » Mon Feb 05, 2018 4:55 pm

Wendy is right, there are various ways to deal with multiple polytopes.

  • You might want to add those, this is what leads to Minkowski sums, or to zonotopes.
  • You might want to multiply them. Usually this is being understood to mean Cartesian products. But there are others too, as Wendy said. And then considering the product of a polytope with itself you'll finally get Pn, where n is some integer. That is you'll stick to numerical powers only.
  • But then you might also consider the geometric power application, as defined by PolyhedronDude, i.e. the thing PQ, where both P and Q are polytopes. - And, as I get it, the whole thread was meant to be considering those.

A further thing is the difference between the tailings "...&#xt" (lace towers = P||Q||R||...) and "...&#zx" (tegum sums = hull of compound from P and Q and ...).

And Übersketch is right too.
  • A square, when situated atop its side, can be represented in brique symmetry o2o2... as x2x.
    Thus a square (as a polygon) of any polytope P = xQoRoS... is nothing but xQoRoS...2xQoRoS...,
    or nothing but the duoprism thereof.
  • A square, when standing on its corner, PolyhedronDude then prefers to speak of a diamond, would be represented in brique symmetry as qo2oq&#zx, where q is the sqrt(2)-sized pseudo line, aka diagonal.
    Thus the diamond of any polytope P = xQoRoS... thus would be the hull of the compound of qQoRoS...2oQoRoS... and oQoRoS...2qQoRoS..., what has been denoted by qoQooRooS...2oqQooRooS...&#zy - for lacing edges of according size y.
    This powerpolytope then happens to be just the according duotegum.
  • And combining those, you'll get the octagon, which in brique symmetry is provided as xw2wx&#zx, where w=q+x.
    Thus the octagon of any polytope P = xQoRoS... then is nothing but wxQooRooS...2xwQooRooS...&#zy - for lacing edges of according size y.
Here y has to be sized accordingly so that the "z" property of that tail will be fulfilled, that is that both layers of the diagram will stick to the same space, or, when speaking in the language of lace prisms, that the according height would be zero.

And, for sure, there are also higher dimensional brique symmetrical exponent polytopes too. You even are allowed to decompose these polytopes, just as the octagon was decomposed accordingly. PolyhedronDude already mentions the decompositions of the icosahedron into 3 orthogonal rectangles or the decomposition of the dodecahedron into 3 orthogonal rectangles plus an additional cube. Etc.

E.g. the icosahedron is describable under brique symmetry as fxo2ofx2xof&#zx, where f is the short-chord of the x-scaled pentagon, aka golden ratio.
Thus the icosahedron of any polytope P = xQoRoS... then is fxoQoooRoooS...2ofxQoooRoooS...2xofQoooRoooS...&#zy.

And then one could dream for according brique-symmetrical descriptions of non-spherical polytopes, like J91 = fxo2ofx2oxF&#zx, where F=f+x. That is, for PJ91.

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