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definition

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In geometry, a duoprism is a polytope resulting from the Cartesian product of two polytopes of lower dimensions.

Such as the product of two edges? —Tamfang (talk) 07:47, 12 September 2008 (UTC)[reply]

Yes, {}x{}={4}, edge x edge = square, but as a special case, with {}n={4,3n-2} in general, so any prismatic polytope can be reduced to at most one {} product, and thus can be considered either a prism or not a prism.
So I'd call for example, {3,3,3}x{} as a generalized prism rather than a duoprism, and {3,3}x{3} as a duoprism, and {3,3}x{3}x{} as a duoprism-prism, and {3,3}x{3}x{3} as a triprism, etc.
So perhaps the definition about should say a product of two polytopes, each dimension two or higher? Tom Ruen (talk) 16:59, 12 September 2008 (UTC)[reply]
works for me —Tamfang (talk) 09:07, 13 September 2008 (UTC)[reply]
OK, I've updated the definition accordingly. Hopefully the wording is clear enough.—Tetracube (talk) 16:48, 13 September 2008 (UTC)[reply]
Oh dear, I just don't get it. Double sharp (talk) 03:43, 10 August 2009 (UTC)[reply]
It is a bit confusing. A prism is a product of a line-segment {} and a polygon {p}, so {}x{p} is a p-gon in a plane extruded into the 3rd dimension into a solid prism. A duoprism is similar except in 4D, you can extrude in 2 dimensions from a plane, so you can extrude a polygon product {p}x{q}! I admit it is confusing since there's no other clear lower dimensional analogy. Tom Ruen (talk) 06:12, 10 August 2009 (UTC)[reply]
Generally, in n dimensions the number of kinds of prism is related to the number of partitions of n. 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1; respectively these five expressions correspond to the nonprismatic polychora (4), the polyhedron prisms (3+1), the duoprisms (2+2), prisms built on polygonal prisms (2+1+1) (equivalent to duoprisms in which one of the factors is a square), and the hypercube (1+1+1+1). —Tamfang (talk) 05:45, 11 August 2009 (UTC)[reply]
What about 5 dimensions and higher? Double sharp (talk) 09:27, 12 August 2009 (UTC)[reply]
Same thing goes. Among the partitions of 5 are 4 + 1 (polychoral prism); 3 + 1 + 1 (polyhedral prism prism); 3 + 2 (polyhedron-polygon duoprism); 2 + 2 + 1 (duoprism prism); 2 + 1 + 1 + 1 (cube-polygon prism). Once you get to 5 dimensions and beyond, the lack of standard terminology makes it difficult to express exactly what you're talking about; writing it out as a Cartesian product helps to nail it down precisely (e.g., square * triangle * tetrahedron = 7-dimensional multiprism).—Tetracube (talk) 21:28, 12 August 2009 (UTC)[reply]
They're listed here for 5D Uniform_polyteron#Uniform_prismatic_forms, and here for 6D 6-polytope#Uniform_prismatic_forms. I tried to list them all (complete up to rank 7, reducing []x[]=[4]) on a userpage User:Tomruen/List of Coxeter groups Tom Ruen (talk) 22:46, 12 August 2009 (UTC)[reply]
Since the tesseract is a cubic prism, and the cube itself is a square prism, is it also a duoprism? Double sharp (talk) 08:50, 13 August 2009 (UTC)[reply]
Yes, a tesseract is a 4-4 duoprism, represented as {4}x{4}. It can also be a cubic prism {4,3}x{}. If you're talking uniform polytopes, {4,3,3}={4,3}x{}={4}x{4}={}x{}x{}x{}. But these lower forms have degrees of freedom, like {}x{} can be a rectangle more generally versus a square {4}. Tom Ruen (talk) 08:59, 13 August 2009 (UTC)[reply]
And {4}x{4} can also be a cuboid instead of a cube, I guess. But I still don't get what it said about prisms and antiprisms. In what way is this polyhedron related to this one , to go down a dimension? Double sharp (talk) 09:07, 13 August 2009 (UTC)[reply]
I'd have to bring in the Wythoff construction to explain fully; it's easier to begin with a decagonal antiprism. [oops!] Then label the vertices alternately – you can do this since all faces are even (10-gons and 4-gons) – remove one set, and take the convex hull of the remaining vertices; put another way, this means inscribing pentagons in the decagons and digons (degenerate polygons with only two vertices) in the squares, and putting triangles in place of the missing vertices. The result isn't uniform, but it can be distorted into uniformity (by shifting the vertices), because there are two degrees of freedom with which to solve two equations (to make three numbers equal).
But if we start from a duoprism – let's say a hexagon × octagon – and do the same alternation, then in general you can't get uniformity, because three degrees of freedom are not enough to assure a solution to five constraints (equating six numbers). —Tamfang (talk) 21:09, 13 August 2009 (UTC)[reply]
And if we start with a pentagonal one instead? Now you can't label the vertices alternately, so what happens now? Professor M. Fiendish, Esq. 02:28, 9 September 2009 (UTC)[reply]
You can alternate the vertices and get a star antiprism. (eh? i thought that would be blue) —Tamfang (talk) 06:39, 30 July 2010 (UTC)[reply]

Uniform polytopes

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This isn't really relevant, but there are articles uniform polyhedron, uniform polychoron and uniform polyteron. Uniform polyexon, uniform polyzetton and uniform polyyotton are redirects. Uniform polyxennon doesn't even exist. Can someone tell me why? Double sharp (talk) 09:02, 13 August 2009 (UTC)[reply]

Apathy? Beyond polychora there's not much novelty to be found. —Tamfang (talk) 21:10, 13 August 2009 (UTC)[reply]
Because no one has added them yet? If you wanted to, you could add them. You'd have to be able to supply enough material to justify devoting an entire article to each subject, though.
And I disagree that beyond polychora not much novelty is to be found: the Gosset uniform polytopes, for example, are of quite some interest, up to 8 dimensions or so. Beyond that, there are certain dimensions with sporadically uniform polytopes besides the usual ones derived from the cube/cross and simplices, corresponding to the symmetry of certain mathematical groups. (IIRC there's one of interest in 36 dimensions, but I forget what it corresponds with.) Besides these, there is also the relatively unexplored area of higher-dimensional Catalan polytopes and other cell-transitive polytopes that are not necessarily uniform.—Tetracube (talk) 22:23, 13 August 2009 (UTC)[reply]
I added a redirect to 10-polytope. Most/all of the contents of 6-polytope to 10-polytope are about regular and uniform polytopes, so there could be 2 sets of articles like for lower dimensions, one focused on uniform figures. I started a division on uniform polyteron vs 5-polytope. I added the 5-simplex family to the uniform polyteron, but never finished the tables for other familes, given on a test page User:Tomruen/uniform_polyteron. It's all rather unstable ground, with a firm theory, but the only reference to the figures comes from Dr. Richard Klitzing's webpages, currently hosted at [1]. They are a bit cryptic, using an ASCII "inline" Coxeter-Dynkin diagram and names by Jonathan Bowers. I'll be going offline for a week. Have fun! Tom Ruen (talk) 02:47, 14 August 2009 (UTC)[reply]
p.s. The infinite (tessellation/honeycombs) forms are fun too, especially 3d space ones. Convex uniform honeycomb, and Convex uniform honeycombs in hyperbolic space. The uniform polychorons can be considered tessellations of the surface of the 3-sphere, so they're all related! I've seen a list of the uniform tetracombs (4D honeycombs), and started my own attempted enumeration but didn't get very far User:Tomruen/Convex_uniform_tetracomb. Tom Ruen (talk) 03:23, 14 August 2009 (UTC)[reply]
It was by contemplating the polychora as tilings of S3 that I learned to understand words like runcinated. —Tamfang (talk) 00:43, 15 August 2009 (UTC)[reply]
Maybe uniform polypeton could be justified, but I don't think going higher than that has much more novelty. Professor M. Fiendish, Esq. 02:30, 9 September 2009 (UTC)[reply]
Uniform polyzetton (8-polytopes) could be defended as useful with the the "exceptional" E8 polytope, and Richard Klitzing lists forms up to 8D at: [2], BUT doesn't enumerate them all, and even some of the nonprismatic uniform polypetons are missing element pages.
Actually, the PRIMARY reason to end at 5-polytopes is Norman Johnson's truncation terminology stops at Sterication which is in his yet unpublished book on the subject (I asked him if he had a higher term!). Jonathan Bowers goes higher, BUT they also work a bit differently on the mixed forms. Anyway, nothing published by Bowers at all. Klitzing's website is the only "public" source at all, referencing the Coxeter-Dynkin diagram and the Bowers short names.
Ring-Interval Johnson Bowers Polytopes
1 Truncated Truncated Regular polygon...
2 Cantellated Rhombated? Uniform polyhedron...
3 Runcinated Prismated? Uniform polychoron...
4 Stericated Cellated? Uniform polyteron...
5 Pentellated Terated Uniform polypeton...

Tom Ruen (talk) 02:52, 9 September 2009 (UTC)[reply]

"Duoprism" term and its definition

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Let me state first that I am an Abstract polytope-ist, where the smallest polytope is the null polytope of rank -1, and all polytopes have a (-1)-face.

There is no reason to exclude lower dimensions. Any product P X Q is polytope when rank(P) < 2 or Rank(Q) < 2, or both. Of course, many of these are trivial - and uninteresting. Will return to that point.

"Multiplying" an n-polytope by an "edge" (1-polytope) gives an (n+1)-prism. Multiplying a prism by a point (0-polytope) leaves it unchanged (or isomorphic, anyway). In both these cases the "m+n rule" works.

Multiply anything by the null polytope gives the null polytope. Here only the m+n rule fails.

It is not good mathematics to arbitrarily restrict concepts and definitions - especially when there is no need to do so.

I strongly recommend that the article be renamed "Product polytope".

I am not sure "prism" means anything other than a product of an edge and another polytope.

Then you can refer to an n-polytope that is the product of a j-pol and a k-pol with j, k > 0 as a non-trivial product, and (e.g.) Olshevskian if j,k >1.

Polytopes that are not trivial products can be defined as "prime". Then it would be true that

A polytope whose number of vertices is prime is a prime polytope.

The converse is false - a hexagon is not a non-trivial product, even though it has 6 vertices. SteveWoolf (talk) 00:52, 29 July 2010 (UTC)[reply]

I have no objection to moving to product polytope, or Conway's proprism. The article was started for the uniform 4-polytopes as polytope products as most studied, and that's what's in the table, but otherwise, I've leaned for wider inclusiveness. BUT I'm also supposed to be on wikibreak again! Tom Ruen (talk) 01:25, 29 July 2010 (UTC)[reply]
I like that term product polytope. Who coined it? —Tamfang (talk) 06:43, 30 July 2010 (UTC)[reply]


P and Q interchangable?

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Just question.

As I understand, q-gonal-p-gonal prism is the same as p-gonal-q-gonal prism. I mean, p and q are interchangeable, they define the same figure. Is it right? If I am right, may be it should be mentioned? For example, in the picture of Duoprisms (p,q = 3 - 8), that pictures with yellow "centers" (p < q) are redundant - the same as p-gonal-q-gonal prism, but different projection.

Bor75 (talk) 08:32, 29 December 2011 (UTC)[reply]

Yes, they are interchangeable, but shown in complementary projection directions. It could be mentioned for clarity. Tom Ruen (talk)
Oh, already says that at the section start "The p-q duoprisms are identical to the q-p duoprisms, but look different because they are projected in the center of different cells." Tom Ruen (talk) 19:44, 29 December 2011 (UTC)[reply]
I see now. Thank you. I didn't notice it because in my comp the site looked messy http://comp.chem.umn.edu/~averkiev/Duoprism.png
I changed it a little bit. I think, it is not the best, but it is better, than it was. Bor75 (talk) 03:10, 30 December 2011 (UTC)[reply]

Duoantiprism s{3,2,3}?

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Can anyone explain why the 3-3 duoantiprism s{3,2,3} can't be uniform? Its vertex figure is a gyrobifastigium, a Johnson solid, so on first inspection, I'd expect uniformity. It has 6 octahedron and 18 tetrahedron cells. Tom Ruen (talk) 22:00, 21 July 2013 (UTC)[reply]

The vertices of the gyrobifastigium do not lie on a sphere. —Tamfang (talk) 01:49, 22 July 2013 (UTC)[reply]
Got it! So you can still contruct it by flattening the Johnson solid gyrobifastigium until all points lie on a sphere, and that'll leave different edge lengths for the triangles. (And same calculation can work for any p,q) Tom Ruen (talk) 02:09, 22 July 2013 (UTC)[reply]

Duoantiprism s{5,2,5/3}?

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Gudap vertfig (front)
Gudap vertfig (back)

Can anyone explain why it works as a uniform polychoron? Double sharp (talk) 04:52, 23 March 2014 (UTC)[reply]

Probably a similar reason why the grand antiprism is the a unique uniform polychoron? Tom Ruen (talk) 05:19, 23 March 2014 (UTC)[reply]
p.s. s{5,2,5/3} isn't a valid symbol by Coxeter's usage of snub as an alternated truncation. h(t0,1,2,3{5,2,5/3}) is the only way I know. s{2,2,2} is also invalid, but s{21,1,1} or s{2[3,3]} work. Tom Ruen (talk) 05:26, 23 March 2014 (UTC)[reply]
Front and back views of vertex figure uploaded, in case it helps answer the question. Double sharp (talk) 16:27, 29 March 2014 (UTC)[reply]
Hm. If that's uniform, then what you get if you remove the "frontmost" vertices of the v.f. and take its convex hull also ought to be uniform (and convex), each vertex having an octahedron surrounded alternately by T and 5-antiprisms. Maybe that can't exist for a topological reason. —Tamfang (talk) 17:06, 29 March 2014 (UTC)[reply]
I tried that in Stella4D. The polychoron begins to generate, but doesn't close up. Double sharp (talk) 06:10, 30 March 2014 (UTC)[reply]
Note new article Great duoantiprism. Tom Ruen (talk) 23:57, 29 March 2014 (UTC)[reply]
My vertex figure
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Alternation of a duoprism?

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If both p and q are even, the resulting duoprism has only even-sided faces, and thus alternatable. Are all the possible alternations, other than the alternated 4-4 duoprism (which is the 16-cell), nonuniform? --Dannyu NDos (talk) 01:29, 12 April 2017 (UTC)[reply]

Ah sorry, the article already states that they are generally nonuniform, but there should be some examples by Schlegel diagram, aren't they? --Dannyu NDos (talk) 01:33, 12 April 2017 (UTC)[reply]

The only convex uniform solution is p=q=4; allowing rationals gives one more, p=10 and q=10/3. The general case cannot be made uniform; these two work only because there is some higher symmetry (the great duoantiprism's vertices form a subset of those of the small stellated 120-cell). Double sharp (talk) 08:02, 7 September 2018 (UTC)[reply]

Infinite duoprisms

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If we were to actually take something like {∞}×{n}, the result must have an infinite number of n-gonal prisms and n apeirogonal prisms. This must therefore be a stack of n-gonal prisms where the rest of 3D space is filled by n apeirogonal prisms, each one connecting along a stack of squares that forms one of the n sides of the prismatic tower, as described by Klitzing on his site. In particular {∞}×{∞} = {4,4} as an infinite stack of apeirogonal prisms, a construction using A1 × A1 symmetry. Double sharp (talk) 07:57, 7 September 2018 (UTC)[reply]

Commons files used on this page or its Wikidata item have been nominated for deletion

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The following Wikimedia Commons files used on this page or its Wikidata item have been nominated for deletion:

Participate in the deletion discussion at the nomination page. —Community Tech bot (talk) 17:23, 23 August 2022 (UTC)[reply]

How would you call the cartesian product of a disk and a polygon?

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A p-gonal-cylindrical prism (or cylindrical-p-gonal prism)? 129.104.241.214 (talk) 18:09, 10 February 2024 (UTC)[reply]

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