degenerate polyhedra in 4-space

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

degenerate polyhedra in 4-space

Postby Adam Lore » Mon May 04, 2009 2:15 am

I'm attempting to make sense of this. I think it might involve using a fourth spatial dimension.

We have a square in 3 dimensions. (let's say a paper model) so one side is blue and the other side is yellow.
(the colors aren't important, just to visualize)..
so we have 2 square faces (one blue and one yellow), 4 vertices, and 4 edges, satisfying Euler's polyhedra formula ( F+V-E=2)
.. so we can consider it a degenerate polyhedron..

Let's call it a planar square dihedron.

Now, we apply a baseless wedge or "roof" (I like to call it a "cune") to both faces of the square, each perpendicular to each other.

Image

[in the same way that we could take a cube and turn it into a dodecahedron by adding "roofs" on each face, like so:

Image

if we ignore the original edges of the dihedron, and focus just on the new edges formed by the cunes, topologically we now have a figure with 9 edges, 7 vertices, and, it would seem, 4 faces.

It makes a lot more sense if you make a model out of string.

2 of the faces can be thought of either as triangles with one extra vertex between one edge,
or a square with four vertices, 3 which meet with 2 other edges (of other faces) and one of which only has two edges..

and, similarly, the other two faces are either quadrilaterals with one extra vertex, or pentagons with one vertex that doesn't meet with other faces.

The important bit is that we have TWO PENTAGONAL FACES

this is where the fourth dimension comes in..

is it possible to flatten or fold this weird shape in the fourth dimension to get a regualr pentagon? (or a pentagonal dihedron?)
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Re: degenerate polyhedra in 4-space

Postby Adam Lore » Mon May 04, 2009 2:27 am

I think this is a more simple way to think of it, but now we end up with 4 pentagons to try to make into 2 (in 3-d) and 1(in 2-d).
(We ignore the outer edges that make the square.)

Image

Each pentagon is folded around the others make a square. Does this become only one or two pentagons, in any sense, in the fourth dimension?

can it be thought of a hyperprism with zero height and depth (?) in the same way we can call a square a prism with zero height?
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Re: degenerate polyhedra in 4-space

Postby wendy » Mon May 04, 2009 12:29 pm

The sort of thing that ye discuss, may or may not exist, but they would involve a pleat-marginoid (which is not generally acceptable).

On the other hand, ye can quite easily construct polychora by flattening polyhedra, to give a front- and back- view of them. You can even construct things that are not renderable in 4d, but are none the less polychora (there are thirteen uniform tri-chora, formed by the thirteen triplets of polyhedra that share the same vertices, edges and sets of hedra.

Presenting 4d polytopes down this way usually leads to having faces (chora) match (eg as ye get by looking at a cube face-first or any polyhedron edge first), or they split giving converging elements. However, the octahedral view of the cube gives a distinct view of every element, can be rendered similarly in three or more dimensions (as zonotopes).

For example, consider the cuboctahedron. You can make two other figures from this, by adding the four crossing hexagons as new hedra. Of course this is not a polyhedron. But taking any two of (triangles) (squares) (hexagons), ye get what is in Bowers notation, CO, Oho and Cho. Since each hedron is now a margin between two faces, the thing closes up as a polychoron. It has exactly three faces (CO, Oho, Cho), and is thus a trichoron. Some of the margin-angles are 180 (hexagons), while others are zero (square, triangle), the latter are 'pleat-margins'.

The bowers 'blend' operation is a matter of finding a missing face (or cell) of such a thing [polycell = closure of solid faces].
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Re: degenerate polyhedra in 4-space

Postby Adam Lore » Fri May 08, 2009 7:40 am

What's a pleat-marginoid? and where can I read about it?
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Re: degenerate polyhedra in 4-space

Postby wendy » Sat May 09, 2009 7:08 am

A margin is the boundary between adjacent faces (eg in 3d, an edge). It's also called a 'ridge', but tilings don't have ridges.

A marginoid is a false-margin: that is, it divides two parts of the same face, for example, by an intersection of surface.

A pleat is a joining of cells (solid surtopes) on the same side. It is usually a margin.

A pleat marginoid, then is a false margin where the cells are on the same side. For example, if you bend a pentagon on one of its diagonals, and continue to regard the join as part of the pentagon, it becomes a marginoid (because the joined elements are the same cell or face), and a pleat (because the cells are on the same side).

The terms are in the polygloss.
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Re: degenerate polyhedra in 4-space

Postby Adam Lore » Sun May 10, 2009 2:53 am

Cool, thanks.

I forgot about the polygloss. I need to catch up.
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Re: degenerate polyhedra in 4-space

Postby pat » Wed May 13, 2009 3:52 pm

Somewhat off-topic Adam, but what did you use to make these diagrams. They are quite attractive.
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Re: degenerate polyhedra in 4-space

Postby Adam Lore » Thu May 14, 2009 5:26 am

Thanks, Pat.. The one with the dodecahedron isn't mine, I just found it online. The other two I made in Paint.
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Re: degenerate polyhedra in 4-space

Postby Keiji » Thu May 14, 2009 7:50 am

Adam Lore wrote:The other two I made in Paint.


In future, change the file type from JPEG to PNG. Microsoft are retarded for making the lossy JPEG format the default.
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Re: degenerate polyhedra in 4-space

Postby Adam Lore » Thu May 14, 2009 8:58 am

Cool, thanks for the tip.

I tried it out and it seems much better.
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