Can anyone explain non-orientability?

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

Can anyone explain non-orientability?

Postby ubersketch » Sun Dec 24, 2017 12:27 am

So, in topology I already know what it is. On a mobius strip, if you walk across one and land back at you're starting position at the first time, you'll be facing the opposite direction which means its non-orientable.
So how does this apply to polytopes? How is a sidhei non-orientable?
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Re: Can anyone explain non-orientability?

Postby Marek14 » Sun Dec 24, 2017 6:07 am

Hm, so far I found this:

https://en.wikipedia.org/wiki/Density_(polytope)

It says that a non-orientable polytope doesn't have a well-defined density.

I'd guess that if you are inside the surface of such polytope, you can move in a way that you end up where you began, but as a mirror image.
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Re: Can anyone explain non-orientability?

Postby ubersketch » Sun Dec 24, 2017 2:31 pm

Marek14 wrote:Hm, so far I found this:

https://en.wikipedia.org/wiki/Density_(polytope)

It says that a non-orientable polytope doesn't have a well-defined density.

I'd guess that if you are inside the surface of such polytope, you can move in a way that you end up where you began, but as a mirror image.

That explains why in Natural Density Filling, non-orientables are filled in mod 2.
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Re: Can anyone explain non-orientability?

Postby Klitzing » Sun Dec 24, 2017 5:17 pm

Just consider the facial planes of the polytope. Provided the polytope has a well-defined center, then these planes usually are off that point. Thus you can clearly say, whether a side is out or in, depending on the respective orientation wrt. that special point. But when a face would happen to run through that point, then this type of orientation breaks down.

The simplest example is the tetrahemihexahedron built from a tetrahedral subset of the faces of an octahedron (which clearly show up an individual orientability) plus the 3 diametral squares (which don't).

A more elaborate definition here consideres the connectivity of the facial planes as well. Think about a small ant crawling apon those facial planes, but which would be unaware of any facial intersection. Then this ant might crawl on the outside of one triangle, then will turn around an edge, thus crawling on the top of a square, tunneling through the square intersections, thereby still remaining on the same side of it. When it comes to the next edge it would have to move on at the inside of the next triangle. Thus when moving on and on, this very ant would travel on either side of every facial plane. That is, the polytope is not orientable, just like the Moebius strip.

Even so the consideration of connectivities in the second case is a bit different from the former case of individual orientabilities, the very argument for an after all cross-over comes down again to those facial planes, which happen to run through the center point.

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Re: Can anyone explain non-orientability?

Postby ubersketch » Sun Dec 24, 2017 5:26 pm

Klitzing wrote:snip
--- rk

Wait so a non-orientable is a polytope where the facets cross through the center?
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Re: Can anyone explain non-orientability?

Postby Klitzing » Sun Dec 24, 2017 5:35 pm

For non-trivially symmetric polytopes this comes down to that.

For fully asymmetric ones there isn't even a well-defined center.
Thus it would be hard to say that the facials would have to run through it.

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Re: Can anyone explain non-orientability?

Postby Mercurial, the Spectre » Sun Dec 24, 2017 8:32 pm

The problem with defining orientability on a particular shape is with the center. It is true that C1 (asymmetric) figures have no well-defined center, except when considering the weight of the vertices that lie on the outer boundary.

But one can put a 2D chiral figure and move it along a shape's surface. If one complete movement involves a transformation of its mirror image, the shape is non-orientable and vice versa.

Visualizing is tricky, but one thing to note that a non-orientable figure involves a surface twist that causes the shape's surface to flip at some point, meaning that the inner boundary and outer boundary of the shape switch sides, such as the Mobius strip and the Klein bottle.
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Re: Can anyone explain non-orientability?

Postby ubersketch » Mon Jan 15, 2018 12:30 am

Mercurial, the Spectre wrote:The problem with defining orientability on a particular shape is with the center. It is true that C1 (asymmetric) figures have no well-defined center, except when considering the weight of the vertices that lie on the outer boundary.

But one can put a 2D chiral figure and move it along a shape's surface. If one complete movement involves a transformation of its mirror image, the shape is non-orientable and vice versa.

Visualizing is tricky, but one thing to note that a non-orientable figure involves a surface twist that causes the shape's surface to flip at some point, meaning that the inner boundary and outer boundary of the shape switch sides, such as the Mobius strip and the Klein bottle.

What I'm getting is that the point where are the axes of symmetry intersect is the center. But what does not having a well-defined density mean?
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Re: Can anyone explain non-orientability?

Postby Klitzing » Mon Jan 15, 2018 6:27 am

If some region locally can be viewed as "in", it usually gets a density increase.
But when - by means of non-orientability - the same region has to be considered "out" (and the opposite one "in"),
you would have to apply an density decrease instead.
And this contradiction just states, that the density is not well-defined.
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Re: Can anyone explain non-orientability?

Postby wendy » Fri Jan 19, 2018 10:44 am

Orientability is to do with surface.

Imagine you were covering a surface with paper that was red on one side, and white on the other. For any complex polyhedron, you will end up with a complete red figure. This is because this surface is orientable.

When surfaces cross, the actual crossing is thought of as not visible on the surface. You can only go from face to face only over an edge. For example, in the five-pointed star, or pentagram, you can't cross across to a different alignment when you come to the edge-crossings in the middle of the figure. The surface is orientable, because you can paint the outside of the edges in red, and you have red and white sides of each line.

With figures like the 'Thah', which is an octahedron, with alternate faces removed, and the three diametric squares added, the surface is unorientable. This means that if you start off with a red facing out on a triangle, the squares would be red facing away. But you won't be able to colour all of the faces in with red on one side, because there is a path that allows you to walk across the surface, and come to the inside of the same place, like a Möbius strip.

Not all three-dimensional polyhedra have non-orientable surfaces. For example, the Cho, formed by the six diametric planes of the cuboctahedron, and the eight triangles of it, can be coated with the paper on one side. It's just that for four of the triangles, the coating is on the inside of the triangle.

Closely related to 'non-orientable', you have the notion of 'containing a volume'. In this exercise, we suppose that we use the red-and-white paper, so that the outside of the enclosed space is red. We can assign the four bounded bits of Thah to be 'enclosed volumes', which means that the triangles are all 'red' and the squares are divided into four diagonally, and the red and white sides of the paper alternate on a side. This is a 'skew marginoid'. It's not a margin in the usual sense (ie a boundary between faces), but one where the in-out direction changes. This does not change the density of the paper.

You can make a 'skew marginoid' from the red-and-white paper, by cutting the sheet in two, and rejoining one of the two sheets upside down. If you hold this line, and rotate the paper around the join (so one side comes up and over the join, and the other goes under it), the red and white halves will still be in the same positions. In the red-and-white square above, you can flip the square around the diagonal, and the red and white quarters will be in the same position.

Both non-orientability and skew marginoids suppose that we are dealing with a surface, that is, an N dimensional something that _divides_ a space of N+1 dimensions. So you can have a left- and right- hand side of a line, but not of a lampost.
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