by **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.

The dream you dream alone is only a dream

the dream we dream together is reality.