mr_e_man wrote:Similarly, we may modify the definition to use bounded regions of subspaces, rather than entire subspaces. Then a non-degenerate polyhedron could have coplanar faces, as long as neither face is completely contained in the other. Actually there are several possible realizations of the incidence relations, using boundaries or closures of sets. But this complicates the study of self-intersecting polytopes. It's easier to use entire subspaces.
Such complications of topology and set theory are described in
Polytopes - Abstract and Real.
If each element of a polytope is realized as an entire subspace, then we can't distinguish between a spherical 90°,90°,90° triangle and a 90°,90°,270° triangle (or 90°,270°,270°, depending on whether these are edge lengths or vertex angles). But these are quite different! One triangle is regular, while the other is only isosceles. Something is missing in my definitions. I think we need some notion of orientation; not necessarily "left vs. right", but rather "out vs. in". See
Filling Polytopes.
Here are some possible notions of orientation for an edge:
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o--------o
o-->-->--o
(-)------(+)
----o--------o----
o--------o
----o o----
I'll define an abstract polytope to be
orientable (in the "left vs. right" sense) when any sequence of flag moves (applying the "dyadic" or "diamond" rule to change one element of the flag), with the same starting and ending flag, has an even number of moves. The polytope is non-orientable when some such sequence has an odd number of moves. A flag move is to be thought of as some kind of reflection, which reverses orientation. In an orientable polytope, the flags can be split into two classes, where all the flags in one class are related by an even number of moves. An
orientation of the polytope is a choice of one such class, declared to be "right-handed".
I'll further define a k-
dyad as four elements: a pair of k-dimensional elements incident with a (k+1) and a (k-1)-dimensional element. So a 0-dyad is essentially an edge, a 1-dyad is essentially an angle in a face, a 2-dyad is essentially a dihedral angle in a cell, and so on.
Now assume for simplicity that the polytope is non-degenerate, though I hope to make sense of degenerate polytopes as well.
In a 0-dyad, the two points divide the line into two regions, which are obviously the edge's "inside" and "outside". (You might see three regions, but here we suppose that the left-outside and right-outside are connected through a point at infinity.) An
insiding of the edge is a choice of one such region, declared to be "inside" in spite of preconceptions. For the ambiguous spherical triangle described above, an insiding of an edge chooses either the 90° arc or the 270° arc.
By default, any Euclidean edge is insided the obvious way, so that it has a finite positive length. We may say that an inside-out edge has negative length, corresponding to the notion that +270° = -90°.
In a 1-dyad, if the two edges are insided, so that they can be represented by rays rather than lines through the vertex, then they divide the plane into two regions. An insiding of the 1-dyad is a choice of one such region. Equivalently, it's a choice between an angle θ and its opposite -θ or 360°-θ.
Insiding is local. In the "hourglass" quadrilateral shown below, if all angles are taken as 45°, then a diagonal edge does not consistently divide the plane into two regions inside and outside the polygon. (At the northwest vertex, the edge's northeast is inside the polygon; but at the southeast vertex, the same edge's southwest is inside the polygon.) In contrast, if the northern angles are 45° while the southern angles are 315°, then any edge consistently divides the plane into inside and outside. In that case we can say that the polygon as a whole is insided. Still it is local to edges; a point in the middle of the apparent triangle in the south cannot be said to be inside or outside the polygon.
- polygonInsiding1.png (9.11 KiB) Viewed 4782 times
- polygonInsiding2.png (15.96 KiB) Viewed 4782 times
(EDIT: A single edge cannot determine that a point is inside a polygon, even if it's convex. So the second half of the second picture is less meaningful than I seem to have thought it was.)
We can define insiding inductively. In a k-dyad, if the two k-faces are insided, so that they can be represented locally by k-halfplanes rather than k-planes at the (k-1)-face, then they divide the (k+1)-plane into two regions. An insiding of the k-dyad is a choice of one such region, or equivalently a choice between opposite angles. This also associates a (k+1)-halfplane to each of the two k-faces locally at this (k-1)-face.
In a (k+1)-polytope, different k-dyads give possibly different (k+1)-halfplanes to each k-face. If the k-dyads are insided consistently so that each k-face gets a unique (k+1)-halfplane, then we can say that the (k+1)-polytope as a whole is insided.
I'm not sure, but I guess that a polytope is insidable if and only if it's orientable.
Anyway, my main idea here is to include a single bit of information (a binary choice) for each dyad in the polytope. I still don't think self-intersecting polytopes should be realized as bounded regions (sets of points) as Johnson and Inchbald suggest, mainly because of the problems of "holes".
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