## Polytope Definition

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

### Polytope Definition

Since it seems everyone has a slightly different definition of a (non-degenerate) polytope, this is a topic to share them in.
I'll start. Under my definition of a polytope, each n-dimensional space contains no more than 1 element, each edge has two vertices, and all facets and vertex figures are also valid polytopes.
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### Re: Polytope Definition

1. Given an n-dimensional element and an n-2-dimensional element there are either 0 or 2 n-1 dimensional elements adjacent to both.
2. No two elements may share all of their vertices.
3. Not a compound, nor has compound elements and element figures.

I also considered adding a fourth restriction.

4. Not a member of a regiment with members which don't satisfy 1.
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ubersketch
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### Re: Polytope Definition

Polytopes are like a cube, for varying meanings of 'like'. There's not much more to say.
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wendy
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### Re: Polytope Definition

A polytope is:
a uniquely closed shape with no curves.
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### Re: Polytope Definition

When I say "it's like a cube", different people will tease out some feature of the cube and generalise it. The resulting mess is that you get figures that are nothing like each other.

If you suppose that it is due to something like a closure of polygons, then something like {6,3} is a polytope. Polytopes intersect with tilings, and it is possible to represent a cube as a spherical tiling {4,3}. But if you insist on 'flat faces', then even something like {8,3} exists with right-angles between its octagons. I mean, this thing actually tiles as {8,3,4}.

Then you have the 'dyadic' relation. If two surtopes of N+1 and N-1 are incident on each other, there are exactly two surtopes N that are incident on both. Spheres and cones have surtopes, but there are no edges incident on the apex, even though the 2d sloping surface is.

Opposite to dyadic, is Sheppard's "complex polytopes", which are described in complex analytical geometry. This is your ordinary geometry, such as a line is $$y=ax+b$$, but all of these numbers are complex. You still get the usual polyhedral stuff, but a line can contain 3, 4, 5, ... vertices at its ends. Ie instead of a line being $$x = y \pm l$$, it becomes $$x = y + l \exp(2\pi j n/v)$$, in other words, a polygon. A polygon is a sparse array of these edges in 4D, and a polyhedron exists in six real dimensions, such as the 3{3}3{3}3, a right-angle thing with 27 vertices.

In the notion of density, a cube is a space with a designated 'inside', and its surface is discrete and thin. That is, there is a definite boundary a line might cross at a point anywhere. This is the definition of 'solid'. Spheres and cones are solid too, but the gaussian cloud $$d = \exp(-r^2)$$ is not.

In the end, it is like shining a torch (flashlight) on something, and saying these are polytopes, and you get the same sorts of arguments when people start testing the boundry between 'blue' and 'green'.
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### Re: Polytope Definition

Mecejide wrote:Since it seems everyone has a slightly different definition of a (non-degenerate) polytope, this is a topic to share them in.
I'll start. Under my definition of a polytope, each n-dimensional space contains no more than 1 element, each edge has two vertices, and all facets and vertex figures are also valid polytopes.

It highlights the inconsistencies between different definitions used by different people, and proposes a (very) general scheme to try to encompass them all. Still, a precise definition that satisfies everyone seems elusive -- it either includes too much, or excludes too much, depending on each person's specific area of interest.
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### Re: Polytope Definition

For me, they’re realizations of abstract polytopes without vertices coinciding in space (and without duplicate elements, of course). I’d be willing to ignore my only constraint if it didn’t lead to Grünbaum’s enormous amount of possibilities (like {6/2} regular polygons, to state the simplest consequence).
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### Re: Polytope Definition

I define a realization, of an abstract polytope A, as a function f that sends each element x in A to some subspace f(x) (an affine subspace in a Euclidean space, or a geodesic sub-manifold in a curved space), where the rank (or abstract dimension) of x is the same as the dimension of f(x). The function must respect the partial order (or incidence relations): If x≤y in A, then f(x) is a subspace of f(y).

I'll say the realization is non-degenerate when the above implication goes both ways: If f(x)⊆f(y), then x≤y in A. This property implies that different vertices must have different locations in space, and that different faces cannot be coplanar, and that a vertex that's not part of a face cannot be in the plane of that face.

A skew polygon only has vertices and edges; its (2D) face is not realized. We may define a k-skeletal n-polytope as an abstract n-polytope which is only partially realized, up to its k-dimensional elements. 0-skeletal and 1-skeletal polytopes are basically the same: If we have the 1-skeleton (vertices and edges realized), then of course we have the 0-skeleton (vertices realized). If we have the 0-skeleton (as well as the abstract structure), then we can construct a 1-skeleton, because there's a line between any two points (though it's not unique if the points coincide). Again I'll define non-degeneracy by the equivalence of x≤y and f(x)⊆f(y), whenever f(x) and f(y) exist; that is, whenever x and y have rank ≤k. So a 1-skeletal polyhedron may have two faces with exactly the same vertices and edges, and still be non-degenerate, if the vertices and edges have the correct incidence relations.

Alternatively, we may modify the definition to allow curved subspaces, so that a skew polygon's face could be realized as, say, a minimal surface. But by default an n-polytope is fully realized (n-skeletal) and has flat faces, so skew polygons are not allowed.

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.

Of course the least face is realized as the empty set, with "dimension" = -1. I suppose there's some connection with projective geometry, subtracting 1 from the dimensions.

I do allow digons, though they're usually degenerate. I don't see why Grünbaum rejects them, especially while allowing edges to coincide otherwise. At first I thought it was because he defines an edge as a pair of vertices (if there's only one pair of vertices, then there's only one edge), but that seems to be contradicted by (P4) in his polyhedron definition, where a pair of vertices may determine two different edges.

Note that a degenerate edge, with length 0, still has a defined direction. (That's another difference from Grünbaum.) Similarly, a face with no area still has a defined plane. For example, the formula for an n-gonal antiprism's dihedral angle is still valid when n=2, giving 125.26° between a triangle and a digon.
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mr_e_man
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### Re: Polytope Definition

What is the purpose of Grünbaum's (P4)? If I'm reading it correctly, it's redundant; any violation of this condition would involve 2-gons or 2-valent vertices, which are already excluded by (P3).
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### Re: Polytope Definition

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:
Code: Select all
    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. (Another example would be a point west of the centre.)
polygonInsiding1.png (9.11 KiB) Viewed 1192 times

polygonInsiding2.png (15.96 KiB) Viewed 1192 times

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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### Re: Polytope Definition

Aha! I should have thought of using vectors. That makes it straightforward to extend to degenerate polytopes. Here's my revised definition of insiding.

In what follows we should assume that the space, and any subspace, is orientable. So this is not relevant to polytopes in elliptic space. (Note that 3D elliptic space is orientable, but 2D elliptic space is not.)

For a k-dyad {w,x,y,z}, where x and y are the k-faces and z is the (k+1)-face, there are two opposite unit vectors normal/perpendicular to f(x) and tangent/parallel to f(z). One such vector can be chosen; call it v(x,z,w). (Remember f is the realization function, which gives a subspace for an element of the polytope.) (In a non-orientable space, it may be possible to slide the vector around on f(x) and end up with the opposite vector, confounding the choice.) The polytope is insided when this choice is made for all dyads, subject to two consistency conditions:

(1) Given x and z, this vector should be the same for all w. So we can write v(x,z,w) = v(x,z). (Ordinarily you can't compare vectors at different locations in a non-Euclidean space. I guess what I really mean is that v(x,z) should be a continuous vector field defined on (and perpendicular to) f(x), including on f(w) for various w≤x.)

(2) Given w and z, the vector for x should be "opposite" the vector for y, in the following sense.

(2a) For a 0-dyad, indeed we just require the vectors to point in opposite directions along the line f(z). Here's a comparison of this notion of insiding to the previous notion, for an edge shrinking to 0 and then negative length:
Code: Select all
      o--------o      ------o=>----<=o--      o-----o         ------o=>-<=o-----      o               ----<=o=>------------o  o-----------    -<=o--o=>---------
The previous insiding is undefined, or at least discontinuous, when the edge is degenerate. But the new insiding behaves nicely regardless of degeneracy.

(2b) For k>0, there's a 2-dimensional subspace normal to f(w) and tangent to f(z), which contains four of our chosen vectors: one pointing from x to z, one from y to z, one from w to x, and one from w to y. We require v(w,x) v(x,z) = - v(w,y) v(y,z), using the geometric product. (It could as well be the wedge product, since v(w,x) and v(x,z) are perpendicular anyway.) What this means is that, if going from x to z (around w) is considered clockwise, then going from y to z should be anticlockwise.
insiding3.png (16.82 KiB) Viewed 1151 times
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### Re: Polytope Definition

Due to condition (1), an insiding vector is assigned to each incident pair of elements, one dimension apart; that is, to each edge in the Hasse diagram. Actually this is only true if each vertex also gets a vector "from the nulloid", which doesn't make much sense unless we use an extra dimension. But we can get the same effect by just putting a + or - sign on each vertex. Then condition (2a) has the same form as (2b), except that (w being the nulloid) v(w,x) and v(w,y) are scalars. (Also, v(x,z) and v(y,z) generally must be slid along the line f(z) before they can be compared.)

Given an insided polytope, we can turn any element x inside-out, and the result is still a valid insided polytope. This is done by negating v(x,z) or v(z,x), for all z incident with x, one dimension higher or lower; that is, by negating all vectors associated with edges in the Hasse diagram connected directly to x. Any insiding can be gotten from any other insiding by doing this repeatedly, varying x. Here's an edge of a square turning inside-out:
insiding4.png (12.27 KiB) Viewed 1062 times

mr_e_man wrote:I'm not sure, but I guess that a polytope is insidable if and only if it's orientable.

In an orientable space, with orientable subspaces, a polytope is insidable if and only if it's orientable. (It's not trivial to prove this, though.) In a non-orientable space, any polytope is non-insidable, but it may or may not be orientable. For example, any polygon is abstractly orientable, and you can put a polygon on an elliptic plane or a Klein bottle, though the extended edge lines can't always be given consistent normal vectors.
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### Re: Polytope Definition

I see that Polyhedron Dude had similar ideas; see the bottom of the page https://www.polytope.net/hedrondude/polygons.htm

The inside-out square, 'asq', has Schlafli symbol {4/3} = {-4}. In general, {n/d} = {n/(d±n)}, because the angle (d/n)*360° is equivalent to (d/n ± 1)*360° = (d ± n)/n*360°.

Without insiding, we can't tell the difference between θ and 360° - θ, so d/n*360° is further equivalent to (1 - d/n)*360° = (n-d)/n*360°, and thus {n/d} is equivalent to {n/(n-d)}. But I won't say they're equal. By default, the Schlafli symbol describes a regular insided polygon.

This helps with constructing non-convex antiprisms. The {5/2} antiprism is different from the {5/3} antiprism, even if we ignore the polyhedron insiding.
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