Polytope Definition

[Mecejide]

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.

[ubersketch]

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.

[wendy]

Polytopes are like a cube, for varying meanings of 'like'. There's not much more to say.

[Mercurial, the Spectre]

A polytope is:
a uniquely closed shape with no curves.

[wendy]

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'.

[quickfur]

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.

I think this article would be relevant: https://sites.math.washington.edu/~grunbaum/Your%20polyhedra-my%20polyhedra.pdf

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.

[URL]

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).

[mr_e_man]

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.

[mr_e_man]

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).

[mr_e_man]

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.

polygonInsiding1.png (9.11 KiB) Viewed 20200 times

polygonInsiding2.png (15.96 KiB) Viewed 20200 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".

[mr_e_man]

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 20159 times

[mr_e_man]

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 20070 times

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.

[mr_e_man]

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.

[mr_e_man]

Here are the insidings of the regular pentagons {5/d}. All vertex-to-edge vectors are shown, but only one edge-to-body vector is shown, to reduce clutter (and to show a pattern of rotation around the rightmost vertex).

Insiding 5.png (40.35 KiB) Viewed 16692 times

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°.

Hmm... What about the edge lengths?

We're used to normalizing things so that the edge length is 1, or 2, or the circumradius is 1, etc. But do we want something with negative edge length to be equivalent to something with positive edge length?

Consider what happens to an ordinary square when it's scaled by a factor varying from positive to negative. By continuity, its insiding vectors should remain unchanged, while the vertices and edges move. The negated square is thus 'adasq'. Likewise, 'asq' negated is 'adesq'. (The results are the same when the vertices and edges are unchanged but the insiding vectors are all negated.)

The edge length of {n/d}, with unit circumradius, is 2*sin(1/2*d/n*360°). This is negated when d is replaced with d±n. So, I think we should say that {4/3} is 'asq', but {4/-1} is 'adesq', and {4/5} is 'adasq'.

Insiding 6.png (42.8 KiB) Viewed 16692 times

[wendy]

Yes, I been playing with vectors on polytopes for years now.

In essence, an 'out vector' is a unit normal. Specifically in one can render it as the gradient of density.

Volume is then the moment of out-vector. That is, you calculate \{\int x\cdot da\}, where x is the corrdinate at the element of surface, da. We have too \{a = \nabla d\}, where d is the density.

Non-orientable polytopes are handled by a skew marginoid. This is a division of the surface that does not appear as a surtope. Instead, as you cross it (walking alonging the surface), the direction of the outvector reverses, while the intensity remains unchanged. This is because the surface tension (the size of the outvector), is not observed, and so must be identical across the skew marginoid.

[mr_e_man]

(The gradient points in the direction of increase, not decrease. So I think it would be "invector", not "outvector". But that word looks like it's related to "invect", having emphasis on the "vect" rather than the "in"....)

The gradient of density has lower-dimensional density as its magnitude, while my vectors all have unit magnitude.

Any dihedron, e.g. {5/2, 2}, has density 0 everywhere. On the other hand, the insiding vectors are still defined (one face points up and the other face points down). So density cannot be used to calculate the vectors, in general. (I guess it works for non-degenerate polytopes.)

Also, given the insiding vectors, any constant can be added to the density. For example, {4/5} could be a '+' shape ('adesq') with density 0, and four corners ('adasq') with density 1, as Polyhedron Dude intended; or it could be a '+' shape with density -1, and four corners with density 0. Furthermore, that depends on the lower-dimensional density; I've assumed that the antidyad has density 0 and 1.

[steelpillow]

Branko Grünbaum used to say ruefully that a polytope is anything you want it to be. In his classic "Convex Polytopes" he discussed several definitions then in use, and plenty more have arisen since - even without considering non-convex cases.

To the classical Greeks it was a solid with flat faces, from Kepler and Euler on it was increasingly regarded as a closed surface or hyper-surface, whose interior was of no consequence. This gave rise to the discipline of topology, in which it is regarded as a closure-finite decomposition of that hypersurface into cells, all meeting at vertices and with simple interiors (i.e. all cells are topological balls).

In the first half of the 20th century, Convexity theory then went wild with unions of half-spaces, or of the interior region within bounding planes; the inside suddenly became all-important again. Complex theory went for what are closer to, and include, configurations, banishing anything to do with finiteness or interiors. Combinatorics, spawned from convexity theory, went for vertex sets. Then, in the second half, the study of more general incidence complexes - a side issue from the topological approach - merged with the combinatorial approach to create a set-theoretic formulation known as abstract polytopes.

In abstract theory, to create a real geometric figure the appropriate abstract polytope must be "realized" by injecting or mapping it into real space. If such a realization is a one-to-one bijection then the realization is "faithful". But if it is a bit of a mess then the realization is "unfaithful". Curiously, abstract theory allows interiors which are not simple balls but twisted things like projective polytopes. I refer to this non-simple property as "anaploid". Attempts to bar them from abstract theory have proved singularly unsuccessful.

In my morphic approach I try to reconcile some of these differences. Beginning with some abstract polytope, I distinguish two stages of realization. In the first, interpretive stage, we decide that we are not dealing with tables, chairs and beer mugs, but with points, line segments and plane regions. We also give the ordering of the abstract elements a direction: this direction determines which of a dual pair of polytopes we are constructing. The result is a topological polytope, having a sort of rubber quasi-form but no fixed measurements, which I call a morphic polytope. In the second step, of concretization, we inject it into a real space and impose decisions about lengths, angles, and the twists which we subject any anaploid interiors to (and simple interiors if we so wish).