Why there are no regular stars beyond 4D

[quickfur]

There are regular star polytopes only up to 4D; beyond that, there are no regular stars and no regular pentagon figures like the 120-cell/600-cell. What is the reason for this? I can understand that there are no convex 5-polytopes with 120-cell or 600-cell facets, as their angle defect is just too small to allow it. But what about star 5-polytopes? Are there any non-convex, non-prismatic 5-polytopes that take, say, the 120-cell as facets? Why can't they be regular? Can they be uniform?

[wendy]

The symmetry of a regular star, like any other regular figure, comprises of intersecting mirrors. For something like {5/2,3,3}, there are other mirrors that cross the cell dividing it into 191 smaller cells, which becomes [5,3,3]. Every platonic star must ultimately then have the same vertices + faces as a non-stary figure. Since none of {3,3,3,3}, {3,3,3,4}, {4,3,3,3} give such stars, there are no stars in 5d or higher.

You can of course, _begin_ to construct any of the possible peice-wise finite stars {5/2,3,3,3}, {5,5/2,3,3}, {5/2,5,5/2,3} and their duals, but these never close. You can even construct something like {5,3,5/2,3}, but its vertex-figure never closes either.

There are none the less, "starry" things. You can replace any octagon, eg x4x, by an octagram x4/3x, and a dodecagon x6x by the corresponding dodecagram x6/5x. One uses such tricks to derive the octagrammy, the d73 stellation of o3x4x3o. Note that x4x need only have an x somewhere on each side of the 4, so something like x3o4o3x has a stellated form x3o4/3o3x. The parity is changed by each 4, octagons arrising only when there are x's of different parity. Something like x4o3o4x does not contain octagons, because tbe x's are the same parity.

[Eric B]

The short answer is that beyond 4-D, the only "families" of objects that continue are the simplex ("triangle family"), the orthotope ("[hyper]cube" family"), the cross-polytope (octahedron family; or with the Greek forms, it would be "diatope"), and the "hypersphere" family (I call it "achanetope").

None of these are stellatable. In 2-D, they are represented by the triangle, square and circle. Only the pentagon and higher are stellatable. Stellation is a result of extending alternate facets together. This is not possible in the square, because the alternating (opposite) sides are parallel. In the triangle, an "alternating" side would actually be the other adjacent side, which on the far end is diverging from the first side! So the tetrahedron, pentachoron, hexatesseron, as well as the cube, tesseract, penteract, etc. will also have alternate facets as parallel or diverging. The 3D pentagon, the dodecahedron, and the 4D pentagon, the hecatonicosachoron, are stellatable. But for some reason, these families do not go past 4D. I guess with more space dimensions, it becomes harder and harder to enclose space in "regular" geometrical figures. You can have any number (above 3) sided regular 2D polygon, but in 3D, you can only have 4, 6, 8, 12, and two of 20 faces. In 4D, you can only have 5, 8, 16, 120, and I think 600. In 5D, it is 6, 10, and I believe the diatope is 32 (IIRC, they double as the dimension goes up, am I right?) And in 6D, it would be 7, 12 and probably 64. So three each in each dimension above 4 (plus the hypersphere), and no representatives of the pentagon, hexagon or higher.

Now, in non-Euclidean geometry, the simplex and orthotopes can be stellated, as the facets are crrved so they they can meet if extended. So that way you would be able to have "stars' in any dimension.

[quickfur]

The short answer is that beyond 4-D, the only "families" of objects that continue are the simplex ("triangle family"), the orthotope ("[hyper]cube" family"), the cross-polytope (octahedron family; or with the Greek forms, it would be "diatope"), and the "hypersphere" family (I call it "achanetope").

Well, I know that. I guess what I'm looking for is some insight into why regular stellated figures must share the vertices/face planes of regular convex figures.

None of these are stellatable. In 2-D, they are represented by the triangle, square and circle.

Actually, this is not accurate. The 2D equivalent of the octahedron is the diamond, which is the same as the square rotated 45 degrees.

Only the pentagon and higher are stellatable. Stellation is a result of extending alternate facets together.

Yes, although one wonders why there are no regular stars that are not a result of extending the convex regular polytopes.

[...]I guess with more space dimensions, it becomes harder and harder to enclose space in "regular" geometrical figures.

This is an interesting area of discussion: why the regular polytopes seem to collapse to only 3 families after 4D. The most superficial reason, of course, is that the 120-cell and 600-cell have very large dihedral angles, and so cannot form the basis of higher regular polytopes (since it must be possible to join three of them together and still have non-zero angle defect in order for them to fold up into 5D: note how there are no regular polyhedra with hexagonal faces for the same reason).

But the deeper question remains: what is it about 5D and beyond that so constrains the type of polytopes that may inhabit it?

You can have any number (above 3) sided regular 2D polygon, but in 3D, you can only have 4, 6, 8, 12, and two of 20 faces. In 4D, you can only have 5, 8, 16, 120, and I think 600.

Don't forget the 24-cell. It's truly a unique polytope, which "almost" spawns another family of higher polytopes; unfortunately, 24-cells tile 4D, and so have an angle defect of 0, thus they cannot fold up into 5D to form a regular 5-polytope.

In 5D, it is 6, 10, and I believe the diatope is 32 (IIRC, they double as the dimension goes up, am I right?) And in 6D, it would be 7, 12 and probably 64. So three each in each dimension above 4 (plus the hypersphere), and no representatives of the pentagon, hexagon or higher.

It is interesting to surmise which polygons exactly correspond with the higher polytopes. I suspect the real answer is, none of them are exact analogs.

One may think, for example, that the n-simplex is the n-dimensional equivalent of the triangle, and in many aspects, this can be construed to be so. However, the n-simplex also has additional properties that make it behave like other polygons, too. For example, sometimes the tetrahedron can play the role of a square (due to having 4 faces). Similarly, the cube sometimes can play the role of a hexagon, in that it has 6 faces, and also has a maximal projection and a maximal intersection in the form of a hexagon (the same goes for the octahedron). The dodecahedron is often thought of as the analog of the pentagon, but in reality, it is an even polytope (the simplices are the only odd regular polytopes above 2D after all), and in many ways behave more like a hexagon (such as in projections of the 120-cell), or a decagon. The icosahedron also has hexagon- and decagon-like properties much more than pentagon-like properties.

Truth be told, analogy only carries so far. Each polytope is unique in its own way, and the polygons are really too simplistic to adequately capture all the beauty and symmetry of the higher polytopes.

Now, in non-Euclidean geometry, the simplex and orthotopes can be stellated, as the facets are crrved so they they can meet if extended. So that way you would be able to have "stars' in any dimension.

Well, that sidesteps the original issue, which is to understand why exactly Euclidean space confines the possibilities in such an intriguing way: an infinitude of regular polygons, yet only 5 regular polyhedra, then a tantalizing 6 regular polychora, before collapsing to a strangely impoverished 3 regular n-polytopes thereafter.

All of these results are obvious from many respects, of course, as one could argue using angle defects, etc.. But it quite intrigues me as to whether there is a deeper underlying cause for this strange pattern of infinity, 5, 6, 3, 3, 3, 3, 3, ... . That anomalous 6 is most fascinating, since if the sequence were doomed to collapse to 3, why does it spike to 6 in 4D?

[Eric B]

The short answer is that beyond 4-D, the only "families" of objects that continue are the simplex ("triangle family"), the orthotope ("[hyper]cube" family"), the cross-polytope (octahedron family; or with the Greek forms, it would be "diatope"), and the "hypersphere" family (I call it "achanetope").

Well, I know that. I guess what I'm looking for is some insight into why regular stellated figures must share the vertices/face planes of regular convex figures.

So what you're talking about is sticking "points" anywhere on the object? Like just pasting triangles on the sides of a square, to make a four pointed star). Stellation is defined as extensions of the "sides" (facets)

None of these are stellatable. In 2-D, they are represented by the triangle, square and circle.

Actually, this is not accurate. The 2D equivalent of the octahedron is the diamond, which is the same as the square rotated 45 degrees.

That I know. So it was represented in the "square".

Only the pentagon and higher are stellatable. Stellation is a result of extending alternate facets together.

Yes, although one wonders why there are no regular stars that are not a result of extending the convex regular polytopes.

What else would a regular star be?

[...]I guess with more space dimensions, it becomes harder and harder to enclose space in "regular" geometrical figures.

This is an interesting area of discussion: why the regular polytopes seem to collapse to only 3 families after 4D. The most superficial reason, of course, is that the 120-cell and 600-cell have very large dihedral angles, and so cannot form the basis of higher regular polytopes (since it must be possible to join three of them together and still have non-zero angle defect in order for them to fold up into 5D: note how there are no regular polyhedra with hexagonal faces for the same reason).

OK, this explain to me. With all of this stuff I know; I never got into "dihedral angles". I remember reading about them, but forgot what they were. "dihedral" would mean two-faced.
I always wondered about hexagonal faces. It seems because they perfectly tile a flat plane. (the three 120° angles fit together to complete the 360°). If you angle them by any amount, they would no longer fit together. Volleyballs use a combination of hexagons and pentagons.

You can have any number (above 3) sided regular 2D polygon, but in 3D, you can only have 4, 6, 8, 12, and two of 20 faces. In 4D, you can only have 5, 8, 16, 120, and I think 600.

Don't forget the 24-cell. It's truly a unique polytope, which "almost" spawns another family of higher polytopes; unfortunately, 24-cells tile 4D, and so have an angle defect of 0, thus they cannot fold up into 5D to form a regular 5-polytope.

OK; thought I might be missing something. Isn't the icosatetrachoron also related to the octahedron?
Also, the other mistake I made is that it is the dodecahedron, not the icosahedron that comes in two forms.
(So, or is it the rhombic dodecahedron the icosatetrahedron is related to ?)

In 5D, it is 6, 10, and I believe the diatope is 32 (IIRC, they double as the dimension goes up, am I right?) And in 6D, it would be 7, 12 and probably 64. So three each in each dimension above 4 (plus the hypersphere), and no representatives of the pentagon, hexagon or higher.

It is interesting to surmise which polygons exactly correspond with the higher polytopes. I suspect the real answer is, none of them are exact analogs.

One may think, for example, that the n-simplex is the n-dimensional equivalent of the triangle, and in many aspects, this can be construed to be so. However, the n-simplex also has additional properties that make it behave like other polygons, too. For example, sometimes the tetrahedron can play the role of a square (due to having 4 faces). Similarly, the cube sometimes can play the role of a hexagon, in that it has 6 faces, and also has a maximal projection and a maximal intersection in the form of a hexagon (the same goes for the octahedron). The dodecahedron is often thought of as the analog of the pentagon, but in reality, it is an even polytope (the simplices are the only odd regular polytopes above 2D after all), and in many ways behave more like a hexagon (such as in projections of the 120-cell), or a decagon. The icosahedron also has hexagon- and decagon-like properties much more than pentagon-like properties.

Truth be told, analogy only carries so far. Each polytope is unique in its own way, and the polygons are really too simplistic to adequately capture all the beauty and symmetry of the higher polytopes.

I judge the analogues by the angles, not the number of facets or shape of projections. Simplexes are always 60°; orthotopes are 90°, and diatopes seem to be 45°. Of course, these objects consist of their lower dimensional counterparts as facets. So the same for the dodecahedron and hecatonicosachoron composed of 108° pentagons, and so on.
So what you say about the cube and the hexagon is true, but globally, it is obvious that the cube is related more to the square. (It's the number of dimensions being projected onto 2D that affect the flattened shape it takes. So one projection of the 4-cube (tesseract) looks like an octagon with an octagram inside of it).

But the deeper question remains: what is it about 5D and beyond that so constrains the type of polytopes that may inhabit it?

Now, in non-Euclidean geometry, the simplex and orthotopes can be stellated, as the facets are curved so they they can meet if extended. So that way you would be able to have "stars' in any dimension.

Well, that sidesteps the original issue, which is to understand why exactly Euclidean space confines the possibilities in such an intriguing way: an infinitude of regular polygons, yet only 5 regular polyhedra, then a tantalizing 6 regular polychora, before collapsing to a strangely impoverished 3 regular n-polytopes thereafter.

All of these results are obvious from many respects, of course, as one could argue using angle defects, etc.. But it quite intrigues me as to whether there is a deeper underlying cause for this strange pattern of infinity, 5, 6, 3, 3, 3, 3, 3, ... . That anomalous 6 is most fascinating, since if the sequence were doomed to collapse to 3, why does it spike to 6 in 4D?

It seems to have to do with the fact that the shapes all become more complex with more dimensions.
Look a the facets of polygons. Identical line segments. Facets are only two points or nullitopes. All you have to do is join them end to end, and any number of them will enclose 2-space dependingon he angle.

With polyhedrons, the facets are now polygons, and they all have their own 2D shape that now has to fit together. So the four sides of right angled squares will fit together to form a cube, but the hexagons' six sides at 120° angles again can only fit together flat, and cannot enclose 3-space. Any number higher than 6 cannot even tesselate on a flat surface! So right away, no [angular] analogues of anything higher than a pentagon can exist in more then 2 dimensions.
In 4-space, you have more possible regular shapes, because of the new "family", allowed by the freedom gained in the new dimension. Just like the diatope (cross-polytope) being introduced in 3D. But now you have even less regular 3-facets to work with. So how many more regular polychora would you think can possibly be made?
So apparently in 5D, all the polytopes and their facets outside the simplex, orthotope and diatope are too complex to fit together to close up the space. The more dimensions, the harder it is to enclose a volume of space. And the more complex the d-1 facets, the harder it is to fit them together. This, making even new "families" employing the added freedom of the higher dimension impossible. (Unless the mathematicians have simply not been able to recognize all the possibilities in the higher spaces). Only the 45, 60 and 90° angles can do it in all dimensions.

[quickfur]

Well, I know that. I guess what I'm looking for is some insight into why regular stellated figures must share the vertices/face planes of regular convex figures.

So what you're talking about is sticking "points" anywhere on the object? Like just pasting triangles on the sides of a square, to make a four pointed star). Stellation is defined as extensions of the "sides" (facets)

The definition of a regular n-polytope is that its facets must be regular (n-1)-polytopes, which must be transitive. Nothing in this definition (at least not directly) says anything about non-convex regular polytopes necessarily sharing the same vertices/face planes as convex regular polytopes. Somewhere between the definition and the derived fact is the insight into why the latter must hold, which is what I'm interested in.

It's not so much about sticking points anywhere, but about why we can't, for example, take a 4D star, and build with it a regular 5D polytope? What is it about Euclidean space that precludes such a figure from closing?

None of these are stellatable. In 2-D, they are represented by the triangle, square and circle.

Actually, this is not accurate. The 2D equivalent of the octahedron is the diamond, which is the same as the square rotated 45 degrees.

That I know. So it was represented in the "square".

Mea culpa. I didn't notice you include the hyperspheres in your list of families.

Only the pentagon and higher are stellatable. Stellation is a result of extending alternate facets together.

Yes, although one wonders why there are no regular stars that are not a result of extending the convex regular polytopes.

What else would a regular star be?

A polytope with transitive regular star-shaped facets. What is it about Euclidean space that necessitates that these star-shaped facets join in such a way so that their vertices match the vertices of a convex regular polytope?

[...]
This is an interesting area of discussion: why the regular polytopes seem to collapse to only 3 families after 4D. The most superficial reason, of course, is that the 120-cell and 600-cell have very large dihedral angles, and so cannot form the basis of higher regular polytopes (since it must be possible to join three of them together and still have non-zero angle defect in order for them to fold up into 5D: note how there are no regular polyhedra with hexagonal faces for the same reason).

OK, this explain to me. With all of this stuff I know; I never got into "dihedral angles". I remember reading about them, but forgot what they were. "dihedral" would mean two-faced.
I always wondered about hexagonal faces. It seems because they perfectly tile a flat plane. (the three 120° angles fit together to complete the 360°). If you angle them by any amount, they would no longer fit together. Volleyballs use a combination of hexagons and pentagons.

The term "dihedral angle" is a misnomer, really. It really should be "difacet angle" or some suitably Latinized/Greekized form thereof. It's the angle between the hyperplanes that two adjacent facets lie in (or, if you want to get formal, the angle between the normals of said hyperplanes). For regular polytopes, it's the same between every pair of adjacent facets, so we may say the polytope has a single, characteristic, dihedral (difacet) angle.

As for volleyballs, the shape is just a truncated icosahedron, suitably expanded onto a spherical surface. It's a coincidence that hexagonal faces result from this truncation, really; in the higher polytopes, truncating analogous facets result in shapes such as, to use 4D as an example, truncated tetrahedra (4 triangles, 4 hexagons), or truncated octahedra (8 hexagons, 6 squares). So in some sense, these polyhedra have some hexagon-like attributes, at least when one considers the analogy of truncation applied to two analogous polytopes.

You can have any number (above 3) sided regular 2D polygon, but in 3D, you can only have 4, 6, 8, 12, and two of 20 faces. In 4D, you can only have 5, 8, 16, 120, and I think 600.

Don't forget the 24-cell. It's truly a unique polytope, which "almost" spawns another family of higher polytopes; unfortunately, 24-cells tile 4D, and so have an angle defect of 0, thus they cannot fold up into 5D to form a regular 5-polytope.

OK; thought I might be missing something. Isn't the icosatetrachoron also related to the octahedron?

Its facets are 24 octahedra. A truly intriguing coincidence in 4D, being both an analog of the cuboctahedron (as a truncated cross polytope) and the rhombic dodecahedron (as a (hyper)cube augmented with square (cubical) pyramids), and on top of this, just happening to be regular because of the way <1,1,1, ...> and <2,0,0, ...> have the same Euclidean length in 4D, an interesting consequence of the fact that 4*1² = 2² = 4.

Also, the other mistake I made is that it is the dodecahedron, not the icosahedron that comes in two forms.
(So, or is it the rhombic dodecahedron the icosatetrahedron is related to ?)

The (pentagonal) dodecahedron and the rhombic dodecahedron have pretty much no relation, except for coincidentally having the same number of faces. The 24-cell is a sort of 4D rhombic dodecahedron, in the sense of being the pyramid-augmentation of the cube, but not in the sense of being the dual of the midpoint between the cube and the cross polytope.

[...]
It is interesting to surmise which polygons exactly correspond with the higher polytopes. I suspect the real answer is, none of them are exact analogs.
[...]
Truth be told, analogy only carries so far. Each polytope is unique in its own way, and the polygons are really too simplistic to adequately capture all the beauty and symmetry of the higher polytopes.

I judge the analogues by the angles, not the number of facets or shape of projections. Simplexes are always 60°; orthotopes are 90°, and diatopes seem to be 45°.

This does not carry to higher dimensions very well, because using this definition, the icosahedron is also 60°, and so is the 600-cell, but they are very different from simplexes.

Always keep in mind that in n dimensions, the facets are (n-1)-dimensional, not 2-dimensional. So a more "faithful" measure, IMHO, is the dihedral angle (or difacet angle, as above). If we use this measure, then we see that both the cube and the cross have dihedral angles of 90°, which hints at their duality. The simplices begin with 60° in 2D, then to 70.5° in 3D. I don't know what happens in 4D and beyond, but it does have an interesting increasing trend, as far as I can see.

But as far as projections are concerned, I wasn't really basing my analogies purely on projections, since that would be quite unreliable, but more on the effect of certain operations such as truncation, which I mentioned above. The reason I brought up projections was because it is in projections that the analogy in the effect of these operations become overt. Regardless, one can examine the analogy between the various truncations of the cube, and the various truncations of the tesseract. The first truncation of the cube yields a polyhedron with octagonal and triangular faces, and the first truncation of the tesseract yields a polychoron with tetrahedral and truncated-cubical cells. So one observes an analogy between triangles and tetrahedra, and between octagons and truncated cubes. The second truncation of the cube yields a cuboctahedron, with squares and triangles for faces; the second truncation of the tesseract yields a polychoron with octahedra and cuboctahedra as cells. So one could draw an analogy between triangles and octahedra, and between squares and cuboctahedra. It quickly becomes obvious that the analogy between the dimensions can only be imperfect, as I alluded to, even though one does see some similarity between a triangle and an octahedron (which has triangular faces), and between a square and a cuboctahedron (the cuboctahedron being a rectified cube, just as the square is a rectified dual square).

One could go farther, and consider, for example, the rhombated cube (the small rhombicuboctahedron) and the cantellated tesseract (same as the cantellated 16-cell). Here, the analogy between cubes and squares, tetrahedra and triangles, is obvious. Then one could consider the handful of distinct 4D analogues of the great rhombicuboctahedron, and observe that the octagons in the great rhombicuboctahedron has analogues in rhombicuboctahedra and even great rhombicuboctahedra, and the hexagons in the same has analogues in octahedra, truncated cubes, and so forth. All of these analogies do have some basis, but again, one is quickly reminded that these analogies are imperfect, and only serve as a crude tool with which to probe into the higher dimensions.

[...]So what you say about the cube and the hexagon is true, but globally, it is obvious that the cube is related more to the square. (It's the number of dimensions being projected onto 2D that affect the flattened shape it takes. So one projection of the 4-cube (tesseract) looks like an octagon with an octagram inside of it).

It is true that the cube is related more to the square, in the sense of both being the measure polytope. It is equally true that the octahedron is also related to the square, being the dual of the cube just as the square is the dual of a square, and having a dihedral angle of 90°, which a square also has, and being the convex hull of equidistant points lying on the positive and negative coordinate axes, just as the square is the same.

One does observe, as Wendy has pointed out elsewhere, that in the lower dimensions, many polytope characteristics are conflated, whereas in the higher dimensions, their differences begin to show up. In 2D, the measure polytope and the cross are conflated, but in 3D onwards, their role becomes distinct. In 3D, the pyramid-augmentation of the cube coincides with the dual of octahedron's rectate, and in 4D this is still true (although the tesseract's rectate is no longer the same as the 16-cell's rectate, unlike the cube's rectate, which is the same as the octahedron's rectate), but in 5D, this is no longer true. In 3D, the alternated cube coincides with the simplex (tetrahedron), but in 4D, the alternated cube becomes the cross polytope instead. In 5D, the alternated cube loses its regularity, and in 6D and above, becomes no longer uniform, and thus one sees a family of polytopes that becomes distinct from the usual categories of regulars and uniforms as one goes up the dimensions.

Perhaps the reason that regular figures become so rare in higher dimensions is due to the divergence of these polytopic characteristics into separate families of polytopes as one goes up the dimensions. In the lower dimensions, many of these characteristics intersect, giving rise to many highly-symmetrical figures, but as one travels up the dimensions, they begin to diverge, so that there are no longer as many points of coincidence. Perhaps this is in part an answer to my question, as quoted below:

But the deeper question remains: what is it about 5D and beyond that so constrains the type of polytopes that may inhabit it?

Nevertheless, one still wonders at the "anomalous 6" in 4D, as I call it:

All of these results are obvious from many respects, of course, as one could argue using angle defects, etc.. But it quite intrigues me as to whether there is a deeper underlying cause for this strange pattern of infinity, 5, 6, 3, 3, 3, 3, 3, ... . That anomalous 6 is most fascinating, since if the sequence were doomed to collapse to 3, why does it spike to 6 in 4D?

It seems to have to do with the fact that the shapes all become more complex with more dimensions.

So it seems on the surface; although one could argue that as far as regular polytopes are concerned, the complexity of the facets are not necessarily such a big hindrance, since for a polytope to be regular, its facets already must be regular, and in fact, identical; so for them to join up with each other is no difficulty. What is difficult, however, is for them not only to match up, but also to close. As Wendy has already pointed out, one could take a few identical regular polytopes (not necessarily convex), and begin to assemble them together (again, not necessarily maintaining convexity). However, something about higher-dimensional Euclidean space dictates that these figures would not close, and so they fail to generate more regular polytopes.

(In fact, one can already see these failures begin in 4D: in 3D, all of the stars generated by pentagonal star polygons do give rise to regular star polyhedra; however, in 4D, not all of the 3D stars (all of which are with icosahedral symmetry, I might add) form closed 4D stars. In 5D, none of the 4D stars form closed 5D stars. So one observes that as the dimension increases, something about the fabric of that space prevents these potential regular figures from closing.)

Look a the facets of polygons. Identical line segments. Facets are only two points or nullitopes. All you have to do is join them end to end, and any number of them will enclose 2-space dependingon he angle.

With polyhedrons, the facets are now polygons, and they all have their own 2D shape that now has to fit together. So the four sides of right angled squares will fit together to form a cube, but the hexagons' six sides at 120° angles again can only fit together flat, and cannot enclose 3-space. Any number higher than 6 cannot even tesselate on a flat surface! So right away, no [angular] analogues of anything higher than a pentagon can exist in more then 2 dimensions.

Ah, but we failed to account for non-convex polytopes here. Remember that the original question concerns non-convex star polytopes; for non-convex things, we are not really that worried about whether or not their angle defect allows them to fold up into space, as one could, conceivably, traverse several multiples of 360° until all the edges finally line up (since we allow self-intersections). In fact, star polygons of high degree have even less digonal angle (e.g., a 12/11 star polygon has a miniscule angle that could conceivably allow these things to easily fold up into 3D: the only problem being that if we attempt to construct a polyhedron with 12/11 stars, their edges would never match up, so the figure never closes. Only the pentagonal stars eventually close up---the question here is, why pentagonal?[i]---remembering that a pentagonal star does [i]not have a 108° angle!).

In 4-space, you have more possible regular shapes, because of the new "family", allowed by the freedom gained in the new dimension. Just like the diatope (cross-polytope) being introduced in 3D. But now you have even less regular 3-facets to work with. So how many more regular polychora would you think can possibly be made?

Actually, I wouldn't say the cross polytope is introduced in 3D: it has always been there, as the dual of the square, which in 2D just happens to be the same as the square itself. The fact that in 3D these two types of polytope diverge is a testament to Wendy's observation that these polytopic characteristics diverge as one goes up the dimensions. In 2D, the measure and the cross coincide, but thereafter, they diverge. (One may note that in 1D, all three families coincide: the simplex is a digon, and so is the measure polytope, and so is the cross. In 2D, the simplex family branches off, and in 3D the cross branches off.)

The regularity of the 24-cell seems all the more a strange coincidence, given these considerations.

So apparently in 5D, all the polytopes and their facets outside the simplex, orthotope and diatope are too complex to fit together to close up the space. The more dimensions, the harder it is to enclose a volume of space. And the more complex the d-1 facets, the harder it is to fit them together. This, making even new "families" employing the added freedom of the higher dimension impossible. (Unless the mathematicians have simply not been able to recognize all the possibilities in the higher spaces). Only the 45, 60 and 90° angles can do it in all dimensions.

I'm not sure where the 45° comes in. As far as I know, there are no polytopes with a 45° angle outside of the 2D stars, neither dihedral nor polygonal.

[Eric B]

The definition of a regular n-polytope is that its facets must be regular (n-1)-polytopes, which must be transitive. Nothing in this definition (at least not directly) says anything about non-convex regular polytopes necessarily sharing the same vertices/face planes as convex regular polytopes. Somewhere between the definition and the derived fact is the insight into why the latter must hold, which is what I'm interested in.

It's not so much about sticking points anywhere, but about why we can't, for example, take a 4D star, and build with it a regular 5D polytope? What is it about Euclidean space that precludes such a figure from closing?

A polytope with transitive regular star-shaped facets. What is it about Euclidean space that necessitates that these star-shaped facets join in such a way so that their vertices match the vertices of a convex regular polytope?

Oh, so you're talking about star shaped facets. I have never studied those. I know I saw some polyhedrons like that on one of the sites that had all those pictures of shapes. Like I think you could get pentagram facets by truncating the vertices of a dodecahedron, or something like that, IIRC.

Anyway, those are going to be even more complex than convex polygon facets, in having all of those additional surfaces (including the concave acute angles) that would need to fit together. Again; I do not know anything about stellation in higher dimensions (except that it is not possible with the simplex and orthotope). So I would not know exactly what it doesn't seem to be possible.

The term "dihedral angle" is a misnomer, really. It really should be "difacet angle" or some suitably Latinized/Greekized form thereof. It's the angle between the hyperplanes that two adjacent facets lie in (or, if you want to get formal, the angle between the normals of said hyperplanes). For regular polytopes, it's the same between every pair of adjacent facets, so we may say the polytope has a single, characteristic, dihedral (difacet) angle.

OK; so basically the angles between the "sides". (Was too lazy to look it up last night). This would also explain why it's "-hedral":
http://mathworld.wolfram.com/DihedralAngle.html

[quickfur]

Oh, so you're talking about star shaped facets. I have never studied those. I know I saw some polyhedrons like that on one of the sites that had all those pictures of shapes. Like I think you could get pentagram facets by truncating the vertices of a dodecahedron, or something like that, IIRC.

You do realize that some of the 3D stars are made of pentagram faces, right? Look up the Kepler-Poinsot polyhedra sometime.

The intriguing thing is that even though we could construct such polyhedra by folding up these pentagram faces, the resulting polyhedron can also be obtained as a stellation of a dodecahedron, or equivalently, a faceting of an icosahedron, which, obviously, has the same vertices and facet planes as the convex polyhedra (the dodecahedron and icosahedron, to be precise). The pressing question is still, why? Why is it that we can't, for example, fold up 12/11 star polygons and have them still close into a polyhedron?

Anyway, those are going to be even more complex than convex polygon facets, in having all of those additional surfaces (including the concave acute angles) that would need to fit together.
[...]

Well, actually, since we are restricting ourselves to regular polytopes, these figures aren't significantly more complex than the figures obtained otherwise. At least, the ones that do close up happen to be the same figures as the ones we obtain by stellating or faceting the convex polytopes. This is at the crux of my question: why is it that allowing starry facets doesn't give us anything more than what we can already obtained by stellation and faceting?

It is even more curious that when we relax the regularity requirement to merely vertex transitivity (uniformity), we do obtain vast numbers of polytopes, of which there are over 1000 in 4D, consisting of both convex and non-convex polyhedra as cells. So it seems that there is something about stars and regularity that makes them incompatible past 4D. (I don't know if there are any uniform 5D stars, but I wouldn't be surprised if there are, since there are so many uniform 4D stars to choose from for facets).

P.S. I accidentally hit "submit" in my previous reply before I was done, so I had to go back to edit it. You might want to re-read it to see the rest of my responses.

[wendy]

There are, in every dimension, uniform figures that are 'starry'. Every figure that contains octagons, either as hedra or as a curcuit of edges (like the octagons in the rhombo-cuboctahedron), leads to a starry figure containing octagrams. This holds in every dimension.

Apart from those figures constructed by Wythoff's mirror-edge construction, there are no known examples of uniform figures in dimensions higher than four. In practice, the list of non-wythoffian uniforms, and laminate tilings, is very short:

3t snub quadlat, snub hexlat
3d antiprisms, snub cube, snub dodecahedron
4d snub 24choron, grand antiprism
4t snub: s3s4o3o3o.

In six, seven and eight dimensions, there is a third branch of the trigonal groups, which yield the non-regular Gossett-Elte figures, and for all dimensions, five and greater, the second branch of the trigonal groups yield figures that are only vaguely derived from the measure polytopes.

The infinite class of laminate tilings is still very small, compared with those derived from Wythoff's construction on the four tiling-symmetries.

Stellations are known in hyperbolic tilings, and star-compounds exist in 2t and 4t as well.

The existance of the four regular polytopes in any dimension, is a direct result of the four products being metric. Two are also coherent.

[Eric B]

Oh, so you're talking about star shaped facets. I have never studied those. I know I saw some polyhedrons like that on one of the sites that had all those pictures of shapes. Like I think you could get pentagram facets by truncating the vertices of a dodecahedron, or something like that, IIRC.

You do realize that some of the 3D stars are made of pentagram faces, right? Look up the Kepler-Poinsot polyhedra sometime.

Yeah; those are what I was thinking of. Particularly the small stellated dodecahedron. Heres one with octagrams: http://en.wikipedia.org/wiki/Image:Grea ... hedron.png

But as far as projections are concerned, I wasn't really basing my analogies purely on projections, since that would be quite unreliable, but more on the effect of certain operations such as truncation, which I mentioned above. The reason I brought up projections was because it is in projections that the analogy in the effect of these operations become overt. Regardless, one can examine the analogy between the various truncations of the cube, and the various truncations of the tesseract. The first truncation of the cube yields a polyhedron with octagonal and triangular faces, and the first truncation of the tesseract yields a polychoron with tetrahedral and truncated-cubical cells. So one observes an analogy between triangles and tetrahedra, and between octagons and truncated cubes. The second truncation of the cube yields a cuboctahedron, with squares and triangles for faces; the second truncation of the tesseract yields a polychoron with octahedra and cuboctahedra as cells. So one could draw an analogy between triangles and octahedra, and between squares and cuboctahedra. It quickly becomes obvious that the analogy between the dimensions can only be imperfect, as I alluded to, even though one does see some similarity between a triangle and an octahedron (which has triangular faces), and between a square and a cuboctahedron (the cuboctahedron being a rectified cube, just as the square is a rectified dual square).

One could go farther, and consider, for example, the rhombated cube (the small rhombicuboctahedron) and the cantellated tesseract (same as the cantellated 16-cell). Here, the analogy between cubes and squares, tetrahedra and triangles, is obvious. Then one could consider the handful of distinct 4D analogues of the great rhombicuboctahedron, and observe that the octagons in the great rhombicuboctahedron has analogues in rhombicuboctahedra and even great rhombicuboctahedra, and the hexagons in the same has analogues in octahedra, truncated cubes, and so forth. All of these analogies do have some basis, but again, one is quickly reminded that these analogies are imperfect, and only serve as a crude tool with which to probe into the higher dimensions.

This does not carry to higher dimensions very well, because using this definition, the icosahedron is also 60°, and so is the 600-cell, but they are very different from simplexes.

Yeah, in that case, it is more than the angle. There is only one simplex, and they happen to always have 60° angles. Anything else in that dimension with the same angle is another family. But 90° is always an orthotope. So that's what I meant by saying I determined it by the angles. I wasn't going by truncations and stuff like that, or even "crossing". So while many other analogies can be drawn, "simplex", orthotope", etc. seem to be the most obvious.

Always keep in mind that in n dimensions, the facets are (n-1)-dimensional, not 2-dimensional. So a more "faithful" measure, IMHO, is the dihedral angle (or difacet angle, as above). If we use this measure, then we see that both the cube and the cross have dihedral angles of 90°, which hints at their duality. The simplices begin with 60° in 2D, then to 70.5° in 3D. I don't know what happens in 4D and beyond, but it does have an interesting increasing trend, as far as I can see.

That means the corner where all three planes and lines meet is not 60°? (I know the dihedral angles are, becuase they are equilateral triangles). At that point, I really do not know as much about 3D or higher geometry, as far as how angles are measured.

It is true that the cube is related more to the square, in the sense of both being the measure polytope. It is equally true that the octahedron is also related to the square, being the dual of the cube just as the square is the dual of a square, and having a dihedral angle of 90°, which a square also has, and being the convex hull of equidistant points lying on the positive and negative coordinate axes, just as the square is the same.

One does observe, as Wendy has pointed out elsewhere, that in the lower dimensions, many polytope characteristics are conflated, whereas in the higher dimensions, their differences begin to show up. In 2D, the measure polytope and the cross are conflated, but in 3D onwards, their role becomes distinct. In 3D, the pyramid-augmentation of the cube coincides with the dual of octahedron's rectate, and in 4D this is still true (although the tesseract's rectate is no longer the same as the 16-cell's rectate, unlike the cube's rectate, which is the same as the octahedron's rectate), but in 5D, this is no longer true. In 3D, the alternated cube coincides with the simplex (tetrahedron), but in 4D, the alternated cube becomes the cross polytope instead. In 5D, the alternated cube loses its regularity, and in 6D and above, becomes no longer uniform, and thus one sees a family of polytopes that becomes distinct from the usual categories of regulars and uniforms as one goes up the dimensions.

Perhaps the reason that regular figures become so rare in higher dimensions is due to the divergence of these polytopic characteristics into separate families of polytopes as one goes up the dimensions. In the lower dimensions, many of these characteristics intersect, giving rise to many highly-symmetrical figures, but as one travels up the dimensions, they begin to diverge, so that there are no longer as many points of coincidence. Perhaps this is in part an answer to my question, as quoted below:

But the deeper question remains: what is it about 5D and beyond that so constrains the type of polytopes that may inhabit it?

Probably so.

Nevertheless, one still wonders at the "anomalous 6" in 4D, as I call it:

All of these results are obvious from many respects, of course, as one could argue using angle defects, etc.. But it quite intrigues me as to whether there is a deeper underlying cause for this strange pattern of infinity, 5, 6, 3, 3, 3, 3, 3, ... . That anomalous 6 is most fascinating, since if the sequence were doomed to collapse to 3, why does it spike to 6 in 4D?

It seems to have to do with the fact that the shapes all become more complex with more dimensions.

So it seems on the surface; although one could argue that as far as regular polytopes are concerned, the complexity of the facets are not necessarily such a big hindrance, since for a polytope to be regular, its facets already must be regular, and in fact, identical; so for them to join up with each other is no difficulty. What is difficult, however, is for them not only to match up, but also to close. As Wendy has already pointed out, one could take a few identical regular polytopes (not necessarily convex), and begin to assemble them together (again, not necessarily maintaining convexity). However, something about higher-dimensional Euclidean space dictates that these figures would not close, and so they fail to generate more regular polytopes.

(In fact, one can already see these failures begin in 4D: in 3D, all of the stars generated by pentagonal star polygons do give rise to regular star polyhedra; however, in 4D, not all of the 3D stars (all of which are with icosahedral symmetry, I might add) form closed 4D stars. In 5D, none of the 4D stars form closed 5D stars. So one observes that as the dimension increases, something about the fabric of that space prevents these potential regular figures from closing.)

The intriguing thing is that even though we could construct such polyhedra by folding up these pentagram faces, the resulting polyhedron can also be obtained as a stellation of a dodecahedron, or equivalently, a faceting of an icosahedron, which, obviously, has the same vertices and facet planes as the convex polyhedra (the dodecahedron and icosahedron, to be precise). The pressing question is still, why? Why is it that we can't, for example, fold up 12/11 star polygons and have them still close into a polyhedron?

Anyway, those are going to be even more complex than convex polygon facets, in having all of those additional surfaces (including the concave acute angles) that would need to fit together.
[...]

Well, actually, since we are restricting ourselves to regular polytopes, these figures aren't significantly more complex than the figures obtained otherwise. At least, the ones that do close up happen to be the same figures as the ones we obtain by stellating or faceting the convex polytopes. This is at the crux of my question: why is it that allowing starry facets doesn't give us anything more than what we can already obtained by stellation and faceting?

It is even more curious that when we relax the regularity requirement to merely vertex transitivity (uniformity), we do obtain vast numbers of polytopes, of which there are over 1000 in 4D, consisting of both convex and non-convex polyhedra as cells. So it seems that there is something about stars and regularity that makes them incompatible past 4D. (I don't know if there are any uniform 5D stars, but I wouldn't be surprised if there are, since there are so many uniform 4D stars to choose from for facets).

I have never studied higher than 4 dimensions, and neither stellations, and objects other than the simplex and orthotope, really, so I would not know. All of that is interesting, however.

Ah, but we failed to account for non-convex polytopes here. Remember that the original question concerns non-convex star polytopes; for non-convex things, we are not really that worried about whether or not their angle defect allows them to fold up into space, as one could, conceivably, traverse several multiples of 360° until all the edges finally line up (since we allow self-intersections). In fact, star polygons of high degree have even less digonal angle (e.g., a 12/11 star polygon has a miniscule angle that could conceivably allow these things to easily fold up into 3D: the only problem being that if we attempt to construct a polyhedron with 12/11 stars, their edges would never match up, so the figure never closes. Only the pentagonal stars eventually close up---the question here is, why pentagonal?[i]---remembering that a pentagonal star does [i]not have a 108° angle!).

What I was getting at, was that I would think it would be even harder to get concave polytopes to fit together because of the "inner" <180° angle, at the point of "concavity", if you will. But then again; I don't know enough about stellations in higher dimensions to say for sure if that would be a barrier.

In 4-space, you have more possible regular shapes, because of the new "family", allowed by the freedom gained in the new dimension. Just like the diatope (cross-polytope) being introduced in 3D. But now you have even less regular 3-facets to work with. So how many more regular polychora would you think can possibly be made?

Actually, I wouldn't say the cross polytope is introduced in 3D: it has always been there, as the dual of the square, which in 2D just happens to be the same as the square itself. The fact that in 3D these two types of polytope diverge is a testament to Wendy's observation that these polytopic characteristics diverge as one goes up the dimensions. In 2D, the measure and the cross coincide, but thereafter, they diverge. (One may note that in 1D, all three families coincide: the simplex is a digon, and so is the measure polytope, and so is the cross. In 2D, the simplex family branches off, and in 3D the cross branches off.)

The regularity of the 24-cell seems all the more a strange coincidence, given these considerations.

Well, what I meant there, was "diverge". A new "family" is in fact being introduced as a separate group as it diverges.

So apparently in 5D, all the polytopes and their facets outside the simplex, orthotope and diatope are too complex to fit together to close up the space. The more dimensions, the harder it is to enclose a volume of space. And the more complex the d-1 facets, the harder it is to fit them together. This, making even new "families" employing the added freedom of the higher dimension impossible. (Unless the mathematicians have simply not been able to recognize all the possibilities in the higher spaces). Only the 45, 60 and 90° angles can do it in all dimensions.

I'm not sure where the 45° comes in. As far as I know, there are no polytopes with a 45° angle outside of the 2D stars, neither dihedral nor polygonal.

I was referring to the cross polytope family, with its right isoceles triangles.

[quickfur]

[...]
I'm not sure where the 45° comes in. As far as I know, there are no polytopes with a 45° angle outside of the 2D stars, neither dihedral nor polygonal.

I was referring to the cross polytope family, with its right isoceles triangles.

A cross polytope does not have any right isoceles triangles. At least, none on its surface. You could get right triangles if you add a point at the origin and consider triangles formed by the origin and the vertices of the cross, I suppose, but this seems a bit arbitrary, since doing the same with the hypercubes also gives a 45° angle, and doing the same with a simplex does not give 60° except in the 2D case.

[Eric B]

Oh; you know what? All this time, I was misled by an illusion of right triangles on the octahedron. I was thinking that the angles at the vertices were the right angles, and the other two, 45°. But then I just realized, all the angles are at vertices!

So what are they, equilateral? For some reason, they do look like isoceles right triangles, where on the tetrahedron and icosahedron, they don't. I guess because when you look straight down on a vertex, it looks like a square crossed with two lines at 45° connecting opposite angles. You forget that when you look at it on an angle and see the global shape, the angles are different as well.

[quickfur]

Oh; you know what? All this time, I was misled by an illusion of right triangles on the octahedron. I was thinking that the angles at the vertices were the right angles, and the other two, 45°. But then I just realized, all the angles are at vertices!

So what are they, equilateral?

Yes, they are equilateral; otherwise, the octahedron would not be a regular figure.

[quickfur]

Just a small update. Having actually read the relevant sections of Coxeter's Regular Polytopes last night , the following insight occurred to me: the reason all regular stars must share the vertices/face planes of a regular convex polytope is that for the star to be regular, it must be transitive on all its elements, including its vertices. But since it is transitive on its elements, the convex hull of its vertices must also be transitive on its elements: it must be a convex regular polytope. So it's not possible for a star to not have the vertices of a convex regular polytope. Being thus constrained, we see that there can be no regular stars in 5D or higher, since the only 3 regular polytopes therein have too small a dihedral angle to admit any stellations/facetings (except for degenerate compounds). So the ultimate cause traces back to angle defect: the 120-cell and 600-cell have too great a dihedral angle, which therefore excludes any regular pentagonal polytopes in 5D and higher.

Having said all that, however, Coxeter points out another tantalizing possibility: semi-regular stars that are transitive on its vertices (or facets), but not necessarily on all other elements. I haven't thought through this very much yet, but intuition would suggest that these would arise as stellations/facetings of the n-dimensional analogues of the Catalan solids. This is very appealing to me, because I have always been very interested in facet-transitive polytopes.