4D Hemipolychora?

[SharkRetriver]

I've been thinking a little about possible hemipolychora. (I think someone probably has all of these listed somewhere )
I think that the simplest one would be the 4D analogue of the tetrahemihex.
Anyway, this is the tetrahemihex: http://en.wikipedia.org/wiki/Tetrahemihexahedron
It has an octahedral envelope with 3 square faces through its equators and 4 triangles in an alternating pattern.
Does this mean that the 4D analogue has a 16-cell envelope with 4 octahedra passing through its equators and 8 alternating tetrahedra?

Of course this is only with the 16-cell envelope. I think that a LOT more exist (3D ones here: http://en.wikipedia.org/wiki/Hemipolyhedron)
Please correct me if anything's wrong there since I'm a lot newer to this than almost all of you I guess...

[quickfur]

Hmm, now that I think of it, since these hemipolychora are uniform, they must be part of Jonathan Bowers' list of non-convex uniform polychora? There are about 1800+ of these known. (Bowers' original list goes up to 8000+, but a lot of it had some strange features like coplanar facets or fissary cases; the amended list is in the 1800's.) I'm pretty sure hemipolychora would be somewhere in that list.

[quickfur]

Aha! Found your polychoron in Bowers' list: http://www.polytope.net/hedrondude/regulars.htm (look near the bottom for "Tho"). Apparently it's non-orientable. Fun, fun. The cells are exactly as we predicted. Interestingly enough, the vertex figure of Tho is Thah (i.e., tetrahemihexahedron). So it's a direct 4D analogue.

[SharkRetriver]

I knew that it was going to be in there somewhere, although I have no idea how to look for one since they're all named their own way...
"tesseractihemioctachoron" Well, at least the full name's provided :3
I also don't know what non-orientable means...chiral?
Anyway, it's time to sharpen some vertex figure and cross section skills...seems I'll need them.

Edit: http://www.polytope.net/hedrondude/icoes.htm There's a whole group of hemis here! Now only if I knew what the acronyms meant. Are CRFs listed on that site?

[wendy]

The class exists in all dimensions, and the class is very common where AB polytopes exist.

This means, for example, the AB polytopes /3.3.3/ and /3.3.4 , as does /3,3,3,3/ and /3,3,3,4 in 5D and so forth. Generally, there are not all that many figures that girth on polyhedra.

[Keiji]

I also don't know what non-orientable means...chiral?

Easiest way to explain this is by example, if you take a Moebius strip (cut out a strip of paper and glue the ends together into a loop, but with a 180 degree twist before you glue it), then you can start on one side of the paper, trace your finger all the way around and get to the point you started at but on the other side. So the Moebius strip only has one side, whereas the un-twisted loop would have two sides.

For polyhedra one side would be the "outside", the other side would be the "inside" (treating it as hollow of course). So, the non-orientability of the tetrahemihexahedron means it only has one side, while a convex polyhedron like the cube would have two sides.

Edit: http://www.polytope.net/hedrondude/icoes.htm There's a whole group of hemis here! Now only if I knew what the acronyms meant. Are CRFs listed on that site?

I don't believe so, if they were, why would we be searching for them?

[wendy]

Bower's site is devoted to finding the uniform star-polyhedra. In other words, these correspond to the 75 uniform starry polytopes in 3d. In four dimensions, there are something like 8200 of them. He brought in some pretty fancy terms to do it: armies and regiments and colonels and so forth.

The CRF figures are different. The distinction for the PG is CUH and CUC, being convex + uniform surhedra, or convex + uniform surchora. Of Bower's site, there are only 55 CUT (convex + uniform surtera), which would classify here. The rest are utterly different, but do include, like the quoted Oho and Tho, a 'hemichora'.

[SharkRetriver]

The class exists in all dimensions, and the class is very common where AB polytopes exist.

This means, for example, the AB polytopes /3.3.3/ and /3.3.4 , as does /3,3,3,3/ and /3,3,3,4 in 5D and so forth. Generally, there are not all that many figures that girth on polyhedra.

What does 3.3.3 stand for here? (I would read it as the vertex figure of a tetrahedron, although that doesn't really make sense here.)

[quickfur]

Bower's site is devoted to finding the uniform star-polyhedra. In other words, these correspond to the 75 uniform starry polytopes in 3d. In four dimensions, there are something like 8200 of them. He brought in some pretty fancy terms to do it: armies and regiments and colonels and so forth. [...]

I think the number has been reduced to 1800+ if one excludes the more exotic ones (he explains this in the "definition" section on his site). He classifies them as "tame", "feral", and "wild" based on what kind of strange properties they exhibit.

[Klitzing]

The class exists in all dimensions, and the class is very common where AB polytopes exist.

This means, for example, the AB polytopes /3.3.3/ and /3.3.4 , as does /3,3,3,3/ and /3,3,3,4 in 5D and so forth. Generally, there are not all that many figures that girth on polyhedra.

What does 3.3.3 stand for here? (I would read it as the vertex figure of a tetrahedron, although that doesn't really make sense here.)

Hehe, this is one of Wendy's cryptix!

On the first run somthing like p.q.r.s etc. in her notation just means the symmetry group o-p-o-q-o-r-o-s-o. I.e. the dot (or comma) represents an unringed node (those at the ends are to be understood in addition). The slash is a representant of a ringed node. So /3.3.3/ just means the Dynkin symbol x3o3o3x. (Beware: sometimes she uses semicolons instead of slashes too.)

But this is only the begin. The next step would be to understand her content. What has a x3o3o3x and a x3o3o4o in common?
This is the true mystery here, which has to be uncovered. :-)

It happens to be an analysis of the rectified cross-polytopes o3x3o3o...o4o : those can be given as a bistratic lace tower o3x3o...o3o || x3o3o...o3x || o3o3o...x3o . In fact, within an other direction they likewise can be given as a similar tower x3o3o...o4o || o3x3o...o4o || x3o3o...o4o . Therefore there can be a hemipolytope be constructed therefrom by using those hemifacets x3o3o...o3x together with those x3o3o...o4o of the other direction. (The hull clearly would be the o3x3o3o...o4o we started with.)

Esp. in 3d we have o3x4o (co) which can be given as x3o || x3x || o3x or likewise as x4o || o4q || x4o. Accordingly there is a polyhedron using the diametral x3x for hemifacets and additionally the facets x4o of the former. That one is cho (cubihemioctahedron).
In 4d we get o3x3o4o (a different description of ico, the 24-cell). It can be displayed as tower as o3x3o || x3o3x || o3x3o or alternatively as x3o4o || o3x4o || x3o4o. (In fact those alternate descriptions happen to be identical stacks of oct || co || oct. This is due to the additional symmetry of the 24-cell.) Accordingly we have a hemipolytope using as the hemifacets those x3o3x and as outer facets x3o4o.
In 5d we use o3x3o3o4o as o3x3o3o || x3o3o3x | o3o3x3o resp. as x3o3o4o || o3x3o4o || x3o3o4o. Therefore there is a hemipolytope using those hemifacets x3o3o3x and those outer x3o3o4o ones.
Etc.

--- rk

[SharkRetriver]

Hehe, this is one of Wendy's cryptix!

On the first run somthing like p.q.r.s etc. in her notation just means the symmetry group o-p-o-q-o-r-o-s-o. I.e. the dot (or comma) represents an unringed node (those at the ends are to be understood in addition). The slash is a representant of a ringed node. So /3.3.3/ just means the Dynkin symbol x3o3o3x. (Beware: sometimes she uses semicolons instead of slashes too.)

But this is only the begin. The next step would be to understand her content. What has a x3o3o3x and a x3o3o4o in common?
This is the true mystery here, which has to be uncovered. :-)

It happens to be an analysis of the rectified cross-polytopes o3x3o3o...o4o : those can be given as a bistratic lace tower o3x3o...o3o || x3o3o...o3x || o3o3o...x3o . In fact, within an other direction they likewise can be given as a similar tower x3o3o...o4o || o3x3o...o4o || x3o3o...o4o . Therefore there can be a hemipolytope be constructed therefrom by using those hemifacets x3o3o...o3x together with those x3o3o...o4o of the other direction. (The hull clearly would be the o3x3o3o...o4o we started with.)

Esp. in 3d we have o3x4o (co) which can be given as x3o || x3x || o3x or likewise as x4o || o4q || x4o. Accordingly there is a polyhedron using the diametral x3x for hemifacets and additionally the facets x4o of the former. That one is cho (cubihemioctahedron).
In 4d we get o3x3o4o (a different description of ico, the 24-cell). It can be displayed as tower as o3x3o || x3o3x || o3x3o or alternatively as x3o4o || o3x4o || x3o4o. (In fact those alternate descriptions happen to be identical stacks of oct || co || oct. This is due to the additional symmetry of the 24-cell.) Accordingly we have a hemipolytope using as the hemifacets those x3o3x and as outer facets x3o4o.
In 5d we use o3x3o3o4o as o3x3o3o || x3o3o3x | o3o3x3o resp. as x3o3o4o || o3x3o4o || x3o3o4o. Therefore there is a hemipolytope using those hemifacets x3o3o3x and those outer x3o3o4o ones.
Etc.

--- rk

I have yet to learn even the simplest of coxeter-dynkin diagrams (e.g. ones with more than one x or one that's non-linear). Also, I've never heard of lace towers, although I've heard of || being used for segmentochora. (Same thing?)
On a side note, what's a simpler definition for a powertope? triangle^octagon led me to a {3}x{8} duoprism.

[Keiji]

On a side note, what's a simpler definition for a powertope? triangle^octagon led me to a {3}x{8} duoprism.

That has nothing to do with CD symbols actually. See powertope and brick product. For triangle^octagon, see triangular octagoltriate - {p}x{q} duoprisms are created by the Cartesian product of a p-gon and a q-gon. Hope that helps

[quickfur]

[...]
I have yet to learn even the simplest of coxeter-dynkin diagrams (e.g. ones with more than one x or one that's non-linear).

I really should make a page to explain this. It's actually very intuitive once you understand how it works (at least, the ringing part is very easy to understand).

Also, I've never heard of lace towers, although I've heard of || being used for segmentochora. (Same thing?)

The || notation is basically to denote the top/bottom layers of a segmentotope (it works for any dimension, not just 4). A square antiprism, for example, can be denoted square || dual_square. If the top and bottom cells of a segmentotope have the same underlying symmetry group, say two members of the cube family of uniform polyhedra, then they form a lace prism. The side cells that connect the top and bottom polyhedra are called lacing cells, and correspond with each node of the CD diagrams of the top/bottom cells.

A lace tower is what you get when you stack a bunch of lace prisms on top of each other. Many lace towers are simply stacks of CRF lace prisms, although there are some cases where the entire tower is CRF but the individual layers may not be. The "ursatopes" we were talking about recently are an example of this. One simple ursatope is the 3D tridiminished icosahedron. Even though you can get this shape by cutting off 3 vertices from the icosahedron, you can also construct it as a lace tower: take a unit triangle, a triangle scaled by the golden ratio, and a dual triangle, and put them in parallel planes. Adjust the relative heights between them until edge lengths become equal between them. The result is the tridiminished icosahedron, which is a lace tower consisting of the above three polygons: triangle, triangle scaled by golden ratio, dual triangle.

A very interesting 4D CRF that Wendy discovered is the so-called "4D teddy", a lace tower consisting of a unit tetrahedron, a tetrahedron scaled by the golden ratio, and a unit octahedron. Its cells are 1 tetrahedron at the top, 4 tridiminished icosahedra around the sides, 1 octahedron at the bottom, and 4 tetrahedra filling the gaps between the octahedron and the tridiminished icosahedra. It's a direct analogue of the tridiminished icosahedron: 1 top triangle --> 1 top tetrahedron; 3 surrounding pentagons --> 4 surrouding tridiminished icosahedra; 1 bottom dual triangle (as a rectified triangle) --> 1 bottom octahedron (as rectified tetrahedron); and 3 lateral triangles around the bottom --> 4 tetrahedra around the bottom.

[wendy]

The notation pqrs or p.q.r.s etc, is the schläfli symbol, standing as a pseudoregular trace, or linearised dynkin-symbol. The / is the ancestral form of the node-marker 'x'. A semicolon is the loop-node 'z'. It's kind of 'ye olde english tea shoppe' thing, it's just that the / notation is twenty years older than the corresponding 'x' notation.

So something like /3.3.3/ is the same as x3o3o3x, represents a motif applied to the 4-simplex group, resulting in the runcinated simplex.

The next bit is 'AB' polytopes. AB polytopes are characterised by planes through them, which can be directed to 'above' and 'below'. For example, the orthotope /3...4 (or x3o..o4o), the 'above' side corresponds to positive coordinate in one axis. The below side is then the negative side. With the /3...3/, there are N+1 planes that cut through the figure. Each of these can be given an 'up' side so that the N+1 up-sides point to the N+1 vertices of a simplex.

The whole point about AB polytopes, is that what's written in the plane, is itself a polytope (of the same kind, but one dimension less). Let's call this C. So what you get is then a figure with faces C, and ones faces that have an even (or odd) number of 'A'.

Code: Select all

            above

       AB  /AA\  AB  /AA\
     \----/----\----/----\
      \BB/  BA  \BB/  BA  \

             below


You can see here that the faces above the line start with an A. Below the line, is a B. For other lines, like the crossing lines, the second line has is either A on one side or B on the other.

You could, for example, have AA = BB = triangles, and AB = BA = squares, will give you the cuboctahedron. The triangles and pentagons of the ID work exactly the same way. In a higher dimension, you can put an A or B in front of these to get a three-letter cluster, for polychora. You then have all eight possibilities, AAA to BBB including the likes of ABA and ABB. Now, you can sort the letters to get AAA = BBB, ABB = AAB etc.

In four dimensions, the orthotope produces simplices only for all combos. For the /3..3/ one has simplex-prisms, where AAA is a tetrahedron, AAB is a triangle-line prism, etc. AAAAA is a 5-simplex, AAABB is a 3-simplex × 2-simplex, and so forth.

What happens, is that the cutting planes are themselves polytopes, of the face-dimension. So they can be used as faces. Cut out those which have either an even number of A's or an odd number of A's. You then get the equal of the Oho or Cho etc, by keeping just odd A's (AB, BA), or even A's (AA, BB). It works in all dimensions.

It's more common with tilings to have AB structures.

/4.3..3.4/ gives n-cubes, say, with the cells centred on the integer-coordinates. One can take a cube and make it AAA... (say, all even) The one directly opposite at the vertex is BBB... (say all odd coordinates) Then the A's and B's correspond to the even and odd numbers, and the B side of any plane is the one that faces the odd numbers.

/3,3.....3: The group An produces this tiling, corresponding to a rhombic of 60°, with additional faces to slice the rhombics into planes, perpendicular to the vertices. Since the rhombic is itself an AB tiling, we just need to give direction to these additional slices to make it work. It does.

/3/3...3: This is the same construction as above, but the additional slicing planes are now halfway between the ones at the top. In other words, we are slicing halfway between vertices. The various AAAA.. is a simplex, BAAAA.. is a truncated simplex, BBAAA.. is a bitruncated simplex, etc.

[wendy]

Dynkin graphs and their motifs.

A dynkin graph can be thought of as a kaleidoscope. You know, those little tubes with mirrors in them, you turn the thing at the bottom to make pretty reflective patterns. The dots and other things we write on them are the motif, the little beads in the bottom of it.

In the dynkin graph, a point is called a 'node', and represents a mirror. A line between two nodes is called a 'branch', and is an angle between two mirrors. When there is no branch between two nodes, the mirrors are at right-angles to each other. So unconnected nodes are mirrors like '+', those connected by an unmarked angle are like a six-pointed asterix, and for marked branches (branches with a number above or on then), an asterix with that number of lines through the point.

This means, that our kaleidoscope at each meeting-point, has 2n regions around each mirror-crossing. This is the deep magic of it. What we're going to do now is build motifs (polytopes) on the kaleidoscope.

Mirror-edge polytopes
-------------------------

The most common construction is wythoff's mirror-edge. A 'mirror-edge' is simply an edge, whose two ends are reflections of each other (ie they are perpendicular to a mirror). These are usually the marked nodes (eg x, /, @ in various ascii graphs). If there is no edge, then the edge is unmarked (eg o, .) Your vertex or starting point, is away from the marked morrors (each node is a mirror), and on the unmarked mirrors (since we're not making an edge here).

Wythoff's construction does not suppose all edges be the same, you can use different lengths and/or colours for edges to each kind of vertex.

As you step through a mirror, you lay down an edge. So if you have a kaleidoscope like x--3--o, you step alternately through mirrors A and B (ie the first and second), laying mirrors whenever there is an x (ie A). You get a triangle. For o--3--x, you get a different triangle, with edges only when you step through a B mirror. For x--3--x, you get edges for both A and B mirrors, which gives a hexagon.

For unconnected mirrors, a single A or B by itself, does not produce a solid region, but putting down edges at each A and B gives ABAB, a rectangle or square.

For polyhedra, the same thing works, you find the three kinds of polygon, by considering pairs of nodes. So, x3o4x has two marked nodes (on mirrors A and C. Mirrors A and B will produce a 3x gon, for the number of x's given: ie a triangle. Mirrors B and C give a 4x gon, on the same condition, so we get squares there. Mirrors A and C are unconnected, so we use '2' as the marking. We get 2x-gon, for the marks on A and C, here a square again.

The connection between two kinds of face, are given by considering the common mirror. So the edge between AC and BC is whatever C is (here marked), so we have an edge between AC and BC. There is no edge between AB and BC, because the common mirror here is B, is unmarked.

For polychora, the same thing happens. You need three mirrors to make the polyhedral faces, and pairs of faces share common elements, according to what shape the shared mirrors make. So x3o4x3o has 4 mirrors, so it's a polychoron or 4d polytope. Removing each of the four nodes individually, gives the kinds of faces, and the pairs of nodes gives the shared margins between them.

So ABC is the rhomboCO we looked at above. ABD gives "x3o o". This thing is simply a triangle. It equates to a triangle-prism of zero height, but because every face using D equates to a line, (ie AD = line, BD = point), the only face produced is by mirrors AB, which means the thing is entirely flat. This is what happens when there are isolated unmarked regions.

ACD gives x x3o. This is pretty much like the previous, but we see that AC gives a square, and AD a line. AC's are connected to AD's by lines, but these are flat and pass the connection straight through as if they were on a mirror. So the faces ACD are connected to other ACD's through the AC wall (which is a square). CD gives a triangle as before,

BCD gives a figure o3x4o. It has only one kind of edge on C. BC drops only 3 edges, so is a triangle. CD drops four edge, so is a square. BD drops nothing, is a point at the centre (of where these mirrors cross). BCD gives a cuboctahedron.

There are more fancy rules around, but this will give you a start.

LACE PRISMS
---------------

The naming-type lace-prism is the polygonal antiprism. Think of it as a drummer's drum, with the top and bottom laced together by the zigzag.

In practice, lace-prisms work the same way, using a kaleidoscope etc, but we have two points and a line segment (lacing), sitting between the mirrors.

For the anti-prism, you have xoPox. The edge go from the left-mirror to the right-mirror, like a ladder sitting between two walls at that angle.

The top face is read by just looking at x.Po. (ie we hide the second vertex entirely). This gives a polygon P. The bottom face .oP.x gives also a polygon. You can even get a point (at the crossing of mirrors), when both nodes are 'o'. This would give a pyramid.

The idea with lace-prisms and lace-towers is that instead of a kaleidoscope of 3 mirrors, you can have different action (polytopes), on different floors, each vertex is connected to the ones above and below by lacing. This allows one to spread the coordinates of the vertices without having to give pages of coordinates.

Hope it helps.

[wendy]

The 4d teddy was one of the very early ones to find. It's essentially the vf of s3s4o3o3o, and i used it as the second part of finding the vertex-angle measure of the 3,3,5. All of the rest come from the tilings, sparce or discrete, they are all peicewise finite.

The 4d teddy can be reassembled to make the vertex-figures of (3,5,5/2), of (3,3,4), of (3,3,5) and 5 (3,3,3). The first two have vertex-angles of 76s and 5s, making all together 81s. The last two give 39s - 5(3,3,3). The value of (3,3,3) comes from the tiling of x3o3o3o3o3o3z, as the margin angle minus 1/5, or 25s 20 f V8 less 24s, or 1s20f108. It's then pretty straignt forward to find (3,3,5) at 33s 15s 60.

The vertex angles of the 4d regular polychora come to (all-space = 120 s, 1s = 120 f, 1f = 120

3,3,3 as 1s 20f 108
3,3,4 as 5s (exactly) = 1/24
4,3,3 as 7s 60 f exactly = 1/16
3,4,3 as 15 s exactly = 1/8
3,3,5 as 33s 15f 60
5,3,3 as 38s 24f (which is just over 1/pi at 38s 23f 80

Some time later, i found that the cubic radian was 1/2pi² or 6s 9f 60, and the tegmic radian at 1s 1f 70, which must be less than the simplex of the same dimension. 3,3,3 is 1.158852 = 1:19.07.

[Klitzing]

Aha! Found your polychoron in Bowers' list: http://www.polytope.net/hedrondude/regulars.htm (look near the bottom for "Tho"). Apparently it's non-orientable. Fun, fun. The cells are exactly as we predicted. Interestingly enough, the vertex figure of Tho is Thah (i.e., tetrahemihexahedron). So it's a direct 4D analogue.

Well, to speak about Jonathans List, its just an enumeration of the uniform polychora, i.e. all vertex surroundings are mapped transiently by some symmetry. This listing is being done by an complete analogue to the one of Coxeter et al. for the 3d uniforms:

• Consider the vertex figures of the (known) convex polyhedra (polychora).
• For each individual such verf consider all possible facetings. For 4d we have additionally: any such faceting face has to be a verf of a uniform polyhedron; for 3d that would reduce to: any such faceting edge has to be secant of some regular polygon.
• For any such faceted verf reconstruct the according uniform polytope.

So your question about hemichora is just the question about verfs which allow for hemifaces. Consider e.g. "tho": its verf is "thah", i.e. has squares for hemifaces. Accordingly the "tho" has diametral "oct"s. Just as thah is a faceting of oct, and oct is the verf of hex, likewise tho itself is a faceting of hex.

Similarily you could consider "ico" (24-cell). Ico's verf is a cube. So you would have to look for faceting faces of the cube, which contain the body-center. An obvious choice would the rectangle of side length 1 resp. sqrt(2). That one in turn is the verf of "co" (cuboctahedron), which clearly is a diametral pseudo cell of "ico". So it just is to ask for a complete vertex figure polyhedron (closed surface) which contains that rectangle, and you could construct the corresponding hemichoron. - Others would be either of the crossed tetragons inscribed into that rectangle. Those are the verfs of "cho" and "oho".

If you would look at his ico regiment page (http://www.polytope.net/hedrondude/icoes.htm), esp. at the displayed verfs, you would read off that nearly all members of that regiment are hemipolychora...

But that shown concept is even more general!

--- rk

[Mecejide]

By my count there are 31 unifrom hemipolychora.