Johnsonian Polytopes

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

Johnsonian Polytopes

Postby wintersolstice on Wed Dec 02, 2009 8:20 pm

I've recently been searching for "Johnsonian Polytopes" (only polychora atm)

Defined as:

1) Strictly-Convex from body to cell
(all the elements from body to cell are Strictly-convex: Convex and all it's angles between facets <180 degrees)

2) Non-Vertex-Transitive
(it's not transitive about it's vertices)

3) Regular Faced
(all it's faces are regular polygona)

I've found about 193 so far and was wondering if anyone would be interested in helping me.

(this definition and name is not official)
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Re: Johnsonian Polytopes

Postby PWrong on Thu Dec 03, 2009 12:17 am

What exactly do you mean by strictly convex? If all the angles are <180 degrees then you can't have any lines or planes in your shape. :\

I was reading the wikipedia article on vertex transitive graphs and I realised I've forgotten everything I know about group theory, and I never knew much about graph theory :(.

It sounds cool anyway. What sort of shapes have you found so far?
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Re: Johnsonian Polytopes

Postby Keiji on Thu Dec 03, 2009 11:01 am

Convex implies that angles are below 180 degrees. "Convex" does not mean "non-self-intersecting".

Sounds like your "Johnsonian Polytopes" are the same shapes FLD can describe.
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Re: Johnsonian Polytopes

Postby wintersolstice on Thu Dec 03, 2009 8:32 pm

PWrong wrote:What exactly do you mean by strictly convex? If all the angles are <180 degrees then you can't have any lines or planes in your shape. :\


It sounds cool anyway. What sort of shapes have you found so far?


21 pyramids, dipyramids, elongated pyramids and elongated dipyramids of 21 polyhedra
92 prisms of the "Johnson solids" (these shapes are based on them)
augmented tesseracts
and a few more
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Re: Johnsonian Polytopes

Postby Keiji on Fri Dec 04, 2009 5:44 am

wintersolstice wrote:Can I ask a question about your comment? :D

How can a shape be convex but self-intersecting?

And I thought convex meant the angles were "less than or equal than 180 degrees" (given that all the diagonals lie within the shape)

These shapes are based on the "Johnson solids"
"Johnsonian" was an name suggested by Bowers, I spoke to him about these shapes. And he seemed to understand the definition I created.

thanks (and sorry if I've upset you!)


Oh my, please don't go sending things that are meant to be forum posts in PMs.

A shape can't be both convex and self-intersecting, but it can be neither. Thus merely saying that a shape is non-self-intersecting does not necessarily mean it is convex.

Also I don't see why any convex polytope would have an 180 degree angle anywhere (that wasn't just an arbitrary point inside a facet).
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Re: Johnsonian Polytopes

Postby PWrong on Fri Dec 04, 2009 7:23 am

I was going by the definition of a strictly convex space. I guess a strictly convex polytope is different.

I still have no idea what vertex transitive means :(
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Re: Johnsonian Polytopes

Postby Keiji on Fri Dec 04, 2009 9:30 am

Vertex-transitive means that you can pick any two vertices in the shape, and the facets joined to them will all be equivalent. So a cuboctahedron for example is vertex-transitive because on every vertex there are 2 triangles and 2 squares.
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Re: Johnsonian Polytopes

Postby wintersolstice on Fri Dec 04, 2009 12:51 pm

Well I read about "Johnson solids" and they were described as "strictly-convex" and I made a deduction about what it meant
But it's not important.

Anyway they're the 4D analog of Johnson solids if anyone wants to look them up :D
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Re: Johnsonian Polytopes

Postby PWrong on Sat Dec 05, 2009 1:18 am

Ok I'm a bit confused. The wiki article says that a square pyramid ad a pentagonal pyramid are Johnson solids, but not a hexagonal pyramid. Surely none of these are vertex transitive?
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Re: Johnsonian Polytopes

Postby wendy on Sat Dec 05, 2009 7:46 am

A Johnson polyhedron is made of regular polygons, and is convex, but is not elsewhere included in platonic/uniform/etc.

The current thinking of uniform is equal edge + vertex transitive + uniform surtopes.

The hexagonal pyramid, or the hexagonal

Uniform johnson polychora, for example, would be all convex polychora made of regular polygons, except for the sixty-seven regulars, and their classes.

There are of course, the 92 prisms of the three-dimensional ones.

One might here include joys like xo3of3ox, a polytope made of 4 tri-diminished icosahedra, five tetrahedra, and an octahedron, the various diminished 500chora, the various augmented tesseracts (which are indeed convex, giving diminished, or all of the segmentotopes enumerated by Richard Klitzing (eg cube || icosahedron), and various sectionings of the 500ch.

Beside this, there are the figures like point | x-diminished icosa | x-diminished icosa | point, and all various sections thereof, for which gives at least four separate figures.
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Re: Johnsonian Polytopes

Postby Keiji on Sat Dec 05, 2009 7:28 pm

Ok I'm a bit confused. The wiki article says that a square pyramid ad a pentagonal pyramid are Johnson solids, but not a hexagonal pyramid. Surely none of these are vertex transitive?


Johnson solids need regular faces, but need not be vertex-transitive. The square and pentagonal pyramids can have regular triangular faces on them, but the hexagonal pyramid cannot because the dihedral angle would be zero, which would make it degenerate. Then it's obvious why n-gonal pyramids with n > 6 are not Johnson solids.

Back to the original point of this topic...

wintersolstice wrote:I've found about 193 so far and was wondering if anyone would be interested in helping me.


Could you list those 193?
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Re: Johnsonian Polytopes

Postby wintersolstice on Sun Dec 06, 2009 4:02 pm

Keijj wrote:Could you list those 193?


I'll have go!

(Elongated)
Tetrahedron, octahedron, icosahedron, cube, triangular prism, pentagonal prism, square antiprism, pentagonal antiprism, snub disphenoid, (bi)(tri) augmented triangular prism, gyroelongated square (di) pyramid, (metabi)(tri) diminished icosahedon, square pyramid, pentagonal (di) pyramid
(di)pyramid
excluding tetrahedron pyramid (regular pentachoron) and octahedron dipyramid (regular hexadecachoron)
82 in total

92 prisms of the Johnson solids

(Parabi) diminished hexacosichoron (2 shapes)(incomplete list)

ortho/gyrobi cube cupola
elongated cube gyrobicupola (gyrate sidpith)
cube cupola
elongated cube cupola (diminished sidpith)

the others are various augmented tesseracts (incomplete list and some don't have names yet)

I hope I've descibed them well enough :D
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Re: Johnsonian Polytopes

Postby wintersolstice on Sun Dec 13, 2009 2:53 pm

I'll have a go at explaining "Strictly convex" Some shapes have facets that lie in the same subplane but instead of joining them together to make a single facet they're kept as seperate facets. I think you call this "co-planar" some polychora have co-planar facets. A convex shape which is not co-planar is called "strictly convex"

I hope that explains it.

Edit: I've just realised if a 4D+ shape is coplanar then it's facets can be joined and it could still have regular faces so it would still be "strictly convex" so only the cells have to be strictly convex.
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Re: Johnsonian Polytopes

Postby PWrong on Mon Dec 14, 2009 2:46 am

I get the idea of strictly convex now, thanks.

It's an impressive list, do you think there are many more? How did you go about counting the ones you've found?
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Re: Johnsonian Polytopes

Postby wendy on Mon Dec 14, 2009 8:11 am

A large number of the Johnson polyhedra are laminate: that is, one can build them by placing successive layers like pyramid+prism+pyramid. We (Richard Klitzing and myself), devised a notion for forming notations, that represent more polytopes by means of "layers of dynkin symbols". For example, the general antiprism, might be considered as xPo || oPx (meaning, xPo on top, and oPx), or xoPox, which represents an array, the n_th layer is represented by layer n.

The tri-diminished dodecahedron, then might be x3o || f3o || o3x, which becomes xfo3oox&#t. (See eg Lace towers in my Polygloss). This occurs as a face of a four-dimensional polytope xf3oox3ooo&#t, a figure bounded by four tri-diminished icosahedra, five tetrahedra, and an octahedron.

Google on Klitzing Segmentotopes , and ye will find a number of interesting things that will help you. Klitzing enumerated the segmentotopes, being polytopes of equal edge, for which all vertices lie in two parallel rings of a sphere. The .pdf file corresponds to a paper listing some 184 segmentatopes or so, many of these are Johnson polychora.

One of the more interesting figures by Klitzing is "cube || icosahedron". The axial symmetry here is 3*2, or pyritohedral, it occurs in the {3,3,5}, where the cube is part of the dodecahedral layer, and the icosahedron is the large icosahedron layer below the mid-ring.

The next step is that ye can stack these, to get various laminatopes. A useful place to start is something like the 500ch {3,3,5}. Its layers give a fruitful source of johnsons, eg oxoo3ooox5ooxo, a polytope which comprises of several rings of the {3,3,5}, the face consist being tetrahedra, 12 diminished icosahedra, and one icosadodecahedron.

This has four bands, which must occur, and variations of putting a fifth band in, to cater for any of 1-12 apiculations (the diminished icosahedra can have a pyramid mounted on it, replacing this face with tetrahedra and a pentagonal pyramid.

You can mount any consecutive number of these bands together, with an prism layer ("elongated"), with a reversal to any earlier point.
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Re: Johnsonian Polytopes

Postby Tamfang on Mon May 24, 2010 3:53 am

Keiji wrote:Vertex-transitive means that you can pick any two vertices in the shape, and the facets joined to them will all be equivalent. So a cuboctahedron for example is vertex-transitive because on every vertex there are 2 triangles and 2 squares.

That is a consequence of transitivity, not the definition; in other words it is a necessary property but not sufficient. The classic counterexample is the pseudorhombicuboctahedron, Johnson #37.

A figure is vertex-transitive if all vertices are equivalent under the figure's isometry group.
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