## Klitzing snubs, multi-snubs

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

### Klitzing snubs, multi-snubs

I was studying this post, and thinking that there should be a more abstract description of the snubbing process, applicable to the symmetry group itself, rather than the polytope.

For ordinary snubs, i.e. alternated omnitruncates, the new symmetry group is simply the rotational subgroup of the original symmetry group, which was generated by reflections. The new group is generated by the same reflections, but the total number of reflections applied must be an even number.

Klitzing wrote:In the example above we considered omnitruncated figures only, i.e. Dynkin symbols with all nodes being ringed, for snubbing. Now we will go on. We not only allow polytopes, the Dynkin symbols of which do use unringed nodes as well, for starting figures. We even would allow to alternate not only vertices, but even higher dimensioanl elements as well, e.g. edges, faces, cells, etc.

The procedure would be as follows. Take any Dynkin diagram of any reflectional symmetry, and decorate the nodes by any choice with ringed nodes, with un-ringed nodes, and with snub nodes. I would use as an example x3o4s, but it would apply generally.

1st step: replace any "s" node by an "x" node (here: x3o4x = sirco). This would be the starting figure.
2nd step: omit any "s" node together with all incident links (here: x3o . = triangle). This is the element to be alternated.
Alternation again takes place as above, just that we dont have pyramids any longer, but more general cupolaic elements, which are to be removed locally in an alternating way. Still there are new facets underneath the omitted elements, which come in as new facets, while the other ones are to be diminished somehow.

In our example the cut off elements are x3o || x3q (triangles atop semiregular hexagons). Thus we omit any second triangle, and replace those by those hexagons. The other triangles would be kept in place. The squares in cubical positions are to be reduced to one of their diagonals, the squares in rhombical positions are completely withdrawn together with those alternated triangles. So we would result with a figure, which is a topological variant of the truncated tetrahedron, in fact it is q3x3o.

In that example, the original symmetry group obviously is o3o4o; an element of this group is a composition of any number of reflections using the three mirrors represented by the three nodes. An element of the new symmetry group is a composition of any number of reflections using the first two mirrors (the 'x' and 'o' nodes) and an even number of reflections using the third mirror (the 's' node).

Let's call the three generating reflections f,g,h (corresponding to 'x','o','s' respectively), so we have f^2 = g^2 = h^2 = (fg)^3 = (gh)^4 = (fh)^2 = 1 (the identity transformation). The original group contains all of the following transformations; the new group contains only those where the number on the right is even:

1; (0 non-snubbed reflections, 0 snubbed reflections)
f, g; (1 non-snubbed, 0 snubbed)
h; (0 non-snubbed, 1 snubbed)
ff=1, fg, gf, gg=1; (2 non-snubbed, 0 snubbed)
fh, gh, hf=fh, hg; (1 non-snubbed, 1 snubbed)
hh=1; (0 non-snubbed, 2 snubbed)
fff=f, ffg=g, fgf, fgg=f, gff=g, gfg, ggf=f, ggg=g; (3, 0)
ffh, fgh, gfh, ggh, fhf, fhg, ghf, ghg, hff, hfg, hgf, hgg; (2, 1)
fhh, ghh, hfh, hgh, hhf, hhg; (1, 2)
hhh; (0, 3)
etc.

It seems that a general snub symmetry group is just all of those transformations where the total number of reflections in snubbed mirrors is an even number. This is clearly a subgroup; it contains the identity (as 0 is even), it's closed under composition (as a sum of even numbers is even), and it's closed under inversion (as reversing a sequence of reflections doesn't change the count of each type). It's another question, whether the subgroup is actually smaller than the original group. Note that a transformation can have many different representations, some with odd numbers and some with even numbers; consider fgfgfg=1, where the left representation has 3 f's and 3 g's, and the right has 0 f's and 0 g's.
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mr_e_man
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### Re: Klitzing snubs, multi-snubs

A multi-snub symmetry group, then, is all of those transformations where the total number of reflections from each snubbed subset is an even number. That is, several sums must all be even, instead of just one sum being even.

The snubbed subsets may even overlap; some nodes could be snubbed twice!

This clearly doesn't depend on the order. (It's just an intersection of subgroups, each being singly-snubbed.) So what is going on here with s3s4s' ?

One answer, of course, is that Klitzing is considering polytopes, whereas I'm considering groups. Another answer is the fixing of edge lengths, which may add symmetry to the polytope or change the faces of the convex hull. Note that the hexagons in tut may be broken into 4 triangles each, making it isomorphic to ike.

In any case, the symmetry group s3s4s' is the intersection of s3s4o (pyritohedral group) and o3o4s (full tetrahedral group), which is isomorphic to s3s3s (chiral tetrahedral group).

What about the notation? The current multi-snub notation doesn't allow overlapping sets of snubbed nodes, and it doesn't describe the edge lengths of non-uniform snubs....
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### Re: Klitzing snubs, multi-snubs

The use of 's' in the CD was simply to enclose the wythoff-snubs into the CD diagram. It is more because there was no satisfactory way of handling rotation groups, and so this fudge was invented. Note that Mrs Stott (who suggested the idea) removes the mirror, leaving the ring to be alternated.

1). There is no provision for overlaping groups, because overlaping groups do not exist. Especially, were such to happen, then one would alternate the other over the common overlap.

2). In a snub, the edge lengths are not driven by the nodes, but pairs of nodes. If there are four circled nodes, there would be six edges, with only four degrees of freedom. Alternated nodes work in all dimensions, but except for the Coxeter-snubs and the Wythoff-snubs, none become regular, and all have more edges than you can set degrees of freedom on.

3). The proper notation for snubs in polyhedra is the conway-thurston notation, which allows one to set each edge of a polygon separately. But you have to deal with 'aspect' as well here. If there is more than three actions in the CT symbol, then those in excess can set aspect. The group 2 2 2 2, for example, has four symbols, or actions, and this group covers the general quadralateral, rotated in each of its four edges. It doesn't have to be a rectangle even.

All the same, for things like the pyritohedral groups and the snub cube etc, the CT can handle different edge lengths, including setting edges so that the incident faces are coplaner and hence can be merged. Thus, we can write for example, the octahedral figures (except tCO), as variations of the snub cube, by setting different combinations of edge to 0, 1 and %. Likewise, the pyritohedral group give six uniforms by setting its three positions to 0, 1, % too.

In the snub group, the base form is Px Qy Rz, where P Q R is the Conway Thurston Orbifold, and x,y,z is settable lengths. In this case, there is no asterix, so the three groups are 'cones', or a simple rotation. The x,y,z are the edges of the triangle adjacent to the polygons P, Q, R. One of these may be a digon, which disappears unless it is followed by %. Since this is a triangle, the rules of triangles (that the maximum is not greater than the other two) apply. If there is no degree of freedom, we can replace something like x:x:o with x:%:o triangle, where the second side is precalculated to equal the first.

Putting a single 'o' causes the triangle to shrink to preserve the 'o' at the vertex. There is than eg an order P rotation of polygon-vertices at that point. So for example 2x 3% 5o gives an icosahedron, literally digons and triangles repeated 5 times.

The xxx combo is x2 x3 x5 or snub dodecahedron. Except in the six special cases, this is the usual form.

With xx%, the side against the percent sign swallows the triangle, and the polygon is doubled. So 2x 3x 5% has decagons, surrounded alternately by digons and triangles. 2% 3x 5x has a square (the digon has swallowed the two triangles), along with triangles and pentagons.

The pyritohedral group P * Q has mark points at Px * yQz . Again we have edges o, x, %, but the results are somewhat different. P is still a cone, meaning a simple rotation group, while Q is a reflection polygon in the form xQo etc. It is possible to drop two edges from the cone-apex to the mirror, so you can have a generalised alternating edges over Q. The Pyritohedral group is 3 * 2, but there are similar groups 3 * 3 and 4 * 2 in euclidean space.

The root form is 3x * y2z, this has eight triangle faces 3x, six rectangles y:z and 12 general trapezia yz&x, that is, top y, bottom z, and slope x. The six rectangles are such that the y-edge of one faces the z-edge of the next.

3x*o2o is an octahedron. The rectangle is shrunk to a 0×0 rectangle, the trapezium is 00&1, ie a digon.
3o*x2% is a cube. The rectangles are 1:%, while the trapezium have no slope length, and so 1=%.
3x*o2% is the cuboctahedron. It has triangles, and a base rectangle o%. The trapezium is %o&1, which gives a right-angle triangle to be joined over the line %

3x*x2o and 3x*o2x are the left and right icosahedra, the triangle is 1, the rectangle is 1:0 (ie a line), and the trapezium is 01&1, ie a triangle.
3x*x2% gives a triangle, and a rectangle x:%. The trapezia are now x%:1, which merge over the % to give octagons. The result is a truncated cube.

3x*x2x is the rCO, with triangles 3x, and rectangles (squares) 1:1. The trapezia are 11&1, which are also squares.
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wendy
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### Re: Klitzing snubs, multi-snubs

wendy wrote:1). There is no provision for overlaping groups, because overlaping groups do not exist. Especially, were such to happen, then one would alternate the other over the common overlap.

I suppose you mean something like 'xor'. I did consider this. What if neither of the subsets contains the other? Consider sPsQo and oPsQs. Either one can be replaced with sPoQs, but they still overlap. (Essentially, we have two equations: x+y=0(mod2), and y+z=0(mod2); either equation can be added to the other and thus replaced with x+z=0(mod2). Here x,y,z are the numbers of times the three mirrors are used.)
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