The symmetry of sadi

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

The symmetry of sadi

Postby student91 » Mon Apr 07, 2014 11:06 am

Since I found a way to express sadi (snub demicube/snub 24-cell) in full demicubic symmetry, I've been wondering if it would be possible that sadi might have full demitesseractic symmetry.
First what wikipedia states: wikipedia says that sadi has the following symmetries: [3+,4,3] (i.e. diminished 24-cell symmetry), [(3,3)+,4] (i.e. diminished 16-cell symmetry) and [31,1,1]+ (i.e. chiral demitesseractic symmetry). Now I think that apart from these symmetries, sadi also has full demicubic symmetry.

The normal construction of sadi from demicubic symmetry is as a snub, i.e. you take the 24-cell: o3x3o*b3o. Truncate it: x3x3x*b3x, and finally snub it: s3s3s*b3s.

A funny property of the 24-cell is that it's self-dual. I think this property can be exploited to give it full demitesseractic symmetry.
when you take the 24-cell: o3x3o*b3o, and first dual it, before you apply the alternation, you get the following:

24-cell: o3x3o*b3o

dual 24-cell: ooq3ooo3oqo3*b3qoo&#zx <- still in perfect demitesseractic symmetry, but oriented differently

truncated dual 24-cell: qAooAq3oooooo3AqqAoo*b3ooAqqA&#zx <- also perfect demitesseractic symmetry, although a bit unwieldy ( A=2sqrt(2) )



This buildup of the truncated dual 24-cell is nicely grouped in pairs: You have qA3oo3Aq*b3oo + oo3oo3qA*b3Aq + Aq3oo3oo*b3qA. When you draw these as diagrams, you can see the pairs look like rotors that have their nodes swapped. You can say one of every pair is "left-turning", and the other one is "right turning" The alternation process is done by taking all things that are turning one way, so you get: q3oo3A*b3o + o3o3q*b3A + A3o3o*b3q. This is still in full demitesseractic symmetry. now the relaxation is done by changing every q in an x and every A in an f. then you get the compound I found in the D4.11/D4.12 threat.

Sadi can't get full 16-cell symmetry this way, nor full 24-cell symmetry, but do you think it is possible that it does have full demitesseractic symmetry? :)
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Re: The symmetry of sadi

Postby wendy » Mon Apr 07, 2014 12:27 pm

The symmetry attached to s3s4o3o is as the Wiki says [3+, 4, 3]. I mean, this was run past me.

The symmetry contains a subgroup 3,3,A, of order 172 (dec 192), the reflective cells are formed by dividing the octahedron of {3,4,3} into eight octants. The octahedra actually have pyritohedral symmetry, which is implemented by allowing the octants to individually rotate around the line perpendicular to its face. This triples the order to 496, (dec 576). This is the highest symmetry common to 3,3,5 (496*25=1,0000, dec 576*25=14400), and the {3,4,3} 496*2=972, dec 576*2=1152.

The icosahedral faces of sadi sit inside the octahedral faces described above. There are tetrahedra sitting at the vertices of the {3,4,3}, and the remaining 96 tetrahedra are triangular pyramids, the sloping sides are shared with the icosahedra. In the vertex-figure 'teddi', one has the three pentagons representing the three icosahedra. The triangle by itself represents the apex of the triangular pyramid. The three triangles sharing two sides with the pentagons, make the remaining vertices of the triangular pyramids (at the base), and the bottom triangle, not connected to the pentagons, is a fully symmetric tetrahedron, which forms at the vertex of the {3,4,3}.

The {3,3,4} has sixteen mirrors, twelve of a group 'h', and four belonging to a rectangular group. These give h=APEC, and r give AC. The sadi, has only twelve mirrors, corresponding to the group 'h'. The 24choron has 24 mirrors, two sets of 'h'.
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Re: The symmetry of sadi

Postby student91 » Mon Apr 07, 2014 7:49 pm

wendy wrote:The symmetry attached to s3s4o3o is as the Wiki says [3+, 4, 3]. I mean, this was run past me.

The symmetry contains a subgroup 3,3,A, of order 172 (dec 192), the reflective cells are formed by dividing the octahedron of {3,4,3} into eight octants. The octahedra actually have pyritohedral symmetry, which is implemented by allowing the octants to individually rotate around the line perpendicular to its face. This triples the order to 496, (dec 576). This is the highest symmetry common to 3,3,5 (496*25=1,0000, dec 576*25=14400), and the {3,4,3} 496*2=972, dec 576*2=1152.
[...]
This symmetry is exactly the symmetry my "union" exibits! If i'm right, those reflective cells have three edges with 6 cells around it, and three edges with 4 cells around. Now this reflective cell is exactly the reflective cell of the demitesseract (if I'm right). Furthermore my "union" exists of three parts, that implement the pyritohedral part. So the symmetry on which the union is based, is known (at least to you), and highly corresponds to demitesseractic symmetry (If I'm right, it does correspond to it, right?). The only question I'm having is, why doe wikipedia not mention this symmetry, and how well known is it? Also, is it valid to say sadi has D4 symmetry, or does this imply the demitesseractic reflective cells to be in a specific orientation? (i.e. in an orientation that would only allow for chiral D4 symmetry?)
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Re: The symmetry of sadi

Postby wendy » Wed Apr 09, 2014 8:33 am

Let's see, we have four nodes, the central one is nil, and the other three are a3o3bAc, gives where 'A' = 2q.

a = qA oo Aq
b = Aq qA oo
c = oo Aq qA

It's an interesting riddle, since the figure is actually comprised of six different elements, aganist o3aAbBc.

1, qAo 2, Aqo, 3, oqA, 4 oAq, 5, Aoq, 6, qoA.

Should need to work through this, but i suspect that there is a 3+ segment works as well.
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