polychoronlover wrote:I noticed that the IncMats website says that o3o3x3x3/2 *a and o3x3x3o3/2 *a lead to 2 superimposed copies of the facetorectified tesseract, firt. However, due to the possibility of demitesseractic symmetry, and the fact that the vertex figure can reduce its symmetry from trigonal prismatic to trigonal pyramidal (see sto and gotto), it appears that o3o3x3x3/2 *a and o3x3x3o3/2 *a form firt, not 2firt.
Confer https://bendwavy.org/klitzing/explain/p ... #xpxqorosa :
Let P=n/d, Q=m/b<>2, R=k/c<>2, S=s/e<>2 and Q<S, then we have as general incidence matrix for xPxQoRoS*a :
- Code: Select all
.... | G/2k | k k | 2k k k | k k 1 1
-----+---------+---------------+-------------------------+----------------------------
x... | 2 | G/4 * | 2 2 0 | 1 2 1 0
.x.. | 2 | * G/4 | 2 0 2 | 2 1 0 1
-----+---------+---------------+-------------------------+----------------------------
xx.. | 2n | n n | G/2n * * | 1 1 0 0
x..o | s | s 0 | * G/2s * | 0 1 1 0
.xo. | m | 0 m | * * G/2m | 1 0 0 1
-----+---------+---------------+-------------------------+----------------------------
xxo. | g(3)/2 | g(3)/4 g(3)/2 | g(3)/2n 0 g(3)/2m | G/g(3) * * *
xx.o | g(2)/2 | g(2)/2 g(2)/4 | g(2)/2n g(2)/2s 0 | * G/g(2) * *
x.oo | g(1)/2k | g(1)/4 0 | 0 g(1)/2s 0 | * * G/g(1) *
.xoo | g(0)/2k | 0 g(0)/4 | 0 0 g(0)/2m | * * * G/g(0)
and the vertex figure would be : x(Q)oRox(S)&#x(2P) = x(Q)Ro || oRx(S)
(where x(Q) means an edge, the size of which is the chord length of the Q-gon; etc.)
Your case then is :
P=3 Q=3/2 R=3 S=3 -> n=3 m=3 k=3 s=3 g(0)=24 µ(0)=3 g(1)=24 µ(1)=1 g(2)=24 µ(2)=1 g(3)=24 µ(3)=3 G=384 µ=8
cells are tet and tut.
And as firt itself has for incidence matrix :
- Code: Select all
32 | 6 | 6 6 | 2 6
---+----+-------+------
2 | 96 | 2 2 | 1 3
---+----+-------+------
3 | 3 | 64 * | 1 1
6 | 6 | * 32 | 0 2
---+----+-------+------
4 | 6 | 4 0 | 16 *
12 | 18 | 4 4 | * 16
it becomes obvious that x3x3/2o3o3*a ought be its double cover.
Same for the symmetrical case, cf. https://bendwavy.org/klitzing/explain/p ... #xpxqoroqa ,
i.e. with P=n/d, Q=m/b<>2, R=k/c<>2 we get for xPxQoRoQ*a :
- Code: Select all
.... | G/2k | 2k | 2k 2k | 2k 2
-------+---------+---------+-----------------+----------------
x... & | 2 | G/2 | 2 2 | 3 1
-------+---------+---------+-----------------+----------------
xx.. | 2n | 2n | G/2n * | 2 0
.xo. & | m | m | * G/m | 1 1
-------+---------+---------+-----------------+----------------
xxo. & | g(2)/2 | 3g(2)/4 | g(2)/2n g(2)/2m | 2G/g(2) *
.xoo & | g(0)/2k | g(0)/4 | 0 g(0)/2m | * 2G/g(0)
where g(0)=g(1) and g(2)=g(3). And the verf becomes x(Q)oRox(Q)&#x(2P) = x(Q)Ro || oRx(Q)
Here your case is
P=3 Q=3 R=3/2 -> n=3 m=3 k=3 g(0)=24 µ(0)=3 g(2)=24 µ(2)=1 G=384 µ=8
but also
P=3 Q=3/2 R=3/2 -> n=3 m=3 k=3 g(0)=24 µ(0)=5 g(2)=24 µ(2)=2 G=384 µ=40
would produce the same one. I.e. x3x3o3/2o3*a = x3x3/2o3/2o3/2*a, both again the double cover of firt.
--- rk