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Which tridental polychora are possible?

Within spherical geometry clearly there is o3o3o *b3o. And moreover any number of those "3" could be replaced by "3/2". This is the demitesseractic symmetry. But the question rises, whether there could be any other tridental graph in spherical geometry. Perhaps one with a leg mark being 5/2, or such the like? (For sure we are looking for non-degenerate tridentals only, i.e. none of the emmanating links from the bifurcation node bears a mark-value 2.) – The answer then will be negative. A proof could run like this:

Consider the set of possible link marks under this restriction of geometry. Possible values are 2, 3, 3/2, 4, 4/2, 4/3, 5, 5/2, 5/3, 5/4 – that's all. (For the matter of research for tridental graphs we further will not distinguish between 4/2 and 2.) Next consider the addition rule for Goursat tetrahedra, reproduced in parts below:

    o               o               o
    r               r1              r2
  p o t     =     p o t     +     x'o y'
  q   s           x   y           q   s
o   u   o       o   u   o       o   u   o

(lines omitted, "o" marking the nodes), where

1/r = 1/r1 + 1/r2
1 = 1/x + 1/x'
1 = 1/y + 1/y'

For tridental outcomes (left hand sides) we have essentially the follwing possibilities:

  1. p = t = u = 2
  2. q = s = u = 2
  3. p = q = r = 2
  4. r = s = t = 2

Cases A and B will ask for further tridental ones for input (right hand side), so will not be useful in this research. For cases C and D we have r = 2, and thus

1/2 = 1/r = 1/r1 + 1/r2, i.e.
r1 = 2 r2 / (r2 - 2)

The first form of this restriction shows that it is symmetrical in r1 and r2. The second one shows that we have r2 > 2 (and thus r1 > 2 as well). – Hence we just have to try any thus allowed value from the above provided link mark set, and to calculate the corresponding value of r1:

r2 = 5/2  →  r1 = 10
r2 = 3    →  r1 =  6
r2 = 4    →  r1 =  4
r2 = 5    →  r1 = 10/3

In here, there is just a single row, where r1 is within that relevant set of possible link marks as well: the single case r1 = r2 = 4. On the other hand, it is known in addition that a link mark 4 and one of the form 5/d cannot be within the same Goursat tetrahedron. Therefore pentachoric (no "4") and hecatonicosachoral ("5/d" exists) symmetries are out of scope from now on.

Finally, as shown explicitly, and being mentioned here, any Goursat tetrahedron of demitesseractic symmetry will be reproduced within tesseractic as well as within icositetrachoric symmetry. Its multiplicity thereby gets doubled resp. multiplied by 6. (Just that some link marks change from being 2 into 4/2.)

Thus it is shown, that the shape of any tridental Goursat tetrahedron is one of those provided within demitesseractic symmetry, that is, it has to be of the form mentioned right within the first paragraph. Q.E.D.

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