Wythoff Polytope

Higher-dimensional geometry (previously "Polyshapes").

Wythoff Polytope

Postby wendy » Wed Feb 12, 2014 11:56 am

I put together some thoughts on http://hddb.teamikaria.com/wiki/Wythoff_polytope but i don't know enough latex yet to get matrices to work properly. Here's what it's supposed to look like.

Code: Select all
<nowiki>
             (  3    2r2    r2   )                        (  2   -r2    0 )
  A_ij  =    ( 2r2    4     2    )          D_ij  = 1/2   ( -r2   2    -1 )
             (  r2    2     2    )                        (  0    -1    2 )

          Stott Matrix                               Dynkin Matrix
</nowiki>


p.s. It was the change in web address, and not the bad weather that i could not get into the page earlier. It's sad when you can look at a town from the other side of the world, and say "yes, that's Dawlish".
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Re: Wythoff Polytope

Postby quickfur » Thu Feb 20, 2014 1:01 am

Could you give an example of how one calculates the Dynkin matrix of, say, the tetrahedron?
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Re: Wythoff Polytope

Postby wendy » Thu Feb 20, 2014 7:03 am

D_ij is always calculated as 1/2 [2 cos( pi/(ij))], where i and j are vectors normal to the reflective planes.

If you draw two lines representing mirrors at angle pi/n, the graph would be a branch o----n----o . The vectors pointing into the region, and perpendicular to them, would point at the complement angle (ie pi - x), for which the cos is -cos(x). The dot product of the vector and itself is 1. But we multiply the whole thing inside by two, and divide the outside by two, simply to save having to remember a different set of numbers.

So, for the simplex A------B------C, the branches AB and BC give -2cos(60), that is, 1. The branch AC gives -2 cos 90, that is 0. The 2 cos(0) for AA or BB or CC gives 2.

Then one gets
Code: Select all

    [  A -1  0 ]
    [ -1  B -1 ]
    [  0 -1  C ]



And then replace A, B, C with +2, and multiply the lot by 1/2. Note that if you are doing this by machine, you can slip the half into the matrix by dividing it all by 2, so the whole array represents the dot product of unit vectors ȧ<sub>i</sub> ̇· a<sub>j</sub>, where the angles between vectors is taken as the supplement (pi- pi/n) of the branches on the graph. The dot of a vector and itself is 1.
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Re: Wythoff Polytope

Postby wendy » Fri Feb 21, 2014 8:56 am

I rewrote this page under the matrices, to show how to calculate the values in general, and what they mean.
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Re: Wythoff Polytope

Postby quickfur » Fri Feb 21, 2014 5:48 pm

I'm still not 100% clear how one goes from these matrices to the actual coordinates of the uniform polytopes. How do you get from the Stott/Dynkin matrices to, say, the vertices of the icosahedron? I know there's an epacs involved somewhere, whereas for the n-cube polytopes, it's apacs, but how does one tell, given some arbitrary Stott/Dynkin matrix, what permutation operators are needed?
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Re: Wythoff Polytope

Postby wendy » Sat Feb 22, 2014 7:23 am

The whole point behind the stott-matrix is to allow one to access vector-dots without having to convert to an orthogonal system. It's not really intended for generating vertex coordinates.

Coordinate functions like EPACS and APACS and EP+C and EPEC represent the four subgroups of the cubic symmetry.

APACS is the full octahedral group. EPACS is the pyratohedral, APECS is the tetrahedral group, EP&CS is the octahedral rotational group (eg snub cube), and EPECS is the tetrahedral rotation group. Stating a point is (1,0,0) EPACS means that it is in one of 24 reflection-rotary cells of the pyritohedral group. You might use this information to reduce it to (1,0,0), (0,1,0), (0,0,1) ACS, which is the ordinary rectangular group.

Well, you have AI and EI, all icosahedral and even-icosahedral (snub dodecahedron). These reduce to 5 points in EPACS and EPECS. A point off a mirror (ie has no 0 in the coordinate), would reduce to 5 points to EP, and then to 15 points for ECS or ACS to expand to 60 or 120 points.

The dynkin matrix, applied at the AI point, allows one to find the coordinates of the vertex in different mirrors. A point (a,b,c) reflected in the rectangular mirror A would give (-a,b,c), ie (a, b, c)-a(2,0,0). Here, the icosahedral coordinate of a point reflected in mirror A is (a,b,c)-a(2,-f,0).
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