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I would like to understand commutation classes of reduced expressions of the longest element in $S_5$ a little better. For this, it makes sense to look at the permutohedron of order 5. Since I am only interested in commutation classes of reduced expressions, the square faces in this 4-polytope are kind of superfluous information (just as for $S_4$ it is enough to think of the octahedron instead of the truncated octahedron).

So my question is, what kind of 4-polytope would I get if I contract all 90 square faces to a single vertex? Is this procedure even well-defined? If so, would the hexagon faces turn all into triangles?

If I should guess, my answer would be the rectified 5-cell, but I don't have much evidence for this.

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    $\begingroup$ For the next permutohedron, the squares are no longer facets, so one cannot just remove the squares by removing the associated inequalities defining the polytope, as was the case in dimension 3. $\endgroup$
    – F. C.
    Commented May 19, 2020 at 16:49
  • $\begingroup$ @F.C. Oh, I see. So, there is no 4-polytope analogue of the polyhedron of degree 5, s.t. "commutative squares" like {12345,21345,12435,21435} are contracted? $\endgroup$
    – Bipolar Minds
    Commented May 19, 2020 at 17:24
  • $\begingroup$ Maybe this helps: the 4-dimensional permutahedron is also known as the omnitruncated 4-simplex which belongs to the class of $A_4$-polytopes. Go trough the list of these polytopes, maybe one of these fits your purpose. The omnitruncated one is the "most truncated version" of all of these, so Yes, you can "untruncate" it in some sense. $\endgroup$
    – M. Winter
    Commented May 27, 2020 at 20:12

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