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The two quaternions $\omega={1\over 2}(-1,1,1,1)$ and $q={1\over 4}(0,2,\sqrt{5}+1,\sqrt{5}-1)$ generate a finite group under multiplication with 120 elements that form the vertices of a 600 cell, when considered as vectors of ${\mathbb R}^4$.

I tried to find a similar pair of quaternions that generate the dual polytope, the 120 cell. Since it is not mentioned on the wiki-page on quaternions and I couldn't find it somewhere else, I assume that it is not possible to find such a pair of quaternionic generators?

Is there an easy explanation, why it works for the 600 cell and not for the 120 cell?

p6majo
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  • How do they generate a finite group? What is the group operation here? – Jaap Scherphuis May 16 '22 at 11:42
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    Good remark! Quaternion multiplication, I'll adjust the question. – p6majo May 16 '22 at 11:54
  • Has it something to do with reflections, that are not provided by the quaternions. In other words, maybe subgroup of B4 of order 120 that is represented by the 600cell is purely rotationally generated. All determinants are 1, where as the subgroup of order 600 is generated by at least one reflection, that cannot be represented by a unit quaternion. – p6majo May 16 '22 at 17:22
  • The finite subgroups of $SO(3)$, and hence $S^3$, are classified. This includes the 600-cell, aka the binary icosahedral group $2I$, but not the 120-cell (which I will call $2D$ for the moment). What I am curious about: $2I$ ought to act on $2D$ by left- (or right-, or bi-) multiplication, which would seem to have $5$ orbits, but that would mean the 120-cell has 5 inscribed 600-cells. But no search results seem to verify that, which I would expect if it were true. Dunno what I'm misunderstanding. – anon May 26 '22 at 22:57
  • I converted the vertices of the 120-cell into quaternions and checked, whether they form a finite group. But they didn't. This procedure worked nicely for the 600-cell. – p6majo May 30 '22 at 18:24

1 Answers1

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Summarizing the discussion in the comments.

The binary icosahedral group $2I$ in the quaternions form a 600-cell with 120 vertices. The dual polychoron, the 120-cell with 600 vertices, is not a group, so it doesn't make sense to talk about generators.

What is true, though, is that there is a free action of $2I$ on it by left- (or right- or two-sided) multiplication which is free with five orbits, realizing the 120-cell as a compound of 5 inscribed 600-cells (see my question about this).

anon
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  • I will accept it as an answer, though it's a bit unsatisfactory just to say that it is not a group, since it's not among the classified finite sub-groups. Tbh, I don't know anything about $2I$. So far I only worked with the coxH4 group with its 14400 elements in its matrix representation. Depending on the starting vector, one can generate the 600cell or the 120cell or any other polytope of the family. Short, I don't have any intuition, why the 600cell is a group and the other one is not. – p6majo Jun 02 '22 at 17:53
  • @p6majo The finite subgroups of the quaternions are in 1-to-1 relation with the finite groups of rotations in 3D. If the vertices of the 120-cell were to form a group, then this group would consist of 600 elements. The corresponding group in 3D would then have 300 elements. But there is no such group in 3D as you might know. – M. Winter Jun 25 '22 at 23:48