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Another question got me thinking. The $600$-cell has $600$ tetrahedral cells, and $120$ vertices which may be viewed as elements of the binary icosahedral group $2I$, a subset of the unit quaternions $S^3$. The dual polychoron is the $120$-cell with $120$ dodecahedral cells and $600$ vertices, which I will call $2D$.

Any two adjacent vertices of the polychora are a minimal positive distance apart (among vertices of the same polychoron), and conversely if the vertices are that distance apart they are adjacent. The normalized average of the vertices of any cell, i.e. that cell's center, yields the vertex of the corresponding dual polychoron. Because of this, multiplying elements of $2D$ by elements of $2I$ should result in elements of $2D$, right? That is, $2I$ acts on $2D$ from the left and the right. This action should be free, which means there must be $5$ orbits.

Each orbit should be a copy of $2I$. Shouldn't this mean $2D$ has five inscribed $2I$s? If this were a fact, I'd expect it to show up in search results, but I don't see it, so I'm wondering if my reasoning is wrong somewhere.

anon
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1 Answers1

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The second nearest neighbour vertices of a chosen dodecahedral vertex form a semiregular hexagon having alternating side lengths of one unit (edge of dodecahedron) and its golden ratio (1.618..: diagonal of pentagon). The edge length of the dodecahedron's vertex-inscribed 5-tetrahedron-compound then would be nothing but the distance of alternating vertices of that hexagon.

This 5-tetrahedron-compound moreover uses the vertices of the dodecahedron just once. And moreover each tetrahedral face (triangle) will be in one-to-one relation to exactly one vertex of the dodecahedron. In fact its center normal does point to it.

Next consider the structure of the 120-cell. Around each edge there are 3 dodecahedra. At a vertex there will be 4. That is any vertex of the 120-cell defines exactly one inscribed triangle underneath this very vertex per incident dodecahedron. All those 4 therefore describe a further tetrahedron around this chosen vertex. Thence those dodecahedron-inscribed tetrahedra all get face connected by those vertex-tetrahedra, and vice versa.

This is how you can construct geometrically your proposed 5-(600-cell)-compound. In fact, it already was known to Coxeter, who described it in his compound notation as {5,3,3}[5{3,3,5}].

--- rk

Dr. Richard Klitzing
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