# Why do some uniform polyhedra have a “conjugate” partner?

While browsing through a list of uniform polynohedra, I noticed that the square of the circumradius $R_m$ of the small snub icosicosidodecahedron ($U_{32}$) with unit edge lengths is, $$R_{32}^2 =\tfrac1{16}\big (13 + 3\sqrt5 + \sqrt{102 + 46\sqrt5}\big)$$

while that for the small retrosnub icosicosidodecahedron ($U_{72}$) is its real Galois conjugate $$R_{72}^2 =\tfrac1{16}\big (13 + 3\sqrt5 - \sqrt{102 + 46\sqrt5}\big)$$

which are the two roots of, $$\frac{(4R-1)^2}{(3R-1)}=4\phi^2$$

Furthermore, both have the same number of faces, edges, and vertices: $$F = 112,\; E = 180,\; V = 60$$

$U_{32}$ and $U_{72}$ pictured below,

Another pair: the great rhombicuboctahedron, $$R_{11}^2 =\tfrac1{4}\big (13+ 6\sqrt2\big)$$ and great truncated cuboctahedron $$R_{20}^2 =\tfrac1{4}\big (13- 6\sqrt2\big)$$ both with, $$F = 26,\; E = 72,\; V = 48$$

Intrigued, I went through the entire list.

Q: Why is it that if squared circumradius $R_m^2$ of a uniform polyhedra is an algebraic number of even degree $k>1$, then its real Galois conjugate is also a squared circumradius $R_n^2$ of another uniform polyhedra? Furthermore, why do both solids have the same number of faces, edges, and vertices?

There are exactly $75$ non-prismatic uniform polyhedra so it was a finite exercise of checking. It is $23+52=75$, with $k=1$, and $k>1$, respectively.

The observation is true for all $k>1$. But why?

• Because of this observation, I found two erroneous entries in Mathworld (for $U_{53}$ and $U_{61}$). Accurate ones can be found in David McCooey's Visual Polyhedra. – Tito Piezas III Dec 10 '17 at 13:28
• Can they be positioned so that you just apply Galois conjugation to the vertices of one polyhedron to get the other? – Noah Snyder Dec 10 '17 at 14:41
• @NoahSnyder: I've afraid I wouldn't know. Some pairs have the same convex hull, but the pair $U_{32}$ and $U_{72}$ do not. The former has a truncated icosahedron, while the latter has a truncated dodecahedron. – Tito Piezas III Dec 10 '17 at 15:31
• Would it be due to the fact that the isommetry group of those polyhedra are actually Galois groups of some field extension whose considered circumradii are elements? – Sylvain JULIEN Dec 10 '17 at 15:42
• If all the vertices have coordinates in a real field K, then the K-linear combos of them will also lie in K, and that means a dense subset of every edge and every face are in K. So you can just apply conjugation to all those points and get a dense subset of the other polyhedron. – Noah Snyder Dec 10 '17 at 16:34

There is nothing particularly mysterious here. Roughly, fix the lattice (incidence relations of all faces) and the edge lengths. Then there are several realizations of this structure, typically a finite number of them, but not always (see here). The space of realizations is a semi-algebraic set, but the degrees of the polynomials are typically very large, leading to a large number of realizations (cf. Theorem 1.4 and §7.4 in this paper). But when the extra symmetry assumptions are added one would expect the degree of polynomials to become small (like 4 and 2 in your examples), leading to your "conjugate" phenomenon.

Note that the paper linked above also explains why the same holds for the circumradius, etc. Hope this helps.

Igor Pak's answer suggests that conjugate polyhedra in the sense of the question should be different realizations of the same abstract polyhedron (defined by a face lattice); so in this answer I'll list those which are abstractly isomorphic. The two main references for this answer are the paper "Closed-Form Expressions for Uniform Polyhedra and Their Duals" by Peter Messer and the page "Polytopes & their Incidence Matrices" by Richard Klitzing.

### Abstract isomorphism

(In earlier revisions of this answer I attempted to say something about abstract isomorphism from the table of vertex cycle symbols in Messer's paper, but as I wasn't able to get very far with that approach, I've deleted it. The first two comments to this answer refer to that text.)

Klitzing's site seems to be unique in that the subpages indicate which sets of polyhedra are abstractly isomorphic, but it's unfortunately a bit tricky to navigate (I could not find an easy to parse list of just the 75 non-prismatic uniform polytopes; the closest was this). With Wikipedia's list of uniform polyhedra, I managed to find each of the polyhedra on his site by googling their names with the keyword "site:bendwavy.org".

Here's a summary:

The conjugate groupings of uniform polyhedra in Tito Pieza III's list correspond to abstract isomorphism classes of uniform polyhedra except that:

• 13 and 14 in his list form one abstract isomorphism class ($U_{33},U_{61},U_{43},U_{42}$ are all abstractly isomorphic),
• 23 ($U_{46}$) is actually abstractly isomorphic to $U_{64}$,
• $U_{30}$ and $U_{47}$ are abstractly isomorphic, but not on the list of conjugates,
• $U_{62}$ and $U_{65}$ are also abstractly isomorphic and not on the list of conjugates,
• and finally, 22 ($U_{12}$) is in an abstract isomorphism class on its own, as are $U_n$ with $n=1, 2,3,4,5,6,7,8,15,16,36,38,41,44,45,56,59,62,65,75$ (the same list as in his comment on that answer, except that 30, 47, 62, 64, 65 are omitted, per above).

(Here's the text document I compiled while searching Klitzing's site that includes the names of the polyhedra in nonsingleton abstract isomorphism classes and links.)

To get from this to the squared circumradii, the results in the paper cited in Igor Pak's answer state that the squared circumradii of two abstractly isomorphic polyhedra should be roots of the same polynomial (actually we require some generalization of those theorems to non-simplicial polyhedra, which might be straightforward).

One thing to keep in mind is that the polynomial equations satisfied by general polynomial invariants (in the sense of Fedorchuk and Pak's paper) could be reducible. So their results alone are not sufficient to show that circumradii of uniform polyhedra that are abstractly isomorphic are actually Galois conjugate.

Indeed, the fact that the abstract isomorphism classes above don't all correspond to the Galois conjugacy classes found previously bear this out!

This is of course a good conceptual framework to think about what's going on, but a more hands-on approach is also possible:

Messer gives explicit expressions for the circumradii in his paper; in section 8, the circumradius ${}_{0}R$ is expressed as $\csc\phi$, where $\phi$ is the angle subtended by a half-edge when viewed from the center of the circumsphere. In section 9, he gives expressions for $\cot\phi$ for the uniform polyhedra for the first 3 of the 4 "forms" mentioned in Tito Pieza III's answer. Appendix A contains polynomials that define $Y=2-\tan^2\phi$ for the fourth "snub" case, and in fact the table given there places the Wythoff symbols together when their $Y$ satisfy the same polynomial. In each case, it should be possible to check the conjugacy from these expressions (after applying some trigonometric identities and rewriting certain cosines of rational multiples of $\pi$ as roots of polynomials).

• Among the $23$ $U_n$ where the squared circumradius has degree $k=1$, the only pairs with the same number of faces, edges, and vertices are the small and great ditrigonal icosidodecahedron ($U_{30},\, U_{47}$) which you mentioned, and the small and great dodecahemicosahedron ($U_{62},\,U_{65}$). – Tito Piezas III Dec 11 '17 at 11:13
• @TitoPiezasIII thanks! I rewrote parts of my answer since that example showed me that my "numerator" criterion wasn't sufficient. – j.c. Dec 11 '17 at 16:04
• I agree with the Galois group comment. There is actually very little work in this direction, such as this paper turpion.org/php/paper.phtml?journal_id=sm&paper_id=798 – Igor Pak Dec 11 '17 at 19:33

(Not an answer, but a long addendum to the question above to prevent clutter.)

Here are the $2+23+1 = 26\,$ "conjugates": as $2$ singletons, $23$ pairs, and $1$ quadruplet. As requested in the comments, the Wythoff symbol is included. The table is divided into symbols of $$\text{Form}\,1:\; p|q\;r\\ \text{Form}\,2:\; p\;q|r\\ \text{Form}\,3:\; p\;q\;r|\\ \text{Form}\,4:\; |p\;q\;r$$ It turns out that if a uniform polyhedron has Wythoff symbol of $\text{Form}\,n$, then its "conjugate" also has symbol of $\text{Form}\,n$. A convenient list of $U_n$ is here.

\begin{array}{|c|c|rl|} \hline \text{Number}&\text{Polyhedron}&\text{Wythoff}&\!\!\!\text{symbol}\\ \hline 1 & U_{22}\;\text{(icosahedron)} & 5&|\ 2\;3\\ 1 & U_{53} & 5/2&|\ 2\;3\\ 2 & U_{23}\;\text{(dodecahedron)} & 3&|\ 2\;5\\ 2 & U_{52} & 3&|\ 2\;5/2\\ 3 & U_{24} & 2&|\ 3\;5\\ 3 & U_{54} & 2&|\ 5/2\;3\\ 4 & U_{34} & 5&|\ 2\; 5/2\\ 4 & U_{35} & 5/2&|\ 2\; 5\\ \hline 5 & U_{10} & 3\;4&|\ 2\\ 5 & U_{17} & 3/2\;4&|\ 2\\ 6 & U_{27} & 3\;5&|\ 2\\ 6 & U_{67} & 5/3\;3&|\ 2\\ 7 & U_{25} & 2\;5&|\ 3\\ 7 & U_{55} & 2\;5/2&|\ 3\\ 8 & U_{31} & 5/2\;3&|\ 3\\ 8 & U_{48} & 3/2\;5&|\ 3\\ 9 & U_{9} & 2\;3&|\ 4\\ 9 & U_{19} & 2\;3&|\ 4/3\\ 10 & U_{13} & 3/2\;4&|\ 4\\ 10 & U_{14} & 3\;4&|\ 4/3\\ 11 & U_{26} & 2\;3&|\ 5\\ 11 & U_{66} & 2\;3&|\ 5/3\\ 12 & U_{37} & 2\;5/2&|\ 5\\ 12 & U_{58} & 2\;5&|\ 5/3\\ 13 & U_{42} & 3\;5&|\ 5/3\\ 13 & U_{43} & 5/3\;3&|\ 5\\ 14 & U_{33} & 3/2\;5&|\ 5\\ 14 & U_{61} & 5/2\;3&|\ 5/3\\ 15 & U_{49} & 3/2\;3&|\ 5\\ 15 & U_{71} & 3/2\;3&|\ 5/3\\ 16 & U_{51} & 5/4\;5&|\ 5\\ 16 & U_{70} & 5/3\;5/2&|\ 5/3\\ \hline 17 & U_{11} & 2\;3\;4&| \\ 17 & U_{20} & 4/3\;2\;3&| \\ 18 & U_{28} & 2\;3\;5&| \\ 18 & U_{68} & 5/3\;2\;3&| \\ 19 & U_{18} & 3/2\;2\;4&| \\ 19 & U_{21} & 4/3\;3/2\;2&| \\ 20 & U_{50} & 3/2\;3\;5&| \\ 20 & U_{63} & 5/3\;5/2\;3&| \\ 21 & U_{39} & 2\;5/2\;5&| \\ 21 & U_{73} & 3/2\;5/3\;2&| \\ \hline \color{blue}{22} & U_{12} & &|\ 2\;3\;4\\ \color{brown}{23} & U_{46} & &|\ 5/3\;3\;5\\ 24 & U_{32} & &|\ 5/2\;3\;3\\ 24 & U_{72} & &|\ 3/2\;3/2\;5/2\\ 25 & U_{40} & &|\ 2\;5/2\;5\\ 25 & U_{60} & &|\ 5/3\;2\;5\\ \color{green}{26} & U_{29} & &|\ 2\;3\;5\\ \color{green}{26} & U_{57} & &|\ 2\;5/2\;3\\ \color{green}{26} & U_{69} & &|\ 5/3\;2\;3\\ \color{green}{26} & U_{74} & &|\ 3/2\;5/3\;2 \\ \hline \end{array}

Notes :

1. All these $52$ $U_n$ have $R_n^2$ that is an algebraic number of degree $k>1$. But the $U_n$ with $|p\;q\;r$ have degree $k\geq3$.
2. The special case of $R_{12}^2$ and $R_{46}^2$ (for the snub cube and snub icosidodecadodecahedron) use cubics with one real root, namely the well-known tribonacci constant and plastic constant.
3. Two pairs of $R_n^2$ are quartic roots.
4. Four $R_n^2$ use the same sextic with four real roots.
• The $23$ $U_n$ with $k=1$ and hence do not appear on the table above are, $$n=1, 2, 3, 4, 5, 6, 7, 8, 15, 16, 30, 36, 38, 41, 44, 45, 47, 56, 59, 62, 64, 65, 75$$ – Tito Piezas III Dec 11 '17 at 5:08
• Thanks! What does the "Number" in the leftmost column correspond to? – j.c. Dec 11 '17 at 5:17
• @j.c.: Its conjugate number. – Tito Piezas III Dec 11 '17 at 5:18
• I guess what I mean to ask is does the conjugate number have any intrinsic meaning, or is it just some arbitrary ordering? – j.c. Dec 11 '17 at 5:20
• @j.c.: Other than binding conjugates together, the ordering is arbitrary. :) – Tito Piezas III Dec 11 '17 at 5:37

The best way to describe Wythoffian polytopes is to use the decorated Coxeter-Dynkin graph. Then consider the being used link marks.

In the context of this question the conjugacy just comes down to replacements of
4 <-> 4/3,
5 <-> 5/3, and
5/2 <-> 5/4.

--- rk