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I know that I will ultimately be using the orbit counting theorem involving $$\frac{1}{|G|}\sum_{g\in G}|\mbox{Fix}_A(g)|$$. Where specifically for the cube, the group is $S_4$, $A=\{\mbox{colorings of edges of cube with two colors}\}$ and each $g$ (or $w$ as expressed below) is a cycle type conjugacy class representative of the axis of symmetries of a cube (for example, $(2,2)$ is the cycle type representative of a $90^{\circ}$ rotation about the axis connecting two opposite centers of faces). I know the following about the size of these conjugacy classes: $$\begin{array}{|c|c|c|c|c|c|} \hline \text{conjugacy class $\mathcal{C}$} & (1,1,1,1) & (2,1,1) & (2,2) & (3,1) & (4)\\ \hline |\mathcal{C}| & 1 & \binom{4}{2}=6 & \binom{3}{1}=3 & \binom{4}{1} 2!=8 & 3!=6\\ \hline \end{array} $$ I also know that $|A|=2^{12}$.
What I am primarily having trouble with is understanding which conjugacy class represents which rotation. Further, then struggling to visualize what those rotations are doing for the edges I am trying to fix. I am trying to fill out the following table in order to calculate the number of distinct ways:

$$\begin{array}{|c|c|c|c|c|c|} \hline \text{conjugacy class $\mathcal{C}$} & (1,1,1,1) & (2,1,1) & (2,2) & (3,1) & (4)\\ \hline |\mathrm{Fix}_A(w)| &2^{12} & & 2^2 & & \\ \hline \end{array} $$

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To answer this question we first need the cycle index of the edge permutation group of the cube $E$ by rotations, which we now compute. There is the identity, which contributes $$a_1^{12}.$$ Rotations by $120$ degrees and $240$ degrees about an axis passing through pairs of opposite vertices contribute $$4\times 2 a_3 ^4.$$ Rotations by $90$ and $270$ and $180$ degrees about an axis passing through the center of opposite faces contribute $$3 \times (2 a_4^3 + a_2^6).$$ Finally rotations by $180$ degrees about an axis passing through midpoints of opposite edges contribute $$6\times a_1^2 a_2^5.$$

This gives the cycle index $$Z(E) = \frac{1}{24} \left(a_1^{12} + 8 a_3^4 + 6 a_4^3 + 3 a_2^6 + 6 a_1^2 a_2^5\right).$$

With this cycle index we obtain by Burnside for edge colorings with $n$ colors the formula

$$\frac{1}{24} \left(n^{12} + 8 n^4 + 6 n^3 + 3 n^6 + 6 n^7\right).$$ This gives the sequence $$1, 218, 22815, 703760, 10194250, 90775566, 576941778, 2863870080,\ldots$$ which points us to OEIS A060530, where we find that indeed we have the right cycle index. For two colors we obtain the value

$$\bbox[5px,border:2px solid #00A000]{ 218.}$$

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  • $\begingroup$ I've never seen the $a_1^{12}$ etc. notation. What does it mean? $\endgroup$
    – Sam
    Dec 6 '19 at 20:37
  • $\begingroup$ Uniquely factorize a permutation into disjoint cycles and represent it by a product of variables $a_l$ where $l$ is the length of a contributing cycle. Here is Wikipedia on cycle indices. For a permutation to fix an assignment of colors it must be constant on the cycles (Burnside). $\endgroup$ Dec 6 '19 at 21:12

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