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I've recently learned about Burnside's Lemma (https://en.wikipedia.org/wiki/Burnside%27s_lemma) and its applications to rotating necklaces, coloring cubes and such, but I fear my understanding of it isn't mature enough and am unable to apply it to the following situation:

Suppose you have a 2x3 matrix, and a set of 3 distinct colors $R, Y, B$. How many non-equivalent ways are there to color the matrix? Note that two matrices $m_1$ and $m_2$ are considered equivalent if you can turn the former into the latter by swapping any rows and/or columns as many times as you want.

So I got started by setting up the Burnside's equation as follows:

\begin{align*}\frac{1}{|G|}\sum_{g \in G} |X^g| &= \frac{1}{H! W!} \sum_{\sigma \in S_H}\sum_{\tau \in S_W} |X^{(\sigma, \tau)}|. \end{align*}

where $|G|$ is the total no. of elements that act to permute the set $X$ of matrices with height and width $H\times W$. Since there are $H!$ ways to permute the rows and $W!$ ways to permute the columns, the total no. of ways to permute the matrix is $|G| = H!W!$. So we can iterate through every way to permute the rows and every way to permute the columns (hence the double summation above), and for each permutation, find out the value of |$X^{(\sigma, \tau)}$| -- which is the number of matrices that are fixed (a.k.a. not changed) when applying permutation $\sigma$ to the rows and $\tau$ to the columns.

For example, say we have a 2x3 grid with the following indices for each cell:

(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)

There would be 2! = 2 ways to permute the rows and 3! = 6 ways to permute the columns, so the Burnside's equation becomes: \begin{align*}\frac{1}{2! 3!} \sum_{\sigma \in S_2}\sum_{\tau \in S_3} |X^{(\sigma, \tau)}| \end{align*}

It is here that I hit a stumbling block. Given we have 3 distinct colors to work with, it isn't clear to me how to count up |$X^{(\sigma, \tau)}$| for each permutation. If someone could show me a step-by-step way to compute the answer for this specific example, I feel I could probably learn to apply it to a more general situation, with arbitrary values of $W$, $H$, and number of colors. Thanks!

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Let me just observe that while rigorous notation is a must for any serious mathematician it can sometimes block the path of the beginning reader. What is happening here is very simple. We need the cycle index of the permutations from $S_2\times S_3$ acting on the slots of the matrix. While the present case can be computed by inspection I will explain the general method.

We require the cycle index $Z(M_{2,3})$ of the pairs $(\sigma,\tau)$ of row and column permutations acting simultaneously on the slots of the matrix ($6$ of them, and with $12$ permutations total). Start from the two cycle indices for the row and column permutations

$$Z(S_2) = \frac{1}{2} a_1^2 + \frac{1}{2} a_2$$

and

$$Z(S_3) =\frac{1}{6} b_1^3 + \frac{1}{2} b_1 b_2 + \frac{1}{3} b_3.$$

The method here is as follows. We draw a diagram of the cycle types of $\sigma$ and $\tau,$ perhaps one beneath the other. Now for the cycle index of the cartesian product of $S_2$ and $S_3$ we must factor the combined action of the two permutations on the slots into cycles. We represent row-column pairs i.e. slots by marking say the row and the column and connecting them with a edge, not to be confused with the directed edges of the two cycles. This edge travels in parallel along the two cycles it is on and returns to its initial position after $\mathrm{lcm}(l_1, l_2)$ steps, where $l_{1,2}$ are the lengths of the cycles. As the pair of cycles contributes $l_1\times l_2$ slot identifiers we get for the contribition to the combined cycle index the term

$$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2/\mathrm{lcm}(l_1, l_2)}.$$

We now do the computation. There are thus six possible combinations of cycle types that combine to form $Z(M_{2,3})$. We process these in turn, leaving the most difficult part for last.

First, combining $a_1^2$ and $b_1^3.$ This fixes all six slots for a contribution of

$$\frac{1}{12} a_1^6.$$

Second, combining $a_1^2$ and $b_3.$ This partitions the pairs into three-cycles for a contribution of

$$\frac{1}{6} a_3^2.$$

Third, combining $a_2$ and $b_1^3.$ Here we have everything on two-cycles to get

$$\frac{1}{12} a_2^3.$$

Fourth, combining $a_2$ and $b_3$. This will produce a six-cycle to get

$$\frac{1}{6} a_6.$$

Now for the tricky part. Fifth, combining $a_1^2$ and $b_1 b_2.$ There are two pairs that are fixed by these, and two pairs on two-cycles and we obtain

$$\frac{1}{4} a_1^2 a_2^2.$$

Finally, sixth, combining $a_2$ and $b_1 b_2.$ We have everything on two-cycles and obtain

$$\frac{1}{4} a_2^3.$$

Adding everything we now have the cycle index

$$Z(M_{2,3}) = \frac{1}{12} a_1^6 + \frac{1}{3} a_2^3 + \frac{1}{6} a_3^2 + \frac{1}{4} a_1^2 a_2^2 + \frac{1}{6} a_6.$$

In order to apply Burnside and ask about colorings with at most $N$ colors we have that the assignment of the colors must be constant on each cycle and we obtain

$$\frac{1}{12} N^6 + \frac{1}{3} N^3 + \frac{1}{6} N^2 + \frac{1}{4} N^4 + \frac{1}{6} N.$$

This yields the sequence

$$M_n = 1, 13, 92, 430, 1505, 4291, 10528, 23052, \ldots$$

which is OEIS A027670. In particular we find there are $92$ colorings using at most three distinct colors. We could apply PET at this point since we have the cycle index. E.g. for three colors we obtain

$$1/12\, \left( R+G+B \right) ^{6} \\ +1/4\, \left( R+G+B \right) ^{2} \left( {B}^{2}+{G}^{2}+{R }^{2} \right) ^{2}+1/6\, \left( {B}^{3}+{G}^{3}+{R}^{3} \right) ^{2} \\ +1/3\, \left( {B}^{2}+ {G}^{2}+{R}^{2} \right) ^{3}+1/6\,{B}^{6} +1/6\,{G}^{6}+1/6\,{R}^{6}$$

which expands to

$${B}^{6}+{B}^{5}G+{B}^{5}R+3\,{B}^{4}{G}^{2}+3\,{B}^{4}GR \\ +3\,{B}^{4}{R}^{2}+3\,{B}^{3}{G}^{3}+6\,{B}^{3}{G}^{2}R \\ +6\,{B}^{3}G{R}^{2}+3\,{B}^{3}{R}^{3}+3\,{B}^{2}{G}^{4} \\ +6\,{B}^{2}{G}^{3}R+11\,{B}^{2}{G}^{2}{R}^{2}+6\,{B}^{2}G{R}^{3} \\ +3\,{B}^{2}{R}^{4}+B{G}^{5}+3\,B{G}^{4}R+6\,B{G}^{3}{R}^{2} \\ +6\,B{G}^{2}{R}^{3}+3\,BG{R}^{4}+B{R}^{5}+{G}^{6}+{G}^{5}R \\ +3\,{G}^{4}{R}^{2}+3\,{G}^{3}{R}^{3}+3\,{G}^{2}{R}^{4} \\+G{R}^{5}+{R}^{6}.$$

The reader might want to verify some of these with pen and paper.

We may also apply inclusion-exclusion to obtain the count of colorings of the matrix with exactly $N$ colors. The nodes of the poset correspond to sets $P\subseteq [N]$ of colors which include colorings that use some subset of these colors, with the top node representing at most $N$ colors, which is the only $P$ that includes colorings with exactly $N$ colors. Colorings with exactly $p \lt N$ colors where $p\ge 1$ are included at all nodes that are supersets of the $p$ colors. We thus obtain for the total weight

$$\sum_{q=p}^N {N-p\choose q-p} (-1)^{N-q} = (-1)^{N-p} \sum_{q=0}^{N-p} {N-p\choose q} (-1)^q = 0$$

since $N-p\ge 1.$ Colorings with less than $N$ colors have total weight of zero in the poset. We thus obtain

$$\sum_{q=1}^N {N\choose q} (-1)^{N-q} M_q.$$

We get a finite sequence since with six slots it is not possible to have a coloring with more than six distinct colors:

$$1, 11, 56, 136, 150, 60$$

The last term represents colorings with exactly six colors. This means all slots in the matrix are distinct. Therefore all orbits have the same size, the number of permutations in the group, which is twelve, and indeed $6!/12 = 60.$ The term for two colors indicates that the two monochrome colorings have been excluded.

This MSE link has the Maple code for the general case.

Addendum. Here is what we mean when we say in the introduction that the cycle index can be computed by inspection. This refers to the isomorphism between $M_{2,3} = S_2\times S_3$ and $D_6$, the dihedral group (reflections and rotations of regular polygons) acting on six slots in this case. The cycle indices $Z(D_p)$ are tabulated and have simple closed forms, consult e.g. Wikipedia. Label the vertices of a hexagon in clockwise order with the labels $(0,0), (1,1), (0,2), (1,0), (0,1)$ and $(1,2).$ Then it is not difficult to see that the rotations of the hexagon are in a bijection with the pairs of cycles from $C_2 \times C_3$ embedded in $M_{2,3}.$ E.g. the rotation that takes the top vertex to its clockwise neighbor corresponds to the two cycles $(0,1)$ and $(0,1,2)$. The reflections in an axis passing through opposite vertices preserve parity (permutation $(0)(1)$ from $S_2$) and fix one element from $0,1,2$ while permuting the other two in two-cycles. The reflections in an axis passing through opposite edges flip parity (permutation $(0,1)$ from $S_2$) and fix one of three elements from $0,1,2$ while exchanging the other two. In this way we have bjectively accounted for all permutations and the proof of the isomorphism is complete.

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  • $\begingroup$ Hi, great explanation, thanks! I have a question. though: how to get from $Z(S_2)$ and $Z(S_3)$ to $Z(M_{2,3})$ other than manually... Unfortunately I can't read Maple code. $\endgroup$ – Vlad Dumitrescu Jan 28 '17 at 16:17
  • $\begingroup$ Oh, I think I got it (it was the "tricky part" that fooled me). $\endgroup$ – Vlad Dumitrescu Jan 28 '17 at 18:56
  • $\begingroup$ Hello again! I think there is an error in the description above. With a slight change in notation: When combining two cycles $a_i^j$ and $a_m^n$, the number of slots are not $ i \times m$, but $i \times j \times m \times n$. So the resulting monome is $a_{lcm(i,m)}^{ijmn/lcm(i,m)}$, which matches the values derived thereafter. Is this correct or am I confused? $\endgroup$ – Vlad Dumitrescu Feb 28 '17 at 20:59
  • $\begingroup$ Yes this is in fact the formula used at the more generic MSE link for this problem in the optimized Maple version, please do consult it. (The first version, which is also available there, processed cycles in pairs of singletons.) I would not call it an error. Let's call it a clarification, which is now documented in the comments. $\endgroup$ – Marko Riedel Feb 28 '17 at 21:10
  • $\begingroup$ I asked to make sure I understood correctly, like I said I'm not really at home with Maple code. Thanks! $\endgroup$ – Vlad Dumitrescu Feb 28 '17 at 22:35

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