Polya

Orbits, Burnside Lemma
Suppose a group G act by permutations on a set A. The subset Ga of M is called the orbit of a. There are three natural properties related directly to the definition of a group.
 * Reflexivity: $$ a \in Ga $$
 * Symmetry: if $$ b \in Ga $$, then $$ a \in Gb $$
 * Transitivity: " if $$ a \in Gb $$ and $$ b \in Gc $$ then $$ a \in Gc $$

These three properties ensure that M is a distinct reunion of orbit. Orbits will have different names in distinct context, like would be equivalence classes, types, patterns, etc.

Take the action of a group G on block of transitivity (orbit)

There are $$degree \times order= 3 \times 6 =18 $$ values in these tables.

Among them, exactly $$ \ order = 6 \ $$ of them are "fixing points" such that $$ga=a$$. This happens because the same number of permutations transports a given $$a$$ to any given $$b$$.

The number of orbits is then $$ \# \ of \ fixing \ points \over order $$

Colored permutations
Take an uncolored permutation, e.g. $$\sigma = (A)(B)(C)(D,E,F)$$. Suppose we can color permutations i.e. elements in the same cycle have the same color.

Then, for each colored $$\sigma_c$$, e.g. $$ \sigma_c = \color {red} (A) \color {green} (B) \color {blue} (C) \color {green} (D, E, F)$$ we associate  a colored partition $$ |\ \color {red}   A}\ |\ {\color {green}  B, D, E, F }\ |\ {\color {blue} C\ | $$  that is fixed by $$\sigma = (A)(B)(C)(D,E,F)$$.

Different colorings of $$\sigma = (A)(B)(C)(D,E,F)$$ produces different colored partitions.

Reversely, given a colored partition eg. $$ |\ \color {red}   A,B\ |\ \color {green}  C, D, E, F \ |$$  we observe that, to be fixed by fixed by $$\sigma$$, the color must be constant over a cycle.

Thus, there is a coloring $$\sigma_c = \color {red} (A)(B) \color {green}(C)(D,E,F)$$ that produces the given partition.

In conclusion, there is a one-to-one correspondence between colorings of a permutation and colored partitions fixed by that permutation.

Take now $$ \sum \ of \ all \ colored \ permutations \ like \ would \ be \ \color {red}c_1 \color {green} c_1 \color {blue}c_1\color {green}c_3 $$ and re-groupe it like

$$ \sum of \ colored  \ permutations \ like \ ( \color{red}c_1 \color{black} + \color{green} c_1 \color{black} + \color {blue} c_1 \color{black} ) \cdot ( \color{red}c_1 \color{black} + \color{green} c_1 \color{black} + \color {blue} c_1 \color{black} ) \cdot ( \color{red}c_1 \color{black} + \color{green} c_1 \color{black} + \color {blue} c_1 \color{black} ) \cdot ( \color{red}c_3 \color{black} + \color{green} c_3 \color{black} + \color {blue} c_3 \color{black} ) $$

We build now a generating polynomial by formally replacing in the above expression colored cycles with "counters"

$$ \sum of \ factors \ like \ ( r + g  +  b ) \cdot ( r + g  +  b ) \cdot ( r + g +  b ) \cdot ( r^3 + g^3  +  b^3 ) $$

the exponents of the "counters" r, g, b are not counting only the length of cycles but also indicate the distribution of colors on the colored cube.

For our cube, the generating polynomial has 28 terms. Suppose we want to know the number of colorings depending on one color - the other not being specified. We take $$g=b=1$$ (the so called zero weight) and we obtain :

$$r^6 + 2r^5 + 6r^4 + 10r^3 + 16r^2 + 12r + 10$$

10 stands here for 10 different colorings using only g and b.

Double permutations
we wonder how many double permutations stabilize

$$ \begin{bmatrix} \color {red} R \\ 1 \color {black} \end{bmatrix} \begin{bmatrix} \color {green} G \\ 2 \color {black} \end{bmatrix} \begin{bmatrix} \color {blue} B \\ 3 \color {black} \end{bmatrix} \begin{bmatrix} 4, 5, 6     \end{bmatrix}                       $$ e.g. $$ \sigma_c =  { \color {red} R \choose \color {black} 1 }{ \color {green} G \choose \color {black} 2} { \color {blue} B \choose \color {black} 3 }( 4, 5, 6 ) $$

For this to happen, the cycle structure of R,G,B must be the same as 1,2,3. To obtain synchronicity, cycles must be mapped onto cycles of the same size.

Let's count how many 3-cycles are in a cube.

$${ c_1^6 + 6 c_1^2 c_4 + 3 c_1^2 c_2^2 + 8 c_3^2 + 6 c_2^3 \over b_1^3 + b_3 + b_3 } = { c_1^6 + 8 c_3^2 \over b_1^3 + b_3 + b_3 } = { c_1c_1c_1c_1c_1c_1 \over b_1b_1b_1} + { 8 c_3^2  \over b_3 + b_3 } $$

$$= 6.5.4 + 8.2 { c_3c_3 \over b_3 } = 120 + 8.2.2 {c_3 \over b_3 }$$

to evaluate b3/c3 we have to observe that a double cycle $$ \begin{bmatrix} ( \color {red} R, \color {green} G, \color {blue} B  ) \\ ( 1,2,3 )     \end{bmatrix}               $$ stabilizes three one-to-one functions

$$ \begin{bmatrix} \color {red} R \\ 1 \color {black} \end{bmatrix} \begin{bmatrix} \color {green} G \\ 2 \color {black} \end{bmatrix} \begin{bmatrix} \color {blue} B \\ 3 \color {black} \end{bmatrix}  $$ and $$ \begin{bmatrix} \color {red} R \\ 2 \color {black} \end{bmatrix} \begin{bmatrix} \color {green} G \\ 3 \color {black} \end{bmatrix} \begin{bmatrix} \color {blue} B \\ 1 \color {black} \end{bmatrix}  $$  and $$ \begin{bmatrix} \color {red} R \\ 3 \color {black} \end{bmatrix} \begin{bmatrix} \color {green} G \\ 1 \color {black} \end{bmatrix} \begin{bmatrix} \color {blue} B \\ 2 \color {black} \end{bmatrix}  $$

Eventually we obtain 216 fixed points for 3.24 functions i.e. 3 orbits.

To establish a formula we remark that a k-cycle $$c_k$$ contributes to a product either with $$\ 1 \ $$ or with $$\ kb_k \ $$, so $$ \ c_k = 1+kb_k \ $$

Eventually we get this nice Bruijn page166

$$P_G \left( \frac {\partial}{\partial x_1}, \frac {\partial}{\partial x_2}, \frac {\partial}{\partial x_2}, \cdots  \right) P_H ( 1 + z_1, 1+ 2z_2, 1 + 3z_3, \cdots) $$ evaluated at

$$ z_1 = z_2 = z_3 = \cdots = 0 $$

Coloring cube sides and vertices - Stirling - exact # of colors
https://math.stackexchange.com/questions/2828924/coloring-cube-sides-and-vertices

$$H_X(N, M) = \frac{N! \times M!}{24} $$ $$ \times \left( \left\{\begin{array}{ll}\ 6 \\ N \end{array} \right\} \left\{\begin{array}{ll}\ 8 \\ M \end{array} \right\} + 6 \left\{\begin{array}{ll}\ 3 \\ N \end{array} \right\} \left\{\begin{array}{ll}\ 2 \\ M \end{array} \right\} + 3 \left\{\begin{array}{ll}\ 4 \\ N \end{array} \right\} \left\{\begin{array}{ll}\ 4 \\ M \end{array} \right\} + 8 \left\{\begin{array}{ll}\ 2 \\ N \end{array} \right\} \left\{\begin{array}{ll}\ 4 \\ M \end{array} \right\} + 6 \left\{\begin{array}{ll}\ 3 \\ N \end{array} \right\} \left\{\begin{array}{ll}\ 4 \\ M \end{array} \right\} \right) $$

$$ \begin {Bmatrix} 6 \\ N \end {Bmatrix} \begin {Bmatrix} 8 \\ M \end {Bmatrix} + 6 \begin {Bmatrix} 3 \\ N \end {Bmatrix} \begin {Bmatrix} 2 \\ M \end {Bmatrix}

$$

The cube

 * $$Z_C(t_1,t_2,t_3,t_4) = \frac{1}{24}\left(t_1^6 + 6 t_1^2 t_4 + 3 t_1^2 t_2^2 + 8 t_3^2 + 6 t_2^3\right)$$



{Cube \over { \{ X,X,X \} } } = a_1{a_1^2 + a_2 \over 2 }. b_1{b_1^2 + b_2 \over 2 } + {a_1^3 + 2 a_3 \over 3}.{a_1^3 +  2 a_3 \over 3} $$