Let's say we have a tetrahedron labelled as such:
We want to find the number of distinct ways to color the vertices and edges, such that 2 vertices are green, 2 vertices are red, 4 edges are black and 2 edges are white. Colorings are considered the same under rotation, but not reflection.
Just by counting I think there are 9. But I want to know how Polya's enumeration theorem is applied in this case.
In practice we can do the following.
1 Get the cycle index for vertices and edges.
| Rotation | Cycle index vertices | Cycle index edges |
|---|---|---|
| $1\times id$ | $x_1^4$ | $y_1^6$ |
| $4 \times 120°$ | $x_1^1x_3^1$ | $y_3^2$ |
| $4 \times 240°$ | $x_1^1x_3^1$ | $y_3^2$ |
| $3\times 180°$ | $x_2^2$ | $y_1^2y_2^2$ |
Where for $\alpha ^i_j$, $j$ represents the cycle length, and $i$ its multiplicity.
2 Multiply out the rows and add $$ \frac{1}{12} (x_1^4y_1^6 + 8\cdot x_1^1x_3^1y_3^2 + 3\cdot x_2^2y_1^2y_2^2) $$ 3 Substitute in the colors
Let's take $(a+b)$ as the generating function for the colors of vertices and $(c+d)$ for edges. Thus we get. $$ \frac{1}{12} ((a+b)^4(c+d)^6 + 8\cdot (a+b)^1(a^3+b^3)^1(c^3+d^3)^2 + 3\cdot (a^2+b^2)^2(c+d)^2(c^2+d^2)^2) $$
4 Find the coefficient of $a^2b^2c^4d^2$, which is 9
The main problem I have with understanding my solution from a group theoretic perspective are points 2 and 3, that is:
- Why is the cycle index for edges and vertices combined the dot product of the two columns?
- Why can we use a separate generating function for $x$ and $y$?