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Pólya enumeration counts edge-set orbits on the 2-subsets of [n]
Statement
Let be the set of two-element subsets of , and let act on by
If
where the sum runs over the -orbits of edge-sets and, for each orbit , denotes any member of (so is well defined), then
Facts & Assumptions
Given: an integer .
A subset is exactly a - colouring of the pair set : colour a pair black when it lies in and white otherwise.
The pair set is finite, with (A finite set with elements has exactly two-element subsets, and ).
Weighted Pólya enumeration evaluates the orbit inventory at the power sums of the colour weights (The weighted pattern inventory is the cycle index evaluated at the power sums).
Proof
By [F1], edge-sets on are exactly colourings of the finite set by the two colours white and black. The induced action of on those colourings is exactly the relabelling action on edge-sets.
Give white weight and black weight . Then the weight of a colouring corresponding to is precisely . The -th power sum of the two colour weights is therefore .
Apply [L2] to the action of on the colourings of . By step 2.1, the resulting pattern inventory is exactly , and by the definition of weight it is also exactly , where for each orbit the symbol denotes any member of .
Depends on
Used by
Dependency tree · two levels
19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Applied Combinatorics, Section 15.5: Applications of Pólya's Enumeration Formula (standard reference, not scraped)