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Pólya enumeration counts edge-set orbits on the 2-subsets of [n]

Statement

Let En=[n]2 be the set of two-element subsets of [n], and let Sn act on En by

σ⋅{i,j}:={σ(i),σ(j)}.

If

Fn(x):=∑Ox∣A∣,

where the sum runs over the Sn-orbits O of edge-sets A⊆En and, for each orbit O, A denotes any member of O (so ∣A∣ is well defined), then

Fn(x)=ZSn↷En(1+x,1+x2,…,1+x∣En∣).

Facts & Assumptions

Given: an integer n≥0.

[F1]

A subset A⊆En is exactly a 0-1 colouring of the pair set En: colour a pair black when it lies in A and white otherwise.

[L1]
[L2]

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

technique · direct
1.1F1L1

By [F1], edge-sets on [n] are exactly colourings of the finite set En by the two colours white and black. The induced action of Sn on those colourings is exactly the relabelling action on edge-sets.

2.1step 1.1algebra

Give white weight 1 and black weight x. Then the weight of a colouring corresponding to A⊆En is precisely x∣A∣. The d-th power sum of the two colour weights is therefore 1d+xd=1+xd.

3.1step 2.1L2∎

Apply [L2] to the action of Sn on the colourings of En. By step 2.1, the resulting pattern inventory is exactly ZSn↷En(1+x,1+x2,…,1+x∣En∣), and by the definition of weight it is also exactly ∑Ox∣A∣=Fn(x), where for each orbit O the symbol A denotes any member of O.

Depends on

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Sources