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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):=OxA,

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

Fn(x)=ZSnEn(1+x,1+x2,,1+xEn).

Facts & Assumptions

Given: an integer n0.

[F1]

A subset AEn 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
1.1

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.

F1L1
2.1

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

step 1.1algebra
3.1

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

step 2.1L2

Depends on

Used by

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