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Composing with a transposition reverses $(- 1)^{inv (σ)}$

Statement

Let $σ \in S_{n}$ and let $τ$ be a transposition. Then

$(- 1)^{inv (σ τ)} = - (- 1)^{inv (σ)}, (- 1)^{inv (τ σ)} = - (- 1)^{inv (σ)} .$

Thus composing on either side with a transposition reverses inversion sign.

Facts & Assumptions

Given: A natural $n$ , a permutation $σ \in S_{n}$ , and a transposition $τ$ ; composition acts from right to left.

[L1]

The inversion number counts pairs $i < j$ whose values occur in decreasing order, and inversion sign is $(- 1)$ raised to that number (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

Proof

technique · direct

1.1

If $s_{a} = (a a + 1)$ is an adjacent transposition, right composition by $s_{a}$ swaps the values of $σ$ in positions $a$ and $a + 1$ . Their mutual pair toggles its inversion status, while for every third position the two affected pairs merely exchange their total contribution. Hence the inversion number changes by an odd number and the inversion sign is negated.

givenL1

2.1

For $a < b$ , the transposition $(a b)$ equals $s_{a} s_{a + 1} \dots s_{b - 2} s_{b - 1} s_{b - 2} \dots s_{a + 1} s_{a}$ , a product of $2 (b - a) - 1$ adjacent transpositions.

step 1.1L1

3.1

Repeatedly applying step 1.1 along the odd-length product in step 2.1 gives the first formula. For the second, $τ σ = σ (σ^{- 1} τ σ)$ and $σ^{- 1} τ σ$ is the transposition obtained by applying $σ^{- 1}$ to the two moved points, so the first formula applied on the right gives the second.

step 2.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Stanford Math 51H, Permutations (standard reference, not scraped)

Composing with a transposition reverses (−1)inv⁡(σ)(−1)inv(σ)