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Under Foata's fundamental transformation, anti-excedances become descents

Statement

Let σ∈Sn and let w=F(σ) be Foata's fundamental transformation. Then a position j is a descent of w if and only if the entry w(j) is an anti-excedance of σ. In particular,

des⁡(w)=∣AExc⁡(σ)∣.

Facts & Assumptions

Given: A permutation σ∈Sn, its standard cycle form

(a1,0 a1,1 … a1,r1−1)⋯(as,0 as,1 … as,rs−1),

with each first entry at,0 the largest in its cycle and a1,0<⋯<as,0, and the one-line word w=F(σ) obtained by deleting the parentheses.

Proof

technique · direct
1.1given

For every cycle and every 0≤j<rt−1, the cycle notation means σ(at,j)=at,j+1. Therefore at,j is an anti-excedance of σ exactly when at,j+1<at,j, which is exactly the condition that the adjacent pair (at,j,at,j+1) contributes a descent in the stripped word w.

1.2given

Across cycle boundaries, the stripped word has no descent: the last entry of cycle t is followed by the first entry at+1,0 of cycle t+1, and the standard cycle form orders these first entries increasingly, so at,rt−1<at+1,0.

2.1step 1.1given

The last entry at,rt−1 of a cycle is never an anti-excedance, because σ(at,rt−1)=at,0 and at,0 is the largest element of the cycle.

3.1step 1.1step 2.1step 1.2∎

By steps 1.1, 2.1 and 1.2, the descents of w occur exactly at the entries of w that are anti-excedances of σ. Counting them gives des⁡(w)=∣AExc⁡(σ)∣.

Depends on

Used by

Dependency tree · two levels

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Sources