P ermutation patterns and the Mbius funtion Vt Jelnek, Eva - PowerPoint PPT Presentation

P ermutation patterns and the M�bius fun tion V�t Jel�nek, Eva Jel�nk ov� and Eina r Steingr�msson in luding some joint w o rk (in p rogress) with Alex Burstein

· · · · · · 2314 123 132 213 231 312 321 12 21 1 The p oset of p ermutations w.r.t. pattern ontainment 1

Computing the M�bius fun tion µ ( • , y ) -1 2 1 0 • -1 -1 -1 • • • 1 • • • The M�bius fun tion is de�ned b y µ ( x, x ) = 1 and • • if x < y 2 � µ ( x, t ) = 0 x � t � y

Some examples 461532 132 2134 35142 41532 51432 12 21 213 123 2413 2431 3142 4132 1432 1 12 132 ∅ µ (12 , 2134) = 1 µ ( ∅ , 132) = 0 µ (132 , 461532) = − 2 3

A p ermutation is de omp osable if it is the dire t sum of t w o o r mo re (nonempt y) p ermutations: • • • • 3 1 2 4 6 5 3 4 1 2 • • • • Inde omp osable • • W e write π = π 1 ⊕ π 2 ⊕ · · · ⊕ π n only if ea h π i is inde omp osable 312465 = 312 ⊕ 1 ⊕ 21 4

Let σ = σ 1 ⊕ σ 2 ⊕ · · · ⊕ σ m and Let π >i = π i +1 ⊕ π i +2 ⊕ · · · ⊕ π n et . fo r π = π 1 ⊕ π 2 ⊕ · · · ⊕ π n First Re urren e: Let ℓ � 0 and k � 1 b e maximal so that π � i , σ >i , σ � i 1 and 1 . Then if ℓ � k − 2 if ℓ = k − 1 σ 1 = σ 2 = · · · = σ ℓ = π 1 = π 2 = · · · = π k = if ℓ � k  0     µ ( σ, π ) = − µ ( σ � k , π >k ) Example:    µ ( σ >k , π >k ) − µ ( σ � k , π >k )  µ (132 , 1237564) = 0 5 µ (132 , 126453) = − µ (21 , 4231) = − 2 µ (132 , 13524) = µ (21 , 2413) − µ (132 , 2413) = 3 − ( − 1) = 4

Main Theo rem: Supp ose π 1 � = 1 . Let k � 1 b e maximal so that π 1 = π 2 = · · · = π k . Then m k Co rolla ry: If σ = a ⊕ b and , where c, d � = 1 , c � = d , � � µ ( σ, π ) = µ ( σ � i , π 1 ) µ ( σ >i , π >j ) then i =1 j =1 Co rolla ry: If σ is inde omp osable (so m = 1 ), then π = c ⊕ d if π = π 1 ⊕ π 1 ⊕ · · · ⊕ π 1 µ ( σ, π ) = µ ( a, c ) · µ ( b, d ) if π = π 1 ⊕ π 1 ⊕ · · · ⊕ π 1 ⊕ 1 ( π 1 � = 1) otherwise • µ ( σ, π ) = µ ( σ, π 1 ) 6 • µ ( σ, π ) = − µ ( σ, π 1 ) • µ ( σ, π ) = 0

A p ermutation is sepa rable if it an b e generated from 1 b y dire t sums and sk ew sums . • • • • • 42351786 = 42351 ⊕ 231 = 4 2 3 5 1 7 8 6 • (3124 ⊖ 1) ⊕ 231 = · · · Sepa rable • A p ermutation is sepa rable if • and only if it avoids the pat- terns 2413 and 3142. • 2 4 1 3 • Nonsepa rable 7 • •

+ + q : p : − − − − 10 + + + + 1 2 3 3 6 9 13 σ = 123 1 2 4 5 7 8 11 12 The sepa rating trees of σ and π π = 3 , 1 , 2 , 6 , 4 , 5 , 9 , 7 , 8 , 10 , 13 , 11 , 12 ( σ and π sepa rable) 8

+ + +1 +1 − − − − − − − − + + + + + + + + + + +1 − 1 − − − − − − − − + + + + + + + + Unpaired o urren es of σ = 123 in π + − 1 − − − − 9 + + + +

Theo rem: If σ and π a re sepa rable p ermutations, then � 1) parity(X) where the sum is over unpaired o urren es of σ in π . Note: This omputes µ ( σ, π ) in p olynomial time, although � µ ( σ, π ) = ( the size of the interval [ σ, π ] ma y gro w exp onentially . X Co rolla ry: If π is sepa rable then where σ ( π ) is the numb er of o urren es of σ in π . In pa rti ula r: If π avoids 132 then | µ ( σ, π ) | � σ ( π ) | µ ( σ, π ) | � σ ( π ) 10

Mo re results - 1 2 k . . . 42 , 135 . . . 2 n − 1 2 n . . . 42) = In pa rti ula r, µ ( σ, π ) is not b ounded � n + k − 1 If π is sepa rable then µ ( 1 , π ) ∈ { 0 , 1 , − 1 } � • µ (135 . . . 2 k n − k If σ is inde omp osable and π = π 1 ⊕ 1 ⊕ π 2 then µ ( σ, π ) = 0 • • • 11

Op en p roblems: When is µ ( σ, π ) = 0 ? When is | µ ( σ, π ) | = σ ( π ) ? What is max { µ ( 1 , π ) : | π | = n } ? • Can w e relate µ ( σ, π ) to the homology of [ σ, π ] ? • Whi h intervals [ σ, π ] a re shellable? • Conje ture: max π ∈ S n | µ ( 1 , π ) | is unb ounded as a fun tion of n • • • 12