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

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Jan 06, 2023 392 likes •531 views

P ermutation patterns and the Mbius funtion Vt Jelnek, Eva Jelnk ov and Eina r Steingrmsson inluding some joint w o rk (in p rogress) with Alex Burstein 2314 123 132 213 231

SLIDE 1 P ermutation patterns and the Mbius fun tion Vt Jelnek, Eva Jelnk

and Eina r Steingrmsson in luding some joint w

(in p rogress) with Alex Burstein

SLIDE 2

· · · 2314 · · ·

123 132 213 231 312 321 12 21 1 The p

p ermutations w.r.t. pattern

ntainment

SLIDE 3 Computing the Mbius fun tion µ(•, y)

1 2

Mbius fun tion is dened b y µ(x, x) = 1 and

µ(x, t) = 0

if x < y 2

SLIDE 4 Some examples 2134 213 123 12

µ(12, 2134) = 1

132 12 21 1

∅ µ(∅, 132) = 0

461532 35142 41532 51432 2413 2431 3142 4132 1432 132

µ(132, 461532) = −2

SLIDE 5 A p ermutation is de omp

sable

if it is the dire t sum

t w

mo re (nonempt y) p ermutations:

1 2 4 6 5

312465 = 312 ⊕ 1 ⊕ 21

3 4 1 2

Inde omp
sable

W e write π = π1 ⊕ π2 ⊕ · · · ⊕ πn

if ea h πi is inde omp

sable

SLIDE 6 Let σ = σ1 ⊕ σ2 ⊕ · · · ⊕ σm and

π = π1 ⊕ π2 ⊕ · · · ⊕ πn

Let π>i = πi+1 ⊕ πi+2 ⊕ · · · ⊕ πn et . fo r

πi, σ>i, σi

First Re urren e: Let ℓ 0 and k 1 b e maximal so that

σ1 = σ2 = · · · = σℓ =

1 and

π1 = π2 = · · · = πk =

1 . Then

µ(σ, π) =

        

if ℓ k − 2

−µ(σk, π>k)

if ℓ = k − 1

µ(σ>k, π>k) − µ(σk, π>k)

if ℓ k Example:

µ(132, 1237564) = 0 µ(132, 126453) = −µ(21, 4231) = −2 µ(132, 13524) = µ(21, 2413)−µ(132, 2413) = 3−(−1) = 4

SLIDE 7 Main Theo rem: Supp

se π1 =

1 . Let k 1 b e maximal so that π1 = π2 = · · · = πk . Then

µ(σ, π) =

µ(σi, π1)µ(σ>i, π>j)

Co rolla ry: If σ = a⊕b and

π = c⊕d

, where c, d = 1, c = d , then

µ(σ, π) = µ(a, c) · µ(b, d)

Co rolla ry: If σ is inde omp

sable

(so m = 1 ), then

µ(σ, π) = µ(σ, π1)

if π = π1 ⊕ π1 ⊕ · · · ⊕ π1

µ(σ, π) = −µ(σ, π1)

if π = π1 ⊕ π1 ⊕ · · · ⊕ π1 ⊕ 1 (π1 = 1)

µ(σ, π) = 0
therwise

SLIDE 8 4 2 3 5 1 7 8 6

Sepa

rable 2 4 1 3

Nonsepa

rable 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 = (3124 ⊖ 1) ⊕ 231 = · · ·

A p ermutation is sepa rable if and

if it avoids the patterns 2413 and 3142. 7

SLIDE 9

− + − − − + + + + + q : p : 1 2 3 1 2 3 4 5 6 7 8 9 10 11 12 13 σ = 123 π = 3, 1, 2, 6, 4, 5, 9, 7, 8, 10, 13, 11, 12

The sepa rating trees

f σ

and π (σ and π sepa rable) 8

SLIDE 10

− + − − − + + + + − + − − − + + + + − + − − − + + + + − + − − − + + + + − + − − − + + + + +1 +1 +1 −1 −1

Unpaired

urren es
f σ = 123

in π 9

SLIDE 11 Theo rem: If σ and π a re sepa rable p ermutations, then

µ(σ, π) =

(

1)parity(X) where the sum is

unpaired

urren es
f σ

in π . Note: This

mputes µ(σ, π)

in p

lynomial

time, although the size

the interval [σ, π] ma y gro w exp

nentially

. Co rolla ry: If π is sepa rable then

|µ(σ, π)| σ(π)

where σ(π) is the numb er

f
urren es
f σ

in π . In pa rti ula r: If π avoids 132 then |µ(σ, π)| σ(π) 10

SLIDE 12 Mo re results

µ(135 . . . 2k
1 2k . . . 42, 135 . . . 2n − 1 2n . . . 42) =

n+k−1

n−k

pa rti ula r, µ(σ, π) is not b

unded
If π

is sepa rable then µ( 1, π) ∈ {0, 1, −1}

If σ

is inde omp

sable

and π = π1 ⊕ 1 ⊕ π2 then µ(σ, π) = 0 11

SLIDE 13 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} ?

w e relate µ(σ, π) to the homology

f [σ, π]

Whi h

intervals [σ, π] a re shellable?

Conje ture: maxπ∈Sn|µ(

1, π)| is unb

unded

as a fun tion