Permutations, sequences, and partially ordered sets

Sergey KitaevReykjavik University

Joint work with

Mireille Bousquet-Mélou Anders Claesson Mark Dukes

Overview of results

Bijections (respecting several statistics) between the following objects

unlabeled (2+2)-free posets on n elements

pattern-avoiding permutations of length n

ascent sequences of length n

linearized chord diagrams with n chords = certain involutions

Closed form for the generating function for these classes of objects

Pudwell’s conjecture (on permutations avoiding 31524) is settled using modified ascent sequences

_ _

Ascent sequencesNumber of ascents in a word: asc(0, 0, 2, 1, 1, 0, 3, 1, 2, 3) = 4

(0,0,2,1,1,0,3,1,2,3) is not an ascent sequence, whereas (0,0,1,0,1,3,0) is.

Unlabeled (2+2)-free posets

A partially ordered set is called (2+2)-free if it contains no induced sub-posets isomorphic to (2+2) =

Such posets arise as interval orders (Fishburn):

P. C. Fishburn, Intransitive indifference with unequal indifference intervals, J. Math.Psych. 7 (1970) 144–149.

bad guy good guy

Unlabeled (2+2)-free posets

Theorem. (Not ours!) A poset is (2+2)-free iff the collection of strict down-sets may be linearly ordered by inclusion.

Unlabeled (2+2)-free posets

How can one decompose a (2+2)-free poset?

Unlabeled (2+2)-free posets

2

Unlabeled (2+2)-free posets

1 1 3

1 0 1

Read labels backwards: (0, 1, 0, 1, 3, 1, 1, 2) – an ascent sequence!

Removing last point gives one extra 0.

Theorem. There is a 1-1 correspondence between unlabeled (2+2)-free posets on n elements and ascent sequences of length n.

(0, 1, 0, 1, 3, 1, 1, 2)

Permutations avoiding

31524 avoids 32541 contains

How can one decompose such permutations?

Permutations avoiding

61832547 avoids . Remove the largest element, 8:

61 32 54 7

61 32 54 72 410 3

8 corresponds to the position labeled 1.

61 32 54 210 3

7 corresponds to the position labeled 3. Etc.

Read obtained labels starting from the recent one, to get (0, 1, 1, 2, 2, 0, 3, 1) – an ascent sequence!

Remove 7:

Theorem. There is a 1-1 correspondence between permutations avoiding on n elements and ascent sequences of length n.

(0, 1, 1, 2, 2, 0, 3, 1)6 1 8 3 2 5 4 7

Restricted permutations versus (2+2)-free posets

Modified ascent sequences

Some statistics preserved under the bijections

(0, 1, 0, 1, 3, 1, 1, 2)

(0, 1, 0, 1, 3, 1, 1, 2 )

(0, 1, 0, 1, 3, 1, 1, 2) 3 1 7 6 4 8 2 5

3 1 7 6 4 8 2 5

3 1 7 6 4 8 2 5

min zeros leftmost decr. run

min maxlevel

last element

0 1 2

label in fron ofmax element

(0, 3, 0, 1, 4, 1, 1, 2)

Level distri-bution

letter distributionin modif. sequence

element distributionbetween act. sites

Some statistics preserved under the bijections

(0, 1, 0, 1, 3, 1, 1, 2)

(0, 1, 0, 1, 3, 1, 1, 2)

(0, 1, 0, 1, 3, 1, 1, 2) 3 1 7 6 4 8 2 5

3 1 7 6 4 8 2 5

3 1 7 6 4 8 2 5

highestlevel

number of ascentsnumber of ascentsin the inverse

2 7 1 5 8 4 3 6

(0, 3, 0, 1, 4, 1, 1, 2)

right-to-left maxright-to-left maxin mod. sequencemax

compo-nents

Components inmodif. sequence components

(0, 3, 0, 1, 4, 1, 1, 2)

RLCD versus unlabeled (2+2)-free posets

RLCD versus unlabeled (2+2)-free posets

Sketch of a proof for surjectivity

(2+2)-free poset interval order

Sketch of a proof for surjectivity

Posets avoiding and

Ascent sequences are restricted as follows:

m-1, where m is the max element here

Catalan many

An open problemFind a bijection between (2+2)-free posets and permutations

avoiding using a graphical way, like one suggested below.

Thank you for your attention!Any questions?