permutations, sequences, and partially ordered sets sergey kitaev reykjavik university joint work...
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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?