On a rank-unimodality conjecture of Morier-Genoud and Ovsienko - PowerPoint PPT Presentation

On a rank-unimodality conjecture of Morier-Genoud and Ovsienko Thomas McConville Kennesaw State University Bruce Sagan Michigan State University www.math.msu.edu/˜sagan Clifford Smyth University of North Carolina, Greensboro University of Minnesota Combinatorics Seminar September 25, 2020

Introduction Fences with long segments Fences with at most three segments Other approaches

Let α = ( a , b , . . . ) be a composition. A fence is a poset F = F ( α ) with elements x 1 , . . . , x n and covers x 1 ✁ x 2 ✁ . . . ✁ x a +1 ✄ x a +2 ✄ . . . ✄ x a + b +1 ✁ x a + b +2 ✁ . . . . Ex. x 3 F (2 , 3 , 1) = x 2 x 4 x 1 x 5 x 7 x 6 The maximal chains of F are called segments . Note that if α = ( α 1 , α 2 , . . . ) then � n = # F ( α ) = 1 + α i . i

Let L = L ( α ) be the distributive lattice of order ideals of F ( α ). These lattices can be used to compute mutations in a cluster algebra on a surface with marked points. Who When What Propp 2005 perfect mathings on snake graphs Yurikusa 2019 perfect matchings of angles Schiffler 2008, 2010 T -paths Schiffler and Thomas 2009 T -paths Propp 2005 lattice paths on snake graphs Claussen 2020 lattice paths of angles Claussen 2020 S -paths

Lattice L ( α ) is ranked with rank function rk I = # I . We let R k ( α ) = { I ∈ L ( α ) | rk I = k } and r k ( α ) = # R k ( α ) . We will also use the rank generating function � r k ( α ) q k . r ( q ; α ) = k This generating function was used by Morier-Genoud and Ovsienko to define q -analogues of rational numbers. Call a sequence a 0 , a 1 , . . . or its generating function unimodal if there is an index m with a 0 ≤ a 1 ≤ . . . ≤ a m ≥ a m +1 ≥ . . . . Conjecture (Morier-Genoud and Ovsienko, 2020) For any α we have that r ( q ; α ) is unimodal. Previous work: Gansner (1982), Munarini and Salvi (2002), Claussen (2020).

Call sequence a 0 , a 1 , . . . , a n symmetric if, for all k ≤ n / 2, a k = a n − k . Call the sequence top heavy or bottom heavy if, for all k ≤ n / 2, a k ≤ a n − k or a k ≥ a n − k , respectively. Call the sequence top interlacing (TI) if a 0 ≤ a n ≤ a 1 ≤ a n − 1 ≤ a 2 ≤ . . . ≤ a ⌈ n / 2 ⌉ or bottom interlacing (BI) if a n ≤ a 0 ≤ a n − 1 ≤ a 1 ≤ a n − 2 ≤ . . . ≤ a ⌊ n / 2 ⌋ . Note that interlacing implies unimodality and heaviness. Conjecture (MSS) Suppose α = ( α 1 , . . . , α s ) . (a) If s is even, then r ( q ; α ) is BI. (b) Suppose s ≥ 3 is odd and let α ′ = ( α 2 , . . . , α s − 1 ) . (i) If α 1 > α s or α 1 < α s then r ( q ; α ) is BI or TI, respectively. (iii) If α 1 = α s then r ( q ; α ) is symmetric, BI, or TI depending on whether r ( q ; α ′ ) is symmetric, TI, or BI, respectively.

A chain decomposition (CD) of a poset P is a partition of P into disjoint saturated chains. If P is ranked then the center of a chain C is cen C = rk (min C ) + rk (max C ) . 2 If rk P = n then a CD is symmetric (SCD) if for all chains C in the CD cen C = n 2 . A CD is top centered (TCD) if for all chains C in the CD cen C = n n + 1 or . 2 2 A bottom centered CD (BCD) has cen C = n / 2 or ( n − 1) / 2 for all chains C . If P has an SCD, TCD, or BCD then its rank sequence is symmetric, top, or bottom interlacing, respectively. Conjecture (MSS) For any α , the lattice L ( α ) admits an SCD, TCD, or BCD consistent with the previous conjecture.

Theorem (MSS) Let α = ( α 1 , . . . , α s ) and suppose that for some t we have � α t > α i . i � = t Then r ( q ; α ) is unimodal. Theorem (MSS) Let α = ( α 1 , . . . , α s ) where for some t � α t = 1 + α i . i � = t If L ( α ) has an SCD, TCD, or BCD then so does L ( β ) where β = ( α 1 , . . . , α t − 1 , α t + a , α t +1 , . . . , α s ) for any a ≥ 0 . Theorem (MMS) If α has at most three parts then L ( α ) has an SCD, TCD, or BCD.

The following recursion also has a version for s even. Lemma Let α = ( α 1 , α 2 , . . . , α s ) . Then for s odd r ( q ; α ) = r ( q ; α 1 , . . . , α s − 1 , α s − 1)+ q α s +1 · r ( q ; α 1 , . . . , α s − 2 , α s − 1 − 1) . Proof. If I ∈ L ( α ) then either x n �∈ I or x n ∈ I where n = # F ( α ). So or I = I = x n x n Using the lemma as well as induction on α 2 + · · · + α s : Theorem We have r ( q ; α ) unimodal if α = ( α 1 , α 2 , . . . , α s ) satisfies α 1 ≥ α 2 + α 3 + · · · + α s . A similar result hold when α s plays the role of α 1 .

For long segments other than the first or last we use: Lemma Suppose α = ( α 1 , . . . , α s ) , n = # F ( α ) , and for some t � α t ≥ 1 + α i . (1) i � = t Let S be the segment of length α t , F ′ = F − S, m = # F ′ ℓ = # L ( F ′ ) and Then the maximum size of a rank of L = L ( α ) is ℓ and this maximum occurs at ranks m + 1 through n − m − 1 . Proof. If I ∈ L ( α ) then I = J ∪ K where J ∈ L ( F ′ ) and K ∈ L ( S ). Since L ( S ) is a chain, given J and rk I , there is ≤ 1 choice for K . This lemma permits us to prove that if (1) holds then L ( α ) is rank unimodal. It also has an SCD, TCD or BCD as long as that is true for the base case of equality in (1).

Claussen proved that if α has at most four parts then L ( α ) is rank unimodal. We are able to prove the stronger SCD, TCD, and BCD conditions when α has at most three parts using a variant of the Greene-Kleitman core. If P is a poset on [ n ] = { 1 , . . . , n } then we can associate with any I ⊆ P a zero-one word w I = w 1 . . . w n which is the indicator function of I . Form the Greene-Kleitman core , GK( w I ), by pairing any w i = 0 with w i +1 = 1, then pair zeros and ones separated only by pairs, etc. Ex. Let F = F (2 , 3 , 1) be labeled starting with the middle segment, then the last, and finally the first, and let I = { 1 , 4 , 5 } . GK( w I ) = ∗ � 0 � w I = 1001100 and 011 ∗ ∗ = { (2 , 5) , (3 , 4) } . 7 F (2 , 3 , 1) = 6 3 5 2 4 1

Let ✂ be the order relation in F ( α ). Call an unpaired f ∈ [ n ] frozen with respect to I if there is ( i , j ) ∈ GK( w I ) with f ✄ i or f ✁ j . We now define the core of w I to be core w I = GK( w I ) ∪ { f | f is frozen with respect to I } . The elements i �∈ core w I are free with respect to I . Given κ = core w I we form a saturated chain C κ by starting with w 0 which has core κ and all free positions equal to zeros. We then change each free zero to a one from left to right. Ex. Let F = F (2 , 3 , 1) be labeled as before I = { 1 , 4 , 5 } . w I = 1001100 and GK( w I ) = { (2 , 5) , (3 , 4) } 7 ✄ 2 ∈ (2 , 5) so 7 frozen and 1 ✁ 4 ∈ (3 , 4) so 1 frozen κ = core w I = { 1 , (2 , 5) , (3 , 4) , 7 } and 6 is free C κ = { 1001100 , 1001110 } Theorem If κ = ( a , b , c ) , a ≥ c, then the C κ form an SCD or TCD of L ( α ) .

(1) An explicit formula. There is an explicit formula for r ( q ; α ) in terms of powers of q and [ n ] q = 1 + q + · · · + q n − 1 . This expression appeared in a more complicated form in Claussen’s thesis. Let Z ( α ) = { z 1 , z 2 , . . . , z u } be the set of maxima of F ( α ) written in order from left to right. Theorem (Claussen) If α = ( α 1 , . . . , α 2 u − 1 ) then � q # Z r ( Z , 1) r ( Z , 2) · · · r ( Z , u ) r ( q ; α ) = Z ⊆ Z ( α ) where, letting α 0 = 1 ,   q α 2 i − 2 + α 2 i − 1 − 1 if z i − 1 , z i ∈ Z,     q α 2 i − 2 [ α 2 i − 1 ] q if z i − 1 ∈ Z and z i �∈ Z, r ( Z , i ) =  q α 2 i − 1 [ α 2 i − 2 ] q if z i − 1 �∈ Z and z i ∈ Z,     1 + q [ α 2 i − 2 ] q [ α 2 i − 1 ] q if z i − 1 , z i �∈ Z.

2. Lexicographic CDs. Suppose P is a poset on [ n ]. We construct the chains C 1 , C 2 , C 3 , . . . of a CD of L = L ( P ) as follows. Suppose C 1 , . . . , C i − 1 have been constructed. Since P = [ n ] as sets, we can consider any ideal I of P as a subset of { 1 , . . . , n } and we will not make any distinction between an ideal and its subset. So given two ideals, we can compare them in the lexicographic order on subsets. Now we form C i by starting with the unique ideal I 0 which has minimum rank and is also lexicographically least among all elements of L ′ = L − ( C 1 ∪ · · · ∪ C i − 1 ). We now consider all ideals of L ′ which cover I 0 and take the lexicographically least of them to be the next element I 1 on C i . We continue in this manner until we come to an ideal which has no cover in L ′ at which point C i terminates. We have the following conjecture which we have verified for all compositions α with � i α i ≤ 6. Conjecture For every α there is a labeling of F ( α ) with [ n ] such that the correponding lexicographic CD of L ( α ) is an SCD, TCD, or BCD.

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