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 x1, . . . , xn and covers x1 ✁ x2 ✁ . . . ✁ xa+1 ✄ xa+2 ✄ . . . ✄ xa+b+1 ✁ xa+b+2 ✁ . . . . Ex. F(2, 3, 1) = x1 x2 x3 x4 x5 x6 x7 The maximal chains of F are called segments. Note that if α = (α1, α2, . . .) then n = #F(α) = 1 +

• i

αi.

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 Rk(α) = {I ∈ L(α) | rk I = k} and rk(α) = #Rk(α). We will also use the rank generating function r(q; α) =

• k

rk(α)qk. This generating function was used by Morier-Genoud and Ovsienko to define q-analogues of rational numbers. Call a sequence a0, a1, . . . or its generating function unimodal if there is an index m with a0 ≤ a1 ≤ . . . ≤ am ≥ am+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 a0, a1, . . . , an symmetric if, for all k ≤ n/2, ak = an−k. Call the sequence top heavy or bottom heavy if, for all k ≤ n/2, ak ≤ an−k

• r

ak ≥ an−k, respectively. Call the sequence top interlacing (TI) if a0 ≤ an ≤ a1 ≤ an−1 ≤ a2 ≤ . . . ≤ a⌈n/2⌉

• r bottom interlacing (BI) if

an ≤ a0 ≤ an−1 ≤ a1 ≤ an−2 ≤ . . . ≤ a⌊n/2⌋. Note that interlacing implies unimodality and heaviness.

Conjecture (MSS)

Suppose α = (α1, . . . , αs). (a) If s is even, then r(q; α) is BI. (b) Suppose s ≥ 3 is odd and let α′ = (α2, . . . , αs−1).

(i) If α1 > αs or α1 < αs then r(q; α) is BI or TI, respectively. (iii) If α1 = αs 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 2

• r

n + 1 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 = (α1, . . . , αs) and suppose that for some t we have αt >

• i=t

αi. Then r(q; α) is unimodal.

Theorem (MSS)

Let α = (α1, . . . , αs) where for some t αt = 1 +

• i=t

αi. If L(α) has an SCD, TCD, or BCD then so does L(β) where β = (α1, . . . , αt−1, αt + a, αt+1, . . . , αs) 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 = (α1, α2, . . . , αs). Then for s odd r(q; α) = r(q; α1, . . . , αs−1, αs−1)+qαs+1·r(q; α1, . . . , αs−2, αs−1−1).

Proof.

If I ∈ L(α) then either xn ∈ I or xn ∈ I where n = #F(α). So I = xn

• r I =

xn Using the lemma as well as induction on α2 + · · · + αs:

Theorem

We have r(q; α) unimodal if α = (α1, α2, . . . , αs) satisfies α1 ≥ α2 + α3 + · · · + αs. A similar result hold when αs plays the role of α1.

For long segments other than the first or last we use:

Lemma

Suppose α = (α1, . . . , αs), n = #F(α), and for some t αt ≥ 1 +

• i=t

αi. (1) Let S be the segment of length αt, F ′ = F − S, m = #F ′ and ℓ = #L(F ′) 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 wI = w1 . . . wn which is the indicator function of I. Form the Greene-Kleitman core, GK(wI), by pairing any wi = 0 with wi+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}. wI = 1001100 and GK(wI) = ∗ 011 ∗ ∗ = {(2, 5), (3, 4)}. F(2, 3, 1) = 5 6 7 3 2 1 4

Let ✂ be the order relation in F(α). Call an unpaired f ∈ [n] frozen with respect to I if there is (i, j) ∈ GK(wI) with f ✄ i

• r

f ✁ j. We now define the core of wI to be core wI = GK(wI) ∪ {f | f is frozen with respect to I}. The elements i ∈ core wI are free with respect to I. Given κ = core wI we form a saturated chain Cκ by starting with w0 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}.

wI = 1001100 and GK(wI) = {(2, 5), (3, 4)} 7 ✄ 2 ∈ (2, 5) so 7 frozen and 1 ✁ 4 ∈ (3, 4) so 1 frozen κ = core wI = {1, (2, 5), (3, 4), 7} and 6 is free Cκ = {1001100, 1001110}

Theorem

If κ = (a, b, c), a ≥ c, then the Cκ 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 + · · · + qn−1. This expression appeared in a more complicated form in Claussen’s

• thesis. Let

Z(α) = {z1, z2, . . . , zu} be the set of maxima of F(α) written in order from left to right.

Theorem (Claussen)

If α = (α1, . . . , α2u−1) then r(q; α) =

• Z⊆Z(α)

q#Zr(Z, 1)r(Z, 2) · · · r(Z, u) where, letting α0 = 1, r(Z, i) =            qα2i−2+α2i−1−1 if zi−1, zi ∈ Z, qα2i−2[α2i−1]q if zi−1 ∈ Z and zi ∈ Z, qα2i−1[α2i−2]q if zi−1 ∈ Z and zi ∈ Z, 1 + q[α2i−2]q[α2i−1]q if zi−1, zi ∈ Z.

• 2. Lexicographic CDs. Suppose P is a poset on [n]. We

construct the chains C1, C2, C3, . . . of a CD of L = L(P) as follows. Suppose C1, . . . , Ci−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

• rder on subsets. Now we form Ci by starting with the unique ideal

I0 which has minimum rank and is also lexicographically least among all elements of L′ = L − (C1 ∪ · · · ∪ Ci−1). We now consider all ideals of L′ which cover I0 and take the lexicographically least of them to be the next element I1 on Ci. We continue in this manner until we come to an ideal which has no cover in L′ at which point Ci terminates. We have the following conjecture which we have verified for all compositions α with

i αi ≤ 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.

THANKS FOR LISTENING!