Steep-bounce zeta map in the parabolic Cataland Wenjie Fang, - - PowerPoint PPT Presentation

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Steep-bounce zeta map in the parabolic Cataland Wenjie Fang, - - PowerPoint PPT Presentation

Two Tamaris Bijections Zeta Discussion Steep-bounce zeta map in the parabolic Cataland Wenjie Fang, Institute of Discrete Mathematics, TU Graz Joint work with Cesar Ceballos and Henri M uhle 11 December 2018, AG Diskrete Mathematik, TU


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Two Tamaris Bijections Zeta Discussion

Steep-bounce zeta map in the parabolic Cataland

Wenjie Fang, Institute of Discrete Mathematics, TU Graz Joint work with Cesar Ceballos and Henri M¨ uhle 11 December 2018, AG Diskrete Mathematik, TU Wien

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Two Tamaris Bijections Zeta Discussion

Tamari lattice, as an order on Dyck paths

Dyck path : n north steps (N) and n east steps (E), above the

  • diagonal. Counted by Catalan numbers
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Two Tamaris Bijections Zeta Discussion

Tamari lattice, as an order on Dyck paths

Covering relation: take a valley •, let be the next point wiht the same distance to the diagonal...

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Two Tamaris Bijections Zeta Discussion

Tamari lattice, as an order on Dyck paths

..., and push the segment to the left. The path gets larger. This gives the Tamari lattice.

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Two Tamaris Bijections Zeta Discussion

ν-Tamari lattice

Generalization with ν an arbitrary directed walk as “diagonal”! Horizontal distance = # east steps until touching the other side of ν

2 1 1 2 1 1 ν 1 2 1 1 p p′ E E ≺ν ν ν 1

ν-Tamari lattice (Pr´ eville-Ratelle and Viennot 2014): Tν with arbitrary ν (called canopy) with steps N, E.

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Why is it important ?

Generalizing a lot of cases (m-Tamari, rational Tamari) Bijective links (non-separable planar maps and related objects) Algebraic aspect (subword complexes, Diagonal coinvariant spaces, etc.)

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Two Tamaris Bijections Zeta Discussion

Tamari lattice, as quotient of the weak order

Sn as a Coxeter group generated by si = (i, i + 1) For w ∈ Sn, ℓ(w) = min. length of factorization of w into si’s. Weak order : w covered by w′ iff w′ = wsi and ℓ(w′) = ℓ(w) + 1

321 312 231 132 213 123 321 312 132 213 123

Sylvester class : permutations with the same binary search tree Only one 231-avoiding in each class. Induced order = Tamari. Works for other types

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Parabolic subgroup and parabolic quotient of Sn

Let α = (α1, . . . , αk) be a composition of n. Parabolic subgroup : Sα1 × · · · × Sαk ⊂ Sn. Generated by si except for i = α1 + α2 + · · · + αj.

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

Parabolic quotient : Sα

n = Sn/(Sα1 × · · · × Sαk).

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

Increasing order in each block

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Two Tamaris Bijections Zeta Discussion

Parabolic permutations avoiding 231

Pattern (α, 231) : three indices i < j < k in three distinct blocks with w(k) < w(i) < w(j), w(k) + 1 = w(i). (α, 231)-avoiding permutations: without (α, 231) patterns

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

n(231) : set of (α, 231)-avoiding permutations

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Two Tamaris Bijections Zeta Discussion

Parabolic Tamari lattice

Parabolic Tamari lattice T α

n = weak order restricted to Sα n(231)

(M¨ uhle and Williams 2018+)

12|34|5 12|35|4 13|24|5 12|45|3 13|25|4 14|23|5 23|14|5 14|35|2 23|15|4 15|23|4 24|13|5 34|25|1 15|24|3 25|13|4 34|12|5 15|34|2 25|14|3 35|12|4 35|24|1 45|12|3 45|13|2 45|23|1

Works for other types!

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Two Tamaris Bijections Zeta Discussion

Parabolic non-crossing partitions

· · · · · · bump {1, 6, 8}, {2, 9}, {3, 7}, {4}, {5} 1 2 3 4 5 6 7 8 9

Parabolic α-partition: a set of bumps, ≤ 1 incoming/outgoing

1 2 3 4 5 6 7 8 9

Parabolic non-crossing α-partition : without bumps crossing

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Two Tamaris Bijections Zeta Discussion

Parabolic non-nesting partitions

Parabolic non-nesting α-partition : no bumps (i, j), (k, ℓ) with i < k < ℓ < j. Encoding with points (i, j)

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

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Parabolic non-nesting partitions

Parabolic non-nesting α-partition : no bumps (i, j), (k, ℓ) with i < k < ℓ < j. Encoding with points (i, j)

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

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Two Tamaris Bijections Zeta Discussion

Parabolic non-nesting partitions

Parabolic non-nesting α-partition : no bumps (i, j), (k, ℓ) with i < k < ℓ < j. Encoding with points (i, j)

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

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Parabolic non-nesting partitions

Parabolic non-nesting α-partition : no bumps (i, j), (k, ℓ) with i < k < ℓ < j. Bounce pair: A Dyck path above a bounce path

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

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Detour to pipe dreams

Hopf algebra on pipe dreams (Bergeron, Ceballos et Pilaud, 2018+).

1 2 3 4 5 6 1 2 3 4 5 6

  • Dim. of homogeneous comps. of a sub-algebra (generated by identities)

= # pipe dreams with an “identity by block” permutation Proposition (Bergeron, Ceballos and Pilaud, 2018+) Pipe dreams whose permutation is an “identity by block” of size n are in bijection with bounce pairs of order n. Already a link to the parabolic Catalan objects!

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Counting and relations ?

All three objects are in bijection (M¨ uhle and Williams), but not easy. Numbers of parabolic Catalan objects of order n: 1, 1, 3, 12, 57, 301, 1707, 10191, 63244, 404503, . . . (OEIS A151498) = certain walks in the quadrant Bijective link? An easier-to-understand structure?

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Marked paths and steep pairs

Walks in the quadrant: {(1, 0), (1, −1), (−1, 1)}, ending with y = 0. Considered in (Bousque-M´ elou and Mishna, 2010) and counted in (Mishna and Rechnitzer, 2009) In bijection with level-marked Dyck paths: level ≤ marking before the point

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Level-marked Dyck paths and steep pairs

Steep pairs : 2 nested Dyck paths, the one above has no EE except at the end Bijection: Path below: path without marking Path above: read the N’s, marked → N, not marked → EN

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Steep-Bounce conjecture

Conjecture (Bergeron, Ceballos and Pilaud 2018+, Conjecture 2.2.8) The following two sets are of the same size: bounce pairs of order n with k blocks; steep pairs of order n with k east steps E on y = n. A proof gives the counting of all these objects (pipe dreams and parabolic Catalan) The cases k = 1, 2, n − 1, n already proved

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A scheme of the bijections

5 3 4 10 1 2 7 6 9 13 14 8 11 12 Ξperm Ξnc Ξbounce Ξdyck Ξsteep

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Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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Two Tamaris Bijections Zeta Discussion

Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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Two Tamaris Bijections Zeta Discussion

Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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Two Tamaris Bijections Zeta Discussion

Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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Two Tamaris Bijections Zeta Discussion

Left-aligned colored trees

T : plane tree with n non-root nodes; α = (α1, . . . , αk) : composition of n Active nodes : not yet colored, but parent is colored or is the root. Coloring algorithm : For i from 1 to k, If there are less than αi active nodes, then fail; Otherwise, color the first αi from left to right with color i.

α = (1, 3, 1, 2, 4, 3) ⊢ 14

When succeeded, it is a left-aligned colored tree (or a LAC tree).

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To permutations

(T, α) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Ξperm(T, α) = 5 | 3 4 10 | 1 | 2 7 | 6 9 13 14 | 8 11 12 ∈ Sα

n(231)

Ξperm

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To parabolic non-crossing partitions

(T, α)

Ξnc LAC tree → partition : flatten the layers Partition → LAC tree : look at the sky

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To bounce pairs

α = (1, 3, 1, 2, 4, 3) ⊢ 14 jk = 1 ak = 5 p = jk − r + 1 = 1 q = jk + ak − s = 4 r = 1 s = 2

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To bounce pairs

jk = 4 ak = 6 p = jk − r + 1 = 4 q = jk + ak − s = 8 r = 1 s = 2 α = (1, 3, 1, 2, 4, 3) ⊢ 14

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To bounce pairs

jk = 4 ak = 6 p = jk − r + 1 = 2 q = jk + ak − s = 6 r = 3 s = 4 α = (1, 3, 1, 2, 4, 3) ⊢ 14

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To bounce pairs

α = (1, 3, 1, 2, 4, 3) ⊢ 14

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To steep pairs

Ξdyck(T, α) Ξsteep(T, α) (T, α)

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Steep-Bounce theorem

Theorem (Ceballos, F., M¨ uhle 2018+) There is a natural bijection Γ between the following two sets: bounce pairs of order n with k blocks; steep pairs of order n with k each steps E on y = n. So we know how to count them!

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A bijection between the two Tamaris

5 3 4 10 1 2 9 6 8 13 14 7 11 12

⋗L ⋖να

5 3 4 10 1 2 7 6 9 13 14 8 11 12

↓ ↓

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One isomorphic to the dual of the other

12|34|5 12|35|4 13|24|5 12|45|3 13|25|4 14|23|5 23|14|5 14|35|2 23|15|4 15|23|4 24|13|5 34|25|1 15|24|3 25|13|4 34|12|5 15|34|2 25|14|3 35|12|4 35|24|1 45|12|3 45|13|2 45|23|1

Theorem (Ceballos, F., M¨ uhle 2018+) The parabolic Tamari lattice indexed by α is isomorphic to the ν-Tamari lattice with ν = N α1Eα1 · · · N αkEαk.

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Detour to q, t-Catalan combinatorics

a(1) = 0 1 2 3 3 3 3 1 a(9) = 2 5 2 area(D) =

i a(i) = 18

dinv(D) = #{(i, j) | i < j, (a(i) = a(j) ∨ a(i) = a(j) + 1} = 17 bounce(D) =

i(i − 1)αi = 7

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A non-trivial symmetry

Theorem (Garsia and Haiman 1996, Haiman 2001) By summing up all Dyck paths of order n, we have

  • D

qarea(D)tbounce(D) =

  • D

qbounce(D)tarea(D). The proof goes by the Hilbert series of the diagonal coinvariant space with two sets of variables. No combinatorial proof! Theorem (Haglund 2008, Proof of Theorem 3.15) There is a bijection ζ on Dyck paths that transfers the pairs of statistics (dinv, area) → (area, bounce).

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Our zeta map

area(D) = 18 bounce(D) = 7 dinv(D) = 18 area(D) = 7 Γ = Ξbounce ◦ Ξ−1

steep

Ξsteep Ξbounce

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Our zeta map, Steep-Bounce version

Theorem (Ceballos, F., M¨ uhle 2018+) There is a natural bijection Γ between the following sets: bounce pairs of order n with k blocks; steep pairs of order n with k east steps E on y = n. ζ = special case of Γ, with steep pairs and bounce pairs constructed in a greedy way A generalization to explore!

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Possible directions

Many questions in enumeration (but possibly very difficult) How are the statistics transferred, and which ones? Action by symmetries? Implication in diagonal coinvariant spaces?

  • etc. ?
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Possible directions

Many questions in enumeration (but possibly very difficult) How are the statistics transferred, and which ones? Action by symmetries? Implication in diagonal coinvariant spaces?

  • etc. ?

Thank you for listening!