Survey

# Download Parallelogram polyominoes, the sandpile model on Km,n, and a q,t

Document related concepts

Transcript

Parallelogram polyominoes, the sandpile model on Km,n , and a q, t-Narayana polynomial. Mark Dukes University of Strathclyde 21 February 2013 * joint work with Y. Le Borgne (Bordeaux) * 1. A pentagram of correspondences (2+2)-free posets on {1, . . . , n} Composition matrices on {1, . . . , n} Labelled bivincular pattern avoiding perms Sn (2|31) M. Dukes (University of Strathclyde) Labelled Stoimenow matchings Labelled Ascent sequences QMUL 2013 1 / 22 1. A pentagram of correspondences 2 4 1 5 6 5 {6} {3} ∅ ∅ ∅ {1,5} ∅ {4} ∅ {2} 33 16 51 45 24 62 M. Dukes (University of Strathclyde) 3 1 5 4 6 2 (0,1,0,1,1,3) h{6},{4},{3},{1,5},{2}i QMUL 2013 1 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 3 0 0 3 0 M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 3 0 0 3 0 M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 3 0 1 4 0 3 0 0 3 0 0 M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 3 0 1 4 0 3 0 0 3 0 0 M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 1 0 1 3 0 1 4 0 2 0 1 3 0 3 0 0 4 0 0 M. Dukes (University of Strathclyde) 0 QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 1 0 1 3 0 1 4 0 2 0 1 3 0 3 0 0 4 0 0 M. Dukes (University of Strathclyde) 0 QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 1 0 1 1 0 1 3 0 1 4 0 2 0 1 2 1 1 3 0 3 0 0 4 0 1 0 1 0 0 * v0 = sink v7 v8 v9 v4 v5 v6 v1 v2 v3 M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 1 0 1 1 0 1 3 0 1 4 0 2 0 1 2 1 1 3 0 3 0 0 4 0 1 0 1 0 * v0 = sink v7 v8 v9 v4 v5 v6 v1 v2 v3 0 (0, 3, 0, 1, 3, 0, 1, 0, 0) → (0, 3, 0, 1, 4, 0, 1, 0, 0) → (0, 4, 0, 2, 0, 1, 1, 1, 0) → (1, 0, 1, 2, 1, 1, 1, 1, 0) M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 2. The sandpile model Some form of grid where grid entries contain grains of sand. A toppling rule for when too many grains occupy a grid entry. 1 0 0 1 0 0 1 1 0 1 1 0 1 3 0 1 4 0 2 0 1 2 1 1 3 0 3 0 0 4 0 1 0 1 0 * v0 = sink v7 v8 v9 v4 v5 v6 v1 v2 v3 0 (0, 3, 0, 1, 3, 0, 1, 0, 0) → (0, 3, 0, 1, 4, 0, 1, 0, 0) → (0, 4, 0, 2, 0, 1, 1, 1, 0) → (1, 0, 1, 2, 1, 1, 1, 1, 0) σ(0, 3, 0, 1, 4, 0, 1, 0, 0) = (1, 0, 1, 2, 1, 1, 1, 1, 0) M. Dukes (University of Strathclyde) QMUL 2013 2 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. M. Dukes (University of Strathclyde) di eij QMUL 2013 = = degree of vertex vi number of edges between vi and vj . 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. M. Dukes (University of Strathclyde) QMUL 2013 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. M. Dukes (University of Strathclyde) QMUL 2013 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. Toppling: vertex vi unstable ⇒ M. Dukes (University of Strathclyde) xi → xi − di xj → xj + eij QMUL 2013 for all j 6= i. 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. Toppling: vertex vi unstable ⇒ xi → xi − di xj → xj + eij vc for all j 6= i. va vz vi v M. Dukes (University of Strathclyde) QMUL b2013 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. Toppling: vertex vi unstable ⇒ xi → xi − di xj → xj + eij vc for all j 6= i. va 5 vz vi v M. Dukes (University of Strathclyde) QMUL b2013 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. Toppling: vertex vi unstable ⇒ xi → xi − di xj → xj + eij vc for all j 6= i. va 9 vz vi v M. Dukes (University of Strathclyde) QMUL b2013 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. Toppling: vertex vi unstable ⇒ xi → xi − di xj → xj + eij vc for all j 6= i. va 9 vz vi v M. Dukes (University of Strathclyde) QMUL b2013 3 / 22 3. The sandpile model on (balanced) graphs Let G = (V , E ) be a graph on V = {v0 , v1 , . . . , vn } where vertex v0 is a sink. di eij = = degree of vertex vi number of edges between vi and vj . A configuration x = (x1 , . . . , xn ) is called stable if xi < di for all 1 ≤ i ≤ n. Otherwise it is called unstable. Toppling: vertex vi unstable ⇒ xi → xi − di xj → xj + eij vc +1 for all j 6= i. +3 va 4 vz vi +1 v M. Dukes (University of Strathclyde) QMUL b2013 3 / 22 4. Recurrent states/configurations All stable states can be classified as either transient or recurrent. M. Dukes (University of Strathclyde) QMUL 2013 4 / 22 4. Recurrent states/configurations All stable states can be classified as either transient or recurrent. 3 1 3 2 1 2 3 1 3 1 0 1 1 0 1 1 0 1 3 1 3 2 1 2 3 1 3 M. Dukes (University of Strathclyde) QMUL 2013 4 / 22 4. Recurrent states/configurations All stable states can be classified as either transient or recurrent. 0 0 0 0 0 0 0 0 0 M. Dukes (University of Strathclyde) ? QMUL 2013 0 0 0 0 0 0 0 0 0 4 / 22 4. Recurrent states/configurations All stable states can be classified as either transient or recurrent. 0 0 0 3 1 3 0 0 0 1 0 1 0 0 0 3 1 3 M. Dukes (University of Strathclyde) is transient; QMUL 2013 is recurrent. 4 / 22 4. Recurrent states/configurations All stable states can be classified as either transient or recurrent. 0 0 0 3 1 3 0 0 0 1 0 1 0 0 0 3 1 3 is transient; is recurrent. Test for Recurrence A stable state x = (x1 , . . . , xn ) on a graph G = (V , E ) is recurrent if and only if σ(x + t) = x where t = (t1 , . . . , tn ) is the vector with ti = e0i . M. Dukes (University of Strathclyde) QMUL 2013 4 / 22 4. Recurrent states/configurations All stable states can be classified as either transient or recurrent. 0 0 0 3 1 3 0 0 0 1 0 1 0 0 0 3 1 3 is transient; is recurrent. Test for Recurrence A stable state x = (x1 , . . . , xn ) on a graph G = (V , E ) is recurrent if and only if σ(x + t) = x where t = (t1 , . . . , tn ) is the vector with ti = e0i . rec(G ) = set of all recurrent states on the graph G M. Dukes (University of Strathclyde) QMUL 2013 4 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 v2 M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 x G = v1 v3 x +t σ(x + t) recurs (0, 0, 0) v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 x x +t (0, 0, 0) (1, 0, 1) σ(x + t) recurs v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 x x +t σ(x + t) (0, 0, 0) (1, 0, 1) (1, 0, 1) recurs v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 x x +t σ(x + t) recurs (0, 0, 0) (1, 0, 1) (1, 0, 1) 5 v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) M. Dukes (University of Strathclyde) QMUL 2013 x x +t σ(x + t) recurs (0, 0, 0) (0, 0, 1) (0, 1, 0) (0, 1, 1) (1, 0, 0) (1, 0, 1) (1, 1, 0) (1, 1, 1) (1, 0, 1) (1, 0, 2) (1, 1, 1) (1, 1, 2) (2, 0, 1) (2, 0, 2) (2, 1, 1) (2, 1, 2) (1, 0, 1) (1, 1, 0) (1, 1, 1) (0, 1, 1) (0, 1, 1) (1, 0, 1) (1, 1, 0) (1, 1, 1) 5 5 5 3 5 3 3 3 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) x x +t σ(x + t) recurs (0, 0, 0) (0, 0, 1) (0, 1, 0) (0, 1, 1) (1, 0, 0) (1, 0, 1) (1, 1, 0) (1, 1, 1) (1, 0, 1) (1, 0, 2) (1, 1, 1) (1, 1, 2) (2, 0, 1) (2, 0, 2) (2, 1, 1) (2, 1, 2) (1, 0, 1) (1, 1, 0) (1, 1, 1) (0, 1, 1) (0, 1, 1) (1, 0, 1) (1, 1, 0) (1, 1, 1) 5 5 5 3 5 3 3 3 rec(G ) = {(0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)} M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 5. Test for recurrence - an example Example v0 G = v1 v3 v2 stable(G ) = {(x1 , x2 , x3 ) : xi ∈ {0, 1}} t = (1, 0, 1) x x +t σ(x + t) recurs (0, 0, 0) (0, 0, 1) (0, 1, 0) (0, 1, 1) (1, 0, 0) (1, 0, 1) (1, 1, 0) (1, 1, 1) (1, 0, 1) (1, 0, 2) (1, 1, 1) (1, 1, 2) (2, 0, 1) (2, 0, 2) (2, 1, 1) (2, 1, 2) (1, 0, 1) (1, 1, 0) (1, 1, 1) (0, 1, 1) (0, 1, 1) (1, 0, 1) (1, 1, 0) (1, 1, 1) 5 5 5 3 5 3 3 3 rec(G ) = {(0, 1, 1), (1, 0, 1), (1, 1, 0), (1, 1, 1)} Example Elements of rec(Kn+1 ) are parking functions of order n. M. Dukes (University of Strathclyde) QMUL 2013 5 / 22 6. The sandpile model on Km,n v0 v1 Determining rec(Km,n ) vm−1 v2 ······ Given x ∈ stable(Km,n ), x ∈ rec(Km,n ) m σ (x + (0, . . . , 0, 1, . . . , 1)) = x M. Dukes (University of Strathclyde) QMUL 2013 ······ vm vm+1 vm+n−1 6 / 22 6. The sandpile model on Km,n v0 v1 Determining rec(Km,n ) vm−1 v2 ······ Given x ∈ stable(Km,n ), x ∈ rec(Km,n ) m σ (x + (0, . . . , 0, 1, . . . , 1)) = x ······ vm vm+1 vm+n−1 Definition incm,n (x) is the rearrangement of x ∈ stable(Km,n ) such that the first m − 1 values are weakly increasing, and the subsequent n values are weakly increasing. Example If x = (2, 1, 2, 0, 0, 2; 6, 1, 5, 1) ∈ stable(K7,4 ) then inc7,4 (x) = (0, 0, 1, 2, 2, 2; 1, 1, 5, 6). M. Dukes (University of Strathclyde) QMUL 2013 6 / 22 7. From stable states to polyominoes Example (1) 2 6 M. Dukes (University of Strathclyde) 1 2 1 0 0 5 QMUL 2013 2 1 7 / 22 7. From stable states to polyominoes Example (1) 2 6 1 2 1 0 0 5 2 1 x = (2, 1, 2, 0, 0, 2, 6, 1, 5, 1) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 2, 2, 2, 1, 1, 5, 6) M. Dukes (University of Strathclyde) QMUL 2013 7 / 22 7. From stable states to polyominoes Example (1) 2 6 1 2 1 0 0 5 2 1 x = (2, 1, 2, 0, 0, 2, 6, 1, 5, 1) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 2, 2, 2, 1, 1, 5, 6) 0 0 1 2 2 2 M. Dukes (University of Strathclyde) QMUL 2013 7 / 22 7. From stable states to polyominoes Example (1) 2 6 1 2 1 0 0 5 2 1 x = (2, 1, 2, 0, 0, 2, 6, 1, 5, 1) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 2, 2, 2, 1, 1, 5, 6) 6 5 1 1 0 0 1 2 2 2 M. Dukes (University of Strathclyde) QMUL 2013 7 / 22 7. From stable states to polyominoes Example (1) 2 1 6 2 1 0 0 5 2 1 x = (2, 1, 2, 0, 0, 2, 6, 1, 5, 1) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 2, 2, 2, 1, 1, 5, 6) ∩ 6 5 1 1 0 0 1 2 2 2 M. Dukes (University of Strathclyde) QMUL 2013 7 / 22 7. From stable states to polyominoes Example (1) 2 1 6 2 1 0 0 5 2 1 x = (2, 1, 2, 0, 0, 2, 6, 1, 5, 1) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 2, 2, 2, 1, 1, 5, 6) ∩ 6 5 1 1 = = fm,n (x) 0 0 1 2 2 2 M. Dukes (University of Strathclyde) QMUL 2013 7 / 22 7. From stable states to polyominoes Example (1) 2 1 6 2 1 0 0 5 2 1 x = (2, 1, 2, 0, 0, 2, 6, 1, 5, 1) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 2, 2, 2, 1, 1, 5, 6) ∩ 6 5 1 1 = = fm,n (x) 0 0 1 2 2 2 x is not recurrent, i.e. x 6∈ rec(K7,4 ) M. Dukes (University of Strathclyde) QMUL 2013 7 / 22 7. From stable states to polyominoes Example (2) 2 4 M. Dukes (University of Strathclyde) 1 0 6 0 1 6 QMUL 2013 1 4 8 / 22 7. From stable states to polyominoes Example (2) 2 4 1 0 6 0 1 6 1 4 x = (2, 1, 0, 0, 1, 1, 4, 6, 6, 4) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 1, 1, 2, 4, 4, 6, 6) M. Dukes (University of Strathclyde) QMUL 2013 8 / 22 7. From stable states to polyominoes Example (2) 2 4 1 0 6 0 1 6 1 4 x = (2, 1, 0, 0, 1, 1, 4, 6, 6, 4) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 1, 1, 2, 4, 4, 6, 6) 0 0 1 1 1 2 M. Dukes (University of Strathclyde) QMUL 2013 8 / 22 7. From stable states to polyominoes Example (2) 2 4 1 0 6 0 1 6 1 4 x = (2, 1, 0, 0, 1, 1, 4, 6, 6, 4) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 1, 1, 2, 4, 4, 6, 6) 6 6 4 4 0 0 1 1 1 2 M. Dukes (University of Strathclyde) QMUL 2013 8 / 22 7. From stable states to polyominoes Example (2) 2 1 4 0 6 0 1 6 1 4 x = (2, 1, 0, 0, 1, 1, 4, 6, 6, 4) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 1, 1, 2, 4, 4, 6, 6) ∩ 6 6 4 4 0 0 1 1 1 2 M. Dukes (University of Strathclyde) QMUL 2013 8 / 22 7. From stable states to polyominoes Example (2) 2 1 4 0 6 0 1 6 1 4 x = (2, 1, 0, 0, 1, 1, 4, 6, 6, 4) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 1, 1, 2, 4, 4, 6, 6) ∩ 6 6 4 4 = = fm,n (x) 0 0 1 1 1 2 M. Dukes (University of Strathclyde) QMUL 2013 8 / 22 7. From stable states to polyominoes Example (2) 2 1 4 0 6 0 1 6 1 4 x = (2, 1, 0, 0, 1, 1, 4, 6, 6, 4) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 1, 1, 1, 2, 4, 4, 6, 6) ∩ 6 6 4 4 = = fm,n (x) 0 0 1 1 1 2 x is not recurrent, i.e. x 6∈ rec(K7,4 ) M. Dukes (University of Strathclyde) QMUL 2013 8 / 22 7. From stable states to polyominoes Example (3) 2 1 3 2 6 0 2 1 2 6 x = (2, 1, 2, 0, 2, 2, 3, 6, 1, 6) ∈ stable(K7,4 ) inc7,4 (x) = (0, 1, 2, 2, 2, 2, 1, 3, 6, 6) ∩ 6 6 3 1 = = fm,n (x) 0 1 2 2 2 2 x is recurrent, i.e. x ∈ rec(K7,4 ) M. Dukes (University of Strathclyde) QMUL 2013 9 / 22 7. From stable states to polyominoes Example (4) 0 3 3 0 6 3 3 3 0 3 x = (0, 3, 0, 3, 3, 0, 3, 6, 3, 3) ∈ stable(K7,4 ) inc7,4 (x) = (0, 0, 0, 3, 3, 3, 3, 3, 3, 6) ∩ 6 3 3 3 = = fm,n (x) 0 0 0 3 3 3 x is recurrent, i.e. x ∈ rec(K7,4 ) M. Dukes (University of Strathclyde) QMUL 2013 10 / 22 8. Parallelogram polyominoes A parallelogram polyomino is a polyomino such that its intersection with every line of slope −1 is a connected segment. Let Para m,n be the set of all parallelogram polyominoes whose bounding rectangle is [0, m] × [0, n]. Example P1 = ∈ Para 7,4 ; M. Dukes (University of Strathclyde) P2 = 6∈ Para 7,4 ; QMUL 2013 P3 = ∈ Ribbon 7,4 . 11 / 22 9. Classifying recurrent states of Km,n by polyominoes Theorem (MD & Le Borgne) Let u = (u1 , . . . , um+n−1 ) ∈ stable(Km,n ). Then u ∈ rec(Km,n ) M. Dukes (University of Strathclyde) ⇐⇒ QMUL 2013 fm,n (u) ∈ Para m,n . 12 / 22 9. Classifying recurrent states of Km,n by polyominoes Theorem (MD & Le Borgne) Let u = (u1 , . . . , um+n−1 ) ∈ stable(Km,n ). Then u ∈ rec(Km,n ) ⇐⇒ fm,n (u) ∈ Para m,n . Corollary The number of ‘different’ recurrent configurations in rec(Km,n ) is Nara(m + n − 1, m) where 1 a a Nara(a, b) = a b b−1 are the Narayana numbers. M. Dukes (University of Strathclyde) QMUL 2013 12 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 2 0 0 2 2 1 3 6 3 M. Dukes (University of Strathclyde) QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 2 0 0 2 2 1 4 7 4 M. Dukes (University of Strathclyde) QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 2 0 0 2 2 1 8 is first to topple, 4 7 4 M. Dukes (University of Strathclyde) QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 3 1 1 3 3 2 8 is first to topple, then 1, 4 and 5, 4 0 4 M. Dukes (University of Strathclyde) QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 0 1 1 0 0 2 8 is first to topple, then 1, 4 and 5, then 7 and 9, 7 3 7 M. Dukes (University of Strathclyde) QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 2 3 3 2 2 4 0 3 0 M. Dukes (University of Strathclyde) 8 is first to topple, then 1, 4 and 5, then 7 and 9, and finally 2, 3 and 6. QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 2 0 0 2 2 1 {2, 3, 6} {1, 4, 5} (P, A, B) = 3 6 {7, 9} 3 M. Dukes (University of Strathclyde) {8} QMUL 2013 13 / 22 10. Topplings of rec(Km,n ) and bounce paths for Para m,n How to distinguish between recurrent configurations which map to the same P ∈ Param,n ? Example 2 0 0 2 2 1 {2, 3, 6} {1, 4, 5} (P, A, B) = 3 6 {8} {7, 9} 3 Elements of rec(Km,n ) are in 1-1 correspondence with parallelogram polyominoes in Para m,n whose ‘bounce path’ is decorated with a set partition. M. Dukes (University of Strathclyde) QMUL 2013 13 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example D= M. Dukes (University of Strathclyde) ∈ D5 QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example (3,3) ∈ D5 D= (1,1) M. Dukes (University of Strathclyde) QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example (3,3) bounce(D) = 1 + 3 = 4 ∈ D5 D= (1,1) M. Dukes (University of Strathclyde) QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example (3,3) bounce(D) = 1 + 3 = 4 ∈ D5 D= (1,1) M. Dukes (University of Strathclyde) QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example bounce(D) = 1 + 3 = 4 area(D) = 3 (3,3) ∈ D5 D= (1,1) M. Dukes (University of Strathclyde) QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example bounce(D) = 1 + 3 = 4 area(D) = 3 (3,3) ∈ D5 D= (1,1) The q, t-Catalan polynomial is defined as the generating function of these two statistics: X Cn (q, t) = q area(D) t bounce(D) . D∈Dn M. Dukes (University of Strathclyde) QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials Let Dn be the set of all Dyck paths of semi-length n. Example bounce(D) = 1 + 3 = 4 area(D) = 3 (3,3) ∈ D5 D= (1,1) The q, t-Catalan polynomial is defined as the generating function of these two statistics: X Cn (q, t) = q area(D) t bounce(D) . D∈Dn e.g. C3 (q, t) = q0t 3 + M. Dukes (University of Strathclyde) q1t 2 + q2t 1 + q1t 1 + q3t 0. QMUL 2013 14 / 22 11. Interlude: q, t-Catalan numbers/polynomials What’s so interesting about these numbers/polynomials? Jim Haglund’s book The q, t-Catalan Numbers and the Space of Diagonal Harmonics: With an Appendix on the Combinatorics of Macdonald Polynomials, AMS Univ. Lect. Series 41, 2008, contains a wealth of information about these polynomials. Related to Macdonald polynomials and the space of diagonal harmonics. Cn (q, t) was shown to be symmetric in q and t by Garsia and Haglund (2001/2002). n q (2) Cn (q, 1/q) is the nth q-Catalan number. Egge, Haglund, Kremer & Killpatrick (2003) asked if the lattice path statistics for Cn (q, t) can be extended, in a way which preserves the rich combinatorial structure, to related combinatorial objects. M. Dukes (University of Strathclyde) QMUL 2013 15 / 22 12. A q, t-Narayana polynomial Given P ∈ Para m,n let area(P) be the number of cells in P. M. Dukes (University of Strathclyde) QMUL 2013 16 / 22 12. A q, t-Narayana polynomial Given P ∈ Para m,n let area(P) be the number of cells in P. Let parabounce(P) be a statistic defined in the following way: 1 2 2 3 P1 = 3 4 4 5 3 1 1 1 ∈ Para 9,7 ; 3 3 P2 = 2 2 1 2 2 1 1 1 1 ∈ Para 8,2 . 5 parabounce(P1 ) = 1 + 1 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 3 + 3 + 4 + 4 +5 + 5 = 41 parabounce(P2 ) = 1 + 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 = 13. M. Dukes (University of Strathclyde) QMUL 2013 16 / 22 12. A q, t-Narayana polynomial Let Fm,n (q, t) be the generating function of the bi-statistic (area, parabounce) on Para m,n : def Fm,n (q, t) = X q area(P) t parabounce(P) . P∈Para m,n Note that Fm,n (1, 1) = Nara(m + n − 1, m). We call Fm,n (q, t) the q, t-Narayana polynomial. q, t-Narayana symmetry conjectures (i) Fm,n (q, t) is symmetric in q and t. (ii) Fm,n (q, t) is symmetric in m and n. M. Dukes (University of Strathclyde) QMUL 2013 17 / 22 13. More on the q, t-Narayana polynomial These two conjectures have recently been solved in Statistics on parallelogram polyominoes and a q, t-analogue of the Narayana numbers J-C. Aval, M. D’Adderio, myself, A. Hicks, Y. Le Borgne arXiv:1301.4803 Theorem Fm,n (q, t) is also the g.f. for the bi-statistic (dinv, area) Theorem Fm,n (q, t) = (qt)m+n−1 h∇em+n−2 , hm−1 hn−1 i where ek and hk are the elementary and homogeneous symmetric functions of degree k, respectively. M. Dukes (University of Strathclyde) QMUL 2013 18 / 22 13. Further exciting developments Combinatorics of Labelled Parallelogram polyominoes J-C. Aval, F. Bergeron, A. Garsia – arXiv:1301.3035 An algebraic look at actions of the symmetric group on labelled parallelogram polyominoes which shows connections to Macdonal polynomials, amongst other things. The sandpile model on Km,n and a Cyclic Lemma J-C. Aval, M. D’Adderio, myself, Y. Le Borgne – preprint 2013 Introduces operators on stable configurations of the sandpile model that lead to an algorithmic bijection between recurrent and parking configurations which preserves their equivalence classes with respect to the sandpile group. Studies them in the special case of the graph Km,n , showing their connection to a generalization of the well known Cyclic Lemma of Dvoretsky and Motzkin. M. Dukes (University of Strathclyde) QMUL 2013 19 / 22 14. What about the original class? The collection of configurations that correspond to the original class, i.e. upper triangular matrices or pattern avoiding permutations is {u ∈ Rec(Dn,n ) : u is minanz and waveu (vn+x ) ≤ waveu (vx ) ∀x ≥ 1} In terms of parallelogram polyominoes, they correspond to ribbon polyominoes which do not pass through the cell [1, 2] × [0, 1]. M. Dukes (University of Strathclyde) QMUL 2013 20 / 22 15. A relation to Haglund’s bounce path Let LowerParan,n−1 be the set of all P ∈ Para n,n−1 whose upper defining Dyck path rests on the main diagonal. 5 4 P= 3 D = dyck(P) = 2 1 0 Let X Sn (q, t) = q area(P) t parabounce(P) . P∈LowerParan,n−1 Theorem Sn (q, t) = (qt)2n Cn−1 (q, t 2 ). M. Dukes (University of Strathclyde) QMUL 2013 21 / 22 16. A conjecture related to walks in the plane A lattice walks conjecture Let an be the number of configurations x ∈ rec(Kn,n ) such that incn,n (x) = x and all x entries are non-zero. Then 2n − 2 2n 1 , an = n−1 n n−2 the number of walks from (0, 0) to (0, 1) that remain in the upper half-plane (y ≥ 0) using 2n − 3 unit steps {n, s, e, w }. Is there a combinatorial explanation for why these configurations are equinumerous with certain walks in the plane? M. Dukes (University of Strathclyde) QMUL 2013 22 / 22