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

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
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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
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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)
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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)
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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
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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
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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)
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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
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=
=
degree of vertex vi
number of edges between
vi and vj .
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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