Combinatorics of the Del-Pezzo 3 Quiver: Aztec Dragons, Castles, …musiker/Beyond.pdf · 2015. 3. 16. · Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond

Combinatorics of the Del-Pezzo 3 Quiver:Aztec Dragons, Castles, and Beyond

Tri Lai (IMA) and Gregg Musiker (University of Minnesota)

AMS Special Session on Integrable Combinatorics

March 15, 2015

http//math.umn.edu/∼musiker/Beyond.pdf

Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 1 / 54

Outline

1 Introducing the del Pezzo 3 Quiver

2 Contours in the Honeycomb Lattice

3 Reimagining the Aztec Castles of Leoni-M-Neel-Turner

4 A Third Direction

5 Further Directions

Thank you to NSF Grants DMS-1067183, DMS-1148634, DMS-1362980,and the Institute for Mathematics and its Applications.

http//math.umn.edu/∼musiker/Beyond.pdf


Motivations beyond studying Bipartite Graphs on a Torus

Clust. Int. System ↔ Poisson Struct. ↔ Newton Polygon ↔ Bip. Graph

In String Theory, referred to as Brane Tilings and appearWhen studying intersections of NS5 and D5-branes or the geometry in

a (3+1)-dimensional N = 1 supersymmetric quiver gauge theory.

In Algebraic Geometry, they are used to

Probe certain toric Calabi-Yau singularities, and relate tonon-commutative crepant resolutions and the 3-dimensional McKaycorrespondence.

Certain examples of path algebras with relations (Jacobian Algebras) (alsoknown as Dimer Algebras or Superpotential Algebras) can be constructedby a quiver and potential (Q0,Q1,Q2) dual to a bipartite graph on asurface.


Brane Tilings (Combinatorially)

Physicists/Geometers use Kasteleyn matrix and weighted enumeration ofperfect matchings of these bipartite graphs to obtain bivariate Laurentpolynomials whose Newton polygon recovers toric data of variety whosecoordinate ring is center of corresponding Jacobian algebra.

Examples:

3

1

33

1

2 2

3

4

,

2 4 6 1 3

2 4

1 3 5

2

6

5 7 6 1

4

7 2 4

5

5 73 6 1

3

7

3

We will instead view such tessellations as doubly-periodic tessellations ofits universal cover, the Euclidean plane.


Brane Tilings from a Quiver Q with Potential W

A Brane Tiling can be associated to a pair (Q,W ), where Q is a quiverand W is a potential (called a superpotential in the physics literature).

A quiver Q is a directed graph where each edge is referred to as an arrow.

A potential W is a linear combination of cyclic paths in Q (possibly aninfinite linear combination).

For combinatorial purposes, we assume other conditions on (Q,W ):

• Each arrow of Q appears in one term of W with a positive sign, andone term with a negative sign.

• The number of terms of W with a positive sign equals the numberwith a negative sign. All coefficients in W are ±1.

Also, #V −#E + #F = 0 where F = {terms in the potential}.

Lastly, by a result of Vitoria, we wish to focus on potentials where nosubpath of length two appears twice.


Brane Tilings from a Quiver Q with Potential W

A Brane Tiling can be associated to a pair (Q,W ), where Q is a quiverand W is a potential (called a superpotential in the physics literature).

A quiver Q is a directed graph where each edge is referred to as an arrow.

A potential W is a linear combination of cyclic paths in Q (possibly aninfinite linear combination).

For combinatorial purposes, we assume other conditions on (Q,W ):

• Each arrow of Q appears in one term of W with a positive sign, andone term with a negative sign.

• The number of terms of W with a positive sign equals the numberwith a negative sign. All coefficients in W are ±1.

Also, #V −#E + #F = 0 where F = {terms in the potential}.

Lastly, by a result of Vitoria, we wish to focus on potentials where nosubpath of length two appears twice.


The Del Pezzo 3 Quiver

We now consider the quiver and potential whose associated Calabi-Yau3-fold is the cone over CP2 blown up at three points.

Its Toric Diagram = {(−1, 1), (0, 1), (1, 0), (1,−1), (0,−1), (−1, 0), (0, 0)}.

QdP3 = Q =

4

6

1

3

5

2

,

W = A16A64A42A25A53A31 + A14A45A51 + A23A36A62

− A16A62A25A51 − A36A64A45A53 − A14A42A23A31.


The Del Pezzo 3 Quiver

We now consider the quiver and potential whose associated Calabi-Yau3-fold is the cone over CP2 blown up at three points.

Its Toric Diagram = {(−1, 1), (0, 1), (1, 0), (1,−1), (0,−1), (−1, 0), (0, 0)}.

QdP3 = Q =

4

6

1

3

5

2

,

W = A16A64A42A25A53A31 + A14A45A51 + A23A36A62

− A16A62A25A51 − A36A64A45A53 − A14A42A23A31.


The Del Pezzo 3 Quiver (continued)

We now unfold Q onto the plane, letting the three positive (resp. negative)terms in W depict clockwise (resp. counter-clockwise) cycles on Q.

Q =

4

6

1

3

5

2

unfolds to Q =

66

6

F

AD

B

A

C

F E F

F

B

D

C

5

3

5

3

3

5

3

1

2

4

4

4

1

2

1

2

4

1

2

W = A16A64A42A25A53A31(A) + A14A45A51 (B) + A23A36A62 (C )

− A16A62A25A51 (D) − A36A64A45A53 (E ) − A14A42A23A31(F ).



Taking the planar dual yields a bipartite graph on a torus (Brane Tiling):

66

6

F

AD

B

A

C

F E F

F

B

D

C

5

3

5

3

3

5

3

1

2

4

4

4

1

2

1

2

4

1

2

Q −→ TQ =

6

4

2

5

3

1

1

3

6

1

6

4

2

5

3

1

3

2

5

2

4

F

AD

B

A

C

F F

F

B

D

E

B

E

C

B

D

C

Negative Term in W ←→ Counter-Clockwise cycle in Q ←→ • in TQPositive Term in W ←→ Clockwise cycle in Q ←→ ◦ in TQ(To obtain Q from TQ , we dualize edges so that white is on the right.)



Summarizing the dP3 Example:

Q

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TQNegative Term in W ←→ Counter-Clockwise cycle in Q ←→ • in TQPositive Term in W ←→ Clockwise cycle in Q ←→ ◦ in TQDoubly periodic Perfect Matchings ↔ points in Toric Diagram.


Quiver Mutation (a.k.a. Seiberg Duality in Physics)

Pick vertex j of Q. (1) For every two path i → j → k, add a new arrowi → k . (2) Reverse all arrows incident to j . (3) Delete any 2-cycles.

Example: Start with dP3 and mutate at vertices 1, 4, and 3 in order.

Together, these are known as the Four Toric Phases of dP3. Note that weonly mutate at vertices with in-degree and out-degree 2.


Toric Phases of dP3 (Models I, II, III, and IV)

Toric mutations take place at vertices with in-degree and out-degree 2.

Compare with Figure 27 of Eager-Franco (“Colored BPS Pyramid PartitionFunctions, Quivers and Cluster Transformations”)

I IVIIIII

6 4 3

211

6

5

4 3

21 2

34

5

6 6

1

2

3

4

5

5


Toric Phases of dP3 (Models I, II, III, and IV)

Toric mutations take place at vertices with in-degree and out-degree 2.

Starting with any of these four models of the dP3 quiver, any sequence oftoric mutations yields a quiver that is graph isomorphic to one of these.

Figure 20 of Eager-Franco (Incidences betweeen these Models):

3

3

2

3

2

11 2

22

3

3

3 3

2

3

1

4

2

422


Cluster Variable Mutation

In addition to the mutation of quivers, there is also a complementarycluster mutation that can be defined.

Cluster mutation yields a sequence of Laurent polynomials inQ(x1, x2, . . . , xn) known as cluster variables.

Given a quiver Q (the potential is irrelevant here) and an initial cluster{x1, . . . , xN}, then mutating at vertex k yields a new cluster variable x ′k

defined by x ′k =

∏k→i∈Q

xi +∏

i→k∈Qxi

/xk .

Example: Mutating the dP3 quiver periodically at 1, 2, 3, 4, 5, 6, 1, 2, . . .yields Laurent polynomials x3x5+x4x6

x1, x4x6+x3x5

x2,

x2x3x25+x1x3x5x6+x2x4x5x6+x1x4x26x1x2x3

,x2x3x25+x1x3x5x6+x2x4x5x6+x1x4x26

x1x2x4,

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x5

, (x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x6

, . . .






defined by x ′k =

∏k→i∈Q

xi +∏

i→k∈Qxi

/xk .


x1, x4x6+x3x5

x2,



x1x2x4,

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x5

, (x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x6

, . . .






defined by x ′k =

∏k→i∈Q

xi +∏

i→k∈Qxi

/xk .


x1, x4x6+x3x5

x2,



x1x2x4,

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x5

, (x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x6

, . . .


Goal: Combinatorial Formula for Cluster Variables

Example from S. Zhang (2012 REU): Periodic mutation1, 2, 3, 4, 5, 6, 1, 2, . . . yields partition functions for Aztec Dragons (asstudied by Ciucu, Cottrell-Young, and Propp) under appropriate weightedenumeration of perfect matchings.

1 x3x5+x4x6x1

2 x4x6+x3x5x2

2

31


42

1


2

3

42

5

1

1 1

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x5

2

3

64

2

5

1

1

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x6


Goal: Combinatorial Formula for Cluster Variables

Example from M. Leoni, S. Neel, and P. Turner (2013 REU):Mutations at antipodal vertices of dP3 quiver yield τ -mutation sequences.Resulting Laurent polynomials correspond to Aztec Castles underappropriate weighted enumeration of perfect matchings.

e.g. 1, 2, 3, 4, 1, 2, 5, 6 yields cluster variable

(x1x22 x

33 x

45 + x32 x

23 x4x

45 + 2x21 x2x

33 x

35 x6 + 4x1x

22 x

23 x4x

35 x6 + 2x32 x3x

24 x

35 x6 + x31 x

33 x

25 x

26

+ 5x21 x2x23 x4x

25 x

26 + 5x1x

22 x3x

24 x

25 x

26 + x32 x

34 x

25 x

26 + 2x31 x

23 x4x5x

36 + 4x21 x2x3x

24 x5x

36

+ 2x1x22 x

34 x5x

36 + x31 x3x

24 x

46 + x21 x2x

34 x

46 )/x

21 x

22 x

23 x

24 x6 =

(x1x3 + x2x4)(x4x6 + x3x5)2(x1x6 + x2x5)

2

x21 x22 x

23 x

24 x6

1

1

13 3

5

4

4 4

2

2


Introducing Contours on the del Pezzo 3 Lattice

We wish to understand combinatorial interpretations for more generaltoric mutation sequences, not necessarily periodic or coming frommutating at antipodes.

To this end, we cut out subgraphs of the dP3 lattice by using six-sidedcontours

64

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1

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a

b

c

d

e

f

indexed as (a, b, c , d , e, f ) with a, b, c , d , e, f ∈ Z.Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 16 / 54


a

b

c

d

e

f

ff

ff

aa aa

a

b

b

b

b b

cc

cc

c

d d

d

e ee

e

e

f

From left to right, the cases where (a, b, c, d , e, f ) =(1) (+,+,+,+,+,+), (2) (+,−,+,+,−,+), (3) (+,−,+, 0,−,+), (4) (+,−,+,+,−,−), (5) (+,+,+,−,+,−)

Sign determines direction of the sides. Magnitude determines length.

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1

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1

6

e=6

3

a=3

b=2

c=4

d=2

e=3

f=3

a=3

c=2

d=2

f=1

a=3

c=4

f=1

d=0

e= -1

b= -4

e= -3

b= -4

a=5

b= -6

c=3d=0

e= -1f= -2

a=5b= -6

c=2

d=2

e= -3

f= -1

a=2

b=1

c=1

d= -3

f= -4

3

42

5



a

b

c

d

e

f

ff

ff

aa aa

a

b

b

b

b b

cc

cc

c

d d

d

e ee

e

e

f

From left to right, the cases where (a, b, c, d , e, f ) =(1) (+,+,+,+,+,+), (2) (+,−,+,+,−,+), (3) (+,−,+, 0,−,+), (4) (+,−,+,+,−,−), (5) (+,+,+,−,+,−)


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4

1

6

e=6

3

a=3

b=2

c=4

d=2

e=3

f=3

a=3

c=2

d=2

f=1

a=3

c=4

f=1

d=0

e= -1

b= -4

e= -3

b= -4

a=5

b= -6

c=3d=0

e= -1f= -2

a=5b= -6

c=2

d=2

e= -3

f= -1

a=2

b=1

c=1

d= -3

f= -4

3

42

5



(1) G(3, 2, 4, 2, 3, 3), (2) G(3,−4, 2, 2,−3, 1), (3) G(3,−4, 4, 0,−1, 1),

(4) G(5,−6, 3, 0,−1,−2), (5) G(5,−6, 2, 2,−3,−1), (6) G(2, 1, 1,−3, 6,−4)

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13

35

35

2

35

2

5

2

5

24

24

6 24

6 24

6 24

64

6

1

4

6

1

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

1

4

1

6

e=6

3

a=3

b=2

c=4

d=2

e=3

f=3

a=3

c=2

d=2

f=1

a=3

c=4

f=1

d=0

e= -1

b= -4

e= -3

b= -4

a=5

b= -6

c=3d=0

e= -1f= -2

a=5b= -6

c=2

d=2

e= -3

f= -1

a=2

b=1

c=1

d= -3

f= -4

3

42

5


Turning a Contour C into the Subgraph G(C)

1) Draw the contour C on top of the dP3 lattice starting from a degree 6white vertex.2) For all sides of “positive” length, we erase all the black vertices.3) For all sides of “negative” length, we erase all the white vertices.For sides of “zero length” (between two sides of positive length), we erasethe white corner or keep it depending on convexity.4) After removing “dangling” edges and their incident faces, remainingsubgraph inside contour is G(C).

e.g.

2

3

6

4

15

4

15

2

3

6 2

3

6

2

3

64

15

2

3

6

4

15

e=-2

d=+3

c=-1

b=-1

a=+2

f=0

2

3

6

4

15

4

15

2

3

6 2

3

6

2

3

64

15

2

3

6

4

15

e=-2

d=+3

c=-1

b=-1

a=+2

f=0

4

1

1

2

2

315

e=-2

d=+3

c=-1

b=-1

a=+2

f=0




e.g.

2

3

6

4

15

4

15

2

3

6 2

3

6

2

3

64

15

2

3

6

4

15

e=-2

d=+3

c=-1

b=-1

a=+2

f=0

2

3

6

4

15

4

15

2

3

6 2

3

6

2

3

64

15

2

3

6

4

15

e=-2

d=+3

c=-1

b=-1

a=+2

f=0

4

1

1

2

2

315

e=-2

d=+3

c=-1

b=-1

a=+2

f=0




e.g.

2

3

6

4

15

4

15

2

3

6 2

3

6

2

3

64

15

2

3

6

4

15

e=-2

d=+3

c=-1

b=-1

a=+2

f=0

2

3

6

4

15

4

15

2

3

6 2

3

6

2

3

64

15

2

3

6

4

15

e=-2

d=+3

c=-1

b=-1

a=+2

f=0

4

1

1

2

2

315

e=-2

d=+3

c=-1

b=-1

a=+2

f=0


Further Examples of Subgraphs G(C) from Contours C1) Draw the contour C on top of the dP3 lattice starting from a degree 6white vertex.2) For all sides of “positive” length, we erase all the black vertices.3) For all sides of “negative” length, we erase all the white vertices.For sides of “zero length” (between two sides of positive length), we erasethe white corner or keep it depending on convexity.4) After removing “dangling” edges and their incident faces, remainingsubgraph inside contour is G(C).

1

4

113

513

5

6

13

5

24

6

13

5

24

13

55

5

24

35

24

6

13

5

24

6

1

46

13

5

24

6

1

46

13

5

24

6

1

4

4

5

24

13

5

2

5

5

24

5

5

24

13

5

2

35

2

35

2

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

4

6

1

4

6

1

4

6

1

4

6

1

4

6

1

4

113

5

6

13

5

6

13

3

4

6

1

4

5

24

5

24

6

13

5

24

6

1

4

6

1

46

13

5

24

6

13

5

24

6

5

24

2

35 1

35 1

35

35 1

5

24

5

24

4

6

1

4

6

1

4

6

13

5

24

6

13

5

24

5

24

5

24

5 13

5 13

5

2

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

35

6

13

5

24

6

13

5

24

13

5 13

5

6

1

4

1

6

1

46

13

5

24

6

1

4

4

5

24

5

24

2

35

2 6

13

5

24

6

5

24

6

13

5

24

6

13

5

24

6

1

4

113

5

6

13

5

6

13

3

35

2

13

5 13

5 1

6

6

13

6

13

35

35

2

35

2

5

2

5

24

24

6 24

6 24

6 24

64

6

1

4

6

1

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

1

4

1

6

e=6

3

a=3

b=2

c=4

d=2

e=3

f=3

a=3

c=2

d=2

f=1

a=3

c=4

f=1

d=0

e= -1

b= -4

e= -3

b= -4

a=5

b= -6

c=3d=0

e= -1f= -2

a=5b= -6

c=2

d=2

e= -3

f= -1 a=2

b=1

c=1

d= -3

f= -4

3

42

5


Aztec Dragons (Ciucu, Cottrell-Young, Propp) Revisited

Dn+1/2 = G(n + 1,−n,−1, n + 2,−n − 1, 0).

Dn = G(n + 1,−n − 1, 1, n,−n, 0).

24

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2 6

13

5

24

35

2

3

6

13

6

13

6

13

5 1

6

1

4

6

1

4

6

1

4

4

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

24

6

13

6

6

13

3

6

133

5

2

6

13

5

24

35

2

5

2

5

24

5

24

44

6

1

4

5

24

5

24

5

2 6

13

5

24

6

13

5

24

6

1

4

1

6

13

5

6

133

5

2

3

6

1

24

6

6

13

6

5

3

3

35

2

6

13

5

24

5

2

1 5

24

44

5

2

5

24

1

3

6

13

5

6

1

4

24

5

5

2

5

42

1

D1/2 1 3/2

2 5/2 3

D D

D D D

a=1

b=0 c= -1

d=2

e= -1

f=0a=2

b= -2

c=1d=1e= -1

f=0

a=2

b= -1

c= -1

d=3

e= -2

f=0

a=3 b= -3

c=1

d=2e= -2

f=0

a=3

b= -2

c= -1

d=4e= -3

f=0

a=4

b= -4

c=1

d=3e= -3

f=0

31

6

3

2


Aztec Castles (Leoni-M-Neel-Turner) Revisited

NE-Aztec Castles γji = G(j ,−i − j , i , j + 1,−i − j − 1, i + 1)

SW-Aztec Castles γ−j−i = G(−j + 1, i + j ,−i ,−j , i + j + 1,−i − 1)

e.g. γ11 = G(1,−2, 1, 2,−3, 2) and γ−3−2 = G(−2, 5,−2,−3, 6,−3).

6

133

5

2

35

2

2 6 24

4

6

1

4

1

6

13

5

6

5

24

6

13

5

2

3

a=1

b= -2

c=1

d=2

e= -3

f=2

4

e=6

d= -3

c= -2

b=5

a= -2

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

3

6

133

5

2

35

2

35

2

35

2

2 6 24

6

5

24

6

13

5

24

5

24

6

13

5

24

5

24

4

6

1

4

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

4

113

513

5

6

13

5

24

6

13

5

24

6

13

5

6

13

5

2

f= -342

53

1

6

4


Aztec Castles (Leoni-M-Neel-Turner) Revisited

NE-Aztec Castles γji = G(j ,−i − j , i , j + 1,−i − j − 1, i + 1)

SW-Aztec Castles γ−j−i = G(−j + 1, i + j ,−i ,−j , i + j + 1,−i − 1)

e.g. γ11 = G(1,−2, 1, 2,−3, 2) and γ−3−2 = G(−2, 5,−2,−3, 6,−3).

1

1

13 3

5

4

4 4

2

2

6

3 52 63

32 6

32 6

2 62 62 62 63

6 2 6 2 6 2 6

6 2 65 1

4

5 1

45 1

4

5 14

5 14

5 14

5 14

5 14

54

6

54

5333

6

5 1 5 1 3 3

5

5

36


Turning Subgraphs into Laurent Polynomials

G −→ cm(G )∑

M = a perfect matching of G

x(M), where

x(M) =∏

edge e∈M1

xixj(for edge e straddling faces i and j),

cm(G ) = the covering monomial of the graph Gn (which records what facelabels are contained in G and along its boundary).

Remark: This is a reformulation of weighting schemes appearing in workssuch as Speyer (“Perfect Matchings and the Octahedron Recurrence”),Goncharov-Kenyon (“Dimers and cluster integrable systems”), and DiFrancesco (“T-systems, networks and dimers”).

Alternative definition of cm(G): We record all face labels inside contourand then divide by face labels straddling dangling edges.


Initial cluster {x1, x2, . . . , x6} in terms of contours

Consider the following six special contours

C1 = (0, 0, 1,−1, 1, 0), C2 = (−1, 1, 0, 0, 0, 1),

C3 = (0, 1,−1, 1, 0, 0), C4 = (1, 0, 0, 0, 1,−1),

C5 = (1,−1, 1, 0, 0, 0), C6 = (0, 0, 0, 1,−1, 1).

4

15

c=+1

e=+1

f=a=b=0

d=-1 2

3

6

f=+1

a=-1b=+1

c=d=e=0

Applying our general algorithm, G(Ci )’s correspond to graphs consisting ofa single edge and a triangle of faces.

Using G −→ cm(G )∑

M = a perfect matching of G x(M), we see

cm(G(C1)) = x1x4x5 and x(M) =1

x4x5, hence G −→ x1x4x5

x4x5= x1

Similar calculations show G(Ci ))←→ xi for i ∈ {1, 2, . . . , 6}.


Initial cluster {x1, x2, . . . , x6} in terms of contours


C1 = (0, 0, 1,−1, 1, 0), C2 = (−1, 1, 0, 0, 0, 1),

C3 = (0, 1,−1, 1, 0, 0), C4 = (1, 0, 0, 0, 1,−1),

C5 = (1,−1, 1, 0, 0, 0), C6 = (0, 0, 0, 1,−1, 1).

4

15

c=+1

e=+1

f=a=b=0

d=-1 2

3

6

f=+1

a=-1b=+1

c=d=e=0

Applying our general algorithm, G(Ci )’s correspond to graphs consisting ofa single edge and a triangle of faces.

Using G −→ cm(G )∑

M = a perfect matching of G x(M), we see

cm(G(C1)) = x1x4x5 and x(M) =1

x4x5, hence G −→ x1x4x5

x4x5= x1

Similar calculations show G(Ci ))←→ xi for i ∈ {1, 2, . . . , 6}.Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 25 / 54

Visualizing Contours as points in Z2


C1 = (0, 0, 1,−1, 1, 0), C2 = (−1, 1, 0, 0, 0, 1),

C3 = (0, 1,−1, 1, 0, 0), C4 = (1, 0, 0, 0, 1,−1),

C5 = (1,−1, 1, 0, 0, 0), C6 = (0, 0, 0, 1,−1, 1).

Note that the six subgraphs corresponding to the initial contours are

G(C1) = σγ−10 , G(C3) = γ0−1, G(C5) = σγ00 ,

G(C2) = γ−10 , G(C4) = σγ0−1, G(C6) = γ00

using γji = G(j ,−i − j , i , j + 1,−i − j − 1, i + 1) for i , j ∈ Z2.

σγji = G(j + 1,−i − j − 1, i + 1, j ,−i − j , i) for i , j ∈ Z2.

Remark: The operation σ appears to be 180◦ rotation. Better to think ofit as σ(a, b, c , d , e, f ) = (a + 1, b − 1, c + 1, d − 1, e + 1, f − 1).



G(C1) = σγ−10 , G(C3) = γ0−1, G(C5) = σγ00 ,

G(C2) = γ−10 , G(C4) = σγ0−1, G(C6) = γ00

(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2

τ3 τ2 τ1τ1τ2τ3

τ3

Dn+1/2 = γn+1−1 = G(n + 1,−n,−1, n + 2,−n − 1, 0).

Dn = σγn0 = G(n + 1,−n − 1, 1, n,−n, 0).

e.g. Dn’s and Dn+1/2 correspond to vertical lines (0, n) and (−1, n + 1).



G(C1) = σγ−10 , G(C3) = γ0−1, G(C5) = σγ00 ,

G(C2) = γ−10 , G(C4) = σγ0−1, G(C6) = γ00

(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2


τ3

Dn+1/2 = γn+1−1 = G(n + 1,−n,−1, n + 2,−n − 1, 0).

Dn = σγn0 = G(n + 1,−n − 1, 1, n,−n, 0).

e.g. Dn’s and Dn+1/2 correspond to vertical lines (0, n) and (−1, n + 1).


Adding a Third Dimension

About the same time the REU 2013 students and I were investigatingAztec Castles, there was independent work of T. Lai on (unweighted)enumeration of tilings of Aztec Dragon Regions.

“New Aspects of Hexagonal Dungeons”, arXiv:1403.4481,

inspired by T. Lai’s work on Blum’s Conjecture with M. Ciucu (JCTA2014, arXiv:1402.7257)

Unifying all these points of view, we define the contours

σkCji = (j+k,−i−j−k , i+k , j+1−k ,−i−j−1+k , i+1−k) for (i , j , k) ∈ Z3.


G(σ0Cji ) = γji , G(σ1Cji ) = σγji , G(σ1C−j−i+1) = γ−j−i , G(σ0C−j−i+1) = σγ−j−i .


Triangles in 45− 45− 90 square lattice ←→ prisms in Z3

σkCji = (j+k,−i−j−k , i+k , j+1−k ,−i−j−1+k, i+1−k) for (i , j , k) ∈ Z3.

(i-1,j) (i,j)

(i,j-1)

1

2

3

4

5

6

(i-1,j) (i.j)

(i-1,j+1)

(i-1,j,k-1)

(i-1,j,k)

(i-1,j+1,k)

(i-1,j+1,k-1)

(i.j.k)

(i.j.k-1)3

2

(i,j-1,k)

(i,j,k)

6

5(i,j,k-1)

1 (i,j-1,k-1)

(i-1,j,k)

4(i-1,j,k-1)

{(0,−1), (−1, 0), (0, 0)} ←→ [σγ−10 , γ−10 , γ0−1, σγ0−1, σγ

00 , γ

00 ]

Thus {(0,−1), (−1, 0), (0, 0)} ←→ [x1, x2, . . . , x6], the initial cluster.Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 29 / 54

Introducing τ -mutation sequences

Start with Model I of the dP3 quiver and the initial cluster {x1, x2, . . . , x6}.

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

42

53

1

6

Let τ1 = µ1 ◦ µ2 ◦ (12), τ2 = µ3 ◦ µ4 ◦ (34), and τ3 = µ5 ◦ µ6 ◦ (56).

Each τi mutates at a vertex of dP3 followed by mutation at its antipode.These mutations commute.

Further on the level of clusters, (τ1τ2)3 = (τ1τ3)3 = (τ2τ3)3 = 1(as discussed with P. Pylyavskyy) and there are no other relations.

Thus 〈τ1, τ2, τ3〉 ∼= A2, considering the action of τi ’s on clusters.

Remark: The permutations are necessary here so that (τiτj)3 does not

reorder clusters as labeled seeds.




64

2

53

1

64

2

53

1

64

2

53

1

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2

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1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

42

53

1

6










64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

42

53

1

6










64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

42

53

1

6






reorder clusters as labeled seeds.Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 30 / 54

Mutating Model I to Model II and back to Model I

By applying τ1 = µ1 ◦ µ2 ◦ (12), τ2 = µ3 ◦ µ4 ◦ (34), or τ3 = µ5 ◦ µ6 ◦ (56),we mutate the quiver:

−→ −→

Claim: Corresponding action on triangular prisms:

32

(i,j-1,k)

(i,j,k)

6

5(i,j,k-1)

1 (i,j-1,k-1)

(i-1,j,k)

4(i-1,j,k-1)

1

23

4

5

6

(i-1,j,k-1)

(i-1,j,k)

(i-1,j+1,k)

(i-1,j+1,k-1)

(i.j.k)

(i.j.k-1)(i-1,j,k)

(i-1,j,k-1)

(i,j,k-1)(i,j,k)

(i,j,k)

(i-1,j+1,k)

36

2

45

1


Defining an action of τ1, τ2, and τ3 on Contours

(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2


τ3

1) Start at the initial triangle in the 45-45-90 Square Lattice.2) Given τ -mutation sequence S , take an alcove walk to the new triangle{(i1, j1), (i2, j2), (i3, j3)}, ordered so that

ir − jr ≡ r mod 3

3) Define six-tuple of contouors CS = [CS1 , CS2 , . . . , CS6 ] by

CS ←→ prism {(i1, j1, 1), (i1, j1, 0), (i2, j2, 0), (i2, j2, 1), (i3, j3, 1), (i3, j3, 0)}.


Defining an action of τ1, τ2, and τ3 on Contours

(i-1,j)

(i-1,j+1)

(i,j+1)

(i,j)

1

3

2

1

32132

(i-1,j+2)

(i-2,j)

1) Start at {(0,−1), (−1, 0), (0, 0)} in the 45-45-90 Square Lattice.2) Given τ -mutation sequence S , take an alcove walk to the new triangle{(i1, j1), (i2, j2), (i3, j3)}, ordered so that





Theorem 1 [Lai-M 2015+]

Theorem (Reformulation of [Leoni-M-Neel-Turner 2014]): LetZS = [z1, z2, . . . , z6] be the cluster obtained after applying τ -mutationsequence S to the initial cluster {x1, x2, . . . , x6}.

Let w(G ) = cm(G )∑

M a perfect matching of G x(M).

Let G(Ci ) be the subgraph cut out by the contour Ci .

Then ZS = [w(G(CS1 ),w(G(CS

2 ), . . . ,w(G(CS6 )].

1) Start at {(0,−1), (−1, 0), (0, 0)} in the 45-45-90 Square Lattice.2) Given τ -mutation sequence S , take an alcove walk to the new triangle{(i1, j1), (i2, j2), (i3, j3)}, ordered so that





Example 1: τ -mutation sequence τ1τ2τ3

(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2


τ3

We start at the initial triangle {(0,−1), (−1, 0), (0, 0)}. Applying theτ -mutation sequence τ1, τ2, τ3 corresponds to the alcove walk

{(0,−1), (−1, 0), (0, 0)} → {(−1, 1), (−1, 0), (0, 0)} → {(−1, 1), (0, 1), (0, 0)} → {(−1, 1), (0, 1), (−1, 2)}

Recall that (i , j)↔ (j ,−i − j , i , j + 1,−i − j − 1, i + 1) and(j + 1,−i − j − 1, i + 1, j ,−i − j , i).

C1 = (0, 0, 1,−1, 1, 0), C2 = (−1, 1, 0, 0, 0, 1), C3 = (0, 1,−1, 1, 0, 0),C4 = (1, 0, 0, 0, 1,−1), C5 = (1,−1, 1, 0, 0, 0), C6 = (0, 0, 0, 1,−1, 1).Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 35 / 54


(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2


τ3


{(0,−1), (−1, 0), (0, 0)} → {(−1, 1), (−1, 0), (0, 0)} → {(−1, 1), (0, 1), (0, 0)} → {(−1, 1), (0, 1), (−1, 2)}


C ′1 = (2,−1, 0, 1, 0,−1),C ′2 = (1, 0,−1, 2,−1, 0),C3 = (0, 1,−1, 1, 0, 0),C4 = (1, 0, 0, 0, 1,−1),C5 = (1,−1, 1, 0, 0, 0),C6 = (0, 0, 0, 1,−1, 1).Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 36 / 54


(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2


τ3


{(0,−1), (−1, 0), (0, 0)} → {(−1, 1), (−1, 0), (0, 0)} → {(−1, 1), (0, 1), (0, 0)} → {(−1, 1), (0, 1), (−1, 2)}


C ′1 = (2,−1, 0, 1, 0,−1),C ′2 = (1, 0,−1, 2,−1, 0),C ′3 = (1,−1, 0, 2,−2, 1),C ′4 = (2,−2, 1, 1,−1, 0),C5 = (1,−1, 1, 0, 0, 0),C6 = (0, 0, 0, 1,−1, 1).Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 37 / 54


(0,0)(-1,0)(-2,0)

(1,0)(2,0)

(-1,1) (0,1) (1,1)(2,1)(-2,1)

(-2,-1)

(-2,-2)

(-1,-1)

(-1,-2)

(0,-1)

(0,-2)

(1,-1)

(1,-2) (2,-2)

(2,-1)

τ3 τ2

τ1

τ1

τ2


τ3


{(0,−1), (−1, 0), (0, 0)} → {(−1, 1), (−1, 0), (0, 0)} → {(−1, 1), (0, 1), (0, 0)} → {(−1, 1), (0, 1), (−1, 2)}


C ′1 = (2,−1, 0, 1, 0,−1),C ′2 = (1, 0,−1, 2,−1, 0),C ′3 = (1,−1, 0, 2,−2, 1),

C ′4 = (2,−2, 1, 1,−1, 0),C ′5 = (3,−2, 0, 2,−1,−1),C ′6 = (2,−1,−1, 3,−2, 0).


Example 1: τ -mutation sequence τ1τ2τ3 = µ1µ2µ3µ4µ5µ6

C ′1 = (2,−1, 0, 1, 0,−1),C ′2 = (1, 0,−1, 2,−1, 0),C ′3 = (1,−1, 0, 2,−2, 1),

C ′4 = (2,−2, 1, 1,−1, 0),C ′5 = (3,−2, 0, 2,−1,−1),C ′6 = (2,−1,−1, 3,−2, 0).

2 x4x6+x3x5x2

1 x3x5+x4x6x1

42

1


2

31


2

3

64

2

5

1

1

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x6

2

3

42

5

1

1 1

(x2x5+x1x6)(x1x3+x2x4)(x3x5+x4x6)2

x21 x22 x3x4x5

24

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2 6

13

5

24

35

2

3

6

13

6

13

6

13

5 1

6

1

4

6

1

4

6

1

4

4

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

24

6

13

6

6

13

3

6

133

5

2

6

13

5

24

35

2

5

2

5

24

5

24

44

6

1

4

5

24

5

24

5

2 6

13

5

24

6

13

5

24

6

1

4

1

6

13

5

6

133

5

2

3

6

1

24

6

6

13

6

5

3

3

35

2

6

13

5

24

5

2

1 5

24

44

5

2

5

24

1

3

6

13

5

6

1

4

24

5

5

2

5

42

1

D1/2 1 3/2

2 5/2 3

D D

D D D

a=1

b=0 c= -1

d=2

e= -1

f=0a=2

b= -2

c=1d=1e= -1

f=0

a=2

b= -1

c= -1

d=3

e= -2

f=0

a=3 b= -3

c=1

d=2e= -2

f=0

a=3

b= -2

c= -1

d=4e= -3

f=0

a=4

b= -4

c=1

d=3e= -3

f=0

31

6

3

2


Beyond Aztec Castles

In previous work of T. Lai, he enumerated (without weights) the numberof perfect matchings in certain graphs cut out by contours as above.

We give a cluster algebraic interpretation of these formulas.

Recall

τ1 = µ1 ◦ µ2 ◦ (12),

τ2 = µ3 ◦ µ4 ◦ (34),

τ3 = µ5 ◦ µ6 ◦ (56)

satisfy τ21 = τ22 = τ23 = 1 and (τiτj)3 = 1 as actions on labeled clusters.

We also consider

τ4 = µ1 ◦ µ4 ◦ µ1 ◦ µ5 ◦ µ1 ◦ (145), and

τ5 = µ2 ◦ µ3 ◦ µ2 ◦ µ6 ◦ µ2 ◦ (236).

Note: unlike mutations in τ1, τ2, and τ3, we do not have commutationhere. However, the dP3 quiver is still sent back to itself after τ4 or τ5.




We give a cluster algebraic interpretation of these formulas. Recall

τ1 = µ1 ◦ µ2 ◦ (12),

τ2 = µ3 ◦ µ4 ◦ (34),

τ3 = µ5 ◦ µ6 ◦ (56)


We also consider

τ4 = µ1 ◦ µ4 ◦ µ1 ◦ µ5 ◦ µ1 ◦ (145), and

τ5 = µ2 ◦ µ3 ◦ µ2 ◦ µ6 ◦ µ2 ◦ (236).

Note: unlike mutations in τ1, τ2, and τ3, we do not have commutationhere. However, the dP3 quiver is still sent back to itself after τ4 or τ5.




We give a cluster algebraic interpretation of these formulas. Recall

τ1 = µ1 ◦ µ2 ◦ (12),

τ2 = µ3 ◦ µ4 ◦ (34),

τ3 = µ5 ◦ µ6 ◦ (56)


We also consider

τ4 = µ1 ◦ µ4 ◦ µ1 ◦ µ5 ◦ µ1 ◦ (145), and

τ5 = µ2 ◦ µ3 ◦ µ2 ◦ µ6 ◦ µ2 ◦ (236).

Note: unlike mutations in τ1, τ2, and τ3, we do not have commutationhere. However, the dP3 quiver is still sent back to itself after τ4 or τ5.Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 40 / 54

Action of τ4 and τ5 on Triangular Prisms

32

(i,j-1,k)

(i,j,k)

6

5(i,j,k-1)

1 (i,j-1,k-1)

(i-1,j,k)

4(i-1,j,k-1)

(i-1,j,k)

(i-1,j,k-1)

(i,j,k-1)(i,j,k)

(i,j,k)

(i-1,j+1,k)

36

2

45

1

(i,j,k)

(i,j−1,k)

(i,j,k)

(i,j−1,k+1)

(i−1,j+1,k)

(i−1,j,k)

(i,j,k−1)

1

2

5

64

3

(i,j−1,k+1)

(i,j,k)

(i,j,k−1)

(i−1,j−1,k)

(i−1,j,k)

2 (i,j−1,k)

6

1

3

(i−1,j,k+1)

(i−1,j,k)(i−1,j−1,k)

(i,j−1,k)

(i,j−1,k+1)

36

5

4

1

2

4

5

(i−1,j,k)

3

5

4

6

2

(i−1,j,k+1)

(i,j−1,k+1)

(i,j,k+1)

(i,j−1,k)

(i,j,k)

1

µ1µ4µ1µ5µ1 lifts layer (k − 1) to (k + 1) and rotates the triangle by (145).


Action of τ4 and τ5 on Clusters

Recall

τ1 = µ1 ◦ µ2 ◦ (12),

τ2 = µ3 ◦ µ4 ◦ (34),

τ3 = µ5 ◦ µ6 ◦ (56)


We also consider

τ4 = µ1 ◦ µ4 ◦ µ1 ◦ µ5 ◦ µ1 ◦ (145), and

τ5 = µ2 ◦ µ3 ◦ µ2 ◦ µ6 ◦ µ2 ◦ (236).

We then get the relations τ24 = τ25 = 1, and τi and τj commute fori ∈ {1, 2, 3} and j ∈ {4, 5}. However, (τ4τ5) has infinite order.

In summary, we have a “book” of affine A2 lattices where we apply τ4 andτ5 in an alternating fashion to get between pages.



Recall

τ1 = µ1 ◦ µ2 ◦ (12),

τ2 = µ3 ◦ µ4 ◦ (34),

τ3 = µ5 ◦ µ6 ◦ (56)


We also consider

τ4 = µ1 ◦ µ4 ◦ µ1 ◦ µ5 ◦ µ1 ◦ (145), and

τ5 = µ2 ◦ µ3 ◦ µ2 ◦ µ6 ◦ µ2 ◦ (236).


In summary, we have a “book” of affine A2 lattices where we apply τ4 andτ5 in an alternating fashion to get between pages.



Recall

τ1 = µ1 ◦ µ2 ◦ (12),

τ2 = µ3 ◦ µ4 ◦ (34),

τ3 = µ5 ◦ µ6 ◦ (56)


We also consider

τ4 = µ1 ◦ µ4 ◦ µ1 ◦ µ5 ◦ µ1 ◦ (145), and

τ5 = µ2 ◦ µ3 ◦ µ2 ◦ µ6 ◦ µ2 ◦ (236).


In summary, we have a “book” of affine A2 lattices where we apply τ4 andτ5 in an alternating fashion to get between pages.Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 42 / 54

Putting it all together

Because of the commutation relations, to get clusters obtained from{x1, x2, . . . , x6} by applying sequences S of τ1, τ2, . . . , τ5, sufficient to letS = S123S45 where S123 is a τ -mutation sequence (as earlier) andS45 = τ4τ5τ4 . . . τ4, τ4τ5τ4 . . . τ5, τ5τ4τ5 . . . τ4, or τ5τ4τ5 . . . τ4.

Let [(i1, j1), (i2, j2), (i3, j3)] denote the triangle reached after applying thealcove walk associated to S123 starting from the initial triangle.

CS = CS123S45 ↔ [(i1, j1, k1), (i1, j1, k2), (i2, j2, k2), (i2, j2, k1), (i3, j3, k1), (i3, j3, k2)]

in Z3 where {k1, k2} = {|S45|, |S45|+ 1} (resp. {−|S45|,−|S45|+ 1}) ifS45 starts with τ5 (resp. τ4). (i.e. τ4τ5 pushes down, τ5τ4 pushes up)Further, we let k1 = ±|S2| if |S2| is odd and k2 = ±|S2| otherwise.

(i , j , k)↔ σkC ji = (j +k,−i− j−k, i +k , j +1−k ,−i− j−1+k, i +1−k).

Then CS is defined as

[σk1Cj1i1 , σk2Cj1i1 , σ

k2Cj2i2 , σk1Cj2i2 , σ

k1Cj3i3 , σk2Cj3i3 ].



Let S = S123S45 and S({x1, x2, . . . , x6}) = {z(S)1 , z(S)2 , . . . , z

(S)6 }. Let

CS = (C ′1,C′2, . . . ,C

′6) be the six-tuple of contours obtained as above.

Theorem 2 (Lai-M 2015+) : As long as the contoursCS = (C ′1,C

′2, . . . ,C

′6) are (without self-intersections), then the resulting

Laurent expansions of the cluster variables ZS = {z(S)1 , z(S)2 , . . . , z

(S)6 }

given byZS = [w(G(C′1),w(G(C′2), . . . ,w(G(C′6)].

Proof (Method): Kuo’s Method of Graphical Condensation for CountingPerfect Matchings. We isolate four vertices {a, b, c , d} (in cyclic order andalternating color) near the boundary of the contour and argue that

w(G )w(G−{a, b, c , d}) = w(G−{a, b})w(G−{c , d})+w(G−{a, d})w(G−{b, c})

corresponds to the cluster mutation and translations in the contours.

Remark: As side lengths change and contour goes from convex toconcave or vice-versa, we have to use different types of Kuo Condensations(Balanced, Unbalanced, Non-alternating).




(S)6 }. Let

CS = (C ′1,C′2, . . . ,C



′2, . . . ,C



(S)6 }





Remark: As side lengths change and contour goes from convex toconcave or vice-versa, we have to use different types of Kuo Condensations(Balanced, Unbalanced, Non-alternating).




(S)6 }. Let

CS = (C ′1,C′2, . . . ,C



′2, . . . ,C



(S)6 }





Remark: As side lengths change and contour goes from convex toconcave or vice-versa, we have to use different types of Kuo Condensations(Balanced, Unbalanced, Non-alternating).Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 44 / 54

Example 2: S = τ1τ2τ3τ1τ2τ3τ2τ1τ4

We reach {(1, 3), (1, 2), (0, 3)} and applying τ4 yields CS = [σ−1C31 , C31 , C

21 , σ

−1C21 , σ−1C30 , C

30 ] =

[(2,−3, 0, 5,−6, 3), (3,−4, 1, 4,−5, 2), (2,−3, 1, 3,−4, 2), (1,−2, 0, 4,−5, 3), (2,−2,−1, 5,−5, 2), (3,−3, 0, 4,−4, 1)].

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

2 6

5

24

5

24

5

24

4

6

1

4

6

1

4

6

1

6

13

6

13

6

13

3

35

2

35

2

35

2

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2

35

2

35

2

35

2

2 6 24

6

5

24

5

24

4

6

1

4

6

1

24

6

6

13

6

13

6

13

6

13

3

4

5

24

6

5

24

6 24

2

35

2 6

13

5

24

6

13

5

24

6

13

5

24

35

2

35

2

3

6

13

6

13

6

13

5 1

6

1

4

6

1

4

4

6

5

24

6 24

6 24

2

35

2 6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2 6

13

5

24

35

2 6

13

5

24

35

2

3

6

13

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13

6

13

6

1

6

1

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24

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24

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6

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24

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6

13

5

24

35

2

35

2

35

2

35

2

2 6 24

6

5

24

4

24

5

24

4

6

1

4

6

1

4

6

1

4

1

6

13

5

6

13

6

13

6

13

3

5

24

4

6

1

4

6

1

6

1

46

13

5

24

6

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6 24

6 24

2

35

2

35

2

35

2

35

2

35

2

3

6

13

6

13

6

13

6

13

6

13

5

24

6

13

5

24

6

13

5

6 2

53

1

5

24

6

13

5

24

c=1

d=4

e= -5

f=2a=2

b= -3

c=1

d=3

e= -4

f= 2

a=1b= -2

c=0

d=4e= -5

f=3a=2

c= -1

b= -2

d=5e= -5

f= 2 a=3 b= -3

c=0

d=4e= -4

f= 1

b=-4a=3f=3

e= -6 d=5

c=0

b= -3a=2



[(0,−2, 1, 3,−5, 4), (−1,−1, 0, 4,−6, 5), (0,−1,−1, 5,−6, 4),

(1,−2, 0, 4,−5, 3), (0,−1, 0, 3,−4, 3), (−1, 0,−1, 4,−5, 4)].

f=3

e= -4

d=3

c=0

b= -1

a=0f=3

e= -5

d=4

c=0

b= -2

a=1

f=4

e= -6 d=5

c= -1

b= -1

a=0f=5

e= -6

d=4

c=0

b= -1

a= -1

f=4

e= -5

d=3

c=1

b= -2

a=0

33

6

13

5

24

6

13

5

24

6

13

5

24

6 24

6 24

2

35

2

35

2

35

2

3

6

13

6

13

6

1

64

35

2

35

2

35

2

2 6 24

6 24

6 24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

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35

2

3

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35

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35

2

35

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35

2 6

13

5

24

6

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24

6

1

6

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24

6

1

4

4

6

5

24

6 24

6 24

2

6

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24

6

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5

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24

6

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24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

6

13

66

1

24

64

6 24

6 24

6 24

2

35

2

35

2

35

2

35

2

35

2 6

13

6

13

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13 a=-1

24

6

13

5

24

6

13

5

24

2

35

2

35

2

35

2

35

2

3

6

13

6

13

6

13

5 1

6

1

4

64

6 24

6 24

6 24

2

6

13

4

66 24

6 24

6 24

6 24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2

35

2

35

f=4

b=0 c= -1

d=4

e= -5

5

2

53 3

1

6

31

6

31

6

1

64

2

53

1

6

42

53

1

6

42

53

1

6

42

53

1

6

2

53

2



[(0,−2, 1, 3,−5, 4), (−1,−1, 0, 4,−6, 5), (0,−1,−1, 5,−6, 4),

(1,−2, 0, 4,−5, 3), (0,−1, 0, 3,−4, 3), (−1, 0,−1, 4,−5, 4)].

f=3

e= -4

d=3

c=0

b= -1

a=0f=3

e= -5

d=4

c=0

b= -2

a=1

f=4

e= -6 d=5

c= -1

b= -1

a=0f=5

e= -6

d=4

c=0

b= -1

a= -1

f=4

e= -5

d=3

c=1

b= -2

a=0

33

6

13

5

24

6

13

5

24

6

13

5

24

6 24

6 24

2

35

2

35

2

35

2

3

6

13

6

13

6

1

64

35

2

35

2

35

2

2 6 24

6 24

6 24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2

3

6

13

6

13

6

13

6

6

6

13

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

6

13

6

13

3

35

2

35

2

35

2

35

2 6

13

5

24

6

13

5

24

6

1

6

13

5

24

6

1

4

4

6

5

24

6 24

6 24

2

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

6

13

66

1

24

64

6 24

6 24

6 24

2

35

2

35

2

35

2

35

2

35

2 6

13

6

13

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13 a=-1

24

6

13

5

24

6

13

5

24

2

35

2

35

2

35

2

35

2

3

6

13

6

13

6

13

5 1

6

1

4

64

6 24

6 24

6 24

2

6

13

4

66 24

6 24

6 24

6 24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

35

2

35

2

35

f=4

b=0 c= -1

d=4

e= -5

5

2

53 3

1

6

31

6

31

6

1

64

2

53

1

6

42

53

1

6

42

53

1

6

42

53

1

6

2

53

2


Example of Kuo Condensations Used in Proof (Balanced)

35

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

1

4

113

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5 13

5 13

5

2 6

13

5

24

6

13

5

24

35

2 6

13

5

24

6

13

5

6

13

6

133

5

2

3

5

24

5

24

5

24

5

24

5

24

5

24

4

6

1

4

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

1

4

113

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5 13

5 13

5

2 6

13

5

24

6

13

5

24

35

2 6

13

5

24

6

13

5

6

13

6

133

5

2

3

5

24

5

24

5

24

5

24

5

24

5

24

4

6

1

4

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

1

4

113

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5 13

5 13

5

2 6

13

5

24

6

13

5

24

35

2 6

13

5

24

6

13

5

6

13

6

133

5

2

3

5

24

5

24

5

24

5

24

5

24

5

24

4

6

1

4

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

1

4

113

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5 13

5 13

5

2 6

13

5

24

6

13

5

24

35

2 6

13

5

24

6

13

5

6

13

6

133

5

2

3

5

24

5

24

5

24

5

24

5

24

5

24

4

6

1

4

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

1

4

113

513

5

6

13

5

2

1

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5 13

5 13

5

2 6

13

5

24

6

13

5

24

35

2 6

13

5

24

6

13

5

6

13

6

133

5

2

3

5

24

5

24

5

24

5

24

5

24

5

24

4

6

1

4

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

13

5

24

6

1

4

113

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5 13

5 13

5

2 6

13

5

24

6

13

5

24

35

2 6

13

5

24

6

13

5

6

13

6

6

4

1

64

2

53

1

64

2

53

1

6

4

1

64

2

53

1

64

2

53

1

6

42

53

1

6

42

1

53

1

6

42

53

1

6

4

1

64

2

53

1

64

2

53

1

6

4

1

64

2

53

1

6

4

1

6

335

2

3

4

42

5

42

5

42

5

42

5

42

5

42

4

5

AB

C

D

AB

C

D

AB

C

D

A

D

B

C


Example of Kuo Condensations Used in Proof (Balanced)

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

46

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

5

24

5

24

5

24

24

4

6

1

4

6

1

4

6

1

24

6

5

24

5

24

4

6

1

4

6

1

4

1

6

4

13

5

24

6

1

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

113

513

513

513

513

513

513

55 13

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

46

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

5

24

5

24

5

24

24

4

6

1

4

6

1

4

6

1

24

6

5

24

5

24

2

4

6

1

4

6

1

46

13

5

24

46

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

113

513

513

513

513

513

513

55 13

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

46

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

5

24

5

24

5

24

24

4

6

1

4

6

1

4

6

1

2

5

46

5

24

5

24

4

6

1

4

6

1

46

13

5

24

24

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

113

513

513

513

513

513

513

55 13

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

46

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

5

24

5

24

5

24

24

4

6

5

1

4

6

1

4

6

1

24

6

5

24

5

24

4

6

1

4

6

1

46

13

5

24

24

4

6

5

6

13

5

24

6

13

5

24

6

13

5

24

6

1

3

5

24

6

13

5

24

113

513

513

513

513

513

513

55 13

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

46

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

5

24

5

1

24

5

24

24

4

6

1

4

6

1

4

6

1

24

6

5

24

5

24

4

6

1

4

6

1

46

13

5

24

4

6

1

4

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

113

513

513

513

513

513

513

55 13

513

5

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

1

4

6

1

4

6

1

46

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

6

13

5

24

5

24

5

24

5

24

5

24

5

24

24

4

6

1

64

2

53

1

64

2

53

1

6

6

13

5

2

42

5

4

31

64

2

53

1

6

42

53

1

6

53

1 53

15 53

1 53

1 53

1 53

1 53

1 53

1 53

1 1

42

53

1

64

2

53

1

64

2

53

1

64

2

53

1

64

2

53

1

6

5

6

4

3

1

A

B

C

D

A

B

C

D

A

B

C

D

A

D

B

C


Example of Kuo Condensations Used (Unbalanced)

1

1

64

1

64

513

51

6

351

264

351

264

3

1

6

351

264

3

1

6

3

3

5

2

3

5

2

3

5

2

3

5

2

3

2 264

5

264

5

24

51

264

3

51

264

351

264

3

51

264

351

264

351

264

351

264

3

3

1

6

35

2

3

5

2

351

264

31

6

3

1513

51

6

351

264

351

264

35

2

3

51

264

35

2

3

2 264

5

264

51

264

351

264

351

264

31

64

1

64

1

64

1

64

5

24

51

264

3

51

264

351

264

3

4

51

264

3

1

64

1

64

51

264

3

1

264

4

5

2

5

24

5

264

264

2

5

2

3

3

5

2

3

4

6

3

1

6

3

51

264

35

2

3

5

2

351

264

351

264

351

6

351

31

1

64

51

264

3

51

264

351

264

3

51

264

351

264

3

3

1

4

5

24

1

64

1

64

1

64

1

64

513

51

6

3

1

6

3

1

6

35

2

4

5

2

3

5

2

3

5

2

3

2 264

5

264

5

24

51

264

3

51

264

3

51

264

351

264

3

51

264

3

51

264

351

264

3

51

264

351

264

3

51

264

3

1

3

6

1

6

35

1

46

2

3

5

2

3

5

2

351

264

31

6

3

4

5

24

51

264

3

5

24

1

64

1

64

1

64

51

264

3

1

42

5

4

1

4

51

264

35

264

2 264

5

2

3

51

264

351

264

351

6

351

31

1

64

51

264

351

264

351

264

351

264

3

5

4

5

3

5

5

4

5

2

5

5

1

4

44

1

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Future Directions

We now understand many sequences inside three-dimensional subspace oftoric mutations (at vertices with 2 in-coming arrows, 2 out-going arrows)for the dP3 Quiver.

Question 1: We only understand when contours have no self-intersections.(Corresponds to the length of S123 being longer enough than S45.)

How can we extend our combinatorial interpretation to cases withself-intersections?

Questions 2: How can this new point of view (via contours) for the dP3quiver better our understanding of Gale-Robinson, Octahedron Recurrence,and other cases of toric mutations?

Concentrating still on the dP3 quiver, there are more toric mutationsequences, but also more relations.


Future Directions







Future Directions







Future Directions







Future Directions







Future Directions


Example: µ1µ4µ1µ4µ1 actually sends labeled cluster back to itself(except with permutation (14)).

Question 3: Accounting for these relations and others, what is left in thespace of toric mutation sequences (for this example)?

Question 4: Do τ1, τ2, . . . , τ5 represent integrable directions of mutationsequences? What about the compositions τ1τ2τ3τ1τ2τ3, τ1τ2τ1τ3, τ4τ5?


Future Directions






Future Directions






Thank You For Listening

• Aztec Castles and the dP3 Quiver (with Megan Leoni, Seth Neel, andPaxton Turner, Journal of Physics A: Math. Theor. 47 474011,arXiv:1308.3926

• In-Jee Jeong, Bipartite Graphs, Quivers, and Cluster Variables, REUReport, http://www.math.umn.edu/∼reiner/REU/Jeong2011.pdf• Sicong Zhang, Cluster Variables and Perfect Matchings of Subgraphs ofthe dP3 Lattice, REU Report,http://www.math.umn.edu/∼reiner/REU/Zhang2012.pdf• Gale-Robinson Sequences and Brane Tilings (with In-Jee Jeong and andSicong Zhang), Discrete Mathematics and Theoretical Computer ScienceProc. AS (2013), 737-748. (Longer version in preparation.)

• Tri Lai, New Aspects of Hexagonal Dungeons, arXiv:1403.4481

• Tri Lai, New regions whose tilings are enumerated by powers of 2 and 3,preprint, http://ima.umn.edu/∼tmlai/Newdungeon.pdf

http//math.umn.edu/∼musiker/Beyond.pdfLai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond Et Tu Brutus 54 / 54

Documents

Combinatorics of the Del-Pezzo 3 Quiver: Aztec Dragons, Castles, …musiker/Beyond.pdf · 2015. 3. 16. · Lai-M (IMA and Univ. Minnesota) The dP3 Quiver: Aztec Castles and Beyond