tiling expression of minors - university of nebraska–lincolntlai3/jtubc.pdf · 2016-11-09 ·...

Tiling expression of minors

Tri Lai

Institute for Mathematics and its ApplicationsMinneapolis, MN 55455

ColloquiumUniversity of British Columbia, Vancourver

February 3, 2016

Tri Lai Tiling expression of minors

Outline

1 Introduction to the Enumeration of Tilings.

2 Electrical networks

3 Tiling expression of minors

4 Future work.


Tilings

A lattice divides the plane into disjoint pieces, called fundamentalregions.

A tile is a union of any two fundamental regions sharing an edge.

A tiling of a region is a covering of the region by tiles so that thereare no gaps or overlaps.

The number of tilings of a 8× 8 chessboard is 12,988,816.


Tilings

We would like to find the number of tilings of certain regions.

Tilings can carry weights, we also care about the weighted sum oftilings:

∑T wt(T ), called the tiling generating function.


Kasteleyn–Temperley–Fisher Theorem

Theorem (Kasteleyn, Temperley and Fisher 1961)

The number of tilings of a 2m × 2n chessboard equals

m∏j=1

n∏k=1

(4 cos2

( jπ

2m + 1

)+ 4 cos2

( kπ

2n + 1

)).


Connections and Applications to Other Fields

Statistical Mechanics: Dimer model, double-dimer model, squareice model, 6-vertex model, fully packed loops configuration...

Probability: Limit shapes and Arctic Curves.

Graph Theory: Bijection between tilings and perfect matchings.

Cluster Algebras: Combinatorial interpretation of cluster variables.

Electrical networks:

. . .

Other topics in combinatorics: Alternating-sign matrices,monotone triangles, plane partitions, lattice paths, symmetricfunctions...


Aztec Diamond

Theorem (Elkies, Kuperberg, Larsen and Propp 1991)

The Aztec diamond of order n has 2n(n+1)/2 domino tilings.

Figure: The Aztec diamond of order 5 and one of its tilings.


An Aztec Temple


MacMahon’s Formula

Theorem (MacMahon 1900)

The number of (lozenge) tilings of a semi-regular hexagon of sidesa, b, c , a, b, c on the triangular lattice is

a∏i=1

b∏j=1

c∏t=1

i + j + t − 1

i + j + t − 2=

H(a) H(b) H(c) H(a + b + c)

H(a + b) H(b + c) H(c + a),

where the hyperfactorial H(n) = 0! · 1! · 2! . . . (n − 1)!.

a=4 b=5

c=6

b=5 a=4

c=6


Connection to Plane Partitions

a=10

b=10

c=10

c=10

a=10

b=10

k

ij

O


MacMahon’s q-formula

Theorem (MacMahon)∑π

qvol(π) =Hq(a) Hq(b) Hq(c) Hq(a + b + c)

Hq(a + b) Hq(b + c) Hq(c + a),

where the sum is taken over all stacks π fitting in an a× b × c box.

q-integer [n]q := 1 + q + q2 + . . .+ qn−1

q-factorial [n]q! := [1]q · [2]q · [3]q . . . [n]q

q-hyperfactorial Hq(n) := [0]q! · [1]q! · [2]q! . . . [n − 1]q!.


Generalizing MacMahon’s Formula

t

m

x

c

y+bm

z

a+b

t+c

y

m

x

z+b+c

a

a

m

y+bm

z

a+b

t+c

y

mx

z+b+c

a

t

m

12

3

4

56

k

ij

O

b

c

ac

x

m b

c

∑stacks

qvolume of the stack=?


Generalizing MacMahon’s Formula

Theorem (L. 2015+)

For non-negative integers x , y , z , t,m, a, b, c∑π

qvol(π)=Hq(∆ + x + y + z + t)

Hq(∆ + x + y + t) Hq(∆ + x + y + z)

×Hq(∆ + x + t) Hq(∆ + x + y) Hq(∆ + y + z) Hq(∆)

Hq(∆ + z + t) Hq(∆ + x) Hq(∆ + y)

×Hq(m + b + c + z + t) Hq(m + a + c + x) Hq(m + a + b + y)

Hq(m + b + y + z) Hq(m + c + x + t)

× Hq(c + x + t) Hq(b + y + z)

Hq(a + c + x) Hq(a + b + y) Hq(b + c + z + t)

× Hq(m)3 Hq(a)2 Hq(b) Hq(c) Hq(x) Hq(y) Hq(z) Hq(t)

Hq(m + a)2 Hq(m + b) Hq(m + c) Hq(x + t) Hq(y + z),

where ∆ = m + a + b + c.


Quasi-hexagon

c=3

a=5

b=3

b=3

a=5

c=3

In 1999, James Propp collected 32 open problems in enumeration oftilings.

Problem 16 on the list asks for the number of tilings of aquasi-hexagon.


Quasi-hexagon

c=3

a=5

b=3

b=3

a=5

c=3

Theorem (L. 2014)

The number of tilings of a quasi-hexagon is a power of 2 times thenumber of tilings of a semi-regular hexagon.


Blum’s Conjecture and Hexagonal Dungeon

b=6

a=2

2a=4

2a=4

b=6

a=2

Theorem (Blum’s (ex-)conjecture)

The number of tilings of the hexagonal dungeon of side-lengths

a, 2a, b, a, 2a, b (b ≥ 2a) is 132a214ba2

2 c.

The conjecture was proven by Ciucu and L. (2014).


Circular Planar Electrical Networks

Study of electrical networks comes from classical physics with thework of Ohm and Kirchhoff more than 100 years ago.

The circular planar electrical networks were studied systematically byColin de Verdiere and Curtis, Ingerman, Mooers, and Morrow.

A number of new properties have been discovered recently.


Circular Planar Electrical Networks

Definition

A circular planar electrical network is a finite graph G = (E ,V )embedded in a disk with a set of distinguished vertices N of V on thecircle, called nodes, and a conductance function wt : E → R+

3

4

7

3

2

5

5

4

1

32

7

6

8 9

2

6

1

2

3

4

5

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7 8

9

12

1

3

4

5

6

13

2


Well-connected networks

A = {a1, a2, . . . , ak} and B = {b1, b2, . . . , b`} are non-interlacedon the circle if we do not have a1 ≤ bj ≤ ak or b1 ≤ ai ≤ b`.

A network G is well-connected if for any pair (A,B) ofnon-interlaced sets with k nodes (1 ≤ k ≤ b n2c) we can find kpairwise vertex-disjoint paths connecting nodes in A to nodes in B.


Well-connected networks

1

8

7

65

4

3

2


Motivational Problem

We would like to test the well-connectivity of givennetworks.


Response Matrix

Associated with a network is a response matrix Λ = (λi,j)1≤i,j≤n,that measures the response of the network to potential applied atthe nodes.

−λi,j is the current that would flow into node j if node i is set toone volt and the remaining nodes are set to zero volts.

Two networks is electrically equivalent of they have the sameresponse matrix.


Electrical Moves

bc/(a+b+c)

a

a

a

b a+

ba b

ab

/(a+

b)

a

bc

ab/(a+

b+c)

ac/(

a+b+c)

Two networks are electrically equivalent if and only if they can beobtained from each other by using the “electrical moves”.


Circular Minors

Arrange the indices 1, 2, . . . , n of a general matrix M =(mi,j

)1≤i,j≤n in

counter-clockwise order around the circle.

A = {a1, a2, . . . , ak} in counter-clockwise order, andB = {b1, b2, . . . , bk} in clockwise order around the circle.

Circular minor

MBA = det

b1 b2 ... bk−1 bk

a1 ma1,b1 ma1,b2 . . . ma1,bk−1ma1,bk

a2 ma2,b1 ma2,b2 . . . ma2,bk−1ma2,bk

......

.... . .

......

ak−1 mak−1,b1 mak−1,b2 . . . mak−1,bk−1mak−1,bk

ak mak ,b1 mak ,b2 . . . mak ,bk−1mak ,bk


Circular Minors: Examples

M =

1 1 4 1 6 1 1 11 0 4 1 1 9 1 10 2 4 1 1 1 1 01 2 0 2 6 0 1 11 0 1 0 1 1 1 00 0 4 0 2 1 0 01 0 1 0 6 0 1 24 0 4 1 6 9 1 15 2 1 0 1 9 1 0

b

1

2

3

12

3

b

b

M1,2,4

9,7,5

= 1

2

34

5

6

7

8

9

a

a

a



M =

b3 b2 b1

a1 1 1 4 1 6 1 1 1a2 1 0 4 1 1 9 1 1

0 2 4 1 1 1 1 0a3 1 2 0 2 6 0 1 1

1 0 1 0 1 1 1 00 0 4 0 2 1 0 01 0 1 0 6 0 1 24 0 4 1 6 9 1 15 2 1 0 1 9 1 0

M9,7,51,2,4 = det

b1 b2 b3

a1 1 1 1a2 1 9 1a3 1 0 2



b

1

2

3

12

3

b

b

M1,2,4

9,7,5

= 1

2

34

5

6

7

8

9

a

a

a

= det

b1 b2 b3

a1 1 1 1a2 1 9 1a3 1 0 2


Test the well-connectivity

(Colin de Verdiere) A network is well-connected if and only if allnon-interlaced circular minors ΛB

A > 0.

Kenyon and Wilson showed a test of the well-connectivity of anetwork by checking the positivity of (only)

(n2

)special circular

minors of the response matrix.


Contiguous Minors

The contiguous minor CONa,b,y (M) := MBA where

A = {a, a + 1, . . . , a + y − 1} and B = {b + y − 1, . . . , b + 1, b}

(c)(a) (b)

1

2

34

567

8

9

11

12

1416

1515

101

2

1614

567

8

13

5

1213

1416

2

1

15

10

17

34

11

67

8

9

11

12

9

43

17

181810

17

13

18

Figure: M12,10,91,2,4 , CON1,6,3(M), CON1,10,3(M).


Central Minors

The central minor CMx,y (M) is the contiguous minorCONa,b,y (M), where

a =

⌊x − y

2

⌋and b =

⌊x − y + n − (n − 1 mod 2)

2

⌋.

a and b are opposite or almost opposite.

1

2

1516

567

8

1413

12

10

4

17

18

3

9

11

Figure: CON1,10,3(M) = CM6,3(M).


Small Central Minors

If 1 ≤ x ≤ n, 1 ≤ y < n/2 or y = n/2 and x + y odd, then we callCMx,y (M) a small central minor of M.

There are total(n2

)small center minors.

x=1 x=2 x=3 x=4 x=5

y=1

y=2

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5

1

2

3

4 5


Small Central Minors

If 1 ≤ x ≤ n, 1 ≤ y < n/2 or y = n/2 and x + y odd, then we callCMx,y (M) a small central minor of M.

There are total(n2

)small center minors.

x=1 x=2 x=3 x=4 x=5 x=6

y=1

y=2

y=3

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6

1

23

4

5 6


Kenyon–Wilson Test

Theorem (Kenyon–Wilson Test)

If(n2

)small central minors of the response matrix are all positive, then

the network is well-connected.


Kenyon–Wilson Theorem

Theorem (Kenyon-Wilson 2014)

1 Each contiguous minor can be written as a Laurent polynomial (withpositive coefficients) in central minors.

2 Moreover, the Laurent polynomial is the generating function ofdomino tilings of a truncated Aztec diamond.


Aztec Diamonds

Definition

Denote by ADhx0,y0 of the Aztec diamond of order h with the center

(x0, y0).

(a) (b)

(0,0) y=0

(c)

(−13,4)

(0,5)

(13,1)

Figure: (a) AD5−13,4. (b) AD4

0,5. (c) AD613,1.


Truncated Aztec Diamonds

Definition

The truncated Aztec diamond TADh,nx0,y0 is the portion of ADh

x0,y0between the lines y = 0 and y = n.

y=8

(a) (b) (c)

(0,0) y=0

(−13,4)

(0,5)

(13,1)

Figure: (a) TAD5,8−13,4. (b) TAD4,8

0,5. (c) TAD6,813,1.


Weight Assignment

Definition

Assign to each lattice point (x , y) a weight vx,y :

vx,y := CMx,y (M) if 0 ≤ y ≤ n.

vx,y = 1 if y < 0 or y > n.


Weight Assignment

Definition

We assign to each domino a weight 1vx1,y1vx2,y2

, where the lattice

points (x1, y1) and (x2, y2) are the centers of the long sides of thedomino.

W(R) :=∑

T wt(T ), where wt(T ) is the product of weights of alldominoes in the tiling T .

(x+1,y)(x,y)

(x,y-1)

(x,y)

1

vx,yvx,y−1

1

vx,yvx+1,y


Weight of Dominoes

Definition

The covering monomial: F(R) :=∏

(x,y) vx,y , taken over all lattice

points (x , y) inside R or on the boundary of R, except for the90◦-corners.

The tiling-polynomial: P(R) := W(R) F(R).


Calculating P(TAD2,131,1 )

(1,1)

(1,0) (2,0)

(0,1)

(0,0)

(-1,1) (3,1)

(2,2)

(1,3)

(2,1)

(0,2) (1,2)

F(

TAD2,131,1

)= v0,0 · v1,0 · v2,0 · v−1,1 · v0,1 · v1,1 · v2,1 · v3,1 · v0,2 · v1,2 · v2,2 · v1,3


Calculating P(TAD2,131,1 )

(1,1)

(1,0)

(0,1)(-1,1) (3,1)

(2,2)

(2,1)

(0,2) (1,2)

(1,1)(0,1)(-1,1)(3,1)

(1,3)

(2,1)

(1,2)

(1,1)(2,0)

(0,1)(-1,1)

(1,3)

(2,1)

(1,2)

(1,1)

(1,0)

(0,1)(-1,1) (3,1)

(1,3)

(2,1)

(1,2)

(1,1)

(0,1)

(0,0)

(3,1)

(1,3)

(2,1)

(0,2) (1,2)

(2,0)

(0,1)

(0,0)

(2,2)

(1,3)

(2,1)

(0,2)

(1,2)

(2,2)

wt(T1) = 1v−1,1v0,1

1v1,0v1,1

1v2,1v3,1

1v0,2v1,2

1v1,2v2,2

F(TAD2,131,1 )wt(T1) =

v0,0v2,0v1,3v1,2

P(TAD2,131,1 ) =

v0,0v2,0v1,3v1,2

+v0,0v1,0v2,0v0,2v2,2

v0,1v1,1v2,1+

v0,0v1,0v0,2v3,1v2,1v0,1

+v0,0v2,0v0,2v2,2

v1,1v1,2+

v1,0v2,0v−1,1v2,2v0,1v2,1

+v1,0v−1,1v1,1v3,1

v0,1v2,1.


Kenyon–Wilson Theorem (cont.)


CONa,b,y (M) = P(TAD|h|,nx−h,y ),

where h is the integer closest to 0 so that

CONa,b+h,y (M) = CMx,y (M).

= ?6

7

8

45

91110

32

1

13

12

Figure: CON1,5,1(M) with n = 13.





h is the integer closest to 0 so that CONa,b+h,y (M) = CMx,y (M).

7

8

9

45

10 1111109

8

7

65 4

6 2

1

13

1212

13

32

1

3

Figure: h = +2, x = 3, y = 1





where h is the integer closest to 0 so that CONa,b+h,y (M) = CMx,y (M).

=1

2

345

6

7

8

911

12

13

10

45

11109

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7

6

45

11109

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45

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6

1

2

3

13

12

1

2

3

13

12

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2

3

13

12

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2

3

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12

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2

3

13

12

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2

3

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12

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2

3

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12

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2

3

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3

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12

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2

3

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12

1

2

3

13

12

1

2

3

13

12

CON1,5,1(M) = P(TAD2,131,1 )





h is the integer closest to 0 so that CONa,b+h,y (M) = CMx,y (M).

=1

2

345

6

7

8

911

12

13

10

45

11109

8

7

6

45

11109

8

7

6

45

11109

8

7

6

45

11109

8

7

6

45

11109

8

7

6

45

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8

7

6

45

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7

6

45

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8

7

6

45

11109

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7

6

45

11109

8

7

6

45

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7

6

45

11109

8

7

6

45

11109

8

7

6

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

1

2

3

13

12

CON1,5,2(M) = P(TAD2,132,2 )


Semicontiguous minors

A semicontiguous minor is a circular minor MBA when exactly one of A

and B is contiguous.

(a) (b)


Structure of Semicontiguous Minors

k

k

k

k

1k1

2

k2

3

k3

4

k4

t1

t2

3t

SMa,b(k1, . . . , ks ; t1, . . . , ts−1)

The h- and x-parameters are that of the truncated Aztec diamondcorresponding to the contiguous minor MB1

A1


Kenyon–Wilson Conjecture

Conjecture (Kenyon-Wilson 2014)

Any semicontiguous minor can be represented as the tiling-polynomialP(R) of some region R on the square lattice.


Main Result

Theorem (L. 2015)

Any semicontiguous minor can be represented as the tiling-polynomialP(R) of some region R on the square lattice.

=1

2

34

5

6

7

8

911

12

13

10

45

11109

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7

6

45

11109

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7

6

45

11109

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45

11109

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Goal

Target

Define a family of regions

Rx,h(k1, . . . , ks ; t1, . . . , ts−1)

corresponding to the minors

SMa,b(k1, . . . , ks ; t1, . . . , ts−1).


Aztec Rectangles

(a) (b)

n=6n=6

m=3 m=4y=0

(0,0)

(−4,2)(8,3)

Figure: (a) AR3,6−4,2. (b) AR4,6

8,3.


Family of Regions Rx ,h(k1, . . . , ks ; t1, . . . , ts−1)

kk k k1

2 3 4

t1 t t2 3

Figure: Z(k1, k2, . . . , ks ; t1, t2, . . . , ts−1)

The infinite extension to the right of Z is denoted by Z+.

The infinite extension to the left of Z is denoted by Z−.


Case 1. h > t + k

ARh−k+k1,h+k1x−h,0

h-k+k

h+k 1

y=0

y=n

1

k = k1 + · · ·+ ks and t = t1 + · · ·+ ts−1


Case 2. h ≤ 0.

AR2k+t−h−k1,k+t−h−k1x−h,0

y=0

k+t-h

-k1

2k+t-h-k1

y=n


Case 3. 0 < h ≤ t + k .

(a) (b) (c)

Figure: AD1 ⊕L

AD2.

AD1 := ADh+k1x−h,0

AD2 := AD2k+t−h−k1−1x−h+t,0


Case 3. 0 < h ≤ t + k .

h+k

2k+t−h−k

1

−1

1

y=0

y=n


Case 3. 0 < h ≤ t + k .

h+k 1 2k+

t−h−k

y=0

y=n

1 −1


Case 3. 0 < h ≤ t + k .

h+k 1

2k+t−h−k

y=n

y=0

−1

1


Main Theorem

Theorem (L. 2015)

SMa,b(k1, . . . , ks ; t1, . . . , ts−1) = P(Rx,h(k1, . . . , ks ; t1, . . . , ts−1))

Idea of the proof.

Prove by induction on s + k + t:

Base case s = 1, following from Kenyon-Wilson’s Theorem.

Induction step: Show that two sides satisfy the same recurrence.

Apply Dodgson condensation to obtain a recurrence onsemicontiguous minors.Apply Kuo condensation to obtain the same recurrence ontiling-polynomials of R-type regions.


Dodgson Condensation

Lemma (Dodgson condensation)

XC SENW

SW

NEX = -X

∣∣∣∣∣∣1 2 34 5 69 7 8

∣∣∣∣∣∣ ∣∣5∣∣ =

∣∣∣∣5 67 8

∣∣∣∣ ∣∣∣∣1 24 5

∣∣∣∣− ∣∣∣∣2 35 6

∣∣∣∣ ∣∣∣∣4 59 7

∣∣∣∣−9 · 5 = (−2) · (−3)− (−3) · (−17)


Dodgson Condensation

Lemma (Dodgson condensation)

XC SENW

SW

NEX = -X

Charles Lutwidge Dodgson (1832–1898).

Who is he?


Alice’s Adventures in Wonderland


Future Work

Open Problem

When can a circular minor be expressed as the tiling-polynomial P(R) ofa region R on the square lattice?

We only need to consider the case of MBA with A and B both

non-contiguous.

y=0

Z1 2Z


The end

Thank you!

Email: [email protected]

Website: http://www.ima.umn.edu/~tmlai/

arXiv: http://arxiv.org/abs/1507.02611


http://www.ima.umn.edu/~tmlai/

http://arxiv.org/abs/1507.02611

tiling expression of minors - university of nebraska–lincolntlai3/jtubc.pdf · 2016-11-09 ·...

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