Tiling expression of minors - University of Nebraska–Lincolntlai3/JTUBC.pdf · 2016-11-09 ·...
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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
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Outline
1 Introduction to the Enumeration of Tilings.
2 Electrical networks
3 Tiling expression of minors
4 Future work.
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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.
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
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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.
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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
)).
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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...
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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.
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An Aztec Temple
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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
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Connection to Plane Partitions
a=10
b=10
c=10
c=10
a=10
b=10
k
ij
O
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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!.
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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=?
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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.
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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.
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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.
Tri Lai Tiling expression of minors
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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.
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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).
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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.
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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
6
7 8
9
12
1
3
4
5
6
13
2
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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.
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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.
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Well-connected networks
1
8
7
65
4
3
2
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Well-connected networks
1
8
7
65
4
3
2
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Well-connected networks
1
8
7
65
4
3
2
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Motivational Problem
We would like to test the well-connectivity of givennetworks.
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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.
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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”.
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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
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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
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Circular Minors: Examples
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
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Circular Minors: Examples
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
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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.
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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.
Tri Lai Tiling expression of minors
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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).
Tri Lai Tiling expression of minors
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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).
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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).
Tri Lai Tiling expression of minors
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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).
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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Kenyon–Wilson Theorem (cont.)
Theorem (Kenyon-Wilson 2014)
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.
Tri Lai Tiling expression of minors
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Kenyon–Wilson Theorem (cont.)
Theorem (Kenyon-Wilson 2014)
CONa,b,y (M) = P(TAD|h|,nx−h,y ),
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
Tri Lai Tiling expression of minors
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Kenyon–Wilson Theorem (cont.)
Theorem (Kenyon-Wilson 2014)
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).
=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
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
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
CON1,5,1(M) = P(TAD2,131,1 )
Tri Lai Tiling expression of minors
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Kenyon–Wilson Theorem (cont.)
Theorem (Kenyon-Wilson 2014)
CONa,b,y (M) = P(TAD|h|,nx−h,y ),
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
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
11109
8
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 )
Tri Lai Tiling expression of minors
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Semicontiguous minors
A semicontiguous minor is a circular minor MBA when exactly one of A
and B is contiguous.
(a) (b)
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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.
Tri Lai Tiling expression of minors
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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
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Tri Lai Tiling expression of minors
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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).
Tri Lai Tiling expression of minors
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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.
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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−.
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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
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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Case 3. 0 < h ≤ t + k .
h+k
2k+t−h−k
1
−1
1
y=0
y=n
Tri Lai Tiling expression of minors
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Case 3. 0 < h ≤ t + k .
h+k 1 2k+
t−h−k
y=0
y=n
1 −1
Tri Lai Tiling expression of minors
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Case 3. 0 < h ≤ t + k .
h+k 1
2k+t−h−k
y=n
y=0
−1
1
Tri Lai Tiling expression of minors
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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.
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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)
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Dodgson Condensation
Lemma (Dodgson condensation)
XC SENW
SW
NEX = -X
Charles Lutwidge Dodgson (1832–1898).
Who is he?
Tri Lai Tiling expression of minors
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Dodgson Condensation
Lemma (Dodgson condensation)
XC SENW
SW
NEX = -X
Charles Lutwidge Dodgson (1832–1898).
Who is he?
Tri Lai Tiling expression of minors
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Alice’s Adventures in Wonderland
Tri Lai Tiling expression of minors
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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
Tri Lai Tiling expression of minors
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The end
Thank you!
Email: [email protected]
Website: http://www.ima.umn.edu/~tmlai/
arXiv: http://arxiv.org/abs/1507.02611
Tri Lai Tiling expression of minors