LOZENGE TILINGS WITH GAPS IN A 90 WEDGE DOMAIN WITH MIXED BOUNDARY - PowerPoint PPT Presentation

LOZENGE TILINGS WITH GAPS IN A 90 ◦ WEDGE DOMAIN WITH MIXED BOUNDARY CONDITIONS Mihai Ciucu Department of Mathematics, Indiana University, Bloomington, Indiana 47405 Research supported in part by NSF grant DMS-1101670.

Correlation in a sea of dimers

[C, ’05–’10] In bulk, for large separations, this is asymptotically 2D electrostatics

What about the interaction with boundary? Two natural types: free boundary constrained boundary

Previous examples

“straight line” constrained boundary

straight line free boundary

60 ◦ angle, constrained boundary

120 ◦ angle, constrained boundary

Current talk: 90 ◦ angle, mixed boundary

� x x O � y+ 1 y+ 1 n n D n,x,y : n = 6 , x = 5 , y = 4 D n,x,y ( α , β ): n = 6 , x = 5 , y = 4 , α = 2 , β = 4

• M f ( D ): # tilings of D with tiles allowed to protrude across free boundary portions • : ω c ( α , β ) (correlation of the gap with the corner): M f ( D n,n, 0 ( α , β )) ω c ( α , β ) := lim M f ( D n,n, 0 (1 , 1)) n →∞

D 10 , 10 , 0 (3 , 4). D 10 , 10 , 0 (1 , 1).

O 3 O 1 O O 2 4 The gap and its three images for α = 3, β = 4

The main result of this talk: Theorem. Let q be a fixed positive rational number. As α and β approach infinity so that α = q β , we have � 16 ∼ 32 d( O 1 , O 2 ) d( O 3 , O 4 ) ω c ( α , β ) ∼ d( O 1 , O 3 ) d( O 1 , O 4 ) d( O 2 , O 3 ) d( O 2 , O 4 ) , � π q 2 + 1 3 π Rq 3 where d is the Euclidean distance.

x y+ 1 i i i 2 i k k 1 1 n D i 1 ,...,i k for n = 6, x = 5, y = 4, k = 4, i 1 = 1, i 2 = 3, i 3 = 5, i 4 = 6. n,x,y It turns out we can reduce to enumerating tilings of such regions.

Great strike of luck: They are given by “round” formulas!

Proposition. For any integers n, x ≥ 0 and y ≥ − 1 , and for any integers 1 ≤ i 1 < · · · < i k ≤ n , we have k � x + y + n + i a � i b − i a � � M f ( D i 1 ,...,i k ) = . n,x,y y + 2 i a y + i b + i a a =1 1 ≤ a<b ≤ k

x x y+ 1 y+ 1 i i i 2 i i i i 2 i k k 1 1 k k 1 1 n n Tilings and paths Starting and ending segments The tilings are in bijection with non-intersecting families of paths of rhombi: • starting points: fixed • ending points: can vary among a specified set

x y+ 1 i i i 2 i k k 1 1 n Regarding the paths of lozenges as lattice paths in Z 2

A result of Stembridge expresses this as a Pfa ffi an. After using some combinatorial identities, this Pfa ffi an can be evaluated explicitly using Schur’s Pfa ffi an Identity: Theorem (Schur’s Pfaffian Identity). Let n be even, and let x 1 , . . . , x n be indeterminates. Then we have � n � x j − x i x j − x i � Pf = . x j + x i x j + x i i,j =1 1 ≤ i<j ≤ n

Generalization of SSC plane partitions, even by even by even case.

Generalization of SSC plane partitions, even by odd by odd case.

Corollary (Generalization of SSC plane partitions). Let n, x ≥ 0 and 1 ≤ k 1 < · · · < k s ≤ n be integers. If k 1 > 1 set t = 0 , otherwise define t by requiring k i − i = 0 , i = 1 , . . . , t , and k t +1 − ( t + 1) > 0 . Let { 1 , . . . , n } \ { k 1 , . . . , k s } = { i 1 , . . . , i n − s } . Then we have: (a) . M − , | ( H 2 n, 2 n, 2 x ( k 1 , . . . , k s )) = M f ( D i 1 ,...,i n − s n,x, 2 t − 1 ) n − s � x + 2 t + n + i a − 1 � i b − i a � � = 2 t + i a + i b − 1 . 2 t + 2 i a − 1 a =1 1 ≤ a<b ≤ n − s (b) . M − , | ( H 2 n +1 , 2 n +1 , 2 x ( k 1 , . . . , k s )) = M f ( D i 1 ,...,i n − s ) n,x, 2 t n − s � x + 2 t + n + i a � i b − i a � � = . 2 t + 2 i a 2 t + i a + i b a =1 1 ≤ a<b ≤ n − s

A limit formula for regions with two dents Proposition. For any fixed integers 1 ≤ i < j , we have � � D [ n ] \{ i,j } M f � = 4 j − i 1 � 2 i − 1 � 1 � 2 j − 1 � n,n, 0 lim . 2 2 i − 2 2 2 j − 2 � j + i i − 1 j − 1 D [ n ] \{ 1 , 2 } n →∞ M f n,n, 0

To finish the proof: • a double sum formula • its asymptotic analysis

v 3 R 2 R ( v R ) 2 3 Changing from ( α , β ) to ( R, v )-coordinates.

A double sum formula Lemma. Write α = 2 v − R , β = R , with R and v non-negative integers. Then we have ω c ( α , β ) = ω c (2 v − R, R ) � R R ( − 1) a + b ( R + a − 1)! ( R + b − 1)! � � � = 4 R � (2 a )! ( R − a )! (2 b )! ( R − b )! � � a =0 b =0 (2 v ′ + 2 a + 1)! (2 v ′ + 2 b + 1)! ( b − a ) 2 � � � , × 2 2(2 v ′ + a + b ) ( v ′ + a )! ( v ′ + a + 1)! ( v ′ + b )! ( v ′ + b + 1)! 2 v ′ + a + b + 2 � where v ′ = 2 v − R − 1 .

D 6 , 6 , 0 (3 , 3; { 1 , 2 , 6 , 8 } ) Paths of lozenges

8 8 6 6 2 2 1211 1 1 6 5 4 3 2 1 6 4 2 1 D 1 , 2 , 4 , 6 Labeling starting and ending points 6 , 6 , 0 ( { 1 , 2 , 6 , 8 } )

Outline of proof of double sum formula • Free boundary is sum over constrained boundaries: � M f ( D n,n, 0 ( α , β )) = M( D n,n, 0 ( α , β ; S )) S ⊂ T | S | = n − 2 • Use Pfa ffi an formula for lattice paths and Laplace expansion to get M( D n,n, 0 ( α , β ; S )) = � � � � ( − 1) a + b ( b − a )( R + a − 1)! ( R + b − 1)! M( D [ n ] \{ 2 v − R + a, 2 v − R + b } � � � ( S )) � n,n, 0 � (2 a )! ( R − a )! (2 b )! ( R − b )! � � 0 ≤ a<b ≤ R � �

• Sum over boundaries to get M f ( D n,n, 0 ( α , β )) = � � � � ( − 1) a + b ( b − a )( R + a − 1)! ( R + b − 1)! M f ( D [ n ] \{ 2 v − R + a, 2 v − R + b } � � � 2 R ) � � n,n, 0 (2 a )! ( R − a )! (2 b )! ( R − b )! � � 0 ≤ a<b ≤ R � � • divide by M f ( D n,n, 0 (1 , 1)), let n → ∞ , and use 2-dent limit formula

Reduction of the double sum to simple sums • The double sum separates if we write � 1 1 x 2 v ′ + a + b +1 dx 2 v ′ + a + b + 2 = 0 • Moment sums ( k ∈ Z , x ∈ [0 , 1]): R T ( k ) ( R, v ; x ) := 1 ( − R ) a ( R ) a (3 / 2) v + a � x � a � a k R (1) a (1 / 2) a (2) v + a 4 a =0

Lemma. We have that ω c (2 v − R, R ) = � 1 � 1 � � � 2 � T (2) ( R, v ′ ; x ) T (0) ( R, v ′ ; x ) x 2 v ′ +1 dx − x 2 v ′ +1 dx � T (1) ( R, v ′ ; x ) � 8 R � , � � � 0 0 where v ′ = 2 v − R − 1 .

The asymptotics of the integrals in the lemma It follows from results in [C, Mem. AMS, 2005] that: � 1 T (2) ( R, v ′ ; x ) T (0) ( R, v ′ ; x ) x 2 v ′ +1 dx 0 � 1 � 2 1 � 1 − x − arctan 1 x � � � x 2 qR cos 2 R arccos 4 − x + π dx ∼ π R � 2 q q 2 + x (4 − x ) 0 4 − x and � 1 � 2 � T (1) ( R, v ′ ; x ) dx ∼ 0 � 1 � � � �� 2 1 1 − x − arctan 1 x � � x 2 qR 1 + cos 2 R arccos 4 − x + π dx π R � 2 q q 2 + x (4 − x ) 0 4 − x

Lemma then implies � � � 1 � � ω c (2 v − R, R ) ∼ 16 1 x 2 qR � � dx � � π � q 2 + x � (4 − x ) � 0 4 − x � � as R and v approach infinity so that 2 v − R = qR .

We have � 1 1 1 1 x 2 qR dx ∼ R, R → ∞ � � q 2 + q 2 + 1 x (4 − x ) 0 3 q 4 − x 3 Then we get 16 1 ω c (2 v − R, R ) ∼ R, � q 2 + 1 3 π q 3 which proves the Theorem.

A general conjecture for regions Ω n on the triangular lattice

The two types of zig-zag corners in Ω n

An example of Ω n

� 2 +1 +1 +1 The corresponding steady state heat flow problem

• O ( n ) , . . . , O ( n ) k : finite unions of unit triangles from the interior of Ω n (the gaps) 1 • for fixed i , O ( n ) ’s are translates of one another for all n ≥ 1 i • O ( n ) shrinks to point a i ∈ Ω in scaling limit, i = 1 , . . . , k i • Ω n → Ω , n → ∞ • E : heat energy when sources/sinks are at positions a 1 , . . . , a k ( n ) ’s be translations of the O ( n ) Conjecture. Let O ′ ’s that shrink to distinct i i points a ′ 1 , . . . , a ′ k ∈ Ω in the scaling limit as n → ∞ . Then M f ( Ω n \ O ( n ) ∪ · · · ∪ O ( n ) k ) → exp( − E ) 1 exp( − E ′ ) , ( n ) ∪ · · · ∪ O ′ ( n ) ) M f ( Ω n \ O ′ 1 k where E ′ is the heat energy of the system obtained from S by moving the point heat sources to positions a ′ 1 , . . . , a ′ k .