Proof of a generalization of Aztec diamond theorem

Transcription

1 Proof of a generalization of Aztec diamond theorem Institute for Mathematics and its Applications University of Minnesota Minneapolis, MN December 5, 2014

2 Aztec Diamond Theorem Definition The Aztec diamond of order n is the collection of unit squares inside the contour x + y = n + 1. Theorem (Elkies, Kuperberg, Larsen and Propp 1991) There are 2 n(n+1)/2 ways to cover the Aztec diamond of order n by dominoes so that there are no gaps or overlaps.

3 Aztec Diamond Theorem Theorem (Elkies Kuperberg Larsen Propp, 1991) There are 2 n(n+1)/2 ways to cover the Aztec diamond of order n by dominoes so that there are no gaps or overlaps. The first four proofs have been given by Elkies, Kuperberg, Larsen and Propp. Many further proofs: Propp 2003 Kuo 2004 Brualdi and Kirkland 2005 Eu and Fu 2005 Bosio and Leeuwen 2013 Fendler and Grieser 2014

4 Terminology Introduction A lattice divides the plane into fundamental regions called cells.

5 Terminology Introduction A lattice divides the plane into fundamental regions called cells. A tile is a union of any two cells sharing an edge.

6 Terminology Introduction A lattice divides the plane into fundamental regions called cells. A tile is a union of any two cells sharing an edge. A tiling of a region is a covering of the region by tiles so that there are no gaps or overlaps.

7 Terminology Introduction A lattice divides the plane into fundamental regions called cells. A tile is a union of any two cells sharing an edge. A tiling of a region is a covering of the region by tiles so that there are no gaps or overlaps. Denote by M(R) the number of tilings of R.

8 Hybrid Domino-Lozenge Tilings and Douglas Theorem n=2 n=4 n=6 Theorem The number of tilings of the region of order 2k is 2 2k(k+1).

9 Hybrid Domino-Lozenge Tilings and Douglas Theorem Are there similar regions on the square lattice with arbitrary diagonals drawn in?

10 Hybrid Domino-Lozenge Tilings and Douglas Theorem Are there similar regions on the square lattice with arbitrary diagonals drawn in? What is the number of tilings of the region?

11 Hybrid Domino-Lozenge Tilings and Douglas Theorem Are there similar regions on the square lattice with arbitrary diagonals drawn in? What is the number of tilings of the region? Is it still a power of 2?

12 Generalized Douglas region D a (d 1, d 2,..., d k ). A l a l D B d1 d2 d3 d4 C Figure : The region D 7 (4, 2, 5, 4). The width is the numbers of squares running from B to C.

13 Generalization of Aztec diamond theorem Theorem (L, 2014) Assume that D a (d 1,..., d k ) has the width w. Assume that the the vertices B and D are on the same level. Then M(D a (d 1,..., d k )) = 2 C w(w+1)/2 (1) where C is the number of black squares and black up-pointing triangles (which we called regular cells).

14 Example Introduction A l a l D B d1 d2 d3 d4 C The region D 7 (4, 2, 5, 4) has the width w = 8 and C = = 65. Therefore M(D 7 (4, 2, 5, 4)) = 2 C w(w+1)/ = 2 2 = 536, 870, 912.

15 Remark Introduction Q: Why did we assume that B and D are on the same level in the main theorem?

16 Remark Introduction Q: Why did we assume that B and D are on the same level in the main theorem? A: Otherwise the numbers black and white cells are not the same, so the region has no tiling.

17 A remark Introduction A A balanced balanced D B D B balanced C balanced C

18 A remark Introduction A D B C

19 Bijection between tilings and perfect matchings Figure : Bijection between tilings of the Aztec diamond of order 5 and perfect matchings of its dual graph. The dual graph of a region R is the graph whose vertices are the cells in R and whose edges connect precisely two adjacent cells. A perfect matching of a graph G is a collection of disjoint edges covering all vertices of G.

20 From tilings to families of non-intersecting paths A l l D B d1 d2 d3 d4 C

21 From tilings to families of non-intersecting paths G G

22 From tilings to families of non-intersecting paths G G

23 From tilings to families of non-intersecting paths G

24 From tilings to families of non-intersecting paths

25 From tilings to families of non-intersecting paths x

26 From tilings to families of non-intersecting paths x

27 Schröder paths Definition A Schröder path is a lattice path: 1 starting and ending on the x-axis; 2 never go below x-axis; 3 using the steps: (1,1) up, (1,-1) down, and (2,0) flat x

28 Large and small Schröder numbers gessel/homepage/slides/schroder.pdf r n = the number of Schröder paths from (0, 0) to (2n, 0).

29 Large and small Schröder numbers gessel/homepage/slides/schroder.pdf r n = the number of Schröder paths from (0, 0) to (2n, 0). s n = the number of Schröder paths from (0, 0) to (2n, 0) with no flat step on the x-axis.

30 Large and small Schröder numbers gessel/homepage/slides/schroder.pdf r n = the number of Schröder paths from (0, 0) to (2n, 0). s n = the number of Schröder paths from (0, 0) to (2n, 0) with no flat step on the x-axis. r n = 2s n

31 Large and small Schröder numbers gessel/homepage/slides/schroder.pdf r n = the number of Schröder paths from (0, 0) to (2n, 0). s n = the number of Schröder paths from (0, 0) to (2n, 0) with no flat step on the x-axis. r n = 2s n R(x) = n=0 r nx n = 1 x 1 6x + x 2 2x

32 Schröder paths with barriers Definition 1 For integers a m > a m 1 >... > a 1 0, define B(a 1, a 2,..., a m ) is the set of length-1 barriers running along the lines y = x + a i (i.e. connecting ( a i + k, k ) and ( a i + k + 1, k )). 2 If a Schröder path P avoids all barriers in B = B(a 1, a 2,..., a m ), we say P is compatible with B

33 Two families of non-intersecting Schröder paths with barriers Definition 1 Let x i be the i-th largest negative odd number in Z\{ a 1,..., a m }. 2 Π n (a 1,..., a m ) set of n-tuples of non-intersecting Schröder paths (π 1,..., π n ), where π i connects A i = (x i, 0) and B i = (2i 1, 0) and is compatible with B(a 1, a 2,..., a m ).

34 Families of non-intersecting paths in D a (d 1, d 2,..., d k ) x Figure : Families of non-intersecting paths in D 7 (4, 2, 5, 4). Π w (a 1,..., a k 1 ) where a i := d k d k i+1 + i 1, for i = 1, 2,..., k 1.

35 Families of non-intersecting paths in D a (d 1, d 2,..., d k ) x Lemma M(D a (d 1,..., d k )) = Π w (a 1,..., a k 1 ), (2) where a i := d k d k i+1 + i 1, for i = 1, 2,..., k 1.

36 Inductive proof on the number layers Theorem M(D a (d 1, d 2,..., d k )) = 2 C w(w+1)/2 (*) If k = 1, the theorem follows from the Aztec diamond theorem. Assume that (*) is true for any generalized Douglas regions with less than k layers (k 2). We need to show for any D a (d 1, d 2,..., d k ). Consider the d k = 2x (the case d k = 2x 1 can be obtained similarly).

37 Reduction using Lindström Gessel Viennot Lemma. Lemma (Linström Gessel Viennot) 1 G is an acyclic digraph. 2 U = {u 1, u 2,..., u n }, V = {v 1, v 2,..., v n } 3 For any 1 i < j n, P : u i v j intersects Q : u j v i. Then a 1,1 a 1,2... a 1,n a 2,1 a 2,2... a 2,n #(τ 1, τ 2,..., τ n ) U V = det..., a n,1 a n,2... a n,n where a i,j is the number of paths from u i to v j.

38 Reduction using Linström Gessel Viennot Lemma. M(D a (d 1,..., d k )) = Π w (a 1,..., a m ) = det(h w (a 1,..., a m )), (3) where r 1,1 r 1,2... r 1,w r 2,1 r 2,2... r 2,w H w := H w (a 1,..., a m ) :=..., (4) r w,1 r w,2... r w,w and where r i,j = the number of Schröder paths connecting A i to B j compatible with B(a 1,..., a m ).

39 Reduction using Linström Gessel Viennot Lemma. A4 A 3 A 2 A 1 B 1 B B B 2 3 4

40 Reduction using Linström Gessel Viennot Lemma. Definition Given B(a 1,..., a m ) 1 A bad level step is a level step on x-axis avoiding all points ( a i, 0). 2 Λ n (a 1,..., a m ) set of n-tuples of non-intersecting Schröder paths (λ 1,..., λ n ), where λ i connects A i = (x i, 0) and B i = (2i 1, 0), is compatible with B(a 1, a 2,..., a m ), and has no bad flat step.

41 Reduction using Linström Gessel Viennot Lemma. Λ w (a 1,..., a m ) = det(q w (a 1,..., a m )), (5) where s 1,1 s 1,2... s 1,w s 2,1 s 2,2... s 2,w Q w := Q w (a 1,..., a m ) :=..., (6) s w,1 s w,2... s w,w and where s i,j = the number of Schröder paths connecting A i to B j, is compatible with B(a 1,..., a m ), and has no bad flat step.

42 Reduction using Linström Gessel Viennot Lemma. A4 A 3 A 2 A 1 B 1 B B B 2 3 4

43 Reduction using Linström Gessel Viennot Lemma. Lemma If a 1 > 1, then r i,j = 2s i,j. det(h w (a 1,..., a m )) = 2 w det(q w (a 1,..., a m ))

44 Reduction using Linström Gessel Viennot Lemma. Lemma If a 1 > 1, then r i,j = 2s i,j. det(h w (a 1,..., a m )) = 2 w det(q w (a 1,..., a m )) Π w (a 1,..., a m ) = 2 w Λ w (a 1,..., a m )

45 Reduction using Linström Gessel Viennot Lemma. Π w (a 1,..., a m ) = 2 w Λ w (a 1,..., a m ) Lemma If a 1 > 1, then Π w 1 (a 1 2,..., a m 2) = Λ w (a 1,..., a m )

46 Reduction using Linström Gessel Viennot Lemma. Π w (a 1,..., a m ) = 2 w Π w 1 (a 1 2,..., a m 2)

47 Reduction using Linström Gessel Viennot Lemma. Π w (a 1,..., a m ) = 2 w Π w 1 (a 1 2,..., a m 2) = 2 x 1 i=0 (w i) Π w x (0, a 2 a 1,..., a m a 1 )

48 Comparing a and x A D B x x+1 C a x w x + 1

49 Case a = x Introduction A D B C

50 Case a = x Introduction A D B C We have d 1 = d 2 =... = d k 1 = 1 and w = k + x 1 so Π w x (0, a 2 a 1..., a m a 1 ) = Π k 1 (0, 2, 4,..., 2(k 1)) = 1.

51 Case a = x Introduction Π k 1 (0, 2, 4,..., 2(k 1)) =

52 Case a = x Introduction M(D a (d 1,..., d k )) = Π w (a 1,..., a m ) = 2 x 1 i=0 (w i). Need to show that x 1 (w i) = C w(w + 1)/2 i=0

53 Case a = x Introduction Need to show that x 1 (w i) = C w(w + 1)/2 i=0 A D B C w x 1 C = (w + 1)x + (w i) i=0

54 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) Π w x (0, a 2 a 1,..., a m a 1 ) Lemma Π w x (0, a 2 a 1,..., a m a 1 ) = Π w x 1 (a 2 a 1 2,..., a m a 1 2)

55 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) Π w x 1 (a 2 a 1 2,..., a m a 1 2) Recall that we have bijection between tilings of D a (d 1,..., d k ) and elements of Π w (a 1, a 2,..., a m ).

56 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) Π w x 1 (a 2 a 1 2,..., a m a 1 2) Recall that we have bijection between tilings of D a (d 1,..., d k ) and elements of Π w (a 1, a 2,..., a m ). Tilings of D := D a x (d 1, d 2,..., d k 1 1) correspond to elements of Π w x 1 (a 2 a 1 2,..., a m a 1 2).

57 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) Π w x 1 (a 2 a 1 2,..., a m a 1 2) Recall that we have bijection between tilings of D a (d 1,..., d k ) and elements of Π w (a 1, a 2,..., a m ). Tilings of D := D a x (d 1, d 2,..., d k 1 1) correspond to elements of Π w x 1 (a 2 a 1 2,..., a m a 1 2). M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) M(D a x (d 1, d 2,..., d k 1 1))

58 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) M(D a x (d 1, d 2,..., d k 1 1))

59 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) 2 C w (w +1)/2

60 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) 2 C w (w +1)/2 w w = x + 1

61 Case a > x Introduction M(D a (d 1,..., d k )) = 2 x 1 i=0 (w i) 2 C w (w +1)/2 w w = x + 1 C C = (w + 1)x + w + x(w x 1)

62 Case a > x Introduction M(D a (d 1,..., d k )) = 2 C w(w+1)/2

63 Questions? Introduction Thank you!