SLC 69, Strobl, Austria

Transcriptie

1 Eric Joint work with Benjamin Young Fakultät für Mathematik Universität Wien SLC 69, Strobl, Austria

2 The Aztec Diamond Aztec diamonds of orders 1, 2, 3 and 4.

3 The Aztec Diamond Aztec diamonds of orders 1, 2, 3 and 4. The diamond of order n can be tiled in 2 n(n+1)/2 ways. Elkies, Kuperberg, Larsen & Propp 1992

4 The Aztec Diamond

5 The number of tilings of an order n Aztec diamond is 2 (n+1 2 ). Jonathan observed that det [( 2i j )] n i,j=1 = 2 (n+1 2 ).

6

7

8 The shuffling algorithm The Arctic Parabola Theorem.

9

10

11

12

13 Introduce a coordinate system: Xj i (t) is the position of the jth particle on level i at time t Note that X i j (t) X i 1 (t) < Xj+1(t) i j

14 Recursion equations X 1 1 (t) = X 1 1 (t 1) + β 1 1(t) X j 1 (t) = X j 1 (t 1) + βj 1 (t) 1{X j 1 (t 1) + βj j 1 1 (t) = X1 (t) + 1} for j 2 X j j (t) = X j j (t 1) + βj j (t) + 1{X j j (t 1) + βj j 1 j (t) = Xj 1 (t)} for j 2 X j i (t) = X j i (t 1) + βj i (t) + 1{X j i (t 1) + βj j 1 i (t) = Xi 1 (t)} 1{X j i (t 1) + βj j 1 i (t) = Xi (t) + 1} for j > i > 1 where all βj i (t) are independent coin flips.

15 Particle dynamics Time shift: x j i (t) = X j i (t j)

16 Half-Aztec diamond

17 The Arctic Parabola Theorem Theorem Consider uniform measure on tilings of the. The region in which the density of particles (i.e. vertical lozenges) is assymptotically non-zero is bounded by a parabola.

18 Proposition (N & Y 2011) The limit shape in the Half-Aztec diamond is the semi-circle. Jockusch, Propp & Shor (1998) Cohn, Kenyon & Propp (2001)

19 The Arctic Parabola Theorem

20 s

21 s Consider n Bernoulli walkers started at 1, 2,..., n, and conditioned to end up at positions y 1,..., y n, at time N conditioned never to intersect. The number of such configurations is given by the Lindström-Gessel-Viennot Theorem as the determinant of [( )] N n M = y j i i,j=1

22 Eynard-Mehta Theorem Theorem (Eynard & Mehta (1998), Borodin & Rains (2005)) The probability that there is a walker at each of (x 1, t 1 ),..., (x k, t k ) is det[k(t i, x i ; t j, x j )] k i,j=1 where ( ) s r K(r, x; s, y) = 1{s > r} y x n + i,j=1 ( N r y i x ) ( ) s [M 1 ] i,j y j

23 [( det N y i j 1{j s} ( n 1 i=1 )] n 1 i,j=1 = N! (y i 1)!(N y i + n)! ) det[f (i, j, s)] n 1 i,j=1 where f (i, j, s) = { (y i j + 1) (y i 1)(N y i + j + 1) (N y i + n), j < s, (y i j) (y i 1)(N y i + j + 2) (N y i + n), j s.

24 Let mean taking the Vandermonde determinant in the variables. For s = 1, sage gave us ( n 2 ) n 1 P n,1 (N, y) = (y) (N + i) n 1 i (y j 1) i=1 j=1

25 Let mean taking the Vandermonde determinant in the variables. For s = 2, sage gave us P 3,2 (N, y) =(N + 1) (y)( 2e 2 (y) + (N + 4)e 1 (y) (3N + 8)) P 4,2 (N, y) =(N + 1) 2 (N + 2) (y)( 3e 3 (y) + (N + 6)e 2 (y) (3N + 12)e 1 (y) + (7N + 24)) P 5,2 (N, y) =(N + 1) 3 (N + 2) 2 (N + 3) (y)( 4e 4 (y) + (N + 8)e 3 (y) (3N + 16)e 2 (y) + (7N + 32)e 1 (y) (15N + 64)) P 6,2 (N, y) =(N + 1) 4 (N + 2) 3 (N + 3) 2 (N + 4) (y)( 5e 5 (y) + (N + 10)e 4 (y) (3N + 20)e 3 (y) + (7N + 40)e 2 (y) (15N + 80)e 1 (y) + (31N + 160))

26 n 2 P n,s (N, y) = (y) (N + r) n 1 r n 1 s 1 s l=0 k=0 j=1 i=0 r=1 j ( 1) n+s+l+j Nk j l e n 1 l (y) s(s 1 j, k i) i!(s 1)! ( ( ) ) d i (s ) 1 (n 1) (n j) (1) dn j where s(n, k) are the Stirling numbers of the first kind.

27 Matrix Inverse Theorem Let Then [M 1 ] i,j = M = [( )] N n. y i j i,j=1 ( j N+n 1 )( N 1+j k k 1 j k ( N+n 1 k=1 y i 1 ) n ) ( 1) k+j l=1, l i k y l y i y l.

28 Proof [MM 1 ] α,γ = = β=1 k=1 n [M] α,β [M 1 ] β,γ β=1 n γ ( ) N + n 1 1 ( N + n 1 ( 1) k+γ y β 1 k 1 ( )( ) n N 1 + γ k N γ k y β α i=1, i β ) k y i y β y i. (2)

29 Lagrange interpolation Let (x 1, y 1 ),..., (x n, y n ) R 2 and let p k (x) = n i=1,i k x x i x k x i. Then n f (x) = y k p k (x) k=1 has the property that f (x i ) = y i for i = 1,..., n.

30 Proof [MM 1 ] α,γ = γ ( )( ) N 1 + β k N ( 1) β+j β k k γ ( ) 0 = = δ α,γ (3) α γ k=1

31 Corollary The correlation functions for the are determinental, with given by K(r, x; s, y) = φ r,s (x, y) n + i,j=1 ( n+1 r 2i x )( s ( 2n 2i 1 y j ) ( 1)k+j+i+n (i 1)!(n i)! ) j k=1 n l=1, l i ( 2n k 1 (k 2l) )( n + j k j k ) where for r s, φ 0 and for r < s, ( ) s r φ r,s (x, y) =. y x

32

33 Thank you four your attention, Young, Domino shuffling on s and Aztec half-diamonds, Electron. J. of Combin. 18 (2011), no. 1., Young, s for the Process, FPSAC 2012 proceedings, arxiv:

34 q-analog Theorem has inverse N = 2 ) [N 1 ] i,j = qnb i ( Bi ] a=0 b=0 [ A+n 1 B i 1 q q [ [ ] ] n A q (B j i ) 2, B j i q i,j=1 n k=1, k i j 1 n 1 [ ] [ ] b n b 1 j 1 a a 1 q B i q B k q (j 1 q 2 )+(a+b)(a j 1) b 1+aA ( 1) b e b (q B 1,..., q B i,..., q B n ).