From alternating sign matrices to Painlevé VI

Transcription

1 From alternating sign matrices to Painlevé VI Hjalmar Rosengren Chalmers University of Technology and University of Gothenburg Nagoya, July 31, 2012 Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

2 Outline 1 Three-coloured chessboards 2 Combinatorial results 3 Symmetric polynomials 4 Painlevé VI 5 Future problems Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

3 Outline 1 Three-coloured chessboards 2 Combinatorial results 3 Symmetric polynomials 4 Painlevé VI 5 Future problems Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

4 Three-coloured chessboards n n Chessboard of size (n + 1) (n + 1). Paint squares with three colours 0, 1, 2 mod 3. Adjacent squares have distinct colour. Domain wall boundary conditions (DWBC). Read entries mod 3. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

5 Three-coloured chessboards n n Chessboard of size (n + 1) (n + 1). Paint squares with three colours 0, 1, 2 mod 3. Adjacent squares have distinct colour. Domain wall boundary conditions (DWBC). Read entries mod 3. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

6 Example When n = 3 there are seven chessboards. 0 = black, 1 = red, 2 = yellow. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

7 Other descriptions Chessboard Alternating sign matrix Ice graph Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

8 Bijection to alternating sign matrices Represent colours (residue classes mod 3) by integers so that neighbours differ by Contract each block [ a c d b ] to (b + c a d)/2 { 1, 0, 1} Gives bijection to n n alternating sign matrices. Non-zero entries in each row and column alternate in sign and add up to 1. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

9 Bijection to alternating sign matrices Represent colours (residue classes mod 3) by integers so that neighbours differ by Contract each block [ a c d b ] to (b + c a d)/2 { 1, 0, 1} Gives bijection to n n alternating sign matrices. Non-zero entries in each row and column alternate in sign and add up to 1. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

10 Bijection to alternating sign matrices Represent colours (residue classes mod 3) by integers so that neighbours differ by Contract each block [ a c d b ] to (b + c a d)/2 { 1, 0, 1} Gives bijection to n n alternating sign matrices. Non-zero entries in each row and column alternate in sign and add up to 1. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

11 Bijection to ice graphs (states of six-vertex model) Put arrows between adjacent squares. Larger entry to the right, 0 < 1 < 2 < Each vertex has two incoming and two outgoing edges. Domain wall boundary conditions. Vertex = Oxygen, Incoming arrow = Hydrogen bond. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

12 Bijection to ice graphs (states of six-vertex model) Put arrows between adjacent squares. Larger entry to the right, 0 < 1 < 2 < Each vertex has two incoming and two outgoing edges. Domain wall boundary conditions. Vertex = Oxygen, Incoming arrow = Hydrogen bond. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

13 Bijection to ice graphs (states of six-vertex model) Put arrows between adjacent squares. Larger entry to the right, 0 < 1 < 2 < 0. Each vertex has two incoming and two outgoing edges. Domain wall boundary conditions. Vertex = Oxygen, Incoming arrow = Hydrogen bond. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

14 Bijection to ice graphs (states of six-vertex model) Put arrows between adjacent squares. Larger entry to the right, 0 < 1 < 2 < 0. Each vertex has two incoming and two outgoing edges. Domain wall boundary conditions. Vertex = Oxygen, Incoming arrow = Hydrogen bond. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

15 Bijection to ice graphs (states of six-vertex model) Put arrows between adjacent squares. Larger entry to the right, 0 < 1 < 2 < 0. Each vertex has two incoming and two outgoing edges. Domain wall boundary conditions. Vertex = Oxygen, Incoming arrow = Hydrogen bond. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

16 Bijection to ice graphs (states of six-vertex model) Put arrows between adjacent squares. Larger entry to the right, 0 < 1 < 2 < 0. Each vertex has two incoming and two outgoing edges. Domain wall boundary conditions. Vertex = Oxygen, Incoming arrow = Hydrogen bond. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

17 Outline 1 Three-coloured chessboards 2 Combinatorial results 3 Symmetric polynomials 4 Painlevé VI 5 Future problems Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

18 Example For n = 5, the number of chessboards with exactly k squares of colour 0 and l squares of colour 2 are as follows. k= l= = Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

19 Example For n = 5, the number of chessboards with exactly k squares of colour 0 and l squares of colour 2 are as follows. k= l= = Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

20 Example For n = 5, the number of chessboards with exactly k squares of colour 0 and l squares of colour 2 are as follows. k= l= = ( 6 k ) Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

21 Example For n = 5, the number of chessboards with exactly k squares of colour 0 and l squares of colour 2 are as follows. k= l= = ( 6 k = ) Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

22 Example For n = 5, the number of chessboards with exactly k squares of colour 0 and l squares of colour 2 are as follows. k= l= = ( ) 6 k = ( ) ( 4 k + 4 ) k 2 Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

23 Example For n = 5, the number of chessboards with exactly k squares of colour 0 and l squares of colour 2 are as follows. k= l= = ( ) 6 k = ( ) ( 4 k + 4 k 2 ) = Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

24 Generating function (partition function) Z n (t 0, t 1, t 2 ) = chessboards of size (n+1) (n+1) # squares coloured 0 # squares coloured 1 # squares coloured 2 t 0 t 1 t 2 Z 5 (t 0, t 1, t 2 ) = t 14 0 t 14 1 t t 13 0 t 14 1 t Partition function for three-colour model with DWBC. Studied by Baxter (1970) for periodic boundary conditions. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

25 Generating function (partition function) Z n (t 0, t 1, t 2 ) = chessboards of size (n+1) (n+1) # squares coloured 0 # squares coloured 1 # squares coloured 2 t 0 t 1 t 2 Z 5 (t 0, t 1, t 2 ) = t 14 0 t 14 1 t t 13 0 t 14 1 t Partition function for three-colour model with DWBC. Studied by Baxter (1970) for periodic boundary conditions. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

26 Generating function (partition function) Z n (t 0, t 1, t 2 ) = chessboards of size (n+1) (n+1) # squares coloured 0 # squares coloured 1 # squares coloured 2 t 0 t 1 t 2 Z 5 (t 0, t 1, t 2 ) = t 14 0 t 14 1 t t 13 0 t 14 1 t Partition function for three-colour model with DWBC. Studied by Baxter (1970) for periodic boundary conditions. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

27 Three questions How many states are there (for fixed n)? What is Z n (1, 1, 1)? How common are the various colors? What is Z n t j (1, 1, 1) = chessboards #squares of colour j? What is the joint distribution of the three colours? What is Z n (t 0, t 1, t 2 )? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

28 Three questions How many states are there (for fixed n)? What is Z n (1, 1, 1)? How common are the various colors? What is Z n t j (1, 1, 1) = chessboards #squares of colour j? What is the joint distribution of the three colours? What is Z n (t 0, t 1, t 2 )? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

29 Three questions How many states are there (for fixed n)? What is Z n (1, 1, 1)? How common are the various colors? What is Z n t j (1, 1, 1) = chessboards #squares of colour j? What is the joint distribution of the three colours? What is Z n (t 0, t 1, t 2 )? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

30 Question 1: Enumeration Alternating sign matrix theorem: #chessboards = 1! 4! 7! (3n 2)! n!(n + 1)!(n + 2)! (2n 1)!. Conjectured by Mills Robbins Rumsey (1983). Proved by Zeilberger (1996). Much simpler proof by Kuperberg (1996), using six-vertex model. We generalize Kuperberg s work using eight-vertex-solid-on-solid (8VSOS) model. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

31 Question 1: Enumeration Alternating sign matrix theorem: #chessboards = 1! 4! 7! (3n 2)! n!(n + 1)!(n + 2)! (2n 1)!. Conjectured by Mills Robbins Rumsey (1983). Proved by Zeilberger (1996). Much simpler proof by Kuperberg (1996), using six-vertex model. We generalize Kuperberg s work using eight-vertex-solid-on-solid (8VSOS) model. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

32 Question 1: Enumeration Alternating sign matrix theorem: #chessboards = 1! 4! 7! (3n 2)! n!(n + 1)!(n + 2)! (2n 1)!. Conjectured by Mills Robbins Rumsey (1983). Proved by Zeilberger (1996). Much simpler proof by Kuperberg (1996), using six-vertex model. We generalize Kuperberg s work using eight-vertex-solid-on-solid (8VSOS) model. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

33 8VSOS model (no details) Introduced by Baxter Generalizes both three-colour model and six-vertex model. Same states, but more general weight function. It gives a nice way to put 2n extra parameters into three-colour model (or 2 extra parameters into six-vertex model). The Yang Baxter equation implies non-trivial and useful symmetries in these extra parameters. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

34 8VSOS model (no details) Introduced by Baxter Generalizes both three-colour model and six-vertex model. Same states, but more general weight function. It gives a nice way to put 2n extra parameters into three-colour model (or 2 extra parameters into six-vertex model). The Yang Baxter equation implies non-trivial and useful symmetries in these extra parameters. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

35 8VSOS model (no details) Introduced by Baxter Generalizes both three-colour model and six-vertex model. Same states, but more general weight function. It gives a nice way to put 2n extra parameters into three-colour model (or 2 extra parameters into six-vertex model). The Yang Baxter equation implies non-trivial and useful symmetries in these extra parameters. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

36 Trigonometric case The 8VSOS model involves elliptic functions. Using the trigonometric limit case, we can prove that, with ω = e 2πi/3, ( ) 1 Z n (1 λ), 1 3 (1 λω), 1 3 (1 λω 2 ) 3 = (1 λω2 ) 2 (1 λω n+1 ) 2 (A n (1 + ω n λ 2 ) + ( 1) n C n ω 2n λ), (1 λ 3 ) n2 +2n+3 where A n are alternating sign matrix numbers and C n = n j=1 (3j 1)(3j 3)! (n + j 1)! count cyclically symmetric plane partitions. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

37 Consequence The case λ = 0 is the ASM Theorem: Z n (1, 1, 1) = A n. Applying / λ gives expressions for the first moments λ=0 #squares of colour j, chessboards which answers Question 2. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

38 Consequence The case λ = 0 is the ASM Theorem: Z n (1, 1, 1) = A n. Applying / λ gives expressions for the first moments λ=0 #squares of colour j, chessboards which answers Question 2. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

39 Answer to Question 2: How common are the colours? Suppose n 0 mod 6 and consider the colour 0 (all other cases are similar). Probability that random square from random chess-board (chosen uniformly) has colour 0 is (3n 1) 9(n + 1) (3n 2) + 4 9(n + 1) 2 = Γ(1/3) 9 Γ(2/3) n 5/ n 2 + O(n 3 ), n. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

40 Answer to Question 2: How common are the colours? Suppose n 0 mod 6 and consider the colour 0 (all other cases are similar). Probability that random square from random chess-board (chosen uniformly) has colour 0 is (3n 1) 9(n + 1) (3n 2) + 4 9(n + 1) 2 = Γ(1/3) 9 Γ(2/3) n 5/ n 2 + O(n 3 ), n. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

41 Answer to Question 2: How common are the colours? Suppose n 0 mod 6 and consider the colour 0 (all other cases are similar). Probability that random square from random chess-board (chosen uniformly) has colour 0 is (3n 1) 9(n + 1) (3n 2) + 4 9(n + 1) 2 = Γ(1/3) 9 Γ(2/3) n 5/ n 2 + O(n 3 ), n. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

42 Question 3: What is Z n in general? Z n can be split as a sum of two parts. Each part is a specialized affine Lie algebra character, and a tau function of Painlevé VI. Moreover, each part satisfies a Toda-type recursion. Before giving some details, we mention one more thing. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

43 Question 3: What is Z n in general? Z n can be split as a sum of two parts. Each part is a specialized affine Lie algebra character, and a tau function of Painlevé VI. Moreover, each part satisfies a Toda-type recursion. Before giving some details, we mention one more thing. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

44 Question 3: What is Z n in general? Z n can be split as a sum of two parts. Each part is a specialized affine Lie algebra character, and a tau function of Painlevé VI. Moreover, each part satisfies a Toda-type recursion. Before giving some details, we mention one more thing. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

45 Free energy Supppose t 0, t 1, t 2 are positive. Conjecture: log Z n (t 0, t 1, t 2 ) lim n n 2 where ζ is determined by = 1 3 log(t 0t 1 t 2 ) + log ( (ζ + 2) 3 4 (2ζ + 1) ζ 1 12 (ζ + 1) 4 3 (t 0 t 1 + t 0 t 2 + t 1 t 2 ) 3 (t 0 t 1 t 2 ) 2 = 2(ζ2 + 4ζ + 1) 3 ζ(ζ + 1) 4, ζ 1. Compare Baxter s formula for periodic boundary conditions: ( ) 1 3 log(t ζ 1 3 (ζ + 1) 4 3 0t 1 t 2 ) + log. (2ζ + 1) 3 2 ), Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

46 Free energy Supppose t 0, t 1, t 2 are positive. Conjecture: log Z n (t 0, t 1, t 2 ) lim n n 2 where ζ is determined by = 1 3 log(t 0t 1 t 2 ) + log ( (ζ + 2) 3 4 (2ζ + 1) ζ 1 12 (ζ + 1) 4 3 (t 0 t 1 + t 0 t 2 + t 1 t 2 ) 3 (t 0 t 1 t 2 ) 2 = 2(ζ2 + 4ζ + 1) 3 ζ(ζ + 1) 4, ζ 1. Compare Baxter s formula for periodic boundary conditions: ( ) 1 3 log(t ζ 1 3 (ζ + 1) 4 3 0t 1 t 2 ) + log. (2ζ + 1) 3 2 ), Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

47 Outline 1 Three-coloured chessboards 2 Combinatorial results 3 Symmetric polynomials 4 Painlevé VI 5 Future problems Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

48 Symmetric polynomials Let S n (x 1,..., x n, y 1,..., y n, z) n i,j=1 = G(x i, y j ) 1 i<j n (x j x i )(y j y i ) det 1 i,j n ( ) F(xi, y j, z), G(x i, y j ) F(x, y, z) = (ζ + 2)xyz ζ(xy + yz + xz + x + y + z) + ζ(2ζ + 1), G(x, y) = (ζ + 2)xy(x + y) ζ(x 2 + y 2 ) 2(ζ 2 + 3ζ + 1)xy + ζ(2ζ + 1)(x + y). This is a symmetric (!) polynomial in all 2n + 1 variables, depending on parameter ζ. It can be identified with a character of A (2) 4n 3 affine Lie algebra. Cauchy-type : all minors have the same form. Very useful! Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

49 Symmetric polynomials Let S n (x 1,..., x n, y 1,..., y n, z) n i,j=1 = G(x i, y j ) 1 i<j n (x j x i )(y j y i ) det 1 i,j n ( ) F(xi, y j, z), G(x i, y j ) F(x, y, z) = (ζ + 2)xyz ζ(xy + yz + xz + x + y + z) + ζ(2ζ + 1), G(x, y) = (ζ + 2)xy(x + y) ζ(x 2 + y 2 ) 2(ζ 2 + 3ζ + 1)xy + ζ(2ζ + 1)(x + y). This is a symmetric (!) polynomial in all 2n + 1 variables, depending on parameter ζ. It can be identified with a character of A (2) 4n 3 affine Lie algebra. Cauchy-type : all minors have the same form. Very useful! Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

50 Symmetric polynomials Let S n (x 1,..., x n, y 1,..., y n, z) n i,j=1 = G(x i, y j ) 1 i<j n (x j x i )(y j y i ) det 1 i,j n ( ) F(xi, y j, z), G(x i, y j ) F(x, y, z) = (ζ + 2)xyz ζ(xy + yz + xz + x + y + z) + ζ(2ζ + 1), G(x, y) = (ζ + 2)xy(x + y) ζ(x 2 + y 2 ) 2(ζ 2 + 3ζ + 1)xy + ζ(2ζ + 1)(x + y). This is a symmetric (!) polynomial in all 2n + 1 variables, depending on parameter ζ. It can be identified with a character of A (2) 4n 3 affine Lie algebra. Cauchy-type : all minors have the same form. Very useful! Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

51 Symmetric polynomials Let S n (x 1,..., x n, y 1,..., y n, z) n i,j=1 = G(x i, y j ) 1 i<j n (x j x i )(y j y i ) det 1 i,j n ( ) F(xi, y j, z), G(x i, y j ) F(x, y, z) = (ζ + 2)xyz ζ(xy + yz + xz + x + y + z) + ζ(2ζ + 1), G(x, y) = (ζ + 2)xy(x + y) ζ(x 2 + y 2 ) 2(ζ 2 + 3ζ + 1)xy + ζ(2ζ + 1)(x + y). This is a symmetric (!) polynomial in all 2n + 1 variables, depending on parameter ζ. It can be identified with a character of A (2) 4n 3 affine Lie algebra. Cauchy-type : all minors have the same form. Very useful! Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

52 Symmetric polynomials Let S n (x 1,..., x n, y 1,..., y n, z) n i,j=1 = G(x i, y j ) 1 i<j n (x j x i )(y j y i ) det 1 i,j n ( ) F(xi, y j, z), G(x i, y j ) F(x, y, z) = (ζ + 2)xyz ζ(xy + yz + xz + x + y + z) + ζ(2ζ + 1), G(x, y) = (ζ + 2)xy(x + y) ζ(x 2 + y 2 ) 2(ζ 2 + 3ζ + 1)xy + ζ(2ζ + 1)(x + y). This is a symmetric (!) polynomial in all 2n + 1 variables, depending on parameter ζ. It can be identified with a character of A (2) 4n 3 affine Lie algebra. Cauchy-type : all minors have the same form. Very useful! Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

53 Relation to three-colour model Let p n (ζ) = elementary factor ( ζ S n 2ζ + 1,..., 2ζ + 1, }{{} ζ + 2,..., n+1 This is a polynomial in ζ of degree n(n + 1)/2. ζ ζ + 2 } {{ } n Result: Z n (t 0, t 1, t 2 ) is a linear combination (with elementary coefficients) of p n 1 (ζ) and p n 1 (1/ζ), where (t 0 t 1 + t 0 t 2 + t 1 t 2 ) 3 (t 0 t 1 t 2 ) 2 = 2(ζ2 + 4ζ + 1) 3 ζ(ζ + 1) 4. ). Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

54 Relation to three-colour model Let p n (ζ) = elementary factor ( ζ S n 2ζ + 1,..., 2ζ + 1, }{{} ζ + 2,..., n+1 This is a polynomial in ζ of degree n(n + 1)/2. ζ ζ + 2 } {{ } n Result: Z n (t 0, t 1, t 2 ) is a linear combination (with elementary coefficients) of p n 1 (ζ) and p n 1 (1/ζ), where (t 0 t 1 + t 0 t 2 + t 1 t 2 ) 3 (t 0 t 1 t 2 ) 2 = 2(ζ2 + 4ζ + 1) 3 ζ(ζ + 1) 4. ). Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

55 More precisely... Suppose n 0 mod 6 and (t 0 t 1 + t 0 t 2 + t 1 t 2 ) 3 (t 0 t 1 t 2 ) 2 = 2(ζ2 + 4ζ + 1) 3 ζ(ζ + 1) 4. Z n (t 0, t 1, t 2 ) = (t 0 t 1 t 2 ) n(n+2) 3 ) n 2 12 ( 2 ζ(ζ + 1) 4 p n 1 (ζ) ζ (t n p n 1 (1/ζ) 0 1 ζ t 0t 1 t 2 (ζ 2 + 4ζ + 1) t 0 t 1 + t 0 t 2 + t 1 t 2 Note that only asymmetry comes from t 0. ) n p n 1 (ζ) ζ 2 2 p n 1 (1/ζ) 1 ζ 2 Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

56 More precisely... Suppose n 0 mod 6 and (t 0 t 1 + t 0 t 2 + t 1 t 2 ) 3 (t 0 t 1 t 2 ) 2 = 2(ζ2 + 4ζ + 1) 3 ζ(ζ + 1) 4. Z n (t 0, t 1, t 2 ) = (t 0 t 1 t 2 ) n(n+2) 3 ) n 2 12 ( 2 ζ(ζ + 1) 4 p n 1 (ζ) ζ (t n p n 1 (1/ζ) 0 1 ζ t 0t 1 t 2 (ζ 2 + 4ζ + 1) t 0 t 1 + t 0 t 2 + t 1 t 2 Note that only asymmetry comes from t 0. ) n p n 1 (ζ) ζ 2 2 p n 1 (1/ζ) 1 ζ 2 Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

57 Table of p n n p n (ζ) ζ ζ ζ 2 + 7ζ (35ζ ζ ζ ζ ζ ζ + 2) (63ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ + 2) Conjecture: The polynomials p n have positive coefficients. Conjecture: The polynomials p n are unimodal. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

58 Table of p n n p n (ζ) ζ ζ ζ 2 + 7ζ (35ζ ζ ζ ζ ζ ζ + 2) (63ζ ζ ζ ζ ζ ζ ζ ζ ζ ζ + 2) Conjecture: The polynomials p n have positive coefficients. Conjecture: The polynomials p n are unimodal. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

59 Plot of the 105 complex zeroes of p 14. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

60 Outline 1 Three-coloured chessboards 2 Combinatorial results 3 Symmetric polynomials 4 Painlevé VI 5 Future problems Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

61 Painlevé VI PVI is the nonlinear ODE (Painlevé, Fuchs, Gambier ) d 2 y dt = 1 ( y + 1 y ( 1 + y t t + 1 t y t y(y 1)(y t) 2t 2 (t 1) 2 ) ( dy dt ) dy dt ) 2 ( α + β t y + γ t 1 t(t 1) + δ 2 (y 1) 2 (y t) 2 Most general second order ODE such that all movable singularities are poles. Special functions of the 21st Century? ). Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

62 Painlevé VI PVI is the nonlinear ODE (Painlevé, Fuchs, Gambier ) d 2 y dt = 1 ( y + 1 y ( 1 + y t t + 1 t y t y(y 1)(y t) 2t 2 (t 1) 2 ) ( dy dt ) dy dt ) 2 ( α + β t y + γ t 1 t(t 1) + δ 2 (y 1) 2 (y t) 2 Most general second order ODE such that all movable singularities are poles. Special functions of the 21st Century? ). Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

63 Bäcklund transformations If y = y(t) solves PVI, then so does t/y, with parameters α β, γ 1 2 δ. Such Bäcklund transformations y F(t, y, y ) generate group (extended affine Weyl group) containing Z 4. Given a seed solution y = y 0000, acting with Z 4 gives new solutions y k1 k 2 k 3 k 4. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

64 Bäcklund transformations If y = y(t) solves PVI, then so does t/y, with parameters α β, γ 1 2 δ. Such Bäcklund transformations y F(t, y, y ) generate group (extended affine Weyl group) containing Z 4. Given a seed solution y = y 0000, acting with Z 4 gives new solutions y k1 k 2 k 3 k 4. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

65 Bäcklund transformations If y = y(t) solves PVI, then so does t/y, with parameters α β, γ 1 2 δ. Such Bäcklund transformations y F(t, y, y ) generate group (extended affine Weyl group) containing Z 4. Given a seed solution y = y 0000, acting with Z 4 gives new solutions y k1 k 2 k 3 k 4. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

66 Picard s seed solution When α = β = γ = 0, δ = 1/2, PVI can be solved in terms of elliptic functions (Picard, 1889). Picard s solutions include the algebraic solution which is parametrized by y 4 4ty 3 + 6ty 2 4ty + t 2 = 0, y = ζ(ζ + 2) ζ(ζ + 2)3, t = 2ζ + 1 (2ζ + 1). 3 If we choose y 0000 = y, what is y k1 k 2 k 3 k 4? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

67 Picard s seed solution When α = β = γ = 0, δ = 1/2, PVI can be solved in terms of elliptic functions (Picard, 1889). Picard s solutions include the algebraic solution which is parametrized by y 4 4ty 3 + 6ty 2 4ty + t 2 = 0, y = ζ(ζ + 2) ζ(ζ + 2)3, t = 2ζ + 1 (2ζ + 1). 3 If we choose y 0000 = y, what is y k1 k 2 k 3 k 4? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

68 Picard s seed solution When α = β = γ = 0, δ = 1/2, PVI can be solved in terms of elliptic functions (Picard, 1889). Picard s solutions include the algebraic solution which is parametrized by y 4 4ty 3 + 6ty 2 4ty + t 2 = 0, y = ζ(ζ + 2) ζ(ζ + 2)3, t = 2ζ + 1 (2ζ + 1). 3 If we choose y 0000 = y, what is y k1 k 2 k 3 k 4? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

69 Example y 1,2, 3,1 = ζ(ζ + 2)(ζ3 + 3ζ 2 + 3ζ + 5)(5ζ ζ 2 + 7ζ + 1) (2ζ + 1)(ζ 3 + 7ζ ζ + 5)(5ζ 3 + 3ζ 2 + 3ζ + 1) The non-trivial factors are called tau functions. Note that 5ζ ζ 2 + 7ζ + 1 = p 2 (ζ). Result: There is a 4-dim cone in Z 4 where tau functions are given by specializations of S n. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

70 Example y 1,2, 3,1 = ζ(ζ + 2)(ζ3 + 3ζ 2 + 3ζ + 5)(5ζ ζ 2 + 7ζ + 1) (2ζ + 1)(ζ 3 + 7ζ ζ + 5)(5ζ 3 + 3ζ 2 + 3ζ + 1) The non-trivial factors are called tau functions. Note that 5ζ ζ 2 + 7ζ + 1 = p 2 (ζ). Result: There is a 4-dim cone in Z 4 where tau functions are given by specializations of S n. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

71 Example y 1,2, 3,1 = ζ(ζ + 2)(ζ3 + 3ζ 2 + 3ζ + 5)(5ζ ζ 2 + 7ζ + 1) (2ζ + 1)(ζ 3 + 7ζ ζ + 5)(5ζ 3 + 3ζ 2 + 3ζ + 1) The non-trivial factors are called tau functions. Note that 5ζ ζ 2 + 7ζ + 1 = p 2 (ζ). Result: There is a 4-dim cone in Z 4 where tau functions are given by specializations of S n. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

72 Tau functions in a cone Let S (k 0,k 1,k 2,k 3 ) ( n (ζ) = S n 2ζ + 1 }{{} k 0, ζ, ζ + 2 }{{} k 1 Then, if k i are non-negative integers with k 0 + k 1 + k 2 + k 3 = 2n 1, ζ(2ζ + 1) ζ + 2 } {{ } k 2, 1 }{{} k 3 ). y k0 +k 2 n+1,k 1 +1,n k 0 k 1 1,n k 1 k 2 1 = elementary factor S(k 0+1,k 1,k 2,k 3 +1) n (ζ)s (k 0,k 1 +1,k 2 +1,k 3 ) n (ζ) S (k 0+1,k 1,k 2 +1,k 3 ) n (ζ)s (k 0,k 1 +1,k 2,k 3 +1) n (ζ), where as before y = y(t), t = ζ(ζ + 2)3 (2ζ + 1) 3. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

73 Tau functions in a cone Let S (k 0,k 1,k 2,k 3 ) ( n (ζ) = S n 2ζ + 1 }{{} k 0, ζ, ζ + 2 }{{} k 1 Then, if k i are non-negative integers with k 0 + k 1 + k 2 + k 3 = 2n 1, ζ(2ζ + 1) ζ + 2 } {{ } k 2, 1 }{{} k 3 ). y k0 +k 2 n+1,k 1 +1,n k 0 k 1 1,n k 1 k 2 1 = elementary factor S(k 0+1,k 1,k 2,k 3 +1) n (ζ)s (k 0,k 1 +1,k 2 +1,k 3 ) n (ζ) S (k 0+1,k 1,k 2 +1,k 3 ) n (ζ)s (k 0,k 1 +1,k 2,k 3 +1) n (ζ), where as before y = y(t), t = ζ(ζ + 2)3 (2ζ + 1) 3. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

74 Consequence: Recursions Tau functions satisfy bilinear recursions. For instance, it follows that p n+1 (ζ)p n 1 (ζ) = A n (ζ)p n (ζ) 2 + B n (ζ)p n (ζ)p n(ζ) with explicit coefficients. + C n (ζ)p n(ζ) 2 + D n (ζ)p n (ζ)p n(ζ). Conjectured by Bazhanov and Mangazeev Gives fast way of computing Z n. Possibly, it can be used to prove our conjectured expression for the free energy. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

75 Consequence: Recursions Tau functions satisfy bilinear recursions. For instance, it follows that p n+1 (ζ)p n 1 (ζ) = A n (ζ)p n (ζ) 2 + B n (ζ)p n (ζ)p n(ζ) with explicit coefficients. + C n (ζ)p n(ζ) 2 + D n (ζ)p n (ζ)p n(ζ). Conjectured by Bazhanov and Mangazeev Gives fast way of computing Z n. Possibly, it can be used to prove our conjectured expression for the free energy. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

76 Consequence: Recursions Tau functions satisfy bilinear recursions. For instance, it follows that p n+1 (ζ)p n 1 (ζ) = A n (ζ)p n (ζ) 2 + B n (ζ)p n (ζ)p n(ζ) with explicit coefficients. + C n (ζ)p n(ζ) 2 + D n (ζ)p n (ζ)p n(ζ). Conjectured by Bazhanov and Mangazeev Gives fast way of computing Z n. Possibly, it can be used to prove our conjectured expression for the free energy. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

77 Outline 1 Three-coloured chessboards 2 Combinatorial results 3 Symmetric polynomials 4 Painlevé VI 5 Future problems Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

78 Questions Symmetry classes of three-coloured chessboards. For the six-vertex model, various classical Lie algebra characters appear (Kuperberg, Okada,... ). For the three-colour model, we expect various affine Lie algebras. Macroscopic boundary effects. Arctic curves. New phenomenon: Some tau functions for Painlevé VI are specialized affine Lie algebra characters, given by Cauchy-type determinant formulas. Does this happen for other solutions to Painlevé equations? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

79 Questions Symmetry classes of three-coloured chessboards. For the six-vertex model, various classical Lie algebra characters appear (Kuperberg, Okada,... ). For the three-colour model, we expect various affine Lie algebras. Macroscopic boundary effects. Arctic curves. New phenomenon: Some tau functions for Painlevé VI are specialized affine Lie algebra characters, given by Cauchy-type determinant formulas. Does this happen for other solutions to Painlevé equations? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

80 Questions Symmetry classes of three-coloured chessboards. For the six-vertex model, various classical Lie algebra characters appear (Kuperberg, Okada,... ). For the three-colour model, we expect various affine Lie algebras. Macroscopic boundary effects. Arctic curves. New phenomenon: Some tau functions for Painlevé VI are specialized affine Lie algebra characters, given by Cauchy-type determinant formulas. Does this happen for other solutions to Painlevé equations? Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

81 Specializations of the symmetric polynomials S n recently appeared (implicitly) in three contexts: Partition function for three-colour model with DWBC (R. 2011). Ground state eigenvalue for the Q-operator of eight-vertex model (Bazhanov and Mangazeev 2005, 2006). Eigenvectors of XYZ Hamiltonian (Razumov and Stroganov 2010, Bazhanov and Mangazeev 2010, Zinn-Justin 2012). Why do the same functions appear in three contexts? Although the physical models are closely related, this is far from clear. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35

82 Specializations of the symmetric polynomials S n recently appeared (implicitly) in three contexts: Partition function for three-colour model with DWBC (R. 2011). Ground state eigenvalue for the Q-operator of eight-vertex model (Bazhanov and Mangazeev 2005, 2006). Eigenvectors of XYZ Hamiltonian (Razumov and Stroganov 2010, Bazhanov and Mangazeev 2010, Zinn-Justin 2012). Why do the same functions appear in three contexts? Although the physical models are closely related, this is far from clear. Hjalmar Rosengren (Chalmers University) Nagoya, July 31, / 35