Towards combinatorics of elliptic lattice models Hjalmar Rosengren - PowerPoint PPT Presentation

Towards combinatorics of elliptic lattice models Hjalmar Rosengren Chalmers University of Technology and University of Gothenburg Firenze, 22 May 2015 Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 1 / 47

A missing big picture Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 2 / 47

Outline Introduction 1 Eigenvectors (Mangazeev & Bazhanov 2010, Razumov & 2 Stroganov 2010, Zinn-Justin 2013) Eigenvalues of Q -operator (Bazhanov & Mangazeev 2005, 3 2006) Three-coloured chessboards (R. 2011) 4 Towards a synthesis (R., to appear) 5 Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 3 / 47

Outline Introduction 1 Eigenvectors (Mangazeev & Bazhanov 2010, Razumov & 2 Stroganov 2010, Zinn-Justin 2013) Eigenvalues of Q -operator (Bazhanov & Mangazeev 2005, 3 2006) Three-coloured chessboards (R. 2011) 4 Towards a synthesis (R., to appear) 5 Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 4 / 47

Main point All three contexts eigenvectors eigenvalues of Q -operator domain wall partition functions lead to polynomials that have positive coefficients are Painlevé tau functions We know what these polynomials "are", but conceptual explanations are still lacking. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 5 / 47

Main point All three contexts eigenvectors eigenvalues of Q -operator domain wall partition functions lead to polynomials that have positive coefficients are Painlevé tau functions We know what these polynomials "are", but conceptual explanations are still lacking. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 5 / 47

Solvable lattice models - XXZ 6 V � Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 6 / 47

Solvable lattice models - XYZ 8 V � . . . . . . . . . . . . . . . . . . . . . . . . . . . . ? ? - XXZ 6 V � Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 6 / 47

Solvable lattice models elliptic SOS � - 8 V � XYZ - . . . . . . . . . . . . . . . . . . . . . . . . . . . . ? ? - XXZ 6 V � Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 6 / 47

Solvable lattice models elliptic SOS ............................... � � - three-colour 8 V � XYZ - . . . . . . . . . . . . . . . . . . . . . . . . . . . . ? ? - XXZ 6 V � Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 6 / 47

Solvable lattice models elliptic SOS ............................... . � . . . . . . . . . . . . � - ? three-colour trig SOS 8 V � XYZ - . . � . . . . . . . . . . . . . . . . . . . . . . . . . . - ? ? 6 V XXZ � - Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 6 / 47

Solvable lattice models elliptic SOS ............................... . � . . . . . . . . . . . . � - ? three-colour trig SOS 8 V � XYZ - . . . . . . . . . . . . . . . . . . . . . . . . . . . . ? ? - XXZ 6 V � elliptic models Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 6 / 47

"Combinatorial" parameter values ∆ = 1 / 2: ASM enumeration, three-colourings etc. ∆ = � 1 / 2: supersymmetry Magic in spectra, Razumov–Stroganov etc. ∆ = 0: free fermions Domino tilings, arctic circle etc. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 7 / 47

"Combinatorial" parameter values ∆ = 1 / 2: ASM enumeration, three-colourings etc. ∆ = � 1 / 2: supersymmetry Magic in spectra, Razumov–Stroganov etc. ∆ = 0: free fermions Domino tilings, arctic circle etc. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 7 / 47

"Combinatorial" parameter values ∆ = 1 / 2: ASM enumeration, three-colourings etc. ∆ = � 1 / 2: supersymmetry Magic in spectra, Razumov–Stroganov etc. ∆ = 0: free fermions Domino tilings, arctic circle etc. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 7 / 47

Outline Introduction 1 Eigenvectors (Mangazeev & Bazhanov 2010, Razumov & 2 Stroganov 2010, Zinn-Justin 2013) Eigenvalues of Q -operator (Bazhanov & Mangazeev 2005, 3 2006) Three-coloured chessboards (R. 2011) 4 Towards a synthesis (R., to appear) 5 Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 8 / 47

XYZ spin chain Hamiltonian acting on ( C 2 ) ⌦ N , X N � � H = � 1 J x σ j x σ j + 1 + J y σ j y σ j + 1 + J z σ j z σ j + 1 ; x y z 2 j = 1 ✓ 0 ◆ ✓ 0 ◆ ✓ 1 ◆ 1 � i 0 σ x = , σ y = , σ z = , 1 0 i 0 0 � 1 Periodic boundary conditions: σ N + 1 = σ 1 . If N is odd and J x J y + J x J z + J y J z = 0 ( ∆ = � 1 / 2) then H has lowest eigenvalue � N 2 ( J x + J y + J z ) . Observed by Stroganov (2001), proved by Hagendorf (2013). Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 9 / 47

XYZ spin chain Hamiltonian acting on ( C 2 ) ⌦ N , X N � � H = � 1 J x σ j x σ j + 1 + J y σ j y σ j + 1 + J z σ j z σ j + 1 ; x y z 2 j = 1 ✓ 0 ◆ ✓ 0 ◆ ✓ 1 ◆ 1 � i 0 σ x = , σ y = , σ z = , 1 0 i 0 0 � 1 Periodic boundary conditions: σ N + 1 = σ 1 . If N is odd and J x J y + J x J z + J y J z = 0 ( ∆ = � 1 / 2) then H has lowest eigenvalue � N 2 ( J x + J y + J z ) . Observed by Stroganov (2001), proved by Hagendorf (2013). Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 9 / 47

Ground state eigenvectors Consider cyclically symmetric eigenvector Ψ in sector e ± ⌦ · · · ⌦ e ± with even number of plus signs. Unique up to normalization. Razumov & Stroganov observed that if J z = ζ 2 � 1 J x = 1 + ζ , J y = 1 � ζ , , 2 then X Ψ = Ψ k 1 ··· k N e k 1 ⌦ · · · ⌦ e k N , k 1 ··· k N 2 {±} where Ψ k 1 ··· k N seem to be polynomials in ζ with positive integer coefficients. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 10 / 47

Ground state eigenvectors Consider cyclically symmetric eigenvector Ψ in sector e ± ⌦ · · · ⌦ e ± with even number of plus signs. Unique up to normalization. Razumov & Stroganov observed that if J z = ζ 2 � 1 J x = 1 + ζ , J y = 1 � ζ , , 2 then X Ψ = Ψ k 1 ··· k N e k 1 ⌦ · · · ⌦ e k N , k 1 ··· k N 2 {±} where Ψ k 1 ··· k N seem to be polynomials in ζ with positive integer coefficients. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 10 / 47

Example: N=7 Ψ � + � + � ++ = 7 + ζ 2 , Ψ �� + � +++ = 3 + 5 ζ 2 , Ψ ��� ++++ = 1 + 5 ζ 2 + 2 ζ 4 , Ψ �� ++ � ++ = 4 + 3 ζ 2 + ζ 4 . All other components are equal to one of these four, up to multiplication by ζ or ζ 2 . Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 11 / 47

Conjectures There are polynomials s n , ¯ s n , given by explicit recursions, such that Ψ �� ··· � = ζ n ( n + 1 ) / 2 s n ( ζ � 2 ) , Ψ + ··· + � = N � 1 ζ n ( n � 1 ) / 2 ¯ s n ( ζ � 2 ) , where N = 2 n + 1. Sum rule X Ψ 2 k 1 ··· k N = ( 4 / 3 ) n ζ n ( n + 1 ) s n ( ζ � 2 ) s � n � 1 ( ζ � 2 ) , k where s n is naturally extended to n < 0. Proved by Zinn-Justin, up to certain conjecture. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 12 / 47

Conjectures There are polynomials s n , ¯ s n , given by explicit recursions, such that Ψ �� ··· � = ζ n ( n + 1 ) / 2 s n ( ζ � 2 ) , Ψ + ··· + � = N � 1 ζ n ( n � 1 ) / 2 ¯ s n ( ζ � 2 ) , where N = 2 n + 1. Sum rule X Ψ 2 k 1 ··· k N = ( 4 / 3 ) n ζ n ( n + 1 ) s n ( ζ � 2 ) s � n � 1 ( ζ � 2 ) , k where s n is naturally extended to n < 0. Proved by Zinn-Justin, up to certain conjecture. Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 12 / 47

More conjectures There are polynomials q n , r n , given by explicit recursions, such that for n even ( N = 2 n + 1) ✓ 1 � ζ ◆ n ( n − 2 ) 2 ( ζ � 1 ) . Ψ � + � + � ··· + � = Const n ( ζ ( 3 + ζ )) r n − 2 q n − 2 4 3 + ζ 2 and for n odd ✓ 1 � ζ ◆ n 2 − 1 4 r n − 1 2 ( ζ � 1 ) . Ψ + � + � ··· + � + = Const n ( ζ ( 3 + ζ )) q n − 3 3 + ζ 2 Factorizations s 2 n + 1 ( y 2 ) = Const n r n ( y ) r n ( � y ) , ✓ y � 1 ◆ s 2 n ( y 2 ) = Const n ( 1 + 3 y ) n ( n + 1 ) r � n � 1 q n � 1 ( y ) . 3 y + 1 Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 13 / 47

More conjectures There are polynomials q n , r n , given by explicit recursions, such that for n even ( N = 2 n + 1) ✓ 1 � ζ ◆ n ( n − 2 ) 2 ( ζ � 1 ) . Ψ � + � + � ··· + � = Const n ( ζ ( 3 + ζ )) r n − 2 q n − 2 4 3 + ζ 2 and for n odd ✓ 1 � ζ ◆ n 2 − 1 4 r n − 1 2 ( ζ � 1 ) . Ψ + � + � ··· + � + = Const n ( ζ ( 3 + ζ )) q n − 3 3 + ζ 2 Factorizations s 2 n + 1 ( y 2 ) = Const n r n ( y ) r n ( � y ) , ✓ y � 1 ◆ s 2 n ( y 2 ) = Const n ( 1 + 3 y ) n ( n + 1 ) r � n � 1 q n � 1 ( y ) . 3 y + 1 Hjalmar Rosengren (Chalmers University) Firenze, 22 May 2015 13 / 47