Sum rule for a mixed boundary qKZ equation Caley Finn May 2015, - PowerPoint PPT Presentation

Sum rule for a mixed boundary qKZ equation Caley Finn May 2015, GGI, Firenze In collaboration with Jan de Gier

Outline 1 Mixed boundary q KZ equation 2 Sum rule 3 Bases of the Hecke algebra

Section 1 Mixed boundary q KZ equation

One boundary Temperley–Lieb algebra • Generators e 0 , . . . e N − 1 • Bulk relations e 2 i = − [2] e i , e i e i ± 1 e i = e i : = − [2] = = i i i i i +1 i − 1 i • Boundary relations e 2 0 = e 0 , e 1 e 0 e 1 = e 1 : = = 0 1 0 0 1 • With t -number [ u ] = t u − t − u t − t − 1 .

Action on Ballot paths • Ballot path: ( α 0 , . . . , α N ) , with α i ≥ 0 , α i +1 − α i = ± 1 , and α N = 0 . • Ballot paths of length N = 3 Ω = α 1 = α 2 = • Example e 2 | Ω � = = = = = | α 2 � • Will take general vector of the form � | Ψ( z 1 , . . . , z N ) � = ψ α ( z 1 , . . . , z N ) | α � α

Mixed boundary q KZ equation • Write q KZ equation in component form. • For 0 ≤ i ≤ N − 1 (bulk and left boundary) � � � � � � ψ α ( z 1 , . . . , z N ) e i | α � = T i ( − 1) ψ α ( z 1 , . . . , z N ) | α � , α α where T i ( u ) are generators of a Baxterized Hecke algebra (will be defined) • Reflection at the right boundary ψ α ( . . . , z N − 1 , z N ) = ψ α ( . . . , z N − 1 , t 3 z − 1 N ) • Relates Temperley–Lieb action on Ballot paths (LHS) to Hecke algebra action on coefficient functions (RHS).

Baxterized Hecke algebra • Bulk generators ( 1 ≤ i ≤ N − 1 ): T i ( u ) = ( tz i − t − 1 z i +1 ) 1 − π i − [ u − 1] , π i : z i ↔ z i +1 z i − z i +1 [ u ] • Boundary generator T 0 ( u ) = k ( z 1 , ζ 1 ) 1 − π 0 π 0 : z 1 ↔ z − 1 − B 0 ( u ) , 1 , z 1 − z − 1 1 and k ( z 1 , ζ 1 ) , B 0 ( u ) are simple functions. • These generators obey Yang–Baxter (bulk) and reflection equations (boundary).

Solution of the q KZ equation Theorem (de Gier, Pyatov, 2010) The solutions of the q KZ equation have a factorised form ր u i,j � ψ α ( z 1 , . . . , z N ) = T i ( u i,j ) ψ Ω i,j The product is constructed using a graphical representation of the Hecke generators u u T 0 ( u ) = , T i ( u ) = . i − 1 ii +1 0 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N )

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0)

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 1 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 • Label remaining tiles by rule u i,j = max { u i +1 ,j − 1 , u i − 1 ,j − 1 } + 1 1 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 • Label remaining tiles by rule u i,j = max { u i +1 ,j − 1 , u i − 1 ,j − 1 } + 1 2 1 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 • Label remaining tiles by rule u i,j = max { u i +1 ,j − 1 , u i − 1 ,j − 1 } + 1 3 3 2 1 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 • Label remaining tiles by rule u i,j = max { u i +1 ,j − 1 , u i − 1 ,j − 1 } + 1 4 3 3 2 1 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 • Label remaining tiles by rule u i,j = max { u i +1 ,j − 1 , u i − 1 ,j − 1 } + 1 5 4 3 3 2 1 1

Factorised solutions • Factorised solution for ψ α ( z 1 , . . . , z N ) • Fill to maximal Ballot path Ω = ( N, N − 1 , . . . , 0) • Label corners with 1 • Label remaining tiles by rule u i,j = max { u i +1 ,j − 1 , u i − 1 ,j − 1 } + 1 5 4 3 3 2 1 1 ψ α = T 0 (1) .T 1 (2) T 0 (3) .T 3 (1) T 2 (3) T 1 (4) T 0 (5) ψ Ω and ψ Ω = ∆ − t ( z 1 , . . . , z N )∆ + t ( z 1 , . . . , z N )

Section 2 Sum rule

Consecutive integer filling • Fill with consecutive integers along rows, e.g. for previous shape tilted by 45 ° ψ 4 , 2 , 1 ( u 1 + 1 , u 2 + 1 , u 3 + 1) u 1 + u 1 +3 u 1 +2 u 1 +1 4 = u 2 + u 2 +1 2 u 3 + 1 • In terms of Hecke generators ψ a 1 ,...,a n ( u 1 + 1 , . . . , u n + 1) = T a n ( u n + 1) . . . T a 1 ( u 1 + 1) ψ Ω where T a ( u + 1) = T a − 1 ( u + 1) . . . T 1 ( u + a − 1) T 0 ( u + a ) • T a ( u + 1) gives a row of length a , numbered from u + 1 .

Staircase diagram • Call the largest such element the staircase diagram : . . . u 1 +1 ψ ¯ a n ( u 1 +1 , . . . , u n +1) = , . . . a 1 ,..., ¯ . . . u n + 1 where n = ⌊ N/ 2 ⌋ , ¯ a i = N − 2 i + 1 . • In terms of Hecke generators ψ ¯ a n ( u 1 + 1 , . . . , u n + 1) a 1 ,..., ¯ = T N − 2 n +1 ( u n + 1) . . . T N − 3 ( u 2 + 1) T N − 1 ( u 1 + 1) ψ Ω .

Generalised sum rule Theorem The staircase diagram has the expansion � ψ ¯ a n ( u 1 + 1 , . . . , u n + 1) = c α ψ α ( z 1 , . . . , z N ) , a 1 ,..., ¯ α where the coefficients c α are non-zero and are monomials in [ u i ] y i = − [ u i + 1] , y i = − B 0 ( u i + 1) . ˜ • At specialization u i = 1 , t = e ± 2 π i / 3 , all coefficients c α = 1 . The sum gives the normalization of Temperley-Lieb loop model ground state vector. The sum has been computed at this point [Zinn-Justin 2007]. • Proof of the sum rule requires expanding staircase diagram in two stages.

First expansion The first stage of the expansion gives the form of the coefficients. Lemma (First expansion) ψ a 1 ,...,a n ( u 1 + 1 , . . . , u n + 1) = T a n ( u n + 1) . . . T a 1 ( u 1 + 1) ψ Ω � = ( T a i (1) + y i T a i − 1 (1) + ˜ y i ) ψ Ω . i = n,n − 1 ,..., 1 where [ u i ] y i = − [ u i + 1] , y i = − B 0 ( u i + 1) . ˜

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes.

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes. y 1 10 9 8 7 6 5 4 3 2 1

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes. y 1 10 9 8 7 6 5 4 3 2 1 1 9 8 7 6 5 4 3 2 1

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes. y 1 10 9 8 7 6 5 4 3 2 1 1 9 8 7 6 5 4 3 2 1 1 7 6 5 4 3 2 1

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes. y 1 10 9 8 7 6 5 4 3 2 1 1 9 8 7 6 5 4 3 2 1 1 7 6 5 4 3 2 1 y 4 ˜

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes. y 1 10 9 8 7 6 5 4 3 2 1 1 9 8 7 6 5 4 3 2 1 1 7 6 5 4 3 2 1 y 4 ˜ y 5 2 1

First expansion terms Procedure to expand ( T a n (1) + y n T a n − 1 (1) + ˜ y n ) . . . ( T a 1 (1) + y 1 T a 1 − 1 (1) + ˜ y 1 ) ψ Ω • Start from the empty outline. • Working from top down, a row may be left empty (factor ˜ y i ), filled one short (factor y i ), or filled completely (no additional factor). • Delete empty rows and boxes. y 1 10 9 8 7 6 5 4 3 2 1 1 9 8 7 6 5 4 3 2 1 1 7 6 5 4 3 2 1 y 4 ˜ y 5 2 1 1 1