A factorization of Temperley–Lieb diagrams Michael Hastings and Sarah Salmon Northern Arizona University Department of Mathematics and Statistics [email protected] [email protected] FAMUS November 22, 2013 M. Hastings & S. Salmon A factorization of TL-diagrams 1 / 23 

Admissible type A Temperley–Lieb diagrams An admissible diagram in type An must satisfy the following requirements: • The diagram starts with a box with n + 1 nodes along the north face and n + 1 nodes along the south face. • Every node must be connected to exactly one other node by a single edge. • The edges cannot cross. • The edges cannot leave the box. M. Hastings & S. Salmon A factorization of TL-diagrams 2 / 23 

Type A Temperley–Lieb diagrams Example Here is an example of an admissible 5-diagram. Here is an example of an admissible 6-diagram. Here is an example that is not an admissible diagram. M. Hastings & S. Salmon A factorization of TL-diagrams 3 / 23 

The type A Temperley–Lieb diagram algebra TL(An ) is the Z[δ]-algebra having (n + 1)-diagrams as a basis. When multiplying diagrams, it is possible to obtain a loop. In this case, we replace each loop with a coefficient δ. = δ M. Hastings & S. Salmon A factorization of TL-diagrams 4 / 23 

Type An simple diagrams We define n simple diagrams as follows: d1 = · · · 1 2 n n + 1 . . . di = · · · · · · 1 i i + 1 n + 1 . . . dn = · · · 1 2 n n + 1 M. Hastings & S. Salmon A factorization of TL-diagrams 5 / 23 

Important relations in type An Theorem TL(An ) satisfies the following: • d2 i = δdi ; • di dj = dj di when |i − j| > 1; • di dj di = di when |i − j| = 1. Theorem The set of simple diagrams generate all admissible diagrams in the Temperley–Lieb algebra of type An . M. Hastings & S. Salmon A factorization of TL-diagrams 6 / 23 

Proof of one relation in type An Proof We see that for |i − j| = 1 (here j = i + 1) di dj di = · · · · · · · · · · · · · · · · · · · · · · · · i i + 1 i + 2 = · · · · · · = di M. Hastings & S. Salmon A factorization of TL-diagrams 7 / 23 

Products of simple diagrams Example Consider the product d1 d3 d2 d4 d3 in type A4 . = M. Hastings & S. Salmon A factorization of TL-diagrams 8 / 23 

Historical context Comments • TL(An ) was discovered in 1971 by Temperley and Lieb as an algebra with abstract generators and a presentation with the relations above. • It first arose in the context of integrable Potts models in statistical mechanics. • As well as having applications in physics, TL(An ) appears in the framework of knot theory, braid groups, Coxeter groups and their corresponding Hecke algebras, and subfactors of von Neumann algebras. • Penrose/Kauffman used a diagram algebra to model TL(An ) in 1971. • In 1987, Vaughan Jones (awarded Fields Medal in 1990) recognized that TL(An ) is isomorphic to a particular quotient of the Hecke algebra of type An (the Coxeter group of type An is the symmetric group, Sn+1 ). M. Hastings & S. Salmon A factorization of TL-diagrams 9 / 23 

Factorization in type An We have discovered an algorithm to reconstruct the factorization given an admissible diagram. ←→ ←→ By our algorithm, the diagram equals d2 d4 d1 d3 d2 . M. Hastings & S. Salmon A factorization of TL-diagrams 10 / 23 

Factorization in type An Let’s verify our calculation. d2 d4 d1 d3 d2 = = M. Hastings & S. Salmon A factorization of TL-diagrams 11 / 23 

Admissible type B Temperley–Lieb diagrams An admissible diagram must satisfy the following requirements: • The diagram starts with a box with n + 1 nodes along the north face and n + 1 nodes along the south face; • Every node must be connected to exactly one other node by a single edge; • The edges cannot cross; • The edges cannot leave the box; • All decorations must be exposed to the west face; • There are a few technical restrictions on what decorations can occur where; • All loops (decorated or not) are replaced with a coefficient, δ; • Decorations are restricted by the relations below. = = = 2 M. Hastings & S. Salmon A factorization of TL-diagrams 12 / 23 

Type B Temperley–Lieb diagrams In type Bn , there are slightly different simple diagrams which generate the admissible diagrams. We define n simple diagrams as follows: d1 = · · · 1 2 n n + 1 . . . di = 1 n + 1 · · · · · · i i + 1 . . . dn = · · · n n + 1 1 2 M. Hastings & S. Salmon A factorization of TL-diagrams 13 / 23 

Type B Temperley–Lieb diagrams Theorem TL(Bn ) satisfies the following: • d2 i = δdi ; • di dj = dj di when |i − j| > 1; • di dj di = di when |i − j| = 1 and i, j = 1; • di dj di dj = 2di dj if {i, j} = {1, 2}. Theorem The set of simple diagrams generate all admissible diagrams in the Temperley–Lieb algebra of type Bn . M. Hastings & S. Salmon A factorization of TL-diagrams 14 / 23 

Proof of one relation in type Bn Proof For i = 1 and j = 2, d1 d2 d1 d2 = · · · · · · · · · · · · = 2 · · · = 2 · · · · · · = 2d1 d2 M. Hastings & S. Salmon A factorization of TL-diagrams 15 / 23 

Admissible diagrams in type Bn Example Here is an example of a product of several simple diagrams in type B4 . d1 d2 d4 d1 d3 d2 = = M. Hastings & S. Salmon A factorization of TL-diagrams 16 / 23 

Factorization of type Bn Example Let’s take the same diagram and work towards the factorization. ←→ ←→ Therefore the original diagram equals d1 d4 d2 d1 d3 d2 . M. Hastings & S. Salmon A factorization of TL-diagrams 17 / 23 

Factorization of type Bn That matches our previous calculation (up to commutation). d1 d4 d2 d1 d3 d2 = = M. Hastings & S. Salmon A factorization of TL-diagrams 18 / 23 

An exception Unlike type A, there is one exception to our algorithm. d2 d1 d2 = · · · · · · · · · = · · · M. Hastings & S. Salmon A factorization of TL-diagrams 19 / 23 

An exception Example There is one case where we must slightly adjust how we factor the diagram. ←→ ←→ This diagram equals d2 d5 d1 d4 d2 . M. Hastings & S. Salmon A factorization of TL-diagrams 20 / 23 

Factorization of type Bn Example Let’s try a larger diagram. ←→ ←→ Therefore, the original diagram equals d1 d4 d8 d10 d3 d5 d9 d2 d4 d6 d1 d3 d5 d7 d2 d4 d6 d8 d1 d3 d5 d2 d4 . M. Hastings & S. Salmon A factorization of TL-diagrams 21 / 23 

Factorization of type Bn Let’s check our calculation: d1 d4 d8 d10 d3 d5 d9 d2 d4 d6 d1 d3 d5 d7 d2 d4 d6 d8 d1 d3 d5 d2 d4 = = M. Hastings & S. Salmon A factorization of TL-diagrams 22 / 23 

Open Questions • There are other types to look at such as C and D. • Here is an example of C. • Will our algorithm work on these other types? • What type of special cases will we run into when checking if our algorithm works? M. Hastings & S. Salmon A factorization of TL-diagrams 23 / 23