A factorization of Temperley–Lieb diagrams, Part 1 Michael Hastings Northern Arizona University Department of Mathematics and Statistics [email protected] SUnMaRC March, 2014 Joint work with Sarah Salmon M. Hastings A factorization of TL-diagrams, Part 1 1 / 14 

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 A factorization of TL-diagrams, Part 1 2 / 14 

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 A factorization of TL-diagrams, Part 1 3 / 14 

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 A factorization of TL-diagrams, Part 1 4 / 14 

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 A factorization of TL-diagrams, Part 1 5 / 14 

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 A factorization of TL-diagrams, Part 1 6 / 14 

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 A factorization of TL-diagrams, Part 1 7 / 14 

Products of simple diagrams Example Consider the product d1 d3 d2 d4 d3 in type A4 . = M. Hastings A factorization of TL-diagrams, Part 1 8 / 14 

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 A factorization of TL-diagrams, Part 1 9 / 14 

Factorization in type An We have discovered an algorithm to reconstruct the factorization given an admissible diagram. ←→ 1 1 2 2 3 ←→ By our algorithm, the diagram equals d2 d4 d1 d3 d2 . M. Hastings A factorization of TL-diagrams, Part 1 10 / 14 

Factorization in type An Let’s verify our calculation. d2 d4 d1 d3 d2 = 1 2 3 = M. Hastings A factorization of TL-diagrams, Part 1 11 / 14 

Factorization in type An An example of our algorithm on a more difficult diagram. 1 1 1 2 2 2 2 2 3 3 3 4 4 5 5 6 6 7 By our algorithm, the diagram equals d2 d6 d10 d1 d3 d5 d9 d11 d4 d8 d10 d7 d9 d6 d8 d5 d7 d6 . M. Hastings A factorization of TL-diagrams, Part 1 12 / 14 

Factorization in type An Let’s verify our calculation: d2 d6 d10 d1 d3 d5 d9 d11 d4 d8 d10 d7 d9 d6 d8 d5 d7 d6 = 1 2 3 4 5 6 7 = M. Hastings A factorization of TL-diagrams, Part 1 13 / 14 

A taste of type Bn Example ←→ M. Hastings A factorization of TL-diagrams, Part 1 14 / 14