Low Dimensional Representations of the Loop Braid Group LB 3

Low Dimensional Representations of the Loop Braid Group LB 3 Liang Chang Texas A&M University UT Dallas, June 1, 2015 Supported by AMS MRC Program Joint work with Paul Bruillard, Cesar Galindo, Seung-Moon Hong, Ian Marshall, Julia Plavnik, Eric Rowell and Michael Sun 1 / 18

Outline Loop braid group LB 3 Irreps of B 3 for dimension d 5 Irreps of LB 3 for dimension d 5 2 / 18

Braid Group B 3 The 3-strand braid group B 3 is a group generated by σ 1 and σ 2 subject to σ 1 σ 2 σ 1 = σ 2 σ 1 σ 2 3 / 18

Loop Braid Group LB 3 The loop braid group LB 3 is defined as the motion group of 3 unknotted and unlinked oriented circles in R 3. 4 / 18

Loop Braid Group LB 3 LB 3 is generated by σ 1, σ 2, s 1 and s 2 subject to Braid relation: σ 1 σ 2 σ 1 = σ 2 σ 1 σ 2 Permutation relation: s 1 s 2 s 1 = s 2 s 1 s 2, s 2 1 = s2 2 = 1 Mixed relation: s 1 s 2 σ 1 = σ 2 s 1 s 2, σ 1 σ 2 s 1 = s 2 σ 1 σ 2 5 / 18

Irreps of B 3 for dimension d 5 Tuba and Wenzl completely classified all irreducible B 3 representations of dimension d 5. For any dim 5 irrep of B 3, there exists a basis, with respect to which A = ρ(σ 1 ) and B = ρ(σ 2 ) act in ordered triangular form. A = λ 1 0.... 0 0 λ d, B = λ d 0 0.... 0 λ 1 where all nonzero entries are rational functions of λ 1,...,λ d and γ := (λ 1 λ d ) 1/d. 6 / 18

Irreps of B 3 for dimension d 5 [ ] [ ] λ d = 2: A = 1 λ 1 λ, B = 2 0 0 λ 2 λ 2 λ 1 λ 1 λ 1 λ 3 λ 1 2 + λ 2 λ 2 d = 3: A = 0 λ 2 λ 2, 0 0 λ 3 λ 3 0 0 B = λ 2 λ 2 0 λ 2 λ 1 λ 3 λ 1 2 λ 2 λ 1 7 / 18

Irreps of B 3 for dimension d 5 λ 1 (1 + D 1 + D 2 )λ 2 (1 + D 1 + D 2 )λ 3 λ 4 0 λ d = 4: A = 2 (1 + D 1 )λ 3 λ 4 0 0 λ 3 λ 4, 0 0 0 λ 4 λ 4 0 0 0 λ B = 3 λ 3 0 0 Dλ 2 (D + 1)λ 2 λ 2 0 D 3 λ 1 (D 3 + D 2 + D)λ 1 (D 2 + D + 1)λ 1 λ 1 where D = λ 2 λ 3 /λ 1 λ 4. 8 / 18

Irreps of B 3 for dimension d 5 λ 1 λ 5 0 0 0 0 0 λ 2 λ 4 0 0 0 d = 5: A = 0 0 λ 3, B = λ 3 0 0 0 0 0 λ 4 λ 2 0 0 0 0 0 λ 5 λ 1 where all nonzero entries are rational functions of λ 1,...,λ 5 and γ := (λ 1 λ 5 ) 1/5. 9 / 18

Goal: For dimension d {2,3,4,5}, classify the irreps of LB 3 over C extended from Tuba-Wenzl representation ρ TW. That is, given A and B in Tuba-Wenzl representation, find matrices S 1 and S 2 such that S 1 S 2 S 1 =S 2 S 1 S 2, S 2 1 = S2 2 = I d S 1 S 2 A =BS 1 S 2, ABS 1 = S 2 AB 10 / 18

Strategy: 1. Find S := S 1 S 2 satisfying S 3 = I d, SA = BS. 2. Solve for S 1 and S 2 from S. "Standard solution": S = cab for (σ 1 σ 2 ) 3 is in the center of B 3. Extending S to a rep of S 3 is not always possible. An matrix S over C with S 3 = I d extends to a rep of S 3 if and only if tr(s) R 11 / 18

Strategy: 1. Find S := S 1 S 2 satisfying S 3 = I d, SA = BS. 2. Solve for S 1 and S 2 from S. "Standard solution": S = cab for (σ 1 σ 2 ) 3 is in the center of B 3. Extending S to a rep of S 3 is not always possible. An matrix S over C with S 3 = I d extends to a rep of S 3 if and only if tr(s) R 11 / 18

Strategy: 1. Find S := S 1 S 2 satisfying S 3 = I d, SA = BS. 2. Solve for S 1 and S 2 from S. "Standard solution": S = cab for (σ 1 σ 2 ) 3 is in the center of B 3. Extending S to a rep of S 3 is not always possible. An matrix S over C with S 3 = I d extends to a rep of S 3 if and only if tr(s) R 11 / 18

Let ω be a third root of unity. Proposition For d {2,3,4,5}, S = ( 1) d 1 (λ 1 λ d ) 2 d AB satisfies S 3 = I d and has eigenvalues ω and ω 2 appearing in pair. S = U I n R... U 1, where R = R [ ] ω 0 0 ω 2. 12 / 18

Proposition The above S induces a representation of the permutation group, that is, there exist S 1 and S 2 s.t. S1 2 = S2 2 = I d and S = S 1 S 2. ε 1... ε n S k = U P φ U 1, 1 k... P φ m k [ ] where ε i {1, 1}, k {1,2} and Pk α = 0 φ 1 ω k φω k for any φ 0. 0 13 / 18

Theorem For each dimension d {2,3,4,5}, every irrep of B 3 extends to an irrep ρ (ε,φ) of LB 3 by the above S 1 and S 2. 14 / 18

Theorem For each dimension d {2,3,4,5}, every irrep of B 3 extends to an irrep ρ (ε,φ) of LB 3 by the above S 1 and S 2. Remark There exists 6-dimensional irrep of B 3 that can not be extended. 14 / 18

Theorem For each dimension d {2,3,4,5}, every irrep of B 3 extends to an irrep ρ (ε,φ) of LB 3 by the above S 1 and S 2. Remark There exists 6-dimensional irrep of B 3 that can not be extended. Conjecture For any dimension, an irreducible B 3 representation can be extended if and only if AB can be normalized to be an S such that S 3 = I and tr(s) R. 14 / 18

Question: Are there "non-standard" solutions for S? 15 / 18

Question: Are there "non-standard" solutions for S? Proposition Suppose matrices A and B corresponds to a Tuba-Wenzl irrep. The equation SA = BS has the general solution for a 0,...,a d 1 C. S = d 1 a n B n AB n=0 15 / 18

λ λ 1 λ 3 1 λ λ 2 2 λ 3 0 0 Example: A = 0 λ 2 λ 2, B = λ 2 λ 2 0 0 0 λ 3 λ 2 λ 1λ 3 λ λ 2 2 λ 1 0 0 S = a 0 AB + a 1 BAB + a 2 B 2 AB = 0 S 2 and (BSA) 2 are skew upper triangular matrices. 6 homogeneous linear eqns of a 0 a 1, a 0 a 2, a 2 1, a 1a 2 and a 2 2 with coefficients in C[λ 1,λ 2,λ 3 ]. a 1 = a 2 = 0 if (λ 1,λ 2,λ 3 ) is not a zero of a set of polynomials in C[λ 1,λ 2,λ 3 ]. 16 / 18

Theorem For each dimension d {2,3,4,5}, there is J d C[λ 1,...,λ d,δ] such that if (λ 1,...,λ d ) / V (J d ), S = ( 1) d 1 (λ 1 λ d ) 2 d AB is the only solution for the loop braid relations. Therefore, for such generic (λ 1,...,λ d ), ρ (ε,φ) are all irreps of LB 3 extended from the given Tuba-Wenzl representation. 17 / 18

Theorem For each dimension d {2,3,4,5}, there is J d C[λ 1,...,λ d,δ] such that if (λ 1,...,λ d ) / V (J d ), S = ( 1) d 1 (λ 1 λ d ) 2 d AB is the only solution for the loop braid relations. Therefore, for such generic (λ 1,...,λ d ), ρ (ε,φ) are all irreps of LB 3 extended from the given Tuba-Wenzl representation. J 2 = /0 J 3 = (λ 1 + λ 2 )(λ 2 + λ 3 )(λ 3 + λ 1 ) If (λ 1,λ 2,λ 3 ) J 3, S = a 0 (t)ab + a 2 (t)b 2 AB. ( J 4 = δ 12 Π 4 n=1 Σ i n (λ i + 1 ) ) λ i J 5 det(m 14 14 (λ 1,...,λ 5,δ)) 17 / 18

Thank You! 18 / 18