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LOW-DIMENSIONAL UNITARY REPRESENTATIONS of B3 1. Introduction Unitary Braid Representations Have Been Constructed in Several

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LOW-DIMENSIONAL UNITARY REPRESENTATIONS of B3 1. Introduction Unitary Braid Representations Have Been Constructed in Several PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Number 9, Pages 2597{2606 S 0002-9939(01)05903-2 Article electronically published on March 15, 2001 LOW-DIMENSIONAL UNITARY REPRESENTATIONS OF B3 IMRE TUBA (Communicated by Stephen D. Smith) Abstract. We characterize all simple unitarizable representations of the braid group B3 on complex vector spaces of dimension d ≤ 5. In particular, we prove that if σ1 and σ2 denote the two generating twists of B3,thenasimple representation ρ : B3 ! GL(V )(fordimV ≤ 5) is unitarizable if and only if the eigenvalues λ1,λ2;::: ;λd of ρ(σ1) are distinct, satisfy jλij =1and (d) ≤ ≤ (d) µ1i > 0for2 i d,wheretheµ1i are functions of the eigenvalues, explicitly described in this paper. 1. Introduction Unitary braid representations have been constructed in several ways using the representation theory of Kac-Moody algebras and quantum groups (see e.g. [1], [2], and [4]), and specializations of the reduced Burau and Gassner representations in [5]. Such representations easily lead to representations of PSL(2; Z)=B3=Z ,where Z is the center of B3,andPSL(2; Z)=SL(2; Z)={1g,where{1g is the center of SL(2; Z). We give a complete classification of simple unitary representations of B3 of dimension d ≤ 5 in this paper. In particular, the unitarizability of a braid rep- resentation depends only on the the eigenvalues λ1,λ2;::: ,λd of the images of the two generating twists of B3. The condition for unitarizability is a set of linear in- equalities in the logarithms of these eigenvalues. In other words, the representation is unitarizable if and only if the (arg λ1; arg λ2;::: ;arg λd) is a point inside a poly- hedron in (R=2π)d, where we give the equations of the hyperplanes that bound this polyhedron. This classification shows that the approaches mentioned previously do not produce all possible unitary braid representations. We obtain representations that seem to be new for d ≥ 3. As any unitary representation of Bn restricts to a unitary representation of B3 in an obvious way, these results may also be useful in classifying such representation of Bn. I would like to thank my advisor, Hans Wenzl, for the ideas he contributed to this paper and Nolan Wallach for his suggestions. Let B3 be generated by σ1 and σ2 with the relation σ1σ2σ1 = σ2σ1σ2.Itis 3 well known that the center of B3 is generated by (σ1σ2) .LetK be any field. If 3 ρ is a simple representation of B3 on a K-vector space V ,thenρ(σ1σ2) must act on V as a scalar δ 2 K.Sinceσ1 and σ2 are conjugates via σ1σ2σ1, their images Received by the editors August 31, 1999 and, in revised form, January 31, 2000. 1991 Mathematics Subject Classification. Primary 20F36, 20C07, 81R10; Secondary 20H20, 16S34. c 2001 American Mathematical Society 2597 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2598 IMRE TUBA A = ρ(σ1)andB = ρ(σ2) have the same eigenvalues λ1,λ2;::: ,λd. We will need the following two results from [3]. Theorem 1.1. 1. Let K be an algebraically closed field, V a d-dimensional K- vector space, and λ1,λ2;::: ,λd 2 K −{0g,whered ≤ 5.Thereexistsa simple representation ρ : B3 ! GL(V ) such that the eigenvalues of A = ρ(σ1) (d) (d) satisfy Qrs =06 for all r =6 s where the polynomials Qrs are as follows: (2) − 2 − 2 Qrs = λr + λrλs λs; (3) 2 2 Qrs =(λr + λsλk)(λs + λrλk) with k =6 r; s. (4) − −1 2 2 Qrs = γ (λr + γ)(λs + γ)(γ + λrλk + λsλl)(γ + λrλl + λsλk) with γ2 = λ ···λ and k; l =6 r; s. 1 4 Y (5) −8 2 2 2 2 2 2 Qrs = γ (γ + λrγ + λr)(γ + λsγ + λs) (γ + λrλk)(γ + λsλk) k=6 r;s 5 with γ = λ1 ···λ5. 2. A simple representation of B3 of dimension d ≤ 5 is uniquely determined up to isomorphism by the eigenvalues of A = ρ(σ1) (for d ≤ 3)andδ,where 3 ρ(σ1σ2) = δ IdV (for d =4; 5). Explicit matrices for A = ρ(σ1)andB = ρ(σ2) are also listed in [3]. (d) The functions Qrs aredefinedin[3]by (d) (d) (d) (d) (d) Pr (B)Ps (A)Pr (B)=Qrs Pr (B); Q (d) − 2πiti where Pr (x)= i=6 r(x λi). Note that substituting λi = e and taking (d) logarithms reduces the problem of finding the zeroes of Qrs to solving a system of linear equations in the ti.(SeeExample4.2.) Proposition 1.2. Let ρ : B3 ! GL(V ) be a simple representation of dimension d ≤ 5. Then the minimal polynomials of A = ρ(σ1) and B = ρ(σ2) are the same as their characteristic polynomials. An immediate consequence of this is Corollary 1.3. If A (or B) is a diagonalizable matrix, then it has distinct eigen- values λ1,λ2;::: ,λd. Proof. Since AQis conjugate to some diagonal matrix D, its minimal polynomial is just p(x)= (x − dj)wherethedj are the distinct diagonal entries of D.By the previous proposition, deg p = d; hence D must have d distinct diagonal entries. Thus all of the diagonal entries of D are distinct. Since we are interested in unitarizable representations, we will let K = C and we will require that jλij =1.Letρ : B3 ! V be a simple d-dimensional represen- tation (d ≤ 5), and A = ρ(σ1), B = ρ(σ2). Any unitarizable complex matrix is diagonalizable, so we can assume that A and B are diagonalizable. So the eigen- values λ1,λ2;:::,λd are distinct by the last corollary. Let δ be the scalar via which 3 3 ρ(σ1σ2) acts on V ,thatis(AB) = δI.DenotetheC-algebra generated by A and B by B.Inotherwords,B = ρ(C B3), where C B3 is the group algebra. Note that B = End(V ) by simplicity. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use LOW-DIMENSIONAL UNITARY REPRESENTATIONS OF B3 2599 The proof proceeds by defining a vector space antihomomorphism { : B!B and proving that it is an algebra antihomorphism and an involution of B in section 2. In section 3, we define a sesquilinear form h:;:i on the ideal I = BeB;1 that is invariant under multiplication by A and B.Weprovethath:;:i is positive definite (d) ≤ ≤ if µ1i > 0for2 i d.Inthiscase,ρ is a unitary representation of B3 on the d-dimensional vector space I.Wealsoprovethatρ is a unitarizable representation (d) ≤ ≤ µ1i > 0for2 i d. In section 4, we give some examples of using the positivity (d) of µ1i . 2. An involution of the image of B3 Let eM;i be the eigenprojection of M to the eigenspace of λi,whereM 2fA; Bg. That is, Y − (d) M λj Pi (M) eM;i = = Q : λi − λj 6 (λi − λj ) j=6 i j=i Note that eA;i and eB;i always exist because the eigenvalues are distinct. Also (d) (d) eM;ieM;j = δij eM;i. Define µij by eB;ieA;jeB;i = µij eB;i.Notethat Q(d) (d) Q ijQ µij = − − : k=6 i(λi λk) k=6 j (λj λk) (d) Lemma 2.1. The µij are real numbers. 6 −1 −1 Proof. For i = j, the proof is by direct computation using λi = λi and γ = γ . For example, for d =5: Q (γ2 + λ γ + λ2)(γ2 + λ γ + λ2) (γ2 + λ λ )(γ2 + λ λ ) (d) i i Q j j Qk=6 i;j i k j k µij = 8 γ 6 (λi − λk) 6 (λj − λk) k=i k=j Q −1 −1 −1 −1 2 2 (γλ +1+γ λi)(γλ +1+γ λj ) 6 (γ + λiλk)(γ + λjλk) = i j k=Qi;j : − −1 − −1 6 − − (1 λjλi )(1 λiλj ) γ k=6 i;j (λi λk)(λj λk) The first of the two quotients is easily seen to be real. For the second quotient, Q ! Q 2 2 −2 −1 −1 −2 −1 −1 6 (γ + λiλk)(γ + λj λk) 6 (γ + λ λ )(γ + λ λ ) k=Qi;j = k=i;jQ i k j k : 6 − − −6 −1 − −1 −1 − −1 γ k=6 i;j (λi λk)(λj λk) γ k=6 i;j (λi λk )(λj λk ) Q 12 3 3 2 Multiply the numerator and the denominator by γ λi λj k=6 i;j λk to see that this is still Q 2 2 6 (γ + λiλk)(γ + λjλk) k=Qi;j : 6 − − γ k=6 i;j (λi λk)(λj λk) License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2600 IMRE TUBA P d For the case i = j,notethat k=1 eA;k = I,so eB;i = eB;iIeB;i Xd = eB;i eA;keB;i k=1 Xd = eB;ieA;keB;i k=1 Xd (d) = µik eB;i: k=1 P P d (d) (d) − (d) Hence k=1 µik =1,andµii =1 k=6 i µik is real. Proposition 2.2. S = feA;ieB;1eA;j j 1 ≤ i; j ≤ d; i =6 jg[feA;i j 1 ≤ i ≤ dg is a basis for the C-vector space B. Proof. Suppose Xd Xd Xd αij eA;ieB;1eA;j + αiieA;i =0: i=1 j=1 i=1 j=6 i Multiply by eA;i both on the left and on the right. The only term of the sum that survives is αiieA;i =0: Let vi be an eigenvector of A corresponding to λi.TheneA;ivi = vi =0,so6 eA;i =0.6 Hence αii =0.
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