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Representations of the Schrödinger-Virasoro algebras1 Junbo Li∗,†), Yucai Su‡) ∗)Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China †)Department of Mathematics, Changshu Institute of Technology, Changshu 215500, China ‡)Department of Mathematics, University of Science and Technology of China, Hefei 230026, China E-mail: sd [email protected], [email protected] Abstract. Inthispaperitisprovedthatanirreducibleweightmodulewithfinite-dimensional 8 0 weight spaces over the Schrödinger-Virasoro algebras is a highest/lowest weight module or 0 2 a uniformly bounded module. Furthermore, indecomposable modules of the intermediate n series over these algebras are completely determined. a J Key words: Schrödinger-Virasoro algebras, modules of the intermediate series, Harish- 5 Chandra modules. 1 MR(2000) Subject Classification: 17B10, 17B65, 17B68. ] A R 1. Introduction . h Let s = 0 or 1 . The Schrödinger-Virasoro algebra L[s] introduced in [1, 2, 3], in the t 2 a context of non-equilibrium statistical physics as a by-product of the computation of n- m point functions that are covariant under the action of the Schrödinger group, is the infinite- [ dimensionalLiealgebrawithC-basis{L , Y , M , c|m, n ∈ Z, p ∈ s+Z}andLiebrackets, 1 m p n v 9 m3 −m 20 [Lm, Lm′] = (m′ −m)Lm′+m +δm,−m′ c, (1.1) 12 2 m . 1 [L , Y ] = (p− )Y , (1.2) m p p+m 0 2 8 0 [Lm, Mn] = nMn+m, (1.3) : v i [Yp, Yp′] = (p′ −p)Mp′+p, (1.4) X ar [Yp, Mn] = [Mn, Mn′] = [L, c] = 0. (1.5) The Lie algebra L[s] contains as subalgebras both the Lie algebra S of invariance of the free Schrödingerequationandthecentralcharge-freeVirasoroalgebraVir,whereS istheinfinite dimensional Lie algebra, called the Schrödinger algebra, with the C-basis {Y , M |n ∈ p n Z, p ∈ s + Z}, and Vir is the Virasoro algebra with the C-basis {L , c|n ∈ Z}. The Lie n algebra L[1] is called the original Schrödinger-Virasoro algebra, while L[0] is called the 2 twisted Schro¨dinger-Virasoro algebra. It is well known that one attempt of introducing two-dimensional conformal field theory is to understand the universal behavior of two-dimensional statistical systems at equilibrium and at the critical temperature. A systematic investigation of the theory of representations 1 Supported by NSF grants 10471091, 10671027 of China, “One Hundred Talents Program” from Uni- versity of Science and Technology of China. Corresponding E-mail: sd [email protected] 1 of the Virasoro algebra in the 80’s led to introduce the unitary minimal models, corresponding to the unitary highest weight representations of the Virasoro algebra with central charge less than one. Miraculously, covariance alone is enough to allow the computation of the n-point functions for these highly constrained physical models. Since both original and twisted Schrödinger-Virasoro Lie algebras are closely related to the Schrödinger Lie algebra and the Virasoro Lie algebra, it is highly expected that they should consequently play an important role akin to that of the Virasoro Lie algebra in two-dimensional equilibrium statistical physics (see, e.g., [1, 2, 3, 4, 6, 10]). Partially due to the above-stated reasons, the Schrödinger-Virasoro Lie algebras have re- cently drawn some attentions in the literature. In particular, the sets of generators provided by the cohomology classes of the cocycles for bothoriginalandtwisted Schrödinger-Virasoro Lie algebras were presented in [6], and the derivation algebra and the automorphism group of the twisted sector were determined in [4]. Furthermore, vertex algebra representations of the Schrödinger-Virasoro Lie algebras out of a charged symplectic boson and a free boson with its associated vertex operators were constructed in [10]. Our motivation in studying the Schrödinger-Virasoro Lie algebras is to have a better understanding of their representations. Let us formulate our main results below. Denote H = span{L ,M ,c}, (1.6) 0 0 which is a maximal torus of L[s] (note that in case of s = 0, since the adjoint operator ad Y0 is not semi-simple, H⊕CY is not a torus but a Cartan subalgebra of L[0]). Denote by H∗ 0 the dual space of H. Definition 1.1 A module V over L[s] is called a (i) Harish-Chandra module if V admits a finite-dimensional weight space decomposition V = ⊕λ∈H∗Vλ, where Vλ = {v ∈ V |xv = λ(x)v, x ∈ H} such that dimVλ < ∞ for all λ ∈ H∗ (in case Vλ 6= 0, we call λ a weight of V); (ii) uniformly bounded module if it is a Harish-Chandra module such that there exists some N > 0 with dimVλ 6 N for all λ ∈ H∗; (iii) module of the intermediate series if V is an indecomposable Harish-Chandra module such that dimVλ 6 1 for all λ ∈ H∗. Remark 1.2 (i) Since the Schrödinger subalgebra S is an ideal of L, if S acts trivially on a module V, then V is simply a module over Vir. Thus in the following, we always suppose that S acts nontrivially on V. (ii) It is well known (see, e.g., [7, 8]) that a module V of the intermediate series over Vir is a quotient of one of the modules A , A(α), B(α) for some a,b,α ∈ C, they all have a,b 2 the basis {x |k ∈ Z} such that c acts trivially, and for n,k ∈ Z, k A : L x = (a+k +bn)x , (1.7) a,b n k k+n A(α) : L x = (k +n)x (k 6= 0), L x = n(n+α)x , (1.8) n k k+n n 0 n B(α) : L x = kx (k 6= −n), L x = −n(n+α)x . (1.9) n k k+n n −n 0 The main results of this paper are presented in the following theorem. Theorem 1.3 (i) An irreducibleHarish-Chandramodule overL[s]is eithera highest/lowest weight module or a uniformly bounded one. (ii) A module of the intermediate series over L[0] is simply a module of the intermediate series over Vir. (iii) A module V of the intermediate series over L[1] such that S acts nontrivially on V is 2 one of the modules A , B , C , D , A (α), A (α), B (α), B (α), C(α,α′), D(β,β′), a,b a,b a a 1 2 1 2 or one of their quotients for a, b, α, α′, β, β′ ∈ C, whose module structures are given as follows (the central element c acts as zero), where k ∈ 1Z, i,n ∈ Z, j,p ∈ 1 +Z, 2 2 A : M x = 0, (1.10) a,b n k Y x = (a+i+2bp)x , Y x = 0, (1.11) p i i+p p j 1 L x = (a+i+bn)x , L x = a+j +(b+ )n x , (1.12) n i i+n n j j+n 2 B : M x = 0, (cid:0) (cid:1) (1.13) a,b n k Y x = 0, Y x = x , (1.14) p i p j j+p 1 L x = (a+i+bn)x , L x = a+j +(b+ )n x , (1.15) n i i+n n j j+n 2 C : M x = 0, (cid:0) (cid:1) (1.16) a n k Y x = (a+i)(a+i+2p)x , Y x = 0, (1.17) p i i+p p j 3 L x = (a+i)x , L x = (a+j + n)x , (1.18) n i i+n n j j+n 2 D : M x = 0, (1.19) a n k Y x = (a+i+p)(a+i−p)x , Y x = 0, (1.20) p i i+n p j 1 L x = (a+i− n)x , L x = (a+j +n)x , (1.21) n i i+n n j j+n 2 A (α) : M x = 0, (1.22) 1 n k 1 Ypxi = (−2 +i+p)xi+p, Ypxj = 0, Lnx21 = n(n+α)x12+n, (1.23) 1 1 1 1 L x = (− +i+ n)x , L x = (− +j +n)x (j 6= ), (1.24) n i i+n n j j+n 2 2 2 2 3 A (α) : M x = 0, (1.25) 2 n k Y x = ix , Y x = 0, L x = −n(n+α)x , (1.26) p i i+p p j n −n 0 1 L x = ix (i 6= −n), L x = (j + n)x , (1.27) n i i+n n j j+n 2 B (α) : M x = 0, (1.28) 1 n k Y x = 0, Y x = x , L x = n(n+α)x , (1.29) p i p j j+p n 0 n 3 L x = (i+n)x (i 6= 0), L x = (j + n)x , (1.30) n i i+n n j j+n 2 B (α) : M x = 0, (1.31) 2 n k Ypxi = 0, Ypxj = xj+p, Lnx1−n = −n(n+α)x1, (1.32) 2 2 1 1 1 1 L x = (− +i− n)x , L x = (− +j)x (j 6= −n), (1.33) n i i+n n j j+n 2 2 2 2 C(α,α′) : M x = −2nα′x , M x = 0 (k 6= −n), (1.34) n −n 0 n k Y x = i(i+2p)x , Y x = α′x , Y x = 0 (j 6= −p), (1.35) p i i+p p −p 0 p j 3 L x = −n(n+α)x , L x = ix (i6=−n), L x = (j+ n)x , (1.36) n −n 0 n i i+n n j j+n 2 1 1 D(β,β′) :Mnx21 =2nβ′xn+21, Mnxk=0(k6=2), Ypx12 =β′xp+21, Ypxj=0(j6=2), (1.37) 1 1 Ypxi = (−2 +i+p)(−2 +i−p)xi+n, Lnx21 = n(n+β)xn+21, (1.38) 1 1 1 1 L x = (− +i− n)x , L x = (− +j +n)x (j 6= ). (1.39) n i i+n n j j+n 2 2 2 2 Throughout the paper, we denote the set of all nonzero integers by Z∗ and that of all positive integers by N. 2. Proof of Theorem 1.3(i) TheproofofTheorem1.3willbedividedbyseverallemmas. LetV beanindecomposable module over L[s]. Since M and c are central elements, whose actions on V must be 0 constants. Thus from (1.6), we can simply regard the weight space as the eigenspace of L , 0 i.e., V = ⊕λ∈CVλ, where Vλ = {v ∈ V |L0v = λv} for λ ∈ C. Lemma 2.1 Fix an a ∈ C such that Va 6= 0. We have V = ⊕n∈1ZVn, where Vn = Va+n = 2 {v ∈ V |L v = (a+n)v} for n ∈ 1Z. 0 2 Proof. For any a ∈ C, denote V(a) = ⊕n∈1ZVa+n. From relations (1.1)–(1.5), one can 2 easily see that V(a) is an L[s]-submodule of V such that V = ⊕a∈C/1ZV(a) is a direct sum 2 of different V(a). Hence V = V(a) for some a ∈ C. (cid:3) 4 Lemma 2.2 Suppose V = ⊕a∈C/1ZV(a) is an irreducible Harish-ChandraL[s]-module with- 2 out highest and lowest weights. Then for any i ∈ Z∗, k ∈ 1Z, 2 L | ⊕L | ⊕Y | ⊕Y | : V → V ⊕V ⊕V ⊕V (2.1) i Vk i+1 Vk i+s Vk i+1+s Vk k k+i k+i+1 k+i+s k+i+1+s is injective. In particular, by taking i = −k (if k ∈ Z) or i = 1−k (if k ∈ 1+Z), we obtain 2 2 that dim V is uniformly bounded. k Proof. Suppose there exists some v ∈ V such that 0 k L v = L v = Y v = Y v = 0. (2.2) i 0 i+1 0 i+s 0 i+1+s 0 Without loss of generality, we can suppose i > 0. Note that when ℓ ≫ 0, we have ℓ = ai+b(i+1), ℓ+s = a′i+b′(i+s), ℓ = a′′(i+s)+b′′(i+1+s), for some a,b,a′,b′,a′′,b′′ ∈ N, from this and (1.1), (1.2) and (1.4), one can easily deduce that L ,Y ,M can be generated by L ,L ,Y ,Y . Therefore there exists some N > 0 ℓ s+ℓ ℓ i i+1 i+s i+1+s such that L v = Y v = M v = 0 for all ℓ ≥ N. ℓ 0 s+ℓ 0 ℓ 0 The rest of the proof is exactly similar to that of [9, Proposition 2.1]. (cid:3) Now Theorem 1.3(i) follows from Lemma 2.1. 3. Proofs of Theorem 1.3(ii) and (iii) From now on, we shall suppose V is a module of the intermediate series over L[s]. As in [7, 8], one sees that c acts trivially on V, so we can omit c in (1.1). First we prove Theorem 1.3(iii), i.e., for the original Schrödinger-Virasoro Lie algebra L[1]. 2 We shall first suppose that dimVk = 1 for all k ∈ 21Z and both V′ = ⊕k∈ZVk and V′′ = ⊕k∈1+ZVk are Vir-modules of the form Aa,b, a,b ∈ C (later on, we shall determine all 2 possible deformations). Therefore, we can choose a basis {x |k ∈ 1Z} of V such that k 2 Y x = f x , (3.1) p k p,k k+p M x = g x , (3.2) n k n,k k+n (a+k +bn)x if k ∈ Z, k+n L x = (3.3) n k ((a+k +b′n)xk+n if k ∈ 21 +Z, for some a, b, b′, f , g ∈ C, where k ∈ 1Z, p ∈ 1 +Z, n ∈ Z. p,k n,k 2 2 Lemma 3.1 b′ = b± 1 or (b,b′) ∈ (0, 3), (3,0), (1,−1), (−1,1) . 2 2 2 2 2 (cid:8) (cid:9) 5 Proof. If f = 0 for all p ∈ 1 + Z, k ∈ 1Z, then from (1.4), we also obtain g = 0 for p,k 2 2 n,k all n ∈ Z, k ∈ 1Z, and so S acts trivially on V. Thus there exists some p ∈ 1 + Z and 2 0 2 k ∈ 1Z such that f 6= 0. Replacing a by a + 1 if necessary (which exchanges V′ and 0 2 p0,k0 2 V′′), we can always suppose f 6= 0 for some p ∈ 1 +Z, k ∈ Z. Then from (1.2), we see p0,k0 0 2 0 that for every p ∈ 1 +Z, 2 f 6= 0 for infinitely many k ∈ Z. (3.4) p,k For any m, n ∈ Z, p ∈ 1 +Z, k ∈ Z, applying 2 m+n n m (p− )[L ,[L ,Y ]] = (p− )(n+p− )[L ,Y ] (3.5) m n p m+n p 2 2 2 to x , using (3.1), (3.3) and comparing the coefficients of x , we obtain k k+p+m+n m+n (p− ) (a+k +p+b′n)(a+k +p+n+b′m)f p,k 2 −(a(cid:16)+k +bn)(a+k +p+n+b′m)f p,k+n −(a+k +bm)(a+k +p+m+b′n)f p,k+m +(a+k +bm)(a+k +m+bn)f p,k+m+n n m (cid:17) = (p− )(n+p− ) (a+k +p+b′m+b′n)f −(a+k +bm+bn)f . (3.6) p,k p,k+m+n 2 2 (cid:16) (cid:17) In the above equation, replacing m, n, k by (i) m, m, k−m, (ii) −m, −m, k+m and (iii) m, −m, k, respectively, we obtain the following three equations: m m (p− )(p+ ) a+k +(2b−1)m 2 2 (cid:16) (cid:0) (cid:1) +(p−m) a+k +(b−1)m (a+k +bm) f p,k+m (cid:17) −2(p−m(cid:0)) a+k +(b−1)m(cid:1) (a+k +p+b′m)f p,k + (p−m)(cid:0)a+k +p+(b′ −(cid:1)1)m (a+k +p+b′m) (cid:16) m (cid:0) m (cid:1) −(p− )(p+ ) a+k +p+(2b′ −1)m f = 0, (3.7) p,k−m 2 2 (cid:17) (cid:0) (cid:1) (p+m) a+k +p−(b′ −1)m (a+k +p−b′m) (cid:16) (cid:0) m m (cid:1) −(p+ )(p− ) a+k +p−(2b′ −1)m f p,k+m 2 2 (cid:17) −2(p+m) a+k −(cid:0)(b−1)m (a+k +p−(cid:1)b′m)f p,k m m (cid:0) (cid:1) + (p+ )(p− ) a+k −(2b−1)m 2 2 (cid:16) (cid:0) (cid:1) +(p+m) a+k −(b−1)m (a+k −bm) f = 0, (3.8) p,k−m (cid:17) (cid:0) (cid:1) 6 (a+k +bm) a+k +p−(b′ −1)m f p,k+m − (a(cid:0)+k +p−b′m) a+k(cid:1)+p+(b′ −1)m (cid:16) m 3m (cid:0) (cid:1) +(p+ )( −p)+(a+k +bm) a+k −(b−1)m f p,k 2 2 (cid:17) +(a+k −bm) a+k +p+(b′ −1)m(cid:0) f = 0. (cid:1) (3.9) p,k−m Regard (3.7)–(3.9) as a system of li(cid:0)near equations on the(cid:1)unknown variables f , f p,k+m p,k and f . Denote by ∆(p,k,m) the coefficient determinant. By (3.4), we see that for any p,k−m pairs (p,m) ∈ (1 +Z)×Z, there exist infinitely many integers k such that ∆(p,k,m) = 0. 2 Since ∆(p,k,m) is a polynomial on p, k, m, we obtain that 1 ∆(p,k,m) = 0 for all p ∈ +Z, k, m ∈ Z. (3.10) 2 It is a little lengthy but straightforward to compute (one can simply use Mathematica to solve a system of linear equations without problem) m6 ∆(p,k,m) = ∆ (∆ (a+k)p+∆ m2 +∆ p2), 0 1 2 3 64 where ∆ = (2b−2b′ −1)(2b−2b′ +1), 0 ∆ = 18(2b+2b′ −3)(2b+2b′ −1), 1 ∆ = 4(b+b′ −1)(−3b−4b2 +4b3 −9b′ +4bb′ +4b2b′ +12b′2 −4bb′2 −4b′3), 2 ∆ = 27−156b+152b2 −16b3 −16b4 −180b′ +328bb′ −80b2b′ 3 −32b3b′ +240b′2 +240b′2 −176bb′2 −112b′3 +32bb′3 +16b′4. It is easy to see that (3.10) holds if and only if ∆ = 0 or ∆ = ∆ = ∆ = 0, 0 1 2 3 if and only if 1 3 3 1 1 b′ = b± or (b, b′) ∈ (0, ), ( , 0), (1, − ) or (− , 1) . (3.11) 2 2 2 2 2 (cid:8) (cid:9) (Note that from ∆ = 0, one obtains that b′ = 3 −b or 1 −b, then from ∆ = ∆ = 0 one 1 2 2 2 3 can solve b.) Thus this lemma follows. (cid:3) By replacing a by a + 1 (then (3.4) does not necessarily hold) if necessary (which ex- 2 changes V′ and V′′), we can always suppose Re(b′) ≥ Re(b) (where Re(b) is the real part of the complex number b). Thus we only need to consider the cases 1 3 1 b′ = b+ or (b,b′) ∈ (0, ), (− ,1) . (3.12) 2 2 2 (cid:8) (cid:9) We shall now consider all the cases given in (3.12) one by one. 7 Case 1 b′ = b+ 1. 2 We need to determine f , g defined in (3.1) and (3.2), where n ∈ Z, p ∈ 1 +Z and p,k n,k 2 k ∈ 1Z. From the relation [Y ,Y ]x = (p−m)M x for m,p ∈ 1 +Z, k ∈ 1Z, we only 2 m p k m+p k 2 2 need to determine f . p,k Lemma 3.2 For any m, k ∈ Z, p ∈ 1 +Z, 2 (a+k +2bp)f = (a+k +m+2bp)f . (3.13) p,k+m p,k Proof. Using (3.7) and (3.8) to cancel f , one can obtain (or simply using Mathematica) p,k−m that (3.13) holds under the following condition ∆ (k,m) = (a+k)(1−4b)+(1+6b)(b−1)p m2 1 (cid:16) (cid:17) +6(a+k)2p+8(a+k)(1−b)p2 +4(1−b)p3 6= 0. (3.14) Next we want to prove that (3.13) holds under the condition ∆ (k) = (a+k)(1−4b)+(1+6b)(b−1)p 6= 0. (3.15) 2 Suppose (3.15) holds. Then ∆ (k + m′,m − m′) is a polynomial on m′ of degree 3 (if 1 b 6= 1) or of degree 2 (if b = 1, in this case the coefficient of m′2 in ∆ (k +m′,m−m′) is 4 4 1 (1+6b)(b−1)p+6p = (6− 5)p 6= 0 since p ∈ 1 +Z), and ∆ (k,m′) is a polynomial on m′ 8 2 1 of degree 2. Thus we can find some m′ such that (a+k +m′ +2bp)∆ (k +m′,m−m′)∆ (k,m′) 6= 0, 1 1 in particular, condition (3.14) holds for the two triples (k+m′,m−m′,p), (k,m′,p). Thus (3.13) holds for these two triples, and one has a+k +2bp (a+k +2bp)fp,k+m = a+k +m′ +2bp(a+(k +m′)+2bp)fp,(k+m′)+(m−m′) a+k +2bp = a+k +m′ +2bp(a+(k +m′)+(m−m′)+2bp)fp,k+m′ a+k +m+2bp = a+k +m′ +2bp(a+k +2bp)fp,k+m′ = (a+k +m+2bp)f . (3.16) p,k Thus (3.13) holds under condition (3.15). Now for any p,k,m, choose m′ such that (a+k +m′ +2bp)∆ (k +m′) 6= 0. (3.17) 2 This follows by noting that if b 6= 1 then ∆ (k+m′) is a polynomial on m′ of degree 1, and 4 2 if b = 1 then ∆ (k + m′) = (1 + 6b)(b − 1)p = −15p 6= 0 (since p ∈ 1 + Z). Now (3.17) 4 2 8 2 8 implies that we have all equalities of (3.16) except the last equality. But the last equality also follows by writing fp,k as fp,(k+m′)+(−m′) and using (3.13) with the pair (k,m) being replaced by (k +m′,−m′) (condition (3.15) holds for this pair). (cid:3) For any m ∈ Z, p ∈ 1+Z, k ∈ 1Z, applying [L ,Y ] = (p−m)Y to x and comparing 2 2 m p 2 p+m k the coefficients of x , one has k+m+p (a+k +bm)f −(a+k +p+b′m)f = m−2pf ifk ∈ Z, p,m+k p,k 2 m+p,k (3.18) ((a+k +b′m)f −(a+k +p+bm)f = m−2pf ifk ∈ 1 +Z. p,m+k p,k 2 m+p,k 2 Taking m = p−n in (3.18) and using b′ = b+ 1, one has 2 p−n p−3n (a+k +bp−bn)f −(a+k +n+bp−bn+ )f = f . (3.19) n,p−n+k n,k p,k 2 2 Using (3.13), one has (a+k +2bn)f = (a+k +p−n+2bn)f . (3.20) n,k+p−n n,k Then using (3.19) together with (3.13) and (3.20), we obtain (p−3n)(a+k +2bp) (a+k +2bn)f −(a+k +m+2bp)f = 0. (3.21) p,k+m n,k (cid:0) (cid:1) Letting m = j −k in (3.21), one has (p−3n)(a+k +2bp) (a+k +2bn)f −(a+j +2bp)f = 0. (3.22) p,j n,k (cid:0) (cid:1) Lemma 3.3 For any k, j ∈ Z, n, p ∈ 1 +Z, one has 2 (a+k +2bn)f = (a+j +2bp)f . (3.23) p,j n,k Furthermore, for any j ∈ Z, p ∈ 1 +Z, f can be written as 2 p,j f = (a+j +2bp)f , (3.24) p,j 0 for some f ∈ C. 0 Proof. Equation (3.24) follows from (3.23) by fixing n ,k such that a+k +bn 6= 0 and 0 0 0 0 letting f = fn0,k0 . It remains to prove (3.23). If (p−3n)(a+k +2bp) 6= 0, then (3.23) 0 a+k0+bn0 follows from (3.22). Next suppose (p−3n)(a+k+2bp) = 0. Choose n ∈ 1 +Z and k ∈ Z 0 2 0 such that (p−3n )(a+k +2bp) 6= 0 and (n−3n )(a+k +2bn) 6= 0. 0 0 0 0 Then by (3.22), (a + k + 2bn )f = (a + j + 2bp)f and also (a + k + 2bn)f = 0 0 p,j n0,k0 n0,k0 (a+k +2bn )f . Thus (3.23) also holds. This proves the lemma. (cid:3) 0 0 n,k As for the case k ∈ 1 +Z, we have the following lemma. 2 9 Lemma 3.4 For any m,k,p ∈ 1 +Z, 2 f = d . (3.25) p,k 0 Proof. First suppose b 6= 0. We want to prove 1 f21,k = d0 for k ∈ 2 +Z. (3.26) Applying [L1,Y21] = 0 to xk for k ∈ 12 +Z and using b′ = b+ 21, we obtain 1 1 (a+k +b+ )f1 −f1 (a+k +b+ ) = 0. (3.27) 2 2,k 2,k+1 2 If a+b ∈/ Z, then a+k+b+ 1 6= 0 for all k ∈ 1 +Z, thus (3.27) implies f1 is a constant, 2 2 2,k denoted by d . Thus (3.26) holds in this case. Assume that a + b ∈ Z. If necessary, by 0 shifting the index k of x by an integer (which does not change V′, V′′), we can suppose k a+b = 0. Then (3.27) shows f1,1 if k > 0, f1 = 2 2 (3.28) ,k 2 ( f1,−1 if k < 0. 2 2 Applying [L−1,Y12] = Y−21 to xk, we obtain f−12,k = (a+k+21−b)f12,k−f21,k−1(a+k−(b+21)), which together with (3.28) gives (using a = −b) f1 1 if k > 1, 2,2 2 f−12,k =  (1−2b)f21,12 +2bf21,−12 if k = 12, (3.29)  f21,−21 if k ≤ −12.    Applying [L1,Y−1] = −Y1 to x1, using (3.28) and (3.29), we obtain 2 2 2 1 1 1 1 (1−2b)f21,12 +2bf12,−21 = −f12,12 = (a+ 2 − 2 +b)f−12,12 −f−21,23(a+ 2 +b+ 2), which implies f1,1 = f1,−1 since a = −b 6= 0. This together with (3.28) gives (3.26). 2 2 2 2 Now for any 2 6= p ∈ 1 +Z, by replacing (n,p) by (p− 1, 1) in (1.2) and applying it to 3 2 2 2 x , one can easily obtain f = d . If p = 2, choosing some n ∈ Z, p ∈ 1 + Z such that k p,k 0 3 2 n+ p = 23 and p − n2 6= 0, we can again obtain f32,k = d0. This proves (3.25) for the case b 6= 0. Now suppose b = 0. Similar to the proof of (3.13), for any m,n ∈ Z, k,p ∈ 1 +Z, one 2 has f = f , (3.30) p,k+m p,k 10