SIMPLICITY OF KAC MODULES FOR THE QUANTUM GENERAL LINEAR SUPERALGEBRA
RANDALL R. HOLMES1 CHAOWEN ZHANG
Abstract. A general necessary and sufficient condition is obtained for a Kac module of the quantum general linear superalgebra to be simple. More explicit conditions are then obtained by considering separately the case where the quantum parameter is not a root of unity and the case where it is a root of unity.
0. Introduction
For fixed positive integers m and n the quantized enveloping superalgebra U = Uq(gl(m, n)) of the general linear Lie superalgebra gl(m, n) is a deformation of the universal enveloping superalgebra of gl(m, n) depending on a parameter q in the defining field F. The superalgebra U, which is defined by R. B. Zhang in [6], is the quantum general linear superalgebra (the term “supergroup” in that paper is changed to “superalgebra” in later work). The superalgebra U has a subalgebra B+ having a quotient U0 isomorphic to the quantized enveloping algebra of the Lie algebra gl(m)⊕gl(n). We can regard a given finite-dimensional 0 + simple U -module L as a B -module and then form the induced U-module K(L) = U⊗B+ L, called a Kac module. Kac modules are of interest because every finite-dimensional simple U-module is a quotient of K(L) for some L. The aim of this paper is to provide necessary and sufficient conditions for K(L) itself to be simple. The study of the simplicity of K(L) was initiated by R. B. Zhang in [6]. Our setting is a bit more general than the setting in that paper. For instance, in [6] the field F is taken to be the field of complex numbers, while our assumptions allow for other fields as well. Also, see Remark 4.4 below. We recover some results in [6] using different methods and providing detailed proofs for some claims that are made in that paper without proofs or with only sketched proofs.
2010 Mathematics Subject Classification. 17B37, 17B50. Key words and phrases. Quantum supergroup, quantum enveloping superalgebra, general linear, repre- sentation, Kac module, simple module. 1Corresponding author: Randall R. Holmes, [email protected], 334.844.6571. The authors thank the referee for a careful reading of the paper and for several useful suggestions. 1 2
The paper is organized as follows. In Section1 the superalgebra U is defined in terms of generators and relations. Section2 gathers together several formulas involving elements of U that will be used throughout the paper. In Section3 some structural features of U are obtained, including a PBW-type basis. In Section4 the Kac module is defined and a general criterion is given for its simplicity. Section5 provides a more explicit simplicity criterion. The remaining sections specialize to the two cases where q is not a root of unity (Section6) and where q is a root of unity (Section7). In both cases, finite-dimensional simple U0-modules are constructed and then explicit criteria are given for simplicity of the corresponding Kac modules.
1. The superalgebra U
Fix positive integers m and n. Put
I0 = {(i, j)|1 ≤ i < j ≤ m or m < i < j ≤ m + n},
I1 = {(i, j)|1 ≤ i ≤ m < j ≤ m + n}, and I = I0 ∪ I1. Let F be a commutative ring with identity and let q be a unit in F such that q − q−1 is also a −1 2 unit. Assume that there exists a ring automorphism Ω of F such that Ω(q) = q and Ω = 1F. (For instance, we could have F = C(q), q an indeterminate, and Ω : x 7→ x (x ∈ C), q 7→ q−1. Or we could have F = C, take q 6= ±1 to be a root of unity, and take Ω to be complex conjugation.) From Section4 on we assume that F is a field, the more general situation being required only for the proof of Theorem 3.1. −1 Denote by X the set of symbols Ei,i+1,Fi,i+1,Kk,Kk with i ∈ [1, m + n) and k ∈ [1, m + n]. (Here and below we use interval notation to denote the set of integers in the indicated range.) Let Uˆ be the free F-algebra on the set X . This algebra has the structure of F-superalgebra (i.e., Z2-graded F-algebra) uniquely determined by assigning degree 1 (odd parity) to the elements Em,m+1 and Fm,m+1 and degree 0 (even parity) to the remaining elements of X . (See [3, Section 1.2] for basic notions regarding superalgebras and Lie superalgebras.)
ηk For k ∈ [1, m + n], put qk = q , where ηk is 1 or −1 according as k ≤ m or k > m.
For a = (i, j) ∈ I, define the elements Ea = Eij and Fa = Fij of Uˆ recursively by
−1 Ea = EikEkj − qk EkjEik, Fa = FkjFik − qkFikFkj, for i < k < j. It is straightforward to check that these definitions are independent of the choice of k and that the elements Ea and Fa have degree 0 if a ∈ I0 and degree 1 if a ∈ I1. SIMPLICITY OF KAC MODULES 3
The supercommutator of homogeneous elements x and y of a superalgebra is given by [x, y] := xy − (−1)x¯y¯yx, wherex ¯ denotes the degree of x.
Denote by U = UF,q the F-algebra with generators X and the relations R given below. (Thus, U is the quotient of Uˆ by the ideal generated by all x − y, with x = y a relation in R.)
−1 (R1) KkKl = KlKk,KkKk = 1, −1 (δkj −δk,j+1) (R2) KkEj,j+1Kk = qk Ej,j+1, −1 −(δkj −δk,j+1) KkFj,j+1Kk = qk Fj,j+1, −1 −1 KiKi+1 − Ki Ki+1 (R3)[ Ei,i+1,Fj,j+1] = δij −1 , qi − qi 2 2 (R4) Em,m+1 = 0,Fm,m+1 = 0, (R5) Ei,i+1Ej,j+1 = Ej,j+1Ei,i+1,Fi,i+1Fj,j+1 = Fj,j+1Fi,i+1 (|i − j| > 1), 2 −1 2 (R6) Ei,i+1Ej,j+1 − (q + q )Ei,i+1Ej,j+1Ei,i+1 + Ej,j+1Ei,i+1 = 0 (|i − j| = 1, i 6= m), 2 −1 2 (R7) Fi,i+1Fj,j+1 − (q + q )Fi,i+1Fj,j+1Fi,i+1 + Fj,j+1Fi,i+1 = 0 (|i − j| = 1, i 6= m), (R8)[ Em−1,m+2,Em,m+1] = 0, [Fm−1,m+2,Fm,m+1] = 0, with i, j ∈ [1, m+n) and k, l ∈ [1, m+n], where δij is the Kronecker delta. (The last relation is omitted if either m = 1 or n = 1.) We use the same letters to denote the images of elements of the generating set X in U under the canonical map Uˆ → U. The superalgebra structure on Uˆ induces a superalgebra structure on U.
The superalgebra U is the quantum general linear superalgebra, denoted Uq(gl(m, n)) in [5],[6]. It is a quantum deformation of the universal enveloping superalgebra of the Lie superalgebra gl(m, n). A bijective Z-linear map f : U → U is an antiautomorphism of U if f(xy) = f(y)f(x) for all x, y ∈ U.
Lemma 1.1 (cf. [6, Appendix A]). The automorphism Ω of F extends to an antiautomor- phism Ω of U with −1 Ω: Ei,i+1 7→ Fi,i+1,Fi,i+1 7→ Ei,i+1,Kj 7→ Kj for all i ∈ [1, m + n), j ∈ [1, m + n]. We have Ω(Ea) = Fa for every a ∈ I.
Proof. Denote by U0 the opposite ring of U regarded as an F-module via cx := Ω(c)x (c ∈ F, x ∈ U). There is a uniquely determined F-algebra homomorphism Uˆ → U0 that maps the 4
−1 elements of X as indicated in the statement and with Kj 7→ Kj. The ideal generated by all x − y with x = y in the relation R is contained in the kernel of this homomorphism, so we get an induced F-algebra homomorphism Ω : U → U0. Regarding Ω as a function 2 Ω: U → U we have Ω = 1U, so Ω is bijective and it is an antiautomorphism extending the given automorphism of F. The final claim follows immediately from the definitions of Ea and Fa.
In [6, Appendix A], R. B. Zhang defines certain automorphisms Ti, i ∈ [1, m + n) \ m, of U. He calls these generalized Lusztig automorphisms since they generalize the automorphisms defined by Lusztig in [4, Theorem 3.1]. We will require only the following properties of these automorphisms. (In the formulas, empty products of automorphisms are regarded as the identity automorphism of U.)
Lemma 1.2 ([6, Appendix A]). Let a = (i, j) ∈ I with j 6= i + 1. If a ∈ I0 let i < k < j be arbitrary; if a ∈ I1, put k = m. We have
j−i+1 −1 −1 −1 (a) Ea = (−1) TiTi+1 ··· Tk−1Tj−1Tj−2 ··· Tk+1(Ek,k+1), j−i+1 −1 −1 −1 (b) Fa = (−1) TiTi+1 ··· Tk−1Tj−1Tj−2 ··· Tk+1(Fk,k+1).
2. Some formulas
We collect here several formulas involving the elements of U that we will require. −1 For a = (i, j) ∈ I, put Ka = KiKj .
Lemma 2.1 (cf. [6, Lemma 2]). For every a = (i, j) ∈ I, k ∈ [1, m + n], and b ∈ I1,
−1 δki−δkj −1 −(δki−δkj ) (a) KkEaKk = qk Ea and KkFaKk = qk Fa, −1 Ka − Ka (b)[ Ea,Fa] = −1 , qi − qi 2 2 (c) Eb = 0 and Fb = 0.
Proof. These identities follow from the defining relations (R2), (R3), and (R4), respectively, using the generalized Lusztig automorphisms and Lemma 1.2. The formulas in the following two lemmas appear in [6, Lemma 2] as entries of a more comprehensive list of formulas. We include here only the formulas that will be required in this paper. Note that additional formulas are easily obtained by applying the antiautomorphism Ω (Lemma 1.1) and using the fact that for x, y ∈ U we have Ω([x, y]) = [Ω(y), Ω(x)]. Lemma 2.2 ([6, Lemma 2]). For every i, j, k ∈ [1, m + n] with i < j < k,
(a)[ Eij,Fjk] = 0 and [Fij,Ejk] = 0, SIMPLICITY OF KAC MODULES 5
F¯ij −1 (b) FikFij = (−1) qi FijFik, ¯ Fjk (c) FikFjk = (−1) qkFjkFik, −1 (d)[ Fik,Eij] = qjFjkKiKj , −1 (e)[ Fij,Eik] = EjkKi Kj, −1 (f)[ Ejk,Fik] = FijKj Kk, −1 −1 (g)[ Eik,Fjk] = qj EijKjKk .
We provide proofs of parts (b) and (c) in the following lemma since our formulas differ slightly from those in [6]. We provide a proof of part (d) as well since the formula conflicts with a formula in [6] (see Remark 2.4 below). Lemma 2.3 (cf. [6, Lemma 2]). For every i, j, k, l ∈ [1, m + n] with i < j < k < l, (a) the products
[Eil,Ejk], [Fil,Fjk], [Eil,Fjk], [Fil,Ejk], [Eij,Ekl], [Fij,Fkl], [Eij,Fkl], [Fij,Ekl] are all zero, −1 −1 (b)[ Eik,Fjl] = (qk − qk)KjKk FklEij, −1 −1 (c)[ Ejl,Fik] = (qk − qk )FijEklKkKj , −1 (d)[ Fjl,Fik] = (qk − qk)FjkFil. ¯ Proof. (b) The operator [Eik, −] is a derivation of U of degree Eik [3, Sections 1.1.2,1.1.4], ¯ ¯ and EikFkl = 0 since (k, l) ∈ I0 if (i, k) ∈ I1, so
[Eik,Fjl] = [Eik,FklFjk − qkFjkFkl] ¯ ¯ EikFkl = (−1) Fkl[Eik,Fjk] − qk[Eik,Fjk]Fkl (Lemma 2.2(a)) −1 −1 −1 −1 = qj FklEijKjKk − qkqj EijKjKk Fkl (Lemma 2.2(g)) −1 −1 = (qk − qk)KjKk FklEij. (Lemma 2.1(a))
(c) This formula follows from part (a) by applying the antiautomorphism Ω. ¯ ¯ ¯ ¯ ¯ (d) By considering cases, it is seen that σ := (−1)FjlFik = (−1)Fjk = (−1)FjkFil . We have
[Fjl,Fik] = FjlFik − σFikFjl
= (FklFjk − qkFjkFkl)Fik − σFik(FklFjk − qkFjkFkl) (defn. of Fjl)
= FklFjkFik − σFikFklFjk − qk(FjkFklFik − σFikFjkFkl) −1 = σqk FklFikFjk − σFikFklFjk − qk(FjkFklFik − qkFjkFikFkl) (Lemma 2.2(c)) −1 = σqk FilFjk − qkFjkFil (defn. of Fil) −1 = (qk − qk)FjkFil. (part (a)) 6
Remark 2.4. In the first line of (4) in [6, Lemma 2] it is asserted that [Fab,Fcd] = −(qb − −1 qb )FcbFad in the situation a < c < b < d, F = C. This appears not to be the case. For example, if m, n = 2 and (a, c, b, d) = (1, 2, 3, 4), Lemma 2.3(d) gives −1 −1 [Fab,Fcd] = [F13,F24] = [F24,F13] = (q − q )F23F14 = (q − q)F14F23 −1 −1 6= −(q − q)F14F23 = −(qb − qb )FcbFad, where the fourth equality uses Lemma 2.3(a) and the inequality uses the fact that q − q−1 is a unit by assumption as well as the fact that F14F23 6= 0 by Theorem 3.1(a) below. For (i, j), (s, t) ∈ I put (i, j) < (s, t) if any of the following holds:
• (i, j) ∈ I0 and (s, t) ∈ I1, • (i, j), (s, t) ∈ I0 and i < s or [i = s and j < t], • (i, j), (s, t) ∈ I1 and j > t or [j = t and i < s]. For a, b ∈ I, we write a ≤ b to mean a < b or a = b and note that ≤ is a total order on the set I. Q Let I ⊆ I. Define FI = a∈I Fa with the factors in the order given by ≤ and where an empty product is defined to be 1. Define EI = Ω(FI ) (= product of the Ea in the reverse order).
For a ∈ I1, we write F≤a to mean FI , where I = {b ∈ I1, b ≤ a}, and similarly for F
For j ∈ (m, m + n], put Cj = {(i, j) | i ∈ [1, m]}. For x, y ∈ U, we write x ∼ y to mean x = uy for some unit u in F.
Lemma 2.5. Let a = (i, j) ∈ I1 and J ⊆ Cj. We have
(a) FaFJ ∼ FJ Fa, (b) If a ∈ J, then FaFJ = 0 and FJ Fa = 0.
Proof. (a) We proceed by induction on |J|. The claim is immediate if |J| = 0, so assume otherwise. Then J has a minimal element b = (k, j) with respect to the order ≤. Putting J 0 = J \ b we have, using Lemma 2.2(c),
FaFJ = FaFbFJ0 ∼ FbFaFJ0 ∼ FbFJ0 Fa = FJ Fa, where the third step uses the induction hypothesis.
(b) Assume that a ∈ J. Putting Ja, we have 2 FaFJ = FaFJa ∼ FJa = 0, SIMPLICITY OF KAC MODULES 7
where we have used part (a) and then Lemma 2.1(c). By part (a), we have FJ Fa = 0 as well.
Lemma 2.6. For every a ∈ I1 and j ∈ (m, m + n] we have FaFCj ∼ FCj Fa. In particular,
FCi FCj ∼ FCj FCi for every i, j ∈ (m, m + n].
Proof. We need only prove the first statement since then the second statement will follow Qi−1 immediately. Let a = (s, t) ∈ I1 and j ∈ (m, m + n]. For i ∈ [1, m + 1], put Pi = k=i+1 Fkj. It suffices to show that
P
Now FsjFit = −FitFsj by Lemma 2.3(a), and FsjP>i = 0 by Lemma 2.5(b), so the second term on the right is zero. Therefore, Equation (1) holds. The remaining case, s < i and t < j, is handled similarly.
Lemma 2.7. For every a, b ∈ I1 with b > a the products FbF>a, F>aFb, FaF Proof. Let a, b ∈ I1 with b > a. Write a = (i, j) and b = (s, t) and put J = {(k, j) | i < k ≤ Q m}. Using Lemma 2.6 we have F>a ∼ l FbFCt = 0 by Lemma 2.5(b), so if t < j the claim follows using Lemma 2.6. Otherwise, since b > a, we have t = j and s > i, so b ∈ J giving FbFJ = 0 by Lemma 2.5(b), and the claim follows once again using Lemma 2.6. Similar arguments handle the remaining claims. Lemma 2.8. For every a ∈ I1, we have F>aFa ∼ F≥a and FaF Q Proof. Let a = (i, j) ∈ I1. Using Lemma 2.6 we have F>a ∼ FJ l Lemma 2.9. For every a ∈ I0 and b ∈ I1, we have 8 (a) EaF≥b = F≥bEa, (b) FaF≤b ∼ F≤bFa. Proof. Let a ∈ I0. We may assume that a = (i, i + 1) for some i ∈ [1, m + n) \ m. (a) We proceed by reverse induction on b ∈ I1. For each b = (s, t) ∈ I1, using Lemmas 2.2(d,f) and 2.1(a) for the first two cases and Lemmas 2.2(a) and 2.3(a) for the third case, we have −1 KiKi+1Fi+1,t, s = i, −1 [Ea,Fb] = [Ei,i+1,Fst] ∼ Ki Ki+1Fsi, t = i + 1, (2) 0, otherwise. If b is the maximal element (m, m + 1) of I1, then neither of the first two cases can occur since i 6= m, so EaF≥b − F≥bEa = [Ea,Fb] = 0 and the claim holds. Let b = (s, t) ∈ I1 and assume that b 6= (m, m + 1). We have EaF≥b = EaFbF>b = [Ea,Fb]F>b + FbEaF>b. (3) In the case s = i of Equation (2), we have (i + 1, t) > (i, t) = b and in the case t = i + 1 we have (s, i) = (s, t − 1) > b. Therefore, by Lemma 2.7, the first term on the right of Equation (3) is zero. Since b is not (m, m + 1) it has an immediate successor c ∈ I1. So EaF≥b = FbEaF>b = FbEaF≥c = FbF≥cEa = F≥bEa, where the third step uses the induction hypothesis. (b) We proceed by induction on b ∈ I1. For each b = (s, t) ∈ I1, using the definition of Fs,i+1 in the first case (see Section1), similarly the definition of Fit in the second case, and finally Lemmas 2.2(b,c) and 2.3(a) in the third case, we get q F F + F , t = i, i st i,i+1 s,i+1 FaFb = Fi,i+1Fst ∼ FstFi,i+1 − Fit, s = i + 1, (4) FstFi,i+1, otherwise. If b is the minimal element (1, m + n) of I1, then neither of the first two cases can occur since i 6= m + n, 0. Thus, FaF≤b = FaFb ∼ FbFa = F≤bFa as claimed. Let b = (s, t) ∈ I1 and assume that b 6= (1, m + n). Then b has an immediate predecessor c ∈ I1. Using the induction hypothesis, we get FaF≤b = FaF≤cFb ∼ F≤cFaFb = F (a) For every a ∈ I1, we have FaFI1 = 0. (b) For every a ∈ I0, we have EaFI1 = FI1 Ea and FaFI1 ∼ FI1 Fa. Proof. (a) Let a ∈ I1. If a is the maximal element (m, m+1) of I1, then FaFI1 = FaF (b) Since FI1 = F≥(1,m+n) and also FI1 = F≤(m,m+1), this follows from Lemma 2.9. r Qr For a positive integer r and an r-tuple x = (x1, x2, . . . , xr) ∈ (I1) put Ex = k=1 Exk and Qr Fx = k=1 Fxk . r Lemma 2.11. For every positive integer r and x ∈ (I1) , (a) Ex is in the F-span of the set {EI | I ⊆ I1}, (b) Fx is in the F-span of the set {FI | I ⊆ I1}. Proof. It suffices to prove part (b) since then part (a) will follow by applying the antiauto- morphism Ω. Let r be a positive integer. We argue by induction using the lexicographical r r order on (I1) . Let x ∈ (I1) . If x1 ≤ x2 ≤ · · · ≤ xr, then either xk = xk+1 for some k, in which case Fx = 0 by Lemma 2.1(c), or the inequalities are strict, in which case Fx = FI , where I = {x1, x2, . . . , xr}, so the claim holds. Therefore, we may assume that xk > xk+1 for some k. Writing xk = (i, j) and xk+1 = (s, t) we have i, s < j, t since (i, j), (s, t) ∈ I1. Additionally, since (i, j) > (s, t) we have j < t or (j = t and i > s). Now ( −1 −FstFij + (qj − qj )FitFsj, if j < t and i < s, Fxk Fxk+1 = FijFst ∼ (6) FstFij, otherwise, where the first case uses Lemma 2.3 part (d) and then part (a) (to switch the order of the factors in the second term), and the second case uses Lemmas 2.2(b) and 2.3(a) if j < t and i ≥ s and Equation 2.2(c) if j = t and i > s. Since (s, t) < (i, j) and, in the first case, (i, t) < (i, j), by using Equation (6) to replace Fxk Fxk+1 in the product Fx we see that Fx is a r linear combination of Fy with y ∈ (I1) and with y strictly less than x in the lexicographical order, so the induction hypothesis applies to complete the proof. 10 3. Structure of U Below, we define sub(super)algebras of U (shown on the left) by giving their generating sets (shown on the right): 0 ±1 U : {Ei,i+1,Fi,i+1,Kj | i ∈ [1, m + n) \ m, j ∈ [1, m + n]}; 00 ±1 U : {Ki | i ∈ [1, m + n]}; 0+ U : {Ei,i+1 | i ∈ [1, m + n) \ m}; 0− U : {Fi,i+1 | i ∈ [1, m + n) \ m}; 1+ U : {Ea | a ∈ I1}; 1− U : {Fa | a ∈ I1}. I ψ Q ψ(a) Let I ⊆ I and ψ ∈ N , where N = {0, 1, 2,... }. Define F = a∈I Fa with the factors in the order given by ≤ and where an empty product is defined to be 1. Define Eψ = Ω(F ψ) ψ(a) (= product of the Ea in the reverse order). Put I0 I1 Ψ0 = N , Ψ1 = {0, 1} and I Ψ = {ψ ∈ N | ψ(I1) ⊆ {0, 1}}. Pm+n Denote by Λ the set {λ = i=1 λiεi | λi ∈ Z}, where εi is the (m + n)-tuple with jth entry λ Q λi δij (Kronecker delta). For λ ∈ Λ, put K = i Ki . The proof of the following theorem is an adaptation of the proof in [5], which handles the special case (F, q) = (C(v), v), where v is an indeterminate. Part (c) of the theorem provides a PBW-type basis for U. This theorem is used throughout the rest of the paper (often without explicit reference). Theorem 3.1. j− ψ j+ ψ (a) For j ∈ {0, 1}, U has F-basis {F }ψ∈Ψj and U has F-basis {E }ψ∈Ψj . The space 00 λ U has F-basis {K }λ∈Λ. (b) The multiplication maps U1− ⊗ U0 ⊗ U1+ → U, U0− ⊗ U00 ⊗ U0+ → U0 are F-isomorphisms. In particular, U = U1−U0U1+, U = U1+U0U1−, U0 = U0−U00U0+. SIMPLICITY OF KAC MODULES 11 (c) The set ψ λ ϕ X = {F K E }ψ,ϕ∈Ψ;λ∈Λ ψ λ ϕ = {FI F K E EJ }I,J⊆I1;ψ,ϕ∈Ψ0;λ∈Λ is an F-basis for U. ψ 1+ ψ Proof. By Lemma 2.11 the set {E | ψ ∈ Ψ1} spans U and the set {F | ψ ∈ Ψ1} spans U1−. The remaining spanning assertions of part (a) follow from the proofs of Lemmas 5.3(b) and 5.14 in [5]. The multiplication maps in part (b) are surjective by the proofs of Lemma 5.3(a), Theorem 5.5, and Theorem 5.16 in [5]. It then follows that the set X in part (c) spans U. Assume that (F, q) = (C(v), v), where C(v) is the field of fractions of the polynomial ring C[v](v an indeterminate) and the assumed automorphism Ω of C(v) is the one uniquely −1 ˙ ψ 1+ determined by v 7→ v . By [5, Corollary 8.11], the subset {E }ψ∈Ψ1 of U is linearly independent over F, where the dot in the notation E˙ ψ indicates that the product is relative to the order of the Ea given in that paper (before Lemma 5.9), which is different from the ψ 1+ order in this paper. From the preceding paragraph, {E }ψ∈Ψ1 spans U over F. Since ψ ψ Ψ1 is finite, it follows that {E }ψ∈Ψ1 is linearly independent over F. Similarly, {F }ψ∈Ψ1 is linearly independent over F. The remaining claims of linear independence in part (a) follow from [5, Corollary 8.11]. In part (b), the first isomorphism is noted in the discussion after Corollary 8.11 in [5], the second is given in [5, Lemma 5.3(a)], and the second equality follows by applying the antiautomorphism Ω of U to the first equality. Finally, part (c) follows from parts (a) and (b). This completes the proof of the theorem in the special case (F, q) = (C(v), v). Let (F, q) be arbitrary once again. Denote by Z[v, v−1] the subring of C(v) generated by the set Z ∪ {v, v−1}. Put −1 B = {f(v)/g(v) | f(v), g(v) ∈ Z[v, v ], g(q) 6= 0} ⊆ C(v). The automorphism Ω of C(v) fixes every integer and hence induces an automorphism of the subring B, which is the automorphism we assume in the definition of UB,v. The map that sends an element of the generating set X ⊆ UB,v to the corresponding element of X ⊆ UC(v),v induces a B-algebra homomorphism ϕ : UB,v → UC(v),v (the codomain regarded as a B-algebra by restriction of scalars). For each x in the set X of part (c), we have ϕ(x) = x. Now X ⊆ UC(v),v is linearly independent over C(v), as was observed above, so it is linearly independent over B as well. Therefore, X ⊆ UB,v is linearly independent over B. Regard F as a B-module via the homomorphism B → F, f(v)/g(v) 7→ f(q)/g(q). The map X → X ⊗ 1, x 7→ x ⊗ 1, induces an F-algebra homomorphism U → UB,v ⊗B F, which sends 12 x ∈ X to x⊗1 ∈ X ⊗1. By the preceding paragraph, X ⊗1 is linearly independent over F, so the subset X of U is linearly independent over F as well and is therefore, in view of the first paragraph, an F-basis for U. This completes the proof of part (c). The linear independence claims of part (a) now follow, and so does injectivity of the maps in part (b). In view of the first paragraph the proof is complete. In addition to the Z2-grading on U in the superalgebra structure, there are two Z-gradings that we will use. P˙ Proposition 3.2. The algebra U has uniquely determined Z-gradings U = iU(i) and P˙ U = iU[i] satisfying the following: ±1 (a) Ea,Fa,Ki ∈ U(0) (a ∈ I0, i ∈ [1, m + n]); Ea ∈ U(1),Fa ∈ U(−1) (a ∈ I1); ±1 (b) Ki ∈ U[0] (i ∈ [1, m + n]); Ea ∈ U[j−i],Fa ∈ U[i−j] (a = (i, j) ∈ I). Proof. (a) The free algebra Uˆ can be given the structure of a Z-graded algebra by assigning to its generators arbitrary degrees. We assign degree 1 to Em,m+1, degree −1 to Fm,m+1, and degree 0 to all of the other elements of X . It is straightforward to check that each difference x−y with x = y a relation in R is homogeneous, so the ideal generated by all such differences is graded and we get a uniquely determined induced grading on the quotient of Uˆ by this ideal, which is U. The properties stated follow directly from the definitions of Ea and Fa with a ∈ I. (b) The proof here is similar to the proof of (a), except that we assign degree 1 to each Ei,i+1, ±1 degree −1 to each Fi,i+1, and degree 0 to each Ki . 4. Kac module We assume from here on that F is a field. The set B+ := U0U1+ is a subsuperalgebra of U. This follows from Lemma 5.15 in [5], which assumes that (F, q) = (C(v), v) but holds in general. By Theorem 3.1, B+ has F-basis consisting of the vectors ψ λ ϕ F K E EJ (ψ, ϕ ∈ Ψ0, λ ∈ Λ,J ⊆ I1). Each such vector is homogeneous of degree |J| using the Z-grading of U in Proposition 3.2(a). + + Pmn + In particular, B is a graded subalgebra of U and, since |I1| = mn, we have B = k=0 B(k). + P + + mn+1 P + + Put N = k>0 B(k). We have (N ) ⊆ k>mn B(k) = 0, so N is a nilpotent ideal of B+. SIMPLICITY OF KAC MODULES 13 Let L be a finite-dimensional simple U0-module. (Here and below “finite-dimensional” means + 0 + “finite-dimensional as a vector space over the field F.”) Since B(0) = U , we have B = U0+˙ N+, so B+/N+ ∼= U0. Therefore, L is a B+-module via the canonical map B+ → U0. The Kac module associated with L is the U-module K(L) = U ⊗B+ L. (Throughout, by “U-module” we mean a module M for U in the usual sense that has a Z2-grading M = M0+˙ M1 with UiMj ⊆ Mi+j for all i, j ∈ Z2. Here, the Z2-grading on K(L) is induced by that on U with the elements of 1 ⊗ L declared to be of degree 0.) Kac modules are of interest for the following reason. Proposition 4.1. Every finite-dimensional simple U-module is a homomorphic image of K(L) for some finite-dimensional simple U0-module L. Proof. Let M be a finite-dimensional simple U-module. Regarded as B+-module, M has a simple submodule L, which is necessarily finite-dimensional. Since N+ is an ideal of B+, it follows that N+L is a B+-submodule of L. If N+L = L, then L = (N+)mn+1L = 0L = 0, a contradiction. Since L is simple, we conclude that N+L = 0, that is, N+ annihilates L. Therefore, L can be regarded as a module for B+/N+ ∼= U0, and it is simple as such. The inclusion map L → M induces a U-homomorphism K(L) = U ⊗B+ L → M, which is surjective since M is simple. Our immediate goal is Theorem 4.3, which gives a general necessary and sufficient condition for K(L) itself to be simple. We need some preparation. Lemma 4.2. Let L be a finite-dimensional simple U0-module. Each nonzero submodule of K(L) contains FI1 ⊗ v for some 0 6= v ∈ L. Proof. Assume to the contrary that there exists a nonzero submodule N of K(L) such that FI1 ⊗ v∈ / N for every nonzero v ∈ L. Let x be a nonzero element of N. By Theorem 3.1 we have K(L) = U1− ⊗ L, so X x = FI ⊗ vI (7) I⊆I1 for some uniquely determined vI ∈ L. By assumption, there exists a minimal t(x) ∈ I1 such that t(x) ∈/ I for some I ⊆ I1 with vI 6= 0. This defines a function t : N \ 0 → I1. There exists x ∈ N \{0} with p = t(x) maximal. Write x as in Equation (7). By the definition of p, we have vI = 0 if I is not in the set Y = {I ⊆ I1 | (< p) ⊆ I}, where (< p) denotes the set of all elements of I1 that are less 14 than p. Moreover, Lemmas 2.8 and 2.1(c) give, for each I ∈ Y, ( 0, p ∈ I, FpFI ⊗ vI = 0 FI∪{p} ⊗ vI , p∈ / I, 0 where vI = cvI for some nonzero c ∈ F (depending on I). Therefore, 0 X X 0 x := Fpx = FpFI ⊗ vI = FI∪{p} ⊗ vI . I∈Y I∈Y p/∈I 0 0 0 Now, vI 6= 0 for some I ∈ Y with p∈ / I, so vI 6= 0, implying x 6= 0. Hence, x ∈ N \{0} 0 and t(x ) > p = t(x), contradicting the definition of x. Therefore, the lemma follows. Denote by X(U00) the set of all algebra homomorphisms U00 → F. Let L be a U00-module and let λ ∈ X(U00). The set 00 Lλ = {x ∈ L | tx = λ(t)x for every t ∈ U } is the weight space of L corresponding to λ, and each element of Lλ is a weight vector of weight λ. By relation (R2), we have U00U0+ = U0+U00 =: B0+, so B0+ is a subalgebra of U and it is 0+ P 0+ Z-graded using the grading of Proposition 3.2(b). Moreover, N := i>0 B[i] is an ideal of B0+ and B0+ = U00+˙ N0+, so B0+/N0+ ∼= U00. Similar statements hold for B0− := U0−U00 0− P 0− and N := i<0 B[i] replacing + by −. Given a B0+-module L (which can be regarded as a U00-module by restriction of scalars) 00 0+ and λ ∈ X(U ), a nonzero weight vector x ∈ Lλ with N x = 0 is a maximal vector of weight λ. For λ ∈ X(U00), denote by S(λ) the class of all finite-dimensional simple U0-modules L that have the property that every simple B0+-submodule of L is one-dimensional and is generated by a maximal vector of weight λ. 00 By Theorem 3.1(c), there are uniquely determined elements uψϕ of U such that X ψ ϕ EI1 FI1 = F uψϕE . ψ,ϕ∈Ψ Every term here is homogeneous with respect to the Z-grading of U given in Proposition 3.2(b). The left-hand side has degree zero, so the sum of the terms of nonzero degree on the right-hand side equals zero. Using uniqueness of expression, it follows that uψϕ = 0 whenever ψ ϕ F uψϕE has nonzero degree, which is the case, in particular, if ψ 6= 0, ϕ = 0, and also if SIMPLICITY OF KAC MODULES 15 ψ = 0, ϕ 6= 0. Therefore, X ψ ϕ EI1 FI1 = u0 + F uψϕE , (8) 06=ψ,ϕ∈Ψ where u0 := u00. Theorem 4.3. Let λ ∈ X(U00) and let L ∈ S(λ). The Kac module K(L) is simple if and only if λ(u0) 6= 0. Proof. Assume that K(L) is simple. Regarded as B0+-module, L has a simple submodule, and since L ∈ S(λ), this simple submodule is spanned by a maximal vector vλ of weight λ. Since FI1 ⊗vλ ∈ K(L) is nonzero, UFI1 ⊗vλ = U(FI1 ⊗vλ) is a nonzero submodule of K(L). This submodule is Z2-graded since the vector FI1 ⊗ vλ is homogeneous. By our assumption, K(L) = UFI1 ⊗ vλ. By Theorem 3.1(a), we have U1− = F1 + N1−, where N1− is the F-linear span of the F ψ 1− 1− with 0 6= ψ ∈ Ψ1. By Lemma 2.10(a), N FI1 = 0, so U FI1 = FFI1 . By Lemmas 2.1(a) 0 0 and 2.10(b) we have U FI1 = FI1 U . Therefore, using Theorem 3.1(b), we get 1+ 0 1− K(L) = UFI1 ⊗ vλ = U U U FI1 ⊗ vλ 1+ 0 1+ = U FI1 ⊗ U vλ = U FI1 ⊗ L. In particular, the set B of all EI FI1 ⊗ vi with I ⊆ I1, 1 ≤ i ≤ s spans K(L) over F, where {v1, . . . , vs} is an F-basis for L. On the other hand, it follows from Theorem 3.1(b,c) that 1− K(L) is isomorphic over F to U ⊗F L, which has basis all FI ⊗ vi with I ⊆ I1, 1 ≤ i ≤ s. Since the cardinality of B is at most the cardinality of this basis, we conclude that B is a basis for K(L). Assuming, without loss of generality, that v1 = vλ, we get 0 6= EI1 FI1 ⊗ vλ = 1 ⊗ u0vλ = λ(u0)(1 ⊗ vλ), so that λ(u0) 6= 0. Now assume that λ(u0) 6= 0. Let N be a nonzero (Z2-graded) submodule of K(L). By 0+ 0+ Lemma 4.2, we have FI1 ⊗ v ∈ N for some 0 6= v ∈ L. Now B v is a nonzero B - submodule of L, so it has a simple B0+-submodule. Since L ∈ S(λ), this submodule equals Fv0 for some maximal vector v0 of weight λ. So 0 0+ 0+ FI1 ⊗ v ∈ FI1 ⊗ B v = B (FI1 ⊗ v) ⊆ N, 0+ 0+ where we have used that FI1 B = B FI1 by Lemmas 2.1(a) and 2.10(b). Hence λ(u0)1 ⊗ 0 0 0 v = EI1 FI1 ⊗ v ∈ N. Since λ(u0) 6= 0 by assumption, it follows that 1 ⊗ v ∈ N. The vector 1 ⊗ v0 generates the U-module K(L), so K(L) ⊆ N, forcing K(L) = N. We conclude that K(L) is simple. Remark 4.4. The Kac module K(L) is studied in [6], but the author considers only L ∈ S(λ) λi with λ satisfying the property that for each i ∈ [1, m + n], λ(Ki) = ±q for some integer λi 16 [6, III(12), p. 1241]. We say that λ is “integral” in this case. Theorem 4.3 does not require this property. Moreover, Corollaries 6.5 and 7.6 below include cases where the weight λ is not integral. In the next section, we give an explicit formula for u0 (Theorem 5.4) and use it to replace the condition λ(u0) 6= 0 in the theorem with a more explicit condition (Corollary 5.5). In the last two sections of the paper, we construct simple U0-modules L to which the theorem can be applied, that is, for which L ∈ S(λ). For this, we specialize to the case where q is not a root of unity (Section6) and then to the case where q is a root of unity (Section7). Before leaving the general situation, though, we record a couple of facts. Lemma 4.5. Let M be a U00-module and let N be a submodule of M. If M is the sum of its weight spaces, then so is N. Proof. The argument in the proof of part (d) of Theorem 20.2 in [1] applies here almost verbatim. Lemma 4.6. Let S be a simple B0+-module and assume that there exists a nonzero weight vector v in S. If N0+ acts nilpotently on S, then S = Fv and v is a maximal vector. Proof. Assume that N0+ acts nilpotently on S. By an argument similar to that in the proof of Proposition 4.1, N0+ annihilates S, so that S can be regarded as a module for the quotient 0+ 0+ ∼ 00 00 B /N = U and it is simple as such. Now Fv is a U -submodule of S, so S = Fv. We 0+ also have N v = 0, so v is a maximal vector. If F is algebraically closed, the assumption in the lemma that there exists a nonzero weight vector in S can be removed since this condition is automatically satisfied. 5. Simplicity criterion 00 In this section we obtain a formula for the element u0 of U (Theorem 5.4) and use it to give a more explicit criterion (Corollary 5.5) for simplicity of the Kac module K(L) than that given in Theorem 4.3. Denote by E the left ideal of U generated by the set {Ea | a ∈ I}. Lemma 5.1. For every a, b ∈ I1 with a ≤ b, we have EaF>b ≡ 0 (mod E). Proof. We proceed by reverse induction on b using the order ≤ on I1. The claim follows immediately if b is the maximal element (m, m + 1) of I1. Let a, b ∈ I1 with a ≤ b and assume that b 6= (m, m + 1). Then b has an immediate successor c ∈ I1. We can write SIMPLICITY OF KAC MODULES 17 a = (i, j) and c = (s, t). By Lemmas 2.2(g,e) and 2.3(c,a), respectively, as well as the fact that i, s < j, t, we have q−1E K K−1, if i < s and t = j, s is s j E K−1K , if s = i and t < j, [E ,F ] = [E ,F ] = tj s t a c ij st (q − q−1)F E K K−1, if s < i and t < j, t t si tj t i 0, if i < s and t < j. Since a ≤ b < c the indicated cases cover all possibilities for the pairs (i, j) and (s, t). It now follows from Lemma 2.1(a) that [Ea,Fc] is an element of the left ideal of U generated by the set {Ed | d ∈ I0}. By Lemma 2.9(a), we get [Ea,Fc]F>c ≡ 0 (mod E), since F>c = F≥d if c has an immediate successor d ∈ I1, and the equivalence holds trivially otherwise. Therefore, EaF>b = EaFcF>c = [Ea,Fc]F>c − FcEaF>c ≡ 0 (mod E), where the last step uses the induction hypothesis. m+n The set {εi | 1 ≤ i ≤ m + n} is a basis for the Z-module Z , where εi is the (m + n)- m+n tuple with jth entry δij (Kronecker delta). There is a unique Z-bilinear form on Z with (εi, εj) = ηiδij for each 1 ≤ i, j ≤ m + n, where ηi is 1 or −1 according as i ≤ m or i > m. For a = (i, j) ∈ I, put εa = εi − εj. Put ! 1 X X ρ = ε − ε . 2 a a a∈I0 a∈I1 Lemma 5.2. For each a = (i, j) ∈ I1, we have X (− εb, εa) = 2m − i − j + 1 = (ρ, εa). b>a L R Proof. For a = (i, j) ∈ I1, denote by na and na the left- and right-hand sides, respectively, of the first equation in the lemma. A straightforward computation shows that for every a = (i, j) ∈ I1 and X ∈ {L, R}, we have ( X X n(i+1,j) + 1, if i < m, na = X (9) n(i,j−1) − 1, if j > m + 1. We proceed by reverse induction using the order ≤ on I1. The greatest element of I1 is L R (m, m + 1) and we have n(m,m+1) = 0 = n(m,m+1). Now let a = (i, j) ∈ I1 and assume that a 6= (m, m + 1). We have that either i < m or j > m + 1, so using Equation (9) and the L R induction hypothesis, we get na = na , so the first equality in the lemma is established. 18 According to [6, Appendix B], m m+n X X 2ρ = (m − n − 2k + 1)εk + (3m + n − 2l + 1)εl. k=1 l=m+1 Therefore, for every a = (i, j) ∈ I1, we have 2(ρ, εa) = (2ρ, εi − εj) = (m − n − 2i + 1) + (3m + n − 2j + 1) = 4m − 2i − 2j + 2, so the second equality in the lemma follows. Lemma 5.3. For each a, b ∈ I, we have (εa,εb) (a) KaEb = q EbKa, −(εa,εb) (b) KaFb = q FbKa. Proof. These formulas follow from Lemma 2.1(a) using the definitions. For a unit X in U and an integer i, put qiX − q−iX−1 [X; i] = . q − q−1 00 The element u0 of U in the following result is defined in Equation (8). Theorem 5.4. We have Y u0 = [Ka;(ρ, εa)]. a∈I1 Proof. For every a ∈ I1, we have (recalling the definition of EI for I ⊆ I given before Lemma 2.5) E≥aF≥a = E>aEaFaF>a K − K−1 = E a a F − E F E F (Lemma 2.1(b)) >a q − q−1 >a >a a a >a −1 −1 −1 ≡ (q − q ) (E>aKaF>a − E>aKa F>a) (Lemma 5.1) −1 −1 (ρ,εa) −(ρ,εa) −1 = (q − q ) (q E>aF>aKa − q E>aF>aKa ) (Lemmas 5.3(b), 5.2) = E>aF>a[Ka;(ρ, εa)], where the congruence is modulo the left ideal E. It follows from Lemma 5.3 that E[Ka;(ρ, εa)] ⊆ E SIMPLICITY OF KAC MODULES 19 for each a ∈ I1. A proof by reverse induction using this fact and the above equivalence gives for each a ∈ I1 Y E≥aF≥a ≡ [Kb;(ρ, εb)] (mod E). b≥a Applying this to the minimal element a = (1, m + n) of I1, we get Y u0 ≡ EI1 FI1 ≡ [Kb;(ρ, εb)] (mod E). b∈I1 By Theorem 3.1, the left ideal E is spanned over F by the set {F ψKλEϕ} with λ ∈ Λ, 00 ψ, ϕ ∈ Ψ, ϕ 6= 0, implying that the intersection U ∩ E is trivial. The theorem follows. Corollary 5.5 (cf. [6, Proposition 4]). Let λ ∈ X(U00) and let L ∈ S(λ). The Kac module K(L) is simple if and only if −1 i+j−2m−1 λ(Ki)λ(Kj) 6= ±q for every (i, j) ∈ I1. Proof. Let a = (i, j) ∈ I1. Putting k = (ρ, εa) we have qkλ(K ) − q−kλ(K )−1 λ([K ;(ρ, ε )]) = a a , a a q − q−1 −1 −k which is nonzero if and only if λ(Ki)λ(Kj) = λ(Ka) 6= ±q . By Lemma 5.2 we have k = (ρ, εa) = 2m − i − j + 1. The claim now follows from Theorems 4.3 and 5.4. 6. Not root of unity case In this section we assume that q is not a root of unity. Furthermore, we assume that the characteristic of F is not 2 in order to use results from [2] where that assumption is made. For each i ∈ [1, m + n) \ m, put ˙ −1 ηi ηi −ηi Ki = (KiKi+1) = Ki Ki+1 , recalling that ηi is 1 or −1 according as i ≤ m or i > m. Denote by U˙ 0 the subalgebra of U0 generated by the set ˙ ˙ −1 {Ei,i+1,Fi,i+1, Ki, Ki | i ∈ [1, m + n) \ m}, ˙ 00 00 ˙ ˙ −1 and by U the subalgebra of U generated by the set {Ki, Ki | i ∈ [1, m + n) \ m}. Denote by Uq(g) the quantized enveloping algebra of the complex semisimple Lie algebra g = slm ⊕ sln as defined in [2, 4.3], where the field k in that text is taken to be F. 20 Let Φ be the root system of g relative to the Cartan subalgebra of g consisting of the diagonal matrices in g. For i ∈ [1, m + n] denote by ei the (m + n) × (m + n)-matrix with 1 in the (i, i)-position and zeros elsewhere. Then Π = {αi | i ∈ [1, m + n) \ m} is a basis for Φ, where ∗ ∗ αi = ei − ei+1. Make these choices in the definition of Uq(g) in [2] and take as scalar product P P ∗ ∗ ∗ on i Rαi the restriction of the scalar product on i Rei determined by (ei , ej ) = δij. Proposition 6.1. (a) The multiplication map U0− ⊗ U˙ 00 ⊗ U0+ → U˙ 0 is an F-isomorphism. In particular, U˙ 0 = U0−U˙ 00U0+. ˙ 0 (b) The algebra U is isomorphic to Uq(g). Proof. (a) This follows from Theorem 3.1. (b) Since the relations are preserved, we get a uniquely determined algebra homomorphism U (g) → U˙ 0 by mapping E 7→ E , F 7→ F , K 7→ K˙ , and K−1 7→ K˙ −1, i ∈ q αi i,i+1 αi i,i+1 αi i αi i [1, m + n) \ m. By part (a) and Theorem 4.21 of [2], this map is an isomorphism. We use the isomorphism in the proof of part (b) of the preceding proposition to identify the ˙ 0 algebras U and Uq(g). For λ, µ ∈ X(U00) denote by λ + µ the element of X(U00) uniquely determined by (λ + 00 µ)(Kk) = λ(Kk)µ(Kk), k ∈ [1, m + n]. Then (X(U ), +) is an abelian group. 00 For a = (i, j) ∈ I0 denote byα ¯a the element of X(U ) uniquely determined byα ¯a(Kk) = ∗ (ek,αa) ∗ ∗ qk , k ∈ [1, m + n], where αa = ei − ej . 00 For µ, λ ∈ X(U ) put µ ≤ λ if there exist nonnegative integers na, a ∈ I0, such that λ − µ = P n α¯ . Then ≤ is a partial order on X(U00). (Antisymmetry follows from the a∈I0 a a assumption that q is not a root of unity.) 0 00 Lemma 6.2 (cf. [2, Prop. 5.1]). Let M be a U -module. For every λ ∈ X(U ) and a ∈ I0, EaMλ ⊆ Mλ+¯αa and FaMλ ⊆ Mλ−α¯a . 00 Proof. Let λ ∈ X(U ) and a = (i, j) ∈ I0. For each x ∈ Mλ and k ∈ [1, m + n], Lemma 2.1(a) and the definitions yield ∗ δki−δkj (ek,αa) KkEax = qk EaKkx = qk λ(Kk)Eax =α ¯a(Kk)λ(Kk)Eax = (λ +α ¯a)(Kk)Eax, and the first claim follows. The second claim is proved similarly. SIMPLICITY OF KAC MODULES 21 00 0 Let λ ∈ X(U ). Denote by Iλ the left ideal of U generated by the set 00 {Ea, t − λ(t)1 | a ∈ I0, t ∈ U }. 0 0 0 Put V (λ) = U /Iλ and denote by x 7→ xˆ the canonical map U → U /Iλ = V (λ). It follows 0−ˆ ψˆ from Theorem 3.1 that V (λ) = U 1 and that V (λ) has F-basis {F 1}ψ∈Ψ0 . By Lemma 6.2 each F ψ1ˆ is a weight vector of weight λ − P ψ(a)¯α =: µ, implying µ ≤ λ with strict a∈I0 a P˙ ˆ inequality if and only if ψ 6= 0. In particular, V (λ) = µ≤λV (λ)µ and V (λ)λ = F1. Since 1ˆ generates the U0-module V (λ), each proper submodule of V (λ) (necessarily the sum of its weight spaces by Lemma 4.5) intersects the one-dimensional space F1ˆ trivially. Therefore, the sum W of all proper submodules of V (λ) is the unique maximal submodule of V (λ) and the quotient L(λ) = V (λ)/W is the unique simple quotient of V (λ). A weight space and a maximal vector in a U˙ 0-module are defined just like those for a U0- module except with U00 replaced by U˙ 00. A weight µ ∈ X(U˙ 00) is dominant if there exist nonnegative integers ni and elements σi of {1, −1} with ˙ ni µ(Ki) = σiq , i ∈ [1, m + n) \ m. For λ ∈ X(U00) denote by λ˙ ∈ X(U˙ 00) the restriction of λ to U˙ 00. Put 00 00 ˙ Xd(U ) = {λ ∈ X(U ) | λ is dominant}. Theorem 6.3. 00 (a) For each λ ∈ Xd(U ), we have L(λ) ∈ S(λ). 00 (b) The set {L(λ)}λ∈Xd(U ) is a complete set of representatives of the isomorphism classes of those finite-dimensional simple U0-modules that are the sum of their weight spaces. 00 Proof. (a) Let λ ∈ Xd(U ). The module V (λ) is the sum of its weight spaces (relative to U00) and W is a U00-submodule of V (λ), so it is also the sum of its weight spaces by Lemma P 4.5. Therefore, L(λ) is the sum of its weight spaces as well. We have L(λ) = µ≤λ L(λ)µ, ˆ with L(λ)λ = Fmλ, where mλ is the image of 1 under the canonical map V (λ) → V (λ)/W = 0+ P L(λ). Let M be a simple B -submodule of L(λ). We have M = µ≤λ Mµ, so there 00 0+ exists a maximal η ∈ X(U ) with Mη 6= 0 and η ≤ λ. Then Mη is a nonzero B - 0 0 submodule of M forcing M = Mη. Now U M is a nonzero U -submodule of L(λ), so 0 0− P L(λ) = U M = U Mη ⊆ µ≤η L(λ)µ. Therefore, η = λ, and M = Mλ ⊆ L(λ)λ = Fmλ, so that M = Fmλ. Moreover, mλ is a maximal vector of weight λ. 22 ˙ 0 We claim that L(λ) is simple when regarded as a U -module. The vector mλ is a maximal vector relative to U˙ 0 of weight λ˙ . Moreover, using that U˙ 0 = U0−U˙ 00U0+ we get ˙ 0 0− 0 U mλ = U mλ = U mλ = L(λ), (10) so we get a U˙ 0-epimorphism ϕ : M(λ˙ ) → L(λ), where M(λ˙ ) is the universal highest weight U˙ 0-module of highest weight λ˙ as defined in [2, 5.5]. Denote by N the image under ϕ of the unique maximal submodule of M(λ˙ ). Then N is the unique maximal U˙ 0-submodule of L(λ) and the quotient L(λ)/N is isomorphic to the simple U˙ 0-module L(λ˙ ) of [2, 5.5]. From the relations (R1) and (R2) we see that U00N is a U˙ 0-submodule of L(λ) and that 00 00 ˙ 00 U Nµ ⊆ (U N)µ for each µ ∈ X(U ). Summing over such µ, we get 00 X˙ 00 X˙ 00 00 U N = U Nµ ⊆ (U N)µ = U N, µ µ 00 00 ˙ 00 so, in fact, U Nµ = (U N)µ for each µ ∈ X(U ). Now L(λ) is a homomorphic image of ˙ ˙ M(λ), as observed above, and M(λ)λ˙ is one-dimensional, so we have L(λ)λ˙ = Fmλ, implying ˙ 0 Nλ˙ = 0 (since N 6= L(λ) and mλ generates L(λ) as U -module by Equation (10)). Thus, 00 00 00 ˙ 0 (U N)λ˙ = U Nλ˙ = 0, implying that U N is a proper U -submodule of L(λ). Hence, N ⊆ U00N ⊆ N, forcing U00N = N. But U00N is a U0-submodule of L(λ), so we conclude that N = 0. Thus, we have U˙ 0-isomorphisms L(λ) ∼= L(λ)/N ∼= L(λ˙ ) and the claim is established. Since λ˙ is dominant, L(λ˙ ) is finite-dimensional [2, 5.10], so L(λ) is finite-dimensional as well. Thus, L(λ) ∈ S(λ). 00 (b) Let λ ∈ Xd(U ). By part (a), the simple module L(λ) is finite-dimensional, and it was observed in the proof of (a) that L(λ) is the sum of its weight spaces. Let M be a finite-dimensional simple U0-module that is the sum of its weight spaces. As B0+-module, M has a simple submodule S. For each λ ∈ X(U00), by Lemma 6.2 we have 0+ P 0+ N Mλ ⊆ µ>λ Mµ, so N acts nilpotently on M and hence on S as well. Now S is the sum of its weight spaces (Lemma 4.5), so it has a nonzero weight vector v. By Lemma 4.6, we have S = Fv, and v is a maximal vector of weight, say, λ ∈ X(U00). The homomorphism U0 → M given by x 7→ xv induces a homomorphism V (λ) → M, which must be surjective since it is nonzero and M is simple. Since V (λ) has unique simple quotient L(λ), we get M ∼= L(λ). The vector v is a maximal vector of weight λ˙ in the U˙ 0-module M, so there exists a U˙ 0-homomorphism M(λ˙ ) → M with image U˙ 0v, where M(λ˙ ) is as in the proof of part (a). It follows that M, as U˙ 0-module, has a subquotient isomorphic to the unique simple quotient L(λ˙ ) of M(λ˙ ). Since M is finite-dimensional, so is L(λ˙ ), implying λ˙ is dominant by 00 [2, 5.10], that is, λ ∈ Xd(U ). SIMPLICITY OF KAC MODULES 23 0 00 ∼ 0 0 Finally, let λ, λ ∈ Xd(U ) and assume that L(λ) = L(λ ). There exists a U -isomorphism 0 0 ϕ : L(λ) → L(λ ). The vector ϕ(mλ) is a maximal vector in L(λ ) of weight λ and Fϕ(mλ) 0+ 0 0 0 0 is a B -submodule of L(λ ). Since L(λ ) ∈ S(λ ) by part (a), it follows that λ = λ . Denote by T the set of all tuples (a, b, τ, y) such that a, b ∈ F \ 0, τ ∈ {−1, 1}m+n, y ∈ Zm+n, with y1 ≥ y2 ≥ · · · ≥ ym = 0, ym+1 ≤ ym+2 ≤ · · · ≤ ym+n = 0. For t = (a, b, τ, y) ∈ T there 00 is a uniquely defined λt ∈ X(U ) with yi λt(Ki) = τiciq , i ∈ [1, m + n], where ci = a if i ≤ m and ci = b if i > m. 00 Proposition 6.4. We have Xd(U ) = {λt | t ∈ T }. 00 00 ˙ Proof. Let λ ∈ Xd(U ), so that λ ∈ X(U ) and λ is dominant. There exist nonnegative integers ni and elements σi of {1, −1} with ˙ ni λ(Ki) = σiq , i ∈ [1, m + n) \ m. Put ym, ym+n = 0 and τm, τm+n = 1. Also put a = λ(Km) ∈ F and b = λ(Km+n) ∈ F and note that a, b 6= 0 since Km and Km+n are units and λ is an algebra homomorphism. Recursively define, for i ∈ [1, m + n) \ m, τi = τi+1σi ∈ {1, −1} and yi = yi+1 + ηini (where ηi is 1 or −1 according as i ≤ m or i > m) and note that t := (a, b, τ, y) ∈ T . We argue by reverse induction on i that λ(Ki) = λt(Ki) for each i ∈ [1, m + n]. The cases i = m, m + n follow immediately from the definitions. For i ∈ [1, m + n) \ m we have ˙ ηi ηini yi+1 yi+1+ηini λ(Ki) = λ((Ki) )λ(Ki+1) = σiq τi+1ci+1q = τi+1σici+1q yi = τiciq = λt(Ki), where the second equality uses the induction hypothesis. Therefore, λ = λt, implying that the first set is contained in the second. Now let t = (a, b, τ, y) ∈ T . For each i ∈ [1, m + n) \ m, put σi = τiτi+1 ∈ {1, −1} and ni = ηi(yi − yi+1), and note that each ni is nonnegative. For each i ∈ [1, m + n) \ m we have ˙ ηi −ηi ηi ηiyi −ηi −ηiyi+1 ni λt(Ki) = λt(Ki) λt(Ki+1) = τici q τi+1ci+1q = σiq . ˙ 00 so λt is dominant, that is, λt ∈ Xd(U ). This completes the proof. 00 Corollary 6.5. Let λ ∈ Xd(U ) so that λ = λt for some t = (a, b, τ, y) ∈ T . (a) If ab−1 ∈/ ±qZ, then K(L(λ)) is simple; −1 e (b) If ab = ±q for some e ∈ Z, then K(L(λ)) is simple if and only if yi − yj 6= i + j − 2m − 1 − e for every (i, j) ∈ I1. 24 −1 −1 yi−yj Proof. First note that for every (i, j) ∈ I1 we have λ(Ki)λ(Kj) = ±ab q . If K(L(λ)) is not simple, then Theorem 6.3(a) and Corollary 5.5 give ab−1 = ±qyj −yi qi+j−2m−1 ∈ ±qZ for some (i, j) ∈ I1, so part (a) follows. Now assume that ab−1 = ±qe for some e ∈ Z. It follows from Theorem 6.3(a), Corollary 5.5, and the equality above that K(L(λ)) is simple if and only if yi−yj −e −1 i+j−2m−1−e q = ±q λ(Ki)λ(Kj) 6= ±q for every (i, j) ∈ I1. Since q is not a root of unity, part (b) follows. 7. Root of unity case In this section we assume that q is a primitive lth root of unity, where l is a fixed integer with l ≥ 3. Denote by Z0 the subalgebra of U0 generated by the elements l l ±l Ea,Fa,Ki , with a ∈ I0, i ∈ [1, m + n]. Proposition 7.1. The subalgebra Z0 is contained in the center of U0. Proof. It suffices to show that each of the indicated generators of Z0 is in the center of U0. 0 Let a ∈ I0. The generalized Lusztig automorphisms restrict to automorphisms of U , so l 0 l by Lemma 1.2 the element Ea equals the image under an automorphism of U of Ei,i+1 for l 0 some i ∈ [1, m + n) \ m. Therefore, to show that Ea is in the center of U it is enough to l show that Ei,i+1 is so. Let j ∈ [1, m + n) \ m and let k ∈ [1, m + n). For every integer s > 1 we have the following formulas: s s(δk,i+1−δki) s • Ei,i+1Kk = qk KkEi,i+1, s s −1 s−1 • Ei,i+1Fj,j+1 = Fj,j+1Ei,i+1 + δij[s][KiKi+1; 1 − s]iEi,i+1, s s−1 s • Ei,i+1Ej,j+1 = [s]Ei,i+1Ej,j+1Ei,i+1 − [s − 1]Ej,j+1Ei,i+1 if |i − j| = 1, s −s −1 t −t −1 where we have used the notations [s] = (q −q )/(q−q ), and [X; t]i = (q X−q X )/(qi− −1 qi ) for an invertible element X of U and an integer t (cf. [2, pp. 5,10]). These formulas are established using induction on s with the aid of the relations (R2), (R3), and (R6), respectively. Since q is an lth root of unity, we have [l] = 0 and [l − 1] = −1. It follows, l 0 using relation (R5) as well, that Ei,i+1 commutes with the generators in the definition of U (see Section3), and it is therefore in the center of U0. SIMPLICITY OF KAC MODULES 25 0 l l Since Ω restricts to an antiautomorphism of U it follows that Fi,i+1 = Ω(Ei,i+1) is in the center of U0. Finally, using relations (R1) and (R2) and the fact that q is an lth root of ±l 0 unity, we see that each Ki , i ∈ [1, m+n] is in the center of U . The proposition follows. For an F-algebra homomorphism χ : Z0 → F, a subalgebra S of U0, and an S-module M, we say that M has character χ if zx = χ(z)x for every z ∈ Z0 ∩ S, x ∈ M. Proposition 7.2. If F is algebraically closed, then every finite-dimensional simple U0- module has character χ for some F-algebra homomorphism χ : Z0 → F. Proof. Assume that F is algebraically closed and let M be a finite-dimensional simple U0- module. As Z0-module, M has a simple submodule, which must equal Fv for some nonzero v ∈ M since Z0 is commutative and F is algebraically closed. For each z ∈ Z0, we have zv = χ(z)v for some χ(z) ∈ F. This defines a map χ : Z0 → F, which is seen to be an F-algebra homomorphism. The set {x ∈ M | zx = χ(z)x for all z ∈ Z0} is a U0-submodule of M, and it is nonzero since it contains v, so it equals M. The statement follows. 0 l We fix an F-algebra homomorphism χ : Z → F and assume throughout that χ(Ea) = 0 and l χ(Fa) = 0 for every a ∈ I0. This assumption corresponds roughly to the restricted case in the study of modules for a restricted Lie algebra. 0+ 00 0+ 0+ P 0+ 0+ Recall from Section4 that B := U U and N := i>0 B[i] , where B[i] denotes the ith component of B0+ relative to the Z-grading of U given in Proposition 3.2(b) (with similar definitions for B0− and N0−). Lemma 7.3. (a) N0+ acts nilpotently on every B0+-module having character χ. (b) N0− acts nilpotently on every B0−-module having character χ. Proof. (a) Let M be a B0+-module having character χ. Put j = l P deg E (with the a∈I0 a ψ degree as in Proposition 3.2(b)). Let ψ ∈ Ψ0 and assume that deg E ≥ j. Then X ψ X ψ(a)(deg Ea) = deg E ≥ j = l(deg Ea). a∈I0 a∈I0 l Each deg Ea is positive, so ψ(a) ≥ l for some a. Since M has character χ and χ(Ea) = 0, we ψ 0+ j P 0+ get E M = 0. Now (N ) is contained in i≥j B[i] and this latter space is spanned by all λ ψ ψ 0+ j K E with λ ∈ Λ, ψ ∈ Ψ0, deg E ≥ j, so (N ) M = 0 and the claim follows. (b) This is proved similarly. 00 00 l l Put Xχ(U ) = {λ ∈ X(U ) | λ(Ki ) = χ(Ki ) for every i ∈ [1, m + n]}. 26 00 0 Let λ ∈ Xχ(U ). Denote by Iλ the left ideal of U generated by the set l 00 {Ea,Fa, t − λ(t)1 | a ∈ I0, t ∈ U }. 0 0 0 Put V (λ) = U /Iλ and denote by x 7→ xˆ the canonical map U → U /Iλ = V (λ). (The notation defined in this section is independent of that in the preceding section.) Proposition 7.4. (a) V (λ) is finite-dimensional over F and it is the sum of its weight spaces. (b) The U0-module V (λ) has character χ. (c) V (λ) = Nˆ 0− +˙ F1ˆ. (d) The U0-module V (λ) has a unique maximal submodule W , and W ⊆ Nˆ 0−. 0− ψ 0 Proof. (a) By Theorem 3.1(a), U has F-basis {F | ψ ∈ Ψ }. For each a ∈ I0, the element l 0 ˆ 0 ˆ 0− Fa of U is central (Proposition 7.1) and is contained in Iλ, so V (λ) = U = U is spanned ˆψ l l over F by the set {F | ψ ∈ Ψ0}, where Ψ0 = {ψ ∈ Ψ0 | 0 ≤ ψ(i) < l for every i}, a finite set. Moreover, each Fˆψ is a weight vector, so V (λ) is the sum of its weight spaces. l l (b) Let a ∈ I0. Since Ea and Fa are both central (Proposition 7.1), it follows from the l definition of V (λ) that these elements both annihilate V (λ). By assumption, χ(Ea) = 0 and l l l l l χ(Fa) = 0, so Eav = χ(Ea)v and Fav = χ(Fa)v for every v ∈ V (λ). Let i ∈ [1, m + n]. For v ∈ V (λ) we have l l l Ki v = λ(Ki )v = χ(Ki )v, l 00 using that Ki is central for the first equality and that λ ∈ Xχ(U ) for the second equality. The claim follows. (c) Using Theorem 3.1, we have U0 = B0−U0+ and B0− = N0− + U00, so V (λ) = Uˆ 0 = ˆ 0− ˆ 0− ˆ B = N +F1. The generators in the definition of the left ideal Iλ are homogeneous relative to the Z-grading (Proposition 3.2(b)), so this ideal is Z-graded, implying that V (λ) inherits ˆ 0− P ˆ a Z-grading as well. We have N ⊆ i<0 V (λ)[i], while F1 ⊆ V (λ)[0], so the sum is direct. (d) Let M be a submodule of V (λ) and assume that M * Nˆ 0−. By part (c) we have xˆ + 1ˆ ∈ M for some x ∈ N0−. By part (b) and Lemma 7.3(b), xk annihilates V (λ) for some positive integer k, implying that y(x + 1) acts as the identity map on V (λ), where Pk−1 i 0 ˆ ˆ ˆ y = i=0 (−x) ∈ U . Therefore, 1 = y(x + 1)1 = y(ˆx + 1) ∈ M and we conclude that M = V (λ). It follows that every proper submodule of V (λ) is contained in Nˆ 0− so that the ˆ 0− sum W of all such is the unique maximal submodule of V (λ). Moreover, W ⊆ N . By the proposition, the U0-module L(λ) = V (λ)/W is the unique simple quotient of V (λ), and it has character χ. Theorem 7.5. SIMPLICITY OF KAC MODULES 27 00 (a) For each λ ∈ Xχ(U ), we have L(λ) ∈ S(λ). 00 (b) The set {L(λ)}λ∈Xχ(U ) is a complete set of representatives of the isomorphism classes of those finite-dimensional simple U0-modules that have character χ and that are the sum of their weight spaces. 00 Proof. (a) Let λ ∈ Xχ(U ). By Proposition 7.4(a), V (λ) is finite-dimensional, implying its quotient L(λ) is as well, and V (λ) is the sum of its weight spaces, implying the same holds for its maximal submodule W (Lemma 4.5) and hence for the quotient V (λ)/W = L(λ). Let S be a simple B0+-submodule of L(λ). Since L(λ) has character χ, so does S. By Lemma 7.3, N0+ acts nilpotently on S. Now S is the sum of its weight spaces, so there exists a nonzero weight vectorv ¯ in S of weight, say, λ0 ∈ X(U00). By Lemma 4.6, S = Fv¯ andv ¯ is a maximal vector. Assume that λ0 6= λ. We havev ¯ = v + W for some v ∈ V (λ), and we may assume that v is a weight vector of weight λ0. Now v ∈ Nˆ 0−+˙ F1ˆ by Proposition 7.4(c) and, since 1ˆ has weight λ 6= λ0, it must be the case that v ∈ Nˆ 0−. By Theorem 3.1, U0 = U0−B0+ = U0−U00 + U0−N0+, and N0+v ⊆ W sincev ¯ = v + W is a maximal vector. So using first that v∈ / W (sincev ¯ 6= 0) and that W is maximal, we get V (λ) = U0v + W ⊆ U0−v + W ⊆ Nˆ 0−, where the last step uses the fact that W ⊆ Nˆ 0− (Proposition 7.4(d)). This contradicts that 1ˆ ∈ V (λ) \ Nˆ 0−. We conclude that λ0 = λ. Therefore, L(λ) ∈ S(λ). 00 0 (b) It has already been observed that for λ ∈ Xχ(U ), the U -module L(λ) is finite- dimensional, is simple, is the sum of its weight spaces, and has character χ. Let M be a finite-dimensional simple U0-module having character χ and assume that M is the sum of its weight spaces. Regarded as B0+-module M has a simple submodule, which, arguing as in the proof of part (a), equals Fv with v a maximal vector of weight λ for some λ ∈ X(U00). 00 0 Since M has character χ, it follows that λ ∈ Xχ(U ). The homomorphism U → M given by x 7→ xv induces a homomorphism V (λ) → M, which must be surjective since it is nonzero and M is simple. Since V (λ) has unique simple quotient L(λ), we get M ∼= L(λ). 00 Finally, the modules L(λ), λ ∈ Xχ(U ) are pairwise nonisomorphic by the argument in the proof of Theorem 6.3(b). 00 λi We say that λ ∈ X(U ) is integral if for each i ∈ [1, m + n] we have λ(Ki) = ±q for some integer λi. Put l0 = l if l is odd and l0 = l/2 if l is even. Recall that the assumption throughout this l l section is that χ(Ea) = 0 and χ(Fa) = 0 for every a ∈ I0. 00 Corollary 7.6 (cf. [6, Corollary, p. 1248]). Let λ ∈ Xχ(U ). 28 −1 e (a) K(L(λ)) is simple if and only if for each (i, j) ∈ I1 with λ(Ki)λ(Kj) = ±q for some integer e, we have e 6≡ i + j − 2m − 1 (mod l0). λi (b) Assume that λ is integral, so that for each i ∈ [1, m + n] we have λ(Ki) = ±q for some integer λi. Then K(L(λ)) is simple if and only if λi − λj 6≡ i + j − 2m − 1 0 (mod l ) for every (i, j) ∈ I1. Proof. (a) By Corollary 5.5, K(L(λ)) is simple if and only if for each (i, j) ∈ I1 with −1 e λ(Ki)λ(Kj) = ±q for some integer e, we have qe 6= ±qi+j−2m−1. Since q is a primitive lth root of unity, the statement holds. −1 λi−λj (b) With the stated assumption, we have λ(Ki)λ(Kj) = ±q for each (i, j) ∈ I1, so this part follows from part (a). References [1] James E. Humphreys. Introduction to Lie algebras and representation theory, volume 9 of Graduate Texts in Mathematics. Springer-Verlag, New York-Berlin, 1978. Second printing, revised. [2] Jens Carsten Jantzen. Lectures on quantum groups, volume 6 of Graduate Studies in Mathematics. Amer- ican Mathematical Society, Providence, RI, 1996. [3] V. G. Kac. Lie superalgebras. Advances in Math., 26(1):8–96, 1977. [4] George Lusztig. Quantum groups at roots of 1. Geom. Dedicata, 35(1-3):89–113, 1990. [5] Chaowen Zhang. On the quantum superalgebra Uq(gl(m, n)) and its representations at roots of 1. Science Press, Beijing, 2016. https://arxiv.org/pdf/1211.4115v3.pdf. [6] R. B. Zhang. Finite-dimensional irreducible representations of the quantum supergroup Uq(gl(m/n)). J. Math. Phys., 34(3):1236–1254, 1993. Randall R. Holmes, Department of Mathematics and Statistics, Auburn University, Auburn AL, 36849, USA, [email protected] Chaowen Zhang, Department of Mathematics, China University of Mining and Technology, Xuzhou, 221116, Jiang Su, P. R. China, chaowen163 [email protected]