Simple Weight Q2(ℂ)-Supermodules

U.U.D.M. Project Report 2009:12

Simple weight q2( )-supermodules Ekaterina Orekhova ℂ

Examensarbete i matematik, 30 hp Handledare och examinator: Volodymyr Mazorchuk Juni 2009

Department of Mathematics Uppsala University

Simple weight q2(C)-supermodules

Ekaterina Orekhova

June 6, 2009

1 Abstract

The first part of this paper gives the definition and basic properties of the queer Lie superalgebra q2(C). This is followed by a complete classification of simple h-supermodules, which leads to a classification of all simple highest and lowest weight q2(C)-supermodules. The paper also describes the structure of all Verma supermodules for q2(C) and gives a classification of finite-dimensional q2(C)-supermodules.

2 Contents

1 Introduction 4 1.1 General deﬁnitions ...... 4 1.2 The queer Lie superalgebra q2(C)...... 5

2 Classiﬁcation of simple h-supermodules 6

3 Properties of weight q-supermodules 11

4 Universal enveloping algebra U(q) 13

5 Simple highest weight q-supermodules 15 5.1 Deﬁnition and properties of Verma supermodules ...... 15 5.2 Structure of typical Verma supermodules ...... 27 5.3 Structure of atypical Verma supermodules ...... 28

6 Finite-Dimensional simple weight q-supermodules 31

7 Simple lowest weight q-supermodules 31

3 1 Introduction

Signiﬁcant study of the characters and blocks of the category O of the queer Lie superalgebra qn(C) has been done by Brundan [Br], Frisk [Fr], Penkov [Pe], Penkov and Serganova [PS1], and Sergeev [Se1]. This paper focuses on the case n = 2 and gives explicit formulas, diagrams, and classiﬁcation results for simple lowest and highest weight q2(C) supermodules.

1.1 General deﬁnitions

Definition 1.0.1. A vector space V over a field k is said to be a super vector space if V is Z/2Z-graded, that is if V = V0 ⊕ V1 where we identify Z/2Z with {0, 1}.A super subspace of V is a super vector space W = W0 + W1 such that Wi is a subspace of Vi for i = 0, 1. A homogeneous vector v ∈ V has degree (parity) 0 or 1, denoted by |v|. Element v is said to be even if |v| = 0 and odd if |v| = 1. The vector space V is said to be even if V = V0 and odd if V = V1. Given two super vector spaces V and W , a linear map f : V → W is said to be homogeneous of degree d ∈ Z/2Z if f(Vi) ⊆ W|i+d| for i = 0, 1. The degree of a homogeneous map f is denoted by |f|. The map f is said to be even if f is homogeneous of degree 0 (that is, |f| = 0) and odd if f is homogeneous of degree 1 (|f| = 1). We define the super vector space Homk(V,W ) as the direct sum of the homogeneous vector spaces

Homk(V,W )d = {f : V → W |f is homogeneous of degree d} where d ∈ Z/2Z. Definition 1.0.2. The super vector spaces and homogeneous mappings defined above form a category. If we restrict the homomorphisms to even mappings, we obtain an abelian category, denoted by sV ec(k). Definition 1.0.3. The parity change functor Π: sV ec(k) → sV ec(k) is defined as follows: for V ∈ sV ec(k) and f : V → W ∈ sV ec(k), we take (Π(V ))d = V|d+1|, where d ∈ Z/2Z, and Π(f) = f, as mappings. Note that the underlying field k is considered to be even and hence Π(k) is an odd super vector space, of dimension one.

Definition 1.0.4. A superalgebra over a field k is a k-algebra A with a direct sum decomposition A = A0 ⊕A1 together with a bilinear multiplication that respects the Z/2Z grading, that is A|i|A|j| ⊆ A|i+j|. Definition 1.0.5. Let A be a superalgebra. A left A-supermodule is a super vector space V which is a left A-module in the usual sense such that AθVτ ⊆ V|θ+τ| for θ, τ ∈ Z/2Z. Right A-supermodules are defined analogously. Definition 1.0.6. A Lie superalgebra g over a field k is a super vector space over k together with the Lie bracket [·, ·]: g × g → g, which satisfies the

4 following: 1. Bilinearity: For all scalars a, b ∈ k and all elements x, y, z ∈ g, [ax + by, z] = a[x, z] + b[y, z] and [z, ax + by] = a[z, x] + b[z, y].

2. Super skew symmetry: For all homogeneous elements x, y ∈ g,

[x, y] = −(−1)|x||y|[y, x].

3. The super Jacobi identity: For all homogeneous elements x, y, z ∈ g,

(−1)|z||x|[x, [y, z]] + (−1)|x||y|[y, [z, x]] + (−1)|y||z|[z, [x, y]] = 0.

Deﬁnition 1.0.7. Let g be a Lie superalgebra. A g-supermodule is a super vector space V together with a map g × V → V such that giVj ⊆ V|i+j| and

[x, y]v = (xy − (−1)|x||y|yx)v for all homogeneous elements x, y ∈ g and v ∈ V .

Deﬁnition 1.0.8. The Lie algebra gl2(C) is the complex vector space a b gl ( ) = : a, b, c, d ∈ 2 C c d C where the Lie bracket [·, ·] is the usual commutator. That is, given two matrices A, B,[A, B] = AB − BA.

1.2 The queer Lie superalgebra q2(C)

Deﬁnition 1.0.9. The queer Lie superalgebra q2(C) consists of block matrices of the form AB BA where A, B ∈ gl2(C), and the supercommutator [·, ·] deﬁned for homogeneous elements X,Y ∈ q2(C) as follows: [X,Y ] := XY − (−1)|X||Y |YX.

The even subspace of q, denoted q0, consists of the block matrices with B = 0. The odd subspace of q, denoted q1, consists of the block matrices with A = 0. From now on, we will denote the queer Lie superalgebra simply by q.

0 Let eij, 1 ≤ i, j ≤ 2, be the basis elements in q0, deﬁned as Eij 0 0 Eij

5 where Eij ∈ gl2(C) is the matrix with entries (δij). Similarly, we deﬁne the 1 basis elements eij, 1 ≤ i, j ≤ 2 in q1. Together, these eight elements form a basis of the whole superalgebra q. In this standard basis, the supercommutator [·, ·] has the following form:

σ θ σ+θ σθ σ+θ [eij, ekl] = δjkeil − (−1) δilekj , (1) where σ, θ ∈ Z/2Z and 1 ≤ i, j, k, l ≤ 2. Remark 1.0.1. Note that the supercommutator [A, B] is the usual commutator AB − BA when both entries are even, or when one is even and the other is odd. However, in the case where both entries are odd the minus sign is negated, and hence the supercommutator becomes AB + BA. This is known as the anti-commutator.

To simplify notation, we will denote the basis elements as follows:

0 0 0 0 e = e12, f = e21, h1 = e11, h2 = e22 1 1 1 1 e = e12, f = e21, h1 = e11, h2 = e22

By a direct calculation, we get the following Cayley table for the supercommutator [·, ·] in this basis:

[·, ·] h1 e f h2 h1 e f h2

h1 0 e −f 0 0 e −f 0 e −e 0 h1 − h2 e −e 0 h1 − h2 e f f −(h1 − h2) 0 −f f −(h1 − h2) 0 −f h2 0 −e f 0 0 −e f 0 (2) h1 0 e −f 0 2h1 e f 0 e −e 0 h1 − h2 e e 0 h1 + h2 e f f −(h1 − h2) 0 −f f h1 + h2 0 f h2 0 −e f 0 0 e f 2h2

Remark 1.0.2. Let A, B ∈ {e, f, h1, h2, e, f, h1, h2}. If the parities of A and B are the same, then [A, B] is even. If the parities of A and B are diﬀerent, then [A, B] is odd.

2 Classiﬁcation of simple h-supermodules

Deﬁnition 2.0.10. The Cartan Lie subsuperalgebra of q, denoted by h, is the linear span of h1, h2, h1, and h2.

Deﬁnition 2.0.11. The Cartan subalgebra of gl2(C), denoted by h0, is the linear span of h1 and h2.

6 ∗ Remark 2.0.3. Let h0 denote the dual space of h0 and let 1, 2 be the dual ∗ basis of h1 and h2. Then any f ∈ h0 can be written as f = a1 + b2 for ∗ ∼ 2 2 some a, b ∈ C. Hence, h0 = C , so we can think of the elements in C as linear functionals on h0, via

(λ1, λ2)(ah1 + bh2) = aλ1 + bλ2

2 for (λ1, λ2) ∈ C and a, b ∈ C.

From the previous section, we know that the basis elements h1, h2, h1, h2 satisfy the following relations:

[·, ·] h1 h2 h1 h2

h1 0 0 0 0 h2 0 0 0 0 (3) h1 0 0 2h1 0 h2 0 0 0 2h2

Remark 2.0.4. Note that h1, h2 commute with all other basis elements 2 and that hi = hi for i = 1, 2. Deﬁnition 2.0.12. An h-supermodule is a super vector space V together with two even linear operators H1,H2 and two odd linear operators H1, H2 that satisfy the relations in Table (3).

Remark 2.0.5. This explicit deﬁnition is equivalent to the general deﬁni- tion of a supermodule given by 1.0.7.

Example 2.0.1. Let V (0, 0) = C and let H1, H2, H1, H2 act as zero on V (0, 0). Then V (0, 0) is a one-dimensional (and hence simple) h-supermodule, called the trivial supermodule.

2 Example 2.0.2. Let (a, b) ∈ C \{0} and define V (a, b) as the two-dimensional super vector space with Cx as the even part and Cy as the odd part. Let H , H act on V (a, b) as the scalars a, b respectively and define √ 1 2 √ √ √ H x = ay, H y = ay, H x = −i by, H y = i bx, where for a complex 1 1 √ 2 2 number z we take z to be the principal square root of z. (That√ is, the √argument of z is restricted to the interval [0, π)). In particular, 1 = 1 and −1 = i.) This defines a simple, two-dimensional h-supermodule, shown in the diagram below:

√ √ H1= a, H2=−i b

H1=a, H2=b • % • H1=a, H2=b (4) 9 e e √ √ H1= a, H2=i b

7 Remark 2.0.6. Note that h1, h2 commute with all four operators in h and hence are elements in Z(h), the center of h. In general, given a ﬁnitely- generated algebra A over C and a simple A-module V , any element in the center of A acts on V as a scalar. Thus, given a simple h-supermodule V , the operators H1,H2 act on V as scalars, say λ1 and λ2 respectively. Deﬁnition 2.0.13. An even (resp. odd) h-supermodule homomorphism (or an h-homomorphism) is an even (resp. odd) linear map Φ : V → W between two h-supermodules such that the following diagram is commutative for all X ∈ {H1,H2, H1, H2}: XV V / V

Φ Φ

XW W / W

In other words, ΦXV = XW Φ for all X ∈ {H1,H2, H1, H2}.

Lemma 2.0.1. Let a, b, c, d ∈ C. The h-supermodules V (a, b) and V (c, d) are isomorphic if and only if (a, b) = (c, d). Proof. (⇐) Clear. (⇒) Note that V (0, 0) is one-dimensional and V (a, b) is two-dimensional for (a, b) 6= (0, 0). Thus, V (0, 0) =∼ V (a, b) implies that a = 0 and b = 0. Now suppose that V (a, b) =∼ V (c, d), where (a, b), (c, d) 6= (0, 0). Let Φ : V (a, b) → V (c, d) be an h-isomorphism. Then given any nonzero element x ∈ V (a, b), we have ΦH1x = H1Φx, which implies that aΦ(x) = cΦ(x). Since Φ is an isomorphism, it is in particular injective and hence Φ(x) 6= 0. Thus, a = c. Similarly, by considering the action of H2 we deduce that b = d.

Lemma 2.0.2. Let V be an h-supermodule. Then the function P : V → Π(V ) which sends v 7→ v is an odd h-supermodule isomorphism.

Proof. Let V be an h-supermodule, V = V0 ⊕ V1. Let W := Π(V ) and for 0 v ∈ V denote the image Πv in W by v . It is clear that H1 and H2 commute with P since these act as scalars. Now let v0 ∈ V0 and let v1 = H1v0 ∈ V1. Then, 0 0 P H1v0 = P v1 = v1 = H1v0 = H1P v0.

Hence, P commutes with H1 on V0. Similarly, P commutes with H1 on V1 and hence on all of V . The proof for H2 is identical. Thus P is an odd h-homomorphism and, in fact, an isomorphism because P is bijective by deﬁnition.

2 ∼ Proposition 2.1. Let (a, b) ∈ C . Then V (a, b) = Π(V (a, b)) if and only if a = 0 and b 6= 0 or a 6= 0 and b = 0.

8 Proof. First, note that since V (0, 0) is an even super vector space and Π(V (0, 0)) is an odd super vector space, these two are not isomorphic. Now, 2 ∼ let (a, b) ∈ C \(0, 0) and suppose V (a, b) = Π(V (a, b)). Let x, y be the basis elements of the even and odd parts of V (a, b) respectively, and y0, x0 be the basis elements of the even and odd parts of Π(V (a, b)) respectively. Then any isomorphism Φ : V (a, b) → Π(V (a, b)) must send x 7→ cy0 and y 7→ dx0 for some nonzero c, d ∈ C. Now, since Φ is an even h-homomorphism, we have ΦH1x = H1Φx, which implies that either a = 0 or c = d. Also,

ΦH2x = H2Φx implies that either b = 0 or c = −d. Suppose a = 0. Then b must be nonzero, and hence c = −d. Thus, since the above conditions are suﬃcient as well as necessary for Φ to be an h-isomorphism, the map Φ : V (0, b) → Π(V (0, b)) sending x 7→ y0 and y 7→ −x0 works. Similarly, if b = 0, then a must be nonzero and hence c = d. So the map Φ : V (a, 0) → Π(V (a, 0)) sending x 7→ y0 and y 7→ x0 is an h-isomorphism.

Deﬁnition 2.1.1. Let V be a h-supermodule. A super subspace W ⊂ V is said to be a subsupermodule of V if it is invariant with respect to the action of H1, H2, H1 and H2, that is

H1W ⊂ W, H2W ⊂ W, H1W ⊂ W, H2W ⊂ W. (5)

Lemma 2.1.1. Let V be a simple h-supermodule and assume that H1,H2 ∼ ∼ act on V as λ1, λ2 respectively. Then, V = V (λ1, λ2) or V = Π(V (λ1, λ2)). Proof. Let V be a simple h-supermodule. Then V is nonzero, so assume ﬁrst that there exists a nonzero vector v ∈ V0. If both H1 and H2 act trivially, then Cv is closed under the actions of all four operators and hence V is a one-dimensional supermodule. Moreover, using the relations we see that

2 0 = H1v = H1v = λ1v which implies that λ1 must be zero. Similarly, we see that λ2 must be zero. Hence, V is isomorphic to V (0, 0). Now suppose that H1v = 0, but H2v is nonzero. Let w = H2v. Then,

2 0 = H1v = H1v = λ1v which implies that λ1 = 0. Also,

H1H2v = −H2H1v = 0

9 Thus, the subsupermodule generated by v is two-dimensional, with basis v for the even part and w for the odd part. It is clear that this is simple, and ∼ hence all of V . It follows easily that V = V (0, λ2). Similarly, we see that ∼ V = V (λ1, 0) when H2v = 0 and H1v 6= 0. Finally, assume that both H1v and H2v are nonzero. Then from the relations (3), it follows that v generates a subsupermodule, call it S, with at most four distinct elements: v, H1v, H2v, and H2H1v. If v and H2H1v are linearly dependent, then so are H1v and H2v, so S is a two-dimensional simple subsupermodule and hence all of V . If v and H2H1v are independent, then so are H1v and H2v. In this case, S is not simple because√ it has a two-dimensional subsupermodule generated by the element z := −λ1λ2v+ H2H1v. But this is a contradiction, since S ⊆ V and V is simple. Thus, 0 0 0 0 V = Cx + Cy , where x = v, y = H1v, and H1v, H2v are dependent. q By comparing H z and H z, we see that we must have H v = − λ1 H v. 1 2 1 λ2 2 Thus, the actions of H1, H2 on V can be summarized as follows: r 0 0 0 0 0 λ2 0 0 p 0 H1x = y , H1y = λ1x , H2x = − − y , H2y = −λ1λ2x . λ1

0 Direct calculation shows that the map Φ : V → V (λ1, λ2) sending x 7→ x 0 √ and y 7→ λ1y is an h-isomorphism. The case where v ∈ V1 is nonzero is completely analogous and, due to Lemma 2.0.2, the supermodules occurring in this case are exactly Π-images of the supermodules in the ﬁrst (even) case.

Theorem 2.1.1. (Classiﬁcation of simple h-supermodules) Let V be a simple h-supermodule and assume that H1,H2 act on V as λ1, λ2 respectively. ∼ ∼ Then, V = V (λ1, λ2) or V = Π(V (λ1, λ2)). Explicitly, if we let x be the basis element of the even subspace of V (possibly zero), we have

(i) If (λ1, λ2) = (0, 0), then ( V (0, 0) , iﬀ V is an even super vector space V =∼ Π(V (0, 0)) , iﬀ V is an odd super vector space

(ii) If (λ1, λ2) 6= (0, 0), then ( √ √ V (λ1, λ2) , iﬀ H1H2x = −i λ1 λ2x V =∼ √ √ Π(V (λ1, λ2)) , iﬀ H1H2x = i λ1 λ2x

∼ Moreover, V (λ1, λ2) = Π(V (λ1, λ2)) if and only if λ1 = 0, λ2 6= 0 or λ1 6= 0, λ2 = 0.

10 ∼ ∼ Proof. By Lemma 2.1.1, either V = V (λ1, λ2) or V = Π(V (λ1, λ2)). Case (i) is clear. Now suppose (λ1, λ2) 6= (0, 0). Then V must be two-dimensional with some basis x for the even subspace. Note that if the right hand side of the statement holds true, then the left clearly follows. (For instance, if √ √ 0 0 H1H2x = −i λ1 λ2x, then the map sending x to x , where x is the basis of the even subspace of V (λ1, λ2), is an h-isomorphism.) Now consider the ∼ reverse direction. Suppose ﬁrst that V = V (λ1, λ2). Then there exists an even h-isomorphism Φ : V → V (λ1, λ2). Since Φ is even, it must map x to some scalar multiple of x0. Without loss of generality, we can assume that this scalar is 1, so that Φ(x) = x0. Now, since Φ is an h-homomorphism, we have the following:

p p 0 Φ(H1H2x) = H1H2(Φx) = −i λ1 λ2x . (6)

Since H1 and H2 are both odd maps and the even subspace of V is one- dimensional, we know that H1H2x must√ √ be some scalar multiple of x. Fi- nally, (6) forces this scalar to be −i λ1 λ2, as required. The case where ∼ V = Π(V (λ1, λ2)) is analogous, which proves (ii). The last claim follows from Proposition 2.1.

Remark 2.1.1. In Theorem 2.1.1, we could instead consider the basis element y of the odd subspace of V . This yields the following classiﬁcation for V in the case (λ1, λ2) 6= (0, 0):

( √ √ V (λ1, λ2) , iﬀ H1H2y = i λ1 λ2y V =∼ √ √ Π(V (λ1, λ2)) , iﬀ H1H2y = −i λ1 λ2y

3 Properties of weight q-supermodules

Definition 3.0.2. A q-supermodule is a super vector space V together with four even linear operators, E, F , H1, H2, and four odd linear operators, E, F , H1, H2, that satisfy the relations in Table (2). Remark 3.0.2. This explicit definition is equivalent to the general definition of a supermodule of a Lie superalgebra given by 1.0.7.

Example 3.0.1. Consider the four-dimensional complex super vector space M with basis v0, v1 for the even part and v0, v1 for the odd part. Deﬁne the

11 eight linear operators required for a q-supermodule as follows:

H1=1,H2=0 v0 l , v0 (7) T _ H1=1,H2=0 > K

E E F E E F F F

H1=0,H2=1 v1 l , v1 H1=0,H2=1 All actions not explicitly shown in diagram (7) are zero and those that are shown have coeﬃcient 1 unless otherwise noted. Direct calculation shows that these linear operators satisfy the relations in Table (2) and hence M is a q-supermodule. In some sense, M is the natural q-supermodule because q is an algebra of 4x4 matrices, which act naturally on 4x1 column vec- 4 tors. The basis elements e1, e2, e3, e4 of C correspond to the basis elements v0, v1, v0, v1 respectively. Deﬁnition 3.0.3. An even q-supermodule homomorphism (or a q-homomorphism) is an even linear map Φ : V → W between two q-supermodules such that the following diagram is commutative for all X ∈ {E,F,H1,H2, E, F, H1, H2}: XV V / V

Φ Φ

XW W / W

In other words, ΦXV = XW Φ for all X ∈ {E,F,H1,H2, E, F, H1, H2}. Deﬁnition 3.0.4. A q-supermodule is said to be a weight supermodule if L V = 2 V , where V is the weight space of V corresponding to λ = λ∈C λ λ (λ1, λ2) deﬁned as follows:

Vλ := {x ∈ V : H1x = λ1x, H2x = λ2x}

For a weight supermodule V , the support of V is the set

2 supp(V ) = {λ ∈ C : Vλ 6= 0}.

A weight is an element λ ∈ supp(V ). A weight vector for λ is any nonzero vector v ∈ Vλ. Deﬁnition 3.0.5. Let V be a q-supermodule. A super subspace W ⊆ V is said to be a subsupermodule of V if it is invariant with respect to the action of X for all X ∈ {E,F,H1,H2, E, F, H1, H2}. That is,

XW ⊆ W ∀X ∈ {E,F,H1,H2, E, F, H1, H2}

12 Lemma 3.0.2. (i) Every subsupermodule of a weight supermodule is a weight supermodule.

(ii) Every quotient of a weight supermodule is a weight supermodule.

(iii) Any direct sum of weight supermodules is a weight supermodule

Proof. Trivial.

Lemma 3.0.3. Let V be a weight q-supermodule and let λ = (λ1, λ2) be a weight. Then  V , if X ∈ {F, F }  (λ1−1,λ2+1) XV(λ1,λ2) ⊆ V(λ1,λ2) , if X ∈ {H1,H2, H1, H2}  V(λ1+1,λ2−1) , if X ∈ {E, E}

Proof. The claim is obvious for H1 and H2. Let v ∈ Vλ and consider Ev.

H1(Ev) = [H1,E]v + EH1v = Ev + λ1(Ev)

H2(Ev) = [H2,E]v + EH2v = −Ev + λ2(Ev)

Hence Ev ∈ V(λ1+1,λ2−1). The other cases are proved similarly.

4 Universal enveloping algebra U(q)

Deﬁnition 4.0.6. Let R be the free associative algebra with generators e, f, h1, h2, e, f, h1, h2. Let I be the ideal of R generated by the relations (2) of q. Then the universal enveloping algebra U(q) of q is the quotient R/I. Abusing notation, we usually identify the elements of R with their images in U(q).

Remark 4.0.3. Note that U(q) has the structure of a superalgebra. We take elements e, f, h1, h2 to be even and elements e, f, h1, h2 to be odd. A general element in U(q) has the form x1x2 . . . xn where xi ∈ {e, f, h1, h2, e, f, h1, h2}. Let m be the number of xi’s which are odd. Then we say that element x1x2 . . . xn is even if m is even, and odd otherwise. Remark 4.0.4. Let A be a superalgebra. Then one can deﬁne on A the structure of a Lie superalgebra by taking the Lie bracket [·, ·] to be the supercommutator [a, b] := ab−(−1)|a||b|ba for homogeneous elements a, b, ∈ A. This Lie superalgebra, denoted A(−), is called the underlying Lie superalgebra of A.

13 Deﬁnition 4.0.7. A Lie superalgebra homomorphism Φ : A → B is a linear map between two Lie superalgebras A and B such that

Φ([x, y]) = [Φ(x), Φ(y)] for all x, y ∈ A. Lemma 4.0.4. (i) There is a unique linear map ε : q → U(q) satisfying

ε(e) = e, ε(f) = f, ε(h1) = h1, ε(h2) = h2,

ε(e) = e, ε(f) = f, ε(h1) = h1, ε(h2) = h2.

(ii) The map ε is a Lie superalgebra homomorphism from q to U(q)(−).

Proof. The statement (i) follows from the fact that {e, f, h1, h2, e, f, h1, h2} is a basis for q. The statement (ii) follows from the relations, the deﬁnition of U(q), and the deﬁnition of the underlying Lie superalgebra.

The following theorems are very useful because they give the connection between U(q)-supermodules and q-supermodules. The proofs are standard and therefore not given here. To see analogous proofs for sl2-modules, see [Ma]. Theorem 4.0.2. (Universal Property of U(q)) Let A be any associative superalgebra and φ : q → A(−) be any homomorphism of Lie superalgebras. Then there exists a unique homomorphism φ : U(q) → A of associative superalgebras such that φ = φ ◦ ε. That is, the following diagram commutes:

φ q / A p7 p p ε p p p p φ U(q)

Proposition 4.1. (Uniqueness of U(q)) Let U(q)0 be another associative algebra such that there exists a ﬁxed homomorphism ε0 : q → (U(q)0)(−) of Lie superalgebras with the universal property as described in Theorem 4.0.2. Then U(q)0 is canonically isomorphic to U(q). Theorem 4.1.1. (i) Let V be a q-supermodule given by the Lie superalgebra homomorphism φ : q → L(V )(−). Then the homomorphism φ : U(q) → L(V ), given by the universal property, endows V with the canonical structure of a U(q)-supermodule. (ii) Let V be a U(q)-supermodule given by the superalgebra homomorphism ψ : U(q) → L(V ). Then the composition ψ ◦ ε : q → L(V )(−) is a Lie superalgebra homomorphism which endows V with the canonical structure of a q-supermodule.

14 (iii) The operations in (i) and (ii) are mutually inverse. Theorem 4.1.2. (Poincar´e-Birkhoﬀ-Witt Theorem) The following set

i j k l θ τ δ σ {f h1h2e e f h1h2 : i, j, k, l ∈ N0, θ, τ, δ, σ ∈ {0, 1}} forms a basis of U(q).

Remark 4.1.1. Note that U(q) = U(n−)⊗C U(h)⊗C U(n+) where n− is the linear span of f and f and n+ is the linear span of e and e. The universal enveloping algebras U(n−) and U(n+) are deﬁned analogously to U(q).

5 Simple highest weight q-supermodules

5.1 Deﬁnition and properties of Verma supermodules

Deﬁnition 5.0.1. If V is a weight q-supermodule and µ = (µ1, µ2) ∈ supp(V ), then µ is said to be a highest weight if (µ1 + 1, µ2 − 1) ∈/ supp(V ). The corresponding weight space Vµ is then referred to as the highest weight space and any nonzero vector in Vµ as a highest weight vector.A q-supermodule generated by a highest weight space is called a highest weight q-supermodule.

Definition 5.0.2. Let B = U(h)⊗CU(n+). Let V be a simple h-supermodule and define n+V = 0. Then the Verma supermodule M(V ) is defined as

M(V ) := U(q) ⊗B V Remark 5.0.2. In the construction of M(V ) above, we start with a simple h-supermodule V and then induce it up to a highest-weight q-supermodule by requiring that the action of e, e on V is trivial.

By deﬁnition, M(V ) is isomorphic to the tensor product U(q)/B ⊗C V as a super vector space. This is in turn isomorphic to the super vector 2 2 3 3 space with basis 1, f, f, f , f f, f , f f,... . Recall that V (λ1, λ2) has two-dimensional basis {x, y} if (λ1, λ2) 6= (0, 0) and one-dimensional basis {x} if (λ1, λ2) = (0, 0). Thus, the Verma supermodule M(V (λ1, λ2)) has basis x, y, fx, fy, fx, fy, ... if (λ1, λ2) 6= (0, 0) and basis x, fx, fx, ffx, ... if (λ1, λ2) = 0. In order to simplify notation and generalize the result, we relabel the basis elements as follows:

v0 = x, v0 = y n vn = f v0 for n = 0, 1, 2,... n vn = f v0 for n = 0, 1, 2,...

wn = fvn−1 for n = 1, 2,...

15 Now, since M(V ) is a q-supermodule, there must exist four even linear operators on M(V ), E,F , H1, H2, and four odd linear operators on M(V ), E,F , H1, H2, that satisfy the relations in Table (2). By using these relations and the deﬁnition of M(V ), we can calculate the explicit actions of these linear operators on the basis elements.

Example 5.0.1. Suppose that V = V (λ1, λ2) where (λ1, λ2) 6= (0, 0). In this example, we show how to calculate the action of E on the basis element vn, for n ∈ N0. By deﬁnition of M(V ), we know that E and E act trivially on v0 and v0. The actions of H1, H2, H1, H2 are given by Diagram (4). In order to calculate the action of E on v1, we use the following process:

Ev1 = EF v0 , by deﬁnition of v1

= [E,F ]v0 + F Ev0

= (H1 − H2)v0 + F Ev0 , using Table (2)

= (λ1 − λ2)v0 , using Diagram (4)

We repeat this process for v2 and v3 to see the pattern, and then use mathematical induction to prove the general result for n ∈ N0. Let x ∈ v, v, w, w. Then, the actions of the linear operators are given by

F (xn) = xn+1 for n ∈ N0

H1(xn) = (λ1 − 1)xn for n ∈ N0 H2(xn) = (λ2 + 1)xn for n ∈ N0

( ( 0, if n = 0 0, if n = 0 E(vn) = E(vn) = anvn−1, if n > 0 anvn−1, if n > 0 where an = n(λ1 − λ2 − n + 1). ( ( αv0, if n = 1 βv0, if n = 1 E(wn) = E(wn) = αvn−1 + bnwn−1, if n > 1 βvn−1 + bnwn−1, if n > 1 √ √ √ √ where bn = (n − 1)(λ1 − λ2 − n), α = λ1 − i λ2, and β = λ1 + i λ2.

F (vn) = wn+1F (wn) = 0 F (vn) = wn+1F (wn) = 0

(√ (√ λ1v0, if n = 0 λ1v0, if n = 0 H1(vn) = √ H1(vn) = √ λ1vn − nwn, if n > 0 λ1vn − nwn, if n > 0

p p H1(wn) = − λ1wn + vn H1(wn) = − λ1wn + vn

16 ( √ ( √ −i λ2v0, if n = 0 i λ2v0, if n = 0 H2(vn) = √ H2(vn) = √ −i λ2vn + nwn, if n > 0 i λ2vn + nwn, if n > 0

p p H2(wn) = −i λ2wn + vn H2(wn) = i λ2wn + vn

  0, if n = 0 0, if n = 0   E(vn) = βv0, if n = 1 E(vn) = αv0, if n = 1   nβvn−1 − n(n − 1)wn−1, if n > 1 nαvn−1 − n(n − 1)wn−1, if n > 1 ( ( cv0, if n = 1 cv0, if n = 1 E(wn) = E(wn) = cvn−1 − α(n − 1)wn−1, if n > 1 cvn−1 − β(n − 1)wn−1, if n > 1 where c = λ1 + λ2.

The result is a rather complicated but symmetric q-supermodule. The following diagram shows the actions of E, F , and F on the basis elements of M(V (λ1, λ2)). Here F is given by the dotted arrows, F is given by the the dashed arrows, and E is given by the solid arrows. Only the coeﬃcients for E are shown.

< b 0 0

v0 v0 B ÀT T j j> \ AA T T j j }} a1 A T T j } a1 α AA T j j }} A TjT }} β A j j T T } v1 w1 tj * w1 v1 B ÀT T I U j j> \ AA T T j j }} a2 A b T T j b } a2 α AA2 T j j 2 }} A TjT }} β A j j T T } v2 w2 tj * w2 v2 B ÀT T I U j j> \ AA T T j j }} a3 A b T T j b } a3 α AA3 T j j 3 }} A TjT }} β A j j T T } v3 w3 tj * w3 v3

......

The next diagram shows the action of H1,H2, H2, and H2. The dotted arrows represent the action of H1 and H2, which is constant on each ”level” of the diagram but shown here only for the basis elements vn, vn, for n ∈ N0. The action of H1 and H2 is given by the solid arrows, shown here without coeﬃcients.

17 ' H1=λ1, H2=λ2 v0 v H1=λ1, H2=λ2 7 g 0 h

$' #$ H1=λ1−1, H2=λ2+1 v1 w1 w v H1=λ1−1, H2=λ2+1 7 cd dg 1 1 h

$' #$ H1=λ1−2, H2=λ2+2 v2 w2 w v H1=λ1−2, H2=λ2+2 7 cd dg 2 2 h

......

Finally, the diagram below shows the action of E on the basis elements, without coeﬃcients.

8 f 0 0 v0 v0 iRjVRVRVV hhlhl54 RRVRVVV hhhlhll RRRVVVV hhhhlll RRR VVVV hhhh lll RRhRhhVhVVlVll hhhh RRRlll VVVV hhhh lll RRR VVVV hhhh lll RR VVVV v1 jVh w1 w1 4 v1 iRRVRVVV dJiRRR ll:5 hhlhl5 RRRVVV JJRRR llltt hhhlhll RRRVVVV JJlllRRRtt hhhhlll RRRlVlVlVJVJththRhRhRlll lllRRhRhthtVhJVJVlVllRRR lll hhhhttRRRlllJJVVVV RRR lllhhhh ttlll RRRJJ VVVVRRR llhlhhh tlll RR VVVRVR v2 jVh w2 w2 4 v2 iRRVRVVV dJiRRR ll:5 hhlhl5 RRRVVV JJRRR llltt hhhlhll RRRVVVV JJlllRRRtt hhhhlll RRRlVlVlVJVJththRhRhRlll lllRRhRhthtVhJVJVlVllRRR lll hhhhttRRRlllJJVVVV RRR llhhhh ttlll RRRJJ VVVVRRR lhlhlhh tlll RR VVVRVR v3 hl w3 w3 v3 ...... Together, these three diagrams give a complete picture of the Verma supermodule M(V (λ1, λ2)), for (λ1, λ2) 6= (0, 0). Below is the corresponding diagram for M(V (0, 0)).

18 8 0 (0,0) v0 7 C

0 (−1,1) $' (−1,1) v1 w1 (−1,1) 7 C g : i (1,1) tt Z −2 tt tt −2 tt −2 tt tt(−2,2) tt $' (−2,2) v2 w2 (−2,2) 7 C g : i (1,1) tt Z −6 tt tt −6 tt −6 tt tt(−3,3) tt $' (−3,3) v3 w (−3,3) 7 g 3 (1,1) i . . . . 2 Lemma 5.0.1. Let (λ1, λ2) ∈ C . Then M(V (λ1, λ2)) has support {(λ1 − n, λ2 + n : n ∈ N0}.

Proof. By construction of M(V (λ1, λ2)) and Lemma 3.0.3.

2 Lemma 5.0.2. Let λ ∈ C . Let N be a subsupermodule of M(V (λ)). Then N is a proper subsupermodule if and only if Nλ = 0.

Proof. Suppose Nλ = 0. Then N is a proper subsupermodule since M(V (λ)) contains a nonzero vector with weight λ, but N does not. Now suppose that Nλ 6= 0. Then there exists a weight vector v ∈ N with weight λ. But v ∈ V (λ) and V (λ) is a simple h-supermodule. Hence, V (λ) ⊆ N. As M(V (λ)) is generated by V (λ), it follows that M(V (λ)) ⊆ N, so that N is not proper. This proves the claim.

2 Lemma 5.0.3. Let λ ∈ C . Then the Verma supermodule M(V (λ)) has a unique maximal proper subsupermodule. Proof. Suppose that V and W are both maximal proper subsupermodules of M(V (λ)). Then by Lemma 5.0.2, Vλ = 0 and Wλ = 0. But then (V +W )λ = 0, so that V + W is also a proper subsupermodule. Since V ⊆ V + W and V is maximal, this implies that V = V + W . Similarly, W = V + W and hence V = W .

2 Deﬁnition 5.0.3. Let λ ∈ C and let N be the unique maximal proper subsupermodule of M(V (λ)), guaranteed to exist by Lemma 5.0.3. Then the quotient M(V (λ))/N is unique and simple and we denote it by L(V (λ)).

19 Corollary 5.0.1. Let V be a simple h-supermodule. Then the Verma supermodule M(V ) is indecomposable.

Proof. This follows directly from Lemma 5.0.3.

Theorem 5.0.3. (Universal property of Verma supermodules)

(i) Let V be a q-supermodule and W be a simple h-supermodule of some weight space Vλ such that E(W ) = E(W ) = 0. Then there exists a q-homomorphism φ : M(W ) → V such that 1 ⊗ w is mapped to w for all w ∈ W .

(ii) Let V be a highest weight q-supermodule generated by a simple highest weight space Vλ. Then V is a quotient of M(Vλ). Proof. The ﬁrst part of the theorem is a standard result for Verma modules. See [Di] for details. The second part follows directly from (i): let V be a highest weight q-supermodule generated by a simple highest weight space Vλ. Then, by (i) there exists a q-homomorphism Φ : M(Vλ) → V mapping ∼ 1 ⊗ Vλ to Vλ. But Vλ generates V , and hence Φ is surjective. Thus, V = M(Vλ)/Ker(Φ) by the First Isomorphism Theorem.

Lemma 5.0.4. Let V be a simple weight q-supermodule with highest weight 2 ∼ λ ∈ C . Then Vλ is a simple h-supermodule. In particular, either Vλ = V (λ) ∼ or Vλ = Π(V (λ)).

Proof. Note that Vλ is an h-supermodule since it is a weight space of M(V (λ)). Now suppose for contradiction that W be a proper, simple h-subsupermodule of Vλ. Then, there exists a q-homomorphism Φ : M(W ) → V by The- orem 5.0.3(i). Moreover, since V is simple, this homomorphism must be surjective. But this is a clear contradiction since any element in Vλ \W does not have a pre-image in M(W ). Thus, Vλ must be simple. In particular, ∼ ∼ Vλ = V (λ) or Vλ = Π(V (λ)) by Theorem 2.1.1.

Lemma 5.0.5. Let V be a simple weight q-supermodule with highest weight 2 λ ∈ C . Then, V is isomorphic to either L(V (λ)) or Π(L(V (λ))). Moreover, ( L(V (λ)) , iﬀ V =∼ V (λ) V ∼ λ = ∼ Π(L(V (λ))) , iﬀ Vλ = Π(V (λ))

Proof. By Lemma 5.0.4, Vλ must be a simple h-supermodule. Thus, by ∼ ∼ ∼ Theorem 2.1.1, either Vλ = V (λ) or Vλ = Π(V (λ)). Suppose Vλ = V (λ). Then, by Theorem 5.0.3, there exists a surjective q-homomorphism φ : ∼ M(V (λ)) V , and hence V = M(V (λ))/Ker(φ). But since V is simple,

20 Ker(φ) must be a maximal subsupermodule of M(V (λ)). Thus, L(V (λ)) = M(V (λ))/Ker(φ) by deﬁnition, and hence V =∼ L(V (λ)). The case where ∼ Vλ = Π(V (λ)) is analogous.

Lemma 5.0.6. Let V be a simple weight q-supermodule with highest weight 2 λ = (λ1, λ2) ∈ C \{0}. Let x be the basis for the even subspace of Vλ and y be the basis for the odd subspace of Vλ. Then, (i) ( √ √ ∼ V (λ) , iﬀ H1H2x = −i λ1 λ2x Vλ = √ √ Π(V (λ)) , iﬀ H1H2x = i λ1 λ2x

(ii) ( √ √ ∼ V (λ) , iﬀ H1H2y = i λ1 λ2y Vλ = √ √ Π(V (λ)) , iﬀ H1H2y = −i λ1 λ2y

Proof. By Lemma 5.0.4, Vλ must be a simple h-supermodule. Now the ﬁrst claim follows from Theorem 2.1.1(ii) and the second claim follows from Remark 2.1.1.

Remark 5.0.3. Let V be a simple weight q-supermodule with highest ∼ ∼ weight (0, 0). Then, by Lemma 5.0.4, either V(0,0) = V (0, 0) or V(0,0) = Π(V (0, 0)). Either way, V(0,0) is one-dimensional and the classiﬁcation is ∼ ∼ as follows: if V(0,0) is an even space, then V(0,0) = V (0, 0) and hence V = ∼ L(V (0, 0)) by Lemma 5.0.5. If V(0,0) is an odd space, then V(0,0) = Π(V (0, 0)) and hence V =∼ Π(L(V (0, 0))).

Theorem 5.0.4. (Classiﬁcation of simple weight q-supermodules with highest weight (λ1, λ2)) Let V be a simple weight q-supermodule with highest 2 weight λ = (λ1, λ2) ∈ C and let x be the basis of the even subspace of Vλ, possibly zero. Then, V is isomorphic to either L(V (λ)) or Π(L(V (λ))). Explicitly,

(i) If (λ1, λ2) = (0, 0), then ( L(V (0, 0)) , iﬀ V is an even super vector space V =∼ (0,0) Π(L(V (0, 0)) , iﬀ V(0,0) is an odd super vector space

(ii) If (λ1, λ2) 6= (0, 0), then ( √ √ L(V (λ)) , iﬀ H1H2x = −i λ1 λ2x V =∼ √ √ Π(L(V (λ))) , iﬀ H1H2x = i λ1 λ2x

21 ∼ ∼ Moreover, V = L(V (λ)) = Π(L(V (λ))) if and only if λ1 = 0, λ2 6= 0 or λ1 6= 0, λ2 = 0. Proof. Cases (i) and (ii) follow directly from Lemma 5.0.5, Lemma 5.0.6, and Remark 5.0.3. The last claim follows from Proposition 2.1.

Remark 5.0.4. In Theorem 5.0.4, we could instead consider the basis element y of the odd subspace of Vλ. This yields the following classiﬁcation for V in the case (λ1, λ2) 6= (0, 0):

( √ √ L(V (λ)) , iﬀ H1H2y = i λ1 λ2y V =∼ √ √ Π(L(V (λ))) , iﬀ H1H2y = −i λ1 λ2y The following proposition helps simplify some of the calculations necessary when applying Theorem 5.0.4 or Remark 5.0.4.

2 Proposition 5.1. Let λ = (λ1, λ2) ∈ C \{0} and let vn, vn, wn, wn, n ∈ N0, be the usual basis for M(V (λ1, λ2)) adjoined with the elements w0, w0, which we take to be zero. Then, p p H1H2vn = (n − i λ1 λ2)vn − nαwn p p H1H2vn = (n + i λ1 λ2)vn − nβwn p p H1H2wn = αvn + (−n + i λ1 λ2)wn p p H1H2wn = βvn − (n + i λ1 λ2)wn

Proof. This follows from the formulas for H1 and H2 given earlier.

Our next aim is to describe the structure of all Verma supermodules, and for this we must consider two separate cases.

2 Deﬁnition 5.1.1. A weight λ = (λ1, λ2) ∈ C is said to be atypical if λ1 + λ2 = 0 and typical otherwise. Lemma 5.1.1. Suppose V is a q-supermodule and v ∈ V is an element such that both E and E act trivially on v. Then, E and E also act trivially on H1v and H2v. Proof. Using the relations, we have

EH1v = H1Ev + [E, H1]v = 0 − Ev = 0

EH1v = −H1Ev + [E, H1]v = 0 + Ev = 0

The proof for H2v is similar.

22 Deﬁnition 5.1.2. Let V be a q-supermodule and v ∈ V be a homogeneous element. Then the elements H1v and H2v are called H-images of v.

2 Remark 5.1.1. Let M(V (λ)) be a Verma supermodule for some λ ∈ C . Then a general element z ∈ M(V (λ)) has form

k=N X z = (rkvk + skwk + tkvk + ukwk) k=0 where N ∈ N0, rk, sk, tk, uk ∈ C, and we take w0 = w0 = 0. If λ = (0, 0), then we also have wn = vn = 0 for n ∈ N0. Now suppose that we impose the condition E(z) = 0. Then by the deﬁnition of E and the fact that L M(V (λ)) = 2 M(V (λ))µ, we get the following two equations: µ∈C

"k=N # X E (rkvk + skwk) = 0 k=0 "k=N # X E (tkvk + ukwk) = 0 k=0 Note that we cannot break this down further and say, for instance, that Pk=N E( k=0 rkvk) = 0 because Evk and Ewk are not necessarily linearly independent. The same holds true for E.

2 Lemma 5.1.2. Let λ = (λ1, λ2) ∈ C .

(i) If λ is typical and λ1 − λ2 ∈/ N, then the only elements in M(V (λ)) on which both E and E act as zero are the highest weight vectors, that is elements of the form av0 + bv0.

(ii) If λ is typical and λ1 −λ2 = n ∈ N, then the only elements in M(V (λ)) on which both E and E act as zero are the highest weight vectors and scalar multiples and H-images of the element x := αvn − nwn.

(iii) If λ is atypical and λ1 − λ2 ∈/ N0, then the only elements in M(V (λ)) on which both E and E act as zero are the highest weight vectors and scalar multiples or H-images of the element z := αv1 − (λ1 − λ2)w1. (iv) If λ = (0, 0), then the only elements in M(V (0, 0)) on which both E and E act as zero are the highest weight vectors and scalar multiples or H-images of the element v1.

(v) If λ is atypical and λ1−λ2 = n ∈ N, then the only elements in M(V (λ)) on which both E and E act as zero are the highest weight vectors, the p n p n element z := v1 − 2 w1, the element x := vn − 2 wn, and linear combinations and H-images of these elements.

23 Proof. By Remark 5.1.1 and Lemma 5.1.1, it is suﬃcient to consider elements of form avm + bwm for some m ∈ N. Thus, let y = avm + bwm be a nonzero element in M(V (λ1, λ2)) such that both E and E act trivially on it. This yields the following equalities:

aamvm−1 + bαvm−1 + bbmwm−1 = 0

amβvm−1 − am(m − 1)wm−1 + bcvm−1 − b(m − 1)αwm−1 = 0

First, assume that λ is typical and λ1 − λ2 ∈/ N. Then linear independence of the basis elements yields the following system of equations:

aam + bα = 0 (I)

bbm = 0 (II) amβ + bc = 0 (III) −am(m − 1) − b(m − 1)α = 0 (IV)

Consider the second equation. Suppose that b is zero. Then the third equation is reduced to anβ = 0. Note that β 6= 0 since β divides the nonzero sum λ1 + λ2. So either a = 0 or m = 0, both of which lead to the conclusion that y is indeed a linear combination of v0 and v0. Now suppose that bm is zero. Then by deﬁnition of bm and the fact that λ1 − λ2 ∈/ N, we must have m = 1. So the system of equations is reduced to the following:

λ − λ α a 1 2 = 0 (8) β c b

In order for this matrix equation to have a nonzero solution, the matrix must have zero determinant. Since c = αβ, we get (λ1 − λ2)c = c. But c is assumed to be nonzero, so λ1 − λ2 = 1. This contradicts our assumption that λ1 − λ2 ∈/ N. Hence, we must have b = 0 and the ﬁrst claim holds. Now suppose λ is typical and λ1 − λ2 = m ∈ N. The case b = 0 is just as above. Suppose bn = 0. Then by deﬁnition of bn, either n = 1 or m = n. If n = 1, then just as above we get the matrix equation (8), which yields m = 1, so m = n in this case. Now suppose m = n 6= 1. Then, we get the following matrix equation:

a α a n = 0 nβ c b

The determinant must be zero for a nontrivial solution to exist, which yields λ1 − λ2 = n. Hence, if λ1 − λ2 = n ∈ N, there exist scalars a, b ∈ C such that E, E act trivially on avm + bwm. There are inﬁnitely many solutions to the matrix equation above, but if we ﬁx a = α, then b = −n, so the element

24 x = αvn − nwn works. All other solutions are scalar multiples of this one so, using Lemma 5.1.1, we conclude that the only elements on which both E and E act trivially are either multiples of x or multiples of H-images of x. Now suppose that (λ1, λ2) is atypical and λ1 − λ2 ∈/ N0. Suppose α = 0. Then the fourth equation reduces to am(m − 1) = 0. So either a = 0 or m = 1. If a = 0, then the second equation yields bm = 0, which in turn forces m = 1 since λ1 − λ2 ∈/ N. If m = 1, then the third equation yields a = 0, since β 6= 0. Thus, in either case we have m = 1 and a = 0. So any multiple of w1 will do, in particular the element z = αv1 − (λ1 − λ2)w1. Now suppose β = 0. Then the second equation yields b = 0 or bm = 0. If b = 0, then the fourth equation forces m = 1. If bm = 0, then we also must have m = 1 since λ1 − λ2 ∈/ N. Hence, the only nontrivial equation is the first one, which reduces to a(λ1 − λ2) + bα = 0. There are infinitely many solutions to this equation, but in particular the coefficients of the element z = αv1 − (λ1 − λ2)w1 work. All other possible elements in this case are simply scalar multiples of z and also scalar multiples of H-images of z, by Lemma 5.1.1. Now suppose that (λ1, λ2) = (0, 0). The even part of M(V (0, 0)) has basis vi, i ∈ N0 and the odd part has basis wj, j ∈ N, so the element we consider is simply y = avm, where a 6= 0. In this case, the equations (I)-(IV) reduce to the following system:

aam = 0 am(m − 1) = 0

Since a 6= 0 and am = −m(m − 1) in this case, there is in fact just one unique equation: n(m − 1) = 0. Thus, either m = 0 or m = 1. Hence, any multiple of v0 or v1 satisﬁes the conditions, and so do H-images of these elements by Lemma 5.1.1. Finally, suppose that λ is atypical√ and λ1 − λ2 = n ∈ N. Then we must n n have (λ1, λ2) = ( 2 , − 2 ), so that α = 2n, β = 0, am = m(n − m + 1), and bm = (m − 1)(n − m). Then equation (III) is satisﬁed and equations (I), (II), and (IV) become

aam + bα = 0 (I)

bbm = 0 (II) −am(m − 1) − b(m − 1)α = 0 (IV)

Consider equation (II). First, suppose that b = 0. Then, a 6= 0 by assumption and hence equation (I) yields am = 0. By definition of am either m = 0 or m = n + 1. In the first case, equation (III) is satisfied, but the second case contradicts equation (III). Thus, we must have m = 0, so that the only

25 solution in this case if of form z = av0, a highest weight vector. Now suppose that bm = 0 in equation (II). Then, by deﬁnition of bm either m = 1 or n = m. Assume m = 1. Then equation (III) is satisﬁed and equation p n p n (I) yields b = − 2 a. Thus, all multiples of the element z := v1 − 2 w1 satisfy the necessary conditions. Now assume n = m. Then equation (I) p n and (III) are identical, and they both give the relation b = − 2 a. Thus, all p n multiples of the element x := vn − 2 wn satisfy the necessary conditions. As in all previous cases, linear combinations of these elements and also of their H-images also work.

Lemma 5.1.3. Let M(V (λ)) be a Verma supermodule and suppose that both E and E act as zero on some nontrivial element x ∈ M(V (λ)), where x is not a highest weight vector. Then x generates a proper subsupermodule of M(V (λ)).

Proof. This follows from Lemma 3.0.3 and Lemma 5.0.2.

Lemma 5.1.4. Suppose V is a q-supermodule and v ∈ V such that E(v) = 0. Then both E and E act trivially on the element Ev.

Proof. The claims follows from the commutativity of E and E and the fact 2 that E = 0.

2 Lemma 5.1.5. Suppose (λ1, λ2) ∈ C . Then, (i) ∼ Π(M(V (λ1, λ2))) = M(Π(V (λ1, λ2))).

(ii) ∼ Π(L(V (λ1, λ2))) = L(Π(V (λ1, λ2))).

Proof. This follows from Lemma 2.0.2 and also the deﬁnitions of M(V (λ1, λ2)) and L(V (λ1, λ2)).

Remark 5.1.2. In the next two sections, we will describe the structure of all Verma supermodules of form M(V (λ1, λ2)). Using Lemma 5.1, we can obtain completely analogous results for Verma supermodules of form Π(M(V (λ1, λ2))).

26 5.2 Structure of typical Verma supermodules Theorem 5.1.1. [Structure of typical Verma supermodules] 2 Suppose (λ1, λ2) ∈ C is typical.

(i) The supermodule M(V (λ1, λ2)) is simple if and only if λ1 − λ2 ∈/ N.

(ii) If λ1 − λ2 = n ∈ N then the supermodule M(V (λ1, λ2)) has a unique proper subsupermodule isomorphic to either L(V (λ1 − n, λ2 + n)) or Π(L(V (λ1 − n, λ2 + n))). The quotient is a simple ﬁnite-dimensional q-supermodule isomorphic to L(V (λ1, λ2)).

Proof. Suppose λ1 − λ2 ∈/ N and let V be a nonzero subsupermodule of M(V (λ1, λ2)). Then there exists a nonzero element z ∈ V of form avn +bwn n for some n ∈ N0 and some a, b ∈ C. If E z 6= 0, then it generates the whole supermodule and hence M = V . Otherwise, let m ∈ {1, 2, . . . , n} be minimal such that Emz = 0. If EEm−1z = 0, then Em−1z is a nonzero element in the linear span of vn−m+1 and wn−m+1 such that both E and E act trivially on it, which contradicts Lemma 5.1.2 since n − m + 1 ≥ 1. So m−1 m−1 EE z must be nonzero. If n = m, then EE z must be a multiple of v0 and hence M = V . Otherwise, if m ≤ n, then EEm−1z is a nonzero element in the linear span of vn−m and wn−m. But by Lemma 5.1.4, both E and E act trivially on this element, which is again a contradiction to Lemma 5.1.2 since (n − m) ≥ 1. Thus, M(V (λ1, λ2)) is simple. (⇒) Now assume λ1 − λ2 = n ∈ N. Then by Lemma 5.1.2, the highest weight supermodule generated by x := αvn − nwn is a unique proper subsupermodule of M(V (λ1, λ2)). This subsupermodule has highest weight (λ1 −n, λ2 +n), where (λ1 −n)−(λ2 +n) = λ1 −λ2 −2n = n−2n = −n∈ / N. Hence it is simple by the above argument. Thus this subsupermodule is isomorphic to either L(V (λ1 − n, λ2 + n)) or Π(L(V (λ1 − n, λ2 + n))) by Theorem 5.0.4. Since this subsupermodule is the unique proper subsupermodule of M(V (λ1, λ2)), it is maximal and hence the quotient is simple and isomorphic to L(V (λ1, λ2)) by deﬁnition. This quotient has basis v0,... , vn, v0,... , vn, w1,... , wn−1, w1, ... , wn−1, and is hence a ﬁnite-dimensional q-supermodule, of dimension 4n.

Remark 5.1.3. Note that due to the nature of complex numbers, it is rather complicated to determine whether the proper subsupermodule referred to in Theorem 5.1.1 is isomorphic to L(V (λ1 − n, λ2 + n)) or Π(L(V (λ1 − n, λ2 + n))). For instance, it is tempting to reason that p p p p λ1 − n λ2 + n = (λ1 − n)(λ2 + n) = λ1λ2 √ √ √ but in fact this logic is wrong because in general ab 6= a b for complex numbers a, b. This makes it diﬃcult to calculate what H1H2x is in general, but fortunately the task becomes easier when λ1, λ2 are known explicitly.

27 5.3 Structure of atypical Verma supermodules 2 Now we examine the Verma supermodule M(V (λ1, λ2)) where (λ1, λ2) ∈ C is atypical. Once again, there are several cases. First, consider the case where λ1 − λ2 ∈/ N0. 2 Theorem 5.1.2. Suppose (λ1, λ2) ∈ C is atypical and λ1 − λ2 ∈/ N0. Then the Verma supermodule M(V (λ1, λ2)) has a unique proper subsupermodule, which is isomorphic to either L(V (λ1 −1, λ2 +1)) or Π(L(V (λ1 −1, λ2 +1))) and has simple quotient L(V (λ1, λ2)).

Proof. By Lemma 5.1.2(iii), the element z := αv1 − (λ1 − λ2)w1 generates a unique proper subsupermodule of M(V (λ1, λ2)), which is simple and has highest weight (λ1 −1, λ2 +1). Hence, this subsupermodule is isomorphic to either L(V (λ1−1, λ2+1)) or Π(L(V (λ1−1, λ2+1))) by Theorem 5.0.4. Since this subsupermodule is the only proper subsupermodule of M(V (λ1, λ2)), it is maximal and hence the quotient is isomorphic to L(V (λ1, λ2)) by deﬁni- tion.

Remark 5.1.4. For the next several theorems,√ we will√ be using the fact that that√ if k is a positive√ real√ number, then −k = i k. This holds because −k has roots i k and −i k, but only the ﬁrst root has argument in [0, π).

Theorem 5.1.3 (Structure of M(V (0, 0))). The Verma supermodule M(V (0, 0)) has a unique proper subsupermodule, which is isomorphic to L(V (−1, 1)) and has one-dimensional even quotient isomorphic to L(V (0, 0)).

Proof. The case where λ1 = λ2 = 0 yields α = β = 0. Note that since V (0, 0) is a one-dimensional even h-supermodule, the Verma supermodule M(V (0, 0)) has basis vn, wm, for n ∈ N0, m ∈ N. Now, by Lemma 5.1.2, the element v1 generates a unique nontrivial proper subsupermodule of M(V (0, 0)), which is simple and has highest weight (−1, 1). Moreover, H1H2v1 = v1, so that this subsupermodule is isomorphic to L(V (−1, 1)) by Theorem 5.0.4(ii). The quotient M(V (0, 0))/L(V (−1, 1)) has basis v0, and is hence an even one-dimensional q-supermodule. Thus, it is isomorphic to L(V (0, 0)) by Theorem 5.0.4(i).

1 1 1 1 Theorem 5.1.4 (Structure of M(V ( 2 , − 2 ))). The Verma supermodule M(V ( 2 , − 2 )) 1 1 has a unique proper subsupermodule, which is isomorphic to L(V (− 2 , 2 )). 1 1 1 1 The quotient T := M(V ( 2 , − 2 ))/L(V (− 2 , 2 )) has a unique proper subsu- 3 3 permodule, isomorphic to Π(L(V (− 2 , 2 ))), with ﬁnite-dimensional quotient 1 1 isomorphic to L(V ( 2 , − 2 )).

28 √ Proof. In this√ case, α = 2 and β = 0. Thus, by Lemma 5.1.2(ii), the 1 1 element z = 2v1 − w1 generates a proper subsupermodule of M( 2 , − 2 ), 1 1 1 which is simple and has highest weight (− 2 , 2 ). Moreover, H1H2z = 2 z, 1 1 so this subsupermodule is isomorphic to L(V (− 2 , 2 )) by Theorem 5.0.4(ii). 1 1 1 1 The quotient T := M(V ( 2 , − 2 ))/L(V (− 2 , 2 )) has basis vn, vn, where n ∈ N0. Now if we look for an even element in the quotient such that E acts trivially on it, we see that the element must be a scalar multiple of v2 since

E(vn) = 0 ⇒ n = 2. (9)

Since E(v2) = 0 in the quotient, we see that T has a subsupermodule generated by the element v2 and this subsupermodule is simple by (9). More- 3 3 over, it has highest weight (− 2 , 2 ) and highest weight vector v2 which 3 satisﬁes H1H2v2 = − 2 v2. Thus, this subsupermodule is isomorphic to 3 3 3 3 Π(L(V (− 2 , 2 ))) by Theorem 5.0.4(ii). The quotient T/Π(L(V (− 2 , 2 ))) has basis v0, v0, v1, v1. It is easy to check that this four-dimensional supermod- 1 1 1 ule is simple, with highest weight ( 2 , − 2 ). Moreover, H1H2v0 = 2 v0, and 1 1 hence this supermodule is isomorphic to L(V ( 2 , − 2 )) by Theorem 5.0.4(ii).

Theorem 5.1.5 (Structure of M(V (1, −1))). The Verma supermodule M(V (1, −1)) has a proper subsupermodule S generated by the element v1 − w1, which has a simple proper subsupermodule generated by v2 − w2 and isomorphic to L(V (−1, 1)). The subquotient S/L(V (−1, 1)) is a one-dimensional even space isomorphic to L(V (0, 0)). The quotient T := M(V (1, −1))/S has a one-dimensional subsupermodule generated by w1 and isomorphic to Π(L(V (0, 0))). The quotient A := T/Π(L(V (0, 0))) has a simple subsupermodule generated by v3 and isomorphic to Π(L(V (−2, 2))). Finally, the quotient A/Π(L(V (−2, 2))) is ﬁnite-dimensional and isomorphic to L(V (1, −1)).

Proof. In this case, we have α = 2 and β = 0. By Lemma 5.1.2(iii), the Verma supermodule M(V (1, −1)) has a proper subsupermodule S generated by element −2v1 + 2w1. For simplicity, we can scale this generator to v1 − w1. Now, again by Lemma 5.1.2(iii), M(V (1, −1)) has a proper, simple subsupermodule generated by element v2 − w2, which is a subsupermodule of S since v2 − w2 ∈ S. The highest weight of this subsupermodule is (−1, 1) and the highest weight vector v2 − w2 satisﬁes H1H2(v2 − w2) = (v2 − w2). Hence, this subsupermodule is isomorphic to L(V (−1, 1)) by Theorem 5.0.4(ii). The subquotient S/L(V (−1, 1)) has basis v1 − w1. Thus, it is a one-dimensional even supermodule with highest weight (0, 0). Thus, it is isomorphic to L(V (0, 0)) by Theorem 5.0.4(i). Now, consider the quotient T := M(V (1, −1))/S. This has basis w1, vj, vj, j ∈ N0, where vj = wj for j ∈ N. Note that Xw1 = 0 in the quotient for X ∈ {E, E,F, F, H1, H2}, so that w1 generates a one-dimensional subsupermodule of T. This subsupermodule has weight (0, 0) and is an odd super

29 vector space. Hence, it is isomorphic to Π(L(V (0, 0))) by Theorem 5.0.4(i). Now, the quotient A := T/Π(L(V (0, 0))) has basis vj, vj, j ∈ N0, where vj = wj for j ∈ N. If A has a proper subsupermodule, then it must be generated by some vi, where i ∈ N. The condition E(vi) = 0 yields i = 3. Thus, the only possibility is a subsupermodule generated by element v3. Di- rect calculation shows that E(v3) = 0in the quotient. Hence, v3 generates a proper subsupermodule of A. This subsupermodule is simple (by previous calculations) and has highest weight (−2, 2). Moreover, the highest weight vector v3 satisﬁes H1H2v3 = −2v3 and hence this subsupermodule is isomorphic to Π(L(V (−2, 2))) by Theorem 5.0.4(ii). Finally, note that the quotient A/Π(L(V (−2, 2))) is ﬁnite-dimensional, with basis vi, vi, where i ∈ {0, 1, 2}, such that vi = wi for i = 1, 2. It is simple by previous calculations and has highest weight (1, −1). Since H1H2v0 = v0, this quotient is isomorphic to L(V (1, −1)) by Theorem 5.0.4(ii).

n n Theorem 5.1.6 (Structure of M(V ( 2 , − 2 )), n ∈ {3, 4,... }). The Verma n n supermodule M(V ( 2 , − 2 )), n ∈ {3, 4,... }, has a proper subsupermodule p n S generated by v1 − 2 w1, which has a simple subsupermodule isomorphic n n n n to L(V (− 2 , 2 )). The quotient S/L(V (− 2 , 2 )) is finite-dimensional and iso- n n n n morphic to Π(L(V ( 2 −1, − 2 +1))). The quotient T := M(V ( 2 , − 2 ))/S has n proper subsupermodule generated by vn+1 and isomorphic to Π(L(V (− 2 − n n n n n 1, 2 +1))). The quotient T/Π(L(V (− 2 −1, 2 +1))) is isomorphic to L(V ( 2 , − 2 )). √ Proof. In this case, we have α = 2n and β = 0. By Lemma 5.1.2(v), the n n Verma supermodule M(V ( 2 , − 2 )) has a proper subsupermodule S gener- p n ated by element z = v1 − 2 w1. Now, again by Lemma 5.1.2(v), S has a p n proper, simple subsupermodule generated by element x = vn− 2 wn, which is a subsupermodule of S since x ∈ S. The highest weight of this simple sub- n n supermodule is (− 2 , 2 ) and the highest weight vector x satisfies H1H2x = n n n 2 x. Hence, this simple subsupermodule is isomorphic to L(V (− 2 , 2 )) by n n Theorem 5.0.4(ii). The subquotient S/L(V (− 2 , 2 )) is finite-dimensional, n n simple (by Lemma 5.1.2), and has highest weight ( 2 − 1, − 2 + 1)). More- n over, the highest weight vector z satisfies H1H2z = −( 2 −1)z, and hence this n n subquotient is isomorphic to Π(L(V ( 2 − 1, − 2 + 1))) by Theorem 5.0.4(ii). n n Denote the quotient M(V ( 2 , − 2 ))/S by T . Then T has basis vi, vj, p n where i, j ∈ N0, such that vi = 2 wi. If this supermodule has a proper subsupermodule, then it must be generated by some vi, where i ∈ N. The condition E(vi) = 0 yields i = n + 1. Thus, the only possibility is a subsupermodule generated by element vn+1. Direct calculation shows that E(vn+1) = 0 in the quotient. Hence, vn+1 generates a proper subsupermodule of T . This subsupermodule is simple (by previous calculations) and n n has highest weight (− 2 − 1, 2 + 1). Moreover, the highest weight vector n vn+1 satisfies H1H2vn+1 = −( 2 + 1)vn+1 and hence this subsupermodule is

30 n n isomorphic to Π(L(V (− 2 − 1, 2 + 1))) by Theorem 5.0.4(ii). Finally, note n n that the quotient T/Π(L(V (− 2 − 1, 2 + 1))) is ﬁnite-dimensional, with ba- p n sis vi, vj, where i, j ∈ {0, 1, 2, ··· n}, such that vi = 2 wi. It is simple by n n n previous calculations and has highest weight ( 2 , − 2 ). Since H1H2v0 = 2 v0, n n this quotient is isomorphic to L(V ( 2 , − 2 )) by Theorem 5.0.4(ii).

6 Finite-Dimensional simple weight q-supermodules

Theorem 6.0.7. Let V be a ﬁnite-dimensional simple weight q-supermodule. Then V is isomorphic to L(V (λ1, λ2)) or Π(L(V (λ1, λ2))) for some (λ1, λ2) ∈ 2 C such that one of the following is satisﬁed:

(i) (λ1, λ2) = (0, 0)

(ii) λ1 − λ2 ∈ N.

Moreover, if λ1 = 0 and λ2 6= 0 or if λ1 6= 0 and λ2 = 0, then ∼ ∼ V = L(V (λ1, λ2)) = Π(L(V (λ1, λ2))).

Proof. Let V be a finite-dimensional simple weight q-supermodule. Then V is in particular a highest weight q-supermodule, with some highest weight 2 (λ1, λ2) ∈ C . Hence V is isomorphic to either L(V (λ1, λ2)) or Π(L(V (λ1λ2))) by Lemma 5.0.5. Then, Theorems 5.1.1, 5.1.2, 5.1.3, 5.1.4, 5.1.5, and 5.1.6 show that L(V (λ1, λ2)) and Π(L(V (λ1, λ2)) are finite-dimensional if and only if (λ1, λ2) = (0, 0) or λ1 − λ2 ∈ N. The first part of the claim fol- ∼ lows. Now, if λ1 = 0 and λ2 6= 0 or if λ1 6= 0 and λ2 = 0, then V (λ1, λ2) = ∼ Π(V (λ1, λ2)) by Proposition 2.1 and hence L(V (λ1, λ2)) = Π(L(V (λ1, λ2))). This finishes the proof.

7 Simple lowest weight q-supermodules

Lowest weight q-supermodules are completely analogous to highest weight q-supermodules.

Deﬁnition 7.0.3. If V is a weight q-supermodule and µ = (µ1, µ2) ∈ supp(V ), then µ is said to be a lowest weight if (µ1−1, µ2+1) ∈/ supp(V ). The corresponding weight space Vµ is then referred to as the lowest weight space and any nonzero vector in Vµ as a lowest weight vector.A q-supermodule generated by a lowest weight space is called a lowest weight q-supermodule.

Remark 7.0.5. Recall that U(q) = U(n−) ⊗C U(h) ⊗C U(n+) where n− is the linear span of f and f and n+ is the linear span of e and e.

31 0 Definition 7.0.4. Let B = U(h)⊗CU(n−). Let V be a simple h-supermodule and define n−V = 0. Then the lowest weight supermodule with lowest weight space V , denoted by M 0(V ), is defined as

0 M (V ) := U(q) ⊗B0 V

Remark 7.0.6. In the construction of M 0(V ) above, we start with a simple h-supermodule V and then induce it up to a lowest-weight q-supermodule by requiring that the action of f and f on V is trivial.

This process of constructing lowest weight q-supermodules is completely analogous to that of constructing highest weight q-supermodules. Thus, 0 0 0 0 0 M (V ) has a basis with elements vn, vn, wm, wm for n ∈ N0, m ∈ N. The 0 0 linear operators F and F act as zero on v0 and v0, and the action of H1, H2, 0 0 H1, and H2 on v0 and v0 is known. Just as for highest weight supermodules, we can use the relations in Table (2) to calculate the explicit actions of all eight linear operators E, F , H1, H2, E, F , H1, and H2 on the basis elements of M 0(V ). The following diagram shows the general actions (without coeﬃcients) 0 of E, E, and F on the basis elements of M (V (λ1, λ2)). Here E is given by the dotted arrows, E is given by the the dashed arrows, and F is given by the solid arrows.

......

v3 w3 w v jT j4 3 3 O } O T T j j O AA O }} j jT T AA }} j j T T AA }} j T T AA ~} j j Ù T T Ò v2 j w2 w2 v2 O O jT T j j4 O A O }} TjTj AA }} j j T T AA }} j j T T AA ~}} j j Ù T T A Ò v1 j w1 w T v jT j4 1 1 O } T T j j AA O }} j jT T AA }} j j T T AA }} j T T AA ~} j j T T Ò v0 j v0 0 0 " | Remark 7.0.7. All of the results concerning the classiﬁcation of Verma supermodules have analogues in the study of lowest weight q-supermodules.

32 Acknowledgments

I am grateful to my advisor Volodymyr Mazorchuk for introducing me to the representation theory of Lie algebras and for suggesting the queer Lie superalgebra as the subject of this paper. He is without a doubt the best advisor a math student could ask for, and the only person I have met so far who can explain the ﬁner points of mathematics in such a way that I both understand them and appreciate their signiﬁcance.

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