Representation Theory of Lie Colour Algebras and Its Connection with Brauer Algebras

Representation theory of Lie colour algebras and its connection with Brauer algebras

Mengyuan Cao

Thesis submitted to the Faculty of Graduate and Postdoctoral Studies in partial fulﬁllment of the requirements for the degree of Master of Science in Mathematics1

Department of Mathematics and Statistics Faculty of Science University of Ottawa

c Mengyuan Cao, Ottawa, Canada, 2018

1The M.Sc. program is a joint program with Carleton University, administered by the Ottawa- Carleton Institute of Mathematics and Statistics Abstract

In this thesis, we study the representation theory of Lie colour algebras. Our strategy follows the work of G. Benkart, C. L. Shader and A. Ram in 1998, which is to use the Brauer algebras which appear as the commutant of the orthosymplectic Lie colour algebra when they act on a k-fold tensor product of the standard representation. We give a general combinatorial construction of highest weight vectors using tableaux, and compute characters of the irreducible summands in some borderline cases. Along the way, we prove the RSK-correspondence for tableaux and the PBW theorem for Lie colour algebras.

ii Dedications

To the ones that I love.

iii Acknowledgements

I oﬀer my most sincere gratitude to my supervisors, Dr. Monica Nevins and Dr. Hadi Salmasian, for their exceptional guidance and immense enthusiasm. I would also like to thank Dr. Alistair Savage and Dr. Yuly Billig for their careful reading, motivating questions and comments, and my family and friends for their support and encouragement.

iv Contents

1 Introduction 1

2 Lie colour algebras and spo(V, β) 4 2.1 Lie colour algebras ...... 5 2.2 The category of modules of Lie colour algebras ...... 9 2.3 The orthosymplectic Lie colour algebra spo(V, β)...... 19 2.4 Homogeneous bases for V , V ∗ and spo(V, β)...... 23 2.5 Roots and root vectors in spo(V, β)...... 32

3 The PBW theorem for Lie colour algebras 40 3.1 Tensor algebras and symmetric algebras ...... 41 3.2 Universal enveloping algebras ...... 45 3.3 A spanning set of U(g)...... 48 3.4 The PBW theorem ...... 52 3.5 Proof of Proposition 3.4.2 ...... 58

4 The Brauer algebra action on V ⊗k 68 4.1 The Brauer algebra ...... 68

v CONTENTS vi

4.2 The commuting action of the Brauer algebra on the spo(V, β)- module V ⊗k ...... 72

5 Highest weight vectors in V ⊗k 80 5.1 (r, s)-hook tableaux and simple tensors ...... 81 5.2 Young symmetrizers ...... 86 5.3 Contraction maps on V ⊗k ...... 88 5.4 Construction of highest weight vectors ...... 93 5.5 An illustration of Theorem 5.4.1 ...... 107 5.5.1 The highest weight vectors ...... 107

5.5.2 Veriﬁcation that v1, v2 and v3 are highest weight vectors . . 109 5.5.3 spo(V, β)-submodules generated by the highest weight vectors 110 5.5.4 Summary: a decomposition of V ⊗ V as spo(V, β)-modules when n = 2, m = 4 and k =2 ...... 113

6 Characters of some spo(V, β)-modules 115 6.1 Schur-Weyl duality-like decomposition of V ⊗k ...... 115

6.2 The Brauer algebra B2(η)...... 118 6.3 Borderline case |n − m| = k =2 ...... 121 6.4 More examples on borderline cases ...... 128 6.5 The characters of W Y (λ) and U Y (λ) ...... 133

A Schur polynomials and the RSK-Correspondence 140 A.1 Symmetric functions and Schur polynomials ...... 141 A.2 Young tableaux ...... 142 A.3 Knuth Equivalence ...... 148 A.4 The RSK-Correspondence ...... 154 CONTENTS vii

A.5 An application of the RSK-Correspondence ...... 157

Bibliography 173 Chapter 1

Introduction

One of the remarkable results in representation theory, now referred to as Schur-Weyl

duality, connects the irreducible representations of the general linear group GLn(C)

to the irreducible representations of the symmetric group Sk. Precisely, let V be an

⊗k n-dimensional vector space. There is an action of GLn(C) on V that commutes

⊗k with the action of Sk on V . Moreover, the action of GLn(C) generates the full

centralizer of the action of Sk, and vice versa. Thus Schur-Weyl duality says that

⊗k there is a multiplicity free decomposition of V as an Sk × GLn(C)-module, given by

M V ⊗k ∼= Sλ ⊗ Eλ. λ

Here the sum is over all partitions λ of k with length at most n, and Sλ and Eλ

are the irreducible representations of Sk and GLn(C) respectively parametrized by λ.

From Schur-Weyl duality, one can compute the characters of these GLn(C)-modules, which turn out to be Schur polynomials. These polynomials are certain symmetric polynomials, named after Issai Schur. Diﬀerent variations of Schur polynomials appear as characters of representations of similar algebraic structures. For example the hook

1 1. INTRODUCTION 2

Schur functions describe the characters of the irreducible representations of the Lie superalgebra gl(m, n). See for example [BR87]. The main objective of this thesis is a generalization of Schur-Weyl duality due to G. Benkart, C. L. Shader and A. Ram to the setting of the orthosymplectic Lie colour algebras, spo(V, β).

The general linear group GLn(C) admits several important subgroups, such as the orthogonal group O(n) and the symplectic group Sp(2n). In order to obtain a Schur-Weyl duality-like theorem in terms of O(n), one has to ﬁnd a larger algebra B whose action generates the full centralizer of the action of O(n). Ideally, this algebra

B should contain Sk as a subalgebra. In 1937, Richard Brauer introduced an algebra

in [Bra37], now referred to as the Brauer algebra Bk(η), where η ∈ C. The Brauer

algebra Bk(n) (respectively Bk(−2m)) plays the same role as C[Sk] in Schur-Weyl

duality when GLn(C) is replaced by O(n) (respectively Sp(2m)). The Brauer algebra plays a principal role in the fundamental book, The Classical Groups,[Wey97] by Hermann Weyl. The outline of the thesis is as follows. In Chapter 2, we study the orthosympletic Lie colour algebra, which is a family of Lie colour algebras which generalizes both the orthogonal and symplectic Lie algebras. In Chapter 3, we prove two versions of the Poincar´e-Birkhoﬀ-Witt(PBW) theorem in the Lie colour algebra setting. In Chapter 4, we discuss the commuting actions of the Brauer algebra and spo(V, β) on a tensor space V ⊗k. In Chapter 5, we construct highest weight vectors of spo(V, β)-submodules of V ⊗k following [BSR98]. In the last chapter, we calculate modules in a borderline case that is a case does not satisfy the hypothesis of [BSR98, Proposition 4.2]. We explore extra examples not covered by [BSR98, Proposition 4.2] or Theorem 5.4.1 which show how the conclusions can fail for a variety of reasons. Our main results are 1. INTRODUCTION 3

Theorem 6.3.2 and Corollary 6.5.9. In Appendix A, we include a brief introduction to Schur polynomials and the RSK correspondence. Our main original contributions in this thesis are

(i) We generalize the proof of the PBW theorem to the Lie colour algebra case. In fact, the PBW theorem for Lie algebras and Lie superalgebras are also deduced from our proof.

(ii) In [BSR98], Benkart et. al introduced a right action of the Brauer algebra

⊗k ⊗k Bk(n − m) on V which commutes with the left action of spo(V, β) on V . We elaborate these commuting actions with detailed examples.

(iii) We give a detailed example in Section 5.5 to ﬁnd the highest weight vectors of V ⊗ V , and then ﬁnd the submodules generated by the highest weight vectors.

(iv) Using this example for intuition, we extend one of the theorems [BSR98, Propo- sition 4.2] to a borderline case, which we do in Theorem 6.3.

(v) We compute the characters of the irreducible summands of V ⊗ V in this borderline case, and use them to show that the submodules we obtained in

Theorem 6.3.2 coincide with those predicted by spo(V, β) × Bk duality.

In the future, it would be interesting to learn more about the combinatorial description of the characters of the spo(V, β)-submodules, in terms of variants of Young tableaux. Another unexplored avenue of research is the relation between the spo(V, β)-submodules generated by the highest weight vectors from Theorem 5.4.1

and the modules predicted by spo(V, β) × Bk duality. It is also interesting to extend [BSR98, Proposition 4.2] to other borderline cases beyond the ones we investigated here. Chapter 2

Lie colour algebras and spo(V, β)

The goal of this chapter is to explore the structure theory of the orthosymplectic Lie colour algebra spo(V, β). In Section 2.1, we give the definition of Lie colour algebras and analyze some of their basic properties. In Section 2.2, we define the modules of Lie colour algebras and give some examples such as the trivial module, the contragredient module and the tensor product of modules. Then we give the definition of g-module morphisms, and provide an important example, called the braiding morphism, in Proposition 2.2.10, which plays a vital role in the subsequent chapters. In Section 2.3, we describe the main object in this thesis, the othosymplectic Lie colour algebra spo(V, β). In the last two sections, Section 2.4 and Section 2.5, we first give homogeneous bases of V , V ∗ and spo(V, β) respectively, and then give an explicit description of the roots and root vectors in spo(V, β). Moreover, we provide a basis of root vectors which will be used in the proof of one of the main theorems in our thesis, Theorem 5.4.1. The material in this chapter comes from [BSR98] but the examples as well as the proofs are ours.

4 2. LIE COLOUR ALGEBRAS AND spo(V, β) 5

As of Section 2.4, in the rest of this thesis F will be assumed algebraically closed and char F 6= 2.

2.1 Lie colour algebras

Definition 2.1.1. Let G be a finite abelian group with identity 1G. A symmetric bicharacter on G over a field F is a map β : G × G → F× such that

(i) β(ab, c) = β(a, c)β(b, c), for all a, b, c ∈ G;

(ii) β(a, bc) = β(a, b)β(a, c), for all a, b, c ∈ G; and

(iii) β(a, b)β(b, a) = 1, for all a, b ∈ G.

Notice that β is called a bicharacter because holding one variable ﬁxed gives a character, which is a 1-dimensional representation in the other variable. In the rest of the thesis, β plays a very important role. We will deﬁne new categories of vector spaces, algebras and maps based on β.

Lemma 2.1.2. Let β be a symmetric bicharacter on an abelian group G. Then we have the following:

(i) β(1G, a) = β(a, 1G) = 1 for all a in G,

(ii) β(a−1, a) = β(a, a−1) = β(a, a)−1 for all a in G,

(iii) β(ab, ab) = β(a, a)β(b, b) for all a, b in G,

(iv) β(a, a) ∈ {−1, 1} for all a in G.

Proof. Let a, b ∈ G. By Deﬁnition 2.1.1, we have 2. LIE COLOUR ALGEBRAS AND spo(V, β) 6

(i) β(b, a) = β(1G, a)β(b, a) which implies that β(1G, a) = 1 for all a in G. Similarly

we have β(a, 1G) = 1 for all a in G.

−1 −1 (ii) Consequently we have 1 = β(a, 1G) = β(a, aa ) = β(a, a)β(a, a ). Hence β(a, a−1) = β(a, a)−1 for all a in G.

(iii) For all a, b in G, we have β(ab, ab) = β(a, a)β(a, b)β(b, a)β(b, b) = β(a, a)β(b, b).

(iv) Since β(a, b)β(b, a) = 1, we have β(a, a)β(a, a) = 1 for all a in G. This implies β(a, a) ∈ {−1, 1} for all a in G.

Example 2.1.3. Let G be an abelian group. Then β : G × G → F∗ such that β(a, b) = 1 for all a, b in G is a symmetric bicharacter.

Example 2.1.4. Let G = Z2 × Z2 = {(0, 0), (1, 0), (0, 1), (1, 1)}. Let a = (a1, a2) and

b = (b1, b2) be in G. The group operation of G is a + b := (a1 + b1, a2 + b2). Let us verify that

a·b β(a, b) = (−1) where a · b = a1b1 + a2b2, a, b ∈ G

is a symmetric bicharacter.

Take a = (a1, a2), b = (b1, b2) and c = (c1, c2) in G. Then we have

β(a + b, c) = (−1)(a+b)·c = (−1)(a1+b1)c1+(a2+b2)c2

= (−1)a1c1+a2c2 (−1)b1c1+b2c2

= β(a, c)β(b, c) which verifies the first condition of Definition 2.1.1. The second condition follows since 2. LIE COLOUR ALGEBRAS AND spo(V, β) 7

a · b = b · a. Now for the third condition, we have

β(a, b)β(b, a) = (−1)a·b(−1)b·a = (−1)2a·b = 1.

Therefore, the β we deﬁned in Example 2.1.4 is a symmetric bicharacter.

Deﬁnition 2.1.5. If an F-vector space V has a direct sum decomposition

M V = Va, a∈G

where each Va is subspace of V indexed by a ∈ G, then V is called a G-graded vector space. Moreover, if v ∈ Va for some a ∈ G, then v is called homogeneous of degree a. We say v has colour a.

Deﬁnition 2.1.6. Given a symmetric bicharacter β on a group G, a Lie colour algebra g is a G-graded vector space M g = ga a∈G together with an F-bilinear map [·, ·]: g × g → g such that

(i) [ga, gb] ⊆ gab, for all a, b ∈ G,

(ii) [x, y] = −β(b, a)[y, x], for x ∈ ga, y ∈ gb,

(iii) For all x ∈ ga, y ∈ gb, and z ∈ gc, the Jacobi identity for Lie colour algebras, given by

β(a, c)[x, [y, z]] + β(c, b)[z, [x, y]] + β(b, a)[y, [z, x]] = 0, (2.1.1)

is satisﬁed. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 8

Lemma 2.1.7. The Lie colour algebra Jacobi identity is equivalent to

[x, [y, z]] = [[x, y], z] + β(b, a)[y, [x, z]], for all x ∈ ga, y ∈ gb and z ∈ g. (2.1.2)

Proof. First suppose that x ∈ ga, y ∈ gb, and z ∈ gc, for some a, b, c ∈ G. We swap z with [x, y] in the second term of (2.1.1) and swap z with x in the third term of (2.1.1). Then (2.1.1) becomes

β(a, c)[x, [y, z]] − β(c, b)β(ab, c)[[x, y], z] − β(b, a)β(a, c)[y, [x, z]] = 0, which can be simpliﬁed to

β(a, c)[x, [y, z]] − β(a, c)[[x, y], z] − β(b, a)β(a, c)[y, [x, z]] = 0.

Dividing both sides of the above equation by β(a, c) and rearranging the equation gives [x, [y, z]] = [[x, y], z] + β(b, a)[y, [x, z]].

The resulting equation is independent of the choice of c. Since (2.1.2) is linear in z, we thus conclude it holds for all z ∈ g.

Example 2.1.8.

(i) Let G = {1G} be the trivial group. Then the Lie colour algebra g is a Lie algebra in the classical sense.

ab (ii) Let G = Z2 = {0, 1}, and β(a, b) = (−1) for all a, b ∈ G. Then the Lie colour algebra g is a Lie superalgebra. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 9

We can classify the elements a in G by the value of β(a, a). According to Lemma 2.1.2, we know that β(a, a) = ±1. Therefore we deﬁne

G(0) := {a ∈ G | β(a, a) = 1} and G(1) := {a ∈ G | β(a, a) = −1}. (2.1.3)

We have G = G(0) ∪ G(1), and G(0) is a subgroup of G. Moreover, given a Lie colour algebra g, we deﬁne M M g(0) = ga and g(1) = ga.

a∈G(0) a∈G(1) This gives a decomposition of Lie colour algebras as vector spaces:

g = g(0) ⊕ g(1).

Remark 2.1.9.

(i) g(0) is a Lie colour subalgebra of g.

(ii) g(1) is not subalgebra of g unless [g(1), g(1)] = 0 .

The ﬁrst part is clear since G(0) is a subgroup of G. Therefore g(0) is closed. For

(ii): take x ∈ ga and y ∈ gb such that β(a, a) = β(b, b) = −1. So x, y ∈ g(1), and

[x, y] ∈ gab. However, β(ab, ab) = β(a, a)β(b, b) = 1. Therefore [x, y] is not in g(1) unless [x, y] = 0.

2.2 The category of modules of Lie colour algebras

In this section, as in classic Lie algebra textbooks, we discuss the concepts of Lie colour algebra modules and module morphisms. Moreover we deﬁne the braid morphism which will be a very important tool in later sections. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 10

Deﬁnition 2.2.1. Let g be a Lie colour algebra. A g-module is a G-graded F-vector L space V = a∈G Va together with a g-action

g × V → V, (x, v) 7→ xv which is a bilinear map satisfying the following properties:

(i) If x ∈ ga and v ∈ Vb, then xv ∈ Vab,

(ii) [x, y]v = x(yv) − β(b, a)y(xv), for all x ∈ ga, y ∈ gb and for all v ∈ V .

A g-module is also called a representation of g over F.

Example 2.2.2. Let V = g, with the action of g on itself given by (x, y) 7→ [x, y] for all x, y ∈ g. Then together with this action, g is a g-module by Deﬁnition 2.1.6 and Lemma 2.1.7. This g-module is called the adjoint representation.

Example 2.2.3. The trivial module of g is the one-dimensional vector space V with

grading V = V1G . The g-action is deﬁned as x · v = 0 for all x ∈ g and v ∈ V .

Deﬁnition 2.2.4. Let V be a g-module over the ﬁeld F. The contragredient module of V is the vector space V ∗ := {f : V → F | f is a linear functional }, such that

∗ ∗ −1 (i) the G-grading is deﬁned by (V )a := {f ∈ V | f(Vb) = 0 if a 6= b },

(ii) the g-action is deﬁned on homogeneous elements by

(xf)(v) = −β(b, a)f(xv)

∗ for all x ∈ ga, f ∈ (V )b and v ∈ V. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 11

Let B = {v1, . . . , vn} be a homogeneous basis of a g-module V . We denote

∗ 1 n ∗ i B = {v , . . . , v } the dual basis of V deﬁned by v (vj) = δi,j for all 1 ≤ i, j ≤ n.

Notice that since B is a homogeneous basis of V = ⊕a∈GVa, it can be partitioned into

i ∗ bases for each Va. Thus by the way we deﬁne v , we conclude that B is a homogeneous basis of V ∗. Then we compute the colour of vi by the following lemma.

i −1 Lemma 2.2.5. Let vi ∈ B have colour c. Then v has colour c .

i i i Proof. Let vi have colour c. Since v is homogeneous, v (vi) = 1 6= 0 and v (vj) = 0

i −1 for any vj not in Vc implies that v has colour c by Deﬁnition 2.2.4.

Now that we have deﬁned g-modules, we must deﬁne the morphisms between them.

Deﬁnition 2.2.6. Let V and W be g-modules. A g-module morphism from V to W is an F-linear map φ : V → W satisfying

(i) φ(xv) = x(φv) for all x ∈ g and all v ∈ V, and

(ii) φ(Va) ⊆ Wa for all a ∈ G.

The set of all g-module morphisms from V to W is denoted Homg(V,W ).

Notice that If a g-module morphism has an inverse which is also a g-module morphism, then we call it g-module isomorphism. Moreover, if V = W in Deﬁnition

2.2.6, then an element φ ∈ Homg(V,V ) is a graded operator on V which commutes with the action of g. One frequently-used g-module example in this thesis is the tensor product given by the following deﬁnition. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 12

Definition 2.2.7. Suppose V and W are two g-modules. By definition V and W are G-graded vector spaces. Then V ⊗ W is again a G-graded vector space with respect to the G-grading M (V ⊗ W )c = Va ⊗ Wb. a,b∈G ab=c We define the g-module structure on V ⊗ W by defining the g-action

x(v ⊗ w) = xv ⊗ w + β(b, a)v ⊗ xw

for all x ∈ ga, v ∈ Vb and w ∈ W .

It is straightforward to verify that Definition 2.2.7 satisfies the conditions of Definition 2.2.1.

Example 2.2.8.

(i) Let V be the trivial module, as deﬁned in Example 2.2.3. Then (V ⊗ W )a = ∼ F ⊗ Wa = Wa for all a ∈ G. In particular, V ⊗ W → W such that a ⊗ w 7→ aw for all a ∈ F is a g-module isomorphism.

¯ ¯ (ii) However, let G = Z2 = {0, 1}. Let V = V0¯ ⊕ V1¯ such that V0¯ = {0} and V1¯ = F.

Let W = W0¯ ⊕ W1¯ such that W0¯ = {0} and W1¯ = W . Let g act trivially on V . Then V ⊗ W is isomorphic to W as vector spaces. However, the map

V1¯ ⊗ W1¯ → W1¯ is not a g-module isomorphism since there does not exist a grading compatible homomorphism.

Lemma 2.2.9. Let M, N, U, V be g-modules, and let f : M → U and g : N → V be two g-module morphisms. Then

f ⊗ g : M ⊗ N → U ⊗ V 2. LIE COLOUR ALGEBRAS AND spo(V, β) 13

m ⊗ n 7→ f(m) ⊗ g(n) is again a g-module morphism.

Proof. We first verify the first condition of Definition 2.2.6. Take x ∈ ga, m ∈ Mb and n ∈ N. Then by Definition 2.2.7, x(m ⊗ n) is equal to

xm ⊗ n + β(b, a)m ⊗ xn. (2.2.1)

Since f(xm) = x(fm) and g(xn) = x(gn), applying f ⊗ g to (2.2.1) gives

f(xm) ⊗ g(n) + β(b, a)f(m) ⊗ g(xn) = x(fm) ⊗ g(n) + β(b, a)f(m) ⊗ x(gn), which is equal to xf(m) ⊗ g(n)

since f and g preserve colours. Secondly, recall from Deﬁnition 2.2.7, we have

M (M ⊗ N)c = Ma ⊗ Nb. a,b∈G ab=c

Then take any arbitrary homogeneous simple tensor ma ⊗ nb ∈ Ma ⊗ Nb. By 2.2.7, we have

(f ⊗ g)(ma ⊗ nb) = f(ma) ⊗ g(nb) ∈ Ua ⊗ Vb ⊆ (U ⊗ V )c.

By linearity, (ii) holds.

Next we give some examples of g-module morphisms which play a vital role in 2. LIE COLOUR ALGEBRAS AND spo(V, β) 14

Chapter 4.

ˇ Proposition 2.2.10. Let M,N be g-modules. Consider the linear map RM,N deﬁned on homogeneous elements m ∈ Ma and n ∈ Nb by

ˇ RM,N : M ⊗ N → N ⊗ M

m ⊗ n 7→ β(b, a)n ⊗ m.

ˇ Then RM,N is a g-module isomorphism called the braiding morphism.

ˇ Proof. First we verify that RM,N is a g-module morphism. The second condition of Definition 2.2.6 is satisfied since m ⊗ n has the same colour as n ⊗ m. For the first

condition, take x ∈ ga, m ∈ Mb and n ∈ Nc. Then we have

ˇ ˇ RM,N x(m ⊗ n) = RM,N (xm ⊗ n + β(b, a)m ⊗ xn) ˇ ˇ = RM,N (xm ⊗ n) + β(b, a)RM,N (m ⊗ xn)

= β(c, ab)n ⊗ xm + β(b, a)β(ac, b)xn ⊗ m

= β(c, a)β(c, b)n ⊗ xm + β(b, a)β(a, b)β(c, b)xn ⊗ m

= β(c, a)β(c, b)n ⊗ xm + β(c, b)xn ⊗ m

= β(c, b)xn ⊗ m + β(c, a)n ⊗ xm

= β(c, b)x(n ⊗ m) ˇ = xRM,N (m ⊗ n).

ˇ Therefore by Deﬁnition 2.2.6, RM,N is a g-module morphism. ˇ2 Moreover, since β(a, b)β(b, a) = 1 for all a, b ∈ G, we have that RM,N = 1M⊗N . ˇ Thus, RM,N is an g-module isomorphism. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 15

ˇ Let V be a g-module. The braid morphism RV,V is an isomorphism of V ⊗ V

deﬁned by swapping two factors, that is, sending va ⊗ vb to β(b, a)vb ⊗ va for all

homogenous elements va, vb ∈ V with colour a and b respectively. ˇ ˇ ⊗k Then for all k ∈ Z≥2, we deﬁne Ri = R(i,i+1) on V by

ˇ ⊗(i−1) ˇ ⊗(k−i−1) Ri = id ⊗ (−RV,V ) ⊗ id . (2.2.2)

ˇ ˇ Notice that since RV,V is a g-module isomorphism, −RV,V is also a g-module isomorphism. The reason we introduce this minus sign is that in Chapter 4, we will deﬁne a right action of the Brauer algebra on V ⊗k. Then the action of the generator

⊗k ˇ ⊗k si on V on the right will be the same as Ri acting on V on the left.

ˇ Lemma 2.2.11. The maps Ri satisfy the same relations as the standard generators of the symmetric group. Namely we have

ˇ2 (i) Ri = 1,

ˇ ˇ ˇ ˇ (ii) RiRj = RjRi if |i − j| ≥ 2,

ˇ ˇ ˇ ˇ ˇ ˇ (iii) RiRi+1Ri = Ri+1RiRi+1.

ˇ Proof. The proof is a straightforward calculation from the deﬁnition of Ri’s and the property of the symmetric bicharacter.

Consequently, for any permutation π, if we choose an expression π = si1 ··· sip as ˇ a product of adjacent transpositions, where sij = (ij, ij + 1), then we can deﬁne Rπ as

ˇ ˇ ˇ ˇ Rπ = Ri1 Ri2 ··· Rip . (2.2.3)

ˇ Lemma 2.2.11 implies that Rπ is independent of the choice of expression. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 16

Lemma 2.2.12. Let V be a g-module with homogeneous basis B = {v1, . . . , vn}. Let V ∗ be the contragredient module of V with homogeneous basis B∗ = {v1, . . . , vn} dual

to B. Let V1G be the trivial g-module. Then the following maps

∗ Pn i (i) the map prV : V1G → V ⊗ V such that 1 7→ i=1 vi ⊗ v , and

∗ i (ii) the evaluation map evV : V ⊗V → V1G such that v ⊗vj 7→ δi,j for all 1 ≤ i, j ≤ n are g-module morphisms.

Remark 2.2.13. Note that the map prV is also called the coevaluation map.

Proof. Note that prV (x1) = prV (0) = 0 for all x ∈ g. In order to prove that

prV satisfies the first condition of Definition 2.2.6, it suffices to show that for all

Pn i homogeneous elements x ∈ g, xprV (1) = x( i=1 vi ⊗ v ) = 0, that is to show Pn i Ω = i=1 vi ⊗ v is a g-invariant tensor.

Let B = {v1, . . . , vn} be a homogeneous basis of V such that for all 1 ≤ i ≤ n,

i −1 vi has colour ai ∈ G. By Lemma 2.2.5, the dual basis vector v has colour ai . Let

i x ∈ g be homogeneous with colour b ∈ G. We then write xvi and xv explicitly. First notice that n X xvi = cikvk, (2.2.4) k=1

for some cik ∈ F. Then for each 1 ≤ j ≤ n, we have

i −1 i (xv )(vj) = −β(ai , b)v (xvj) n −1 X i = −β(ai , b) cj`v (v`) `=1

−1 = −β(ai , b)cji. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 17

Thus we deduce that n i X −1 j xv = −β(ci , a)cjiv . (2.2.5) j=1 Therefore we have

n ! n n X i X i X i x vi ⊗ v = xvi ⊗ v + β(ai, b)vi ⊗ xv i=1 i=1 i=1 n n n n X X i X X −1 j = cikvk ⊗ v + −β(ai, b)β(ai , b)cjivi ⊗ v i=1 k=1 i=1 j=1 n n n n X X i X X j = cikvk ⊗ v − cjivi ⊗ v = 0. i=1 k=1 i=1 j=1

Thus follows from the fact that both 1 and Ω have colour 1, prV is a g-module

morphism. The proof of evV being a g-module morphism is similar.

Notice that the maps in Lemma 2.2.12 are independent of the choice of basis. See for example [Kas95, Section II.3].

Now let V and W be two g-modules. We notice that prV ⊗W can be obtained by the following composition of the maps:

prV ∗ ∼ ∗ id⊗prW ⊗id ∗ ∗ V1G −−→ V ⊗ V = V ⊗ V1G ⊗ V −−−−−−→ V ⊗ W ⊗ W ⊗ V .

Similarly the map evV ⊗W can be obtained by

∗ ∗ ∗ id⊗evV ⊗id ∗ ∼ ∗ evW W ⊗ V ⊗ V ⊗ W −−−−−−→ W ⊗ V1G ⊗ W = W ⊗ W −−→ V1G.

Thus inductively, we have the following lemma.

Lemma 2.2.14. Let V be a g-module. Let B = {v1, . . . , vn} be a homogeneous basis

1 n i of V . Let {v , . . . , v } be the dual of B such that v (vj) = δi,j for all 1 ≤ i, j ≤ n. Let 2. LIE COLOUR ALGEBRAS AND spo(V, β) 18

k ≥ 1. Then both of the following maps are g-module morphisms

⊗k ∗ ⊗k (i) pr˜ k : V1G → V ⊗ (V ) such that

X ik i1 1 7→ vi1 ⊗ · · · ⊗ vik ⊗ v ⊗ · · · ⊗ v , and

1≤i1,...,ik≤n

∗ ⊗k ⊗k (ii) ev˜ k :(V ) ⊗ V → V1G such that

k i1 ik Y ev˜ k(v ⊗ · · · ⊗ v ⊗ vjk ⊗ · · · ⊗ vj1 ) 7→ δik,jk ··· δi1,j1 = δi`,j` , `=1

extended by linearity.

Now let us talk about the category of g-modules.

Definition 2.2.15. Let g be a finite dimensional Lie colour algebra over a field F. We denote the category of g-modules by C. Then the morphisms are the g-module morphisms.

In particular, for any two objects V and W in C, by Deﬁnition 2.2.7, V ⊗ W is a g-module. Therefore, V ⊗ W is in C. With this extra tensor structure, C becomes a monoidal category. ∼ ∼ We showed there was an identity object V1G such that V ⊗ V1G = V1G ⊗ V = V for all V ∈ C. Also, Proposition 2.2.10 provides us an g-module isomorphism to swap V ⊗ W to W ⊗ V with some β factor generated. With this morphism, C becomes a braided monoidal category. Furthermore, for any object V in C, the dual V ∗ of V is still an g-module. With

∗ the morphisms deﬁned in Lemma 2.2.12, V1G → V ⊗ V , the category C becomes a braided rigid monoidal category of g-modules. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 19

2.3 The orthosymplectic Lie colour algebra spo(V, β)

In this section, we ﬁrst construct the general linear Lie colour algebra, gl(V, β). Then we provide a particular subalgebra of gl(V, β), the orthosymplectic Lie colour algebra spo(V, β). The rest of this paper will mainly focus on this special case.

Let V be a graded vector space over a ﬁeld F, and let G be an abelian group. Let β be a symmetric bicharacter of G. Let End(V ) be the vector space of all F-linear maps from V to V . We let

gl(V, β)a := {x ∈ End(V ) | xVb ⊆ Vab for all b ∈ G}. (2.3.1)

Deﬁnition 2.3.1. The general linear Lie colour algebra is the vector space

M gl(V, β) = gl(V, β)a a∈G equipped with the Lie colour algebra bracket deﬁned by

[x, y] = xy − β(b, a)yx

for all x ∈ gl(V, β)a, y ∈ gl(V, β)b.

Remark 2.3.2. Notice that as a vector space, gl(V, β) is isomorphic to End(V ).

Lemma 2.3.3. Equipped with the Lie colour bracket deﬁned above, gl(V, β) is a Lie colour algebra.

Proof. We need to show that gl(V, β) satisﬁes Deﬁnition 2.1.6.

(i) Let x ∈ gl(V, β)a, y ∈ gl(V, β)b and v ∈ Vc for some a, b and c ∈ G. Then we 2. LIE COLOUR ALGEBRAS AND spo(V, β) 20

have [x, y]v = xyv − β(b, a)yxv.

For the ﬁrst term, we have

xyv ∈ xyVc ⊆ xVbc ⊆ Vabc

and similarly for the second term we have

β(b, a)yxv ∈ yxVc ⊆ yVac ⊆ Vbac = Vabc

since G is abelian. So we have [x, y]Vc ⊆ Vabc holds for all c ∈ G. So [x, y] ∈ Vab.

Therefore we have shown that [ga, gb] ⊆ gab.

(ii) Take x ∈ gl(V, β)a and y ∈ gl(V, β)b. Then we have

−β(b, a)[y, x] = −β(b, a)yx + β(b, a)β(a, b)xy = [x, y].

(iii) The Lie colour Jacobi identity can be proved by a straightforward calculation by using the relations β(a, b)β(b, a) = 1 and β(ab, c) = β(a, c)β(b, c) for all a, b, c ∈ G.

Remark 2.3.4. We adopt the convention that if v ∈ Va and u ∈ Vb, then we write

β(v, u) = β(a, b), and gl(V, β)v = gl(V, β)a.

That is, we use a vector itself to denote its colour and by extension, we write v−1 for a−1. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 21

We next construct a homogeneous basis for gl(V, β) and discuss the colours of its basis vectors.

Let B = {v1, . . . , vn} be a homogeneous basis of V . Recall that the elementary

matrix Evivj in End(V ) is deﬁned by Evivj vk = δjkvi for all 1 ≤ i, j, k ≤ n, where δjk is the Kronecker delta.

Lemma 2.3.5. The set {Evivj | 1 ≤ i, j ≤ n} forms a homogeneous basis for gl(V, β).

Proof. By Remark 2.3.2, gl(V, β) is isomorphic to End(V ) as vector space. Thus the

set {Evivj | 1 ≤ i, j ≤ n} forms a basis for gl(V, β). Then since Evivj sends vi to vj

−1 for all 1 ≤ i, j ≤ n, by (2.3.1) it has colour viv . Therefore Ev v ∈ gl(V, β) −1 and j i j vivj thus it is homogeneous.

Now we begin to construct the orthosymplectic Lie colour algebras.

Deﬁnition 2.3.6. Let V be a G-graded vector space. An F-bilinear map

h·, ·i : V × V → F is called a β-skew-symmetric bilinear form if it satisﬁes:

(i) h·, ·i is nondegenerate;

−1 (ii) hVa,Vbi = 0 whenever b 6= a ; and

(iii) hv, wi + β(b, a)hw, vi = 0, for all v ∈ Va, w ∈ Vb.

Remark 2.3.7. From part (ii) of Deﬁnition 2.3.6, it follows that part (iii) holds trivially for b 6= a−1. If however, b = a−1, we have β(b, a) = β(a−1, a) = β(a, a)−1 =

−1 ±1. Therefore for all v ∈ Va and w ∈ Wa−1 , hv, wi + β(a , a)hw, vi = 0 implies hv, wi = ±hw, vi. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 22

We explore this form further in Section 2.4.

Deﬁnition 2.3.8. The orthosymplectic Lie colour algebra

M spo(V, β) = spo(V, β)a a∈G is the subalgebra of the Lie colour algebra gl(V, β), where for each a ∈ G, the graded component spo(V, β)a is deﬁned as

spo(V, β)a = {x ∈ gl(V, β)a | hxw, vi+β(b, a)hw, xvi = 0, ∀w ∈ Vb, v ∈ V }. (2.3.2)

Lemma 2.3.9. Equipped with the Lie colour bracket deﬁned above, spo(V, β) is a Lie colour algebra.

Proof. Since for all a ∈ G, spo(V, β)a ⊆ gl(V, β)a. we have spo(V, β) ⊆ gl(V, β) as G-graded vector spaces. Next we prove that spo(V, β) is closed under the Lie colour bracket. It suﬃces to show that for homogeneous elements x and y in spo(V, β), we have

[x, y] ∈ spo(V, β)xy. Therefore it is enough to show that

h[x, y]w, vi = −β(w, xy)hw, [x, y]vi, for all v ∈ V.

The method is to expand [x, y] = xy − β(y, x)yx ﬁrst, and use the linearity of h·, ·i. Therefore h[x, y]w, vi = hxyw, vi − β(y, x)hyxw, vi. (2.3.3) 2. LIE COLOUR ALGEBRAS AND spo(V, β) 23

The ﬁrst term of (2.3.3) can be written as

hxyw, vi = −β(yw, x)hyw, xvi

= β(yw, x)β(w, y)hw, yxvi

= β(w, xy)β(y, x)hw, yxvi

= β(w, xy)hw, β(y, x)yxvi. (2.3.4)

Similarly, the second term of (2.3.3) is −β(y, x)β(xw, y)β(w, x)hw, xyvi which can be simpliﬁed to − β(w, xy)hw, xyvi. (2.3.5)

Thus by adding (2.3.4) and (2.3.5) it follows that (2.3.5) is equal to

β(w, xy)hw, (β(y, x)yx − xy)vi which is −β(w, xy)hw, [x, y]vi as we claimed.

2.4 Homogeneous bases for V , V ∗ and spo(V, β)

In this section, starting with a G-graded vector space carrying a β-skew-symmetric invariant bilinear form, we give homogeneous bases for a G-graded vector space V , its contragredient V ∗, and spo(V, β). Recall that from (2.1.3), we deﬁne

M M V(0) = Va and V(1) = Va.

a∈G(0) a∈G(1)

Lemma 2.4.1. Let h·, ·i be as in Deﬁnition 2.3.6. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 24

(i) The restricted form h·, ·i : V(0) ×V(0) → F is a (nondegenerate) symplectic bilinear form.

(ii) The restricted form h·, ·i : V(1) × V(1) → F is a nondegenerate and symmetric bilinear form.

−1 Proof. Let a, b ∈ G. If ab 6= 1, then hVa,Vbi = 0. So we may assume that b = a . Now

−1 −1 −1 −1 if a ∈ G(i) for i = 0, 1, then by Lemma 2.1.2, β(a , a ) = β(a, a ) = β(a, a) = ±1

−1 implies a ∈ G(i). Thus for all v ∈ Va and w ∈ Va−1 , we have

hv, wi = −β(a−1, a)hw, vi = ±hw, vi.

Thus h·, ·i : V(0) × V(0) → F is skew-symmetric, and h·, ·i : V(1) × V(1) → F is symmetric. Next we prove non-degeneracy. By Deﬁnition 2.3.6, h·, ·i : V × V → F is a nondegenerate form. Then for each nonzero v in V , there exists some w in V such that P hv, wi 6= 0. Write w = b∈G wb as a sum of homogeneous vectors. By Deﬁnition

−1 2.3.6 (ii), hv, wi = hv, wbi where b = a . Without loss of generality, we can replace

−1 −1 w with wb. Now if v ∈ V(i) for i = 0, 1, then ±1 = β(v, v) = β(w , w ) = β(w, w).

Therefore w is in V(i) as well. Therefore h·, ·i : V(i) × V(i) → F is a non-degenerate form for i = 0, 1.

Let us ﬁx our notation. From now on, we ﬁx

dim V(0) = m and dim V(1) = n.

Moreover since h·, ·i : V(0) × V(0) → F is symplectic, we have m = 2r for some r ∈ Z≥0.

But V(1) can have odd or even dimension. We deﬁne s ∈ Z≥0 by declaring n = 2s or n = 2s + 1. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 25

We claim that there exists a homogeneous basis

∗ ∗ B(0) = {t1, t1, . . . , tr, tr}

of V(0) which decomposes V(0) into an orthogonal direct sum of hyperbolic planes

∗ Hi = Span{ti, ti }. Namely, let t ∈ V(0) be homogeneous and nonzero. Then as in the

0 0 proof of Lemma 2.4.1, there exists a homogeneous t ∈ V(0) such that ht, t i= 6 0. Also since char F 6= 2, we can assume ht, ti = ht0, t0i = 0. Thus, replacing t0 by a scalar multiple if necessary, this gives a hyperbolic pair {t, t0}. Repeating this process on the orthogonal complement of Span{t, t0} completes the proof.

If F is algebraically closed, by a similar argument, there exists a homogeneous basis

∗ ∗ B(1) = {u1, u1, . . . , us, us, (us+1)}

∗ of V(1) where each {ui, ui } is a hyperbolic pair. The vector us+1 is only included if the dimension of V(1) is odd, and we put a bracket around such u2s+1 to indicate this.

∗ In this case, we have us+1 = us+1. When F is not algebraically closed, we add the additional hypothesis that V(1) admits a basis B(1) of the above form. Moreover, we make the convention that we extend the deﬁnition of * such that v∗∗ = v for all v ∈ B. Therefore we have the following homogeneous basis for V :

∗ ∗ ∗ ∗ B = B(0) ∪ B(1) = {t1, t1, . . . , tr, tr, u1, u1, . . . , us, us, (us+1)}. (2.4.1)

Remark 2.4.2. In what follows, we frequently need the basis vectors with no ∗ signs. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 26

Therefore for future reference, we let

0 0 0 B = {t1, ··· , tr, u1, ··· , us, (us+1)} = B(0) ∪ B(1)

0 0 where B(0) = {t1, ··· , tr} and B(1) = {u1, ··· , us, (us+1)}.

Since h·, ·i is a nondegenerate form, it induces an isomorphism F : V → V ∗ by v 7→ hv, ·i for all v ∈ V . Next we prove in fact F is a g-module isomorphism between V and its contragredient.

Lemma 2.4.3. The map

F : V → V ∗

v 7→ hv, ·i is a g-module isomorphism.

Proof. It suﬃces to prove that for all homogeneous x ∈ g and homogeneous v, w ∈ V we have F (xv)(w) = (xF (v))(w). Since F (v) ∈ V ∗, by Deﬁnition 2.2.4, we have

(xF (v))(w) = −β(F (v), x)F (v)(xw) = −β(F (v), x)hv, xwi.

By (2.3.2), we have the relation

hxv, wi + β(F (v), x)hv, xwi = 0 which implies that

F (xv)(w) = hxv, wi = −β(F (v), x)hv, xwi = (xF (v))(w) 2. LIE COLOUR ALGEBRAS AND spo(V, β) 27

for all w ∈ V . Note that F is bijective, and let F −1 be the inverse linear map of F . Let w ∈ V ∗ and notice that F (x(F −1w)) = (xF )(F −1w) = xw, thus applying F −1 to both sides we get (xF −1)(w) = F −1(xw) for all x ∈ g and w ∈ V ∗. Therefore F −1 is a g-module morphism. Thus, F is a g-module isomorphism.

Lemma 2.4.4. With the above setting, if vi ∈ B has colour c, then

(i) F (vi) has colour c,

(ii) F −1(vi) has colour c−1.

Proof. The result is immediate by using the facts that both F and F −1 are g-module

i morphisms and v has opposite colour as vi.

We recall the following fact. Let B = {v1, . . . , vn} be an ordered basis of (V, h·, ·i).

1 n ∗ i Then denote by {v , . . . , v } the dual basis of V deﬁned by v (vj) = δij for all

∗ 1 ≤ i, j ≤ n. With respect to these bases, the matrix FB of the map F : V → V such

−1 that v 7→ hv, ·i is given by (FB)i,j = Fi,j = hvi, vji. We follow [BSR98] to use Fi,j to −1 denote the (i, j) entry of the inverse matrix FB . Now based on the homogeneous basis B in (2.4.1), we can ﬁnd the explicit matrix

FB. By the calculation above (2.4.1), for i = 1, . . . , r and j = 1, . . . , s, (s + 1), we have

F ∗ = 1,F ∗ = −1,F ∗ = 1,F ∗ = 1, and F 0 = 0 otherwise. (2.4.2) ti,ti ti ,ti uj ,uj uj ,uj v,v

−1 The inverse of the matrix FB is the matrix FB with

−1 −1 −1 −1 F ∗ = −1,F ∗ = 1,F ∗ = 1,F ∗ = 1, and Fv,v0 = 0 otherwise. (2.4.3) ti,ti ti ,ti uj ,uj uj ,uj 2. LIE COLOUR ALGEBRAS AND spo(V, β) 28

0 ∗ Lemma 2.4.5. For all v, w ∈ B , we have Fv∗,w = −β(w, v )Fw,v∗ .

Proof. Since all v, w ∈ B are homogeneous, we have

∗ ∗ ∗ ∗ Fv∗,w = hv , wi = −β(w, v )hw, v i = −β(w, v )Fw,v∗ .

Theorem 2.4.6. The following vectors form a basis for spo(V, β), as v and w range over B0 as indicated:

∗ 0 (i) Yv∗,w = Ev∗,w + β(v, v)β(wv , w)Ew∗,v, where v 6= w if v, w ∈ B(1);

∗ 0 (ii) Yv,w∗ = Ev,w∗ + β(v , w)Ew,v∗ , where v 6= w if v, w ∈ B(1);

(iii) Yv,w = Ev,w − β(w, w)β(v, w)Ew∗,v∗ .

Proof. First, it is straightforward that the vectors in Theorem 2.4.6 are linearly independent. Next, we first show that those vectors are in spo(V, β). Then we show that they span spo(V, β). It suffices to find all the homogeneous vectors x ∈ gl(V, β) that satisfy (2.3.2). Now take a homogeneous x ∈ gl(V, β). We have the relation hxv, wi + β(v, x)hv, xwi = 0, which is equivalent to

t t t v x FBw + β(v, x)v FBxw = 0.

First, letting v and w run over all pairs of vectors in the set B0, we can reduce the above matrix equation into equations of the form:

t (x )v,w∗ Fw∗,w + β(v, x)Fv,v∗ xv∗,w = 0. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 29

After taking the transpose, we have

xw∗,vFw∗,w + β(v, x)Fv,v∗ xv∗,w = 0. (2.4.4)

∗ By Lemma 2.4.5, we can replace Fw∗,w by −β(w, w )Fw,w∗ , and the equation becomes

∗ −xw∗,vβ(w, w )Fw,w∗ + β(v, x)Fv,v∗ xv∗,w = 0, which is equivalent to

∗ −1 Fw,w∗ xw∗,v = β(w, w ) β(v, x)Fv,v∗ xv∗,w.

0 ∗ −1 By using the fact that Fv,v∗ = 1 for all v ∈ B , and the fact that β(w, w ) = β(w, w) for all w ∈ B0, we have

xw∗,v = β(w, w)β(v, x)xv∗,w. (2.4.5)

If xv∗,w =6 0, then since x is homogeneous in gl(V, β), it must have the same colour as

∗ ∗ ∗ ∗ Ev∗,w which is v w . Therefore (2.4.5) becomes xw∗,v = β(w, w)β(v, v w )xv∗,w, which (by using the fact β(v, v∗) = β(v, v) for all v ∈ B0 and the fact β(v∗, w) = β(v, w∗)) is

∗ xw∗,v = β(w, w)β(v, v)β(v , w)xv∗,w, (2.4.6)

which also holds if xv∗,w = 0. Similarly, for all v, w ∈ B0, we have the relation

∗ xw,v∗ = β(v , w)xvw∗ , (2.4.7) 2. LIE COLOUR ALGEBRAS AND spo(V, β) 30

and for all v, w ∈ B0, we have

xw∗,v∗ = −β(w, w)β(v, w)xv,w. (2.4.8)

0 Note that when v = w in B(1), β(v, v) = −1. Therefore in these cases (2.4.7) and

(2.4.6) imply xv,v∗ = xv∗,v = 0. Thus x is a homogeneous element of spo(V, β) if and only if its entries satisfy (2.4.6), (2.4.7) and (2.4.8). Comparing these conditions with the list of vectors in the statement of the proposition yields:

∗ (i) Yv∗,w = Ev∗,w + β(v, v)β(wv , w)Ew∗,v ∈ spo(V, β)v∗w∗ , where v =6 w if v, w ∈

0 B(1);

∗ 0 (ii) Yv,w∗ = Ev,w∗ + β(v , w)Ew,v∗ ∈ spo(V, β)vw, where v 6= w if v, w ∈ B(1);

(iii) Yv,w = Ev,w − β(w, w)β(v, w)Ew∗,v∗ ∈ spo(V, β)vw∗ which proves that the vectors in Theorem 2.4.6 are all in spo(V, β). Now in order to show that these vectors span spo(V, β), we ﬁrst give an order to the basis B. Namely, we write

∗ ∗ ∗ ∗ B = {t1 < t1 < . . . < tr < tr < u1 < u! < . . . < us < us < (us+1)}.

Take arbitrary x ∈ spo(V, β). We let xv,w be the (v, w)-entry for all v, w ∈ B. Then P we write x = v,w∈B xv,wEv,w which can be regrouped as

X X x = (xv,wEv,w + xw∗,v∗ Ew∗,v∗ ) + (xv∗,wEv∗,w + xw∗,vEw∗,v) v,w∈B0 v

X X + (xv,w∗ Ev,w∗ + xw,v∗ Ew,v∗ ) + (xv∗,vEv∗,v + xv,v∗ Ev,v∗ ). (2.4.9) v

Applying the relation (2.4.8) to the ﬁrst term yields

X X xv,w (Ev,w − β(w, w)β(v, w)Ew∗,v∗ ) = xv,wYv,w. v,w∈B0 v,w∈B0

Similarly, applying relation (2.4.6) and (2.4.7) to the second and third terms of (2.4.9) yields X X xv∗,wYv∗,w and xv,w∗ Yv,w∗ v

respectively. For the fourth term of (2.4.9), notice that Yv,v∗ = 2Ev,v∗ and Yv∗v = 2Ev∗,v. Thus the fourth term of (2.4.9) is

X 1 (xv∗,vYv∗,v + xv,v∗ Yv,v∗ ). 0 2 v∈B(0)

Therefore we can write any arbitrary x ∈ spo(V, β) as

X X X 1 X x = x Y + x ∗ Y ∗ + x ∗ Y ∗ + (x ∗ Y ∗ +x ∗ Y ∗ ). v,w v,w v ,w v ,w v,w v,w 2 v ,v v ,v v,v v,v v,w∈B0 v

Thus, the vectors in Theorem 2.4.6 span spo(V, β), and in turn, they form a homogeneous basis of spo(V, β).

0 Remark 2.4.7. When v = w ∈ B , the vector Yv,w simpliﬁes to Hv = Ev,v − Ev∗,v∗ . Notice that these elements always have colour 1, regardless of the grading on V . 2. LIE COLOUR ALGEBRAS AND spo(V, β) 32

2.5 Roots and root vectors in spo(V, β)

In this section, let g = spo(V, β). We compute the weights of each basis vector in (2.4.1) of V under the standard action of g. Then we construct a root system of g with respect to the homogeneous basis in Theorem 2.4.6. The ﬁrst step is to ﬁnd a Cartan subalgebra of g and its dual. Recall that we have a homogeneous basis

∗ ∗ ∗ ∗ B = {t1, t1, . . . , tr, tr, u1, u1, . . . , us, us, (us+1)}

0 of V , and the corresponding subset B = {t1, . . . , tr, u1, . . . , us, (us+1)} of V .

Deﬁnition 2.5.1. Let Hv = Ev,v − Ev∗,v∗ as in Remark 2.4.7. We call

0 h = SpanF{Hv | v ∈ B }. the standard Cartan subalgebra of g.

Lemma 2.5.2. The set h is an abelian subalgebra of g.

Proof. Take Hv and Hw in h. We have [Hv,Hw] = HvHw − β(Hw,Hv)HwHv which

is HvHw − HwHv since Hv has colour 1G. Also since Hv is diagonal matrix for all

0 0 v ∈ B , Hv commutes with Hw. Therefore we have [Hv,Hw] = 0 for all v ∈ B . Thus h is an abelian subalgebra.

Deﬁnition 2.5.3. An g-module W has a weight space decomposition with respect to h∗ if M W = Wα, α∈h∗ 2. LIE COLOUR ALGEBRAS AND spo(V, β) 33

where Wα = {w ∈ W | Hw = α(H)w for all H ∈ h}. We call α a weight if the corresponding weight space Wα is nonzero, and the nonzero elements of a given weight space are called weight vectors.

Deﬁnition 2.5.4. The dual basis of h is the set

∗ 0 {αv ∈ h | αv(Hw) = δvw for all v, w ∈ B }. (2.5.1)

Let us prove that the αv’s in (2.5.1) are weights of the standard representation of spo(V, β) on V .

Lemma 2.5.5. Each basis vector w in B is a weight vector. For all w ∈ B0, the

∗ weight of w is αw and the weight of w is −αw. In particular, the weight of us+1 ∈ B is 0.

0 Proof. Let v, w ∈ B and take Hv = Ev,v − Ev∗,v∗ in h. Then we have

(Ev,v − Ev∗,v∗ )w = Ev,vw = δw,vw = αw(Hv)w.

Therefore the weight of w is αw. On the other hand, we have

∗ ∗ ∗ ∗ ∗ (Ev,v − Ev∗,v∗ )w = −Ev∗,v∗ w = −δw∗,v∗ w = −δw,vw = −αw(Hv)w .

∗ Thus w has weight −αw.

Lemma 2.5.6. Let v = v1 ⊗ · · · ⊗ vk with each vi ∈ B. Then the weight of v is given by k X αv = αvi . i=1 2. LIE COLOUR ALGEBRAS AND spo(V, β) 34

Proof. Each H ∈ h acts on v1 ⊗ · · · ⊗ vk diagonally as per Deﬁnition 2.2.7. Since the

colour of H is 1, we have β(H, vi) = 1 for all vi ∈ B. Thus we have

k X H · (v1 ⊗ · · · ⊗ vk) = v1 ⊗ · · · ⊗ v`−1 ⊗ (H · v`) ⊗ v`+1 ⊗ · · · ⊗ vk `=1 k X = v1 ⊗ · · · ⊗ v`−1 ⊗ αv` (H)v` ⊗ v`+1 ⊗ · · · ⊗ vk `=1 k X = αv` (H)(v1 ⊗ · · · ⊗ vk). `=1

Therefore we have shown that the tensor v1 ⊗ · · · ⊗ vk is a weight vector with weight Pk i=1 αvi .

Next we provide a root system of g. First notice that the basis vectors deﬁned in Theorem 2.4.6 are all weight vectors under the adjoint representation.

Lemma 2.5.7. For vectors u, v, w in B0, we have the following:

(i) [Hu,Yv,w∗ ] = (αv + αw)(Hu)Yv,w∗ ,

(ii) [Hu,Yv∗,w] = (−αv − αw)(Hu)Yv∗,w,

(iii) [Hu,Yv,w] = (αv − αw)(Hu)Yv,w.

Proof. We only prove (i), and the rest can be calculated similarly. Since Yv,w∗ =

∗ 0 Ev,w∗ + β(v , w)Ew,v∗ , notice that Ev,w∗ Eu,u = 0 = Eu∗,u∗ Ev,w∗ for all u, v, w ∈ B .

Thus we have [Hu,Yv,w∗ ] = Eu,uYv,w∗ − Yv,w∗ Eu∗,u∗ . Therefore

[Hu,Yv,w∗ ] = Eu,uYv,w∗ − Yv,w∗ Eu∗,u∗

∗ ∗ = (Eu,uEv,w∗ + β(v , w)Eu,uEw,v∗ ) − (−Ev,w∗ Eu∗,u∗ − β(v , w)Ew,v∗ Eu∗,u∗ )

∗ ∗ = δuvEu,w∗ + δu,wβ(v , w)Eu,v∗ + δwuEv,u∗ + δuvβ(v , w)Ew,u∗ 2. LIE COLOUR ALGEBRAS AND spo(V, β) 35

∗ ∗ = δuv(Eu,w∗ + β(v , w)Ew,u∗ ) + δuw(Ev,u∗ + β(v , w)Eu,v∗ )

= (αv + αw)(Hu)(Yv,w∗ )

We now give deﬁnitions of roots and the root space decomposition of g.

Deﬁnition 2.5.8. Let h = g0 be the Cartan subalgebra of g introduced in Deﬁnition 2.5.1. The root space decomposition of g relative to h is given by

M g = h ⊕ gα α∈Φ

∗ where Φ := {α ∈ h \{0} | gα 6= 0} where gα := {x ∈ g | [H, x] = α(H)x, ∀H ∈ h}.

We call α a root if α ∈ Φ and call the associated gα a root space.

0 All the vectors Hv lie in h = g0. Moreover, by Lemma 2.5.7, for all v, w ∈ B

∗ ∗ the basis vectors Yv,w , Yv ,w and Yv,w are in the distinct nonzero root spaces gαv+αw ,

g−αv−αw and gαv−αw respectively. Therefore we can conclude that the root spaces are each one-dimensional.

Remark 2.5.9. From the subscripts of our notation, we can read both the colour of

a root vector, and the value of its corresponding root. For example, Yv,w∗ has colour

−1 −1 ∗ vw and it lies in the space gαv+αw . Yv ,w has colour v w , and it is in the root space

−1 g−αv−αw . Yv,w has colour vw , and it is in the root space gαv−αw .

In summary, as shorthand notation, we set εi = αti and δj = αuj for 1 ≤ i ≤ r and 1 ≤ j ≤ s. The roots of spo(V, β), are:

(i) When n = 2s + 1,

Φ0 = {±(εi ± εj), ±2εi, ±(δk ± δl), ±δk | 1 ≤ i =6 j ≤ r, 1 ≤ k 6= l ≤ s}

Φ1 = {±(εi ± δj), ±εi | 1 ≤ i ≤ r, 1 ≤ j ≤ s}; 2. LIE COLOUR ALGEBRAS AND spo(V, β) 36

(ii) When n = 2s

Φ0 = {±(εi ± εj), ±2εi, ±(δk ± δl) | 1 ≤ i 6= j ≤ r, 1 ≤ k 6= l ≤ s}

Φ1 = {±(εi ± δj) | 1 ≤ i ≤ r, 1 ≤ j ≤ s}.

We denote the set of roots of spo(V, β) by Φ = Φ0 ∪ Φ1. The root system we deﬁned is not like the usual root system of Lie algebras. For example, when n = 2s + 1, we

can have both ±εi and ±2εi as roots. However Φ0 and Φ1 are the set of even and odd roots for the Lie superalgebra spo(2r|2s + 1), see for example [FSS00, Table 6 and 9].

Lemma 2.5.10. The union of the following sets

(i) When n = 2s + 1,

+ Φ0 = {εi ± εj, 2ε1, 2εj, δk ± δl, δ1, δl | 1 ≤ i < j ≤ r, 1 ≤ k < l ≤ s}

+ Φ1 = {εi ± δj, εi | 1 ≤ i ≤ r, 1 ≤ j ≤ s};

(ii) When n = 2s,

+ Φ0 = {εi ± εj, 2ε1, 2εj, δk ± δl | 1 ≤ i < j ≤ r, 1 ≤ k < l ≤ s}

+ Φ1 = {εi ± δj | 1 ≤ i ≤ r, 1 ≤ j ≤ s}. are positive roots of spo(V, β), denoted by Φ+.

Proof. We see directly that Φ = Φ+ ∪ −Φ+, and that if two roots in Φ+ are such that their sum is a root, then the sum is in Φ+. Hence Φ+ is a positive root system.

Deﬁnition 2.5.11. Given a set of positive roots, the corresponding set of simple roots ∆ is a basis of V such that each α ∈ Φ+ is a nonnegative integral linear combination of elements in ∆. 2. LIE COLOUR ALGEBRAS AND spo(V, β) 37

Lemma 2.5.12. With respect to the positive root system above, if r = 1, s = 1 and n = 2s, the set of simple roots ∆ is given by

∆ = {ε1 + δ1, ε1 − δ1}.

Otherwise, the set of simple roots ∆ = {γ1, . . . , γr+s} is given by

γi = εi − εi+1, 1 ≤ i ≤ r − 1, γr+j = δj − δj+1, 1 ≤ j ≤ s − 1    εr if n = 1,  δs if n = 2s + 1, γr = γr+s =  εr − δ1 otherwise,  δs−1 + δs if n = 2s,

where we adopt the convention that εi = δi = 0 if i ≤ 0.

Proof. It can be checked that ∆ satisﬁes Deﬁnition 2.5.11.

Deﬁnition 2.5.13. The set of simple root vectors ∆Y is denoted {Yγ|γ ∈ ∆}. If

r = 1, s = 1 and n = 2s, then ∆Y is the set

∗ {E + β(t , u )E ∗ ∗ ,E ∗ + β(t , u )E ∗ }. t1,u1 1 1 u1,t1 t1,u1 1 1 u1,t1

Otherwise, the Yγ’s are explicitly given by:

(i) Y = E − β(t , t )E ∗ ∗ , where 1 ≤ i ≤ r − 1, γi ti,ti+1 i i+1 ti+1,ti   ∗  Etr,u1 + β(tr, u1)Eu1,tr , if n = 1, (ii) Yγr = ∗  E + β(t , u )E ∗ ∗ , otherwise,  tr,u1 r 1 u1,tr

(iii) Y = E + β(u , u )E ∗ ∗ , where 1 ≤ j ≤ s − 1, γr+j uj ,uj+1 j j+1 uj+1,uj   ∗  Eus,us+1 + β(us, us+1)Eus+1,us , if n = 2s + 1, (iv) Yγr+s = ∗  E ∗ + β(u , u )E ∗ , if n = 2s.  us,us−1 s s−1 us−1,us 2. LIE COLOUR ALGEBRAS AND spo(V, β) 38

Lemma 2.5.14. Let v be a weight vector with weight α. Let Y be a simple root vector in ∆Y with weight λ. Then the weight of Y v is α + λ.

Proof. Notice that since β(H, ·) = 1, we have

H(Y v) = [H,Y ]v + Y (Hv) = λ(H)Y v + Y α(H)v, which is (λ + α)(H)Y v.

Example 2.5.15. Take r = 1, s = 2, m = 2r, n = 2s. Then V has the basis

∗ ∗ ∗ B = {t1, t1, u1, u1, u2, u2}.

An arbitrary H in h is of the form H = a1Ht1 + a2Hu1 + a3Hu2 , which in matrix form is given with respect to the basis B by the following diagonal matrix:

  a1      −a   1       a2  H =      −a2       a   3    −a3 where the oﬀ-diagonal entries are all 0’s.

In the following matrix, we label each Evw by the unique root, such that the 2. LIE COLOUR ALGEBRAS AND spo(V, β) 39

corresponding root space has a nonzero projection onto Span{Evw}.

  0 ε1 + ε1 ε1 − δ1 ε1 + δ1 ε1 − δ2 ε1 + δ2     −ε − ε 0 −ε − δ δ − ε −ε − δ −ε + δ   1 1 1 1 1 1 1 2 1 2      δ1 − ε1 δ1 + ε1 0 0 δ1 − δ2 δ1 + δ2    .   −δ1 − ε1 ε1 − δ1 0 0 −δ1 − δ2 −δ1 + δ2      δ − ε δ + ε δ − δ δ + δ 0 0   2 1 2 1 2 1 2 1    −δ2 − ε1 −δ2 + ε1 −δ2 − δ1 −δ2 + δ1 0 0

The set of simple roots in this case is {ε1 − δ1, δ1 − δ2, δ1 + δ2}.

Finally we give the deﬁnition of highest weight vectors.

Deﬁnition 2.5.16. Let W be an spo(V, β)-module. Let g+ be the span of the positive root vectors. Then a vector w in W is a highest weight vector of weight λ ∈ h∗ if the following holds:

(i) Xw = 0, for all X ∈ g+,

(ii) Hw = λ(H)w for all H ∈ h. Chapter 3

The Poincar´e-Birkhoﬀ-Witt Theorem for Lie Colour Algebras

In this chapter, our goal is to state and prove the Poincaré-Birkhoff-Witt(PBW) theorem for Lie colour algebras. The statement and proof of the PBW theorem for Lie algebras in the usual sense can be obtained from our proof in this chapter by choosing β to be the trivial bicharacter of the trivial group. Our proof is inspired by [Hum72, Chapter 17] and [Car05, Chapter 9]. In Section 3.1, we give the definition of tensor algebra and symmetric algebra for any vector spaces, and briefly discuss their grading. In Section 3.2, we provide the definition of the universal enveloping algebra U(g) of any arbitrary Lie colour algebra g. Then in Section 3.3, we provide a spanning set of U(g), which will be proved to be a basis of U(g) in Section 3.4. This existence of this basis is known as one of the versions of the PBW theorem. Thereafter, assuming Proposition 3.4.2, we prove another version of the PBW theorem which states that the associated graded algebra of U(g) is isomorphic to the symmetric algebra. Moreover we give an application of

40 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 41

the PBW theorem at the end of this section. We give the technical details of the proof of Proposition 3.4.2 in Section 3.5.

3.1 Tensor algebras and symmetric algebras

Unlike Chapter 2, in this chapter, g always denotes an arbitrary Lie colour algebra

over a ﬁeld F. Nevertheless, all of the deﬁnitions in this section make sense if g is just a G-graded vector space.

th Deﬁnition 3.1.1. For any m ∈ Z+, the m tensor product of g is deﬁned to be the vector space T m(g) = g ⊗ g · · · ⊗ g | {z } m times that is, T m(g) consists of all linear combinations of m-tensors on g.

Note that by convention, T 0(g) = F.

Deﬁnition 3.1.2. The tensor algebra T (g) of g is an associative algebra with unity,

m deﬁned to be the direct sum of T (g) for all m ∈ Z≥0 :

∞ M i T (g) = T (g) = F ⊕ g ⊕ (g ⊗ g) ⊕ · · · . i=0

The multiplication on homogeneous generators of T (g) is deﬁned by the rule:

k+m (v1 ⊗ · · · ⊗ vk) · (w1 ⊗ · · · ⊗ wm) = v1 ⊗ · · · ⊗ vk ⊗ w1 ⊗ · · · ⊗ wm ∈ T (g)

where k, m ∈ Z+, v1, . . . , vk, w1, . . . , wm ∈ g. Note that with this multiplication rule, T (g) is a graded algebra. 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 42

Deﬁnition 3.1.3. We say a G-graded algebra A is colour-commutative if for all homogeneous elements x, y in A, we have xy = β(y, x)yx .

The tensor algebra is not colour-commutative if dim(g) > 1. We construct a new algebra called the β-symmetric algebra, in which the multiplication is colour- commutative.

Deﬁnition 3.1.4. The β-symmetric algebra of g is deﬁned to be the quotient algebra

T (g) S(g) = I where I is the two sided ideal generated by x ⊗ y − β(y, x)y ⊗ x for all homogeneous x, y in g.

Let us discuss more about I and S(g). We ﬁrst show that I is a homogeneous ideal. Since I is a two-sided ideal generated by x ⊗ y − β(y, x)y ⊗ x, I is spanned as a vector space by elements of the form v · (x ⊗ y − β(y, x)y ⊗ x) · w for some

i homogeneous x, y in g and v, w ∈ T (g). Moreover, since T (g) = ⊕i≥0T (g) is a direct sum of homogeneous components, we can view I as the span of elements of the form

m n v · (x ⊗ y − β(y, x)y ⊗ x) · w, for some v ∈ T (g) and w ∈ T (g) where m, n ∈ Z+. Therefore, in this case we have

v · (x ⊗ y − β(y, x)y ⊗ x) · w ∈ I ∩ T m+n+2(g).

Therefore, since I is spanned by homogeneous elements, it is a homogeneous ideal and it admits a grading of the form

M I = Ij where Ij = I ∩ T j(g). j≥0 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 43

Therefore S(g) is also a graded algebra, with grading

M S(g) = Si(g) i≥0

i i i T (g) T (g) where S (g) = Ii = I ∩ T i(g) for all i ≥ 0.

Lemma 3.1.5. Let y1, . . . , yn ∈ g be homogenous elements. Let π ∈ Sn. Then there

× is a scalar βπ(y1, . . . , yn) ∈ F such that in S(g), we have

y1 ⊗ · · · ⊗ yn + I = βπ(y1, . . . , yn)yπ(1) ⊗ · · · ⊗ yπ(n) + I.

Proof. By the discussion above, for homogeneous elements x, y ∈ g, v ∈ T m(g) and w ∈ T n(g), we have

v ⊗ x ⊗ y ⊗ w = β(y, x)v ⊗ y ⊗ x ⊗ w (3.1.1)

in S(g). Therefore the result follows from the fact that the scalar βπ is obtained by writing π as a product of adjacent transpositions, and inductively applying relation (3.1.1).

Remark 3.1.6. Indeed the function βπ(y1, . . . , yn) depends only on the colours of

y1, . . . , yn.

Deﬁnition 3.1.7. Let

m m M i M i Tm = T (g) and Sm = S (g) i=0 i=0 be the mth ﬁltration subspaces of T (g) and S(g) respectively. 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 44

Now for any G-graded vector space V , we give a basis for S(g). We ﬁrst need the following lemma.

Lemma 3.1.8. Let V be a G-graded vector space with G-graded decomposition V =

U ⊕ W . Then S(U ⊕ W ) ∼= S(U) ⊗ S(W ).

Proof. Let us consider Diagram (3.1.2).

V = U ⊕ W S(U ⊕ W )

ϕ (3.1.2) ρ S(U) ⊗ S(W ) where ρ(u) = u ⊗ 1 and ρ(w) = 1 ⊗ w for all u ∈ U and w ∈ W. Then ρ extends to a map ϕ : S(U ⊕ W ) → S(U) ⊗ S(W ). Here we use the universal property of the symmetric algebra. See for example [DF04, Chapter 11, Theorem 34]. It is trivial that ϕ is surjective. Now let φ be the embedding map from S(U)⊗S(W ) into S(U ⊕W ) generated by the inclusions S(U) ,→ S(U ⊕W ) and S(W ) ,→ S(U ⊕W ).

Then φ ◦ ϕ|V = id which implies that ϕ is injective. L Corollary 3.1.9. Let V be a G-graded vector space such that V = a∈G Va where N each Va is one-dimensional vector space. Then S(V ) = a∈G S(Va).

Remark 3.1.10. By Corollary 3.1.9, we can think of an element of S(V ) as a product

of elements of S(Va).

Proposition 3.1.11. Let V be a G-graded vector space. Let {x1, . . . , xk} be a homogeneous basis of V . Then the set

i1 ik {x1 ··· xk |ij ∈ Z≥0 if β(xj, xj) = 1, ij ∈ {0, 1} if β(xj, xj) = −1} (3.1.3) is a basis of S(V ). 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 45

N Proof. By Corollary 3.1.9, we have S(V ) = a∈G S(Va). Since each Va is one- dimensional for all a ∈ G, we have

T (Va) S(Va) = hx ⊗ x − β(x, x)x ⊗ xi

for all x ∈ Va. Therefore, if β(x, x) = β(a, a) = 1, we have S(Va) = T (Va) and if

T (Va) β(x, x) = −1, we have S(Va) = hx ⊗ xi. Thus the result follows by Remark 3.1.10.

3.2 Universal enveloping algebras

Now we have enough tools to introduce the universal enveloping algebra. Recall that

g denotes a Lie colour algebra over a ﬁeld F.

Deﬁnition 3.2.1. The β-universal enveloping algebra of g is deﬁned as

T (g) U(g) = J where J is the ideal generated by x ⊗ y − β(y, x)y ⊗ x − [x, y] for all homogeneous x, y ∈ g.

Deﬁnition 3.2.2. Let σ be the composite linear map

T (g) g ,→ T (g) → J = U(g) such that σ(x) = x + J for all x ∈ g. 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 46

Note that for all homogeneous x, y ∈ g, we have

σ([x, y]) = [x, y] + J

= x ⊗ y − β(y, x)y ⊗ x + J (3.2.1)

= σ(x)σ(y) − β(y, x)σ(y)σ(x).

Since J is not a homogeneous ideal, U(g) is not automatically graded in accordance with T (g). But later in this chapter, we will consider a natural ﬁltration of U(g) and construct the associated graded algebra of U(g), gr(U(g)). Moreover the PBW

theorem (Theorem (3.4.6)) states gr(U(g)) ∼= S(g). We can also characterize the universal enveloping algebra by the following property, which says that representations of the non-associative, but ﬁnite-dimensional, algebra g are in bijective correspondence with representations of the associative, but inﬁnite- dimensional, algebra U(g).

Proposition 3.2.3. [Universal property] Let A be an associative algebra with unity over F, and let ρ : g → A be an F-linear map such that ρ([x, y]) = ρ(x)ρ(y) − β(y, x)ρ(y)ρ(x). Then there exists a unique homomorphism of algebras ϕ : U(g) → A (sending 1 to 1), such that the following diagram commutes:

g σ U(g)

ϕ (3.2.2) ρ A where σ is the map in (3.2.1).

Remark 3.2.4. We can turn any associative algebra A into a Lie colour algebra by equipping A with the bracket [a, b] = ab − β(y, x)ba for all a, b ∈ A. Then ρ becomes 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 47

a Lie colour algebra homomorphism.

Proof of Proposition 3.2.3. We ﬁrst deﬁne an algebra homomorphism ϕ0 : T (g) → A by

0 ϕ (x1 ⊗ · · · ⊗ xk) = ρ(x1) ··· ρ(xk).

Since ρ is linear, the map ϕ0 is well-deﬁned, and it is straightforward to check ϕ0 is a homomorphism. In particular, for all homogeneous x, y in g, we have

ϕ0(x ⊗ y − β(y, x)y ⊗ x − [x, y]) = ρ(x)ρ(y) − β(y, x)ρ(y)ρ(x) − ρ([x, y]), which is 0 by the hypothesis of ρ. Thus ϕ0 factors through to an algebra homomorphism ϕ : U(g) → A such that the diagram (3.2.2) commutes. To prove uniqueness, let ϕ0 : U(g) → A be another algebra homomorphism satisfying our assumption. Then for any x ∈ g, we have ϕ0(σ(x)) = ρ(x) = ϕ(σ(x)). Thus ϕ0 = ϕ on U(g) since σ(g) generates U(g).

Proposition 3.2.5. Let (ρ, V ) be an g-module. Let v be an element of V . Then the subspace U(g) · v is a g-submodule of V.

Proof. It is enough to show that for all x ∈ g and w ∈ U(g)·v we have ρ(x)w ∈ U(g)·v. We prove this proposition by using the universal property of U(g). In Proposition 3.2.3, let A = gl(V, β), which is an associative algebra. Then ρ : g → gl(V, β) is a map satisfying the hypotheses of Proposition 3.2.3. There is an algebra homomorphism ρ˜ from U(g) to End(V, β) such that ρ˜ ◦ σ = ρ. For each w ∈ U(g) · v, we have w =ρ ˜(u)(v) for some u ∈ U(g). Thus for all x ∈ g, we have

ρ(x)w =ρ ˜(σ(x))˜ρ(u)(v) 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 48

=ρ ˜(σ(x)u)(v) ∈ U(g) · v, which completes the proof.

3.3 A spanning set of U(g)

In order to understand more about the structure of the universal enveloping algebra, we give a spanning set of U(g) in this section. Then we state and prove the Poincaré- Birkhoff-Witt (PBW) theorem in the next section. First, we need to fix some general notation.

Let {x1, . . . , xn} be a homogeneous basis of g. We write σ(xi) = Xi in U(g) for

all i in {1, . . . , n}. Given a sequence J = (j1, ··· , jq), the length of J is deﬁned by

`(J) = q. Moreover we say J is weakly increasing if ji ≤ jk whenever 1 ≤ i < k ≤ q.

Let XJ = Xj1 ··· Xjq ∈ U(g) for any sequence J.

Deﬁnition 3.3.1. Let Up(g) be the subspace of U(g) spanned by the elements of the

form XJ , where J is a sequence with `(J) ≤ p.

Remark 3.3.2. Notice that the subspaces Up(g) give a ﬁltration of U(g). That is,

they give a chain of subspaces of U(g) such that C = U0(g) ⊆ U1(g) ⊆ U2(g) ⊆ · · · S and U(g) = p≥0 Up(g).

Next we will use the following lemma to show that in fact Up(g) in Deﬁnition

3.3.1 is spanned by elements of the form XJ with J weakly increasing and `(J) ≤ p.

Lemma 3.3.3. Let y1, . . . , yp be homogeneous elements of g. Let π be a permutation of {1, . . . , p}. Then

σ(y1) ··· σ(yp) − βπ(y1, . . . , yp)σ(yπ(1)) ··· σ(yπ(p)) ∈ Up−1(g) (3.3.1) 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 49

where βπ is as deﬁned in Lemma 3.1.5.

Proof. Suppose π = (i, i + 1) for some i in {1, . . . , p − 1}. Let Y be the term

σ(y1) ··· σ(yi)σ(yi+1) ··· σ(yp) − β(yi+1, yi)σ(y1) ··· σ(yi+1)σ(yi) ··· σ(yp). which is

Y = σ(y1) ··· σ(yi−1) σ(yi)σ(yi+1) − β(yi+1, yi)σ(yi+1)σ(yi) σ(yi+2) ··· σ(yp).

By the deﬁnition of Lie colour algebra bracket, it becomes

Y = σ(y1) ··· σ(yi−1) [σ(yi), σ(yi+1)] σ(yi+2) ··· σ(yp) which by the deﬁnition of U(g) is

σ(y1) ··· σ(yi−1)σ([yi, yi+1])σ(yi+2) ··· σ(yp). (3.3.2)

Since [yi, yi+1] ∈ g, there are p − 1 terms in (3.3.2). Thus, (3.3.2) is an element

of Up−1(g). Note that the scalar β(yi+1, yi) coincides with βπ(yi+1, yi) deﬁned in Lemma 3.1.5. Since we have proved that the statement of Lemma 3.3.3 holds for any adjacent transpositions (i, i + 1), the statement also holds for any permutation π of {1, . . . , p}.

Proposition 3.3.4. Let Cp = {XJ | J weakly increasing , `(J) ≤ p}. Then the subspace Up(g) is spanned by Cp.

Proof. We prove this proposition by induction. The base case when p = 1 is trivial.

Suppose that the statement holds for p = n − 1. By Deﬁnition 3.3.1, Un(g) is spanned 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 50

by XJ = σ(xj1 ) ··· σ(xjm ) over all possible sequences J = (j1, . . . , jm) such that m ≤ n.

Now ﬁx such an XJ with `(J) ≤ n. We prove that XJ is a linear combination

of XL’s in Cn. Let π be a permutation such that π(j1) ≤ π(j2) ≤ · · · ≤ π(jm). Let

Jπ = (π(j1), ··· , π(jm)). Then by Lemma 3.3.3, there is a nonzero scalar βπ and an

element R of Un−1(g) such that

XJ = βπXJπ + R,

where XJπ ∈ Cn. By induction, R can be written as a linear combination of elements

in Cn−1. Thus the result follows.

Now we ﬁnd a subset of Cn, which also spans Un(g). Let J be a weakly increasing sequence satisfying

each index j such that β(xj, xj) = −1 occurs at most once in J. (3.3.3)

Proposition 3.3.5. Let

Bp = {XJ | J weakly increasing , `(J) ≤ p and J satisﬁes condition (3.3.3)}.

Then Bp is a spanning set of Up(g).

Proof. We prove this proposition by induction. The base case when p = 1 is trivial.

Suppose that the statement holds for p = n − 1. First notice that if XJ ∈ Cn, then 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 51

since J is weakly increasing, we have

m i1 im X XJ = Xi ··· Xm , such that iq ≤ n. k=1

We now show that XJ is a linear combination of elements in Bp. Assume that

there exists an Xq in XJ such that β(Xq,Xq) = β(xq, xq) = −1 and the multiplicity

iq of Xq satisﬁes iq ≥ 2.

2 1 By the fact β(Xq,Xq) = −1 we have Xq = 2 [Xq,Xq]. Also notice that since Pn [xq, xq] ∈ g, we have [xq, xq] = i=1 cixi for some scalar ci ∈ F. Thus [Xq,Xq] = Pn Pn 2 1 Pn σ([xq, xq]) = i=1 ciσ(xi) = i=1 ciXi. This implies that Xq = 2 i=1 ciXi. Therefore iq 1 Pn iq/2 we deduce that if iq is even, then Xq = 2 i=1 ciXi . Thus we have

iq/2 iq/2 n ! 1 i X i X = Xi1 ··· X q−1 c X X q+1 ··· Xim , J 2 1 q−1 i i q+1 m i=1 where the sum of powers is less than n. Thus by induction hypothesis, we deduce that

XJ is a linear combination of elements in Bn.

Similarly, if iq is odd, then

iq−1 iq−1 n ! 2 2 1 i X i X = Xi1 ··· X q−1 c X X X q+1 ··· Xim , J 2 1 q−1 i i q q+1 m i=1 which by induction is also a linear combination of elements in Bn.

S Corollary 3.3.6. Let B = p≥0 Bp. Then B spans the universal enveloping algebra U(g). 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 52

3.4 The PBW theorem

In this section, we will ﬁrst show that the elements in B are linearly independent. Therefore they form a basis for U(g). Then we state and prove the PBW theorem.

Let {x1, . . . , xn} be a homogeneous basis of g. For all 1 ≤ i ≤ n, let zi = xi +I be

the image of xi under the canonical map g → S(g). Let J = (j1, . . . , jq) be a weakly

increasing sequence satisfying condition (3.3.3). We deﬁne zJ := zj1 ··· zjq ∈ S(g). Let

Bm = {zJ | J weakly increasing and `(J) ≤ m such that J satisﬁes condition (3.3.3)}. S Then by Proposition 3.1.11, Bm is a basis of Sm and B = m≥0 Bm is a basis of S(g).

Thus if we can ﬁnd a linear map that sends each XJ = Xj1 ··· Xjq in B to zJ in B, we will be able to conclude that the XJ ’s are linearly independent. So far, all we know about U(g) is that for any associative algebra A, the diagram

g σ U(g)

ϕ ρ A

commutes (see Proposition 3.2.3). The most straightforward idea is to construct a Lie colour algebra homomorphism ρ from g to A = S(g) first. However, if A = S(g), ρ being a Lie colour algebra homomorphism implies that ρ([x, y]) = ρ(x)ρ(y) − β(y, x)ρ(y)ρ(x) = 0 in S(g) since S(g) is colour-commutative. Hence the Lie colour bracket [ρ(x), ρ(y)] will always vanish in A, which makes us not able to analyze any structure of the Lie colour bracket. Therefore instead of pursuing the above unsuccessful idea, we will use a variation of it. That is we construct a representation from g to End(S(g)). Equivalently, we define a g-module structure on S(g) with action ρ : g × S(g) → S(g). In order to construct such an action, we inductively define a family of bilinear 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 53

maps ρm : g × Sm → Sm+1 such that ρm+1 = ρm for all m. Then we can extend g×Sm

all of the ρm to a ρ deﬁned on g × S(g). We give the following deﬁnition in order to introduce a natural action of g on S(g).

Deﬁnition 3.4.1. Given a weakly increasing sequence J = (j1, ··· , jm) and i ∈ Z≥0, we write i ≤ J if i ≤ j1 or J is the empty sequence.

Our idea is as follows. If there are some xi ∈ g, zJ ∈ Bm, then the sequence (i, J)

is again weakly increasing and we get another basis element zizJ ∈ Bm+1. Therefore we shall deﬁne

ρm(xi)zJ = zizJ , if i ≤ J and zJ ∈ Bm. (3.4.1)

Moreover, if i 6≤ J, all we can expect is ρm(xi)zJ ∈ Sm+1. Therefore we should declare that

ρm(xi)zJ − zizJ ∈ Sk when `(J) = k ≤ m. (3.4.2)

Also, ρm should satisfy the Jacobi identity for all m ≥ 0. Therefore we want the relation

ρm(xi)ρm(xj)zL = β(xj, xi)ρm(xj)ρm(xi)zL + ρm([xi, xj])zL, `(L) ≤ m − 1 (3.4.3) to hold. In summary, if there is a g-module homomorphism from g to End(S(g)), then it must satisfy (3.4.1), (3.4.2) and (3.4.3). In Section 3.5, we will prove the following, even stronger, result.

Proposition 3.4.2. There exists a unique linear map ρm : g → Hom(Sm,Sm+1) for each m ≥ 0 such that (3.4.1), (3.4.2) and (3.4.3) hold. 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 54

Assuming this proposition for now, let us deduce a few consequences. Let ρ be the

unique map extending the ρm’s. Let (ρ, S(g)) be a g-module. Therefore by the universal enveloping property, there is a corresponding homomorphism ρ˜ : U(g) → End(S(g)) such thatρ ˜ ◦ σ = ρ.

Corollary 3.4.3. Let J be a weakly increasing sequence satisfying condition (3.3.3). Then we have

ρ˜(XJ ) · 1 = zJ .

Proof. Let J = (j1, . . . , jq) be a weakly increasing sequence. Then

ρ˜(XJ ) · 1 =ρ ˜(Xj1 ) ··· ρ˜(Xjq ) · 1

=ρ ˜(σ(xj1 )) ··· ρ˜(σ(xjq )) · 1

= ρ(xj1 ) ··· ρ(xjq ) · 1

= zj1 ··· zjq = zJ , which completes the proof.

Now together with this g-module structure and Proposition 3.2.3, we can prove the following Theorem.

Theorem 3.4.4 (PBW Theorem, ﬁrst form). Let g be a Lie colour algebra with ordered homogeneous basis {x1, . . . , xn}. Then a basis for the associated universal enveloping algebra U(g) is given by

k1 kn B = {X1 ··· Xn | ki ∈ Z≥0 if β(Xi,Xi) = 1 and ki ∈ {0, 1} if β(Xi,Xi) = −1}.

Proof. We ﬁrst show that set B = ∪p≥0Bp is a basis of U(g). Let ρ : g → gl(S(g)) 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 55

be the Lie colour algebra homomorphism guaranteed by Proposition 3.4.2. Let

J = (j1, . . . , jq) be a weakly increasing sequence satisfying (3.3.3). By Corollary 3.4.3, we have

ρ˜(XJ ) · 1 = zJ .

Therefore the linear map

ϕ : U(g) → S(g),XJ 7→ ρ˜(XJ ) · 1

maps B bijectively onto the set B which is a basis for S(g). Therefore B is a linearly independent set. Since B spans U(g) by Proposition 3.3.4, B is a basis of U(g). The result follows by noticing that

B = {XJ | J weakly increasing and zJ satisﬁes condition (3.3.3)}

k1 kn = {X1 ··· Xn | ki ∈ Z≥0 if β(Xi,Xi) = 1 and ki ∈ {0, 1} if β(Xi,Xi) = −1}.

From the preceding discussion, we observe that there are profound connections between U(g) and S(g). However they are not isomorphic to each other since U(g) is not necessarily colour-commutative and ϕ is not an algebra isomorphism. Our next goal is to construct the associated graded algebra of U(g), gr(U(g)). Then we prove that there is an isomorphism of algebras between gr(U(g)) and S(g). By Remark 3.3.1, we have ﬁltration of U(g). Then we can construct the associated graded algebra gr(U(g)) of U(g) as

M gr(U(g)) = Gm m≥0 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 56 where Gm = Um(g) and G0 = . The multiplication is deﬁned by Um−1(g) F

(xm + Um−1(g))(xn + Un−1(g)) = xmxn + Un+m−1(g)

for any xm ∈ Um(g) and xn ∈ Un(g).

Lemma 3.4.5. With the multiplication deﬁned above, gr(U(g)) is a colour-commutative algebra.

Proof. Note that gr(U(g)) is generated as an algebra by G1 ∼= U(g) ∼= g. Therefore it suﬃces to show that any two elements in G1 colour-commute in gr(U(g)).

1 Take x + U0(g) and y + U0(g) in G such that x and y are homogeneous. Then we have

2 (x + U0(g))(y + U0(g)) = xy + U1(g) ∈ G . (3.4.4)

Whereas we have

2 β(y, x)(y + U0(g))(x + U0(g)) = β(y, x)yx + U1(g) ∈ G . (3.4.5)

Since xy − β(y, x)yx = [x, y] ∈ U1(g), the two classes (3.4.4) and (3.4.5) are equal.

Theorem 3.4.6 (PBW theorem, second form). Let g be a Lie colour algebra over a ∼ ﬁeld F. Then we have an isomorphism of algebras gr(U(g)) = S(g).

Proof. Let {x1, . . . , xn} be an ordered homogeneous basis of g. By Theorem 3.4.4, the

k1 kn Pn elements X1 ··· Xn such that j=1 kj ≤ m form a basis for Um(g). Therefore the

k1 kn Pn Um(g) elements X ··· X +Um−1(g) such that kj = m form a basis for . 1 n j=1 Um−1(g) This gives us a linear isomorphism between the graded pieces α : Gm → Sm such that

k1 kn k1 kn α : X1 ··· Xn + Um−1(g) 7→ z1 ··· zn , 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 57

where k1 + ··· + kn = m. Now we have

k1 kn r1 rn k1 kn r1 rn (X1 ··· Xn +Um−1(g))(X1 ··· Xn +Un−1(g)) = X1 ··· Xn X1 ··· Xn +Un+m−1(g).

Since gr(U(g)) is colour-commutative (Lemma 3.4.5), we can gather the X1,...,Xn together. Thus the above element can be rearranged as

k1+r1 kn+rn βπX1 ··· Xn + Un+m−1(g),

where βπ is as deﬁned in Lemma 3.1.5, which is the same scalar generated as we identify

k1 kn r1 rn k1+r1 kn+rn z1 ··· zn z1 ··· zn with βπz1 ··· zn in S(g). It follows that α(uv) = α(u)α(v) for all u, v ∈ gr(U(g)). Thus we have an algebra isomorphism gr(U(g)) ∼= S(g).

Now we give an application of the PBW theorem. Note that although the following discussion works for any Lie colour algebra, we state it in the context of spo(V, β) in order to avoid re-deﬁning related concepts.

Let h be a Cartan subalgebra of spo(V, β). Let g+ be the span of positive root vectors. Let Φ− be the set of negative root vectors. Then recall that from Deﬁnition 2.5.16, if W is an spo(V, β)-module, a vector w ∈ W is called a highest weight vector

∗ of weight λ ∈ h if Hw = λ(H)w for all H ∈ h and Xw = 0 for all X ∈ g+.

Corollary 3.4.7. In the setting above, let w be a highest weight vector of W . The submodule U(g) · w is spanned by the vectors

+ {Yi1 Yi2 ··· Yik · w | k ∈ N, ij ∈ −Φ for each 1 ≤ j ≤ k}.

Proof. Let g = spo(V, β). Notice that by the PBW theorem (Theorem 3.4.6), by giving an order to the basis of g, we obtain an ordered basis of U(g). Then without 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 58

loss of generality, the basis vectors of U(g) have the form of a product of negative roots vectors, H ∈ h, and positive root vectors. Since w is a highest weight vector, any positive root vector acts on w as 0, and H acts on it as scalar. Then the result follows since all that remains is to only consider a product of negative root vectors acting on w.

3.5 Proof of Proposition 3.4.2

In the previous section, we proved the PBW theorem, modulo Proposition 3.4.2 which is about the existence of a certain g-module structure on S(g). A terse proof Proposition 3.4.2 for Lie algebras in the classical sense can be found in [Hum72, Section 17.4]. In this section, in the more general setting of Lie colour algebras, we elaborate the proof of this proposition by induction. In order to apply induction, we relabel the

three conditions which ρm : g → Hom(Sm,Sm+1) should satisfy by the following:

(Am) ρm(xi)zJ = zizJ for all J with `(J) ≤ m and all i ≤ J,

(Bm) ρm(xi)zJ − zizJ ∈ Sk for all J with `(J) = k ≤ m and all 1 ≤ i ≤ n,

(Cm) ρm(xi)ρm(xj)zL = β(xj, xi)ρm(xj)ρm(xi)zL +ρm([xi, xj])zL for all L with `(L) ≤ m − 1 and all 1 ≤ i, j ≤ n.

Our goal is to deﬁne an action of g on S(g). It is enough to deﬁne the action on the

basis. Recall that from Section 3.4, Bm = {zJ | J weakly increasing and `(J) ≤ m} S is a basis of Sm and B = m≥0 Bm is a basis of S(g).

Lemma 3.5.1. Let m ≥ 1. If there exists a linear map ρm−1 which satisﬁes (Am−1),

(Bm−1) and (Cm−1), then there exists at most one linear map ρm extending ρm−1 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 59

which satisﬁes (Am), (Bm) and (Cm). Moreover for a weakly increasing sequence

J = (j1, . . . , jm), the map ρm must be recursively deﬁned by

  zizJ , if i ≤ J ρm(xi)zJ = (3.5.1)   0 zizJ + β(xj1 , xi)ρm−1(xj1 )w + ρm−1([xi, xj1 ])zJ , otherwise

0 where J = (j2, . . . , jm) and w = ρm−1(xi)zJ0 − zizJ0 ∈ Sm−1.

Proof. When m = 0, then in order to satisfy condition (A0), there is only one way to

deﬁne ρ0, which is

ρ0(xi)1 = zi

and conditions (B0) and (C0) are satisﬁed automatically. So ρ0 exists and is unique.

Now suppose that we have a uniquely determined ρm−1 satisfying conditions

(Am−1), (Bm−1) and (Cm−1). Then our goal is to construct ρm satisfying (Am),

(Bm) and (Cm) and extending ρm−1. In particular, we need to deﬁne the action

of ρm(xi) on zJ for all xi running over our basis of g and J running over all weakly increasing sequence of length `(J) ≤ m.

If i ≤ J, then we have to deﬁne ρm(xi)zJ = zizJ in order to satisfy condition

0 0 (Am). If i 6≤ J, then we have j1 < i. Then we set J = (j1,J ) and note that j1 ≤ J . Then

0 0 ρm(xi)zJ = ρm(xi)zj1 zJ = ρm(xi)ρm−1(xj1 )zJ

where the second equality uses the fact zJ0 ∈ Bm−1 and the induction hypothesis

(Am−1). Also since ρm−1 = ρm and in order to satisfy (Cm), ρm(xi)ρm−1(xj )zJ0 g×Sm−1 1 must be equal to 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 60

0 0 β(xj1 , xi)ρm(xj1 )ρm−1(xi)zJ + ρm−1([xi, xj1 ])zJ . (3.5.2)

The ﬁrst term of 3.5.2 can be simpliﬁed. First notice that (Bm−1) implies

w = ρm−1(xi)zJ0 − zizJ0 ∈ Sm−1.

Therefore by linearity we have

0 β(xj1 , xi)ρm(xj1 )ρm−1(xi)zJ

0 = β(xj1 , xi)ρm(xj1 )(zizJ + w)

= β(xj , xi)(ρm(xj )zizJ0 + ρm−1(xj )w) by ρm−1 = ρm 1 1 1 g×Sm−1

0 = β(xj1 , xi)(zj1 zizJ + ρm−1(xj1 )w) by j1 < i and (Am)

= β(xj1 , xi)(β(xi, xj1 )zizJ + ρm−1(xj1 )w) by the colour-commutativity of S(g)

which is zizJ + β(xj1 , xi)ρm−1(xj1 )w. Therefore in this case, if ρm extends ρm−1 and

satisﬁes (Am), we must have

0 ρm(xi)zJ = zizJ + β(xj1 , xi)ρm−1(xj1 )w + ρm−1([xi, xj1 ])zJ .

Lemma 3.5.1 gives us a recursive formula for a function ρm extending ρm−1, and

by construction it satisﬁes (Am). In the next two lemmas, we prove it satisﬁes (Bm)

and (Cm).

From now on we write j instead of j1 for simplicity. But otherwise retain the notation of Lemma 3.5.1.

Lemma 3.5.2. The map that is uniquely deﬁned by (3.5.1) satisﬁes (Bm). 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 61

Proof. Let xi be in the basis of g, zJ ∈ Bk for some k ≤ m, and i 6< J. We need to

prove that ρm(xi)zJ − zizJ ∈ Sk. By the formula (3.5.1), we have

ρm(xi)zJ = zizJ + β(xj, xi)ρm−1(xj)w + ρm−1([xi, xj])zJ0

where w ∈ Sk−1 and thus, β(xj, xi)ρm−1(xj)w ∈ Sk. Similarly, zJ0 ∈ Sk−1 implies

ρm−1([xi, xj])zJ0 in Sk. Therefore this proves ρm(xi)xJ − zizJ ∈ Sk so that (Bm) is satisﬁed.

Now we are going prove that ρm satisﬁes (Cm) by case analysis.

Lemma 3.5.3. Let L be a weakly increasing sequence with `(L) = m − 1. If either of the two conditions

(i) i ≥ j and j ≤ L, or

(ii) i ≤ j and i ≤ L

hold, then ρm also satisﬁes (Cm) for these values of i and j.

Proof. Suppose i ≥ j and j ≤ L. Then by (Am−1), we know that ρm(xj)zL =

ρm−1(xj)zL = zjzL. Therefore we apply the formula (3.5.1) to ρm(xi)zjzL to obtain

ρm(xi)zjzL = zizjzL + β(xj, xi)ρm−1(xj)w + ρm−1([xi, xj])zL (3.5.3)

where w = ρm−1(xi)zL − zizL ∈ Sm−1 in this case. Since w ∈ Sm−1, we can write

ρm−1(xj)w as ρm(xj)w. Then we expand it as follows

ρm(xj)w = ρm(xj)(ρm−1(xi)zL − zizL)

= ρm(xj)ρm−1(xi)zL − ρm(xj)zizL by linearity 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 62

= ρm(xj)ρm−1(xi)zL − zjzizL since j ≤ i and j ≤ L

= ρm(xj)ρm−1(xi)zL − β(xi, xj)zizjzL.

Since zL ∈ Sm−1, we can replace ρm−1 by ρm in this last term. Therefore if we

substitute β(xj, xi)ρm(xj)w by β(xj, xi)ρm(xj)ρm(xi)zL − zizjzL in 3.5.3, we conclude

ρm(xi)ρm(xj)zL = β(xj, xi)ρm(xj)ρm(xi)zL + ρm([xi, xj])zL, (3.5.4) as required. Now we assume that i ≤ j and i ≤ L. Then by a similar calculation in case (ii), we should conclude that

ρm(xj)ρm(xi)zL = β(xi, xj)ρm(xi)ρm(xj)zL + ρm([xj, xi])zL. (3.5.5)

After rearranging (3.5.5), we get

β(xi, xj)ρm(xi)ρm(xj)zL = ρm(xj)ρm(xi)zL − ρm([xj, xi])zL;

since [xj, xi] = −β(xi, xj)[xi, xj] and ρm is linear in each variable, this is equivalent to

β(xi, xj)ρm(xi)ρm(xj)zL = ρm(xj)ρm(xi)zL + β(xi, xj)ρm([xi, xj])zL.

Multiplying by β(xj, xi) on both sides gives

ρm(xi)ρm(xj)zL = β(xj, xi)ρm(xj)ρm(xi)zL + ρm([xi, xj])zL.

This proves that the condition (Cm) holds in case (ii). 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 63

Now the only remaining case left to prove is when neither i ≤ L nor j ≤ L, in which case we must have that L = (`, L0) with ` < i, j.

Lemma 3.5.4. Let L be a weakly increasing sequence with `(L) = m − 1. Let i, j be such that neither i ≤ L nor j ≤ L. Then ρm satisﬁes (Cm) for these values of i and j.

By looking at (Cm), our strategy to prove this lemma is using Lemma 3.5.3

and the induction hypothesis to rewrite ρm(xi)ρm(xj)zL and β(xj, xi)ρm(xj)ρm(xi)zL

explicitly. Then we prove that the diﬀerence between these two terms is ρm([xi, xj])zL.

To save some space, we write β(i, j) = β(xi, xj) for all xi, xj ∈ g. Also since

[xi, xj] has colour equal to the product of colours of xi and xj, we write ij for this

colour. Moreover, instead of ρm(xi)zL, we write xi · zL in the proof.

0 Proof. First we consider the term ρm(xi)ρm(xj)zL = xi · xj · zL. Since ` ≤ L = (`, L ), we have zL = x` · zL0 . Therefore

xj · zL = xj · x` · zL0 (3.5.6) where j > ` and ` ≤ L0. Thus the right hand side of (3.5.6) satisﬁes our assumption of Lemma 3.5.3 (i). Therefore we have

xj · zL = β(`, j)x` · xj · zL0 + [xj, x`] · zL0 .

Therefore by linearity we have

xi · xj · zL = β(`, j)xi · x` · xj · zL0 + xi · [xj, x`] · zL0 . (3.5.7)

Now we expand (3.5.7) even further. 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 64

0 For the ﬁrst term, since L has length ≤ m − 2, by (Bm), we can write xj · zL0 =

zjzL0 + w for some w ∈ Sm−2. Therefore we can write β(`, j)xi · x` · xj · zL0 as

β(`, j)(xi · x` · zjzL0 + xi · x` · w). (3.5.8)

Now since ` < i, j and ` ≤ L0, the ﬁrst term of (3.5.8) satisﬁes the condition of Lemma

3.5.3 (i). Therefore we can expand it according to (Cm). Moreover, since w ∈ Sm−2, we can also expand the second term of (3.5.8) by the induction hypothesis. Therefore we can expand β(`, j)xi · x` · xj · zL0 by considering xj · zL0 as one term. Thus we have:

β(`, j)xi · x` · xj · zL0 = β(`, j) β(`, i)x` · xi · xj · zL0 + [xi, x`] · xj · zL0 .

Now for the second term of (3.5.7). Since L0 has length ≤ m − 2, by the induction hypothesis, we can expand it as

β(j`, i)[xj, x`] · xi · zL0 + [xi, [xj, x`]] · zL0 .

Therefore, in summary, we can expand (3.5.7) as

β(`, i)β(`, j)x` · xi · xj · zL0 + β(`, j)[xi, x`] · xj · zL0

+ β(j`, i)[xj, x`] · xi · zL0 + [xi, [xj, x`]] · zL0 (3.5.9) where we have underlined some terms for easy reference. Similarly as before, if we

swap the roles of i and j, we conclude that ρm(xj)ρm(xi)zL becomes

β(`, i)β(`, j)x` · xj · xi · zL0 + β(`, i)[xj, x`] · xi · zL0 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 65

+ β(i`, j)[xi, x`] · xj · zL0 + [xj, [xi, x`]] · zL0 .

Therefore β(j, i)ρm(xj)ρm(xi)zL is

β(j, i)β(`, i)β(`, j)x` · xj · xi · zL0 + β(j, i)β(`, i)[xj, x`] · xi · zL0

+ β(j, i)β(i`, j)[xi, x`] · xj · zL0 + β(j, i)[xj, [xi, x`]] · zL0 . (3.5.10)

Subtract (3.5.10) from (3.5.9); the underlined terms will be cancelled, and the diﬀerence is:

β(`, i)β(`, j)x` · xi · xj · zL0 − β(j, i)β(`, i)β(`, j)x` · xj · xi · zL0

+ [xi, [xj, x`]] · zL0 − β(j, i)[xj, [xi, x`]] · zL0 (3.5.11)

The ﬁrst two terms of (3.5.11) can be combined into one term as

β(`, i)β(`, j)x` · xi · xj · zL0 − β(j, i)xj · xi · zL0 .

Moreover since `(L0) ≤ m − 2, by induction hypothesis, the underlined term can be

simpliﬁed further as [xi, xj] · zL0 using (Cm). Therefore we have the reduced form

β(`, i)β(`, j)x` · [xi, xj] · zL0 . (3.5.12)

Using (Cm) again, (3.5.12) is equivalent to

β(`, i)β(`, j)β(ij, `)[xi, xj] · x` · zL0 + β(`, i)β(`, j)[x`, [xi, xj]] · zL0 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 66 which (by the fact that β(`, i)β(`, j)β(ij, `) = 1) is

[xi, xj] · x` · zL0 + β(`, i)β(`, j)[x`, [xi, xj]] · zL0 . (3.5.13)

Thus by replacing the ﬁrst two terms of (3.5.11) by (3.5.13), (3.5.11) becomes

[xi, xj] · x` · zL0 + [xi, [xj, x`]] · zL0 + β(`, i)β(`, j)[x`, [xi, xj]] · zL0 − β(j, i)[xj, [xi, x`]] · zL0 . (3.5.14) Now based on the Jacobi identity, we need to do some manipulations of β to show that the sum of the last three terms in (3.5.14) is 0. More precisely, we show that

[xi, [xj, x`]] + β(`, i)β(`, j)[x`, [xi, xj]] − β(j, i)[xj, [xi, x`]] (3.5.15)

is 0. By Lemma 2.1.7, we have a diﬀerent version of the Jacobi identity, namely in our case, we have

[xi, [xj, x`]] − [[xi, xj], x`] − β(j, i)[xj, [xi, x`]] = 0. (3.5.16)

Notice that the ﬁrst and third term in (3.5.15) and (3.5.16) are identical. Then if we swap [xi, xj] with x` in −[[xi, xj], x`] (the second term of (3.5.16)), we obtain

β(`, ij)[x`, [xi, xj]] which is exactly the same as the second term of (3.5.15). Thus, we

conclude that (3.5.15) = 0. Therefore (3.5.14) becomes [xi, xj] · x` · zL0 = [xi, xj] · zL.

This implies that (3.5.11) = [xi, xj] · zL which is the diﬀerence between xi · xi · zL and

β(j, i)xj · xi · L. Thus we have

xi · xi · zL − β(j, i)xj · xi · zL = [xi, xj] · zL 3. THE PBW THEOREM FOR LIE COLOUR ALGEBRAS 67

which is equivalent to (Cm). This completes the proof.

In Lemma 3.5.1, we constructed the map ρm : g → Hom(Sm,Sm+1) and proved

it satisﬁes (Am). By Lemma 3.5.2, we proved ρm satisﬁes (Bm). By Lemma 3.5.3

and 3.5.4, we proved that ρm satisﬁes (Cm). Thus by the lemmas in this section, we proved Proposition 3.4.2, which guarantees the validities of the two versions of the PBW theorem, Theorem 3.4.4 and 3.4.6. Chapter 4

The Brauer algebra action on V ⊗k

The Brauer algebra was introduced by Richard Brauer in [Bra37]. Brauer’s motivation was to obtain the centralizer of the action of the orthogonal group on tensor powers of its standard module. This idea has also been shown to work when the orthogonal group is replaced by the symplectic group (for example, see [BR99]). In Section 4.1, we discuss the Brauer algebra and in Section 4.2, following [BSR98], we deﬁne an action of the Brauer algebra on tensor powers of the standard spo(V, β)-module, which commutes with the action of spo(V, β).

4.1 The Brauer algebra

In this section, for any positive integer k, we first give the definition of k-diagrams. These k-diagrams will span the Brauer algebra. In particular, we describe how to compose k-diagrams, resulting in the algebra structure. Moreover, we discuss a special subset of the k-diagrams so that we can define a subalgebra of the Brauer algebra which is isomorphic to the group algebra of Sk.

68 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 69

Deﬁnition 4.1.1. Let k ∈ Z+.A k-diagram is a diagram with two parallel rows with k vertices in each row, and k edges such that each vertex is linked to precisely one edge. We number the bottom vertices from left to right by 1, 2, ··· , k and the top vertices from right to left by k + 1, k + 2, ··· , 2k. Note that loops are not allowed in a k-diagram.

The set of all k-diagrams is denoted by Dk.

Example 4.1.2. An example of a k-diagram is

14 13 12 11 10 9 8

∈ D7

1 2 3 4 5 6 7

(2k)! Remark 4.1.3. Let k ∈ Z+. There are (2k − 1) · (2k − 3) ··· 5 · 3 · 1 = 2kk! k-diagrams since we have 2k − 1 possible ways to connect the ﬁrst vertex with another vertex, then we have 2k − 3 possibilities to join the next unconnected vertex and so on.

Next we give an operation on Dk so that the vector space spanned by a basis

corresponding to Dk becomes an algebra. This algebra is called the Brauer algebra.

Let F be a ﬁeld of characteristic 0 and let η ∈ F. If d1, d2 ∈ Dk, then we compose d1

and d2 by putting d1 on top of d2 identifying i and 2k − i + 1 in d1 and d2 respectively for all 1 ≤ i ≤ k. Then after removing the middle line of vertices and any loops, we

c have another k-diagram d and we deﬁne d1d2 = η d where c is the number of loops.

Example 4.1.4. Let d1 be the k-diagram we deﬁned in Example 4.1.2 and

d = 2 . 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 70

Then

2 d1d2 = = η .

Deﬁnition 4.1.5. The Brauer algebra Bk(η) is the F-span of the k-diagrams in Dk with multiplication induced by the composition of k-diagrams.

(2k)! By Remark 4.1.3, the dimension of Bk(η) is 2kk! .

Proposition 4.1.6. The Brauer algebra Bk(η) is generated by the k-diagrams ei and

si, for 1 ≤ i ≤ k − 1, where

...... ei = ...... i i + 1

and ...... si = ...... i i + 1 , modulo the following relations.

2 2 si = 1, ei = ηei, eisi = siei = ei, 1 ≤ i ≤ k − 1,

sisj = sjsi, siej = ejsi, eiej = ejei, |i − j| ≥ 1,

sisi+1si = si+1sisi+1, eiei+1ei = ei, ei+1eiei+1 = ei+1, 1 ≤ i ≤ k − 2,

siei+1si = si+1ei, ei+1eisi+1 = ei+1si, 1 ≤ i ≤ k − 2.

Proof. A proof can be found in [Wen88, Section 2]. 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 71

Now our next plan is to give another presentation of Bk(η) so that we can form a permutation corresponding to each k-diagram. For each k-diagram d, we can represent

d as a sequence of pairs of linked vertices such that d = {(`1, r1), (`2, r2),..., (`k, rk)} where `i < ri for all `i, ri ∈ {1,..., 2k}.

Example 4.1.7. Let d be given in Example 4.1.2. We have

d = {(1, 3), (2, 5), (4, 9), (6, 10), (7, 8), (11, 13), (12, 14)}.

There is a way to construct a permutation of {1,..., 2k} using d ∈ Dk: let

d = {(`1, r1), (`2, r2),..., (`k, rk)} ∈ Dk with `1 < ··· < `k. We construct the

permutation πd associated with d by letting

  1 2 ··· k k + 1 k + 2 ··· 2k   πd =   ∈ S2k. `1 `2 ··· `k rk rk−1 ··· r1

Example 4.1.8. For the diagram e1 ∈ D2, we have e1 = {(1, 2), (3, 4)}, and the

corresponding πe1 is

  1 2 3 4 π =   = (234) = (23)(34). e1   1 3 4 2

Example 4.1.9. For the diagram s1 ∈ D2, we have s1 = {(1, 3), (2, 4)}, and the

corresponding πs is 1   1 2 3 4 π =   = (34). s1   1 2 4 3

Let d ∈ Dk be any diagram such that each vertex in the lower row is connected 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 72

to one in the upper row. Then d = {(1, r1), (2, r2),..., (k, rk)}. Therefore the corresponding permutation is

  1 ··· k k + 1 ··· 2k   πd =   , 1 ··· k rk ··· r1 which is a permutation of {k + 1,..., 2k}. For example,

πsi = (2k − i, 2k − i + 1). (4.1.1)

Remark 4.1.10. Identifying {k + 1,..., 2k} with {1, . . . , k} by the transformation

i 7→ 2k − i + 1, we see that si goes to the transposition (i, i + 1). In general, any

d ∈ Bk(η) with only vertical edges can be written as si1 ··· sin , for some n and

some ij ∈ {1, . . . , k − 1}. Therefore, we can identify such d with the permutation

σ = (i1, i1 + 1) ··· (in, in + 1) = si1 ··· sin ∈ Sk, where sij now denotes also the simple

transpositions (ij, ij + 1). In this way, we identify the subalgebra of Bk(η) generated

by the si with the group algebra F[Sk]. Note that this is the same identiﬁcation as obtained by viewing the diagrams with only vertical edges as permutations of {1, . . . , k} in the natural way.

4.2 The commuting action of the Brauer algebra

on the spo(V, β)-module V ⊗k

In this section, let g = spo(V, β). Let B = {v1, . . . , vn} be the ordered basis given in

1 n (2.4.1). Denote the dual basis by {v , . . . , v }. For k ∈ Z≥0, we deﬁne an action of Dk on V ⊗k, which is provided in [BSR98]. This action is designed to make the generator 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 73

⊗k ⊗(k−2) si act by permutations, and the image of ei on V isomorphic to V . In order ˇ to deﬁne this action, we need the g-module morphisms Rπ in (2.2.3), pr˜ k and ev˜ k in Lemma 2.2.14. Also recall that from Section 2.3, we have the matrix form of our g-module isomorphism F : V → V ∗. Then we have the following deﬁnition.

Deﬁnition 4.2.1. Let d be a k-diagram in Dk, we deﬁne a map

⊗k ⊗k Ψ: Dk → Homg(V ,V )

d 7→ Ψd

where Ψd is the composition

⊗k ⊗k ∼ ⊗k pr˜ k⊗id ⊗k ∗ ⊗k ⊗k Ψd : V = V1G ⊗ V −−−−−→ V ⊗ (V ) ⊗ V

id⊗k⊗(F −1)⊗k⊗id⊗k −−−−−−−−−−−−→ V ⊗2k ⊗ V ⊗k

Rˇ ⊗id⊗k −−−−−−→πd V ⊗2k ⊗ V ⊗k

⊗k ⊗k ⊗k −−−−−−−−−−→id ⊗F ⊗id V ⊗k ⊗ (V ∗)⊗k ⊗ V ⊗k

⊗k id ⊗ev˜ k ⊗k ∼ ⊗k −−−−−→ V ⊗ V1G = V .

Notice that Ψd is a g-module morphism for all d ∈ Dk since the map Ψd is a composite of g-module morphisms.

Next we illustrate Deﬁnition 4.2.1 by evaluating Ψ on the generators of B2(η).

We have the following proposition. Note that we write the action of Bk(η) on the right, and we will justify this convention later.

Proposition 4.2.2. Let vi1 ⊗ vi2 ∈ V ⊗ V with vi1 , vi2 ∈ B. Then 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 74

(i) the action of e1 on vi1 ⊗ vi2 is given by

n X (v ⊗ v )Ψ = F F −1 v ⊗ v , and i1 i2 e1 i1,i2 j1,j2 j1 j2 j1,j2

(ii) the action of s1 on vi1 ⊗ vi2 is given by

(vi1 ⊗ vi2 )Ψs1 = −β(vi2 , vi1 )vi2 ⊗ vi1 .

ˇ ˇ ˇ Proof. Let d = e1 ∈ D2. Then by Example 4.1.8, we have Rπd = R(23)R(34). Notice ˇ ˇ ⊗k ˇ ˇ that when we apply R(23)R(34) on V , we ﬁrst apply R(34), and then apply R(23). By

−1 i Lemma 2.4.4, F (v ) and vi have opposite colour. Again by Remark 2.3.4, we use a vector to represent its only colour and use the inverse v−1 to represent the opposite

colour. Then the sequence of maps Ψe1 does the following:

(vi1 ⊗ vi2 ) →1 ⊗ vi1 ⊗ vi2 n=1 pr˜ ⊗id⊗2 k X j2 j1 −−−−−→ vj1 ⊗ vj2 ⊗ v ⊗ v ⊗ vi1 ⊗ vi2

j1,j2 id⊗2⊗(F −1)⊗2⊗id⊗2 X −1 j2 −1 j1 −−−−−−−−−−−−→ vj1 ⊗ vj2 ⊗ F (v ) ⊗ F (v ) ⊗ vi1 ⊗ vi2

j1,j2 ˇ ⊗2 R(34)⊗id X −−−−−−→ −β(v−1, v−1)v ⊗ v ⊗ F −1(vj1 ) ⊗ F −1(vj2 ) ⊗ v ⊗ v (*) j1 j2 j1 j2 i1 i2 j1,j2 ˇ ⊗2 R(23)⊗id X −−−−−−→ β(v−1, v−1)β(v−1, v )v ⊗ F −1(vj1 ) ⊗ v ⊗ F −1(vj2 ) ⊗ v ⊗ v j1 j2 j1 j2 j1 j2 i1 i2 j1,j2 id⊗2⊗F ⊗2⊗id⊗2 X −−−−−−−−−→ β(v−1, 1)v ⊗ F −1(vj1 ) ⊗ F (v ) ⊗ vj2 ⊗ v ⊗ v . (4.2.1) j1 j1 j2 i1 i2 j1,j2

Notice that β(v−1, 1) = 1 for any v ∈ B, and we will rewrite (4.2.1) using the matrix j1 j1 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 75

form of F and F −1 to complete the calculation. Therefore, we have

n X X X v ⊗ F −1 v ⊗ F vq ⊗ vj2 ⊗ v ⊗ v j1 j1,p p j2,q i1 i2 j1,j2 p=1 q=1 n n ⊗2 id ⊗ev˜ k X X X −−−−−→ v ⊗ F −1 v vj2 (v ) F vq(v ) j1 j1,p p i1 j2,q i2 j1,j2 p=1 q=1 n n X X X = v ⊗ F −1 v δ F δ j1 j1,p p j2,i1 j2,q q,i2 j1,j2 p=1 q=1 n X X X = v ⊗ F −1 v δ F . j1 j1,p p j2,i1 j2,i2 j1 j2 p=1

Consider the underlined part. Since δj2,i1 is nonzero only for j2 = i1, the sum over j2 reduces to j2 = i1. Thus we have

n X X X v ⊗ F −1 v F = F F −1 v ⊗ v j1 j1,p p i1,i2 i1,i2 j1,p j1 p j1 p=1 j1,p X = F F −1 v ⊗ v . i1,i2 j1,j2 j1 j2 j1,j2

Therefore we have (v ⊗ v )Ψ = F P F −1 v ⊗ v . i1 i2 e1 i1,i2 j1,j2 j1,j2 j1 j2

Now let d = s1. In Example 4.1.9, we veriﬁed that πu(s1) = (34). The ﬁrst steps

up until (∗) of the evaluation of (vi1 ⊗ vi2 )Ψs1 is identical to that of (vi1 ⊗ vi2 )Ψe1 . Therefore we have

X (v ⊗ v ) → −β(v−1, v−1)v ⊗ v ⊗ F −1(vj1 ) ⊗ F −1(vj2 ) ⊗ v ⊗ v (*) i1 i2 j1 j2 j1 j2 i1 i2 j1,j2

⊗2 ⊗2 ⊗2 id ⊗F ⊗id X j1 j2 −−−−−−−−−→ −β(vj1 , vj2 )vj1 ⊗ vj2 ⊗ v ⊗ v ⊗ vi1 ⊗ vi2

j1,j2

⊗2 id ⊗ev˜ k X −−−−−→ −β(vj1 , vj2 )vj1 ⊗ vj2 δj1,i2 δj2,i1

j1,j2 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 76

= − β(vi2 , vi1 )vi2 ⊗ vi1 .

Thus (vi1 ⊗ vi2 )Ψs1 = −β(vi2 , vi1 )vi2 ⊗ vi1 .

Corollary 4.2.3. For all k ∈ Z>0 and 1 ≤ i ≤ k − 1, we have

⊗(i−1) ⊗(k−i−1) Ψei = id ⊗ Ψe1 ⊗ id , and

⊗(i−1) ⊗(k−i−1) Ψsi = id ⊗ Ψs1 ⊗ id .

Proof. Although the proof is a time consuming process, it is similar to the proof of Proposition 4.2.2.

By Remark 4.1.10, we identiﬁed the diagram si with the permutation (i, i + 1).

Notice that for a simple tensor v = v1 ⊗ · · · ⊗ vk, we have

ˇ vΨsi = −β(vi+1, vi)v1 ⊗ · · · vi−1 ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vk = Riv.

In fact, for each d ∈ Dk with only vertical edges, by Remark 4.1.10, we view d

as a permutation in Sk. Thus, if we write the permutation corresponding to d as

⊗k π = si1 ··· sin as a product of simple transpositions, then for all v = v1 ⊗· · ·⊗vk ∈ V , we have vπ = vΨ = vΨ ··· Ψ = Rˇ ··· Rˇ v. (4.2.2) d si1 sin in i1

Moreover, we have

vπ = (v1 ⊗ · · · ⊗ vk)π = βπvπ(1) ⊗ · · · ⊗ vπ(k), 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 77

where βπ is the scalar deﬁned in Lemma 3.1.5.

We will discuss more about the maps Ψei in Section 5.2.

Recall n = dim(V(0)) and m = dim(V(1)), and set η = n − m. We have seen for

simple reﬂections that vΨd1d2 = vΨd1 Ψd2 . This justiﬁes the reason we write the action of the Brauer algebra as a right action. It also suggests, as in the following theorem,

op that our action extends not to Bk(n − m) but to Bk(n − m) , where the algebra

operation satisﬁes d1 · d2 := d2d1.

op ⊗k ⊗k Theorem 4.2.4. The map Ψ: Bk(n − m) → Homg(V ,V ) is a homomorphism of algebras.

Proof. First notice that since the k-diagrams span Bk(n − m) (and hence span Bk(n −

op op m) ), Deﬁnition 4.2.1 deﬁnes Ψb for any b ∈ Bk(n − m) by linearity. Thus Ψb ∈

⊗k ⊗k op Homg(V ,V ) for all b ∈ Bk(n − m) . Next, in order to prove that it is an algebra homomorphism, one needs to verify that Ψ preserves each of the relations in Proposition 4.1.6. Most of the symmetric relations are proved in [BSR98]. We

will prove one extra relation Ψs1 Ψe2 Ψe1 = Ψs2 Ψe1 (corresponding to the relation

s1e2e1 = s2e1 in Bk(n − m) for all k > 2) since this relation is neither symmetric

op (and thus distinguishes between Bk(n − m) and Bk(n − m) ) nor straightforward to deduce.

For all k ≥ 3, by Corollary 4.2.3, since both Ψs1 Ψe2 Ψe1 and Ψs2 Ψe1 only act on the ﬁrst three components of a tensor in V ⊗k, it suﬃces to only consider a simple

⊗3 tensor vi1 ⊗ vi2 ⊗ vi3 ∈ V with vi1 , vi2 , vi3 ∈ B. We prove that

(vi1 ⊗ vi2 ⊗ vi3 )Ψs1 Ψe2 Ψe1 = (vi1 ⊗ vi2 ⊗ vi3 )Ψs2 Ψe1 . (4.2.3) 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 78

Recall that we use the vector itself to denote its colour. By Proposition 4.2.2, the right hand side of (4.2.3) is

(vi1 ⊗ vi2 ⊗ vi3 )Ψs2 Ψe1 = −β(vi3 , vi2 )(vi1 ⊗ vi3 ⊗ vi2 )Ψe1 X = − β(v , v )F F −1 v ⊗ v ⊗ v . (4.2.4) i3 i2 i1,i3 j1,j3 j1 j3 i2 j1,j3

The left hand side of (4.2.3) is

(vi1 ⊗ vi2 ⊗ vi3 )Ψs1 Ψe2 Ψe1 = − β(vi2 , vi1 )(vi2 ⊗ vi1 ⊗ vi3 )Ψe2 Ψe1 ! X = − β(v , v ) F F −1 v ⊗ v ⊗ v Ψ , i2 i1 i1,i3 `1,`3 i2 `1 `3 e1 `1,`3 which is

! X X − β(v , v ) F F −1 F F −1 v ⊗ v ⊗ v . (4.2.5) i2 i1 i1,i3 `1,`3 i2,`1 h1,h2 h1 h2 `3 `1,`3 h1,h2

Now let us simplify (4.2.5). Consider the underlined term F −1 F . Since `1,`3 i2,`1 F −1 =6 0 if and only if v∗ = v , and F 6= 0 if and only if v∗ = v , we deduce `1,`3 `1 `3 i2,`1 `1 i2 that this term is nonzero if and only if v = v∗ = v . Therefore the underlined term `3 `1 i2 −1 is F ∗ F`∗,` which is 1 for all v` ∈ B by the explicit matrix form of F we found in `1,`1 1 1 1 (2.4.2) and (2.4.3). Thus (4.2.5) becomes

X − β(v , v )F F −1 v ⊗ v ⊗ v . (4.2.6) i2 i1 i1,i3 h1,h2 h1 h2 i2 h1,h2

∗ Notice that Fi1,i3 =6 0 if and only if vi1 = vi3 , therefore by using the fact that 4. THE BRAUER ALGEBRA ACTION ON V ⊗k 79

∗ β(vi2 , vi3 ) = β(vi3 , vi2 ), (4.2.6) becomes

X −β(v , v )F F −1 v ⊗ v ⊗ v . i3 i2 i1,i3 h1,h2 h1 h2 i2 h1,h2

By re-indexing hi to ji for all i = 1, 2, we conclude that the two sides of (4.2.3) are equal.

Remark 4.2.5. Notice that if we wrote the action of the Brauer algebra on the left,

the relation Ψs1 Ψe2 Ψe1 (vi1 ⊗ vi2 ⊗ vi3 ) = Ψs2 Ψe1 (vi1 ⊗ vi2 ⊗ vi3 ) would not hold.

⊗k Theorem 4.2.4 implies that for all w ∈ V , for all X ∈ g, and for all b ∈ Bk(n−m), we have

(ρ(X)w)Ψb = ρ(X)(wΨb),

⊗k where ρ denotes the g-module structure on V . Thus the right action of Bk(n − m) on V ⊗k commutes with the left action of spo(V, β) on V ⊗k.

Remark 4.2.6. The Brauer action can be deﬁned more systematically using the machinery of Brauer categories, introduced for example in [LZ16]. The advantage of the more elementary approach that we develop in this chapter, is that one obtains an explicit method to combinatorially compute the action of any element of the Brauer algebra. Chapter 5

Highest weight vectors in V ⊗k

In this chapter we construct highest weight vectors for tensor powers of the standard representation of spo(V, β). The idea is to construct certain highest weight vectors for harmonic tensor spaces and to use the Brauer algebra. This technique was originally used in the purely even cases corresponding to orthogonal and symplectic Lie algebras; for a readable exposition, see [GW09]. This technique is generalized to the orthosymplectic Lie colour algebra spo(V, β) by [BSR98]. In Section 5.1, we introduce (r, s)-hook tableaux and their associated simple tensors following [BSR98]. In Section 5.2 and 5.3, we introduce Young symmetrizers and contractions which are realized by elements of the Brauer algebra. We illustrate these with detailed examples. In Section 5.4, we will use these new concepts to construct highest weight vectors for

⊗k V where V is the standard spo(V, β)-module and k ∈ Z≥0. Furthermore, we analyze and justify how this construction works. We conclude in Section 5.5.3 with our own explicit example in V ⊗ V , and construct submodules generated by the highest weight vectors we found.

80 5. HIGHEST WEIGHT VECTORS IN V ⊗k 81

5.1 (r, s)-hook tableaux and simple tensors

As demonstrated in [GW09, Chapter 9], there is a way to construct simple tensors vT in V ⊗k from Young tableaux T . For the deﬁnition of Young tableaux, patitions and related concepts, please see Appendix A. The work of [BSR98] generalized this idea to form simple tensors from (r, s)-hook shape tableaux. Let g = spo(V, β). We construct simple tensors from the subset

0 B = {t1, . . . , tr, u1, . . . , us, (us+1)}

0 of the basis B. The set of ti comes from the set B(0), and the set of uj comes from

0 the the set B(1). Therefore we will put all ti’s together to form a smaller tableau

0 T(0) within T , and similarly, we will gather all uj’s from B to form another smaller

tableau T(1) within T . In order to make this discussion more precise, we ﬁrst give the following deﬁnitions.

Deﬁnition 5.1.1. Let λ = (λ1, . . . , λ`) be a partition of k ∈ Z≥0. We say λ is a

(r, s)-hook partition if λr+1 ≤ s. A standard tableau T of shape λ is called an (r, s)-hook tableau.

Note that any tableaux which can be ﬁtted in Figure 5.1 is called an (r, s)-hook tableau.

··· r rows

Figure 5.1: (r, s)-hook shape tableau frame 5. HIGHEST WEIGHT VECTORS IN V ⊗k 82

From now on, we let K = {1, . . . , k} for some k ∈ Z≥0. We denote the set of all

(r, s)-hook tableaux with entries from K by Γr,s(K).

Example 5.1.2. Let λ = (17, 2, 2). Then λ is a (1, 3)-hook partition since λ1+1 = 2 ≤ 3.

Deﬁnition 5.1.3. Let λ = (λ1, . . . , λ`) ` k. Let Y (λ) be the Young diagram of λ. 0 ˜ ˜ Then the transpose of λ, denoted λ = (λ1,..., λn) is a partition of k whose parts are

0 the lengths of the columns of Y (λ). In particular, Y (λ ) has ` columns and n = λ1 ˜ rows, and λj = Card{i | λi − j ≥ 0}, for all 1 ≤ j ≤ n.

Example 5.1.4. Let λ = (5, 2, 2, 1). Then λ0 = (4, 3, 1, 1, 1). The corresponding Young diagrams are given by

Y (λ) = ,Y (λ0) = .

Deﬁnition 5.1.5. Given a tableau T of shape λ, the transpose of T , denoted T 0 is the tableau of shape λ0 in which the entries of the ith row of T are the entries of the ith column of T 0.

Note that if T is a standard Young tableau of shape λ, then T 0 is still a standard Young tableau of shape λ0.

Example 5.1.6. Let T = 1 3 6 8 9 . Then T 0 = 1 2 5 10 . 2 4 3 4 7 5 7 6 10 8 9 5. HIGHEST WEIGHT VECTORS IN V ⊗k 83

Now for each (r, s)-hook tableau T , we separate it into two diﬀerent tableaux

T(0) and T(1): we take the ﬁrst r rows to form a smaller tableau T(0). Then we take

the transpose of the rest of T to form another smaller tableau T(1) which will have at most s rows.

Formally, let λ = (λ1, . . . , λ`) be an (r, s)-hook partition, and let T be a

standard tableau of shape λ. The subtableau T(0) corresponds to the partition

λ(0) = (λ1, . . . , λr). We form a smaller partition (λr+1, . . . , λ`) by removing λ(0) from 0 ˜ ˜ λ. Then λ(1) = (λr+1, . . . , λ`) = (λr+1,..., λr+s) gives a new partition. The trans-

pose of the subtableau of T obtained by removing T(0) is called T(1). Note that T(1)

has shape λ(1). Note we start labelling the indices of λ(1) from r + 1 because λ(1) is constructed starting from the (r + 1)th row of T . By convention, if ` ≤ r, then ˜ ˜ λr+1 = ··· = λr+s = 0.

Example 5.1.7. Let T = 1 3 6 8 9 . This is a (2, 3)-hook tableau and we have 2 4 5 7 10

T(0) = 1 3 6 8 9 and T(1) = 5 10 . 2 4 7

The corresponding partitions are λ(0) = (5, 2) and λ(1) = (2, 1).

Now for each (r, s)-hook tableau, we construct a simple tensor with factors from B0.

Deﬁnition 5.1.8. Let T ∈ Γr,s(K). Then let vT = v1 ⊗ · · · ⊗ vk be the simple tensor

⊗k in V where vi is deﬁned by

 th  tj if i is in the j row of T(0), vi = th  uj if i is in the j row of T(1). 5. HIGHEST WEIGHT VECTORS IN V ⊗k 84

Example 5.1.9. In Example 5.1.7, the corresponding vT is given by

vT = t1 ⊗ t2 ⊗ t1 ⊗ t2 ⊗ u1 ⊗ t1 ⊗ u2 ⊗ t1 ⊗ t1 ⊗ u1.

Lemma 5.1.10. Let λ = (λ1, . . . , λ`) be an (r, s)-hook partition. Let T be an (r, s)- hook tableau of shape λ in Γr,s(K). Let vT = v1 ⊗ · · · ⊗ vk be the simple tensor as in

0 Deﬁnition 5.1.8 with vi ∈ B for all 1 ≤ i ≤ k. Then the weight of vT is given by

˜ ˜ λ1ε1 + ··· + λrεr + λr+1δ1 + ··· + λr+sδs.

Proof. By Deﬁnition 5.1.8, the number of ti in vT is λi for all 1 ≤ i ≤ r. The number ˜ of uj in vT is λr+j for all 1 ≤ j ≤ s. Therefore the result follows from Lemma 2.5.5.

Example 5.1.11. Let T be a (2, 3)-hook tableau of shape λ = (5, 2, 2, 1) as in

Example 5.1.7. We have λ(0) = (5, 2) and λ(1) = (2, 1). Then the corresponding vT has weight 5ε1 + 2ε2 + 2δ1 + δ2.

⊗k 0 Lemma 5.1.12. Let v = v1 ⊗ · · · ⊗ vk be a simple tensor in V with vi ∈ B for all

1 ≤ i ≤ k. Let π be an element of Sk. Then

(i) if there exists j ∈ {1, . . . , r} such that for all i, i0 ∈ K, π(i) = i0 and i 6= i0 ˇ implies vi = vi0 = tj, then Rπv = sgn(π)v, and

(ii) if there exists j ∈ {1, . . . , s} such that for all i, i0 ∈ K, π(i) = i0 and i 6= i0 ˇ implies vi = vi0 = uj, then Rπv = v.

We will say that such a π permutes tj with tj or uj with uj respectively.

Proof. Since every permutation can be written as a product of transpositions, we prove this lemma for transpositions (i, j) by induction on j − i. Note that for transpositions, ˇ the conclusions of the lemma are Rπv = −β(vi, vi)v in both cases. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 85

The base case is j − i = 1. Suppose that π = (i, i + 1), and vi = vi+1. Then by ˇ (2.2.2) and Proposition 2.2.10, Riv = −β(vi+1, vi)v = −β(vi, vi)v. Now suppose that the result holds when π = (i, j) with j − i ≥ 1. We prove the result holds for π = (i, j + 1).

Let π = (i, j + 1) and suppose vi = vj+1. We have π = (j, j + 1)(i, j)(j, j + 1). Then by (2.2.3), we have ˇ ˇ ˇ ˇ Rπv = RjR(i,j)Rjv.

For easy reference, we use boldface to indicate the factors in v that have been swapped in each step. Therefore we have

ˇ Rjv = −β(vj+1, vj)v1 ⊗ · · · ⊗ vj−1 ⊗ vj+1 ⊗ vj ⊗ vj+2 ⊗ · · · ⊗ vk. (5.1.1)

Using the inductive hypothesis, since the entries in the ith and jth positions are now equal, we have

ˇ ˇ RjR(i,j)v = β(vi, vi)β(vj+1, vj)v1 ⊗ · · · ⊗ vi−1 ⊗ vj+1 ⊗ vi+1

⊗ · · · ⊗ vj−1 ⊗ vi ⊗ vj ⊗ vj+2 ⊗ · · · ⊗ vk. (5.1.2)

ˇ Applying Rj to (5.1.2) will permute vi past vj. Thus we have

ˇ Rπv = −β(vi, vi)β(vj+1, vj)β(vj, vi)v1 ⊗ · · · ⊗ vi−1 ⊗ vj+1 ⊗ vi+1

⊗ · · · ⊗ vj−1 ⊗ vj ⊗ vi ⊗ vj+2 ⊗ · · · ⊗ vk. (5.1.3)

Since vi = vj+1, the simple tensor in (5.1.3) is equal to v. We compute the 5. HIGHEST WEIGHT VECTORS IN V ⊗k 86

coeﬃcient

−β(vi, vi)β(vj+1, vj)β(vj, vi) = −β(vi, vi)β(vi, vj)β(vj, vi) = −β(vi, vi).

This completes the proof of the lemma.

5.2 Young symmetrizers

Let us ﬁrst brieﬂy recall the role of Young symmetrizers in the classical representation theory.

Let Sk be the permutation group on k letters. Then when we consider the

representation theory of the symmetric group Sk for some k ∈ Z≥0, an operator

called a Young symmetrizer is used to construct irreducible Sk-modules which can be parametrized by partitions of k.

Deﬁnition 5.2.1. Let λ be a partition of k. Let T be a standard Young tableau of shape λ. The row group RT is the subgroup of Sk preserving the rows of T . The column group CT is the subgroup of Sk preserving the columns of T .

Example 5.2.2. Suppose that T = 1 2 . 3

Then RT = {id, (12)} and CT = {id, (13)}, where we use id to denote the identity

element in Sk.

Definition 5.2.3. Let F be a field. Let F[Sk] be the group algebra of Sk. Let T be a standard Young tableau of shape λ ` k. The normalized Young symmetrizer is defined 5. HIGHEST WEIGHT VECTORS IN V ⊗k 87 to be X yT = h(λ) sgn(ψ)ψγ ∈ F[Sk] ψ∈RT γ∈CT

+ where h(λ) ∈ Q is chosen so that yT is an idempotent.

dimSλ Remark 5.2.4. In [GW09], Goodman and Wallach prove that h(λ)= k! where λ S is the irreducible Sk-module corresponding to the partition λ. In particular the dimension of Sλ is the number of standard Young tableaux of shape λ with entries from K.

Example 5.2.5. Let T be the tableau in Example 5.2.2. There are only two standard

2 1 Young tableaux of shape (2, 1). So by Remark 5.2.4, we have h(λ) = 3! = 3 . Therefore we have 1 y = (id + (13) − (12) − (12)(13)). T 3

Recall that the group algebra F[Sk] is a subalgebra of the Brauer algebra. Thus, the Young symmetrizer will be acting on V ⊗k on the right, so the following is with respect to a right action. We ﬁrst prove a useful lemma.

Lemma 5.2.6. Let T be a standard tableau with entries from a subset L ⊆ K. Suppose

σ σ ∈ Sk is such that σ(i) < σ(j) whenever i < j and i, j ∈ L. Deﬁne T to be the tableau obtained from T by applying σ to each entry of T . Then T σ is also a standard

−1 Young tableau, with entries from σ(L). Moreover, σ yT σ = yT σ .

σ Proof. It is clear that T is a standard Young tableau. For ψ ∈ RT we can write

ψ = ψr1 ··· ψr` , where ψrj is a permutation which only permutes entries in row j of T . Then we have

−1 −1 −1 σ ψσ = (σ ψr1 σ) ··· (σ ψr` σ). 5. HIGHEST WEIGHT VECTORS IN V ⊗k 88

−1 Therefore it suﬃces to prove that σ ψrj σ is a permutation which only permutes the entries of row j of T σ for all 1 ≤ j ≤ `.

Let S = {i1, . . . , im} be the set of entries of row j of T . Then the entries of row

σ j of T are Sσ = {σ(i1), . . . , σ(im)}. From the deﬁnition of the right action, we know

−1 that σ ψrj σ permutes the elements Sσ the same way that ψrj permutes the entries of

−1 the S. It follows that the map ψ 7→ σ ψσ is a bijection from RT onto RT σ . Similarly,

−1 the map γ 7→ σ γσ is a bijection from CT to CT σ .Therefore by Deﬁnition 5.2.3, we have

−1 −1 X σ yT σ = σ h(λ) sgn(ψ)ψγσ

ψ∈RT γ∈CT X = h(λ) sgn(ψ)(σ−1ψσ)(σ−1γσ)

ψ∈RT γ∈CT X = h(λ) sgn(ψσ)ψσγσ σ ψ ∈RT σ σ γ ∈CT σ

= yT σ .

5.3 Contraction maps on V ⊗k

In classical representation theory, let W be a representation of a Lie algebra l. Then for all 1 ≤ i, j ≤ k, contractions Ci,j are linear maps from W ⊗k to W ⊗(k−2) which contract the ith and jth factors of the tensor into a scalar. For example, C1,2(a ⊗ b ⊗ c ⊗ d) = ω(a, b)c ⊗ d where ω(a, b) is a scalar valued bilinear map depending on a and b. Thus, if dim(W ) = n, then dim(Ker(Ci,j))=nk −nk−2. Typically, when we have a contraction map from V ⊗k to V ⊗(k−2), we tensor an invariant tensor to get a map from V ⊗k to V ⊗k. Some of the facts concerning contraction maps are discussed in [BBL90] and 5. HIGHEST WEIGHT VECTORS IN V ⊗k 89

[Bro55]. The vectors in W ⊗k which are annihilated by any composition of contraction maps Ci,j were ﬁrst called traceless tensors in [Wey97]. The set of all such tensors is called the harmonic space of k-tensors in [GW09], which is an l-submodule of W ⊗k. In [BBL90], Benkart et. al. described an algorithm to ﬁnd highest weight vectors for all classical Lie algebras, which is then generalized to the algorithm given in [BSR98] for Lie colour algebras.

⊗k ⊗k We begin by constructing contraction maps Ci,j : V → V .

Deﬁnition 5.3.1. Let ci,j be the diagram ......

ci,j = ...... i j

⊗k in the Brauer algebra Bk(n − m). Then the contraction map Ci,j ∈ End(V ) is given

by Ci,j = Ψci,j .

Since for all i 6= j, ci,j = cj,i in the Brauer algebra, we have that Ci,j = Cj,i. Thus, we can restrict ourselves to contraction maps Ci,j such that 1 ≤ i < j ≤ k.

Example 5.3.2. Since ci,i+1 = ei, we have Ci,i+1 = Ψei .

Lemma 5.3.3. Let ci,j be the element of the Brauer algebra Bk(η) as deﬁned in

−1 Deﬁnition 5.3.1. Let σ be a permutation in Sk. Then we have σ ci,jσ = cσ(i),σ(j).

−1 Proof. Let us consider the product of σ ci,jσ in Bk(η) which we visualize in Figure 5.2. Then the result follows.

⊗k Lemma 5.3.4. Let Ci,j be a contraction map on V , where 1 ≤ i < j ≤ k. Then

−1 for any σ ∈ Sk, we have σ Ci,jσ = Cσ(i),σ(j). 5. HIGHEST WEIGHT VECTORS IN V ⊗k 90

σ−1(j) σ−1(i) ...... σ−1 = ...... i j ......

ci,j = ...... i j

i j ...... σ = ...... σ(i) σ(j) which is ......

= cσ(i),σ(j) ...... σ(i) σ(j) .

−1 Figure 5.2: The composition σ ci,jσ as diagrams in the Brauer algebra, and the resulting diagram cσ(i),σ(j). We suppress most of the vertical edges in both σ and σ−1. We only keep the vertical edges linking i, j with σ(i), σ(j) respectively in σ, and the vertical edges linking i, j with σ−1(i), σ−1(j) respectively in σ−1, and we label both top and bottom rows by 1, . . . , k for simplicity.

op ⊗k ⊗k Proof. By Theorem 4.2.4, Ψ is a homomorphism from Bk(n−m) to Homg(V ,V ). Thus, by Lemma 5.3.3, we have (as a right action)

−1 −1 −1 Cσ(i),σ(j) = Ψcσ(i),σ(j) = Ψσ Ψci,j Ψσ = σ Ψci,j σ = σ Ci,jσ

as claimed.

⊗k Lemma 5.3.5. Let V be an spo(V, β)-module. The image of C1,2 = Ψe1 in V is isomorphic to V ⊗(k−2). 5. HIGHEST WEIGHT VECTORS IN V ⊗k 91

⊗2 Proof. It is enough to prove that the image of C1,2 in V is isomorphic to the trivial module of spo(V, β). Apply Corollary 4.2.2 to our homogenous basis B, we have that for any u, v ∈ B,

X −1 (u ⊗ v)Ψe1 = Fu,v Fs,t s ⊗ t = Ω. (5.3.1) s,t∈B

Recall that from Lemma 2.2.12, we have a g-module morphism prV such that in our case,

Pn+m i if we relabel the homogeneous basis B = {v1, . . . , vn+m}, then prV (1) = i=1 vi ⊗ v is a g-invariant tensor. Also recall from Lemma 2.4.3, F −1 : V ∗ → V is a g-module isomorphism. Then we have

m+n −1 X −1 i (id ⊗ F ) ◦ prV (1) = vi ⊗ F (v ) i=1 m+n X −1 = Fi,j vi ⊗ vj = Ω. i=1

−1 −1 Thus Ω is a g invariant tensor since x(id ⊗ F )(prV (1)) = (id ⊗ F )(xprV (1)) = 0

for all x ∈ g. Therefore that the image of C1,2 on V ⊗ V is isomorphic to the trivial

⊗k ∼ ⊗(k−2) module and in turn, we conclude that V C1,2 = V .

−1 Notice that the values of Fs,t are given explicitly in (2.4.3), and we conclude that (5.3.1) is r s X ∗ ∗ X ∗ ∗ Ω = (ti ⊗ ti − ti ⊗ ti ) + (uj ⊗ uj + uj ⊗ uj). (5.3.2) i=1 j=1

−1 More generally, using the relation Ci,j = σ C1,2σ for some permutation σ ∈ Sk, we

ﬁnd the following. Let w = w1 ⊗ · · · ⊗ wk with w` ∈ B, then for 1 ≤ i < j ≤ k, the 5. HIGHEST WEIGHT VECTORS IN V ⊗k 92

vector wCi,j is a nonzero scalar multiple of

X −1 Fwi,wj Fs,t w1 ⊗ · · · ⊗ wi−1 ⊗ s ⊗ wi+1 ⊗ · · · ⊗ wj−1 ⊗ t ⊗ wj+1 ⊗ · · · ⊗ wk. (5.3.3) s,t∈B

⊗k ∼ ⊗(k−2) In particular, this implies that V Ci,j = V , for all 1 ≤ i < j ≤ k. As in the classical sense, we want to compose multiple contraction maps.

Deﬁnition 5.3.6. Let p = {p1, . . . , pj} and q = {q1, . . . , qj} be two ordered disjoint subsets of K such that p` < q` for all 1 ≤ ` ≤ j. Then we deﬁne

Cp,q := Cp1,q1 ··· Cpj ,qj .

We also call Cp,q a contraction map.

We pair each p` ∈ p and q` ∈ q together to deﬁne a set of pairs

(p, q) = {(p1, q1),..., (pj, qj) | p` < q` for 1 ≤ ` ≤ j}. (5.3.4)

For each j = 1,..., bk/2c, we deﬁne the set of all such (p, q)’s by

P(j) := {(p, q) | the cardinality of p and q are both j}, and the set of all possible indices of contraction maps on V ⊗k is given by

[k/2] [ P = P(j). (5.3.5) j=0

We give the following deﬁnitions of harmonic tensor spaces before we construct highest weight vectors. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 93

⊗k ⊗k Deﬁnition 5.3.7. The harmonic tensor space H(V ,Cp,q) of V is the kernel of

⊗k the contraction map Cp,q acting on V .

⊗k Lemma 5.3.8. The harmonic tensor space H(V ,Cp,q) is an spo(V, β)-submodule of V ⊗k.

⊗k Proof. Let g = spo(V, β). We have that v ∈ H(V ,Cp,q) if and only if vCp,q = 0. Since

⊗k the action of g = spo(V, β) on V commutes with the action of Bk(n−m), we have for

⊗k all X ∈ g,(Xv)Cp,q = X(vCp,q) = X(0) = 0, which implies g · v ⊆ H(V ,Cp,q).

∗ Example 5.3.9. Suppose that r = 1, s = 0 and B = {t1, t1}. Consider the spo(V, β)-

⊗2 ∗ module V = Span{v ⊗ w | v, w ∈ B}. Using (5.3.1), u ⊗ v ∈ Ker(C1,2) if u 6= v .

⊗2 With a little more work, we can see that H(V ,C1,2) is the span of

∗ ∗ ∗ ∗ {t1 ⊗ t1, t1 ⊗ t1 + t1 ⊗ t1, t1 ⊗ t1}.

Deﬁnition 5.3.10. Let s and t be disjoint subsets of K − (p ∪ q). Then the harmonic

⊗k ⊗k tensor space H(V Cp,q,Cs,t) is the kernel of the contraction map Cs,t : V Cp,q →

⊗k V Cp,q.

⊗k Notice that H(V Cp,q,Cs,t) is also an spo(V, β)-submodule.

5.4 Construction of highest weight vectors

In this section, we construct highest weight vectors of several submodules of the spo(V, β)-module V ⊗k following the algorithm presented by Benkart et. al. in [BSR98]. This result is stated in Theorem 5.4.1, and proven in the course of several lemmas.

Let us ﬁrst motivate the choice of w = wT,p,q appearing in the theorem. The idea

⊗k is to choose some special simple tensor w = w1 ⊗ · · · ⊗ wk ∈ V , and then transform 5. HIGHEST WEIGHT VECTORS IN V ⊗k 94

it into a highest weight vector by applying contraction maps and normalized Young symmetrizers on the right. Let us ﬁrst investigate the contraction maps. By Lemma 5.3.4 and Lemma 5.3.5,

Ci,j will contract two factors wi and wj of w into a g-invariant tensor, namely Fwi,wj Ω.

∗ Since wi, wj ∈ B, we have Fwi,wj 6= 0 if and only if wi = wj . Therefore, wCi,j is

∗ ∗ nonzero if and only if wi = wj . Thus, we can start with vectors wi = t1 and wj = t1 in the position that we will apply contractions, which will be i ∈ p and j ∈ q. Now for i 6∈ (p ∪ q), we can construct an (r, s)-hook tableau with entries from

K − (p ∪ q). Then we choose each wi by using Deﬁnition 5.1.8 for all i ∈ K − (p ∪ q).

Theorem 5.4.1. Let λ be an (r, s)-hook partition of k − 2j, for all 0 ≤ j ≤ [k/2].

Let (p, q) ∈ P(j) and ﬁx T ∈ Γr,s(K − (p ∪ q)) of shape λ. Let T(0) and T(1) be the ˜ ˜ corresponding subtableaux of shape λ(0) = (λ1, ··· , λr) and λ(1) = (λr+1, ··· , λr+s) respectively. Let wT,p,q = w1 ⊗ · · · ⊗ wk be the simple tensor deﬁned by

  t if i ∈ p,  1   ∗  t1 if i ∈ q, wi =  t if i ∈ K − (p ∪ q) and i is in `th row of T ,  ` (0)   th  u` if i ∈ K − (p ∪ q) and i is in ` row of T(1).

⊗k Then v = wT,p,qCp,qyT is a highest weight vector of V . For any s, t ⊂ K − (p ∪ q)

⊗k such that s ∩ t = ∅, v is an element of H(V Cp,q,Cs,t)yT .

Moreover, if n = 2s and if there are s rows in T(1), then there is extra highest weight vector v = wT,p,qCp,qyT , where wT,p,q is the simple tensor obtained from wT,p,q

∗ by replacing each occurrence of us with us.

We will prove Theorem 5.4.1 by means of a sequence of lemmas. For simplicity, 5. HIGHEST WEIGHT VECTORS IN V ⊗k 95

we write w for wT,p,q, w for wT,p,q and wt(u) for the weight of a vector u. We begin by determining the weights of v and v.

Lemma 5.4.2. Let w and Cp,q be deﬁned as in Theorem 5.4.1. Then

(i) wCp,q is annihilated by all Cs,t such that s, t ∈ K − (p ∪ q),

(ii) the weight of v equals wt(w) which is

˜ ˜ λ1ε1 + ··· + λrεr + λr+1δ1 + ··· + λr+sδs, and

(iii) the weight of v is ˜ wt(w) − 2λr+sδs. (5.4.1)

Proof.

(i) If s and t are both in K − (p ∪ q), then the corresponding ws and wt will be in

0 the set B = {t1, . . . , tr, u1, . . . , us, (us+1)}. Therefore Cs,t will contract ws and

0 wt into Fws,wt Ω which is 0 since Fws,wt = 0 for all ws, wt ∈ B .

∗ (ii) By the construction of w, the numbers of t1’s and t1’s coming from p and q

∗ are equal in w, and we have wt(t1) = −wt(t1). Therefore the weight of w is

determined by its factors wi such that i ∈ K − (p ∪ q). Therefore Lemma 5.1.10 implies that

˜ ˜ wt(w) = λ1ε1 + ··· + λrεr + λr+1δ1 + ··· + λr+sδs. (5.4.2)

∗ Moreover, Cp,q will contract pairs of t1 and t1 of w into a g-invariant tensor

which also has weight 0, we have wt(wCp,q) = wt(w). Therefore the equation

wt(v) = wt(wCp,qyT ) = wt(w) holds since yT preserves weights. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 96

(iii) The result follows from (ii) and the fact that w is obtained from w by changing

∗ us to us.

Now we verify that v =6 0. This follows from Lemma 5.4.3 and Proposition 5.4.4. The same process is also valid for the vector v. First, let E be the subspace of V ⊗k spanned by the simple tensors of the form

w1 ⊗ · · · ⊗ wk such that wj ∈ B for all 1 ≤ j ≤ k, and wi 6= wi for some 1 ≤ i ≤ k .

Clearly w 6∈ E. Then we identify a key property of the action of yT on wCp,q by the following lemma.

Lemma 5.4.3. Let ψ ∈ RT and σ ∈ CT . Then w sgn(ψ)ψσ = cw for some scalar c

if and only if ψ ∈ RT(0) and σ ∈ RT(1) . In this case, c = 1.

th Proof. Take ψ ∈ RT(0) . Then ψ permutes elements within the i row of T(0) for

th 1 ≤ i ≤ r. By the deﬁnition of w, the elements in the i row of T(0) will be recorded

as ti in w. Therefore ψ will only permute ti with ti in w. By Lemma 5.1.12, we have wψ = sgn(ψ)w.

Now take σ ∈ CT . If σ ∈ RT(1) , then σ will only permute uj with uj for all 1 ≤ j ≤ s. Again by Lemma 5.1.12, we have wσ = w. Thus we have w sgn(ψ)ψσ = w

if ψ ∈ RT(0) and σ ∈ RT(1) . In order to prove the other direction is true, it is enough to prove the following statement:

“If either σ 6∈ RT(1) or ψ 6∈ RT(0) , then wψσ ∈ E.”

If σ 6∈ RT(1) , without loss of generality, we may assume that σ ∈ CT \RT(1) . Then

th there exists i1 appearing in the j row of T(0) such that σ(i1) 6= i1. We have wi1 = tj.

If σ(i1) appears in T(0), then σ(i1) is in a diﬀerent row than i1. So wσ(i1) 6= wi1 = tj.

Otherwise, wσ(i1) = u` for some `, and again wi1 6= wσ(i1). Moreover, for any ψ ∈ RT , 5. HIGHEST WEIGHT VECTORS IN V ⊗k 97

th the vector in the i1 position of wψ is still tj. This implies that the vector in the

th i1 position of wψσ is not tj. So wψσ ∈ E. This argument can be visualized by the

following ﬁgure; notice in this case, i1 and σ(i1) are in the same column but diﬀerent rows. ···

i1 ···

σ(i1) ··· ···

Now let us consider that ψ 6∈ RT(0) . Then without loss of generality, we assume

that ψ ∈ RT \RT(0) . Then there exists i1 6= i2 in the same row of T but not in T(0)

such that ψ(i1) = i2. If i` is in column j` of T for ` = 1, 2, then wi1 = uj1 =6 uj2 = wi2 .

th Therefore the vector in the i1 position of wψ is wψ(i1) = wi2 = uj2 . Now for any σ ∈ CT ,

σ(i2) is still in column j2 of T . If σ(i2) appears in T(0), then wσψ(i1) = wσ(i2) = tj for

some j. Otherwise, σ(i2) appears in T(1). Then it will also be in row j2 of T(1), so

th wσψ(i1) = uj2 . In either case, the vector in the i1 position of wψσ is not uj1 = wi1 .

Thus if σ ∈ CT but ψ ∈ RT \RT(0) , then wψσ ∈ E. This argument can be visualized

by the following ﬁgure; notice that in our case, we draw the transpose of T(1), and σ

permutes entries in the same column of i2. ψ

...... i . i . 1 .2 T 0 . . (1) . . σ 5. HIGHEST WEIGHT VECTORS IN V ⊗k 98

Proposition 5.4.4. The vector wCp,q can be written as βp,qw + R, where R ∈ E and

βp,q 6= 0.

Proof. We prove this proposition by ﬁrst writing wCi,j explicitly for some 1 ≤ i < j ≤

k. Take w = w1 ⊗ · · · ⊗ wk which is constructed according to Theorem 5.4.1. Then

∗ for each pair (i, j) ∈ (p, q), we have wi = t1 and wj = t1.

Therefore by (5.3.3) and the fact that F ∗ = 1, we have (up to a nonzero scalar t1,t1 multiple, indicated by ∼)

X −1 wCi,j ∼ Fu,v w1 ⊗· · ·⊗wi−1 ⊗u⊗wi+1 ⊗· · ·⊗wj−1 ⊗v ⊗wj+1 ⊗· · ·⊗wk. (5.4.3) u,v∈B

∗ −1 Note that if u = t1 and v = t1, we have Fu,v = −1. Thus, wCi,j ∼ w + Ri,j, where

X −1 Ri,j = Fu,v w1 ⊗ · · · ⊗ wi−1 ⊗ u ⊗ wi+1 ⊗ · · · ⊗ wj−1 ⊗ v ⊗ wj+1 ⊗ · · · ⊗ wk. u,v∈B ∗ u6=t1,v6=t1

th Note that the vector in the i position in w is t1 but the vector in the same position

in any simple summand in Ri,j cannot be t1. Thus Ri,j ∈ E.

If we apply a contractions Cs,t with 1 ≤ s < t ≤ k and {s, t} 6= {i, j} on

th th Ri,j, the vectors in the i and j positions of Ri,j will remain unchanged. Thus

every simple tensor appearing in the expression for Ri,jCs,t is an element of E. Since

wCs,t ∼ w + Rs,t, it follows by induction that wCp,q ∼ w + R, where R ∈ E.

Corollary 5.4.5. If we write v = wCp,qyT as a linear combination of simple tensors in the basis B, then the coeﬃcient of w in v is given by β h(λ) R R . In p,q T(0) T(1) particular, v 6= 0. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 99

Proof. By Proposition 5.4.4, we have wCp,q = βp,qw + R, where R ∈ E. Then we

consider the action of normalized Young symmetrizer yT on w and R separately. We

ﬁrst show that RyT is in E.

By the deﬁnition of yT and the way we construct w (Theorem 5.4.1), yT only

permutes the vectors in R at the positions which will not be contracted by Cp,q. For

each i ∈ p, wi = t1. By the proof of Proposition 5.4.4, for each simple tensor in R,

th there exists i ∈ p such that the vector in its i position is not t1. Therefore every

summand of RyT is an element of E, which implies RyT ∈ E. Moreover, we have

X wyT = h(λ) w sgn(ψ)ψσ.

ψ∈RT ,σ∈CT

By Lemma 5.4.3, w sgn(ψ)ψσ = w if and only if ψ ∈ RT(0) and σ ∈ RT(1) . This implies

that if either ψ 6∈ RT(0) or σ 6∈ RT(1) , then w sgn(ψ)ψσ ∈ E. Thus we can write wyT as a linear combination of w and R00 ∈ E. Namely, we have

X 00 wyT = h(λ) w + R . ψ∈R T(0) σ∈R T(1)

Therefore

X 00 wCp,qyT = βp,qh(λ) w + R + RyT , ψ∈R T(0) σ∈R T(1)

where R00, Ry ∈ E. Thus, the coeﬃcient of w in v is given by β h(λ) R R T p,q T(0) T(1) which is not 0. Therefore v is not a zero vector. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 100

Our ultimate goal is to prove that v and v are highest weight vectors. By the previous discussion we have proved that they are nonzero weight vector with weight as given in (5.4.2) and (5.4.1) respectively. Therefore it remains to check whether the simple root vectors Y deﬁned in Deﬁnition 2.5.13 acting on v and v will give 0.

To do so, we use the following lemmas which helps us to view v = wT,p,qCp,qyT more concretely. The same process also works for v as we outline at the end of this section.

Deﬁnition 5.4.6. For all 1 ≤ j ≤ bk/2c, we let Co,e = C1,2C3,4 ··· C2j−1,2j.

Lemma 5.4.7. Let σ ∈ Sk be the permutation such that

(i) σ(pt) = 2t − 1 and σ(qt) = 2t for all 1 ≤ t ≤ j, and

(ii) σ(i1) > σ(i2) for all i1 > i2 and i1, i2 ∈ K − (p ∪ q).

−1 Then for any contraction maps Cp,q = Cp1,q1 ...Cpj ,qj , we have σCo,eσ = Cp,q.

Proof. The result follows from the fact Co,e = C1,2C3,4 ··· C2j−1,2j and Lemma 5.3.4

σ Lemma 5.4.8. Let σ ∈ Sk be the permutation deﬁned in Lemma 5.4.7. Let T be the standard tableau deﬁned in Lemma 5.2.6 with entries in K − (p ∪ q). We have

wT,p,qσ = βσ(w1, . . . , wk)wT σ,o,e,

where βσ(w1, . . . , wk) is the same scalar deﬁned in Lemma 3.1.5.

Proof. By the way we deﬁned σ, we have

wT,p,qσ = (w1 ⊗ · · · ⊗ wk)σ

= βσ(w1, . . . , wk)wσ(1) ⊗ · · · ⊗ wσ(k). 5. HIGHEST WEIGHT VECTORS IN V ⊗k 101

Let vi = wσ(i) for all 1 ≤ i ≤ k. In particular, we have

  t if σ(i) ∈ p,  1   ∗  t1 if σ(i) ∈ q, wσ(i) =  t if σ(i) is in the `th row of T ,  ` (0)   th  u` if σ(i) is in the ` th row of T(1).

Recalling the deﬁnition of T σ from Lemma 5.2.6, we deduce that this is equivalent to

  t if i ∈ o,  1   ∗  t1 if i ∈ e, vi =  t if i is in the `th row of T σ ,  ` (0)   th σ  u` if i is in the ` row of T(1).

Since this deﬁnes wT σ,o,e, we have wT,p,qσ = βσ(w1, . . . , wk)wT σ,o,e.

For simplicity, we write βσ for βσ(w1, . . . , wk).

Corollary 5.4.9. Retain the notation in Lemma 5.4.7 and 5.4.8. We have

−1 wT,p,qCp,qyT = βσwT σ,o,eCo,eyT σ σ .

Proof. First notice that wT,p,qCp,qyT is equal to

−1 −1 −1 (wT,p,qσ)(σ Cp,qσ)(σ yT σ)σ . (5.4.4)

Then by Lemma 5.4.8, 5.4.7 and 5.2.6, (5.4.4) becomes

−1 βσwT σ,o,eCo,eyT σ σ , 5. HIGHEST WEIGHT VECTORS IN V ⊗k 102 which completes the proof.

Corollary 5.4.10. The vector wT,p,qCp,qyT is equal to

−1 βσ(Ω ⊗ · · · ⊗ Ω ⊗ v2j+1 ⊗ · · · ⊗ vk)yT σ σ (5.4.5)

where Ω is the invariant tensor deﬁned in (5.3.2), and v` = wσ(`).

Proof. Notice that

∗ ∗ wT σ,o,e = t1 ⊗ t1 ⊗ · · · ⊗ t1 ⊗ t1 ⊗ v2j+1 ⊗ · · · ⊗ vk, (5.4.6)

∗ where the ﬁrst 2j terms of (5.4.6) are j pairs of t1 ⊗ t1. Therefore by Lemma 5.3.5

Co,e acting on (5.4.6) will contract the ﬁrst 2j terms into j times of Ω. Thus the result follows from Corollary 5.4.9.

Now in order to prove that v = wT,p,qCp,qyT is a highest weight vector, it suﬃces to prove that the simple root vectors Y act on (5.4.5) as 0. Since Ω is a g-invariant tensor, Y acts on each factor Ω as 0. Therefore let

θY,i = Ω ⊗ · · · ⊗ Ω ⊗ v2j+1 ⊗ · · · ⊗ (Y v2j+i) ⊗ · · · ⊗ vk. (5.4.7)

We have k−2j X Y · (Ω ⊗ · · · ⊗ Ω ⊗ v2j+1 ⊗ · · · ⊗ vk) = cY,iθY,i i=1

for some scalars cY,i.

0 Recall that the vi’s are all in B . Each simple root vector (Deﬁnition 2.5.13) acts by zero on all but one vector in B0. We record how the simple root vectors act on b ∈ B0 in Table 5.1. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 103

0 Y ∈ ∆Y b ∈ B Y b Conditions

Yγi ti+1 ti 1 ≤ i ≤ r − 1

Yγr u1 tr

Yγi+j uj+1 uj 1 ≤ j ≤ s − 1

Yγr+s us+1 us n = 2s + 1 Y v 0 for all other cases

Table 5.1: The action of simple root vectors from Deﬁnition 2.5.13 on v ∈ B0

Therefore, for all pairs (Y, v2j+i) in the ﬁfth row of Table 5.1, we have θY,i = 0.

Now we claim that θY,iyT σ = 0 for all Y ∈ ∆Y and for all 1 ≤ i ≤ 2k − j. From Table

5.1, we observe that Y acting on v2j+i will either lower the subscript by one, or change

u1 into tr. Let us ﬁrst consider an example.

Example 5.4.11. Let Y = E − β(t , t )E ∗ ∗ . Then Y acting on Ω ⊗ t ⊗ t ⊗ t t2,t3 2 3 t3,t2 3 1 3 gives

∗ Ω ⊗ t2 ⊗ t1 ⊗ t3 + β(t3t1, t2t3)Ω ⊗ t3 ⊗ t1 ⊗ t2.

Then θY,1 = Ω ⊗ t2 ⊗ t1 ⊗ t3, θY,2 = 0 and θY,3 = Ω ⊗ t3 ⊗ t1 ⊗ t2.

Lemma 5.4.12. Let θ = θY,i 6= 0 for some Y = Yγi ∈ ∆Y and such that v2j+i = t`+1.

Then for all ψ ∈ RT σ there exists some permutation π in CT σ such that θψπ = −θψ.

∗ ∗ Proof. By the construction of wσ = wT σ,o,e = t1 ⊗ t1 ⊗ · · · ⊗ t1 ⊗ t1 ⊗ v2j+1 ⊗ · · · ⊗ vk, we know that for all i ∈ {1, . . . , k − 2j}, if v2j+i = t`+1 for some 1 ≤ ` ≤ r − 1, then

th σ 2j + i is in the (` + 1) row of T(0). In particular, the tensor wσ looks like

· · · ⊗ t`+1 ⊗ · · · (5.4.8) |{z} (2j+i)th position and so the (` + 1)th row of the tableau T σ looks like 5. HIGHEST WEIGHT VECTORS IN V ⊗k 104

. .

row ` + 1 −→ ··· 2j+i ··· . . By hypothesis θ has the form

· · · ⊗ t` ⊗ · · · . |{z} (2j+i)th position

First suppose that ψ ∈ RT σ is such that ψ(2j + i) = 2j + i. Let i1 be the value in the box directly above 2j + i in T σ, as in: . .

row ` −→ i1

row ` + 1 −→ ··· 2j+i ··· . .

σ Therefore we have vi1 = t`. Hence the permutation (i1, 2j + i) ∈ CT will permute tj with tj in θψ. Thus we have θψ(i1, 2j + i) = −θψ by Lemma 5.1.12.

Next, suppose instead that ψ ∈ RT σ such that ψ(2j + i) = i2 for some i2 6= 2j + i in the (` + 1)th row of T σ. Then θψ is the tensor

· · · ⊗ t` ⊗ · · · ⊗ t`+1 ⊗ · · · . |{z} th |{z}th i2 position (2j+i) position

σ Let i3 be the entry of the tableau T in row ` directly above i2, as in the diagram: . .

row ` −→ i3

row ` + 1 −→ ··· i2 ··· . . 5. HIGHEST WEIGHT VECTORS IN V ⊗k 105

σ In this case, we have vi3 = t`. Hence the permutation (i2, i3) ∈ CT will permute the

corresponding t` and t` in θψ. Therefore, we have θψ(i2, i3) = −θψ. Thus (i2, i3) is the desired permutation.

We have proved that if Y acting on v2j+i = t`+1 for some ` lowers the index

` + 1 to `, then for all ψ ∈ RT σ , there exists some permutation π ∈ CT σ such that

θψπ = −θψ. Note that the case when Y changes u1 into tr can be proved similarly.

Lemma 5.4.13. Retain the set up in Lemma 5.4.12, we have θyT σ = 0.

Proof. By Lemma 5.4.12, for all ψ ∈ RT σ , there exists some π ∈ CT σ such that

θψπ = −θψ. Thus suppose ψ ∈ RT σ , there exists a choice of π ∈ CT σ such that

X X X sgn(ψ)(θψ)γ = sgn(ψ)(θψ)πγ = − sgn(ψ)(θψ)γ,

γ∈CT σ γ∈CT σ γ∈CT σ

P which implies that sgn(ψ)(θψ)γ = 0 for each ψ ∈ R σ and therefore θy σ = γ∈CT σ T T 0.

Now let us consider the case when Y changes u`+1 into u` for all 1 ≤ ` ≤ s − 1.

Lemma 5.4.14. Let θ = θY,i 6= 0 for some Y = Yγr+` ∈ ∆Y and such that v2j+i = u`+1.

Then there exists some permutation π ∈ RT σ such that θπ = θ and sgn(π) = −1.

Proof. The proof is similar to the proof of Lemma 5.4.13. Noticing that if v2j+i = u`+1

th σ 0 for some 1 ≤ ` ≤ s − 1, then 2j + 1 is in the (` + 1) column of (T(1)) (or equivalently,

th σ the (` + 1) row of T(1)). Therefore there exists a permutation π ∈ RT σ such that π permutes 2j + i and the entry in the box directly to the left of the box of 2j + i,

th σ 0 say i1. Note that vi1 = u` since i1 is in the ` column of (T(1)) . Moreover, since

th Yγr+` v2j+i = u`, the (2j + i) position of θ is also u`. In this way, we simply choose 5. HIGHEST WEIGHT VECTORS IN V ⊗k 106

π = (i1, 2j + i), and thus, sgn(π) = −1. Then π permutes u` with u` in θ, which implies θπ = θ.

Lemma 5.4.15. Retain the set-up in Lemma 5.4.14, we have θyT σ = 0.

Proof. By Lemma 5.4.14, there exists some permutation π ∈ RT σ such that θπ = θ and sgn(π) = −1. Therefore we have

X X X sgn(ψ)θψ = sgn(ψ)sgn(π)(θπ)ψ = − sgn(ψ)(θψ)

ψ∈RT σ ψ∈RT σ ψ∈RT σ

P which implies that sgn(ψ)θψ = 0 and therefore θy σ = 0. ψ∈RT σ T

Proposition 5.4.16. The vector wT,p,qCp,qyT we deﬁned in Theorem 5.4.1 is a highest weight vector.

Proof. By Lemma 5.4.13 and 5.4.15, every nonzero θY,i will be transformed into 0

by yT σ . Thus Y wT σo,eCo,eyT σ = 0, and in turn, Y wT,p,qCp,qyT = 0 which implies that

wT,p,qCp,qyT is a highest weight vector.

Notice that the whole process (from Deﬁnition 5.4.6 to Proposition 5.4.16) of proving v is a highest weight vector is also valid for proving v is a highest weight vector. The only change we need to make is to notice that we have one extra case

∗ ∗ in Table 5.1, that is, if n = 2s, T(1) has s rows, and b = us, then Yγr+s us = us−1. Therefore we have one extra case to analyze, that is, the simple root vector acting on

∗ v2j+i changes us to us−1. However, the proof of this case is the same as the one we

∗ provided in the proof of Lemma 5.4.14 by changing u`+1 into us and changing u` into

us−1 in Lemma 5.4.14. Thus, we conclude that the vectors produced by Theorem 5.4.1 are highest weight vectors. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 107

5.5 An illustration of Theorem 5.4.1

We have proved that the vectors in Theorem 5.4.1 are highest weight vectors. Next, we illustrate how to ﬁnd these vectors in Section 5.5.1 by expanding Example 2.5.15. Then we verify these vectors are highest weight vectors in Section 5.5.2. In Section 5.5.3, we ﬁnd spo(V, β)-submodules generated by these highest weight vectors. Then we deduce that that V ⊗ V can be written as a direct sum of these submodules. Moreover, we will extend the result in this section to a more generalized version in Section 6.3.

5.5.1 The highest weight vectors

Retain the setup in Example 2.5.15. Let V = V(0) ⊕ V(1) be the standard spo(V, β)-

module such that dim(V(0)) = 2r = 2 and dim(V(1)) = 2s = 4. Thus V has a

∗ ∗ ∗ homogeneous basis B = {t1, t1, u1, u1, u2, u2}. Elements of the Cartan subalgebra of g

⊗k ⊗2 are given by H = diag(a1, −a1, a2, −a2, a3, −a3). Furthermore, we consider V = V .

In order to find the vectors v = wT,p,qCp,qyT , we first find all possible (p, q) ∈ P(j). Since k = 2, we have 0 ≤ j ≤ 1 = bk/2c. Therefore, by (5.3.5), the set of all indices of all possible contractions are

1 [ P = P(j) = P(0) ∪ P(1) = {( , )} ∪ {(1, 2)} = {( , ), (1, 2)}, j=0 where we use to denote the empty sequence in order to distinguish from the empty tableau ∅.

Let us ﬁrst consider (p, q) = (1, 2). We have Cp,q = C1,2. The only possible 5. HIGHEST WEIGHT VECTORS IN V ⊗k 108

(1, 2)-hook tableau T in Γ1,2( ) is the empty tableau ∅. In this case we have

T = ∅ with T(0) = ∅ and T(1) = ∅.

∗ Since 1 ∈ p and 2 ∈ q, we have w∅,1,2 = t1 ⊗ t1. Additionally, we have yT = id since T = ∅. Therefore the highest weight vector in this case is

∗ v1 = (t1 ⊗ t1)C1,2id. (5.5.1)

In Proposition 4.2.2, we already computed that

∗ ∗ ∗ ∗ ∗ ∗ v1 = −t1 ⊗ t1 + t1 ⊗ t1 + u1 ⊗ u1 + u1 ⊗ u1 + u2 ⊗ u2 + u2 ⊗ u2 = Ω. (v1)

Next we consider the case when (p, q) = ( , ). We have Cp,q = C , = Ψid. In addition, there are two possible (1, 2)-hook tableaux in Γ1,2(K). In particular, we have

T = 1 with T(0) = 1 and T(1) = 2 , or (5.5.2) 2

T = 1 2 with T(0) = 1 2 and T(1) = ∅. (5.5.3)

1 Let T be given as in (5.5.2). We have yT = 2 (id + (12)). In addition, since 1 is in st st the 1 row of T(0), and 2 is in the 1 row of T(1), we have wT,∅,∅ = t1 ⊗ u1. Thus the highest weight vector in this case is

1 v = (t ⊗ u )Ψ (id + (12)). 2 2 1 1 id

Then we write v2 explicitly. By Remark 4.1.10, the action of id + (12) on V ⊗ V is 5. HIGHEST WEIGHT VECTORS IN V ⊗k 109

given by the map Ψid + Ψs1 . Therefore we have

1 1 v = (t ⊗ u )Ψ + (t ⊗ u )Ψ 2 2 1 1 id 2 1 1 s1 which (by Proposition 4.2.2) is

1 1 v = (t ⊗ u ) − β(u , t )(u ⊗ t ). (v ) 2 2 1 1 2 1 1 1 1 2

1 Finally, let T be deﬁned as in (5.5.3). We have yT = 2 (id − (12)). Moreover, since st both 1 and 2 are in the 1 row of T(0), we obtain wT,∅,∅ = t1 ⊗ t1. Therefore in this case, the highest weight vector is

1 v = (t ⊗ t )Ψ (id − (12)) 3 2 1 1 id 1 = (t ⊗ t − (−β(t , t )t ⊗ t )) 2 1 1 1 1 1 1

= t1 ⊗ t1. (v3)

5.5.2 Veriﬁcation that v1, v2 and v3 are highest weight vectors

Recall that in order to check if v1, v2 and v3 are highest weight vectors, it suﬃces to check that H acts on these three vectors by scalars, and then verify that for all positive root vectors X of spo(V, β), X acts on these three vectors as 0.

The vector v1 is a highest weight vector with weight 0 since we already proved that it is an spo(V, β)-invariant vector in Lemma 5.3.5. Notice that the possible weights of vectors in V ⊗ V are in the set

Φ = {±2ε1, ±ε1 ± δ1, ±ε1 ± δ2, ±2δ1, ±2δ2}. 5. HIGHEST WEIGHT VECTORS IN V ⊗k 110

Therefore by Lemma 2.5.14, in order to prove that v2 is a highest weight vector,

it suﬃces to only consider the simple root vectors Y ∈ ∆Y such that wt(Y v2) ∈ Φ.

Since wt(v2) = ε1 + δ1, the only Y that satisﬁes this condition is

Y = E + β(t , u )E ∗ ∗ . t1,u1 1 1 u1,t1

First notice that E acting on t gives 0, and E ∗ ∗ acting on either t or u gives t1,u1 1 u1,t1 1 1 0. Thus we have

∗ Y v2 = β(t1, t1u1)t1 ⊗ Et1,u1 u1 − β(u1, t1)Et1,u1 u1 ⊗ t1

∗ = β(t1, t1)β(t1, u1)t1 ⊗ t1 − β(u1, t1)t1 ⊗ t1

∗ = β(t1, u1)t1 ⊗ t1 − β(u1, t1)t1 ⊗ t1

∗ which is 0 by the fact that β(t1, u1) = β(u1, t1). Therefore v2 is a highest weight vector.

By a similar but easier argument, v3 is also a highest weight vector.

5.5.3 spo(V, β)-submodules generated by the highest weight

vectors

We apply Corollary 3.4.7 to ﬁnd the corresponding spo(V, β)-submodules generated by the highest weight vectors we found. We ﬁrst summarize the possible negative root vectors in spo(V, β) with the corresponding weights in Table 6.1.

Using Corollary 3.4.7, we apply the negative root vectors in Table 6.1 to v1, v2 and

v3 respectively to ﬁnd a for the submodules generated by v1, v2 and v3 respectively.

First of all, we know that v1 generates the trivial module, denoted as W∅. Then

let us consider the case which the highest weight vector is v2. We use Table 5.3 to 5. HIGHEST WEIGHT VECTORS IN V ⊗k 111

notation negative root vector Y E ∗ −2ε1 t1,t1 Y E − β(u , t )E ∗ ∗ δ1−ε1 u1,t1 1 1 t1,u1 ∗ Y E ∗ − β(u , t )E ∗ ∗ −δ1−ε1 u1,t1 1 1 t1,u1 Y E − β(u , t )E ∗ ∗ δ2−ε1 u2,t1 2 1 t1,u2 ∗ Y E ∗ − β(u , t )E ∗ −δ2−ε1 u2,t1 2 1 t1,u2 Y E − β(u , u )E ∗ ∗ −δ1+δ2 u2,u1 2 1 u1,u2 ∗ Y E ∗ − β(u , u )E ∗ −δ1−δ2 u1,u2 1 1 u2,u1 Table 5.2: Negative root vectors of spo(V, β) with m = 2 and n = 4.

record a basis of U(g)·v2= W , and the weights of these basis vectors.

weights actions vectors ε1 + δ1 v2 t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1 ∗ ∗ ∗ ε1 − δ1 Y−δ1+δ2 · (Y−δ1−δ2 · v2) t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1 ε1 + δ2 Y−δ1+δ2 · v2 t1 ⊗ u2 − β(u2, t1)u2 ⊗ t1 ∗ ∗ ∗ ε1 − δ2 Y−δ1−δ2 · v2 t1 ⊗ u2 − β(u2, t1)u2 ⊗ t1 ∗ ∗ ∗ −ε1 + δ1 Y−2ε1 · v2 t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1 ∗ ∗ ∗ ∗ ∗ ∗ −ε1 − δ1 Y−δ1−δ2 · (Y−δ1−ε1 · (Yδ2−ε1 · v2)) t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1 ∗ ∗ ∗ −ε1 + δ2 Y−δ1−ε1 · (Yδ2−ε1 · v2) t1 ⊗ u2 − β(u2, t1)u2 ⊗ t1 ∗ ∗ ∗ ∗ ∗ ∗ −ε1 − δ2 Y−δ1−ε1 · (Y−δ2−ε1 · v2) t1 ⊗ u2 − β(u2, t1)u2 ⊗ t1 δ1 + δ2 Yδ2−ε1 · v2 u1 ⊗ u2 − β(u2, u1)u2 ⊗ u1 ∗ ∗ ∗ δ1 − δ2 Y−δ2−ε1 · v2 u1 ⊗ u2 − β(u2, u1)u2 ⊗ u1 ∗ ∗ ∗ −δ1 + δ2 Y−δ1+δ2 · (Y−δ1−ε1 · v2) u1 ⊗ u2 − β(u2, u1)u2 ⊗ u1 ∗ ∗ ∗ ∗ ∗ ∗ −δ1 − δ2 Y−δ1−δ2 · (Y−δ1−ε1 · v2) u1 ⊗ u2 − β(u2, u1)u2 ⊗ u1 2δ1 Yδ1−ε1 · v2 u1 ⊗ u1 ∗ ∗ −2δ1 Y−δ1−δ2 · (Y−δ1+δ2 · (Y−δ1−ε1 · v2)) u1 ⊗ u1 2δ2 Y−δ1+δ2 · (Yδ2−ε1 · v2) u2 ⊗ u2 ∗ ∗ −2δ2 Y−δ1−δ2 · (Y−δ2−ε1 · v2) u2 ⊗ u2 ∗ ∗ ∗ ∗ 0 Y−δ1−δ2 · (Yδ2−ε1 · v2) u1 ⊗ u1 + u1 ⊗ u1 − u2 ⊗ u2 − u2 ⊗ u2 ∗ ∗ ∗ ∗ 0 Y−δ1−ε1 · v2 −t1 ⊗ t1 + t1 ⊗ t1 + u1 ⊗ u1 + u1 ⊗ u1

Table 5.3: A basis of spo(V, β)-submodule generated by the highest weight vector v2. The third column records all possible vectors obtained by acting negative root vectors on v2. The ﬁrst column records the corresponding weights of the vectors in the third column, and the second column records how the negative root vectors act on v2 in order to produce the corresponding vectors in the third column.

It is routine to verify that the vectors we found in Table 5.3 are linearly independent. We have also veriﬁed that for each w ∈ W , there exists X ∈ U(g) such that Xw is the highest weight vector, whence W is irreducible. Moreover, by counting the number of vectors in Table 5.4, we have dim(W ) = 17.

We then draw Figure 5.3, which represents a three-dimensional space with ε1-axis,

δ1-axis and δ2-axis so that we can see the symmetry of the weights. In particular, 5. HIGHEST WEIGHT VECTORS IN V ⊗k 112 we use two circles around the origin (0, 0, 0) to indicate that there are 2 linearly independent zero weight vectors.

ε1

ε1 − δ1

ε1 − δ2

ε1 − δ2 ε1 + δ1 −2δ1 −δ1 − δ2 −2δ2 0 −δ1 + δ2 δ1 − δ2

2δ2 δ1 + δ2 δ2 2δ1 δ1 −ε1 − δ1 −ε1 − δ2

−ε1 + δ1 −ε1 + δ2

Figure 5.3: Weights in the submodule generated by v2, where diﬀerent colours represent diﬀerent hyperplanes.

Finally let us consider the case of v3. Again Table 5.4 records a basis for

U(g)v3= W , the weights of these basis vectors, and the actions of negative root vectors on v3 in order to produce the resulting basis vectors. Figure 5.4 shows the weights of the basis vectors of the spo(V, β)-module W . 5. HIGHEST WEIGHT VECTORS IN V ⊗k 113

weights actions vectors 2ε1 v3 t1 ⊗ t1 ∗ ∗ −2ε1 Y−2ε1 · Y−2ε1 · v3 t1 ⊗ t1 ε1 + δ1 Yδ1−ε1 · v3 t1 ⊗ u1 + β(u1, t1)u1 ⊗ t1 ∗ ∗ ∗ ε1 − δ1 Y−δ1−ε1 · v3 t1 ⊗ u1 + β(u1, t1)u1 ⊗ t1 ε1 + δ2 Yδ2−ε1 · v3 t1 ⊗ u2 + β(u2, t1)u2 ⊗ t1 ∗ ∗ ∗ ε1 − δ2 Y−δ2−ε1 · v3 t1 ⊗ u2 + β(u2, t1)u2 ⊗ t1 ∗ ∗ ∗ −ε1 + δ1 Yδ1−ε1 · (Y−2ε1 · v3) t1 ⊗ u1 + β(u1, t1)u1 ⊗ t1 ∗ ∗ ∗ ∗ ∗ ∗ −ε1 − δ1 Y−δ1−ε1 · (Y−2ε1 · v3) t1 ⊗ u1 + β(u1, t1)u1 ⊗ t1 ∗ ∗ ∗ −ε1 + δ2 Yδ2−ε1 · (Y−2ε1 · v3) t1 ⊗ u2 + β(u2, t1)u2 ⊗ t1 ∗ ∗ ∗ ∗ ∗ ∗ −ε1 − δ2 Y−δ2−ε1 · (Y−2ε1 · v3) t1 ⊗ u2 + β(u2, t1)u2 ⊗ t1 δ1 + δ2 Yδ2−ε1 · (Yδ1−ε1 · v3) u1 ⊗ u2 + β(u2, u1)u2 ⊗ u1 ∗ ∗ ∗ δ1 − δ2 Y−δ2−ε1 · (Y−δ1−ε1 · v3) u1 ⊗ u2 + β(u2, u1)u2 ⊗ u1 ∗ ∗ ∗ −δ1 + δ2 Y−δ1+δ2 · (Y−δ1−ε1 · (Yδ1−ε1 · v3)) u1 ⊗ u2 + β(u2, u1)u2 ⊗ u1 ∗ ∗ ∗ ∗ ∗ ∗ −δ1 − δ2 Y−δ1−δ2 · (Y−δ1−ε1 · (Yδ1−ε1 · v3) u1 ⊗ u2 + β(u2, u1)u2 ⊗ u1 ∗ ∗ 0 Y−2ε1 · v3 t1 ⊗ t1 + t1 ⊗ t1 ∗ ∗ ∗ ∗ 0 Y−δ1−ε1 · (Yδ1−ε1 · v3) u1 ⊗ u1 − u1 ⊗ u1 − t1 ⊗ t1 − t1 ⊗ t1 ∗ ∗ ∗ ∗ 0 Y−δ2−ε1 · (Yδ2−ε1 · v3) u2 ⊗ u2 − u2 ⊗ u2 − t1 ⊗ t1 − t1 ⊗ t1

Table 5.4: the spo(V, β)-submodule generated by the highest weight vector v3

5.5.4 Summary: a decomposition of V ⊗V as spo(V, β)-modules

when n = 2, m = 4 and k = 2

From the calculations, we see that W∅ is the trivial modules. Moreoever, W is spanned by the vectors which are linear combination of vectors of the form

x ⊗ y − β(y, x)y ⊗ x

for all x, y ∈ B, and W is spanned by the vectors of the form

x ⊗ y + β(y, x)y ⊗ x

for all x, y ∈ B. Thus W∅, W and W have pairwise trivial intersection. The

dimension of W∅, W and W are 1, 18 and 17 respectively. Therefore we have V ⊗ V decomposes into a direct sum of irreducible spo(V, β)-modules

∼ V ⊗ V = W∅ ⊕ W ⊕ W , 5. HIGHEST WEIGHT VECTORS IN V ⊗k 114

ε1

2ε1

ε1 − δ1

ε1 − δ2

ε1 − δ2 ε1 + δ1

−δ1 − δ2 0 −δ1 + δ2 δ1 − δ2

δ1 + δ2 δ2

δ1 −ε1 − δ1 −ε1 − δ2

−ε1 + δ1 −ε1 + δ2

−2ε1

Figure 5.4: Weights of the basis vectors of the submodule generated by v3.

We will extend the discussion in this section further to a more general case when |n − m| = k = 2. We will analyze this decomposition further with respect to the Brauer algebra action in Section 6.1. Chapter 6

Formulas for the characters of spo(V, β)-modules in some borderline cases

In this chapter, we first state the spo(V, β) × Bk decomposition provided in [BSR98]. Then we examine reducibility properties of some borderline cases thoroughly. We then compute the characters of the irreducible summands. Besides the results stated in Section 6.1, Definition 6.5.3 and Definition 6.5.5, the results in this chapter are our own.

6.1 Schur-Weyl duality-like decomposition of V ⊗k

By [AB95, Section 13, Theorem 18], a ﬁnite dimensional algebra A over C is semisimple if it is isomorphic to a direct sum of matrix algebras over C. In [BSR98, Section 4],

Benkart et. al. stated that the Brauer algebra Bk(n − m) is semisimple if |n − m| > k.

115 6. CHARACTERS OF SOME spo(V, β)-MODULES 116

When the Brauer algebra Bk(n − m) is semisimple, by the work of [Wen88], it can be decomposed into a direct sum of matrix algebras such that the simple summands are indexed by partitions. In particular, there exists an isomorphism of algebras

M Θ: Bk(n − m) → Mdλ (F), (6.1.1) λ∈Pk where

Pk := {λ ` k − 2h | h = 0, 1,..., bk/2c}, (6.1.2)

and Mdλ (F) is the set of dλ × dλ matrix with entries in F for some nonnegative integers

dλ depending on λ ∈ Pk.

th For each λ ∈ Pk, and 1 ≤ i, j ≤ dλ, let Eλ,i,j be the matrix in the λ block of L M ( ) with 1 at the (i, j) entry and 0 elsewhere. We denote e = Θ−1(E ). λ∈Pk dλ F λ,i,j λ,i,j

Then eλ,i,j is an element in Bk(n − m). We construct a spo(V, β)-module by the following lemma.

Lemma 6.1.1 ([BSR98, Equation (4.1)]). Let V ⊗k be the tensor product of standard

0 spo(V, β)-modules. Let λ ∈ Pk, and let λ be the transpose of λ. Then

Y (λ0) ⊗k U := V eλ,i,j is an spo(V, β)-module.

Proof. Take x ∈ spo(V, β). Since the action of the Brauer algebra Bk(n−m) commutes with the action of spo(V, β) on V ⊗k, we have

Y (λ0) ⊗k ⊗k ⊗k Y (λ0) xU = x(V eλ,i,j) = (xV )eλ,i,j ⊆ V eλ,i,j = U . 6. CHARACTERS OF SOME spo(V, β)-MODULES 117

Therefore U Y (λ0) is an spo(V, β)-module.

Note that U Y (λ0) can be the trivial vector space. We will give an example such that U Y (λ0) = {0} in Example 6.4.1.

Theorem 6.1.2 ([BSR98, Proposition 4.2]). Let |n − m| > k. Then we have

(i) The spo(V, β)-module U Y (λ0) is independent of the choice of i, j.

(ii) As an spo(V, β) × Bk(n − m) module, we have

⊗k ∼ M Y (λ0) V = U ⊗ Bλ,

λ∈Pk

where Bλ is the irreducible Bk(n − m)-module labelled by the partition λ ∈ Pk.

However, Benkart et. al. did not claim or show anything about the irreducibility of U Y (λ0). In fact in [BSR98, Section 0.2(d)], they mention that U Y (λ0) is in general not necessarily irreducible. Moreover, the relation between U Y (λ0) and the highest weight vectors produced in Theorem 5.4.1 is not clear from [BSR98]. They did not indicate whether or not

0 the highest weight vectors wT,p,qCp,qyT , with T of shape λ lie in the appropriate submodules U Y (λ0). In the next section, we extend the discussion in Section 5.5 to a more general case where |n − m| = k = 2 for all possible n, m > 1. Then we verify that in this special case, Theorem 6.1.2 still holds, and relate the highest weight modules from Theorem 5.4.1 to these U Y (λ)’s. 6. CHARACTERS OF SOME spo(V, β)-MODULES 118

6.2 The Brauer algebra B2(η)

In this section, we ﬁrst show that B2(η) is semisimple whenever η 6= 0 by showing

that B2(η) is abelian and there are exactly three irreducible 1-dimensional modules.

We then explicitly show that B2(0) is not semisimple by giving an indecomposable

but not irreducible module. We give a matrix realization of B2(η) for all η.

Recall from Deﬁnition 4.1.5 that for all η ∈ F, B2(η) has dimension 3. Let ι be

the identity element in B2(η). Then we have

B2(η) = span{ι, e1, s1}.

Also recall from Proposition 4.1.6, the relations determining B2(η) are given by

e1s1 = s1e1 = e1, s1s1 = ι, e1e1 = ηe1.

We deduce that B2(η) is an abelian algebra. Therefore B2(η) is semisimple if and only if it has exactly three (irreducible) 1-dimensional modules.

Now suppose that η 6= 0. We ﬁnd all three 1-dimensional B2(η)-modules. Let

(φ, V ) be an irreducible B2(η)-module. Then since B2(η) is abelian, we can check

that both the image and kernel of φ(e1) are invariant under B2(η). Thus, since φ is

irreducible, we deduce that either φ(e1) is surjective or φ(e1) is the zero map.

(i) If φ(e1) is surjective, then every element of V can be written as ve1 for some

v ∈ V . Thus, the relation e1s1 = e1 implies that s1 acts by 1. The relation

2 e1 = ηe1 implies that e1 acts by η. Moreover, since ι is the identity diagram, ι 6. CHARACTERS OF SOME spo(V, β)-MODULES 119

acts by 1. Thus as an irreducible B2(η)-module, we have

vι = v, vs1 = v, ve1 = ηe1. (6.2.1)

(ii) Now if φ(e1) is the zero map, the only case we need to consider is how s1 acts.

2 By the relation s1 = ι, we deduce that s1 acts by either 1 or −1. Thus we have

the following irreducible B2(η)-modules.

vι = v, vs1 = v, ve1 = 0. (6.2.2)

vι = v, vs1 = −v, ve1 = 0. (6.2.3)

Notice that the modules in (6.2.2) and (6.2.3) are the same as the trivial module

(sometimes denoted as S ) and the sign module (sometimes denoted as S ) of S2 respectively. We denote the modules we deﬁned in (6.2.1), (6.2.2) and (6.2.3) as

B∅,B and B (6.2.4)

respectively.

Since B2(η) is semisimple for all η 6= 0, it can be decomposed into a direct sum of these three 1-dimensional subalgebras. Let M be the set of 3 × 3 diagonal matrices. 6. CHARACTERS OF SOME spo(V, β)-MODULES 120

Using the modules we found above, the map Θ : B2(η) → M such that

      1 η 1             Θ(ι) =  1  , Θ(e1) =  0  , Θ(s1) =  1              1 0 −1 is an algebra isomorphism.

1 In particular, using this matrix realization, we determine that e∅,1,1 = η e1, 1 1 1 e(2),1,1 = − η e1 + 2 (s1 + id) and e(1,1),1,1 = 2 (id − s1).

Now we consider the Brauer algebra B2(0) which is a 3-dimensional but not semisimple algebra. In this case, (6.2.3) still gives an irreducible 1-dimensional module, B . However, (6.2.1) and (6.2.2) are the same. In fact, we have a 2-dimensional

indecomposable B2(0)-module, which we will denote B∅0 with basis {v, w} such that

vι = v, vs1 = v, ve1 = w.

wι = w, ws1 = w, we1 = 0. (6.2.5)

Note that B∅0 contains B∅ = B as a submodule. The corresponding isomorphism

Θ: B2(0) → M is given by

      1 0 1 1             Θ(ι) =  1  , Θ(e1) =  0  , Θ(s1) =  1  .             1 0 −1 6. CHARACTERS OF SOME spo(V, β)-MODULES 121

6.3 The decomposition of V ⊗ V

We are now ready to extend the discussion in Section 5.5. Let g = spo(V, β) and

V = V(0) ⊕ V(1). Let dim(V(0)) = m and dim(V(1)) = n such that |n − m| = 2 = k. Note that this is a case which is not covered by the hypothesis of Theorem 6.1.2. We modify our calculation in Section 5.5 to ﬁnd the highest weight vectors in V ⊗ V and the submodules generated by them. We prove that we can decompose V ⊗ V into a

direct sum of three irreducible spo(V, β) × B2(n − m)-submodules when |n − m| = 2 and n, m > 1.

Theorem 6.3.1. Let r, s ≥ 1. Let m = 2r and n = 2s be such that |n − m| = 2. Then there are three linearly independent highest weight vectors in V ⊗ V .

Proof. First notice that if r = 1, then s = 2. We have already discussed this case in Section 5.5. Therefore we can restrict ourselves to consider the case when r > 1. Since k = 2, the set of all possible indices (p, q) of contraction maps is given by

P = {( , ), (1, 2)}.

By a similar argument as in Section 5.5, when (p, q) = (1, 2), the highest weight vector is given by

r s X ∗ ∗ X ∗ ∗ v1 = (ti ⊗ ti − ti ⊗ ti ) + (uj ⊗ uj + uj ⊗ uj) = Ω. (6.3.1) i=1 j=1

Let us now consider (p, q) = ( , ). We have Cp,q = Ψid. Moreover, the set of 6. CHARACTERS OF SOME spo(V, β)-MODULES 122

possible T ’s in Γr,s({1, 2}) is

T = 1 with T(0) = 1 and T(1) = ∅, or (6.3.2) 2 2

T = 1 2 with T(0) = 1 2 and T(1) = ∅. (6.3.3)

Let T be deﬁned in (6.3.2). In this case, since 1 is in the ﬁrst row of T(0), and 2 is in

the second row of T(0), we have wT, , = t1 ⊗ t2. Therefore the highest weight vector is (t1 ⊗ t2)Ψid(Ψid + Ψs1 ) which is

t1 ⊗ t2 − β(t2, t1)t2 ⊗ t1.

Similarly, for T deﬁned in (6.3.3), the highest weight vector is t1 ⊗ t1. In summary, if r = 1, we have three linearly independent highest weight vectors as we discussed in Section 5.5. If r > 1, the linearly independent highest weight vectors are given by

v1 = Ω, (v1)

v2 = t1 ⊗ t2 − β(t2, t1)t2 ⊗ t1, (v2)

v3 = t1 ⊗ t1. (v3)

Thus in either case, we have three independent highest weight vectors in total.

Similarly as in Section 5.5, we generate bases for the corresponding modules via

∅ an application of the PBW theorem (Corollary 3.4.7). Let us deﬁne W = U(g)v1,

W = U(g)v2 and W = U(g)v3. We ﬁrst summarize the list of all of the negative root vectors in Table 6.1. Then 6. CHARACTERS OF SOME spo(V, β)-MODULES 123 we use extra tables to record the actions of the negative root vectors on each of the highest weight vectors, and the weights of the resulting vectors.

notation negative root vector conditions Y−2ε E ∗ 1 ≤ i ≤ r i ti ,ti Y−ε +ε Et ,t − β(tj , ti)E ∗ ∗ 1 ≤ i < j ≤ r i j j i ti ,tj ∗ Y−ε −ε E ∗ + β(t , ti)E ∗ 1 ≤ i < j ≤ r i j ti ,tj j tj ,ti Y Eu ,t + β(uj , ti)E ∗ ∗ 1 ≤ i ≤ r, 1 ≤ j ≤ s −εi+δj j i ti ,uj ∗ Y E ∗ − β(u , ti)E ∗ 1 ≤ i ≤ r, 1 ≤ j ≤ s −εi−δj uj ,ti j ti ,uj Y−δ +δ Eu ,u + β(u`, uk)E ∗ ∗ 1 ≤ k < ` ≤ s k ` ` k uk,u` ∗ Y−δ −δ E ∗ + β(u , u`)E ∗ 1 ≤ k < ` ≤ s k ` uk,u` k u` ,uk Table 6.1: The negative root vectors when m = 2r > 2 and n = 2s ≥ 2. Notice that if s = 1, then we do not have the last two rows of this table.

∅ First, since Ω is spo(V, β)-invariant, v1 generates the trivial module, W , and we have

dim(W ∅) = 1. (6.3.4)

Let us now consider the case when the highest weight vector is v2. We group diﬀerent weights vectors by diﬀerent tables (Table 6.2 to 6.6).

weights actions vectors conditions ε1 + ε2 v2 t1 ⊗ t2 − β(t2, t1)t2 ⊗ t1

εi + εj Y−ε1+εi · Y−ε2+εj · v2 tj ⊗ ti − β(ti, tj )ti ⊗ tj 1 < j ≤ r ε1 + εj Y−ε2+εj · v2 tj ⊗ t1 − β(t1, tj )t1 ⊗ tj 1 < j ≤ r ∗ ∗ ∗ ε1 − εj Y−ε2−εj · v2 tj ⊗ t1 − β(t1, tj )t1 ⊗ tj 1 < j ≤ r ∗ ∗ ∗ εi − εj Y−ε1+εi · Y−ε2−εj · v2 tj ⊗ ti − β(ti, tj )ti ⊗ tj 1 < i < j ≤ r ∗ ∗ ∗ −ε1 + εj Y−2ε1 · Y−ε2+εj · v2 tj ⊗ t1 − β(t1, tj )t1 ⊗ tj 1 < j ≤ r ∗ ∗ ∗ −εi + εj Y−ε1+εi · Y−2ε1 · Y−ε2+εj · v2 tj ⊗ ti − β(ti , tj )ti ⊗ tj 1 ≤ i < j ≤ r ∗ ∗ ∗ ∗ ∗ ∗ −ε1 − εj Y−2ε1 · Y−ε2−εj · v2 tj ⊗ t1 − β(t1, tj )t1 ⊗ tj 1 < j ≤ r ∗ ∗ ∗ ∗ ∗ ∗ −εi − εj Y−ε1−εi · Y−ε2−εj · v2 tj ⊗ ti − β(ti , tj )ti ⊗ tj 1 < i < j ≤ r

Table 6.2: The vectors of weights ±εi ± εj for all 1 ≤ i < j ≤ r in U(g)v2.

weights actions vectors conditions

2δ` Y−ε2+δl · Y−ε1+δl · v2 u` ⊗ u` 1 ≤ ` ≤ s ∗ ∗ −2δ` Y−ε2−δl · Y−ε1−δl · v2 u` ⊗ u` 1 ≤ ` ≤ s

Table 6.3: The vectors of weights ±2δ` for all 1 ≤ ` ≤ r in U(g)v2. 6. CHARACTERS OF SOME spo(V, β)-MODULES 124

weights actions vectors conditions

ε1 + δ` Y−ε2+δl · v2 t1 ⊗ u` − β(u`, t1)u` ⊗ t1 1 ≤ i ≤ r, 1 ≤ ` ≤ s εi + δ` Y−ε1+εi · Y−ε2+δl · v2 ti ⊗ u` − β(u`, ti)u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ ε1 − δ` Y−ε2−δl · v2 t1 ⊗ u` − β(u` , t1)u` ⊗ t1 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ εi + δ` Y−ε1+εi · Y−ε2−δl · v2 ti ⊗ u` − β(u` , ti)u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ −εi + δ` Y−εi−δl · Y−ε2+δl · Y−ε1+δl · v2 ti ⊗ u` − β(u`, ti )u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ ∗ ∗ ∗ −εi − δ` Y−εi+δl · Y−ε2−δl · Y−ε1−δl · v2 ti ⊗ u` − β(u` , ti )u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s

Table 6.4: The vectors of weights ±εi ± δ` for all 1 ≤ i ≤ r, 1 ≤ ` ≤ s in U(g)v2.

weights actions vectors conditions

δk + δ` Y−εi+δk · Y−ε1+εi · Y−ε2+δl · v2 uk ⊗ u` − β(u`, uk)u` ⊗ uk 1 ≤ k < ` ≤ s ∗ ∗ ∗ ∗ ∗ ∗ −δk − δ` Y−ε2−δk · Y−ε1−δl · v2 uk ⊗ u` − β(u` , uk)u` ⊗ uk 1 ≤ k < ` ≤ s ∗ ∗ ∗ −δk + δ` Y−εi−δk · Y−ε1+εi · Y−ε2+δl · v2 uk ⊗ u` − β(u`, uk)u` ⊗ uk 1 ≤ k 6= ` ≤ s ∗ ∗ ∗ δk − δ` Y−ε1+δk · Y−ε2−δl · v2 uk ⊗ u` − β(u` , uk)u` ⊗ uk 1 ≤ k 6= ` ≤ s

Table 6.5: The vectors of weights ±δk ± δ` for all 1 ≤ k < ` ≤ r in U(g)v2.

weights actions vectors conditions 1 ≤ i ≤ r, ∗ ∗ ∗ ∗ 0 Y−εi−δl · Y−ε1+εi · Y−ε2+δl · v2 u` ⊗ u` + u` ⊗ u` − ti ⊗ ti + ti ⊗ ti 1 ≤ ` ≤ s

∗ ∗ ∗ ∗ 0 Y−δk−δl · Y−εi+δk · Y−ε1+εi · Y−ε2+δl · v2 u` ⊗ u` + u` ⊗ u` − uk ⊗ uk − uk ⊗ uk 1 ≤ k < ` ≤ s

Table 6.6: The vectors of weight 0 in U(g)v2.

If we apply negative root vectors to the vectors in Table 6.2 to 6.6, we will get

another vectors in these tables or 0. Thus we deduce that U(g)v2 is spanned by the vactors in Table 6.2 to 6.6. Moreover we deduce that each nonzero weight space has dimension 1. The vectors of weight 0 in Table 6.6 are linearly dependent, but we ﬁnd a linearly independent subset and conclude that the 0 weight space is spanned by the vectors

∗ ∗ ∗ ∗ u` ⊗ u` + u` ⊗ u` − u1 ⊗ u1 − u1 ⊗ u1, (6.3.5) for 1 ≤ ` ≤ s and the vectors

∗ ∗ ∗ ∗ − ti ⊗ ti + ti ⊗ ti + u1 ⊗ u1 + u1 ⊗ u1 (6.3.6)

for 1 ≤ i ≤ r. Therefore the 0 weight space has dimension r + s − 1. By summarizing the weights appearing in Tables 6.2 to 6.6, we ﬁnd that the nonzero weights are 6. CHARACTERS OF SOME spo(V, β)-MODULES 125

{±2εi | 1 ≤ i ≤ r} ∪ {±εi ± εj | 1 ≤ i < j ≤ r}

∪ {±δk ± δ` | 1 ≤ k, ` ≤ s} ∪ {±εi ± δ` | 1 ≤ i ≤ r, 1 ≤ ` ≤ s}. (6.3.7)

Thus by counting, we have

n − m dim(W ) = m2 + mn + n2 + − 1. (6.3.8) 2

Finally, let us consider the highest weight vector v3. In summary, we put the necessary information in Tables 6.7 to 6.11.

weights actions vectors conditions 2ε1 v1 t1 ⊗ t1 ∗ ∗ −2ε1 Y−2ε1 · Y−2ε1 · v1 t1 ⊗ t1 2εi Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ ti 1 < i ≤ r ∗ ∗ −2εi Y−ε1+εi · Y−ε1−εi · v1 ti ⊗ ti 1 < i ≤ r

Table 6.7: The vectors of weights ±2εi for all 1 ≤ i ≤ r in U(g)v3.

weights actions vectors conditions

εi + εj Y−εi+εj · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ tj + β(tj , ti)ti ⊗ tj 1 ≤ i < j ≤ r ∗ ∗ ∗ εi − εj Y−εi−εj · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ tj + β(tj , ti)tj ⊗ ti 1 ≤ i < j ≤ r ∗ ∗ ∗ −εi + εj Y−εi+εj · Y−2εi · Y−ε1+εi · v1 ti ⊗ tj + β(tj , ti )tj ⊗ ti 1 ≤ i < j ≤ r ∗ ∗ ∗ ∗ ∗ ∗ −εi − εj Y−εi−εj · Y−2εi · Y−ε1+εi · v1 ti ⊗ tj + β(tj , ti )tj ⊗ ti 1 ≤ i < j ≤ r

Table 6.8: The vectors of weights ±εi ± εj for all 1 ≤ i < j ≤ r in U(g)v3.

weights actions vectors conditions

εi + δ` Y−εi+δl · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ u` + β(u`, ti)u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ εi − δ` Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ u` + β(u` , ti)u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ −εi + δ` Y−εi+δl · Y−2εi · Y−ε1+εi · v1 ti ⊗ u` + β(u`, ti )u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s ∗ ∗ ∗ ∗ ∗ ∗ −εi − δ` Y−εi−δl · Y−2εi · Y−ε1+εi · v1 ti ⊗ u` + β(u` , ti )u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ s

Table 6.9: The vectors of weights ±εi ± δj for all 1 ≤ i ≤ r, 1 ≤ ` ≤ s in U(g)v3.

weights actions vectors conditions

δk + δ` Y−εi+δk · Y−εi+δl · Y−ε1+εi · Y−ε1+εi · v1 uk ⊗ u` + β(u`, uk)u` ⊗ uk 1 ≤ k < l ≤ s ∗ ∗ ∗ δk − δ` Y−εi+δk · Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 uk ⊗ u` + β(u` , uk)u` ⊗ uk 1 ≤ k < l ≤ s ∗ ∗ ∗ −δk + δ` Y−δk+δl · Y−εi+δl · Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 uk ⊗ u` + β(u`, uk)u` ⊗ uk 1 ≤ k < l ≤ s ∗ ∗ ∗ ∗ ∗ ∗ −δk − δ` Y−δk−δl · Y−εi+δl · Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 uk ⊗ u` + β(u` , uk)u` ⊗ uk 1 ≤ k < l ≤ s

Table 6.10: The vectors of weights ±δk ± δ` for all 1 ≤ i < j ≤ r in U(g)v3. 6. CHARACTERS OF SOME spo(V, β)-MODULES 126

weights actions vectors conditions ∗ ∗ 0 Y−2εi · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ ti + ti ⊗ ti 1 ≤ i ≤ r ∗ ∗ ∗ ∗ 0 Y−εi+δl · Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 u` ⊗ u` − u` ⊗ u` − ti ⊗ ti − ti ⊗ ti 1 ≤ ` ≤ s

Table 6.11: The vectors of weight 0 in U(g)v3.

From the above calculation, each non-zero weight space is spanned by one vector, and the zero weight space is spanned by

∗ ∗ ∗ ∗ {ti ⊗ ti + ti ⊗ ti , u` ⊗ u` − u` ⊗ u` | 1 ≤ i ≤ r, 1 ≤ l ≤ s} . (6.3.9)

Therefore the dimension of the zero weight space is r + s. The set of nonzero weights is

{±2δ` | 1 ≤ ` ≤ s} ∪ {±εi ± εj | 1 ≤ i < j ≤ r}

∪ {±δk ± δ` | 1 ≤ k, ` ≤ s} ∪ {±εi ± δ` | 1 ≤ i ≤ r, 1 ≤ ` ≤ s}. (6.3.10)

Thus we have n − m dim(W ) = m2 + mn + n2 − . (6.3.11) 2

Notice that W ∅, W and W have pairwise trivial intersection, and the sum of their dimensions (6.3.4)+(6.3.8)+(6.3.11) is (m + n)2 which is dim(V ⊗ V ). Therefore

V ⊗ V = W ∅ ⊕ W ⊕ W . Again as in Section 5.5, in each of the cases, we have checked that for each

w ∈ W Y (λ) with λ ∈ {∅, (1, 1), (2)}, there exists X ∈ U(g) such that Xw is the highest weight vector, whence these modules are irreducible. Since this gives a decomposition of V ⊗ V into a direct sum of inequivalent spo(V, β)-submodules, each

of the summands must be B2(n − m) invariant.

Next we determine the action of the Brauer algebra B2(n − m) on each of these 6. CHARACTERS OF SOME spo(V, β)-MODULES 127

summands by applying the generators to the basis vectors of W ∅,W and W respectively. First notice that the basis vectors of W are linear combinations of vectors of the form

ωx,y = x ⊗ y + β(y, x)y ⊗ x for all x, y ∈ B.

Since ι always acts by 1, it suﬃces to check how e1 and s1 act on the spanning vectors.

First consider ωx,ys1 = ωx,yΨs1 . We have

ωx,ys1 = −β(y ⊗ x)y ⊗ x − β(y, x)β(x, y)x ⊗ y = −ωx,y.

Then notice that each of the nonzero weight spaces has basis ωx,y for a particular

∗ choice of x, y ∈ B and x 6= y , and in this case we have ωx,ye1 = 0. Now for the basis vectors in the zero weight space, a basis is given in (6.3.9). In this case, we can also check that

∗ ∗ (t ⊗ t + t ⊗ t )e = F ∗ Ω + F ∗ Ω = 0, 1 1 1 1 1 t1,t1 t1,t1

and similarly

∗ ∗ (u1 ⊗ u1 − u1 ⊗ u1)e1 = Ω − Ω = 0.

Thus, in this case, the Brauer algebra B2(n − m) acts on W as a direct sum of modules B .

∅ Similarly, the action of B2(n − m) on W is by B∅, and on W is by B . Thus, we have the following theorem.

Theorem 6.3.2. Let r, s ≥ 1 and |n − m| = k = 2. As an spo(V, β) × B2(n − m)- 6. CHARACTERS OF SOME spo(V, β)-MODULES 128 module, we can decompose V ⊗ V as

M Y (λ0) V ⊗ V = W ⊗ BY (λ).

λ∈P2

Proof. We proved that V ⊗ V = W ∅ ⊕ W ⊕ W as spo(V, β)-modules. By the

discussion above, as an spo(V, β) × B2(n − m)-module, this is exactly

∅ V ⊗ V = (W ⊗ B∅) ⊕ (W ⊗ B ) ⊕ (W ⊗ B ).

This is consistent with the Schur-Weyl duality-like decomposition. Note also that

the parametrizing set of B2(n − m)-modules is P2 = {λ ` 2 − 2h | h = 0, 1} =

{∅, (1, 1), (2)}.

We have shown that when |m − n| = k = 2 and r, s ≥ 1, we can decompose

V ⊗ V into a direct sum of three irreducible spo(V, β) × B2(n − m)-submodules, and that the highest weight vectors are those given by Benkart et. al.’s formula (Theorem 5.4.1). This extends [BSR98, Proposition 4.2].

6.4 More examples on borderline cases

In this section, we give examples to analyze Theorem 5.4.1 more by considering some additional borderline cases. In particular, in Example 6.4.1, we show that if m = 2r = 2, n = 2s = 0 and k = 2, we can decompose V ⊗ V into a direct sum of

only two irreducible spo(V, β) × B2(n − m)-modules. In Example 6.4.2, under the assumption m = 2r = 0, n = 2s = 2 and k = 2, we show that Theorem 5.4.1 does not always give the full list of highest weight vectors. In Example 6.4.3, we show 6. CHARACTERS OF SOME spo(V, β)-MODULES 129

that when m = n = 2 and k = 2, we cannot decompose V ⊗ V into a direct sum of spo(V, β)-modules. We omit most of the calculations and only give the necessary information.

Example 6.4.1. Assume that m = 2r = 2, n = 2s = 0 and k = 2. In this case the

only (1, 0)-hook tableaux are T = ∅ and T = . Notice that, for T = , we have

T(0) = and T(1) = ∅.

Thus by Theorem 5.4.1, we have two highest weight vectors

∗ ∗ v1 = Ω = −t1 ⊗ t1 + t1 ⊗ t1 and v2 = t1 ⊗ t1.

∅ Then v1 generates the trivial module W , and the submodule W = U(g)v2 (which is irreducible) of V ⊗ V has a basis

∗ ∗ ∗ ∗ {t1 ⊗ t1, t1 ⊗ t1 + t1 ⊗ t1, t1 ⊗ t1}.

⊗2 Notice that in this case, W coincides with H(V ,C1,2) in Example 5.3.9.

We calculate the B2(−2) action on these modules to deduce that V ⊗V decomposes

into irreducible spo(V, β) × B2(−2) modules as

∅ V ⊗ V = (W ⊗ B∅) ⊕ (W ⊗ B ),

where B∅ and B are given by (6.2.1) and (6.2.3) respectively.

Notice that in Example 6.4.1, we only have two summands. By Theorem 6.1.1,

⊗2 the third summand would be U = V e(2),i,j, corresponding to the Brauer module 6. CHARACTERS OF SOME spo(V, β)-MODULES 130

B , but we can calculate that U is the trivial vector space.

Example 6.4.2. Assume that m = 2r = 0, n = 2s = 2 and k = 2. In this case the

only (0, 1)-hook tableaux are T = ∅ and T = . Notice that, for T = , we have

T(0) = ∅ and T(1) = .

Since in this case n = 2s and there is s = 1 row of T(1), by Theorem 5.4.1 we have three highest weight vectors

∗ ∗ v1 = Ω = u1 ⊗ u1 + u1 ⊗ u1,

v2 = u1 ⊗ u1, and

∗ ∗ v2 = u1 ⊗ u1.

Each of the above highest weight vectors generates a 1-dimensional submodule. Let

∅ W = U(g)v1, W1 = U(g)v2 and W2 = U(g)v2. Therefore, the highest weight vectors produced in Theorem 5.4.1 only generate a 3-dimensional submodule of V ⊗ V .

Now let us decompose V ⊗ V under the action of B2(2). We calculate that

∅ (V ⊗ V )e∅,1,1 = Span{Ω} = W , (6.4.1)

∗ ∗ (V ⊗ V )e(2),1,1 = Span{u1 ⊗ u1, u1 ⊗ u1} = W1 ⊕ W2 , (6.4.2)

∗ ∗ (V ⊗ V )e(1,1),1,1 = Span{u1 ⊗ u1 − u1 ⊗ u1}. (6.4.3)

∗ ∗ Thus let v3 = u1 ⊗ u1 − u1 ⊗ u1. In fact, spo(V, β) is just the orthogonal Lie colour algebra of rank 1, which is a colour-commutative algebra. Thus every weight vector,

in particular v3, is a highest weight vector. Let W = U(g)v3. 6. CHARACTERS OF SOME spo(V, β)-MODULES 131

We have shown that V ⊗ V decomposes into spo(V, β) × B2(n − m) modules as

∼ ∅ V ⊗ V = (W ⊗ B∅) ⊕ ((W1 ⊕ W2 ) ⊗ B ) ⊕ (W ⊗ B ).

Example 6.4.2 shows that the highest weight vectors found in Theorem 5.4.1 are not always all of the highest weight vectors. Moreover, in this example, U = W1 ⊕W2 is not irreducible.

Example 6.4.3. Consider m = 2r = 2, n = 2s = 2 and k = 2. This is a case not covered by the hypothesis of Theorem 6.1.2 (since B2(0) is not semisimple). Therefore the conclusion is not expected to hold.

In this case, the (1, 1)-hook tableaux are T1 = ∅, T2 = and T3 = .

For T1 = ∅, the highest weight vector is

∗ ∗ ∗ ∗ Ω = −t1 ⊗ t1 + t1 ⊗ t1 + u1 ⊗ u1 + u1 ⊗ u1, and it generates the trivial submodule W ∅.

For T2 = , the highest weight vector is t1 ⊗ t1. The submodule W =

U(g)(t1 ⊗ t1) is 8 dimensional, and its basis is recorded in Table 6.12.

basis vectors weights t1 ⊗ t1 2ε1 ∗ ∗ t1 ⊗ t1 −2ε1 u1 ⊗ t1 + β(t1, u1)t1 ⊗ u1 ε1 + δ1 ∗ ∗ ∗ u1 ⊗ t1 + β(t1, u1)t1 ⊗ u1 ε1 − δ1 ∗ ∗ ∗ u1 ⊗ t1 + β(t1, u1)t1 ⊗ u1 −ε1 + δ1 ∗ ∗ ∗ ∗ ∗ ∗ u1 ⊗ t1 + β(t1, u1)t1 ⊗ u1 −ε1 − δ1 ∗ ∗ u1 ⊗ u1 − u1 ⊗ u1 0 ∗ ∗ t1 ⊗ t1 + t1 ⊗ t1 0

Table 6.12: A basis for U(g)(t1 ⊗ t1). 6. CHARACTERS OF SOME spo(V, β)-MODULES 132

For T3 = , we have two diﬀerent highest weight vectors

v1 = t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1 (6.4.4)

∗ ∗ ∗ v1 = t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1. (6.4.5)

The submodule W1 generated by (6.4.4) is 4 dimensional, and it has basis

∗ ∗ ∗ {t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1, u1 ⊗ t1 − β(t1, u1)t1 ⊗ u1, u1 ⊗ u1, Ω}.

The submodule W2 generated by (6.4.5) is also 4 dimensional, and it has basis

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ {t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1, u1 ⊗ t1 − β(t1, u1)t1 ⊗ u1, u1 ⊗ u1, Ω}.

Notice that spo(V, β)-modules W1 and W2 are both 4-dimensional, but have diﬀerent weights, so they are not isomorphic.

∅ Notice also that since Ω appears in three diﬀerent submodules (W ,W1 and

W2 ), we do not have a decomposition of V ⊗ V into a direct sum of irreducible highest weight modules. The sum of the modules we found is 15-dimensional, and there is no 1-dimensional invariant complement.

Now let us consider the Brauer action. We calculate that B2(0) acts on W as

a direct sum of modules B . On the subspace spanned by W1 and W2 , the Brauer

1 ∗ ∗ algebra acts by B∅ = B . However if we set v = 2 (u1 ⊗ u1 + u1 ⊗ u1) and w = Ω,

then Span{v, w} is a B2(0)-submodule isomorphic to B∅0 . Therefore V ⊗ V is a sum

of B2(0)-submodules, but the summands are not necessarily spo(V, β)-invariant. 6. CHARACTERS OF SOME spo(V, β)-MODULES 133

6.5 The characters of W Y (λ) and U Y (λ)

In this section, let r > 0, s ≥ 1 and |n − m| = k = 2. We compute the characters of

W ∅, W and W . Then we compare our results with the character formula for U Y (λ) given in [BSR98, Theorem 4.24(e)].

∗ ∗ ∗ ∗ Let B = {t1, t1, . . . , tr, tr, u1, u1, . . . , us, us, (us+1)} be a homogeneous basis of V as in (2.4.1). We deﬁne a corresponding ordered set of commuting variables

Z = {z < z ∗ < ··· < z < z ∗ < z < z ∗ < ··· < z < z ∗ < (z )} (6.5.1) t1 t1 tr tr u1 u1 us us us+1

such that zbzb∗ = 1 for all b ∈ B.

Deﬁnition 6.5.1. Let W be an irreducible spo(V, β)-submodule of V ⊗k. The character of W , χ(W ), is a polynomial in Z with the property that the coeﬃcient of zi1 . . . zim+n b1 bm+n Pm+n Pm+n is the dimension of the j=1 ijwt(bj)-weight space, where p=1 ip = k and bj ∈ B.

Lemma 6.5.2. Let r > 0, s ≥ 0 and |n − m| = k = 2. The characters of W ∅, W and W from Theorem 6.3.2 are given explicitly by

(i) χ(W ∅) = 1.

X 2 X (ii) χ(W ) = zb + zbzb0 + (r + s). 0 b∈B(0) b,b ∈B zb

X 2 X (iii) χ(W ) = zb + zbzb0 + (r + s − 1). 0 b∈B(1) b,b ∈B zb

Let W = W . Then by (6.3.9), the nonzero weight spaces all have dimension 1, and the zero weight space has dimension r + s. Moreover, the nonzero weights in W are given by the following set

{±2εi | 1 ≤ i ≤ r} ∪ {±εi ± εj | 1 ≤ i < j ≤ r}

∪ {±δi ± δj | 1 ≤ i, j ≤ s} ∪ {±εi ± δj | 1 ≤ i ≤ r, 1 ≤ j ≤ s} (6.5.2)

Thus, the character χ(W ) is

X 2 X X X zb + zbzb0 + zbzb0 + zbzb0 + r + s (6.5.3) 0 0 b∈B(0) b,b ∈B(0) b,b ∈B(1) b∈B(0) 0 zb

X 2 X which is zb + zbzb0 + (r + s) as we claimed. 0 b∈B(0) b,b ∈B zb

We now cite [BSR98, Theorem 4.24(e)] to give a combinatorial description of the character of spo(V, β)-submodules by using the hook Schur functions. First we need to deﬁne bi-tableaux.

Deﬁnition 6.5.3. Let λ be a partition. Let Z be deﬁned in (6.5.1). A bi-tableau T of shape λ with entries from Z is a tableau T such that

(i) the part of T ﬁlled with zt’s and zt∗ ’s is a semi-standard Young tableau of shape µ ⊆ λ.

(ii) The zu’s and zu∗ ’s are weakly increasing along columns and strictly increasing along rows. 6. CHARACTERS OF SOME spo(V, β)-MODULES 135

The set of all bi-tableaux of shape λ with entries from Z is denoted as T (Z, λ).

Example 6.5.4. z z z ∗ is a bi-tableau of shape λ = (3, 2, 1). t1 t1 t1

zu1 zu2

zu1

Deﬁnition 6.5.5. Let T be a bi-tableau of shape λ. Let zT be the product of all entries of T , which is a monomial in Z. The hook Schur function shλ(Z) of shape λ is deﬁned as

X T shλ(Z) = z . (6.5.4) T ∈T (Z,λ)

By convention, if λ is a partition of 0, we deﬁne shλ(Z) = 1. By convention, we

deﬁne sh(−n)(Z) = 0 for all n > 0.

∗ ∗ Notice that in Deﬁnition 6.5.5, if B = {t1, t1, . . . , tr, tr} and we do not assume

0 zbzb∗ = 1 for any b ∈ B , then shλ(Z) is a Schur polynomial as deﬁned in Deﬁnition A.5.5.

Example 6.5.6. Let λ = (1). Then

X shλ(Z) = zb. b∈B

Theorem 6.5.7 ([BSR98] Theorem 4.24(e)). Let U Y (λ) be the spo(V, β)-submodule deﬁned in Lemma 6.1.1 The character of U Y (λ) is given by

1 χ(U Y (λ)) = det sh (Z) + sh (Z). (6.5.5) 2 (λi−i−j+2) (λi−i+j)

Now we compare the character we found in Lemma 6.5.2 with the character computed based on Theorem 6.5.7. 6. CHARACTERS OF SOME spo(V, β)-MODULES 136

Proposition 6.5.8. Let r > 0 and s ≥ 0. Let |n − m| = 2 = k. Let λ be an (r, s)-hook ( ) tableau in the set ∅, , . The character of U Y (λ) coincides with the character of W Y (λ).

Proof. Let us ﬁrst consider λ = (1, 1). We have Y (λ) = . Therefore we have

  1 sh(1)(Z) + sh(1)(Z) sh(0)(Z) + sh(2)(Z) χ(U Y (λ)) = det   2   sh(0)(Z) + sh(0)(Z) sh(−1)(Z) + sh(1)(Z)

which is (by the fact that sh0(Z) = 1 and sh−1(Z) = 0) equal to

  1 2sh(1)(Z) 1 + sh(2)(Z) det   2   2 sh(1)(Z) which is also equal to

2 sh(1)(Z) − sh(2)(Z) − 1. (6.5.6)

P 2 Now the ﬁrst term of (6.5.6) is ( b∈B zb) which can be rewritten as

2 X 2 X sh(1)(Z) = zb + 2zbzb0 . (6.5.7) b∈B b,b0∈B zb

The second term of (6.5.6) is

X 2 X sh(2)(Z) = zb + zbzb0 . (6.5.8) 0 b∈B(0) b,b ∈B zb

Therefore 6.5.6 becomes

X 2 X X 2 X zb + 2zbzb0 − ( zb + zbzb0 ) − 1 0 0 b∈B b,b ∈B b∈B(0) b,b ∈B zb

X 2 X zb + zbzb0 − 1. (6.5.9) 0 b∈B(1) b,b ∈B zb

∗ 0 For second term of (6.5.9), if b = b , we have zbzb0 = 1. In addition, we have r + s pairs of such b and b0. Therefore (6.5.9) becomes

X 2 X zb + zbzb0 + (r + s − 1) 0 b∈B(1) b,b ∈B zb

  1 sh(2)(Z) + sh(2)(Z) sh(1)(Z) + sh(−1)(Z) χ(U Y (λ)) = det   2   sh(−1)(Z) + sh(−1)(Z) sh(−2)(Z) + sh(0)(Z) which is   1 2sh(2)(Z) sh(1)(Z)   det   = sh(2)(Z). 2 0 1

By a similar argument, we have

X 2 X sh(2)(Z) = zb + zbzb0 + (r + s) = χ(W ) 0 b∈B(0) b,b ∈B zb

as we claimed.

Although [BSR98] did not provide any evidence that the submodules generated by highest weight vectors are the same as U Y (λ), we proved this is true in the case |n − m| = k = 2 and r > 0. Thus we have the following corollary.

Corollary 6.5.9. Let m = 2r and n = 2s with r > 0, s ≥ 0 and |n − m| = 2 = k. Then

(i) there are exactly three highest weight vectors (two if s = 0),

∅ (ii) V ⊗ V = (W ⊗ B∅) ⊕ (W ⊗ B ) ⊕ (W ⊗ B ) is a direct sum of irreducible

spo(V, β) × B2(2)-submodules (set W = {0} if s = 0), and

(iii) for all (r, s)-hook partition, λ ∈ {∅, (2), (1, 1)} we have χ(W Y (λ)) = χ(U Y (λ)), and in particular, we have W Y (λ) ∼= U Y (λ).

In fact we have a similar result for r = 0 and n = 2s = 2.

Corollary 6.5.10. Let m = 0, n = 2 and k = 2. Then

(i) every nonzero vector in V ⊗ V is a highest weight vector;

∅ (ii) there is a decomposition V ⊗ V = (W ⊗ B∅) ⊕ (W ⊗ B ) ⊕ (W ⊗ B ) into

a direct sum of spo(V, β) × B2(2)-submodules, where

W := W1 ⊕ W2 ,

and

(iii) for all (r, s)-hook partition, λ ∈ {∅, (2), (1, 1)} we have χ(W Y (λ)) = χ(U Y (λ)), and in particular, we have W Y (λ) ∼= U Y (λ). 6. CHARACTERS OF SOME spo(V, β)-MODULES 139

We have extended Theorem 6.1.2 to a borderline case when |n − m| = k = 2. Also, we have proved that in this case the spo(V, β)-submodule U Y (λ) coincides with the

Y (λ) spo(V, β)-submodule W generated by highest weight vector wT,p,qCp,qyT . However, the relation between U Y (λ) and W Y (λ) in general case remains unclear. Moreover, in the future, we hope to study another combinatorial description of the character of U Y (λ), which can be derived from the formula given in [BSR98, Theorem 5.1]: X χ(U Y (λ)) = zT , (6.5.10) T where T runs over all spo(m, n)-tableaux of shape λ, and the spo(m, n)-tableau is a restricted version of Deﬁnition 6.5.3. Appendix A

Schur polynomials and the RSK-Correspondence

Schur polynomials and Young tableaux play vital roles in the representation theory of the general linear groups and the symmetric groups, and they are the fundamental tools we used in our thesis. Since they are also interesting in their own right, we provide this appendix to give a closer look at these concepts. In this appendix, we first give a definition of Schur polynomials derived from a quotient of determinants (Definition A.1.3). We then define Young tableaux and describe the RSK-correspondence in the following sections. Using this, we give a combinatorial definition of Schur polynomials (see Definition A.5.5), and we prove that the two definitions of Schur polynomials are equivalent. Then we use this new definition to give a different proof of the fact that Schur polynomials are symmetric.

140 A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 141

A.1 Symmetric functions and Schur polynomials

In this section, we briefly discuss the definition of symmetric polynomials and in turn, we give an important example of symmetric polynomials, Schur polynomials. In representation theory, Schur polynomials describe the characters of finite-dimensional irreducible representations of the general linear groups. See for example [Sag01]. Further identities and combinatorial properties about Schur polynomials can be found in [Mac15].

Let C[x1, . . . xn] be the ring of complex n-variable polynomials. Then consider

the action of the symmetric group Sn on C[x1, . . . xn] given by permuting variables. We say a polynomial is symmetric if it is invariant under this action.

In order to deﬁne Schur polynomials, we need the Vandermonde matrix, Vn, with

j−1 entries Vi,j = xi for all 1 ≤ i, j ≤ n. The determinant of Vn is given by

Y ∆ = (xi − xj) 1≤i

Deﬁnition A.1.1. By a partition of n, we mean a weakly decreasing sequence of Pk non-negative integers λ = (λ1, . . . , λk) satisfying |λ| = i λi = n. We write λ ` n if |λ| = n. Moreover, we deﬁne the length of λ, `(λ) to be the index of the last nonzero entry of λ.

Example A.1.2. λ = (λ1, λ2, λ3, λ4, λ5) = (4, 3, 2, 0, 0) is a partition of 9 with `(λ) = 3.

Deﬁnition A.1.3. Let λ be a partition of length ≤ n. We deﬁne the Schur polynomial A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 142

in C[x1, . . . xn] by

λi+n−i det xj s = s (x , ··· , x ) = 1≤i,j≤n . λ λ 1 n ∆

Example A.1.4. Let λ = (2) and consider sλ = s(2)(x1, x2, x3). Then we have

  x4 x4 x4  1 2 3  . 2 2 2 sλ = det x x x  (x1 −x2)(x1 −x3)(x2 −x3) = x +x +x +x1x2 +x1x3 +x2x3.  1 2 3 1 2 3   1 1 1

Proposition A.1.5. The Schur polynomial sλ(x1, ··· , xn) is symmetric.

Y Proof. Let ∆ = (x − x ) and q(x , . . . , x ) = detxλi+n−i . Then j i 1 n j 1≤i,j≤n 1≤i

q(x1, . . . , xi, . . . , xj, . . . , xn) = −q(x1, . . . , xj, . . . , xi, . . . , xn)

which implies q(x1, ··· , xn)|xi=xj = 0. Therefore we have (xi − xj) divides q.

By the above argument the product of the factors (xi − xj) divides q. It follows that ∆ divides q, so that q/∆ is a polynomial. Moreover, since ∆ is also an alternating polynomial, q/∆ is a symmetric polynomial.

A.2 Young tableaux

In this section, we ﬁrst give the deﬁnition of Young tableaux. Then we provide an algorithm to insert a letter into a tableau (Algorithm 1) with its reverse algorithm A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 143

(Algorithm 2). With these algorithms, we deﬁne an operation on Young tableaux, called a product. Then we give the deﬁnition of words in some alphabet, and give an algorithm to transform a word into a Young tableau. At the end this section, we deduce that the tableau product corresponds to the concatenation of words. We synthesized this material from [Ful97].

Deﬁnition A.2.1. Given a partition λ = (λ1, ··· , λk), the Young diagram of shape

λ, denoted Y (λ), is a top-left-aligned diagram with at most k rows of boxes, and λi boxes in row number i.

Example A.2.2. The Young diagram for λ = (4, 3, 2, 0, 0) ` 9 is

An alphabet A is a well ordered set. From now on, unless stated otherwise, we

use the set of positive integers Z>0 as an alphabet.

Deﬁnition A.2.3. Let λ be a partition of n. A semi-standard Young tableau is a Young diagram of shape λ with entries from A in each box such that the entries are:

(i) weakly increasing along each row from left to right;

(ii) strictly increasing along each column from top to bottom.

We call a Young tableau a standard Young tableau if it is a semi-standard Young tableau with strictly increasing entries in each row.

We use ∅ to denote the diagram or tableau associated to the empty partition, λ = (0). A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 144

Example A.2.4. Let λ = (4, 3, 2). Then

(i) 1 1 2 2 is a semi-standard Young tableau of shape λ. 2 2 3 3 4

(ii) 1 2 3 4 is a standard Young tableau of shape λ. 5 6 7 9 10 Given a tableau T with n boxes, and a new entry x in A, in order to construct a new tableau from T and x, we introduce the Row-Insertion Algorithm (See Algorithm 1). The resulting new tableau is denoted T ← x and consists of n + 1 boxes, and the entries of T ← x are x together with all of the entries of T .

Algorithm 1 Row-insertion Algorithm 1: Given a tableau T and an entry x. Set Entry := x, Row Number := 1 2: while Entry 6= ∅, do 3: if Entry is greater than or equal to all the entries in the Row Number row of T , then put a new box at the end of the Row Number row with Entry in it. 4: Entry := ∅ 5: else Find the left-most entry, y, in the Row Number row that is strictly larger than Entry, and replace y by Entry. 6: Entry := y 7: end if 8: Row Number := Row Number + 1 9: end while

Example A.2.5. Let x = 1 and row insert x into the tableau

1 1 2 2 2 2 3 3 4

Using Algorithm 1, 1 will replace the ﬁrst 2 in the ﬁrst row, and 2 then replaces 3 in the second row, which replaces 4 in the third row. Then 4 will generate a new row at A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 145

the bottom of the diagram. We illustrate this process as the following sequence of tableau insertions.

1 1 2 2 ← 1 1 1 1 2 1 1 2 2 2 2 3 2 2 3 ← 2 2 2 2 3 4 3 4 , 3 4 , ← 3

1 1 2 2 2 2 2 3 3 4 .

Lemma A.2.6. If T is a semi-standard Young tableau and x is an element in A, then T ← x is again a semi-standard Young tableau.

Proof. T ← x is clearly weakly increasing along rows by the way we insert x. We then prove that T ← x is strictly increasing along columns. Now suppose that some element y is replaced by x. We have y > x. Then we denote the entry immediately below y (if it exists) by z. Then we have z > y > x. We claim that y will not replace entries which are in a position on the right side of z. This is because if y replaces some entry on the right of z, we should have z ≤ y, which is a contradiction. From the argument above, y replaces an entry below and to the left of the position of x, say u. Then u ≤ x < y. Thus the resulting tableau is still column strict. Hence T ← x is semi-standard by induction.

Deﬁnition A.2.7. Given two tableaux T and U, the product tableau T • U is obtained by row-inserting the entries of U into T from bottom to top and from left to right. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 146

Example A.2.8. Let T be a tableau. Then T • x z = T ← y ← x ← z. y For future reference, we note that Algorithm 1 is reversible if we know the resulting tableau and the position of the added box.

Theorem A.2.9. Suppose that U is a tableau such that U = T ← x for some tableau T and positive integer x. Then T and x are uniquely determined from U and the position of the added box to T in U. Furthermore, there is an algorithm for computing T from U and the position of the added box.

Algorithm 2 is the algorithm promised in Theorem A.2.9. Since it precisely reverses Algorithm 1, the theorem follows.

Algorithm 2 Backwards Row-Insertion Algorithm 1: Given the entry in the added box, Entry := y, and the row number of y, Row Number := i. 2: while Row Number > 0 do Find the right-most entry in row Row Number which is strictly less than Entry, say z, replace z by Entry. 3: Entry := z 4: Row Number := Row Number − 1 5: end while 6: Output: x := Entry.

Next we are going to give the deﬁnition of words, and then relate words to tableaux.

Deﬁnition A.2.10. A word is a ﬁnite sequence of elements of A. We usually express words using concatenation.

Example A.2.11. w = bac is a word in the alphabet {a, b, c, . . .}.

Deﬁnition A.2.12. Let A be an alphabet. Given a word w = x1x2 ··· xn in A, the A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 147

tableau of w is the tableau obtained by row inserting the letters of w from x1 to xn :

T (w) = x1 ← x2) ← x3) ← · · · ← xn).

0 Lemma A.2.13. Given two arbitrary words w and w = x1 . . . xn, then

0 0 T (ww ) = T (w) ← w := (T (w) ← x1) ← · · · ← xn.

This lemma follows directly from the deﬁnition.

Deﬁnition A.2.14. Given a semi-standard Young tableau T , we can construct the corresponding word of T , denoted as W(T ), by reading the entries of T from bottom to top and from left to right.

Example A.2.15. The word of the tableau T in Example A.2.4 is 342231122.

A tableau T can also be recovered from its word W(T ): simply break the word wherever one entry is strictly greater than the next, and the pieces are the rows of T (W(T )), read from bottom to top, left to right.

Example A.2.16. In alphabet Z>0, the word 4332221122 breaks into 4 | 33 | 222 | 1122 and the corresponding tableau is the one we obtained in Example A.2.5.

Deﬁnition A.2.17. We call a word w a tableau word if w = W(T ) for some tableau T .

Lemma A.2.18. Let T and U be two tableaux. The product of two tableaux T • U is equivalent to row-inserting W(U) into T from left to right.

Proof. The result follows from the deﬁnition of T • U and the deﬁnition of W(U). A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 148

A.3 Knuth Equivalence

In this section, an equivalence relation between words called Knuth equivalence will be introduced and in turn, we show that two words are Knuth equivalent if and only if they have the same tableaux. Moreover, we prove that the product of tableaux (Deﬁnition A.2.7) is associative. The materials are mainly from [Ful97], but the important proofs of Proposition A.3.6, A.3.9 and A.3.10 are our own.

Deﬁnition A.3.1. There are two rules of words in an ordered alphabet, named elementary Knuth transformations, which are:

y z x 7→ y x z if x < y ≤ z (K0)

x z y 7→ z x y if x ≤ y < z (K00)

Note that K0 and K00 are not inverse to each other.

Deﬁnition A.3.2. We say two words are Knuth equivalent if they can be transformed into each other by a sequence of elementary Knuth transformations and their inverses. k If words w and w0 are Knuth equivalent, then we write w ≡ w0.

Example A.3.3. 12124 is Knuth equivalent to 21124 by using K0 on the ﬁrst three letters of 12124.

Proposition A.3.4. Consider the tableau T ← x. We have

k W(T ← x) ≡ W(T )x.

Proof. Notice that an elementary Knuth transformation is always going to interchange the largest and smallest in three successive letters when the largest letter is in the A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 149

middle position. The row-insertion of an element x into a tableau T is equivalent to the following process:

We ﬁrst try to test x against the last entry of the ﬁrst row of T , say zq. If x ≥ zq,

then we put a new box at the end of the ﬁrst row with x in it. If x < zq and if the

entry zq−1 immediate before zq is also strictly larger than x, then we swap x and zq.

0 Notice that in this case, we have x < zq−1 ≤ zq which is the condition of K . We

repeat the process by testing the relation among x, zq−1 and zq−2 until we reach some

0 0 yp and x in the ﬁrst row such that yp is the entry immediate before x and the relation

0 x < yp no longer holds, which means we have yp ≤ x < x . Moreover, this process can be viewed as:

0 0 ypx z1 ··· zq−1zqx 7→ ypx z1 ··· zq−1xzq (x < zq−1 ≤ zq)

0 7→ ypx z1 ··· zq−2xzq−1zq (x < zq−2 ≤ zq−1)

···

0 7→ ypx z1xz2 ··· zq (x < z1 ≤ z2)

0 0 7→ ypx xz1 ··· zq (x < x ≤ z1).

0 0 Each of the above transformation is K . Now we have yp ≤ x < x which is the

00 0 0 condition of K . Then we can swap x with yp, and continue the process with x replaced by x. Thus by repeated application of K00 we have:

0 0 0 y1 ··· ypx xz1 7→ y1 ··· yp−1x ypxz1 (yp ≤ x < x )

0 0 7→ y1 ··· yp−2x yp−1ypxz1 (yp−1 ≤ yp < x )

···

0 0 7→ y1x y2y3 ··· ypxz1 (y2 ≤ y3 < x ) A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 150

0 0 7→ x y1y2y3 ··· ypxz1 (y1 ≤ y2 < x ).

Each of the transformations is K00, and this process indicates that x0 is now eﬀectively in the second row of the tableau. We can now continue by testing x0 with the entries in the second row of T , and ﬁnish the row-insertion process once a box has to be added. Therefore, the row-insertion process is about swapping consecutive three letters k by using the rules K0 and K00 accordingly. Thus we have W(T ← x) ≡ W(T )x.

k Corollary A.3.5. For all words w, WT (w) ≡ w.

Proof. Let w = w1 ··· wr, where w1, . . . , wr are in the alphabet A. Then we have

T (w) = ∅ ← w1 ← · · · ← wr. Therefore by using Lemma A.3.4, we have

k k k W(T (w)) ≡ W(∅ ← w1 ← · · · ← wr−1)wr ≡ · · · ≡ w1 ··· wr.

Proposition A.3.6. For any two words w1 and w2, we have if T (w1) = T (w2), then k w1 ≡ w2.

k Proof. By Corollary A.3.5, we know w ≡ W T (w) . Therefore, if T (w1) = T (w2), then k k w1 ≡ W T (w1) = W T (w2) ≡ w2.

k Hence we have w1 ≡ w2.

Corollary A.3.7. For any two words w and w0, we have

T (ww0) = T (w) • T (w0). A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 151

Proof. The result follows from Corollary A.3.5 and Proposition A.3.6.

Next we will prove that if two words are Knuth equivalent, then they have the same tableau. Let us ﬁrst consider two straightforward examples.

Lemma A.3.8. Let x, y, z be characters in an alphabet A. Then

(i) if w1 = xzy and w2 = zxy, with x ≤ y < z, we have

T (w1) = T (w2) = x y , and z

(ii) if w1 = yzx and w2 = yxz, with x < y ≤ z, we have

T (w1) = T (w2) = x z . y

Proof. The result follows by using the row-insertion algorithm on w1 and w2.

00 Notice that in the ﬁrst and second case, w1 and w2 are Knuth equivalent by K and K0 respectively.

k Proposition A.3.9. For two words w1 and w2, if w1 ≡ w2, then T (w1) = T (w2).

Proof. Without loss of generality, we assume that there is only a single Knuth trans- k formation, for example wzxyw0 ≡ wxzyw0. Then by Lemma A.2.13, it suﬃces to show T (wxzy) = T (wzxy). Therefore we assume that the Knuth transformation happens at the end of a word. Then we proceed by induction on the number of rows of T (w). k That is we show that for all xzy ≡ zxy with x ≤ y < z, we have T (wxzy) = T (wzxy), k and for all yzx ≡ yxz with x < y ≤ z, we have T (wyzx) = T (wyxz). The base case

(when T (w) = ∅) was proven in Lemma A.3.8. Suppose that if T (w) has n − 1 rows, A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 152

the two implications still hold. Now we prove that if T (w) has n rows, the equality still holds. Our strategy is the following: we insert x, y, z in the given order into the first row of T (w) and keep track of what is replaced. Then we show that the first row of T (wzxy) is the same as the first row of T (wxzy) (similar for T (wyzx) and T (wyxz)). The letters which have been displaced from the first row of T (w), are therefore inserted into the second row of T (w), or equivalently, are inserted into the tableau formed from removing the first row of T (w), denoted as T 0, which has n − 1 rows. Thus once we show that sequences of letters replaced are Knuth equivalent, the result follows by induction hypothesis. Let the first row of T (w) be

x1 x2 xd

for some d ∈ N. For the purpose of the following argument, we make the convention

that if r > d, then xr does not exist, and should be ignored. If z replaces such an xr, this means z is appended to the end of the row. Let us ﬁrst consider that xzy and zxy are Knuth equivalent by K00. That is, we have x ≤ y < z. Then in this case, we have the following two scenarios . Scenario 1: There exists some r with 2 ≤ r ≤ d + 1 such that

xr−1 ≤ x < z < xr ≤ xr+1. (A.3.1)

Then inserting zxy gives the following: z replaces xr, then x replaces z, and then y

replaces xr+1. Therefore the box of xr is now contains x, and the box of xr+1 contains

y. The replaced letters (in order) are xrzxr+1. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 153

On the other hand, inserting xzy gives the following: x replaces xr, then z replaces

xr+1, and then y replaces z. The box of xr contains x, and the box of xr+1 contains y.

The replaced letters (in order) are xrxr+1z. From the above discussion, the ﬁrst row of T (wzxy) is the same as the ﬁrst

row of T (wxzy). Moreover, by (A.3.1), we have z < xr ≤ xr+1 which implies

k 0 xrzxr+1 ≡ xrxr+1z. Thus, since we are now inserting xrzxr+1 and xrxr+1z into T separately, we conclude that T (wzxy) = T (wxzy) by induction hypothesis. Scenario 2: There exists r, s with 1 ≤ r < s ≤ d such that

xr−1 ≤ x < xr ≤ z < xs. (A.3.2)

Then inserting xzy gives the following: x replaces xr, then z replaces xs, then y will

0 0 replaces some y where y belongs to a box to right of the position of xr. Therefore

0 0 the replaced letters are xrxsy and xr+1 ≤ y ≤ z < xs.

Inserting zxy gives the following: z replaces xs, then x replaces xr, then y will

0 0 replace the same y as we discussed above. Therefore the replaced letters are xsxry In this case, it is straightforward that the ﬁrst row of T (wzxy) is the same as the

0 0 k 0 ﬁrst row of T (wxzy). By (A.3.2), xr ≤ y < xs implies that xrxsy ≡ xsxry . Thus the result follows by induction hypothesis. The case when xzy and zxy are Knuth equivalent by K0 can be proved similarly.

We proved this proposition when w1 and w2 can be changed by applying a sequence which consists of K0 or K00 only, and this implies that it is also true for a sequence of K0 and K00 and their inverses. Thus we conclude that for any Knuth equivalent words, they have the same tableaux.

Proposition A.3.10. The tableau product of Deﬁnition A.2.7 is associative. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 154

Proof. Let U and V be two tableaux. Then by Lemma A.2.18,

U • V = U ← W(V ) = ∅ ← W(U) ← W(V ) = T (W(U)W(V )).

Therefore by Corollary A.3.5 and Proposition A.3.6, we have

W(U • V ) ≡ WT (W(U)W(V )) ≡ W(U)W(V ).

Let W be another tableau. Then we have

k W(U • V ) • W ≡ W(U • V )W(W )

k ≡ W(U)W(V )W(W )

k ≡ W(U)W(V • W )

k ≡ WU • (V • W ).

Then by Proposition A.3.9, we have U • (V • W ) = (U • V ) • W.

A.4 The RSK-Correspondence

In this section, following [Ful97, Chapter 4], we construct a matrix with nonnegative integer entries from an ordered two-rowed array, and thereafter we obtain a one-to-one correspondence between a matrix and a pair of tableaux (P,Q), which is known as the Robinson-Schensted-Knuth correspondence, also referred to as the RSK-Corres- pondence. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 155

Deﬁnition A.4.1. An ordered two-rowed array is a 2 × n matrix

  u u . . . u  1 2 r A =   v1 v2 . . . vr such that

ui ≤ uj, ∀1 ≤ i < j ≤ r, and ui = uj implies vi ≤ vj, ∀1 ≤ i < j ≤ n.

Deﬁnition A.4.2. We associate an m × n matrix M to an ordered two-rowed array   i   such that the (i, j) entry of M is the number of times of   occurs in the array. j

  1 1 1 2 2 3 3 3 3   Example A.4.3. For the two-rowed array  , the corre- 1 2 2 1 2 1 1 1 2 sponding matrix is   1 2     M = 1 1 .     3 1

  u u . . . u  1 2 r Next given an arbitrary two-rowed array,   , we construct a v1 v2 . . . vr pair of tableaux (T,Q) of the same shape by using the row-insertion algorithm:

Start with T1 = v1 , and Q1 = u1 . In order to construct (T2,Q2) from (T1,Q1),

row-insert v2 in T1, getting T2. Then add a box to Q1 in the position of the new box

in T2, and place u2 in it. Repeat the process of inserting (uk, vk) for all 2 ≤ k ≤ r until there are no more unused entries in the two-rowed array. We call Q the recording tableau because we use it to record the process of A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 156

row-insertion.

Example A.4.4. Using the two-rowed array in Example A.4.3, we obtain a pair of tableaux as:

    (T,Q) =  1 1 1 1 1 2 , 1 1 1 2 3 3  . 2 2 2 2 3 3

For any matrix M with nonnegative integer entries, we have an associated two- rowed array A. We write the corresponding pair of tableaux of M as (T (M),Q(M)). Viewing the second row of A as a word w, we deduce that T (M) = T (w).

Corollary A.4.5. Let M be an n × m matrix with nonnegative integer entries. Write   M  1 M =   where M1 consists of the ﬁrst k rows of M for all 1 ≤ k ≤ n, and M2 M2 consists of the rest rows. Then T (M) = T (M1) • T (M2).

Proof. The result follows by the construction of a two-rowed array from M, M1, M2 and Corollary A.3.7, we have

Remark A.4.6. Let A be an two-rowed array with ﬁrst row A1 and second row A2.

Let M be the associated m × n matrix of A, with entries Mi,j for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Then

(i) the row sum of the ith row of M is the same as the number of occurrences of i

in A1, and thus is the same as the number of occurrences of i in Q.

(ii) The sum of the entries in column j of M is the same as the number of occurrences

of j in A2, which is the same as the number of occurrence of j in T . A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 157

Similar to the preceding algorithms, this transformation from an ordered two- rowed array to a pair of tableaux is also reversible. Given a pair of tableaux (P,Q),

in order to construct (Pk−1,Qk−1) from (Pk,Qk), we do the following:

(i) First look for the largest number in Qk and ﬁnd the right-most, top-most box with this largest number in it.

(ii) Then ﬁnd the corresponding box in Pk and apply the backwards row-insert

algorithm (Algorithm 2) to Pk with that box. The resulting tableau is Pk−1, and

th the element is removed from the top row of Pk is the k element of the second row of the two-rowed array.

(iii) Remove the box we found in Qk to form Qk−1, and the box removed contains the kth element of the ﬁrst row of the two-rowed array.

By repeating the process until there are no boxes left in the tableaux, we get a two-rowed array. By the preceding discussion in this section, we have the following theorem:

Theorem A.4.7 (the RSK-correspondence). With the above setting, there is a one- to-one correspondence between matrices with nonnegative integer entries and pairs of tableaux.

A.5 An application of the RSK-Correspondence

In this section, we give an application of the RSK-correspondence, which implies

that a certain n-variable polynomial Schλ(x1, x2, ··· , xn) (see Deﬁnition A.5.5) is symmetric. Then at the end of this section, we generalize the proof provided in [Pro89] A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 158

to show that for each λ this polynomial is equal to the Schur polynomial we deﬁned in Deﬁnition A.1.3.

For a tableau T with entries from alphabet {1, . . . , n}, let mi be the number of

occurrences of i in T . Let m(T ) = (m1, . . . , mn). We call m(T ) the type of T . Note

that m` can be 0 for some ` ∈ {1, . . . , n}.

Example A.5.1. Let T = 1 1 2 2 . Then the type of T is m(T ) = (2, 4, 2, 0, 1). 2 2 3 3 5

Proposition A.5.2. Given a tableau T0 of shape λ, then the number of tableaux T of shape λ having m(T ) = m(T0) is the same as the number of tableaux T of shape λ with m(T ) = (mσ(1), . . . , mσ(n)), where σ is a permutation of {1, . . . , n}.

Remark A.5.3. Fix any tableau T0 of shape λ. Then by Theorem A.4.7, for any tableaux Q of shape λ, there is a one-to-one correspondence between the pairs of

tableaux (T0,Q) and matrices M such that T (M) = T0 . From Remark A.4.6 it follows that Proposition A.5.2 is equivalent to showing that the two sets

E1 = {M : T (M) = T0 and M has row sums m1, m2, ··· , mn}

and

E2 = {M : T (M) = T0 and M has row sums mσ(1), mσ(2), ··· , mσ(n)} have the same cardinality.

Before we prove Proposition A.5.2, we need the following lemma. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 159

Lemma A.5.4. Let C be a matrix with only two rows with integer entries and corresponding pair of tableaux T (C),Q(C). Then there is a unique two-rowed matrix C˜ with corresponding pair of tableaux T (C˜),Q(C˜) such that C˜ has swapped row sums and T (C˜) = T (C).

Proof. By Remark A.4.6, since C has only two rows, the row sum of the ith row is the number of occurrences of i in Q(C) for i = 1, 2. Thus Q(C) only has 1 and 2 as entries. Then by Remark A.5.3, it suﬃces to prove that with a particular Q(C) and T (C) = T (C˜), there is only one corresponding possibility for Q(C˜). Now suppose that Q(C) is given by:

1 ··· 1 1 ··· 1 2 ··· 2

2 ··· 2 s t

r which means the row sum for C is r + s and r + t. Therefore let C˜ have row sum r + t and r + s. Thus Q(C˜) must be

1 ··· 1 1 ··· 1 2 ··· 2

2 ··· 2 t s

Since there are (r + t) 1’s and (r + s) 2’s in Q(C˜), this is the only way to get such a semi-standard Young tableau. Thus Q(C˜) is uniquely determined.

Proof of Proposition A.5.2. In order to prove Proposition A.5.2 holds for all permutations, it suﬃces to prove it when σ is the adjacent transposition of (k, k + 1). Given A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 160

A in E1, we can write A as a block matrix

  B     A = C     D where B consists of the ﬁrst k − 1 rows, C consists of next two rows and D consists of the rest. By Corollary A.4.5, we have

T (A) = T (B) • T (C) • T (D).

Then by Lemma A.5.4, if the matrix C has row sums mk and mk−1, then there is ˜ ˜ a unique corresponding C with row sums mk−1 and mk such that T (C) = T (C).   B   ˜   Therefore, for each A in E1, there is a unique corresponding matrix A = C˜in     D

E2. Thus the sets E1 and E2 are in bijection, which proves the proposition by Remark A.4.6.

Now we introduce a new polynomial deﬁned in terms of semi-standard Young tableaux and in turn, prove it is a symmetric polynomial by Proposition A.5.2.

Deﬁnition A.5.5. Let λ be a partition of any nonnegative integer. We deﬁne an

n-variable polynomial Schλ associated to λ by

X m(T ) X m1 mn Schλ(x1, x2, ··· , xn) = x = x1 ··· xn . T T A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 161 where the summation is over all semi-standard Young tableaux T of shape λ with entries from the alphabet {1, . . . , n}, and (m1, . . . , mn) = m(T ).

Example A.5.6. Let n = 3, λ = (2). Then the possible tableaux are

1 1 , 1 2 , 1 3 , 2 2 , 2 3 , 3 3 ,

and

2 2 2 Sch(2)(x1, x2, x3) = x1 + x2 + x3 + x1x2 + x1x3 + x2x3.

Remark A.5.7. By Deﬁnition A.5.5, the coeﬃcients of Schλ(x1, x2, ··· , xn) are always integers.

Proposition A.5.8. The polynomial Schλ deﬁned in Deﬁnition A.5.5 is symmetric.

m1 mn Proof. By Proposition A.5.2, the coeﬃcient of x1 ··· xn is the same as the coeﬃcient

mσ(1) mσ(n) of x1 ··· xn for all permutation σ of {1, . . . , n}. Thus the result follows.

Recall the Schur polynomial sλ(x1, . . . , xn) in Deﬁnition A.1.3. It is clear that the polynomials in Example A.1.4 and A.5.6 are identical. Thus, in this case, we have

s(2)(x1, x2, x3) = Sch(2)(x1, x2, x3).

Next we prove that for any arbitrary partition λ, and n ∈ Z≥0, we have

sλ(x1, . . . , xn) = Schλ(x1, ··· , xn).

We ﬁrst need the following lemma. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 162

Lemma A.5.9. We have

X |λ|−|µ| sλ(x1, . . . xn) = sµ(x1, . . . , xn−1)xn µ

P where |λ| = i λi and the sum is over all partitions µ such that

λ1 ≥ µ1 ≥ λ2 ≥ µ2 ≥ · · · ≥ µn−1 ≥ λn.

Proof. Using Deﬁnition A.1.3 of sλ(x1, . . . , xn), we see that

λ1+n−1 λ1+n−1 λ1+n−1 x1 x2 ··· xn−1 1

λ2+n−2 λ2+n−2 λ2+n−2 x1 x2 ··· xn−1 1

......

λn−1−1 λn−1−1 λn−1−1 x1 x2 ··· xn−1 1

λn λn λn x1 x2 ··· xn−1 1 sλ(x1, . . . , xn−1, 1) = . n−1 n−1 n−1 x1 x2 ··· xn−1 1

n−2 n−2 n−2 x1 x2 ··· xn−1 1

......

x1 x2 ··· xn−1 1

1 1 ··· 1 1

Then if we subtract the last column from each of the other columns and keep the last A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 163

column unchanged, the determinant does not change, and we have:

λ1+n−1 λ1+n−1 λ1+n−1 x1 − 1 x2 − 1 ··· xn−1 − 1 1

λ2+n−2 λ2+n−2 λ2+n−2 x1 − 1 x2 − 1 ··· xn−1 − 1 1

......

λn−1−1 λn−1−1 λn−1−1 x1 − 1 x2 − 1 ··· xn−1 − 1 1

λn λn λn x1 − 1 x2 − 1 ··· xn−1 − 1 1 sλ(x1, . . . , xn−1, 1) = . n−1 n−1 n−1 x1 − 1 x2 − 1 ··· xn−1 − 1 1

n−2 n−2 n−2 x1 − 1 x2 − 1 ··· xn−1 − 1 1

......

x1 − 1 x2 − 1 ··· xn−1 − 1 1

0 0 ··· 0 1

th For all 1 ≤ i ≤ n − 1, we factor out xi − 1 from the i column of each of the determinants in the numerator and denominator to get the following expression for the right hand side of the above. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 164

λ1+n−2 λ1+n−3 λ1+n−2 λ1+n−3 x1 + x1 + ··· + x1 + 1 ··· xn−1 + xn−1 + ··· + xn−1 + 1 1

λ2+n−3 λ2+n−4 λ2+n−3 λ2+n−4 x1 + x1 + ··· + x1 + 1 ··· xn−1 + xn−1 + ··· + xn−1 + 1 1

......

λn−1−2 λn−1−3 λn−1−2 λn−1−3 x1 + x1 + ··· + x1 + 1 ··· xn−1 + xn−1 + ··· + xn−1 + 1 1

λn−1 λn−2 λn−1 λn−2 x1 + x1 + ··· + x1 + 1 ··· xn−1 + xn−1 + ··· + xn−1 + 1 1 . n−2 n−3 n−2 n−3 x1 + x1 + ··· + x1 + 1 ··· xn−1 + xn−1 + ··· + x2 + 1 1

n−3 n−4 n−3 n−4 x1 + x1 + ··· + x1 + 1 ··· xn−1 + xn−1 + ··· + xn−1 + 1 1

......

1 ··· 1 1

0 ··· 0 1

Recursively from i = 1 to n − 1, subtracting the (i + 1)th row from the ith and keeping A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 165

the last row unchanged, the above expression becomes equal to:

λ1+n−2 λ2+n−2 λ1+n−2 λ2+n−2 x1 + ··· + x1 ··· xn−1 + ··· + xn−1 0

λ +n−3 λ +n−3 λ +n−3 λ +n−3 x 2 + ··· + x 3 ··· x 2 + ··· + x 3 0 1 1 n−1 n−1 ......

λi+n−(i+1) λi+1+n−(i+1) λi+n−(i+1) λi+1+n−(i+1) x1 + ··· + x1 ··· xn−1 + ··· + xn−1 0

......

xλn−1−2 + ··· + xλn ··· xλn−1−2 + ··· + xλn 0 1 1 n−1 n−1

λn−1 λn−1 x1 + ··· + x1 + 1 ··· xn−1 + ··· + xn−1 + 1 1 . n−2 n−2 x1 ··· xn−1 0

n−3 n−3 x1 ··· xn−1 0

......

1 1 ··· 0

0 0 ··· 1

Then we use cofactor expansion for the numerator and denominator along the last A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 166

column to get

λ1+n−2 λ2+n−2 λ1+n−2 λ2+n−2 x1 + ··· + x1 ··· xn−1 + ··· + xn−1

xλ2+n−3 + ··· + xλ3+n−3 ··· xλ2+n−3 + ··· + xλ3+n−3 1 1 n−1 n−1

......

λi+n−(i+1) λi+1+n−(i+1) λi+n−(i+1) λi+1+n−(i+1) x1 + ··· + x1 ··· xn−1 + ··· + xn−1

......

λn−1−2 λn λn−1−2 λn x1 + ··· + x1 ··· xn−1 + ··· + xn−1 . n−2 n−2 x1 ··· xn−1

xn−3 ··· xn−3 1 n−1 ......

0 0 x1 ··· xn−1

Notice that the (i, j) entry of the matrix in the numerator is of the form

λi+1 X µi+(n−1)−i xj , µi=λi A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 167 where 1 ≤ j ≤ n − 1. Thus preceding quotient is

λ2 λ2 X µ1+n−2 X µ1+n−2 x1 ··· xn−1

µ1=λ1 µ1=λ1 ......

λi+1 λi+1

X µi+(n−1)−i X µi+(n−1)−i x1 ··· xn−1

µi=λi µi=λi ......

λn λn X µ X µ x n−1 ··· x n−1 1 n−1 µn−1=λn−1 µn−1=λn−1 . n−2 n−2 x1 ··· xn−1

xn−3 ··· xn−3 1 n−1 ......

0 0 x1 ··· xn−1

Thus by using the fact that the determinant is linear in each row, we conclude P that the preceding quotient of determinants is µ sµ(x1, ··· , xn−1), for all partition

µ such that λ1 ≥ µ1 ≥ λ2 ≥ µ2 ≥ · · · ≥ µn−1 ≥ λn. P In order to homogenize the sum µ sµ(x1, . . . , xn−1) to recover sλ(x1, . . . , xn), we |λ|−|µ| should multiply each of the above terms by xn . Therefore we have

X |λ|−|µ| sλ(x1, ··· xn) = sµ(x1, ··· , xn−1)xn µ

where |λ| = λ1 + ··· + λn and the sum is over all partitions µ such that

λ1 ≥ µ1 ≥ λ2 ≥ µ2 ≥ · · · ≥ µn−1 ≥ λn. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 168

Proposition A.5.10. Deﬁnition A.1.3 is equivalent to Deﬁnition A.5.5. In other words,

sλ(x1, ··· xn) = Schλ(x1, ··· xn).

Proof. It suﬃces to prove that we can express sλ(x1, . . . , xn) as a sum of monomials, and each monomial corresponds to a semi-standard Young tableau of shape λ.

t1 tn Also notice that by Deﬁnition A.5.5, the power of xn in a monomial x1 ··· xn corresponds to the number of n’s in the corresponding semi-standard Young tableau. By removing the n’s, we get a new monomial associated to some partition µ. Therefore we will construct a collection of tableaux by repeatly using Lemma A.5.9. Start with a shape λ0 = λ and apply Lemma A.5.9 once. For each resulting λ1 = µ, we construct a tableau by placing an n in each box where the boxes are in λ0 but not in λ1. Then for each λ1 = µ we obtained, apply Lemma A.5.9 again. We therefore place n − 1 in each box which is outside of the resulting shape λ2 but inside λ1. Now we have constructed some tableaux with empty boxes and boxes with n and n − 1. Repeat this procedure n − 2 more times with n − 2’s, n − 3’s, ··· and 1s. After

n steps we are left with λn = ∅, so we can no longer apply Lemma A.5.9 again. We

ﬁnish the procedure by noticing that s∅(x1, . . . , xn) = 1.

Therefore, we have expressed sλ(x1, ··· xn, ) as a sum of monomials which are indexed by a set of tableaux. Moreover these tableaux we constructed are semi-standard. Because ﬁrst of all the entries obviously weakly increase across the rows and down the columns. Also

k+1 k k λi ≥ λi+1 and λn−k+1 = 0 imply that two n − k’s will not be placed in the same column at step k = 0, 1, ··· , n − 1. This implies that the entries are strictly increasing along columns. A. SCHUR POLYNOMIALS AND THE RSK-CORRESPONDENCE 169

Thus we have assigned a semi-standard Young tableau of shape λ to each sum-

mands appearing in sλ(x1, . . . , xn). Also on the other hand, by the way we construct

induction, we are able to assign a summand in sλ(x1, . . . , xn) to each semi-standard

Young tableaux of shape λ appearing in Schλ(x1, . . . , xn). Thus there is a bijection

between sλ(x1, . . . , xn) and Schλ(x1, . . . , xn), which completes the proof. Bibliography

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