representation theory of lie colour algebras and its ... · in this thesis, we study the...

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 partialfulfillment of the requirements for the degree of

Master of Science in Mathematics1

Department of Mathematics and StatisticsFaculty 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 UY (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

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

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

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

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

V ⊗k ∼=⊕λ

Sλ ⊗ Eλ.

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

Poincare-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

Bk(n −m) on V ⊗k which commutes with the left action of spo(V, β) on V ⊗k.

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 charF 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


(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.

(ii) Consequently we have 1 = β(a, 1G) = β(a, aa−1) = β(a, a)β(a, a−1). 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

β(a, b) = (−1)a·b 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


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

V =⊕a∈G

Va,

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

g =⊕a∈G

ga

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.


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.

(ii) Let G = Z2 = 0, 1, and β(a, b) = (−1)ab for all a, b ∈ G. Then the Lie colour

algebra g is a Lie superalgebra.


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

g(0) =⊕a∈G(0)

ga and g(1) =⊕a∈G(1)

ga.

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.


Definition 2.2.1. Let g be a Lie colour algebra. A g-module is a G-graded F-vector

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

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

(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.


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

B∗ = v1, . . . , vn the dual basis of V ∗ defined by vi(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

bases for each Va. Thus by the way we define vi, we conclude that B∗ is a homogeneous

basis of V ∗. Then we compute the colour of vi by the following lemma.

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

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

for any vj not in Vc implies that vi has colour c−1 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.


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

(V ⊗W )c =⊕a,b∈Gab=c

Va ⊗Wb.

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


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

x(f(m)⊗ g(n)

)since f and g preserve colours.

Secondly, recall from Definition 2.2.7, we have

(M ⊗N)c =⊕a,b∈Gab=c

Ma ⊗Nb.

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


Chapter 4.

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

RM,N : M ⊗N → N ⊗M

m⊗ n 7→ β(b, a)n⊗m.

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

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

RM,N

(x(m⊗ n)

)= RM,N(xm⊗ n+ β(b, a)m⊗ xn)

= 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)

= xRM,N(m⊗ n).

Therefore by Definition 2.2.6, RM,N is a g-module morphism.

Moreover, since β(a, b)β(b, a) = 1 for all a, b ∈ G, we have that R2M,N = 1M⊗N .

Thus, RM,N is an g-module isomorphism.


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.

Then for all k ∈ Z≥2, we define Ri = R(i,i+1) on V ⊗k by

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

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

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

Lemma 2.2.11. The maps Ri satisfy the same relations as the standard generators

of the symmetric group. Namely we have

(i) R2i = 1,

(ii) RiRj = RjRi if |i− j| ≥ 2,

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

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

a product of adjacent transpositions, where sij = (ij, ij + 1), then we can define Rπ as

Rπ = Ri1Ri2 · · · Rip . (2.2.3)

Lemma 2.2.11 implies that Rπ is independent of the choice of expression.


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

(i) the map prV : V1G → V ⊗ V ∗ such that 1 7→∑n

i=1 vi ⊗ vi, and

(ii) the evaluation map evV : V ∗⊗V → V1G such that vi⊗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

homogeneous elements x ∈ g, xprV (1) = x(∑n

i=1 vi ⊗ vi) = 0, that is to show

Ω =∑n

i=1 vi ⊗ vi is a g-invariant tensor.

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

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

x ∈ g be homogeneous with colour b ∈ G. We then write xvi and xvi explicitly. First

notice that

xvi =n∑k=1

cikvk, (2.2.4)

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

(xvi)(vj) = −β(a−1i , b)vi(xvj)

= −β(a−1i , b)

n∑`=1

cj`vi(v`)

= −β(a−1i , b)cji.


Thus we deduce that

xvi =n∑j=1

−β(c−1i , a)cjiv

j. (2.2.5)

Therefore we have

x

(n∑i=1

vi ⊗ vi)

=n∑i=1

xvi ⊗ vi +n∑i=1

β(ai, b)vi ⊗ xvi

=n∑i=1

n∑k=1

cikvk ⊗ vi +n∑i=1

n∑j=1

−β(ai, b)β(a−1i , b)cjivi ⊗ vj

=n∑i=1

n∑k=1

cikvk ⊗ vi −n∑i=1

n∑j=1

cjivi ⊗ vj = 0.

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:

V1G

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

Similarly the map evV⊗W can be obtained by

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

of V . Let v1, . . . , vn be the dual of B such that vi(vj) = δi,j for all 1 ≤ i, j ≤ n. Let


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

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

1 7→∑

1≤i1,...,ik≤n

vi1 ⊗ · · · ⊗ vik ⊗ vik ⊗ · · · ⊗ vi1 , and

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

evk(vi1 ⊗ · · · ⊗ vik ⊗ vjk ⊗ · · · ⊗ vj1) 7→ δik,jk · · · δi1,j1 =

k∏`=1

δi`,j` ,

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.


2.3 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

gl(V, β) =⊕a∈G

gl(V, β)a

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


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.


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 Evivjvk = 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

for all 1 ≤ i, j ≤ n, by (2.3.1) it has colour viv−1j . Therefore Evivj ∈ gl(V, β)viv−1

jand

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

〈·, ·〉 : V × V → F

is called a β-skew-symmetric bilinear form if it satisfies:

(i) 〈·, ·〉 is nondegenerate;

(ii) 〈Va, Vb〉 = 0 whenever b 6= a−1; and

(iii) 〈v, w〉+ β(b, a)〈w, v〉 = 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. Therefore for all v ∈ Va and w ∈ Wa−1 , 〈v, w〉 + β(a−1, a)〈w, v〉 = 0 implies

〈v, w〉 = ±〈w, v〉.


We explore this form further in Section 2.4.

Definition 2.3.8. The orthosymplectic Lie colour algebra

spo(V, β) =⊕a∈G

spo(V, β)a

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 | 〈xw, v〉+β(b, a)〈w, xv〉 = 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

〈[x, y]w, v〉 = −β(w, xy)〈w, [x, y]v〉, for all v ∈ V.

The method is to expand [x, y] = xy − β(y, x)yx first, and use the linearity of 〈·, ·〉.

Therefore

〈[x, y]w, v〉 = 〈xyw, v〉 − β(y, x)〈yxw, v〉. (2.3.3)


The first term of (2.3.3) can be written as

〈xyw, v〉 = −β(yw, x)〈yw, xv〉

= β(yw, x)β(w, y)〈w, yxv〉

= β(w, xy)β(y, x)〈w, yxv〉

= β(w, xy)〈w, β(y, x)yxv〉. (2.3.4)

Similarly, the second term of (2.3.3) is −β(y, x)β(xw, y)β(w, x)〈w, xyv〉 which

can be simplified to

− β(w, xy)〈w, xyv〉. (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)〈w, (β(y, x)yx− xy)v〉

which is −β(w, xy)〈w, [x, y]v〉 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

V(0) =⊕a∈G(0)

Va and V(1) =⊕a∈G(1)

Va.

Lemma 2.4.1. Let 〈·, ·〉 be as in Definition 2.3.6.


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

form.

(ii) The restricted form 〈·, ·〉 : V(1) × V(1) → F is a nondegenerate and symmetric

bilinear form.

Proof. Let a, b ∈ G. If ab 6= 1, then 〈Va, Vb〉 = 0. So we may assume that b = a−1. Now

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

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

〈v, w〉 = −β(a−1, a)〈w, v〉 = ±〈w, v〉.

Thus 〈·, ·〉 : V(0)×V(0) → F is skew-symmetric, and 〈·, ·〉 : V(1)×V(1) → F is symmetric.

Next we prove non-degeneracy. By Definition 2.3.6, 〈·, ·〉 : V × V → F is a non-

degenerate form. Then for each nonzero v in V , there exists some w in V such that

〈v, w〉 6= 0. Write w =∑

b∈Gwb as a sum of homogeneous vectors. By Definition

2.3.6 (ii), 〈v, w〉 = 〈v, wb〉 where b = a−1. Without loss of generality, we can replace

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

Therefore w is in V(i) as well. Therefore 〈·, ·〉 : 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 〈·, ·〉 : 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.


We claim that there exists a homogeneous basis

B(0) = t1, t∗1, . . . , tr, t∗r

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

Hi = Spanti, t∗i . Namely, let t ∈ V(0) be homogeneous and nonzero. Then as in the

proof of Lemma 2.4.1, there exists a homogeneous t′ ∈ V(0) such that 〈t, t′〉 6= 0. Also

since charF 6= 2, we can assume 〈t, t〉 = 〈t′, t′〉 = 0. Thus, replacing t′ by a scalar

multiple if necessary, this gives a hyperbolic pair t, t′. Repeating this process on

the orthogonal complement of Spant, t′ completes the proof.

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

basis

B(1) = u1, u∗1, . . . , us, u

∗s, (us+1)

of V(1) where each ui, u∗i 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 u∗s+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, t∗1, . . . , tr, t∗r, u1, u∗1, . . . , us, u

∗s, (us+1). (2.4.1)

Remark 2.4.2. In what follows, we frequently need the basis vectors with no ∗ signs.


Therefore for future reference, we let

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

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

Since 〈·, ·〉 is a nondegenerate form, it induces an isomorphism F : V → V ∗ by

v 7→ 〈v, ·〉 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→ 〈v, ·〉

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)〈v, xw〉.

By (2.3.2), we have the relation

〈xv, w〉+ β(F (v), x)〈v, xw〉 = 0

which implies that

F (xv)(w) = 〈xv, w〉 = −β(F (v), x)〈v, xw〉 = (xF (v))(w)


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

morphisms and vi has opposite colour as vi.

We recall the following fact. Let B = v1, . . . , vn be an ordered basis of (V, 〈·, ·〉).

Then denote by v1, . . . , vn the dual basis of V ∗ defined by vi(vj) = δij for all

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

that v 7→ 〈v, ·〉 is given by (FB)i,j = Fi,j = 〈vi, vj〉. We follow [BSR98] to use F−1i,j to

denote the (i, j) entry of the inverse matrix F−1B .

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

Fti,t∗i = 1, Ft∗i ,ti = −1, Fuj ,u∗j = 1, Fu∗j ,uj = 1, and Fv,v′ = 0 otherwise. (2.4.2)

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

F−1ti,t∗i

= −1, F−1t∗i ,ti

= 1, F−1uj ,u∗j

= 1, F−1u∗j ,uj

= 1, and Fv,v′ = 0 otherwise. (2.4.3)


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 = 〈v∗, w〉 = −β(w, v∗)〈w, v∗〉 = −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 B′ as indicated:

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

(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

〈xv, w〉+ β(v, x)〈v, xw〉 = 0, which is equivalent to

vtxtFBw + β(v, x)vtFBxw = 0.

First, letting v and w run over all pairs of vectors in the set B′, we can reduce the

above matrix equation into equations of the form:

(xt)v,w∗Fw∗,w + β(v, x)Fv,v∗xv∗,w = 0.


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

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

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

for all w ∈ B′, 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 ∈ B′ 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 ∈ B′, we have the relation

xw,v∗ = β(v∗, w)xvw∗ , (2.4.7)


and for all v, w ∈ B′, we have

xw∗,v∗ = −β(w,w)β(v, w)xv,w. (2.4.8)

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 ∈

B′(1);

(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 < t∗1 < . . . < tr < t∗r < u1 < u∗! < . . . < us < u∗s < (us+1).

Take arbitrary x ∈ spo(V, β). We let xv,w be the (v, w)-entry for all v, w ∈ B. Then

we write x =∑

v,w∈B xv,wEv,w which can be regrouped as

x =∑v,w∈B′

(xv,wEv,w + xw∗,v∗Ew∗,v∗) +∑v<wv,w∈B′

(xv∗,wEv∗,w + xw∗,vEw∗,v)


+∑v<wv,w∈B′

(xv,w∗Ev,w∗ + xw,v∗Ew,v∗) +∑v∈B′

(0)

(xv∗,vEv∗,v + xv,v∗Ev,v∗). (2.4.9)

Applying the relation (2.4.8) to the first term yields

∑v,w∈B′

xv,w (Ev,w − β(w,w)β(v, w)Ew∗,v∗) =∑v,w∈B′

xv,wYv,w.

Similarly, applying relation (2.4.6) and (2.4.7) to the second and third terms of (2.4.9)

yields ∑v<wv,w∈B′

xv∗,wYv∗,w and∑v<wv,w∈B′

xv,w∗Yv,w∗

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

∑v∈B′

(0)

1

2(xv∗,vYv∗,v + xv,v∗Yv,v∗).

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

x =∑v,w∈B′

xv,wYv,w+∑v<wv,w∈B′

xv∗,wYv∗,w+∑v<wv,w∈B′

xv,w∗Yv,w∗+1

2

∑v∈B′

(0)

(xv∗,vYv∗,v+xv,v∗Yv,v∗).

Thus, the vectors in Theorem 2.4.6 span spo(V, β), and in turn, they form a homoge-

neous basis of spo(V, β).

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 .


2.5 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, t∗1, . . . , tr, t∗r, u1, u∗1, . . . , us, u

∗s, (us+1)

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

h = SpanFHv | 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

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

W =⊕α∈h∗

Wα,


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

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 ∈ B′, the

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

is 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

αv =k∑i=1

αvi .


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

H · (v1 ⊗ · · · ⊗ vk) =k∑`=1

v1 ⊗ · · · ⊗ v`−1 ⊗ (H · v`)⊗ v`+1 ⊗ · · · ⊗ vk

=k∑`=1

v1 ⊗ · · · ⊗ v`−1 ⊗ αv`(H)v` ⊗ v`+1 ⊗ · · · ⊗ vk

=k∑`=1

αv`(H)(v1 ⊗ · · · ⊗ vk).

Therefore we have shown that the tensor v1 ⊗ · · · ⊗ vk is a weight vector with weight∑ki=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 B′, 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∗ =

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∗


= 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

g = h⊕⊕α∈Φ

gα

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.

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

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

g−αv−αw . Yv,w has colour vw−1, 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;


(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,

Φ+0 = εi ± εj, 2ε1, 2εj, δk ± δl, δ1, δl | 1 ≤ i < j ≤ r, 1 ≤ k < l ≤ s

Φ+1 = εi ± δj, εi | 1 ≤ i ≤ r, 1 ≤ j ≤ s;

(ii) When n = 2s,

Φ+0 = εi ± εj, 2ε1, 2εj, δk ± δl | 1 ≤ i < j ≤ r, 1 ≤ k < l ≤ s

Φ+1 = 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 ∆.


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

γr =

εr if n = 1,

εr − δ1 otherwise,γr+s =

δs if n = 2s+ 1,

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

Et1,u1 + β(t1, u1)Eu∗1,t∗1 , Et1,u∗1 + β(t∗1, u1)Eu1,t∗1.

Otherwise, the Yγ’s are explicitly given by:

(i) Yγi = Eti,ti+1− β(ti, ti+1)Et∗i+1,t

∗i, where 1 ≤ i ≤ r − 1,

(ii) Yγr =

Etr,u1 + β(tr, u1)Eu1,t∗r , if n = 1,

Etr,u1 + β(tr, u∗1)Eu∗1,t∗r , otherwise,

(iii) Yγr+j = Euj ,uj+1+ β(uj, uj+1)Eu∗j+1,u

∗j, where 1 ≤ j ≤ s− 1,

(iv) Yγr+s =

Eus,us+1 + β(us, us+1)Eus+1,u∗s , if n = 2s+ 1,

Eus,u∗s−1+ β(u∗s, us−1)Eus−1,u∗s , if n = 2s.


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, t∗1, u1, u∗1, u2, u

∗2.

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:

H =

a1

−a1

a2

−a2

a3

−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


corresponding root space has a nonzero projection onto SpanEvw.

0 ε1 + ε1 ε1 − δ1 ε1 + δ1 ε1 − δ2 ε1 + δ2

−ε1 − ε1 0 −ε1 − δ1 δ1 − ε1 −ε1 − δ2 −ε1 + δ2

δ1 − ε1 δ1 + ε1 0 0 δ1 − δ2 δ1 + δ2

−δ1 − ε1 ε1 − δ1 0 0 −δ1 − δ2 −δ1 + δ2

δ2 − ε1 δ2 + ε1 δ2 − δ1 δ2 + δ1 0 0

−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 Poincare-Birkhoff-Witt

Theorem for Lie Colour Algebras

In this chapter, our goal is to state and prove the Poincare-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.

Definition 3.1.1. For any m ∈ Z+, the mth tensor product of g is defined to be the

vector space

Tm(g) = g⊗ g · · · ⊗ g︸︷︷︸m times

that is, Tm(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,

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

T (g) =∞⊕i=0

T i(g) = F⊕ g⊕ (g⊗ g)⊕ · · · .

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

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

where k,m ∈ Z+, v1, . . . , vk, w1, . . . , wm ∈ g. Note that with this multiplication rule,

T (g) is a graded algebra.


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

S(g) = T (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

homogeneous x, y in g and v, w ∈ T (g). Moreover, since T (g) = ⊕i≥0Ti(g) is a direct

sum of homogeneous components, we can view I as the span of elements of the form

v · (x ⊗ y − β(y, x)y ⊗ x) · w, for some v ∈ Tm(g) and w ∈ T n(g) where m,n ∈ Z+.

Therefore, in this case we have

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

Therefore, since I is spanned by homogeneous elements, it is a homogeneous ideal

and it admits a grading of the form

I =⊕j≥0

Ij where Ij = I ∩ T j(g).


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

S(g) =⊕i≥0

Si(g)

where Si(g) = T i(g)I i = T i(g)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 ∈ Tm(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

Tm =m⊕i=0

T i(g) and Sm =m⊕i=0

Si(g)

be the mth filtration subspaces of T (g) and S(g) respectively.


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 )

S(U)⊗ S(W )

ρϕ (3.1.2)

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.

Corollary 3.1.9. Let V be a G-graded vector space such that V =⊕

a∈G Va where

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 homo-

geneous basis of V . Then the set

xi11 · · ·xikk |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 ).


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

S(Va) = T (Va)〈x⊗ x− β(x, x)x⊗ x〉

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

β(x, x) = −1, we have S(Va) = T (Va)〈x⊗ x〉. 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

U(g) = T (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

g → T (g)→ T (g)J = U(g)

such that σ(x) = x+ J for all x ∈ g.


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 U(g)

A

σ

ρϕ (3.2.2)

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


a Lie colour algebra homomorphism.

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

by

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

Since ρ is linear, the map ϕ′ is well-defined, and it is straightforward to check ϕ′ is a

homomorphism. In particular, for all homogeneous x, y in g, we have

ϕ′(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 ϕ′ factors through to an algebra homomorphism

ϕ : U(g)→ A such that the diagram (3.2.2) commutes.

To prove uniqueness, let ϕ′ : U(g) → A be another algebra homomorphism

satisfying our assumption. Then for any x ∈ g, we have ϕ′(σ(x)) = ρ(x) = ϕ(σ(x)).

Thus ϕ′ = 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)


= 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 Poincare-

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) ⊆ · · ·

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)


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


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


since J is weakly increasing, we have

XJ = X i1i · · ·X im

m , such thatm∑k=1

iq ≤ n.

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.

By the fact β(Xq, Xq) = −1 we have X2q = 1

2[Xq, Xq]. Also notice that since

[xq, xq] ∈ g, we have [xq, xq] =∑n

i=1 cixi for some scalar ci ∈ F. Thus [Xq, Xq] =

σ([xq, xq]) =∑n

i=1 ciσ(xi) =∑n

i=1 ciXi. This implies that X2q = 1

2

∑ni=1 ciXi. Therefore

we deduce that if iq is even, then Xiqq =

(12

∑ni=1 ciXi

)iq/2. Thus we have

XJ =

(1

2

)iq/2X i1

1 · · ·Xiq−1

q−1

(n∑i=1

ciXi

)iq/2

Xiq+1

q+1 · · ·X imm ,

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

XJ =

(1

2

) iq−1

2

X i11 · · ·X

iq−1

q−1

(n∑i=1

ciXi

) iq−1

2

XqXiq+1

q+1 · · ·X imm ,

which by induction is also a linear combination of elements in Bn.

Corollary 3.3.6. Let B =⋃p≥0 Bp. Then B spans the universal enveloping algebra

U(g).


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).

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 U(g)

A

σ

ρϕ

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


maps ρm : g× Sm → Sm+1 such that ρm+1

∣∣g×Sm

= ρm for all m. Then we can extend

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.


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 that ρ 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 ,


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

B = Xk11 · · ·Xkn

n | 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))


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)

= Xk11 · · ·Xkn

n | 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

gr(U(g)) =⊕m≥0

Gm


where Gm = Um(g)Um−1(g) and G0 = F. The multiplication is defined by

(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)).

Take x+ U0(g) and y + U0(g) in G1 such that x and y are homogeneous. Then

we have

(x+ U0(g))(y + U0(g)) = xy + U1(g) ∈ G2. (3.4.4)

Whereas we have

β(y, x)(y + U0(g))(x+ U0(g)) = β(y, x)yx+ U1(g) ∈ G2. (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

elements Xk11 · · ·Xkn

n such that∑n

j=1 kj ≤ m form a basis for Um(g). Therefore the

elements Xk11 · · ·Xkn

n +Um−1(g) such that∑n

j=1 kj = m form a basis for Um(g)Um−1(g).

This gives us a linear isomorphism between the graded pieces α : Gm → Sm such

that

α : Xk11 · · ·Xkn

n + Um−1(g) 7→ zk11 · · · zknn ,


where k1 + · · ·+ kn = m. Now we have

(Xk11 · · ·Xkn

n +Um−1(g))(Xr11 · · ·Xrn

n +Un−1(g)) = Xk11 · · ·Xkn

n Xr11 · · ·Xrn

n +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

βπXk1+r11 · · ·Xkn+rn

n + Un+m−1(g),

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

zk11 · · · zknn zr11 · · · zrnn with βπz

k1+r11 · · · zkn+rn

n 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

Yi1Yi2 · · ·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


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

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


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

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

ρm(xi)zJ =

zizJ , if i ≤ J

zizJ + β(xj1 , xi)ρm−1(xj1)w + ρm−1([xi, xj1 ])zJ ′ , otherwise

(3.5.1)

where J ′ = (j2, . . . , jm) and w = ρm−1(xi)zJ ′ − zizJ ′ ∈ 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

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

Then

ρm(xi)zJ = ρm(xi)zj1zJ ′ = ρm(xi)ρm−1(xj1)zJ ′

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

(Am−1). Also since ρm−1 = ρm∣∣g×Sm−1

and in order to satisfy (Cm), ρm(xi)ρm−1(xj1)zJ ′

must be equal to


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)zJ ′ − zizJ ′ ∈ Sm−1.

Therefore by linearity we have

β(xj1 , xi)ρm(xj1)ρm−1(xi)zJ ′

= β(xj1 , xi)ρm(xj1)(zizJ ′ + w)

= β(xj1 , xi)(ρm(xj1)zizJ ′ + ρm−1(xj1)w) by ρm−1 = ρm∣∣g×Sm−1

= β(xj1 , xi)(zj1zizJ ′ + 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

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).


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])zJ ′

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

ρm−1([xi, xj])zJ ′ 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


= 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).


Now the only remaining case left to prove is when neither i ≤ L nor j ≤ L, in

which case we must have that L = (`, L′) 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.

Proof. First we consider the term ρm(xi)ρm(xj)zL = xi · xj · zL. Since ` ≤ L = (`, L′),

we have zL = x` · zL′ . Therefore

xj · zL = xj · x` · zL′ (3.5.6)

where j > ` and ` ≤ L′. 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 · zL′ + [xj, x`] · zL′ .

Therefore by linearity we have

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

Now we expand (3.5.7) even further.


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

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

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

Now since ` < i, j and ` ≤ L′, 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 · zL′ by considering xj · zL′ as one term. Thus we have:

β(`, j)xi · x` · xj · zL′ = β(`, j)(β(`, i)x` · xi · xj · zL′ + [xi, x`] · xj · zL′

).

Now for the second term of (3.5.7). Since L′ has length ≤ m−2, by the induction

hypothesis, we can expand it as

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

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

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

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


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

Therefore β(j, i)ρm(xj)ρm(xi)zL is

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

+ β(j, i)β(i`, j)[xi, x`] · xj · zL′ + β(j, i)[xj, [xi, x`]] · zL′ . (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 · zL′ − β(j, i)β(`, i)β(`, j)x` · xj · xi · zL′

+ [xi, [xj, x`]] · zL′ − β(j, i)[xj, [xi, x`]] · zL′ (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 · zL′ − β(j, i)xj · xi · zL′

).

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

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

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

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

β(`, i)β(`, j)β(ij, `)[xi, xj] · x` · zL′ + β(`, i)β(`, j)[x`, [xi, xj]] · zL′


which (by the fact that β(`, i)β(`, j)β(ij, `) = 1) is

[xi, xj] · x` · zL′ + β(`, i)β(`, j)[x`, [xi, xj]] · zL′ . (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` ·zL′+[xi, [xj, x`]] ·zL′+β(`, i)β(`, j)[x`, [xi, xj]] ·zL′−β(j, i)[xj, [xi, x`]] ·zL′ .

(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` · zL′ = [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


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

1 2 3 4 5 6 7

891011121314

∈ D7

Remark 4.1.3. Let k ∈ Z+. There are (2k−1) · (2k−3) · · · 5 ·3 ·1 = (2k)!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

have another k-diagram d and we define d1d2 = ηcd 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

d2 =.


Then

η2=d1d2 =

.

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.

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

i i+ 1. . . . . .

. . .. . .ei =

and

i i+ 1. . . . . .

. . .. . .si =

,

modulo the following relations.

s2i = 1, e2

i = 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].


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

πd =

1 2 · · · k k + 1 k + 2 · · · 2k

`1 `2 · · · `k rk rk−1 · · · r1

∈ S2k.

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

corresponding πe1 is

πe1 =

1 2 3 4

1 3 4 2

= (234) = (23)(34).

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

corresponding πs1 is

πs1 =

1 2 3 4

1 2 4 3

= (34).

Let d ∈ Dk be any diagram such that each vertex in the lower row is connected


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

corresponding permutation is

πd =

1 · · · k k + 1 · · · 2k

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

(2.4.1). Denote the dual basis by v1, . . . , vn. 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


si act by permutations, and the image of ei on V ⊗k isomorphic to V ⊗(k−2). In order

to define this action, we need the g-module morphisms Rπ in (2.2.3), prk and evk 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

Ψ : Dk → Homg(V⊗k, V ⊗k)

d 7→ Ψd

where Ψd is the composition

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

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

Rπd⊗id⊗k

−−−−−−→ V ⊗2k ⊗ V ⊗k

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

id⊗k⊗evk−−−−−→ V ⊗k ⊗ V1G∼= V ⊗k.

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


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

(vi1 ⊗ vi2)Ψe1 = Fi1,i2

n∑j1,j2

F−1j1,j2

vj1 ⊗ vj2 , and

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

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

Proof. Let d = e1 ∈ D2. Then by Example 4.1.8, we have Rπd = R(23)R(34). Notice

that when we apply R(23)R(34) on V ⊗k, we first apply R(34), and then apply R(23). By

Lemma 2.4.4, F−1(vi) 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

prk⊗id⊗2

−−−−−→n=1∑j1,j2

vj1 ⊗ vj2 ⊗ vj2 ⊗ vj1 ⊗ vi1 ⊗ vi2

id⊗2⊗(F−1)⊗2⊗id⊗2

−−−−−−−−−−−−→∑j1,j2

vj1 ⊗ vj2 ⊗ F−1(vj2)⊗ F−1(vj1)⊗ vi1 ⊗ vi2

R(34)⊗id⊗2

−−−−−−→∑j1,j2

−β(v−1j1, v−1j2

)vj1 ⊗ vj2 ⊗ F−1(vj1)⊗ F−1(vj2)⊗ vi1 ⊗ vi2 (*)

R(23)⊗id⊗2

−−−−−−→∑j1,j2

β(v−1j1, v−1j2

)β(v−1j1, vj2)vj1 ⊗ F−1(vj1)⊗ vj2 ⊗ F−1(vj2)⊗ vi1 ⊗ vi2

id⊗2⊗F⊗2⊗id⊗2

−−−−−−−−−→∑j1,j2

β(v−1j1, 1)vj1 ⊗ F−1(vj1)⊗ F (vj2)⊗ vj2 ⊗ vi1 ⊗ vi2 . (4.2.1)

Notice that β(v−1j1, 1) = 1 for any vj1 ∈ B, and we will rewrite (4.2.1) using the matrix


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

∑j1,j2

vj1 ⊗∑p=1

F−1j1,pvp ⊗

n∑q=1

Fj2,qvq ⊗ vj2 ⊗ vi1 ⊗ vi2

id⊗2⊗evk−−−−−→∑j1,j2

vj1 ⊗n∑p=1

F−1j1,pvp

(vj2(vi1)

n∑q=1

Fj2,qvq(vi2)

)=∑j1,j2

vj1 ⊗n∑p=1

F−1j1,pvp

(δj2,i1

n∑q=1

Fj2,qδq,i2

)=∑j1

∑j2

(vj1 ⊗

n∑p=1

F−1j1,pvpδj2,i1Fj2,i2

).

Consider the underlined part. Since δj2,i1 is nonzero only for j2 = i1, the sum over j2

reduces to j2 = i1. Thus we have

∑j1

vj1 ⊗n∑p=1

F−1j1,pvpFi1,i2 = Fi1,i2

∑j1,p

F−1j1,pvj1 ⊗ vp

= Fi1,i2∑j1,j2

F−1j1,j2

vj1 ⊗ vj2 .

Therefore we have (vi1 ⊗ vi2)Ψe1 = Fi1,i2∑

j1,j2F−1j1,j2

vj1 ⊗ vj2 .

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

(vi1 ⊗ vi2)→∑j1,j2

−β(v−1j1, v−1j2

)vj1 ⊗ vj2 ⊗ F−1(vj1)⊗ F−1(vj2)⊗ vi1 ⊗ vi2 (*)

id⊗2⊗F⊗2⊗id⊗2

−−−−−−−−−→∑j1,j2

−β(vj1 , vj2)vj1 ⊗ vj2 ⊗ vj1 ⊗ vj2 ⊗ vi1 ⊗ vi2

id⊗2⊗evk−−−−−→∑j1,j2

−β(vj1 , vj2)vj1 ⊗ vj2δj1,i2δj2,i1


=− 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

Ψei = id⊗(i−1) ⊗Ψe1 ⊗ id⊗(k−i−1),

and

Ψsi = id⊗(i−1) ⊗Ψs1 ⊗ id⊗(k−i−1).

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

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

π = si1 · · · sin as a product of simple transpositions, then for all v = v1⊗· · ·⊗vk ∈ V ⊗k,

we have

vπ = vΨd = vΨsi1· · ·Ψsin

= Rin · · · Ri1v. (4.2.2)

Moreover, we have

vπ = (v1 ⊗ · · · ⊗ vk)π = βπvπ(1) ⊗ · · · ⊗ vπ(k),


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 = vΨd1Ψd2 . This justifies the reason we write the action

of the Brauer algebra as a right action. It also suggests, as in the following theorem,

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

operation satisfies d1 · d2 := d2d1.

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

of algebras.

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

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

Homg(V⊗k, V ⊗k) for all b ∈ Bk(n − m)op. 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 Ψs1Ψe2Ψe1 = Ψs2Ψe1 (corresponding to the relation

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

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

deduce.

For all k ≥ 3, by Corollary 4.2.3, since both Ψs1Ψe2Ψe1 and Ψs2Ψe1 only act on

the first three components of a tensor in V ⊗k, it suffices to only consider a simple

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

(vi1 ⊗ vi2 ⊗ vi3)Ψs1Ψe2Ψe1 = (vi1 ⊗ vi2 ⊗ vi3)Ψs2Ψe1 . (4.2.3)


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)Ψs2Ψe1 = −β(vi3 , vi2)(vi1 ⊗ vi3 ⊗ vi2)Ψe1

=− β(vi3 , vi2)Fi1,i3∑j1,j3

F−1j1,j3

vj1 ⊗ vj3 ⊗ vi2 . (4.2.4)

The left hand side of (4.2.3) is

(vi1 ⊗ vi2 ⊗ vi3)Ψs1Ψe2Ψe1 =− β(vi2 , vi1)(vi2 ⊗ vi1 ⊗ vi3)Ψe2Ψe1

=− β(vi2 , vi1)

(Fi1,i3

∑`1,`3

F−1`1,`3

vi2 ⊗ v`1 ⊗ v`3

)Ψe1 ,

which is

− β(vi2 , vi1)

(Fi1,i3

∑`1,`3

F−1`1,`3

Fi2,`1∑h1,h2

F−1h1,h2

vh1 ⊗ vh2 ⊗ v`3

). (4.2.5)

Now let us simplify (4.2.5). Consider the underlined term F−1`1,`3

Fi2,`1 . Since

F−1`1,`36= 0 if and only if v∗`1 = v`3 , and Fi2,`1 6= 0 if and only if v∗`1 = vi2 , we deduce

that this term is nonzero if and only if v`3 = v∗`1 = vi2 . Therefore the underlined term

is F−1`1,`∗1

F`∗1,`1 which is 1 for all v`1 ∈ B by the explicit matrix form of F we found in

(2.4.2) and (2.4.3).

Thus (4.2.5) becomes

− β(vi2 , vi1)Fi1,i3∑h1,h2

F−1h1,h2

vh1 ⊗ vh2 ⊗ vi2 . (4.2.6)

Notice that Fi1,i3 6= 0 if and only if vi1 = v∗i3 , therefore by using the fact that


β(vi2 , v∗i3

) = β(vi3 , vi2), (4.2.6) becomes

−β(vi3 , vi2)Fi1,i3∑h1,h2

F−1h1,h2

vh1 ⊗ vh2 ⊗ vi2 .

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 Ψs1Ψe2Ψe1(vi1 ⊗ vi2 ⊗ vi3) = Ψs2Ψe1(vi1 ⊗ vi2 ⊗ vi3) would not hold.

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

we have

(ρ(X)w)Ψb = ρ(X)(wΨb),

where ρ denotes the g-module structure on V ⊗k. 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

V ⊗k 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

B′ = t1, . . . , tr, u1, . . . , us, (us+1)

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

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

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


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.

Then the transpose of λ, denoted λ′ = (λ1, . . . , λn) is a partition of k whose parts are

the lengths of the columns of Y (λ). In particular, Y (λ′) has ` columns and n = λ1

rows, and λj = Cardi | λi − j ≥ 0, for all 1 ≤ j ≤ n.

Example 5.1.4. Let λ = (5, 2, 2, 1). Then λ′ = (4, 3, 1, 1, 1). The corresponding

Young diagrams are given by

Y (λ) = , Y (λ′) = .

Definition 5.1.5. Given a tableau T of shape λ, the transpose of T , denoted T ′ is

the tableau of shape λ′ in which the entries of the ith row of T are the entries of the

ith column of T ′.

Note that if T is a standard Young tableau of shape λ, then T ′ is still a standard

Young tableau of shape λ′.

Example 5.1.6. Let T = 1 3 6 8 92 45 710

. Then T ′ = 1 2 5 103 4 7689

.


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

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 92 45 710

. This is a (2, 3)-hook tableau and we have

T(0) = 1 3 6 8 92 4

and T(1) = 5 107

.

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

B′.

Definition 5.1.8. Let T ∈ Γr,s(K). Then let vT = v1 ⊗ · · · ⊗ vk be the simple tensor

in V ⊗k where vi is defined by

vi =

tj if i is in the jth row of T(0),

uj if i is in the jth row of T(1).


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

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.

Lemma 5.1.12. Let v = v1 ⊗ · · · ⊗ vk be a simple tensor in V ⊗k 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, i′ ∈ K, π(i) = i′ and i 6= i′

implies vi = vi′ = tj, then Rπv = sgn(π)v, and

(ii) if there exists j ∈ 1, . . . , s such that for all i, i′ ∈ K, π(i) = i′ and i 6= i′

implies vi = vi′ = 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,

the conclusions of the lemma are Rπv = −β(vi, vi)v in both cases.


The base case is j − i = 1. Suppose that π = (i, i+ 1), and vi = vi+1. Then by

(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

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

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

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)

Applying Rj to (5.1.2) will permute vi past vj. Thus we have

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


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 23

.

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


to be

yT = h(λ)∑ψ∈RTγ∈CT

sgn(ψ)ψγ ∈ F[Sk]

where h(λ) ∈ Q+ is chosen so that yT is an idempotent.

Remark 5.2.4. In [GW09], Goodman and Wallach prove that h(λ)=dimSλ

k!where

Sλ 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

Young tableaux of shape (2, 1). So by Remark 5.2.4, we have h(λ) = 23!

= 13. Therefore

we have

yT =1

3(id + (13)− (12)− (12)(13)).

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 ∈ Sk is such that σ(i) < σ(j) whenever i < j and i, j ∈ L. Define T σ to be the

tableau obtained from T by applying σ to each entry of T . Then T σ is also a standard

Young tableau, with entries from σ(L). Moreover, σ−1yTσ = yTσ .

Proof. It is clear that T σ 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

σ−1ψσ = (σ−1ψr1σ) · · · (σ−1ψr`σ).


Therefore it suffices to prove that σ−1ψrjσ 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

j of T σ are Sσ = σ(i1), . . . , σ(im). From the definition of the right action, we know

that σ−1ψrjσ permutes the elements Sσ the same way that ψrj permutes the entries of

the S. It follows that the map ψ 7→ σ−1ψσ is a bijection from RT onto RTσ . Similarly,

the map γ 7→ σ−1γσ is a bijection from CT to CTσ .Therefore by Definition 5.2.3, we

have

σ−1yTσ = σ−1h(λ)∑ψ∈RTγ∈CT

sgn(ψ)ψγσ

= h(λ)∑ψ∈RTγ∈CT

sgn(ψ)(σ−1ψσ)(σ−1γσ)

= h(λ)∑

ψσ∈RTσγσ∈CTσ

sgn(ψσ)ψσγσ

= yTσ .

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


[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.

We begin by constructing contraction maps Ci,j : V ⊗k → V ⊗k.

Definition 5.3.1. Let ci,j be the diagram

i j. . . . . .

. . .

. . .

. . . . . .

ci,j =

in the Brauer algebra Bk(n−m). Then the contraction map Ci,j ∈ End(V ⊗k) 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

Definition 5.3.1. Let σ be a permutation in Sk. Then we have σ−1ci,jσ = cσ(i),σ(j).

Proof. Let us consider the product of σ−1ci,jσ in Bk(η) which we visualize in Figure

5.2. Then the result follows.

Lemma 5.3.4. Let Ci,j be a contraction map on V ⊗k, where 1 ≤ i < j ≤ k. Then

for any σ ∈ Sk, we have σ−1Ci,jσ = Cσ(i),σ(j).


i j

σ−1(i)σ−1(j)

. . . . . .

. . .

. . .

. . . . . .

σ−1 =

i j. . . . . .

. . .

. . .

. . . . . .

ci,j =

σ(i) σ(j)

ji

. . . . . .

. . .

. . .

. . . . . .

σ =

which is

σ(i) σ(j)

. . . . . .

. . .

. . .

. . . . . .

= cσ(i),σ(j)

.

Figure 5.2: The composition σ−1ci,jσ as diagrams in the Brauer algebra, andthe resulting diagram cσ(i),σ(j). We suppress most of the vertical edges in bothσ and σ−1. We only keep the vertical edges linking i, j with σ(i), σ(j) respec-tively in σ, and the vertical edges linking i, j with σ−1(i), σ−1(j) respectivelyin σ−1, and we label both top and bottom rows by 1, . . . , k for simplicity.

Proof. By Theorem 4.2.4, Ψ is a homomorphism from Bk(n−m)op to Homg(V⊗k, V ⊗k).

Thus, by Lemma 5.3.3, we have (as a right action)

Cσ(i),σ(j) = Ψcσ(i),σ(j) = Ψσ−1Ψci,jΨσ = σ−1Ψci,jσ = σ−1Ci,jσ

as claimed.

Lemma 5.3.5. Let V be an spo(V, β)-module. The image of C1,2 = Ψe1 in V ⊗k is

isomorphic to V ⊗(k−2).


Proof. It is enough to prove that the image of C1,2 in V ⊗2 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,

(u⊗ v)Ψe1 = Fu,v∑s,t∈B

F−1s,t s⊗ t = Ω. (5.3.1)

Recall that from Lemma 2.2.12, we have a g-module morphism prV such that in our case,

if we relabel the homogeneous basis B = v1, . . . , vn+m, then prV (1) =∑n+m

i=1 vi ⊗ vi

is a g-invariant tensor. Also recall from Lemma 2.4.3, F−1 : V ∗ → V is a g-module

isomorphism. Then we have

(id⊗ F−1) prV (1) =m+n∑i=1

vi ⊗ F−1(vi)

=m+n∑i=1

F−1i,j vi ⊗ vj = Ω.

Thus Ω is a g invariant tensor since x(id⊗ F−1)(prV (1)) = (id⊗ F−1)(xprV (1)) = 0

for all x ∈ g. Therefore that the image of C1,2 on V ⊗ V is isomorphic to the trivial

module and in turn, we conclude that V ⊗kC1,2∼= V ⊗(k−2).

Notice that the values of F−1s,t are given explicitly in (2.4.3), and we conclude that

(5.3.1) is

Ω =r∑i=1

(t∗i ⊗ ti − ti ⊗ t∗i ) +s∑j=1

(uj ⊗ u∗j + u∗j ⊗ uj). (5.3.2)

More generally, using the relation Ci,j = σ−1C1,2σ for some permutation σ ∈ Sk, we

find the following. Let w = w1 ⊗ · · · ⊗ wk with w` ∈ B, then for 1 ≤ i < j ≤ k, the


vector wCi,j is a nonzero scalar multiple of

Fwi,wj∑s,t∈B

F−1s,t w1⊗ · · · ⊗wi−1⊗ s⊗wi+1⊗ · · · ⊗wj−1⊗ t⊗wj+1⊗ · · · ⊗wk. (5.3.3)

In particular, this implies that V ⊗kCi,j ∼= V ⊗(k−2), 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

P =

[k/2]⋃j=0

P(j). (5.3.5)

We give the following definitions of harmonic tensor spaces before we construct

highest weight vectors.


Definition 5.3.7. The harmonic tensor space H(V ⊗k, Cp,q) of V ⊗k is the kernel of

the contraction map Cp,q acting on V ⊗k.

Lemma 5.3.8. The harmonic tensor space H(V ⊗k, Cp,q) is an spo(V, β)-submodule

of V ⊗k.

Proof. Let g = spo(V, β). We have that v ∈ H(V ⊗k, Cp,q) if and only if vCp,q = 0. Since

the action of g = spo(V, β) on V ⊗k commutes with the action of Bk(n−m), we have for

all X ∈ g, (Xv)Cp,q = X(vCp,q) = X(0) = 0, which implies g · v ⊆ H(V ⊗k, Cp,q).

Example 5.3.9. Suppose that r = 1, s = 0 and B = t1, t∗1. Consider the spo(V, β)-

module V ⊗2 = Spanv ⊗ w | v, w ∈ B. Using (5.3.1), u ⊗ v ∈ Ker(C1,2) if u 6= v∗.

With a little more work, we can see that H(V ⊗2, C1,2) is the span of

t1 ⊗ t1, t1 ⊗ t∗1 + t∗1 ⊗ t1, t∗1 ⊗ t∗1.

Definition 5.3.10. Let s and t be disjoint subsets of K− (p∪ q). Then the harmonic

tensor space H(V ⊗kCp,q, Cs,t) is the kernel of the contraction map Cs,t : V ⊗kCp,q →

V ⊗kCp,q.

Notice that H(V ⊗kCp,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

is to choose some special simple tensor w = w1 ⊗ · · · ⊗ wk ∈ V ⊗k, and then transform


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,wjΩ.

Since wi, wj ∈ B, we have Fwi,wj 6= 0 if and only if wi = w∗j . Therefore, wCi,j is

nonzero if and only if wi = w∗j . Thus, we can start with vectors wi = t1 and wj = t∗1

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

wi =

t1 if i ∈ p,

t∗1 if i ∈ q,

t` if i ∈ K − (p ∪ q) and i is in `th row of T(0),

u` if i ∈ K − (p ∪ q) and i is in `th row of T(1).

Then v = wT,p,qCp,qyT is a highest weight vector of V ⊗k. For any s, t ⊂ K − (p ∪ q)

such that s ∩ t = ∅, v is an element of H(V ⊗kCp,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 u∗s.

We will prove Theorem 5.4.1 by means of a sequence of lemmas. For simplicity,


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

the set B′ = t1, . . . , tr, u1, . . . , us, (us+1). Therefore Cs,t will contract ws and

wt into Fws,wtΩ which is 0 since Fws,wt = 0 for all ws, wt ∈ B′.

(ii) By the construction of w, the numbers of t1’s and t∗1’s coming from p and q

are equal in w, and we have wt(t1) = −wt(t∗1). 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 t∗1 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.


(iii) The result follows from (ii) and the fact that w is obtained from w by changing

us to u∗s.

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.

Proof. Take ψ ∈ RT(0) . Then ψ permutes elements within the ith row of T(0) for

1 ≤ i ≤ r. By the definition of w, the elements in the ith 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

there exists i1 appearing in the jth 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 ,


the vector in the ith1 position of wψ is still tj. This implies that the vector in the

ith1 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 .

Therefore the vector in the ith1 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

wσψ(i1) = uj2 . In either case, the vector in the ith1 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.

...

......

......

i1 i2

...

...T ′(1)

σ

ψ


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 = t∗1.

Therefore by (5.3.3) and the fact that Ft1,t∗1 = 1, we have (up to a nonzero scalar

multiple, indicated by ∼)

wCi,j ∼∑u,v∈B

F−1u,vw1⊗· · ·⊗wi−1⊗u⊗wi+1⊗· · ·⊗wj−1⊗v⊗wj+1⊗· · ·⊗wk. (5.4.3)

Note that if u = t1 and v = t∗1, we have F−1u,v = −1. Thus, wCi,j ∼ w+Ri,j , where

Ri,j =∑u,v∈B

u6=t1,v 6=t∗1

F−1u,vw1 ⊗ · · · ⊗ wi−1 ⊗ u⊗ wi+1 ⊗ · · · ⊗ wj−1 ⊗ v ⊗ wj+1 ⊗ · · · ⊗ wk.

Note that the vector in the ith 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

Ri,j, the vectors in the ith and jth 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 βp,qh(λ)∣∣∣RT(0)

∣∣∣∣∣∣RT(1)

∣∣∣. In

particular, v 6= 0.


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,

there exists i ∈ p such that the vector in its ith position is not t1. Therefore every

summand of RyT is an element of E , which implies RyT ∈ E .

Moreover, we have

wyT = h(λ)∑

ψ∈RT ,σ∈CT

w sgn(ψ)ψσ.

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 R′′ ∈ E . Namely, we have

wyT = h(λ)∑

ψ∈RT(0)σ∈RT(1)

w +R′′.

Therefore

wCp,qyT = βp,qh(λ)∑

ψ∈RT(0)σ∈RT(1)

w +R′′ +RyT ,

where R′′, RyT ∈ E . Thus, the coefficient of w in v is given by βp,qh(λ)∣∣∣RT(0)

∣∣∣∣∣∣RT(1)

∣∣∣which is not 0. Therefore v is not a zero vector.


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).

Then for any contraction maps Cp,q = Cp1,q1 . . . Cpj ,qj , we have σCo,eσ−1 = Cp,q.

Proof. The result follows from the fact Co,e = C1,2C3,4 · · ·C2j−1,2j and Lemma 5.3.4

Lemma 5.4.8. Let σ ∈ Sk be the permutation defined in Lemma 5.4.7. Let T σ be the

standard tableau defined in Lemma 5.2.6 with entries in K − (p ∪ q). We have

wT,p,qσ = βσ(w1, . . . , wk)wTσ ,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).


Let vi = wσ(i) for all 1 ≤ i ≤ k. In particular, we have

wσ(i) =

t1 if σ(i) ∈ p,

t∗1 if σ(i) ∈ q,

t` if σ(i) is in the `th row of T(0),

u` if σ(i) is in the `thth row of T(1).

Recalling the definition of T σ from Lemma 5.2.6, we deduce that this is equivalent to

vi =

t1 if i ∈ o,

t∗1 if i ∈ e,

t` if i is in the `th row of T σ(0),

u` if i is in the `th row of T σ(1).

Since this defines wTσ ,o,e, we have wT,p,qσ = βσ(w1, . . . , wk)wTσ ,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

wT,p,qCp,qyT = βσwTσ ,o,eCo,eyTσσ−1.

Proof. First notice that wT,p,qCp,qyT is equal to

(wT,p,qσ)(σ−1Cp,qσ)(σ−1yTσ)σ−1. (5.4.4)

Then by Lemma 5.4.8, 5.4.7 and 5.2.6, (5.4.4) becomes

βσwTσ ,o,eCo,eyTσσ−1,



Corollary 5.4.10. The vector wT,p,qCp,qyT is equal to

βσ(Ω⊗ · · · ⊗ Ω⊗ v2j+1 ⊗ · · · ⊗ vk)yTσσ−1 (5.4.5)

where Ω is the invariant tensor defined in (5.3.2), and v` = wσ(`).

Proof. Notice that

wTσ ,o,e = t1 ⊗ t∗1 ⊗ · · · ⊗ t1 ⊗ t∗1 ⊗ v2j+1 ⊗ · · · ⊗ vk, (5.4.6)

where the first 2j terms of (5.4.6) are j pairs of t1 ⊗ t∗1. 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

Y · (Ω⊗ · · · ⊗ Ω⊗ v2j+1 ⊗ · · · ⊗ vk) =

k−2j∑i=1

cY,iθY,i

for some scalars cY,i.

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 B′. We record how the simple root vectors act on

b ∈ B′ in Table 5.1.


Y ∈ ∆Y b ∈ B′ Y b ConditionsYγi ti+1 ti 1 ≤ i ≤ r − 1Yγr u1 trYγi+j uj+1 uj 1 ≤ j ≤ s− 1Yγr+s us+1 us n = 2s+ 1Y v 0 for all other cases

Table 5.1: The action of simple root vectors from Definition 2.5.13 on v ∈ B′

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,iyTσ = 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 = Et2,t3 − β(t2, t3)Et∗3,t∗2 . Then Y acting on Ω⊗ t3 ⊗ t1 ⊗ t3

gives

Ω⊗ t2 ⊗ t1 ⊗ t3 + β(t3t1, t2t∗3)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 ∈ RTσ there exists some permutation π in CTσ such that θψπ = −θψ.

Proof. By the construction of wσ = wTσ ,o,e = t1 ⊗ t∗1 ⊗ · · · ⊗ t1 ⊗ t∗1 ⊗ v2j+1 ⊗ · · · ⊗ vk,

we know that for all i ∈ 1, . . . , k − 2j, if v2j+i = t`+1 for some 1 ≤ ` ≤ r − 1, then

2j + i is in the (`+ 1)th row of T σ(0). In particular, the tensor wσ looks like

· · · ⊗ t`+1︸︷︷︸(2j+i)th position

⊗ · · · (5.4.8)

and so the (`+ 1)th row of the tableau T σ looks like


...

row `+ 1 −→ · · · 2j+i · · ·...

By hypothesis θ has the form

· · · ⊗ t`︸︷︷︸(2j+i)th position

⊗ · · · .

First suppose that ψ ∈ RTσ 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 · · ·...

Therefore we have vi1 = t`. Hence the permutation (i1, 2j + i) ∈ CTσ 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 ∈ RTσ 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`︸︷︷︸ith2 position

⊗ · · · ⊗ t`+1︸︷︷︸(2j+i)th position

⊗ · · · .

Let i3 be the entry of the tableau T σ in row ` directly above i2, as in the diagram:

...

row ` −→ i3

row `+ 1 −→ · · · i2 · · ·...


In this case, we have vi3 = t`. Hence the permutation (i2, i3) ∈ CTσ 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 ∈ RTσ , there exists some permutation π ∈ CTσ 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 θyTσ = 0.

Proof. By Lemma 5.4.12, for all ψ ∈ RTσ , there exists some π ∈ CTσ such that

θψπ = −θψ. Thus suppose ψ ∈ RTσ , there exists a choice of π ∈ CTσ such that

∑γ∈CTσ

sgn(ψ)(θψ)γ =∑γ∈CTσ

sgn(ψ)(θψ)πγ = −∑γ∈CTσ

sgn(ψ)(θψ)γ,

which implies that∑

γ∈CTσ sgn(ψ)(θψ)γ = 0 for each ψ ∈ RTσ and therefore θyTσ =

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 ∈ RTσ 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

for some 1 ≤ ` ≤ s− 1, then 2j+ 1 is in the (`+ 1)th column of (T σ(1))′ (or equivalently,

the (` + 1)th row of T σ(1)). Therefore there exists a permutation π ∈ RTσ such that

π permutes 2j + i and the entry in the box directly to the left of the box of 2j + i,

say i1. Note that vi1 = u` since i1 is in the `th column of (T σ(1))′. Moreover, since

Yγr+`v2j+i = u`, the (2j + i)th position of θ is also u`. In this way, we simply choose


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 θyTσ = 0.

Proof. By Lemma 5.4.14, there exists some permutation π ∈ RTσ such that θπ = θ

and sgn(π) = −1. Therefore we have

∑ψ∈RTσ

sgn(ψ)θψ =∑

ψ∈RTσ

sgn(ψ)sgn(π)(θπ)ψ = −∑

ψ∈RTσ

sgn(ψ)(θψ)

which implies that∑

ψ∈RTσ sgn(ψ)θψ = 0 and therefore θyTσ = 0.

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 yTσ . Thus Y wTσo,eCo,eyTσ = 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 = u∗s, then Yγr+su∗s = us−1.

Therefore we have one extra case to analyze, that is, the simple root vector acting on

v2j+i changes u∗s 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 u∗s 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.


5.5 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, t∗1, u1, u∗1, u2, u

∗2. Elements of the Cartan subalgebra of g

are given by H = diag(a1,−a1, a2,−a2, a3,−a3). Furthermore, we consider V ⊗k = V ⊗2.

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

P =1⋃j=0

P(j) = P(0) ∪ P(1) = (,) ∪ (1, 2) = (,), (1, 2),

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


(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 ⊗ t∗1. Additionally, we have yT = id since

T = ∅. Therefore the highest weight vector in this case is

v1 = (t1 ⊗ t∗1)C1,2id. (5.5.1)

In Proposition 4.2.2, we already computed that

v1 = −t1 ⊗ t∗1 + t∗1 ⊗ t1 + u1 ⊗ u∗1 + u∗1 ⊗ u1 + u2 ⊗ u∗2 + u∗2 ⊗ 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 = 12

with T(0) = 1 and T(1) = 2 , or (5.5.2)

T = 1 2 with T(0) = 1 2 and T(1) = ∅. (5.5.3)

Let T be given as in (5.5.2). We have yT = 12(id + (12)). In addition, since 1 is in

the 1st row of T(0), and 2 is in the 1st row of T(1), we have wT,∅,∅ = t1 ⊗ u1. Thus the

highest weight vector in this case is

v2 =1

2(t1 ⊗ u1)Ψid(id + (12)).

Then we write v2 explicitly. By Remark 4.1.10, the action of id + (12) on V ⊗ V is


given by the map Ψid + Ψs1 . Therefore we have

v2 =1

2(t1 ⊗ u1)Ψid +

1

2(t1 ⊗ u1)Ψs1

which (by Proposition 4.2.2) is

v2 =1

2(t1 ⊗ u1)− 1

2β(u1, t1)(u1 ⊗ t1). (v2)

Finally, let T be defined as in (5.5.3). We have yT = 12(id− (12)). Moreover, since

both 1 and 2 are in the 1st row of T(0), we obtain wT,∅,∅ = t1 ⊗ t1. Therefore in this

case, the highest weight vector is

v3 =1

2(t1 ⊗ t1)Ψid(id− (12))

=1

2(t1 ⊗ t1 − (−β(t1, t1)t1 ⊗ t1))

= 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.


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 = Et1,u1 + β(t1, u1)Eu∗1,t∗1 .

First notice that Et1,u1 acting on t1 gives 0, and Eu∗1,t∗1 acting on either t1 or u1 gives

0. Thus we have

Y v2 = β(t1, t1u∗1)t1 ⊗ Et1,u1u1 − β(u1, t1)Et1,u1u1 ⊗ t1

= β(t1, t1)β(t1, u∗1)t1 ⊗ t1 − β(u1, t1)t1 ⊗ t1

= β(t1, u∗1)t1 ⊗ t1 − β(u1, t1)t1 ⊗ t1

which is 0 by the fact that β(t1, u∗1) = 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


notation negative root vectorY−2ε1 Et∗1,t1Yδ1−ε1 Eu1,t1 − β(u1, t1)Et∗1,u∗1Y−δ1−ε1 Eu∗1,t1 − β(u∗1, t1)Et∗1,u∗1Yδ2−ε1 Eu2,t1 − β(u2, t1)Et∗1,u∗2Y−δ2−ε1 Eu∗2,t1 − β(u∗2, t1)Et∗1,u2Y−δ1+δ2 Eu2,u1 − β(u2, u1)Eu∗1,u∗2Y−δ1−δ2 Eu∗1,u2 − β(u∗1, u1)Eu∗2,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 vectorsε1 + δ1 v2 t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1ε1 − δ1 Y−δ1+δ2 · (Y−δ1−δ2 · v2) t1 ⊗ u∗1 − β(u∗1, t1)u∗1 ⊗ t1ε1 + δ2 Y−δ1+δ2 · v2 t1 ⊗ u2 − β(u2, t1)u2 ⊗ t1ε1 − δ2 Y−δ1−δ2 · v2 t1 ⊗ u∗2 − β(u∗2, t1)u∗2 ⊗ t1−ε1 + δ1 Y−2ε1 · v2 t∗1 ⊗ u1 − β(u1, t∗1)u1 ⊗ t∗1−ε1 − δ1 Y−δ1−δ2 · (Y−δ1−ε1 · (Yδ2−ε1 · v2)) t∗1 ⊗ u∗1 − β(u∗1, t

∗1)u∗1 ⊗ t∗1

−ε1 + δ2 Y−δ1−ε1 · (Yδ2−ε1 · v2) t∗1 ⊗ u2 − β(u2, t∗1)u2 ⊗ t∗1−ε1 − δ2 Y−δ1−ε1 · (Y−δ2−ε1 · v2) t∗1 ⊗ u∗2 − β(u∗2, t

∗1)u∗2 ⊗ t∗1

δ1 + δ2 Yδ2−ε1 · v2 u1 ⊗ u2 − β(u2, u1)u2 ⊗ u1δ1 − δ2 Y−δ2−ε1 · v2 u1 ⊗ u∗2 − β(u∗2, u1)u∗2 ⊗ u1−δ1 + δ2 Y−δ1+δ2 · (Y−δ1−ε1 · v2) u∗1 ⊗ u2 − β(u2, u∗1)u2 ⊗ u∗1−δ1 − δ2 Y−δ1−δ2 · (Y−δ1−ε1 · v2) u∗1 ⊗ u∗2 − β(u∗2, u

∗1)u∗2 ⊗ u∗1

2δ1 Yδ1−ε1 · v2 u1 ⊗ u1−2δ1 Y−δ1−δ2 · (Y−δ1+δ2 · (Y−δ1−ε1 · v2)) u∗1 ⊗ u∗12δ2 Y−δ1+δ2 · (Yδ2−ε1 · v2) u2 ⊗ u2−2δ2 Y−δ1−δ2 · (Y−δ2−ε1 · v2) u∗2 ⊗ u∗2

0 Y−δ1−δ2 · (Yδ2−ε1 · v2) u∗1 ⊗ u1 + u1 ⊗ u∗1 − u2 ⊗ u∗2 − u∗2 ⊗ u20 Y−δ1−ε1 · v2 −t1 ⊗ t∗1 + t∗1 ⊗ t1 + u∗1 ⊗ u1 + u1 ⊗ u∗1

Table 5.3: A basis of spo(V, β)-submodule generated by the highest weight vector v2. Thethird 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, andthe second column records how the negative root vectors act on v2 in order to produce thecorresponding vectors in the third column.

It is routine to verify that the vectors we found in Table 5.3 are linearly indepen-

dent. 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,


we use two circles around the origin (0, 0, 0) to indicate that there are 2 linearly

independent zero weight vectors.

δ1

δ2

ε1

2δ1

−2δ1

2δ2

−2δ2

ε1 − δ1

ε1 + δ1

ε1 − δ2

ε1 − δ2

−ε1 − δ1

−ε1 + δ1

−ε1 − δ2

−ε1 + δ2

δ1 + δ2

δ1 − δ2

−δ1 + δ2

−δ1 − δ2

0

Figure 5.3: Weights in the submodule generated by v2, where different coloursrepresent 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 .


weights actions vectors2ε1 v3 t1 ⊗ t1−2ε1 Y−2ε1 · Y−2ε1 · v3 t∗1 ⊗ t∗1ε1 + δ1 Yδ1−ε1 · v3 t1 ⊗ u1 + β(u1, t1)u1 ⊗ t1ε1 − δ1 Y−δ1−ε1 · v3 t1 ⊗ u∗1 + β(u∗1, t1)u∗1 ⊗ t1ε1 + δ2 Yδ2−ε1 · v3 t1 ⊗ u2 + β(u2, t1)u2 ⊗ t1ε1 − δ2 Y−δ2−ε1 · v3 t1 ⊗ u∗2 + β(u∗2, t1)u∗2 ⊗ t1−ε1 + δ1 Yδ1−ε1 · (Y−2ε1 · v3) t∗1 ⊗ u1 + β(u1, t∗1)u1 ⊗ t∗1−ε1 − δ1 Y−δ1−ε1 · (Y−2ε1 · v3) t∗1 ⊗ u∗1 + β(u∗1, t

∗1)u∗1 ⊗ t∗1

−ε1 + δ2 Yδ2−ε1 · (Y−2ε1 · v3) t∗1 ⊗ u2 + β(u2, t∗1)u2 ⊗ t∗1−ε1 − δ2 Y−δ2−ε1 · (Y−2ε1 · v3) t∗1 ⊗ u∗2 + β(u∗2, t

∗1)u∗2 ⊗ t∗1

δ1 + δ2 Yδ2−ε1 · (Yδ1−ε1 · v3) u1 ⊗ u2 + β(u2, u1)u2 ⊗ u1δ1 − δ2 Y−δ2−ε1 · (Y−δ1−ε1 · v3) u1 ⊗ u∗2 + β(u∗2, u1)u∗2 ⊗ u1−δ1 + δ2 Y−δ1+δ2 · (Y−δ1−ε1 · (Yδ1−ε1 · v3)) u∗1 ⊗ u2 + β(u2, u∗1)u2 ⊗ u∗1−δ1 − δ2 Y−δ1−δ2 · (Y−δ1−ε1 · (Yδ1−ε1 · v3) u∗1 ⊗ u∗2 + β(u∗2, u

∗1)u∗2 ⊗ u∗1

0 Y−2ε1 · v3 t∗1 ⊗ t1 + t1 ⊗ t∗10 Y−δ1−ε1 · (Yδ1−ε1 · v3) u∗1 ⊗ u1 − u1 ⊗ u∗1 − t1 ⊗ t∗1 − t∗1 ⊗ t10 Y−δ2−ε1 · (Yδ2−ε1 · v3) u∗2 ⊗ u2 − u2 ⊗ u∗2 − t1 ⊗ t∗1 − t∗1 ⊗ 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 ,


δ1

δ2

ε1

2ε1

−2ε1

ε1 − δ1

ε1 + δ1

ε1 − δ2

ε1 − δ2

−ε1 − δ1

−ε1 + δ1

−ε1 − δ2

−ε1 + δ2

δ1 + δ2

δ1 − δ2

−δ1 + δ2

−δ1 − δ2

0

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

Θ : Bk(n−m)→⊕λ∈Pk

Mdλ(F), (6.1.1)

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.

For each λ ∈ Pk, and 1 ≤ i, j ≤ dλ, let Eλ,i,j be the matrix in the λth block of⊕λ∈PkMdλ(F) with 1 at the (i, j) entry and 0 elsewhere. We denote eλ,i,j = Θ−1(Eλ,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

spo(V, β)-modules. Let λ ∈ Pk, and let λ′ be the transpose of λ. Then

UY (λ′) := V ⊗keλ,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

xUY (λ′) = x(V ⊗keλ,i,j) = (xV ⊗k)eλ,i,j ⊆ V ⊗keλ,i,j = UY (λ′).


Therefore UY (λ′) is an spo(V, β)-module.

Note that UY (λ′) can be the trivial vector space. We will give an example such

that UY (λ′) = 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 UY (λ′) is independent of the choice of i, j.

(ii) As an spo(V, β)×Bk(n−m) module, we have

V ⊗k ∼=⊕λ∈Pk

UY (λ′) ⊗Bλ,

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 UY (λ′). In fact in [BSR98, Section 0.2(d)], they mention that UY (λ′) is in general

not necessarily irreducible.

Moreover, the relation between UY (λ′) and the highest weight vectors produced

in Theorem 5.4.1 is not clear from [BSR98]. They did not indicate whether or not

the highest weight vectors wT,p,qCp,qyT , with T of shape λ′ lie in the appropriate

submodules UY (λ′).

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 UY (λ)’s.


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) = spanι, 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

e21 = ηe1 implies that e1 acts by η. Moreover, since ι is the identity diagram, ι


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.

By the relation s21 = 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.


Using the modules we found above, the map Θ : B2(η)→M such that

Θ(ι) =

1

1

1

,Θ(e1) =

η

0

0

,Θ(s1) =

1

1

−1

is an algebra isomorphism.

In particular, using this matrix realization, we determine that e∅,1,1 = 1ηe1,

e(2),1,1 = − 1ηe1 + 1

2(s1 + id) and e(1,1),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∅′ with basis v, w such that

vι = v, vs1 = v, ve1 = w.

wι = w, ws1 = w, we1 = 0. (6.2.5)

Note that B∅′ contains B∅ = B as a submodule. The corresponding isomorphism

Θ : B2(0)→M is given by

Θ(ι) =

1

1

1

,Θ(e1) =

0 1

0

0

,Θ(s1) =

1

1

−1

.


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

v1 =r∑i=1

(t∗i ⊗ ti − ti ⊗ t∗i ) +s∑j=1

(uj ⊗ u∗j + u∗j ⊗ uj) = Ω. (6.3.1)

Let us now consider (p, q) = (,). We have Cp,q = Ψid. Moreover, the set of


possible T ’s in Γr,s(1, 2) is

T = 12

with T(0) = 12

and T(1) = ∅, or (6.3.2)

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


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 conditionsY−2εi Et∗i ,ti

1 ≤ i ≤ rY−εi+εj Etj ,ti − β(tj , ti)Et∗i ,t

∗j

1 ≤ i < j ≤ rY−εi−εj Et∗i ,tj

+ β(t∗j , ti)Et∗j ,ti 1 ≤ i < j ≤ rY−εi+δj Euj ,ti + β(uj , ti)Et∗i ,u

∗j

1 ≤ i ≤ r, 1 ≤ j ≤ sY−εi−δj Eu∗

j ,ti− β(u∗j , ti)Et∗i ,uj

1 ≤ i ≤ r, 1 ≤ j ≤ sY−δk+δ` Eu`,uk + β(u`, uk)Eu∗

k,u∗

`1 ≤ k < ` ≤ s

Y−δk−δ` Eu∗k,u`

+ β(u∗k, u`)Eu∗`,uk

1 ≤ k < ` ≤ s

Table 6.1: The negative root vectors when m = 2r > 2 and n = 2s ≥ 2. Notice that ifs = 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 conditionsε1 + ε2 v2 t1 ⊗ t2 − β(t2, t1)t2 ⊗ t1εi + εj Y−ε1+εi · Y−ε2+εj · v2 tj ⊗ ti − β(ti, tj)ti ⊗ tj 1 < j ≤ rε1 + εj Y−ε2+εj · v2 tj ⊗ t1 − β(t1, tj)t1 ⊗ tj 1 < j ≤ rε1 − εj Y−ε2−εj · v2 t∗j ⊗ t1 − β(t1, t∗j )t1 ⊗ t∗j 1 < j ≤ rεi − εj Y−ε1+εi · Y−ε2−εj · v2 t∗j ⊗ ti − β(ti, t

∗j )ti ⊗ t∗j 1 < i < j ≤ r

−ε1 + εj Y−2ε1 · Y−ε2+εj · v2 tj ⊗ t∗1 − β(t∗1, tj)t∗1 ⊗ tj 1 < j ≤ r

−εi + εj Y−ε1+εi · Y−2ε1 · Y−ε2+εj · v2 tj ⊗ t∗i − β(t∗i , tj)t∗i ⊗ tj 1 ≤ i < j ≤ r

−ε1 − εj Y−2ε1 · Y−ε2−εj · v2 t∗j ⊗ t∗1 − β(t∗1, t∗j )t∗1 ⊗ t∗j 1 < j ≤ r

−εi − εj Y−ε1−εi · Y−ε2−εj · v2 t∗j ⊗ t∗i − β(t∗i , t∗j )t∗i ⊗ t∗j 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 conditions2δ` 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.


weights actions vectors conditionsε1 + δ` Y−ε2+δl · v2 t1 ⊗ u` − β(u`, t1)u` ⊗ t1 1 ≤ i ≤ r, 1 ≤ ` ≤ sεi + δ` Y−ε1+εi · Y−ε2+δl · v2 ti ⊗ u` − β(u`, ti)u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ sε1 − δ` Y−ε2−δl · v2 t1 ⊗ u∗` − β(u∗` , t1)u∗` ⊗ t1 1 ≤ i ≤ r, 1 ≤ ` ≤ sεi + 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 t∗i ⊗ u` − β(u`, t∗i )u` ⊗ t∗i 1 ≤ i ≤ r, 1 ≤ ` ≤ s

−εi − δ` Y−εi+δl · Y−ε2−δl · Y−ε1−δl · v2 t∗i ⊗ u∗` − β(u∗` , t∗i )u∗` ⊗ t

∗i 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 conditionsδk + 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 u∗k ⊗ u

∗` − β(u∗` , u

∗k)u∗` ⊗ u

∗k 1 ≤ k < ` ≤ s

−δk + δ` Y−εi−δk · Y−ε1+εi · Y−ε2+δl · v2 u∗k ⊗ u` − β(u`, u∗k)u` ⊗ u∗k 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

0 Y−εi−δl · Y−ε1+εi · Y−ε2+δl · v2 u∗` ⊗ u` + u` ⊗ u∗` − ti ⊗ t∗i + t∗i ⊗ ti

1 ≤ i ≤ r,1 ≤ ` ≤ s

0 Y−δk−δl · Y−εi+δk · Y−ε1+εi · Y−ε2+δl · v2 u∗` ⊗ u` + u` ⊗ u∗` − u∗k ⊗ uk − uk ⊗ u

∗k 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∗` − u∗1 ⊗ u1 − u1 ⊗ u∗1, (6.3.5)

for 1 ≤ ` ≤ s and the vectors

− ti ⊗ t∗i + t∗i ⊗ ti + u∗1 ⊗ u1 + u1 ⊗ u∗1 (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


±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

dim(W ) = m2 +mn+ n2 +n−m

2− 1. (6.3.8)

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 conditions2ε1 v1 t1 ⊗ t1−2ε1 Y−2ε1 · Y−2ε1 · v1 t∗1 ⊗ t∗12εi Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ ti 1 < i ≤ r−2εi Y−ε1+εi · Y−ε1−εi · v1 t∗i ⊗ t∗i 1 < i ≤ r

Table 6.7: The vectors of weights ±2εi for all 1 ≤ i ≤ r in U(g)v3.

weights actions vectors conditionsεi + εj Y−εi+εj · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ tj + β(tj , ti)ti ⊗ tj 1 ≤ i < j ≤ rεi − εj Y−εi−εj · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ t∗j + β(t∗j , ti)t

∗j ⊗ ti 1 ≤ i < j ≤ r

−εi + εj Y−εi+εj · Y−2εi · Y−ε1+εi · v1 t∗i ⊗ tj + β(tj , t∗i )tj ⊗ t∗i 1 ≤ i < j ≤ r

−εi − εj Y−εi−εj · Y−2εi · Y−ε1+εi · v1 t∗i ⊗ t∗j + β(t∗j , t∗i )t∗j ⊗ t∗i 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 conditionsεi + δ` Y−εi+δl · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ u` + β(u`, ti)u` ⊗ ti 1 ≤ i ≤ r, 1 ≤ ` ≤ sεi − 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 t∗i ⊗ u` + β(u`, t∗i )u` ⊗ t∗i 1 ≤ i ≤ r, 1 ≤ ` ≤ s

−εi − δ` Y−εi−δl · Y−2εi · Y−ε1+εi · v1 t∗i ⊗ u∗` + β(u∗` , t∗i )u∗` ⊗ t

∗i 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 conditionsδk + δ` Y−εi+δk · Y−εi+δl · Y−ε1+εi · Y−ε1+εi · v1 uk ⊗ u` + β(u`, uk)u` ⊗ uk 1 ≤ k < l ≤ sδk − 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 u∗k ⊗ u` + β(u`, u

∗k)u` ⊗ u∗k 1 ≤ k < l ≤ s

−δk − δ` Y−δk−δl · Y−εi+δl · Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 u∗k ⊗ u∗` + β(u∗` , u

∗k)u∗` ⊗ u

∗k 1 ≤ k < l ≤ s

Table 6.10: The vectors of weights ±δk ± δ` for all 1 ≤ i < j ≤ r in U(g)v3.


weights actions vectors conditions0 Y−2εi · Y−ε1+εi · Y−ε1+εi · v1 ti ⊗ t∗i + t∗i ⊗ ti 1 ≤ i ≤ r0 Y−εi+δl · Y−εi−δl · Y−ε1+εi · Y−ε1+εi · v1 u∗` ⊗ u` − u` ⊗ u

∗` − ti ⊗ t

∗i − t∗i ⊗ 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 ⊗ t∗i + ti ⊗ t∗i , 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

dim(W ) = m2 +mn+ n2 − n−m2

. (6.3.11)

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


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

(t1 ⊗ t∗1 + t∗1 ⊗ t1)e1 = Ft1,t∗1Ω + Ft∗1,t1Ω = 0,

and similarly

(u1 ⊗ u∗1 − u∗1 ⊗ 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)-


module, we can decompose V ⊗ V as

V ⊗ V =⊕λ∈P2

W Y (λ′) ⊗BY (λ).

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


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 ⊗ t∗1 + t∗1 ⊗ 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 ⊗ t∗1 + t∗1 ⊗ t1, t∗1 ⊗ t∗1.

Notice that in this case, W coincides with H(V ⊗2, 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,

the third summand would be U = V ⊗2e(2),i,j, corresponding to the Brauer module


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 ⊗ u∗1 + u∗1 ⊗ u1,

v2 = u1 ⊗ u1, and

v2 = u∗1 ⊗ u∗1.

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 = SpanΩ = W∅, (6.4.1)

(V ⊗ V )e(2),1,1 = Spanu1 ⊗ u1, u∗1 ⊗ u∗1 = W1 ⊕W2 , (6.4.2)

(V ⊗ V )e(1,1),1,1 = Spanu1 ⊗ u∗1 − u∗1 ⊗ u1. (6.4.3)

Thus let v3 = u1 ⊗ u∗1 − u∗1 ⊗ 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.


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 ⊗ t∗1 + t∗1 ⊗ t1 + u1 ⊗ u∗1 + u∗1 ⊗ 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 weightst1 ⊗ t1 2ε1t∗1 ⊗ t∗1 −2ε1

u1 ⊗ t1 + β(t1, u1)t1 ⊗ u1 ε1 + δ1u∗1 ⊗ t1 + β(t1, u

∗1)t1 ⊗ u∗1 ε1 − δ1

u1 ⊗ t∗1 + β(t∗1, u1)t∗1 ⊗ u1 −ε1 + δ1u∗1 ⊗ t∗1 + β(t∗1, u

∗1)t∗1 ⊗ u∗1 −ε1 − δ1

u∗1 ⊗ u1 − u1 ⊗ u∗1 0t∗1 ⊗ t1 + t1 ⊗ t∗1 0

Table 6.12: A basis for U(g)(t1 ⊗ t1).


For T3 = , we have two different highest weight vectors

v1 = t1 ⊗ u1 − β(u1, t1)u1 ⊗ t1 (6.4.4)

v1 = t1 ⊗ u∗1 − β(u∗1, t1)u∗1 ⊗ 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 ⊗ t∗1 − β(t∗1, u1)t∗1 ⊗ u1, u1 ⊗ u1, Ω.

The submodule W2 generated by (6.4.5) is also 4 dimensional, and it has basis

t1 ⊗ u∗1 − β(u∗1, t1)u∗1 ⊗ t1, u∗1 ⊗ t∗1 − β(t∗1, u∗1)t∗1 ⊗ u∗1, u∗1 ⊗ u∗1, 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

algebra acts by B∅ = B . However if we set v = 12(u1 ⊗ u∗1 + u∗1 ⊗ u1) and w = Ω,

then Spanv, w is a B2(0)-submodule isomorphic to B∅′ . Therefore V ⊗ V is a sum

of B2(0)-submodules, but the summands are not necessarily spo(V, β)-invariant.


6.5 The characters of W Y (λ) and UY (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 UY (λ)

given in [BSR98, Theorem 4.24(e)].

Let B = t1, t∗1, . . . , tr, t∗r, u1, u∗1, . . . , us, u

∗s, (us+1) be a homogeneous basis of V

as in (2.4.1). We define a corresponding ordered set of commuting variables

Z = zt1 < zt∗1 < · · · < ztr < zt∗r < zu1 < zu∗1 < · · · < zus < zu∗s < (zus+1) (6.5.1)

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 zi1b1 . . . zim+n

bm+n

is the dimension of the∑m+n

j=1 ijwt(bj)-weight space, where∑m+n

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.

(ii) χ(W ) =∑b∈B(0)

z2b +

∑b,b′∈Bzb<zb′zbzb′ 6=1

zbzb′ + (r + s).

(iii) χ(W ) =∑b∈B(1)

z2b +


zbzb′ + (r + s− 1).

Proof. Since W∅ is the trivial module, the first case follows. We prove the second

case.


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

∑b∈B(0)

z2b +

∑b,b′∈B(0)zb<zb′zbzb′ 6=1

zbzb′ +∑

b,b′∈B(1)zb<zb′zbzb′ 6=1

zbzb′ +∑b∈B(0)

b′∈B(1)

zbzb′ + r + s (6.5.3)

which is∑b∈B(0)

z2b +


zbzb′ + (r + s) as we claimed.

The last case can be proved similarly by changing the first term of (6.5.3) into∑b∈B(1)

z2b , and by changing r + s into r + s− 1.

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.


The set of all bi-tableaux of shape λ with entries from Z is denoted as T (Z, λ).

Example 6.5.4. zt1 zt1 zt∗1

zu1 zu2

zu1

is a bi-tableau of shape λ = (3, 2, 1).

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

shλ(Z) =∑

T∈T (Z,λ)

zT . (6.5.4)

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, t∗1, . . . , tr, t∗r and we do not assume

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

shλ(Z) =∑b∈B

zb.

Theorem 6.5.7 ([BSR98] Theorem 4.24(e)). Let UY (λ) be the spo(V, β)-submodule

defined in Lemma 6.1.1 The character of UY (λ) is given by

χ(UY (λ)) =1

2det

(sh(λi−i−j+2)(Z) + sh(λi−i+j)(Z).

)(6.5.5)

Now we compare the character we found in Lemma 6.5.2 with the character

computed based on Theorem 6.5.7.


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 UY (λ) coincides with the character

of W Y (λ).

Proof. Let us first consider λ = (1, 1). We have Y (λ) = . Therefore we have

χ(UY (λ)) =1

2det

sh(1)(Z) + sh(1)(Z) sh(0)(Z) + sh(2)(Z)

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

2det

2sh(1)(Z) 1 + sh(2)(Z)

2 sh(1)(Z)

which is also equal to

sh2(1)(Z)− sh(2)(Z)− 1. (6.5.6)

Now the first term of (6.5.6) is (∑

b∈B zb)2 which can be rewritten as

sh2(1)(Z) =

∑b∈B

z2b +

∑b,b′∈Bzb<zb′

2zbzb′ . (6.5.7)

The second term of (6.5.6) is

sh(2)(Z) =∑b∈B(0)

z2b +


zbzb′ . (6.5.8)


Therefore 6.5.6 becomes

∑b∈B

z2b +


2zbzb′ − (∑b∈B(0)

z2b +


zbzb′)− 1

which is ∑b∈B(1)

z2b +


zbzb′ − 1. (6.5.9)

For second term of (6.5.9), if b∗ = b′, we have zbzb′ = 1. In addition, we have r + s

pairs of such b and b′. Therefore (6.5.9) becomes

∑b∈B(1)

z2b +


zbzb′ + (r + s− 1)

which is χ(W ) in Lemma 6.5.2.

Now let us consider λ = (2). We have Y (λ) = . Therefore we have

χ(UY (λ)) =1

2det

sh(2)(Z) + sh(2)(Z) sh(1)(Z) + sh(−1)(Z)

sh(−1)(Z) + sh(−1)(Z) sh(−2)(Z) + sh(0)(Z)

which is

1

2det

2sh(2)(Z) sh(1)(Z)

0 1

= sh(2)(Z).

By a similar argument, we have

sh(2)(Z) =∑b∈B(0)

z2b +


zbzb′ + (r + s) = χ(W )


as we claimed.

Although [BSR98] did not provide any evidence that the submodules generated

by highest weight vectors are the same as UY (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(UY (λ)),

and in particular, we have W Y (λ) ∼= UY (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(UY (λ)),

and in particular, we have W Y (λ) ∼= UY (λ).


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 UY (λ) coincides with the

spo(V, β)-submodule W Y (λ) generated by highest weight vector wT,p,qCp,qyT . However,

the relation between UY (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 UY (λ), which can be derived from the formula given in [BSR98,

Theorem 5.1]:

χ(UY (λ)) =∑T

zT , (6.5.10)

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

entries Vi,j = xj−1i for all 1 ≤ i, j ≤ n. The determinant of Vn is given by

∆ =∏

1≤i<j≤n

(xi − xj)

Definition A.1.1. By a partition of n, we mean a weakly decreasing sequence of

non-negative integers λ = (λ1, . . . , λk) satisfying |λ| =∑k

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


in C[x1, . . . xn] by

sλ = sλ(x1, · · · , xn) =det(xλi+n−ij

)1≤i,j≤n

∆.

Example A.1.4. Let λ = (2) and consider sλ = s(2)(x1, x2, x3). Then we have

sλ = det

x4

1 x42 x4

3

x1 x2 x3

1 1 1

/

(x1−x2)(x1−x3)(x2−x3) = x21+x2

2+x23+x1x2+x1x3+x2x3.

Proposition A.1.5. The Schur polynomial sλ(x1, · · · , xn) is symmetric.

Proof. Let ∆ =∏

1≤i<j≤n

(xj − xi) and q(x1, . . . , xn) = det(xλi+n−ij

)1≤i,j≤n. Then

q(x1, . . . , xn) is an alternating polynomial since it is a determinant. That is, we

have

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 polyno-

mial.

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


(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).


Example A.2.4. Let λ = (4, 3, 2). Then

(i) 1 1 2 22 2 33 4

is a semi-standard Young tableau of shape λ.

(ii) 1 2 3 45 6 79 10

is a standard Young tableau of shape λ.

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 := 12: while Entry 6= ∅, do3: if Entry is greater than or equal to all the entries in the Row Number row ofT , 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 := y7: end if8: Row Number := Row Number + 19: 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


the bottom of the diagram. We illustrate this process as the following sequence of

tableau insertions.

1 1 2 2

2 2 3

3 4

← 1

,

1 1 1 2

2 2 3

3 4

← 2

,

1 1 2 2

2 2 2

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.


Example A.2.8. Let T be a tableau. Then T • x zy

= T ← y ← x← z.

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 Numberwhich is strictly less than Entry, say z, replace z by Entry.

3: Entry := z4: Row Number := Row Number − 15: end while6: 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


tableau of w is the tableau obtained by row inserting the letters of w from x1 to xn :

T (w) =((

x1 ← x2))← x3)

)← · · · ← xn).

Lemma A.2.13. Given two arbitrary words w and w′ = x1 . . . xn, then

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.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 (K ′)

x z y 7→ z x y if x ≤ y < z (K ′′)

Note that K ′ and K ′′ 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.

If words w and w′ are Knuth equivalent, then we write wk≡ w′.

Example A.3.3. 12124 is Knuth equivalent to 21124 by using K ′ on the first three

letters of 12124.

Proposition A.3.4. Consider the tableau T ← x. We have

W(T ← x)k≡ 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


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.

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

yp and x′ in the first row such that yp is the entry immediate before x′ and the relation

x < yp no longer holds, which means we have yp ≤ x < x′. Moreover, this process can

be viewed as:

ypx′z1 · · · zq−1zqx 7→ ypx

′z1 · · · zq−1xzq (x < zq−1 ≤ zq)

7→ ypx′z1 · · · zq−2xzq−1zq (x < zq−2 ≤ zq−1)

· · ·

7→ ypx′z1xz2 · · · zq (x < z1 ≤ z2)

7→ ypx′xz1 · · · zq (x < x′ ≤ z1).

Each of the above transformation is K ′. Now we have yp ≤ x < x′ which is the

condition of K ′′. Then we can swap x′ with yp, and continue the process with x′

replaced by x. Thus by repeated application of K ′′ we have:

y1 · · · ypx′xz1 7→ y1 · · · yp−1x′ypxz1 (yp ≤ x < x′)

7→ y1 · · · yp−2x′yp−1ypxz1 (yp−1 ≤ yp < x′)

· · ·

7→ y1x′y2y3 · · · ypxz1 (y2 ≤ y3 < x′)


7→ x′y1y2y3 · · · ypxz1 (y1 ≤ y2 < x′).

Each of the transformations is K ′′, and this process indicates that x′ is now effectively

in the second row of the tableau. We can now continue by testing x′ 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

by using the rules K ′ and K ′′ accordingly. Thus we have W(T ← x)k≡ W(T )x.

Corollary A.3.5. For all words w, W(T (w)

) k≡ 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

W(T (w))k≡ W(∅← w1 ← · · · ← wr−1)wr

k≡ · · · k≡ w1 · · ·wr.

Proposition A.3.6. For any two words w1 and w2, we have if T (w1) = T (w2), then

w1k≡ w2.

Proof. By Corollary A.3.5, we know wk≡ W

(T (w)

). Therefore, if T (w1) = T (w2),

then

w1k≡ W

(T (w1)

)=W

(T (w2)

) k≡ w2.

Hence we have w1k≡ w2.

Corollary A.3.7. For any two words w and w′, we have

T (ww′) = T (w) • T (w′).


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 yz

, and

(ii) if w1 = yzx and w2 = yxz, with x < y ≤ z, we have

T (w1) = T (w2) = x zy

.

Proof. The result follows by using the row-insertion algorithm on w1 and w2.

Notice that in the first and second case, w1 and w2 are Knuth equivalent by K ′′

and K ′ respectively.

Proposition A.3.9. For two words w1 and w2, if w1k≡ w2, then T (w1) = T (w2).

Proof. Without loss of generality, we assume that there is only a single Knuth trans-

formation, for example wzxyw′k≡ wxzyw′. 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).

That is we show that for all xzyk≡ zxy with x ≤ y < z, we have T (wxzy) = T (wzxy),

and for all yzxk≡ 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,


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 ′, 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 K ′′. 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.


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

xrzxr+1k≡ 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

replaces some y′ where y′ belongs to a box to right of the position of xr. Therefore

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

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

first row of T (wxzy). By (A.3.2), xr ≤ y′ < xs implies that xrxsy′ k≡ xsxry

′. Thus

the result follows by induction hypothesis.

The case when xzy and zxy are Knuth equivalent by K ′ can be proved similarly.

We proved this proposition when w1 and w2 can be changed by applying a sequence

which consists of K ′ or K ′′ only, and this implies that it is also true for a sequence

of K ′ and K ′′ 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.


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 ) ≡ W(T (W(U)W(V ))

)≡ W(U)W(V ).

Let W be another tableau. Then we have

W((U • V ) •W

) k≡ W(U • V )W(W )

k≡ W(U)W(V )W(W )

k≡ W(U)W(V •W )

k≡ W(U • (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.


Definition A.4.1. An ordered two-rowed array is a 2× n matrix

A =

u1 u2 . . . ur

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

such that the (i, j) entry of M is the number of times of

ij

occurs in the array.

Example A.4.3. For the two-rowed array

1 1 1 2 2 3 3 3 3

1 2 2 1 2 1 1 1 2

, the corre-

sponding matrix is

M =

1 2

1 1

3 1

.

Next given an arbitrary two-rowed array,

u1 u2 . . . ur

v1 v2 . . . vr

, we construct a

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


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

2 2 2

, 1 1 1 2 3 3

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 =

M1

M2

where M1 consists of the first k rows of M for all 1 ≤ k ≤ n, and 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 ofM 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 .


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

the element is removed from the top row of Pk is the kth 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]


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

2 2 3

3 5

. Then the type of T is m(T ) = (2, 4, 2, 0, 1).

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.


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

r

s t

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

r

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 permuta-

tions, it suffices to prove it when σ is the adjacent transposition of (k, k + 1). Given


A in E1, we can write A as a block matrix

A =

B

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).

Therefore, for each A in E1, there is a unique corresponding matrix A =

B

C

D

in

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

Schλ(x1, x2, · · · , xn) =∑T

xm(T ) =∑T

xm11 · · ·xmnn .


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

Sch(2)(x1, x2, x3) = x21 + x2

2 + x23 + 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.

Proof. By Proposition A.5.2, the coefficient of xm11 · · ·xmnn is the same as the coefficient

of xmσ(1)1 · · ·xmσ(n)n 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.


Lemma A.5.9. We have

sλ(x1, . . . xn) =∑µ

sµ(x1, . . . , xn−1)x|λ|−|µ|n

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

sλ(x1, . . . , xn−1, 1) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xλ1+n−11 xλ1+n−1

2 · · · xλ1+n−1n−1 1

xλ2+n−21 xλ2+n−2

2 · · · xλ2+n−2n−1 1

......

. . ....

...

xλn−1−11 x

λn−1−12 · · · x

λn−1−1n−1 1

xλn1 xλn2 · · · xλnn−1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xn−11 xn−1

2 · · · xn−1n−1 1

xn−21 xn−2

2 · · · xn−2n−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


column unchanged, the determinant does not change, and we have:

sλ(x1, . . . , xn−1, 1) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xλ1+n−11 − 1 xλ1+n−1

2 − 1 · · · xλ1+n−1n−1 − 1 1

xλ2+n−21 − 1 xλ2+n−2

2 − 1 · · · xλ2+n−2n−1 − 1 1

......

. . ....

...

xλn−1−11 − 1 x

λn−1−12 − 1 · · · x

λn−1−1n−1 − 1 1

xλn1 − 1 xλn2 − 1 · · · xλnn−1 − 1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xn−11 − 1 xn−1

2 − 1 · · · xn−1n−1 − 1 1

xn−21 − 1 xn−2

2 − 1 · · · xn−2n−1 − 1 1

......

. . ....

...

x1 − 1 x2 − 1 · · · xn−1 − 1 1

0 0 · · · 0 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

For all 1 ≤ i ≤ n − 1, we factor out xi − 1 from the ith column of each of the

determinants in the numerator and denominator to get the following expression for

the right hand side of the above.


∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xλ1+n−21 + xλ1+n−3

1 + · · ·+ x1 + 1 · · · xλ1+n−2n−1 + xλ1+n−3

n−1 + · · ·+ xn−1 + 1 1

xλ2+n−31 + xλ2+n−4

1 + · · ·+ x1 + 1 · · · xλ2+n−3n−1 + xλ2+n−4

n−1 + · · ·+ xn−1 + 1 1

.... . .

......

xλn−1−21 + x

λn−1−31 + · · ·+ x1 + 1 · · · x

λn−1−2n−1 + x

λn−1−3n−1 + · · ·+ xn−1 + 1 1

xλn−11 + xλn−2

1 + · · ·+ x1 + 1 · · · xλn−1n−1 + xλn−2

n−1 + · · ·+ xn−1 + 1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xn−21 + xn−3

1 + · · ·+ x1 + 1 · · · xn−2n−1 + xn−3

n−1 + · · ·+ x2 + 1 1

xn−31 + xn−4

1 + · · ·+ x1 + 1 · · · xn−3n−1 + xn−4

n−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


the last row unchanged, the above expression becomes equal to:

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xλ1+n−21 + · · ·+ xλ2+n−2

1 · · · xλ1+n−2n−1 + · · ·+ xλ2+n−2

n−1 0

xλ2+n−31 + · · ·+ xλ3+n−3

1 · · · xλ2+n−3n−1 + · · ·+ xλ3+n−3

n−1 0

.... . .

......

xλi+n−(i+1)1 + · · ·+ x

λi+1+n−(i+1)1 · · · x

λi+n−(i+1)n−1 + · · ·+ x

λi+1+n−(i+1)n−1 0

.... . .

......

xλn−1−21 + · · ·+ xλn1 · · · x

λn−1−2n−1 + · · ·+ xλnn−1 0

xλn−11 + · · ·+ x1 + 1 · · · xλn−1

n−1 + · · ·+ xn−1 + 1 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xn−21 · · · xn−2

n−1 0

xn−31 · · · xn−3

n−1 0

......

. . ....

1 1 · · · 0

0 0 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

Then we use cofactor expansion for the numerator and denominator along the last


column to get

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xλ1+n−21 + · · ·+ xλ2+n−2

1 · · · xλ1+n−2n−1 + · · ·+ xλ2+n−2

n−1

xλ2+n−31 + · · ·+ xλ3+n−3

1 · · · xλ2+n−3n−1 + · · ·+ xλ3+n−3

n−1

.... . .

...

xλi+n−(i+1)1 + · · ·+ x

λi+1+n−(i+1)1 · · · x

λi+n−(i+1)n−1 + · · ·+ x

λi+1+n−(i+1)n−1

.... . .

...

xλn−1−21 + · · ·+ xλn1 · · · x

λn−1−2n−1 + · · ·+ xλnn−1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xn−21 · · · xn−2

n−1

xn−31 · · · xn−3

n−1

.... . .

...

x01 · · · x0

n−1

∣∣∣∣∣∣∣∣∣∣∣∣∣

.

Notice that the (i, j) entry of the matrix in the numerator is of the form

λi+1∑µi=λi

xµi+(n−1)−ij ,


where 1 ≤ j ≤ n− 1. Thus preceding quotient is

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

λ2∑µ1=λ1

xµ1+n−21 · · ·

λ2∑µ1=λ1

xµ1+n−2n−1

.... . .

...λi+1∑µi=λi

xµi+(n−1)−i1 · · ·

λi+1∑µi=λi

xµi+(n−1)−in−1

.... . .

...λn∑

µn−1=λn−1

xµn−1

1 · · ·λn∑

µn−1=λn−1

xµn−1

n−1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

xn−21 · · · xn−2

n−1

xn−31 · · · xn−3

n−1

.... . .

...

x01 · · · x0

n−1

∣∣∣∣∣∣∣∣∣∣∣∣∣

.

Thus by using the fact that the determinant is linear in each row, we conclude

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.

In order to homogenize the sum∑

µ sµ(x1, . . . , xn−1) to recover sλ(x1, . . . , xn), we

should multiply each of the above terms by x|λ|−|µ|n . Therefore we have

sλ(x1, · · ·xn) =∑µ

sµ(x1, · · · , xn−1)x|λ|−|µ|n

where |λ| = λ1 + · · ·+ λn and the sum is over all partitions µ such that

λ1 ≥ µ1 ≥ λ2 ≥ µ2 ≥ · · · ≥ µn−1 ≥ λn.


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.

Also notice that by Definition A.5.5, the power of xn in a monomial xt11 · · ·xtnn

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

λk+1i ≥ λki+1 and λkn−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.


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.

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