Length two extensions of modules for the Witt algebra

Length two extensions of modules for the

Witt algebra

by

Kathlyn Dykes

A thesis submitted to

the Faculty of Graduate and Postdoctoral Affairs

in partial fulfillment of the requirements

for the degree of Master of Science

in Mathematics

Carleton University

Ottawa, Ontario

c©2015, Kathlyn Dykes

Abstract

In this thesis, we analyse length two extensions of tensor modules for the Witt

algebra. In 1992, a classification of these modules was found by Martin and Piard

in [14], though no explicit form of the extensions were given. In this thesis, we es-

tablish an explicit classification of these modules using a different approach. As we

will show, each module extension is classified by a 1-cocycle from the cohomology

of the Witt algbera with coefficients in the module of the space of homomor-

phisms between the two tensor modules of interest. To use this, we first extended

our module to a module that has a compatible action of the commutative algebra

of Laurent polynomials in one variable. In this setting, we are able to determine

the possible structure of a 1-cocycle and from here, we will be able to directly

compute all possible 1-cocycles.

ii

Contents

Abstract ii

1 Introduction 1

2 Background 4

2.1 The Witt algebra and its irreducible modules . . . . . . . . . . . . . 4

2.2 The A-cover of a W1-module . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Cohomology of Lie algebras . . . . . . . . . . . . . . . . . . . . . . 10

3 Cocycle Functions 13

3.1 Parameters of the module extension . . . . . . . . . . . . . . . . . . 13

3.2 A new module extension . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3 Submodules of Hom(A, T (α, γ)) with finite dimensional weight spaces 23

3.4 Finding an appropriate basis for M . . . . . . . . . . . . . . . . . . 28

4 Polynomial Cocycles 39

4.1 General results for polynomial cocycles . . . . . . . . . . . . . . . . 39

4.2 Cases when n ≤ 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.2.1 Cocycles of degree 1 . . . . . . . . . . . . . . . . . . . . . . 47



iii

CONTENTS iv


4.3 Cases when n ≥ 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54




4.3.4 Cocycles of degree greater than or equal to 8 . . . . . . . . . 61

4.4 Dual Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5 Delta Cocycles 65

5.1 Conditions for delta functions . . . . . . . . . . . . . . . . . . . . . 66

6 Special case of β = 1 70

6.1 Conditions on non-polynomial cocycles . . . . . . . . . . . . . . . . 70

6.2 Cocycles with a factor of m−1 . . . . . . . . . . . . . . . . . . . . . 73

6.3 Delta cocycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

7 Conclusion 79

References 85

List of Tables

7.1 Polynomial cocycles in MN . . . . . . . . . . . . . . . . . . . . . . . 80

7.2 Polynomial cocycles in M . . . . . . . . . . . . . . . . . . . . . . . 81

7.3 Non-polynomial cocycles . . . . . . . . . . . . . . . . . . . . . . . . 82

v

Chapter 1

Introduction

This thesis is concerned with classifying all length two extensions of tensor modules

of the Witt algebra, W1. This is a question that has been studied before by Martin

and Piard in their 1992 paper [14]. The classification is given here explicitly, which

is something that has been lacking in previous works. By comparison, our results

are similar to the classification of the cohomology of the Lie algebra of vector fields

on a line given by Feigin and Fuks in their 1982 paper [8].

The Witt algebra can be described in a few different ways. It is the Lie algebra

of derivations of Laurent polynomials in one variable. It is also the complexification

of the Lie algebra of real vector fields on the circle, or can be described as the Lie

algebra of the group of (orientation preserving) diffeomorphisms of the circle [12].

In conformal field theory, the Witt algebra is described as the Lie algebra of

the conformal group of the complex plane, where the conformal group is the group

of transformations that preserve the angle between any two vectors. In quantum

field theory, the Virasoro algebra, the central extension of the Witt algebra, is

considered instead [10].

The importance of the Witt algebra is strongly linked to its central exten-

1

CHAPTER 1. INTRODUCTION 2

sion. The Virasoro algebra has many important roles in the area of mathematical

physics. From conformal field theory to soliton theory to string theory, repre-

sentations of the Virasoro algebra are studied extensively [12]. In this paper, we

will find module extensions of the Witt algebra that admit delta-cocycles. These

cocycles can be used to construct Lie algebras that contain the Virasoro algebra

as a subalgebra. One of these algebras, the W (2, 2) algebra, can be built from

the semidirect product of Virasoro algebra and its adjoint representation. This

algebra has a very different representation theory than that of the Virasoro al-

gebra; the highest weight modules for W (2, 2) admit vertex operator realizations

[17]. Similarly, we can define the twisted Heisenberg-Virasoro algebra which has

applications in the study of the representations of the toroidal Lie algebra [18].

In his 1992 paper [15], Mathieu gave a complete classification of irreducible

Virasoro modules with finite dimensional weight spaces. Consequently, as any

Witt algebra module is a Virasoro module where the central element acts trivially,

his results are also applicable to Witt algebra modules. As was proved in [15], every

irreducible W1-module with finite-dimensional weight spaces is either a highest

weight modules, a lowest weight module, a tensor module or quotient of a tensor

module. In this thesis, we consider only tensor modules of the Witt algebra.

The problem of classifying length two extensions of modules can be solved

with the help of cohomology. We will make use of the objects of cocycles and

coboundaries, which will in turn completely determine the module extensions.

In this paper, the method used to obtain the classification of these module

extensions M , begins with moving up to the A-cover M . Though it turns out

that this space is too big, the idea is to obtain a classification in the context of

AW1-modules, which is a simpler problem computationally than in the W1-module

setting. We obtain a surjection from the space where the classification is easily

CHAPTER 1. INTRODUCTION 3

calculated back to the original module, which will give a classification of these

extensions.

These results are important in two ways. First, no explicit classification of

these modules had been previously presented. Another aspect of these results is

the method used to obtain this classification. Instead of working in the modules

of the Witt algebra, we lift the module into the context of AW1-modules, which

are modules of the Witt algebra which have a compatible action with the com-

mutative algebra of Laurent polynomials in one variable. In fact, we can extend

this definition to define an AnWn-module [5]. The methods used in this work will

extend to the case of Wn, giving us somewhere to start in the classification of

length two extensions of Wn-modules.

This thesis is organized as follows: In the Chapter 2, we will introduce the

main areas of study that will be used throughout the remaining chapters. This

will include a description of the Witt algebra and the module extensions we are

interested in. We will introduce the notion of an A-cover of a module of the Witt

algebra and present enough cohomology of Lie algebras to serve our purposes. In

Chapter 3, the structure of module extensions is shown to be completely deter-

mined by 1-cocycles. These functions are studied in a general setting and the

conditions necessary for cocycles to be polynomial are found. In Chapter 4, we

classify all polynomial cocycles. In Chapter 5, we classify all delta function cocy-

cles. In Chapter 6, we look at the special case of β = 1, where β is one of the

parameters of the module extension. Finally, we end with a summary of all the

results.

Chapter 2

Background

In this section, we will discuss three topics that are used in the body of this

thesis. Lie algebra modules are the objects we are concerning ourselves with, in

particular Witt algebra modules. We will introduce the tensor modules of the Witt

algebra and then we will introduce the length two module extensions that will be

investigated. We will briefly discuss the notion of an A-cover of a W1-module,

which will be a crucial tool in our investigation of the structure of these modules.

At the end of this chapter, we briefly discuss enough cohomology theory to be of

use. We will also introduce the notations that will be used throughout this thesis.

2.1 The Witt algebra and its irreducible

modules

In general, we can define a commutative associative algebra An for n ∈ N by

An = C[t±11 , t±12 , . . . , t±1n ]

4

CHAPTER 2. BACKGROUND 5

which is the algebra of Laurent polynomials in n variables with the regular poly-

nomial multiplication.

A derivation of an algebra L is defined to be a map d : L → L such that

d(xy) = d(x)y + xd(y) for all x, y ∈ L. Notice this is very similar to the product

rule on derivatives of functions. The most basic example of a derivation of a Lie

algebra is the adjoint representation defined by adx : L→ L where adx(y) = [x, y]

for all y ∈ L.

If we look at the set of all derivations of An, we obtain a Lie algebra, denoted

by Wn. By introducing di = ti∂∂ti

for i = 1, . . . , n, then this Lie algebra is spanned

by the vectors{tk11 . . . tknn di|k1, . . . , kn ∈ Z, i = 1, . . . , n

}[5].

The Lie algebra we are interested in is the Witt algebra and is denoted by W1.

As the notation indicates, the Witt algebra is the case of Wn when n = 1. By

simplifying the notations, we denote A1 by A = C[t±1] so that the Witt algebra

is the Lie algebra of derivations of A. In this case, we only have d1 = t ddt

so that

W1 is spanned by the basis vectors{ek = tk+1 d

dt|k ∈ Z

}.

The bracket of W1 is given by

ek · es − es · ek = [ek, es] = (s− k)ek+s (2.1.1)

where the product on the left is the associative product tk+1 ddtts+1 d

dtof derivations.

As discussed in the introduction, the main interest in the Witt algebra is its

central extension. The Virasoro algebra is defined by W1⊕Cc, where c commutes

with everything and is called the central element. By using the notation that

δj+k,0 denotes the function

δj+k,0 =

1, j = −k

0, j 6= −k


we can write the bracket of the Virasoro algebra as

[ek, es] =(s− k)es+k + δs+k,0

(k3 − k

12

)c

[c, ek] =0

We will now introduce the types of modules that we will concern ourselves

with. In particular, we will only work with modules that have finite dimensional

weight spaces. We first need to introduce the notion of weight spaces.

The normalizer of a subalgebra K of a Lie algebra L is the subalgebra NL(K)

defined by

NL(K) = {x ∈ L|∀k ∈ K, [x, k] ∈ K}

The descending central series of a Lie algebra is defined by the subalgebras Ln

where L0 = L, Ln = [L,Ln−1]. If there exists some k ∈ Z such that Lk = {0},

then this Lie algebra is called nilpotent.

Definition 2.1.1. A Cartan subalgebra H of a Lie algebra L is a nilpotent sub-

algebra satisfying NL(H) = H.

We consider Cartan subalgebras with an additional property of being MAD

(maximal abelian diagonalizable) subalgebras. These subalgebras are abelian and

act diagonally on L, and are not strictly contained in any subalgebra with these

two properties. In W1, Ce0 is the MAD subalgebra [5]

Definition 2.1.2. A module V of a Lie algebra L with Cartan subalgebra H is a

weight module if there exist weights λ ∈ H∗ such that V =⊕

λ Vλ, where

Vλ = {v ∈ V |∀h ∈ H, h.v = λ(h)v}

In W1, we can simplify this definition of a weight module using that H = Ce0.


Definition 2.1.3. A W1-module V is a weight module if there exist weights λ ∈ C

such that V = ⊕λVλ where the weight space of weight λ is given by

Vλ = {v ∈ V |e0.v = λv}

Definition 2.1.4. A weight module is called cuspidal if the dimensions of all

weight spaces are bounded by a common constant.

In the Wn case, we can consider tensor modules which are parameterized by

a complex vector γ ∈ Cn and a finite-dimensional gln-module as given in [5]. In

the 1-dimensional case, gl1 = C so that tensor modules are parameterized by two

constants, γ, α ∈ C.

The structure of these tensor W1-modules, denoted by T (α, γ), are given by

T (α, γ) =⊕

m∈γ+Z

Cvm

with W1 action

ek.vm = (m+ αk)vm+k

Notice that e0vk = kvk, so these modules have 1-dimensional weight spaces Cvk

with weights k ∈ γ + Z. As the dimension of all weight spaces are bounded by 1,

tensor W1-modules are in fact cuspidal modules.

One thing to remark before proceeding further is that T (γ, α) ∼= T (γ′, α) if

γ − γ′ ∈ Z, as stated in Remark 1.1 of [12]. Therefore, we will always assume

that 0 ≤ Re(γ) < 1. Consequently, we will treat the cases of γ ∈ Z and γ = 0 as

interchangeable.

These tensor modules are irreducible except in two special cases: α = 0, γ ∈ Z

and α = 1, γ ∈ Z by Proposition 1.1 of [12]. For convenience, we will denote by


D(0, 0) = span{v0}, T ◦(1, 0) = span{vm|m ∈ Z,m 6= 0}. D(0, 0) is a submodule

of T (0, 0) and T ◦(1, 0) is a submodule of T (1, 0), as shown below.

ekv0 = 0, ∀k ∈ Z in T (0, 0)

ekv−k = 0, ∀k ∈ Z in T (1, 0)

In this thesis, we will concern ourselves with weight W1-modules M which

possess the short exact sequence:

0→ T (α, γ)→M → T (β, γ′)→ 0

In this notation, each arrow represents a homomorphism so that M has a sub-

module T (α, γ) with the quotient M/T (α, γ) ∼= T (β, γ′). This extension is also

taken to be a weight extension, meaning that the weight space Mλ is given by the

exact sequence

0→ T (α, γ)λ →Mλ → T (β, γ′)λ → 0

One final remark about these tensor modules is that every irreducible cuspidal

module of W1 is isomorphic to either a tensor modules, D(0, 0) or T ◦(1, 0), as

the only cuspidal highest and lowest weight modules are trivial (Corollary III.3 of

[13]). We will consider only tensor modules extensions, but the cases of D(0, 0)

and T ◦(1, 0) will come out of these.


2.2 The A-cover of a W1-module

Another tool that will be needed in this paper is the idea of an A-cover of a

W1-module. First, notice that W1 is an A-module with the A-action:

tkem = em+k

and A is a W1-module with the W1-action:

ektm = mtm+k

This action can be extended linearly to all f ∈ A and all x ∈ W1.

If V is both an A-module and a W1-module, then the A-action and the W1-

action is compatible if

(xf)v = x(fv)− f(xv), for all x ∈ W1, f ∈ A, v ∈ V (2.2.1)

If V has a compatible A-action and W1 action, it is called an AW1-module [5].

The space of HomC(A, V ) is an example of AW1-module [5], with the actions

(xψ)(f) =x(ψ(f))− ψ(x(f)) (2.2.2)

(gψ)(f) =ψ(gf), for ψ ∈ Hom(A, V ), x ∈ W1, f, g ∈ A (2.2.3)

We now introduce a submodule of Hom(A, V ) which will be of use to us

throughout the next chapter.

Definition 2.2.1. If V is a W1-module, the A-cover V is defined as

V = span{φ(x, u)|x ∈ W1, u ∈ V } ⊂ Hom(A, V )


where φ(x, u)(f) = (fx)(u) for all f ∈ A.

The A-cover is an AW1-submodule of Hom(A, V ). By Proposition 4.5 of [5], if

V is a weight module, then so is V . In fact, if V is a cuspidal module, then so if

V , as shown in Theorem 4.10 of [5]. The W1-action and A-action on the A-cover

is

yφ(x, u) =φ([y, x], u) + φ(x, yu) (2.2.4)

gφ(x, u) =φ(gx, u) (2.2.5)

for x, y ∈ W1, u ∈ V, g ∈ A.

From V , we can always go back to the space V via the map π : V → V where

π(φ(x, u)) = φ(x, u)(1) = x.u. Notice this map is only surjective if W1V = V

(Proposiiton 4.5 of [5]).

2.3 Cohomology of Lie algebras

For this section, suppose that L is a Lie algebra and V is an L-module.

Definition 2.3.1. An n-dimensional cochain of L with coefficients in V is a skew-

symmetric linear map L⊗n → V . The space of all cochains is denoted by Cn(L, V ).

Equivalently, Cn(L, V ) = Hom(∧nL, V ) as the wedge product is skew-symmetric.

These spaces Cn(L, V ) can be related to each other using the following maps:

dn : Cn(L, V )→ Cn+1(L, V )


where, for ϕ ∈ Cn(L, V ), g1, . . . , gn+1 ∈ L, the map dn is defined as

dnϕ(g1, . . . , gn+1) =∑

1≤i≤n+1

(−1)igiϕ(g1, . . . , gi−1, gi+1, . . . , gn+1)

+∑

1≤i<j≤n+1

(−1)i+j−1ϕ([gi, gj], g1, . . . , gi−1, gi+1, . . . , gj−1, gj+1, . . . , gn+1)

(2.3.1)

For n < 0, Cn(L, V ) = 0 and dn = 0. In fact, dn+1 ◦ dn = 0 (page 15 of [9]) so

that the set of Cn(L, V ) together with the maps dn is an algebraic complex. We

now have the tools to define the cohomology of the Lie algebra L.

Definition 2.3.2. The cohomology of L with coefficients in V , denoted byH∗(L, V ),

is the space H∗(L, V ) =⊕∞

n=0Hn(L, V ), where

Hn(L, V ) = Ker(dn)/Im(dn−1)

An element in the kernel of the dn map is called an n-cocyle. An element in the

image of the dn−1 map is called an n-coboundary.

The space HomC(T (β, γ′), T (α, γ)) is a W1-module with module action given

by

(ek.x)(w) = ek.(x(w))− x(ek.w)

As we are considering graded modules, we will instead look at the graded version of

HomC(T (β, γ′), T (α, γ)). It is the space of the set of maps ϕ ∈ HomC(T (β, γ′), T (α, γ))

such that ϕ(wm) ∈ Cvm+k for all m ∈ γ′+Z, or elements that shift the grading of

the module T (α, γ) by k. This is still a module with W1-action as defined above.

We will consider the cohomology of W1 with values in the graded module

HomC(T (β, γ′), T (α, γ)). Again, as we are working with graded spaces, we will

consider the cohomology that is compatible with the Z-grading of this module.

Thus, we only consider n-cochains from ∧nL → V that preserve the Z-grading.


Then

C0(W1,Hom(T (β, γ′), T (α, γ))) = Hom(C,Hom(T (β, γ′), T (α, γ)))

∼= Hom(T (β, γ′), T (α, γ))

C1(W1,Hom(T (β, γ′), T (α, γ))) = Hom(W1,Hom(T (β, γ′), T (α, γ)))

∼= Hom(W1 ⊗ T (β, γ′), T (α, γ))

1-cocycles are in the kernel of d1 and 1-coboundaries are in the image of d0 and

d1 ◦ d0 = 0 as shown below for ϕ ∈ Hom(C,Hom(T (β, γ′), T (α, γ))), g1, g2 ∈ W1.

(d1 ◦ d0(ϕ))(g1, g2) = d1(d0(ϕ))(g1, g2)

= g2d0(ϕ)(g1)− g1d0(ϕ)(g2) + d0(ϕ)([g1, g2])

= −g2g1ϕ+ g1g2ϕ− [g1, g2]ϕ

= ([g2, g1]− g2g1 + g1g2)ϕ

= 0

These concepts will be useful in our efforts to understand and classify length

two extensions of the tensor modules of the Witt algebra.

Chapter 3

Cocycle Functions

In this section, we look at the module extension

0→ T (α, γ)→M → T (β, γ′)→ 0

where T (α, γ) is spanned by vectors vm and T (β, γ′) is spanned by vectors wm.

In this way, M is spanned by basis vectors vm and wm, where these wm ∈ M are

mapped to wm ∈ T (β, γ′) under the surjection from M to T (β, γ′).

We will introduce how we can use the notion of a cocycle to classify the module

extensions. Then we will use the A-cover of our module M to find an appropriate

basis of M so that the corresponding cocycles will be polynomial, except in a few

special cases.

3.1 Parameters of the module extension

Since T (α, γ) is a W1-submodule of M , then the W1 action is, by definition,

ek.vm = (m+ αk)vm+k, for k ∈ Z,m ∈ γ + Z

13

CHAPTER 3. COCYCLE FUNCTIONS 14

The W1-action on the wm basis vectors is defined as

ek.wm = (m+ βk)wm+k + τ(k,m)vm+k, for k ∈ Z,m ∈ γ′ + Z (3.1.1)

where τ(k,m) is some function in k and m.

Here, we can see that M is parameterized by two objects: the function τ(k,m)

and the complex number γ. First, let us look at the relation between γ and γ′.

Lemma 3.1.1. If γ + Z 6= γ′ + Z, then M = T (α, γ)⊕

T (β, γ′).

Proof. The W1-action is given by

ekwm = (m+ αk)wm+k + τ(k,m)vm+k

but since m+ k /∈ γ + Z, τ(k,m) must be zero.

As long as γ 6= γ′, the extension will be trivial in the sense that τ(k,m) is

the zero function. Since we are interested in non-trivial extensions we will only

consider the case that γ + Z = γ′ + Z. If γ + Z = γ′ + Z, then T (β, γ′) ∼= T (β, γ)

so that we may assume γ = γ′ for the rest of this thesis.

Now, let us look at the conditions on τ(k,m). M is a W1-module and it follows

that the W1-action must be a module action.

[ek, es]wm =(ekes − esek)wm

(s− k)ek+swm =(m+ βs)ekwm+s + ekτ(s,m)vm+s

− (m+ βs)eswm+k − esτ(k,m)vm+k

=⇒ (s− k)(m+ β(k + s))wm+k+s =(m+ βs)(m+ s+ βk)wm+s+k

− (m+ βk)(m+ k + βs)wm+k+s

=⇒ (s− k)τ(k + s,m)vm+s+k =(m+ βs)τ(k,m+ s)vm+k+s


− (m+ s+ αk)τ(s,m)vm+s+k

+ (m+ βs)τ(s,m+ k)vm+s+k

− (m+ k + αs)τ(k,m)vm+k+s

Therefore the following condition on τ ensures that M is a W1-module.

(s− k)τ(k + s,m) =(m+ βs)τ(k,m+ s)− (m+ βk)τ(s,m+ k)

+ (m+ s+ αk)τ(s,m)− (m+ k + αs)τ(k,m)

(3.1.2)

Here we see that since these τ -functions define the W1-action, they define the

module. Hence a classification of these τ -functions will give an explicit classifi-

cation of W1-modules of this type. These τ -functions also have a cohomological

interpretation.

For a Lie algebra L, let M be the L-module extension of the L-modules

(M1, ρ1), (M2, ρ2) given by

0→M1 →M →M2 → 0

If we non-canonically identify M2 with a subspace in M , the module action on M

can be given by

x.v = ρ1(x)v

x.w = ρ2(x)w + τ(x,w)

where v ∈M1, w ∈M2 and τ is a map L⊗M2 →M1.

Lemma 3.1.2. The isomorphism classes of these module extensions M are in

one-to-one correspondence to the cohomology, H1(L,HomC(M2,M1)).


Proof. By viewing the cohomology of L with the module Hom(M2,M1) as in

Chapter 2, C1(L,Hom(M2,M1)) ∼= Hom(L⊗M2,M1) and 1-cocycles are elements

in kerd1, where d1 is defined as in (2.3.1) as

d1 : C1(L,Hom(M2,M1))→ C2(L,Hom(M2,M1))

Suppose that ψ ∈ kerd1. Then

0 = (d1ψ)(x, y)

= ψ([x, y])− xψ(y) + yψ(x)

= ψ([x, y])w − (xψ(y))w + (yψ(x))w

= ψ([x, y])w − x(ψ(y)w) + ψ(y)xw + y(ψ(x)w)− ψ(x)yw

= ψ([x, y], w)− x(ψ(y, w)) + ψ(y, xw) + y(ψ(x,w))− ψ(x, yw)

for all x, y ∈ L and w ∈M2.

Then the module action on M gives that

[x, y]w = x(yw)− y(xw)

[x, y]w + τ([x, y], w) = x(yw + τ(y, w)− y(xw + τ(x,w))

[x, y]w + τ([x, y], w) = x(yw) + τ(x, yw) + xτ(y, w)− y(xw)− τ(y, xw)− yτ(x,w))

=⇒ τ([x, y], w) = τ(x, yw) + xτ(y, w)− τ(y, xw)− yτ(x,w))

Thus τ gives an extension of modules if and only if τ is a 1-cocycle.

By (2.3.1), 1-coboundaries are elements in the image of the map

d0 : C0(L,Hom(M2,M1))→ C1(L,Hom(M2,M1))


where C0 ∼= Hom(M2,M1) and C1 ∼= Hom(L⊗M2,M1). Suppose that ϕ ∈ Im(d0)

so that ϕ : L⊗M2 → M1. Then there exists ϕ ∈ Hom(M2,M1), the preimage of

ϕ under d0. The action of d0 is given in (2.3.1) so that

(d0ϕ)(x) = −xϕ

for all x ∈ L. Thus coboundaries in the space of H1(L,HomC(M2,M1)) are the

functions −xϕ where ϕ ∈ Hom(L⊗M2,M1).

Let us discuss when we have equivalent extensions. Equivalent extensions

correspond to different liftings of M2 into M :

w′ =w + ϕ(w)

where ϕ : M2 →M1.

The L action on w′ becomes:

x.w′ =ρ2(x)w + τ(x,w) + xϕ(w)

=ρ2(x)w + τ(x,w) + (xϕ)x+ ϕ(xw)

=ρ2(x)w′ + τ(x,w) + (xϕ)(w)

As defined above, for x ∈ W1, v ∈ M1, w in the preimage of M2 in M the

W1-action is

x.v =ρ1(x)v

x.w =ρ2(x)w + τ(x,w)

Thus, if there exists a change of basis such that the new 1-cocycle is zero,


then τ(x,w) = −(xϕ)w = −d0ϕ(x,w), i.e. τ is a 1-coboundary. From here, we

observe that two extensions are equivalent if the difference of their cocycles is a

coboundary.

Thus, for L = W1,M1 = T (α, γ),M2 = T (β, γ), τ(k,m) is a 1-cocycle of the

cohomology of W1 with the module Hom(T (β, γ), T (α, γ)). As the extension is

trivial if τ(k,m) = 0 then any cocycle equivalent to the zero cocycle will yield a

trivial extension. This leads to a natural concept of a trivial cocycle.

Definition 3.1.3. A 1-cocycle is trivial if it is a 1-coboundary.

This condition turns out to be equivalent to saying that two τ -functions are

equivalent if for some change of basis, we can obtain one τ -function from the other

one. Since we are interested in modules and not cocycles themselves, a change of

basis will not change our module and thus τ -functions that can be obtained from

a change of basis should be seen as equivalent.

3.2 A new module extension

Our final goal in this section is to obtain a basis for M such that our cocycles will

be polynomials in almost all cases. By this, we mean we want to find wk such that

wk → wk under the map from M → T (β, γ) which will admit polynomial cocycles.

The first thing to do is to lift our short exact sequence, or module extension, into

the setting of AW1-modules. To do this, we make use of the A-cover M of M .

From Theorem 4.10 of [5], the A-cover of M is cuspidal, so M has finite di-

mensional weight spaces. We have the map π : M → M such that π(φ(x, u)) =

φ(x, u)(1) = x.u. Define

M(α, γ) =

{∑i

φ(xi, ui)|∀f ∈ A,∑i

(fxi)ui ∈ T (α, γ)

}.


Lemma 3.2.1. M/M(α, γ) ∼= T (β, γ)

Proof. Consider π : M → T (β, γ) by π(φ(x, u)) = φ(x, u), where u is the image

of u under the map that sends M onto T (β, γ). Clearly, this map is surjective as

M = T (β, γ).

Then for an arbitrary∑

i φ(xi, ui) ∈ M ,

∑i

φ(xi, ui) ∈ kerπ ⇐⇒∑i

φ(xi, ui) = 0, in T (β, γ)

⇐⇒ ∀f ∈ A,∑i

φ(xi, ui)(f) = 0

⇐⇒ ∀f ∈ A,∑i

(fxi)ui = 0

⇐⇒ ∀f ∈ A,∑i

f(xi)u = 0

⇐⇒ ∀f ∈ A,∑i

(fxi)ui ∈ T (α, γ)

⇐⇒∑i

φ(xi, ui) ∈ M(α, γ)

So kerπ = M(α, γ).

Lemma 3.2.2. π is an homomorphism of AW1-modules.

Proof. As M → T (β, γ) is a homomorphism of W1-modules, then it is enough to

show that π is a homomorphism of A-modules.

Since tsφ(ek, u)(tm) = ek+s+mu = φ(tsek, u)(tm), it follows from (2.2.5) that

(fφ)(x, u) = φ(fx, u) for all f ∈ A, x ∈ W1, u ∈M . Then:

fπ(φ(x, u)) = fφ(x, u)

= φ(fx, u)

= π(φ(fx, u))

= π(fφ(x, u)


so that the A-action is preserved by π.

It follows from this Lemma that M(α, γ) is an AW1-module as it is the kernel

of a homomorphism of AW1-modules. We obtain the short exact sequence

0→ M(α, γ)→ M → T (β, γ)→ 0

We want to be able to relate this module back to T (β, γ) instead of T (β, γ) and

this new module extension will turn out to be too large. Instead, we will find a

submodule M of M that will admit a short exact sequence with T (β, γ) ⊂ T (β, γ).

The next two lemmas will help us relate T (β, γ) to T (β, γ). Define εj and ηj

in Hom(A, T (β, γ)) by

εj(tm) = wm+j

ηj(tm) = (m+ j)wm+j

Lemma 3.2.3. εj ∈ T (β, γ) for all j ∈ γ + Z, β 6= 1 and ηj ∈ T (β, γ) for all

j ∈ γ + Z, β 6= 0.

Proof. For β 6= 1, consider

1

1− β(φ(e0, wj)− φ(e1, wj−1)) ∈ T (β, γ)

where wj, wj−1 ∈ T (β, γ), j ∈ γ + Z.

Note that:

1

1− β(φ(e0, wj)− φ(e1, wj−1)) (tm) =

1

1− β((tme0).wj − (tme1).wj−1)

=1

1− β(em.wj − em+1.wj−1)


=1

1− β(j + βm− j + 1− β(m+ 1))wj+m

=1

1− β(1− β)wj+m

= wj+m

So that, as long as β 6= 1, εj ∈ T (β, γ) for all j ∈ γ + Z.

For ηj, first notice that the action of φ(ek, wj) on tm is given by:

ψ(ek, wj)(tm) = βηj+k(t

m) + (1− β)jεj+k(tm) (3.2.1)

Thus for β 6= 0, ηj =1

β(φ(ek, wj−k)− (1− β)(j − k)εj) and so ηj ∈ T (β, γ) as

long as β 6= 0.

Remark 3.2.4. The equation (3.2.1) tells us that T (β, γ) is spanned by the vectors

εj, ηj for j ∈ γ+Z when β 6= 1, and β 6= 0. In the special cases of β = 0 or β = 1,

T (β, γ) is spanned by vectors εj or ηj respectively.

Lemma 3.2.5. Let β 6= 1. The subspace of T (β, γ) spanned by {εi|i ∈ γ + Z} is

isomorphic to T (β, γ).

Proof. The action of W1 on εm is given by:

ek.εm = (m+ βk)εm+k

so that {εi|i ∈ γ + Z} is a submodule of T (β, γ).

Then, as long as β 6= 1, the map εi → wi is surjective and thus the claim

follows.

Consequently, we will denote span{εj|j ∈ Z} by T (β, γ), and view T (β, γ) as

a subspace of T (β, γ) when β 6= 1.


Now we define M = π−1(T (β, γ)). As T (β, γ) contains the zero element of

T (β, γ), M(α, γ) ⊆ M . Also T (β, γ) is an AW1-submodule of T (β, γ) so that, as

M is the homomorphic preimage of an AW1-module, it is itself an AW1-module.

From here, we have the first half of our short exact sequence. The only thing we

need is to find the module M/M(α, γ).

Lemma 3.2.6. π : M →M is surjective when β 6= 1 and (α 6= 1 or γ /∈ Z).

Proof. Notice that π : T (α, γ) → T (α, γ) will be surjective if π(T (α, γ)) =

W1T (α, γ) = T (α, γ) (by proposition 4.5 in [5]). This happens if and only if

α 6= 1 or γ /∈ Z:

For k 6= 0, e0.vk = kvk =⇒ 1

ke0.vk = vk

For k = 0, es.v−s = (−s+ αs)v0 =⇒ v0 ∈ Imπ ⇐⇒ α 6= 1 or γ /∈ Z

Then π : T (α, γ) → T (α, γ) is surjective unless α = 1 and γ ∈ Z so by

extending this map, M(α, γ) → T (α, γ) is also surjective. It is left to show that

π : M/M(α, γ)→ T (β, γ) is surjective.

By Lemma 3.2.1, π : M → T (β, γ) is surjective. As T (β, γ) ⊆ T (β, γ) for β 6= 1

and M = π−1(T (β, γ)), then it follows that π : M → T (β, γ) is surjective.

When α = 1, π will be surjective onto the extension given by the submodule

of T (1, 0):

0→ T ◦(1, 0)→M ′ → T (β, 0)

As we will see at the end of the chapter, these extensions will can be easily extended

to the original extension M .


Now, for β 6= 1, we have the short exact sequence

0→ M(α, γ)→ M → T (β, γ)→ 0

Here, all these modules have are cuspidal; this follows as M(α, γ) and M are

AW1-submodules of M . By definition, T (β, γ) has 1-dimensional weight spaces.

3.3 Submodules of Hom(A, T (α, γ)) with finite

dimensional weight spaces

The next proposition will be very important in showing that for a general case,

1-cocycles are polynomial functions. First, introduce the maps in Hom(A, T (α, γ))

for k ∈ γ + Z, i ∈ Z+,

θ(i)k : tm → (m+ k)i

i!vm+k

and

δk : tm →

vm+k, m = −k

0, m 6= −k

Remark 3.3.1. When α = β, θ(0)k = εk and θ

(1)k = ηk. We make the distinction

between these functions for convenience of notation later on.

Proposition 3.3.2. Any AW1-submodule in Hom(A,T(α, γ)) with finite dimen-

sional weight spaces is contained in a submodule spanned by:

1.{θ(0)k , · · · , θ(N)

k |k ∈ γ + Z}

for some N ∈ N, when α 6= 0

2.{θ(0)k , · · · , θ(N)

k , δk|k ∈ γ + Z}

for some N ∈ N, when α = 0

Proof. Suppose that ϕ ∈ Hom(A, T (α, γ)). Without loss of generality, suppose


that ϕ is an element of weight k and ϕ(tm) = amvm+k. We will show that the

function a(m) = am is a polynomial in m when α 6= 0.

First, we consider t−ieiϕ ∈ Hom(A, T (α, γ)). In this notation, t−ieiϕ =

t−i(eiϕ) which is not the same as (t−iei)ϕ. Since eiϕ is an element with weight

k + i, and t−i is an element of weight −i, t−ieiϕ is still an element of weight k in

Hom(A, T (α, γ)). Then (t−ieiϕ)(tm) = bmvm+k = b(m)vm+k. The functions a and

b are connected by the following relation.

bmvm+k = (t−ieiϕ)(tm)

= (eiϕ)(tm−i)

= ei(ϕ(tm−i))− ϕ(eitm−i)

= ei(am−ivm−i+k)− ϕ((m− i)tm−i−1+i+1

)= am−i(eivm−i+k)− (m− i)ϕ(tm)

= (m− i+ k − αi)am−ivm+k − (m− i)amvm+k

=⇒ bm = (m− i+ k − αi)am−i − (m− i)am

So that

b(m) = m(a(m− i)− a(m)) + (k − i− αi)a(m− i) + ia(m)

Define an action of t−iei on a as

(t−ieia)(m) = m(a(m− i)− a(m)) + (k − i− αi)a(m− i) + ia(m)

Now define:

zn =n+1∑i=0

(−1)i(n+ 1

i

)ti−1e1−i

yn =n∑i=0

(−1)i(n

i

)tie−i


Lemma 3.3.3. zn = zn−1 − yn for all n ∈ N.

Proof.

zn−1 − yn =n∑i=0

(−1)i(n

i

)ti−1e1−i −

n∑i=0

(−1)i(n

i

)tie−i

= t−1e1 +n−1∑i=0

(−1)i+1

(n

i+ 1

)tie−i +

n−1∑i=0

(−1)i+1

(n

i

)tie−i + (−1)n+1tne−n

= t−1e1 +n−1∑i=0

(−1)i+1

(n+ 1

i+ 1

)tie−i + (−1)n+1tne−n

=n+1∑i=0

(−1)i(n+ 1

i

)ti−1e1−i

= zn

We can give an explicit form of zn and yn by the following

(zna)(m) =n+1∑i=0

(−1)i(n+ 1

i

)(m(a(m+ i− 1)− a(m))

+ (k − 1 + i− α + αi)a(m+ i− 1) + (1− i)a(m))

(yna)(m) =n∑i=0

(−1)i(n

i

)(m(a(m+ i)− a(m))

+ (k + i+ αi)a(m+ i)− ia(m))

Since∑j

i=0(−1)i(ji

)= 0 and

∑ji=0(−1)i

(ji

)i = 0, this simplifies to

(zna)(m) =n+1∑i=0

(−1)i(n+ 1

i

)(m+ k + (α + 1)(i− 1))a(m+ i− 1)

(yna)(m) =n∑i=0

(−1)i(n

i

)(m+ k + i+ αi) a(m+ i)


Observe that a is a function from γ + Z to C. By assumption, ϕ in contained

in an AW1-module with finite dimensional weight spaces.

By Lemma 4 of [3], [z−1, zn] = −nzn for all n ∈ N, which means that zn will

be an eigenvector of adz−1. As zn ∈ M`×` for some ` ∈ N, adz−1 ∈ M`2×`2(C).

Thus adz−1 has at most `2 unique eigenvalues, so that zn can only be non-zero for

finitely many n ∈ N.

Let n = max{n ∈ N|zn 6= 0} if this set is non-empty, and set n = 1 if the set is

empty. Then zn = 0 for all n > n, then zn+1 is also zero for all n > n, so zn− zn+1

is zero for all n > n. Thus yn+1 is also zero. By shifting m to m− 1, we observe

(yn+1a)(m) =n+1∑i=0

(−1)i(n+ 1

i

)(m+ k + (α + 1)i− 1) a(m+ i− 1) (3.3.1)

and yn+1 is still the zero function. So

(0 · a)(m) =((zn − yn+1) · a)(m)

=n+1∑i=0

(−1)i(n+ 1

i

)(ma(m+ i− 1) + (k − 1 + i− α + αi)a(m+ i− 1)

−ma(m+ i− 1)− (k + i+ αi− 1)a(m+ i− 1))

=n+1∑i=0

(−1)i(n+ 1

i

)(k − 1 + i− α + αi− k + i+ αi− 1) a(m+ i− 1)

=−n+1∑i=0

(−1)i(n+ 1

i

)αa(m+ i− 1)

So, as long as α 6= 0,

0 =n+1∑i=0

(−1)i(n+ 1

i

)a(m+ i− 1) (3.3.2)

for all n > n and hence {am} satisfy recurrence relations.


This in turn tells us that each a is a polynomial in m by the use of Lemma 3

in [3], so that a(m) is in the span (m+ k)i for i = 1, . . . , N . Then it follows that

ϕ is in the span of{θ(0)m , · · · , θ(N)

m

}for some N ∈ N, where N = deg(a).

If α = 0, then

(zna)(m) =n+1∑i=0

(−1)i(n+ 1

i

)(m+ i− 1 + k)a(m+ i− 1)

Set d(m) = (m+ k)a(m) so that

(zna)(m) =n+1∑i=0

(−1)i(n+ 1

i

)d(m+ i− 1)

By the previous argument, d is a polynomial in m with a root at −k. So there

exists a polynomial g(m) such that d(m) = (m + k)g(m). Thus, a(m) and g(m)

agree on every integer except −k so that

a(m) = g(m) + cδm,−k (3.3.3)

for some c ∈ C, where g(m) = d(m)m+k

is a polynomial in m.

This suggests that it is possible that a could have a delta-function component.

It needs to be shown that if ϕ(tm) = δm,−kvm+k, then ϕ is contained in some

AW1-module with finite dimensional weight spaces, and so this case exists.

Define fk(tm) = δm,−kvm+k. Then

fk(tm) =

v0 ,m = −k

0 ,m 6= −k

It is enough to show that {fk|k ∈ γ+Z} as an AW1-module has finite dimensional

weight spaces.


The A action on fk:

(t`fk)(tm) = fk(t

m+`)

= δm+`,−kv0

= δm,−k−`v0

= fk+`(tm)

So t`fk = fk+` is the A action for all ` ∈ Z.

The W1 action on fs:

(e`fk)(tm) = e`(fk(t

m))− fk(e`tm)

= δm,−ke`v0 − fk(mtm+l`)

= −mδm+`,−kv0

= (k + `)δm,−k−`v0

= fk+`v0

So e`fk = (k + `)fk+` is the W1 action for all ` ∈ Z.

This shows us that this space has 1-dimensional weight spaces and hence, finite

dimensional weight spaces.

3.4 Finding an appropriate basis for M

From here, we would like to obtain a basis of M that will admit polynomial

cocycles. In particular, we would like to make use of the last proposition to show

that for β 6= 1, we can obtain a basis of M so that all possible 1-cocycles are of


the form described in the past proposition.

Let L be the Lie algebra spanned by the elements zi ddz

where i ∈ Z. Denote

L+ to be the subalgebra spanned by the elements zi ddz

where i ∈ Z and i ≥ 1.

Theorem 3.4.1 (Theorem 4.11 of [5]). If V is a cuspidal AW1-module with weights

in γ + Z for some γ ∈ C, then there exists a finite dimensional module (U, ρ) of

L+ such that

V ∼= A⊗ U

and with W1-action given by

ek(tm ⊗ u) = (m+ γ)tm+k ⊗ u+

∞∑i=1

ki

i!tm+k ⊗ ρ

(zid

dz

)u

for all k ∈ Z,m ∈ Z, u ∈ U .

We will assume for the rest of the chapter that β 6= 1. We will make use of this

theorem first on T (β, γ) and later on M . As shown in the last section, T (β, γ) is

spanned by the basis vectors εj, ηj for all j ∈ γ + Z when β 6= 0 and spanned just

by εj for β = 0. By computing the action using (2.2.2), the W1-action on these

elements is given by:

ekεj = (j + βk)εj+k

ekηj = (j + k(β − 1))ηj+k − k2(β − 1)εj+k

If εj = tj ⊗ ε, ηj = tj ⊗ η, then T (β, γ) ∼= A ⊗ U ′, where U ′ = 〈ε, η〉. By the

previous theorem, the W1 action of A⊗ U ′ is given by

ek(tj ⊗ u) = tj+k ⊗

(ju+

∞∑i=1

ki

i!ρ

(zid

dz

)u

)


so it is possible to derive the representation ρ:

ρ

(zd

dz

)ε = βε, ρ

(zid

dz

)ε = 0, ∀i ≥ 2

ρ

(zd

dz

)η = (β − 1)η, ρ

(z2d

dz

)η = −2(β − 1)ε

ρ

(zid

dz

)η = 0, ∀i ≥ 3

We can then write ρ as a matrix as we know its action on the basis of T (β, γ).

ρ

(zd

dz

)=

β 0

0 β − 1

, ρ

(z2d

dz

)=

0 −2(β − 1)

0 0

ρ

(zid

dz

)= 0, ∀i ≥ 3.

Define σ0 ∈ M such that

σ0 =1

1− β(φ(e0, wγ)− φ(e1, wγ−1)) (3.4.1)

where wγ−1, wγ ∈ M , the preimages of the basis vectors wγ−1, wγ ∈ T (β, γ) as at

the start of this chapter. Then

π(σ0) =1

1− β(φ(e0, wγ)− φ(e1, wγ−1))

=εγ

Thus σ0 ∈ π−1(T (β, γ)) = M .

Let σm = tmσ0 and define wm = π(σm) ∈M . It follows that

wm = π(σm)


= σm(1)

= (tmσ0)(1)

= σ0(tm)

As we have already concluded, M is a submodule of a cuspidal AW1-module,

and thus itself a cuspidal AW1-module. Thus, we can write M = A⊗ Mγ, and by

Theorem 3.4.1 the γ-weight space Mγ admits the action of L+. As shown below,

σ0 ∈ Mγ so that σ0 = 1⊗ σ.

e0σ0(tm) = e0(σ(tm))− σ0(e0tm)

=1

β − 1e0.(em.wγ − em+1wγ−1))−mσ0(tm)

= γσ0(tm)

Then

ekσ0 =ek(1⊗ σ) = tk ⊗

(∞∑i=1

ki

i!ρ

(zid

dz

)σ

)

=tk ⊗ (ku1 + k2u2 + · · ·+ knun)

(3.4.2)

for all k ∈ Z where ui =1

i!ρ

(zid

dz

)σ. As σ ∈ Mγ, each ui ∈ Mγ. Then

ekσ0 ∈ Mk+γ.

Then

ekwm = ek(σ0(tm))

= (ekσ0)(tm) + σ0(ekt

m)

= (tk ⊗ (ku1 + k2u2 + · · ·+ knun))(tm) + σ0(mtk+m)

= (1⊗ (ku1 + k2u2 + · · ·+ knun))(tm+k) +mσ0(tk+m)

= (m+ βk)wk+m + ζ(k,m)wk+m + τ(k,m)vk+m


We would like to show that ζ is the zero function and

ekwm = (m+ βk)wk+m + τ(k,m)vk+m

where τ(k,m) is a polynomial function in k and m.

Recall that π : M → T (β, γ) is a homomorphism with ker(π) = M(α, γ), and

M/M(α, γ) ∼= T (β, γ). As was proved before, π is an AW1-module homomorphism

from M ∼= A⊗ Mγ → A⊗ T (β, γ)γ∼= T (β, γ).

From here our main motivation is to make use of both Proposition 3.3.2

and equation (3.4.2). The problem is that the first result is in the context of

Hom(A, T (α, γ)) where as the second result is in the context of L+-modules. The

next two lemmas will give us a way to relate AW1-modules to L+-modules.

Lemma 3.4.2. Suppose ϕ : A⊗X → A⊗Y is a homomorphism of AW1-modules

for some L +-modules X, Y . Then the restriction map ϕ′ : 1 ⊗X → 1 ⊗ Y is a

homomorphism of L+-modules.

Proof. First notice that since ϕ is an A-module homomorphism and a W1-module

homomorphism,

ϕ(tm ⊗ x) = tm · ϕ(1⊗ x) = tm ⊗ ϕ′(x)

ϕ(ek(tm ⊗ x)) = ek(ϕ(tm ⊗ x)) (3.4.3)

From Theorem 3.4.1,

ek(tm ⊗ x) = tm+k ⊗

(mx+

∞∑i=0

ki

i!ρ

(zid

dz

)x

)


so that the left-hand side of (3.4.3) becomes

ϕ

(tm+k ⊗

(mx+

∞∑i=0

ki

i!ρ

(zid

dz

)x

))= tm+k ⊗ ϕ′

(mx+

∞∑i=0

ki

i!ρ

(zid

dz

)x

)

The right-hand side of (3.4.3) becomes

ek(tm ⊗ ϕ′(x)) = tm+k ⊗

(mϕ′(x) +

∞∑i=0

ki

i!ρ

(zid

dz

)ϕ′(x)

)

Then the following is true for all k ∈ Z:

ϕ′

((∞∑i=0

ki

i!ρ

(zid

dz

))x

)=

((∞∑i=0

ki

i!ρ

(zid

dz

))ϕ′(x)

)(3.4.4)

These will be polynomial functions since for all x ∈ L+, there exists n ∈ N

such that ρ(zi ddz

)x = 0. Since these functions are equal on all integer values then

necessarily they must be equal as polynomial functions. Therefore,

ϕ′(ρ

(zid

dz

)x

)= ρ

(zid

dz

)ϕ′(x),∀i ∈ Z,∀x ∈ X

Thus, ϕ′ is a homomorphism of L+-modules.

Now, by defining π′ : 1 ⊗ Mγ → 1 ⊗ T (β, γ)γ, by the above results, π′ is a

homomorphism of L+-modules so that π′ρ = ρ′ for the action ρ of L+ on U . π′

is surjective and we will now determine the kernel.

Lemma 3.4.3. M(α, γ) = A⊗ ker(π′)

Proof. Since M(α, γ) is the kernel of a homomorphism of AW1-modules, then it

is itself an AW1-module. Therefore, M(α, γ) ∼= A⊗ V for some L+-module V .


Take tm ⊗ v ∈ M(α, γ). Then π(tm ⊗ v) = 0 since M(α, γ) = ker(π). So

0 = π(tm ⊗ v)

= tm ⊗ π′(v)

=⇒ π′(v) = 0

so that V ⊆ ker(π′) which implies that A⊗ v ⊆ A⊗ ker(π′).

Take u ∈ ker(π′). Then π′(u) = 0 so that

π(tm ⊗ u) = tm ⊗ π′(u)

= tm ⊗ 0

= 0

so that A⊗ ker(π′) = A⊗ V = M(α, γ).

Using the map π′, we will be able to show that all but one of the terms in the

W1-action of ek on σ0 given in (3.4.2) will lie in a finite dimensional module of

M(α, γ). Proposition 3.3.2 will then come in handy to prove what kind of function

our cocycles could be.

Since 1⊗ π′(σ) = π(σ0) = εγ = 1⊗ ε, we see that π′(σ) = ε so that

π(ekσ0) = tk ⊗ π′(∞∑i=1

ki

i!ρ

(zid

dz

)ε

)

= tk ⊗

(∞∑i=1

ki

i!ρ

(zid

dz

)π′(ε)

)

= tk ⊗

(∞∑i=1

ki

i!ρ

(zid

dz

)ε

)

= tk ⊗ (βkε)


By looking at the formula (3.4.2),

tk ⊗ k(βε) = tk ⊗ (kπ′(u1) + k2π′(u2) + · · ·+ knπ′(un))

so that k2π′(u2) + · · · + knπ′(un) = 0 for all k ∈ Z. Since {k2, k3, . . . , kn} are

linearly independent, this implies that π′(ui) = 0 for 2 ≤ i ≤ n. But then

{ui|2 ≤ i ≤ n} ⊂ ker(π′) so that tk ⊗ ui ∈ M(α, γ) for 2 ≤ i ≤ n.

Equation (3.4.2) is reduced to

tk ⊗ k(βε) =tk ⊗ (kπ′(ui))

tk ⊗ 0 =tk ⊗ k(π′(ui)− βε)

We can conclude that tk⊗k(π′(u1)−βε) = tk⊗kπ′(ω) = 0 for some ω ∈ ker(π′).

Thus tk ⊗ ω ∈ M(α, γ) and π′(u1) = βε+ π′(ω).

Now,

π(ekσ0) = tk ⊗ (kβε+ kπ′(ω) + k2π′(u2) + · · ·+ knπ′(un))

Since tk ⊗ (kω + k2u2 + · · · + knun) ⊂ Mk+γ and are contained in M(α, γ),

they are contained in M(α, γ)k+γ. As M(α, γ) ⊂ Hom(A, T (α, γ)), we may apply

Proposition 3.3.2 to conclude that tk ⊗ (kω + k2u2 + · · · + knun) can be written

in the form τ(k) where τ(k) is a linear combination of{θ(0)k , . . . , θ

(N)k , δk

}. Notice

that τk ⊆ M(α, γ)k+γ.

Finally, we can say something about the W1-action on wm.

ekwm = ekπ(σm)

= ek(σm(1))


= ekσ0(tm)

= (ekσ0)(tm) + σ0(ekt

m)

= tk ⊗ (ku1 + k2u2 + · · ·+ knun)(tm) +mwk+m

= tk ⊗ kβε(tm) + τ(k)(tm) +mwk+m

= (m+ βk)wk+m + τ(k)(tm)

Notice that as for all k,m ∈ Z,

τ(k)(tm) = tk ⊗ (kω + k2u2 + · · ·+ knun)(tm)

= tk+m ⊗ (kω + k2u2 + · · ·+ knun)(1)

= π(tm+k ⊗ (kω + k2u2 + · · ·+ knun))

so that τ(k)(tm) ∈Mk+m+γ. Also τ(k) =∑N

i=0 ciθ(i)k so that

τ(k)(tm) =N∑i=0

ciθ(i)k+γ(t

m)

=N∑i=0

ci(k +m+ γ)i

i!vm+k+γ

=N∑i=0

ciθ(i)k+m+γ(t

0)

=N∑i=0

ciπ (θk+m+γ)

so that we may think of τ(k)(tm) as the image under π of a polynomial τ(k,m) in

k and θm+k+γ functions. Thus,

π(τ(k,m)) = τ(k,m)vm+k+γ


where τ(k,m) is a polynomial in k and m. We may shift m ∈ Z to m ∈ γ + Z to

simplify our notation.

Now we apply the surjective map from M to T (β, γ) which we denote by v → v

for v ∈M, v ∈ T (β, γ). Since τ(k,m) ∈ M(α, γ), using that π(M(α, γ)) ⊂ T (α, γ)

and π(T (β, γ)) ⊆ T (β, γ) we obtain

ekwm = (m+ βk)wm+k + π(τ(k,m))

= kβwk+m +mwk+m

This shows that ζ(k,m) = 0, so that ekwm = (m+βk)wk+m+τ(k,m)vk+m. We

can identify T (β, γ) in M by using basis vectors εm = tm⊗σ so that εm(1) = wm.

Then we find that the W1-action on M is given by

ekεm = (m+ βk)εk+m + τ(k,m)

where τ(k,m) is in a submodule in Hom(A, T (α, γ)) of the form in Proposition

3.3.2. Thus when β 6= 1, τ(k,m) is a polynomial in k and θk+m with the possibility

of a δk+m-function component when α = 0.

This is enough to show that our cocycles are polynomial. By Proposition

3.3.2, we know that τ(k) is contained in the submodule {θ(0)k , . . . , θ(N)k , δn} for

some N . Thus, M(α, γ) is contained in a module MN(α, γ) where MN(α, γ) is

the submodule spanned by {θ(0)k , . . . , θ(N)k , δn}. As M(α, γ) ⊆ MN(α, γ), we may

extend our module M to MN which is given by the extension

0→ MN(α, γ)→ MN → T (β, γ)→ 0

This will be convenient in the next section when we consider τ cocycles as we may


assume that θ(i)k is contained in our submodule for each i ∈ N. As M ⊂ MN , then

π : MN → M will be surjective for β 6= 1 and (α 6= 1 or γ /∈ Z), which will allow

us reclaim the cocycles from the original extension of M . Also, as θ(0)0 ∈ MN ,

θ(0)0 (t0) = v0 so that π : MN(α, γ) → T (α, γ) is surjective even when α = 1 and

γ = 0.

Chapter 4

Polynomial Cocycles

From the previous section, we found that for α 6= 0, β 6= 1, every 1-cocycle is

a polynomial function so that we can assume τ(k,m) =∑n

i=0 cikimn−i for some

ci ∈ C. First, we will work with τ(k,m) and then consider the homomorphism

π : MN →M to obtain the functions τ(k,m). In this section, we will derive some

general results about the coefficients ci of τ(k,m), then we will look at each case

of n ∈ N+ separately to find a classification of all polynomial 1-cocycles.

4.1 General results for polynomial cocycles

The W1-action on εm is given by

ek.εm = (m+ βk)εk+m +n∑i=0

s∑`=0

cik`θ

(i)k+m

where k, n, s ∈ Z and m ∈ γ + Z. The space of polynomials C[k,m] admits a

Z-grading, where homogeneous elements of degree n consist of monomials whose

powers of k and m sum to n. Using (3.1.2), we can see that each homogeneous

component of the polynomial τ(k,m) must independently satisfy the cocycle con-

39

CHAPTER 4. POLYNOMIAL COCYCLES 40

dition. Similarly, this is true for the coboundary condition given in Chapter 3.

For the cocycle τ(k,m), we may introduce an analogous idea of a homogeneous

element. A homogeneous element of degree n will consist of monomials k`θ(i)k+m

for which `+ i = n. In this way, we obtain a Z-grading on the cohomology space.

Again, each homogeneous component of τ(k,m) will independently satisfy (3.1.2)

and (4.1.7), so that it is enough to consider homogeneous cocycles

ek.εm = (m+ βk)εm+k +n∑i=0

cikn−iθ

(i)m+k (4.1.1)

By Theorem 3.4.1,

ek.εm = mεm+k +∞∑i=1

ki

i!

(ρ

(zid

dz

)ε

)(m+k)

(4.1.2)

ek.θ(p)m = mθ

(p)m+k +

∞∑i=1

ki

i!

(ρ

(zid

dz

)θ(p))

(m+k)

(4.1.3)

Remark 4.1.1. By comparing powers of k in (4.1.1) and (4.1.2), cn = 0 for every

m ∈ γ + Z.

Lemma 4.1.2. The W1-action on θ(p)j is given by

ek.θ(p)j =

p−1∑i=0

(−1)p−i

(p+ 1− i)!((p+ 1− i)α− (p+ 1)) kp+1−iθ

(i)j+k

+ (j + (α− p)k)θ(p)j+k

(4.1.4)

Proof. Using the W1-action defined on Hom(A, T (α, γ)) in Chapter 2,

(ekθ

(p)j

)(tm) =ek

(θ(p)j (tm)

)− θ(p)j (ekt

m)

=(m+ j)p

p!ek.vm+j −mθ(p)j (tm+k)

=(m+ j)p

p!(m+ j + αk)vm+j+k −m

(m+ j + k)p

p!vm+j+k


Then, by applying (4.1.4) to tm, we would like to show that this expression is

equal to (m+j)p

p!(m+j+αk)vm+j+k−m (m+j+k)p

p!vm+j+k. Let D denote the difference

of these two expressions. Then

D =

p−1∑i=0

(−1)p−i

(p+ 1− i)!((p+ 1− i)α− (p+ 1)) kp+1−i (m+ j + k)i

i!

+ (j + (α− p)k)(m+ j + k)p

p!− (m+ j)p

p!(m+ j + αk) +m

(m+ j + k)p

p!

p!D =

p−1∑i=0

(−1)p−i((

p

i

)α−

(p+ 1

i

))kp+1−i(m+ j + k)i

+ (j +m+ (α− p)k)(m+ j + k)p − (m+ j)p(m+ j + αk)

p!D =

p∑i=0

(−1)p−i((

p

i

)α−

(p+ 1

i

))kp+1−i(m+ j + k)i

− (α− (p+ 1))k(j + k +m)p + (j +m+ (α− p)k)(m+ j + k)p

− (m+ j)p(m+ j + αk)

p!D =k

p∑i=0

(−k)p−i(p

i

)α(−k)p−i(m+ j + k)i − (m+ j)p(m+ j + αk)

+

p∑i=0

(p+ 1

i

)(−k)p+1−i(m+ j + k)i + (j + k +m)p+1

p!D =k(m+ j)p − (m+ j)p(m+ j + αk) +

p+1∑i=0

(p+ 1

i

)(−k)p+1−i(m+ j + k)i

− (m+ j + k)p+1 + (j + k +m)p+1

p!D =(m+ j)p+1 − (m+ j + k)p+1 + (j + k +m)p+1 − (m+ j)p+1

p!D =0

Therefore, these expressions are equal.

Recall that a trivial cocycle is a cocycle in the equivalence class of the zero

function. In Chapter 3, we derived that a 1-coboundary is of the form −(ekϕ)(wi)


for ϕ(wi) = g(i)vi, wi ∈ M2, vi ∈ M1. For τ(k,m), ϕ ∈ Hom(T (β, γ), MN(α, γ))

such that ϕ(εi) = g(i)θ(p)i , so that

τ(k,m) =− (ekϕ)(εm)

=− ek.ϕ(εm) + ϕ(ek.εm)

=− g(m)ek.θ(p)m + (m+ βk)ϕ(εm+k)

=(m+ βk)g(m+ k)θ(p)m+k − g(m)ek.θ

(p)m

=m(g(m+ k)− g(m))θ(p)m+k + (βkg(m+ k)− (α− p)kg(m))θ

(p)m+k

−p−1∑i=0

(−1)p−i

(p+ 1− i)!((p+ 1− i)α− (p+ 1)) g(m)kp+1−iθ

(i)m+k

Generally, g(m) may have few restrictions. In particular though, we would like

to find when our polynomial cocycles are coboundaries. Thus, we would like to find

coboundaries that are homogeneous polynomials in k and θ(i)m+k. By comparing the

coefficient of θ(0)m+k for α 6= 1 on both sides of the above equality, this coefficient

must be a polynomial in k and independent of m. Thus for p ≥ 1, g(m) must be

a constant. If α = 1 and p ≥ 2, we may consider the coefficient of θ(1)m+k to come

to the same conclusion.

We still need to consider the case of p = 0 and (p = 1 and α = 1). The

coefficient of θ(p) is given by

(m+ βk)g(m+ k)− (m+ (α− p)k)g(m) (4.1.5)

This must be a polynomial in k. If we set m = γ then

(γ + βk)g(γ + k)− (γ + (α− p)k)g(γ)


As g(γ) is a constant, then (γ + βk)g(γ + k) must be a polynomial in k. Then

g(x) = q1(x)(βx−(1−β)γ) for some polynomial q1(x). If we set m = γ + 1, then

(γ + 1 + βk)g(γ + 1 + k)− (γ + 1 + (α− p)k)g(γ + 1)

As g(γ + 1) is a constant, then the first term must be polynomial in k. Then

g(x) = q2(x)(βx+(1−β)(γ+1))

for some polynomial in k. Thus,

q1(x)

(βx− (1− β)γ)=

q2(x)

(βx+ (1− β)(γ + 1))

so that either (βx−(1−β)γ) = (βx+(1−β)(γ+1)) otherwise g(x) is a polynomial

function in x. Let us look at the first case a bit further.

If these two factors are equal, then β = 1 and for some polynomial q(x) and

some constant c,

g(x) =

q(x)

x, x 6= 0

c, x = 0

so that

(γ + k)g(γ + k)− (γ + (α− p)k)g(γ) = q(γ + k)− (γ + (α− p)k)

γq(γ)

and thus α = p to obtain a polynomial coboundary. Notice for p = 1, α = 1 this is

already true and for p = 0, this requires that α = 0 for a non-polynomial function

g(x). This reduces (4.1.5) to

(m+ k)g(m+ k)−mg(m) = q(m+ k)− q(m) + δm,0q(m)− δm+k,0q(m+ k)

for (α = 0, p = 0) and (α = 1, p = 1) so that (4.1.5) is a polynomial function in m


and k with a possible difference of delta functions.

Here, we conclude that there are two different types of non-trivial coboundaries

that arise:

1. When g(m) is a polynomial function in m, it must be a constant. This

admits coboundaries that are homogeneous polynomials in k and θ.

2. When both α = 0 and β = 1, the coboundaries that arise are given by

q(m+ k)− q(m) + δm,0q(m)− δm+k,0q(m+ k)

which admits the delta-function coboundary

δm,0 − δm+k,0 (4.1.6)

Therefore for strictly homogeneous polynomial coboundaries, τ ∼ 0 if there

exists h ∈ C such that

τ(k, j) =h

p−1∑i=0

(−1)p−i

(p+ 1− i)!((p+ 1− i)α− (p+ 1)) kp+1−iθ

(i)j+k

+ hk(α− β − p)θ(p)j+k

(4.1.7)

We can obtain some more information about the representation ρ, which will

be a useful our calculations. From (4.1.1) - (4.1.4), we can derive the formulae:

ρ

(zd

dz

)θ(p) = (α− p)θ(p), (4.1.8)

ρ

(zid

dz

)θ(p) = (−1)i−1 (iα− (p+ 1)) θ(p+1−i), i ≥ 2 (4.1.9)

ρ

(zd

dz

)ε = βε+ cn−1θ

(n−1), (4.1.10)

ρ

(zid

dz

)ε = i!cn−iθ

(n−i), i ≥ 2 (4.1.11)


Also, since ρ is a representation,

[ρ

(zid

dz

), ρ

(zjd

dz

)]u = (j − i)ρ

(zi+j−1

d

dz

)u (4.1.12)

The next two results will determine the conditions on the coefficients ci using

(4.1.12).

Lemma 4.1.3. For all n ∈ N+, and for i, j ∈ Z+,

1. For i ≥ 2,

(α− β − (n− 1))cn−i = (−1)i−1(iα− n)cn−1 (4.1.13)

2. For i, j ≥ 2, i+ j ≤ n− 1, and i 6= j

(i+ j − 1)!(j − i)cn−i−j+1 =j!(−1)i−1(iα− (n− j + 1))cn−j

− i!(−1)j−1(jα− (n− i+ 1))cn−i

(4.1.14)

Proof. For part (1), set j = 1, u = ε so by (4.1.12),

LHS =ρ

(zid

dz

)ρ

(zd

dz

)ε− ρ

(zd

dz

)ρ

(zid

dz

)ε

=ρ

(zid

dz

)(βε+ cn−1θ

(n−1))− ρ(z ddz

)(i!cn−iθ

(n−i))=i!(−1)i−1 (iα− n) cn−1θ

(n−i) − i!(α− (n− i))cn−iθ(n−i)

+ βi!cn−iθ(n−i)

RHS =(1− i)i!cn−iθ(n−i)

=⇒ (α− β − (n− 1))cn−i = (−1)i−1 (iα− n) cn−1


For part (2), by using for i, j ≥ 2 in (4.1.12),

LHS =ρ

(zid

dz

)ρ

(zjd

dz

)ε− ρ

(zjd

dz

)ρ

(zid

dz

)ε

=ρ

(zid

dz

)(j!cn−jθ

(n−j))− ρ(zj ddz

)(i!cn−iθ

(n−i))=j!(−1)i−1 (iα− (n− j + 1)) cn−jθ

(n−i+1−j)

RHS =(i+ j − 1)!(j − i)cn−i−j+1

=⇒ (i+ j − 1)!(j − i)cn−i−j+1 = (−1)i−1j! (iα− (n− j + 1)) cn−j

− (−1)j−1i! (jα− (n− i+ 1)) cn−i

Theorem 4.1.4. For non-trivial 1-cocycles with n ≥ 3, cn−1 = 0 and α−β = n−1.

Proof. Suppose n ≥ 3 and suppose cn−1 is non-zero. Then by (4.1.13),

i!(α− β − (n− 1))cn−i 6= 0

for all i, and this (α− β − (n− 1)) is necessarily non-zero. Thus

cn−i =(−1)i−1(iα− n)

i!(α− β − (n− 1))cn−1

So that the W1 action on εm is

ek.εm =(m+ βk)εm+k + kcn−1θ(n−1)m +

n−2∑i=0

cikn−iθ(i)m

=(m+ βk)εm+k + kcn−1θ(n−1)m +

n∑i=2

cn−ikiθ(n−i)m


=(m+ βk)εm+k + kcn−1θ(n−1)m +

n∑i=2

(−1)i−1(iα− n)

i!(α− β − (n− 1))cn−1k

iθ(n−i)m

=(m+ βk)εm+k + kcn−1θ(n−1)m

+cn−1

(α− β − (n− 1))

n−2∑i=0

(−1)n−i−1((n− i)α− n)

(n− i)!kn−iθ(i)m

Take h =cn−1

α− β − (n− 1)∈ C. Then this 1-cocycle is equivalent to the trivial

1-cocycle. For non-trivial cocycles, cn−1 must be zero, so α− β − (n− 1) must be

zero as well if τ is non-trivial.

From this section, we obtain two crucial conditions for non-trivial 1-cocycles

of degree greater than two: cn−1 = 0, and α− β = n− 1.

4.2 Cases when n ≤ 4

When n ≤ 4, there does not exist positive integers i and j satisfying i ≥ 2, j ≥ 2

and i 6= j such that n − i − j + 1 is still positive. This implies that Part (2) of

Lemma 4.1.3 cannot be used. Instead, each case must be carefully looked at.

4.2.1 Cocycles of degree 1

When n = 1, we cannot use Part (1) of Lemma 4.1.3, since there does not exist

i ≥ 2 such that n − 2 ≥ 0. Equality (4.1.1) does give us the possible form of the

W1 action:

ek.εm = (m+ βk)εm+k + c0kθ(0)m+k

From this equation, we can derive the representation ρ:

ρ

(zd

dz

)=

α c0

0 β

, ρ

(zid

dz

)= 0, for i ≥ 2


For valid cocycles, ρ must satisfy the condition (4.1.12) but as[ρ(z ddz

), ρ(z ddz

)]=

0, this tells us that there are no restrictions on c0 as any value will give a valid

representation.

For trivial cocycles, by (4.1.7) h ∈ C must exist such that

c0kθ(0)m+k = hk(α− β)θ

(0)m+k

If α 6= β, then h =c0

α− β. When α = β, the equivalence relationship is simply

zero, so only cocycles strictly equal can be equivalent. Thus for any non-zero c0,

the cocycle is non-trivial, which gives an equivalence class of kθ(0).


When n = 2, Part (1) of Lemma 4.1.3 applies and we obtain the single equation

for i = 2:

i!(α− β − (n− 1))cn−i = (−1)i−1(iα− n)cn−1

By the proof of Theorem 4.1.4, if α− β − (n− 1) 6= 0, this cocycle will be trivial.

Thus, α − β = 1 and we get two cases when the above equation is equal to zero:

cn−1 = 0, or (iα− n) = 2α− 2 = 0, which is simply the case of α = 1.

Case of cn−1 = 0

When cn−1 = c1 = 0, by (4.1.1),

ek.εm = (m+ βk)εm+k + c0k2θ

(0)m+k


We can explicitly write out the representation ρ:

ρ

(zd

dz

)=

α 0 0

0 α− 1 0

0 0 β

, ρ

(z2d

dz

)=

0 −2α + 2 2c0

0 0 0

0 0 0

ρ

(zid

dz

)= 0, for i ≥ 3.

Since ρ is a representation, (4.1.12) holds for all positive integers i and j. When

i = 1, j = 2 we obtain that the following must hold:

0 −2α(α− 1) 2αc0

0 0 0

0 0 0

−

0 −2(α− 1)2 2βc0

0 0 0

0 0 0

=

0 −2α + 2 2c0

0 0 0

0 0 0

This is true for every α, β, c0 such that α− β = 1. The trivial cocycle is given by

(4.1.7) when p = n− 1 = 1, or

− (α− 1)k2θ(0)m+k (4.2.1)

so if α 6= 1, every cocycle is trivial by setting h =c0

α− 1. Therefore, from this

case we obtain an equivalence class of k2θ(0) for α = 1 and β = 0.

Case of α = 1

When α = 1, then β = 0. By (4.1.1),

ek.εm = mεm+k + c0k2θ

(0)m+k + c1kθ

(1)m+k


We again write out the representation ρ:

ρ

(zd

dz

)=

1 0 0

0 0 c1

0 0 0

, ρ

(z2d

dz

)=

0 0 2c0

0 0 0

0 0 0

ρ

(zid

dz

)= 0, for i ≥ 3.

Since ρ is a representation, (4.1.12) holds for all positive integers i and j. In

particular, when i = 1, j = 2,

0 0 2αc0

0 0 0

0 0 0

−

0 0 0

0 0 0

0 0 0

=

0 0 2c0

0 0 0

0 0 0

so that there are no restrictions on c0 of c1. The trivial cocycle is given by (4.2.1),

which is exactly zero when α = 1. This gives us an equivalence class of k2θ(0) +

kθ(1), which can be reduced to kθ(1) for α = 1 and β = 0 by using the previous

case.


For the case of n = 3, we can now make use of Theorem 4.1.4. For non-trivial

cocycles, α− β = 2, and c2 = 0. Then, by (4.1.1),

ek.εm = (m+ (α− 2)k)εm+k + c0k3θ

(0)m+k + c1k

2θ(1)m+k


We explicitly write out the representation ρ:

ρ

(zd

dz

)=

α 0 0 0

0 α− 1 0 0

0 0 α− 2 0

0 0 0 α− 2

ρ

(z2d

dz

)=

0 −2α + 2 0 0

0 0 −2α + 3 2c1

0 0 0 0

0 0 0 0

ρ

(z3d

dz

)=

0 0 3α− 3 6c0

0 0 0 0

0 0 0 0

0 0 0 0

ρ

(zid

dz

)= 0, for i ≥ 4.

Since ρ is a representation, (4.1.12) holds for all positive integers i and j. The

only non-trivial equations will be obtained from i = 1, j = 2 and i = 1, j = 3.

For any c0 and c1, (4.1.12) will hold so that any values of c0 and c1 will give valid

cocycles. All we need to do is find an element in the 1-dimensional non-trivial

solution space to find the equivalence class of non-trivial cocycles. The trivial

cocycle is given by (4.1.7) and is as follows:

(α− 1)k3θ(0)k+m − (2α− 3)k2θ

(1)k+m

For c0 = 1, and c2 = −2, the cocycle is not trivial. Thus we obtain an equivalence

class of k3θ(0) − 2k2θ(1).



The case of n = 4 is very similar to the previous case. For non-trivial cocycles,

α− β = 3, and c3 = 0. Then by (4.1.1),

ek.εm = (m+ (α− 3)k)εm+k + c0k4θ

(0)m+k + c1k

3θ(1)m+k + c2k

2θ(2)m+k

The matrix form of the representation ρ can be given by:

ρ

(zd

dz

)=

α 0 0 0 0

0 α− 1 0 0 0

0 0 α− 2 0 0

0 0 0 α− 3 0

0 0 0 0 α− 3

ρ

(z2d

dz

)=

0 −2α + 2 0 0 0

0 0 −2α + 3 0 0

0 0 0 −2α + 4 2c2

0 0 0 0 0

0 0 0 0 0

ρ

(z3d

dz

)=

0 0 3α− 3 0 0

0 0 0 3α− 4 6c1

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0


ρ

(z4d

dz

)=

0 0 0 −4α + 4 24c0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

ρ

(zid

dz

)= 0, for i ≥ 5.

which imply the conditions:

For i = 1, j = 2 : 2(α− β − 3)c2 = 0

For i = 1, j = 3 : 6(α− β − 3)c1 = 0

For i = 1, j = 4 : 24(α− β − 3)c0 = 0

For i = 2, j = 3 : 12(α− 1)c1 + 6(α− 1)c2 + 24c0 = 0

As α−β = 3, the first three conditions are trivial and by solving the last equation,

we obtain that cocycles are of the form

−(α− 1)

4(2c1 + c2)k

4θ(0)m+k + c1k

3θ(1)m+k + c2k

2θ(2)m+k

The 1-dimensional trivial solution space is given by (4.1.7) for p = n − 1 = 2, or

by the equivalence class of

(α− 1)k4θ(0)m+k − (3α− 4)k3θ

(1)m+k + 6(α− 2)k2θ

(2)m+k

Then the 1-dimensional non-trivial solution space is spanned by the cocycle where

2c1 = −c2. Thus, we obtain the equivalence class of k3θ(1) − 2k2θ(2).


4.3 Cases when n ≥ 5

As soon as n ≥ 5, Part (2) of Lemma 4.1.3 can be used. If we set i = 2, then

(j + 1)!(2− j)cn−j−1 − 2(−1)j−1(jα− n+ 1)cn−2 − j!(2α− n+ j − 1)cn−j = 0

for j = 3, 4, . . . , n− 2 so that we will obtain n− 3 formulas for appropriately large

n. If we set i = 3,

(j + 2)!(3− j)cn−j−2 − 6(−1)j−1(jα− n+ 2)cn−3 + j!(3α− n+ j − 1)cn−j = 0

for j = 4, 5, . . . , n − 2 so that we obtain n − 5 formulas for appropriately large

n. From this, we obtain the following coefficient matrix, where each column j

corresponds to the coefficient cj−1.

0 0 0 · · · 0 0 0 a1,n−4 a1,n−3 a1,n−2

0 0 0 · · · 0 0 a2,n−5 a2,n−4 0 a2,n−2

0 0 0 · · · 0 a3,n−6 a3,n−5 0 0 a3,n−2...

......

......

......

......

0 an−4,1 an−4,2 · · · 0 0 0 0 0 an−4,n−2

an−3,0 an−3,1 0 · · · 0 0 0 0 0 an−3,n−2

0 0 0 · · · 0 b1,n−6 0 b1,n−4 b1,n−3 0

0 0 0 · · · b2,n−7 0 b2,n−5 0 b2,n−3 0

......

......

......

......

...

0 0 bn−7,2 · · · 0 0 0 0 bn−7,n−3 0

0 bn−6,1 0 · · · 0 0 0 0 bn−6,n−3 0

bn−5,0 0 bn−6,2 · · · 0 0 0 0 bn−5,n−3 0


where, for 1 ≤ i ≤ n− 3,

ai,n−2 = 2(−1)i+2((i+ 2)α− (n− 1))

ai,n−2−i = −(i+ 2)!(2α− (n− i− 1))

ai,n−3−i = −i(i+ 3)!

and for 1 ≤ i ≤ n− 5,

bi,n−3 = 6(−1)i+3((i+ 3)α− (n− 2))

bi,n−3−i = (i+ 3)!(3α− (n− i− 2))

bi,n−5−i = −i(i+ 5)!

This matrix encodes a system of (n− 3) + (n− 5) = 2n− 8 equations in n− 1

variables, our n−1 coefficients. We are guaranteed to always have a 1-dimensional

trivial solution space, which means that the rank of our matrix is at most n− 2.

If there exists a non-trivial solution, then the rank would have to be less than or

equal to n− 3.


To demonstrate this, we can further reduce this matrix to

0 0 0 · · · 0 0 0 a1,n−4 a1,n−3 a1,n−2

0 0 0 · · · 0 0 a2,n−5 0 a2,n−3 a′2,n−2

0 0 0 · · · 0 a3,n−6 0 0 a4,n−3 a′3,n−2...

......

......

......

......

0 an−4,1 0 · · · 0 0 0 0 an−4,n−3 a′n−4,n−2

an−3,0 0 0 · · · 0 0 0 0 an−3,n−3 a′n−3,n−2

0 0 0 · · · 0 b1,n−6 0 b1,n−4 b1,n−3 0

0 0 0 · · · b2,n−7 0 b2,n−5 0 b2,n−3 0

......

......

......

......

...

0 0 bn−7,2 · · · 0 0 0 0 bn−7,n−3 0

0 bn−6,1 0 · · · 0 0 0 0 bn−6,n−3 0

bn−5,0 0 bn−6,2 · · · 0 0 0 0 bn−5,n−3 0

where, 2 ≤ i ≤ n− 3

a1,n−2 =− 2(3α− (n− 1))

a1,n−3 =− 6(2α− (n− 2))

a1,n−4 =− 24

a′i,n−2 =2(−1)i

(i− 1)!

([i+1∑j=3

(j − 3)!(jα− (n− 1))i+1∏`=j

(2α− (n− `))

]

+ (i− 1)!((i+ 2)α− (n− 1))

)

ai,n−3 =6(−1)i

(i− 1)!

i∏j=1

(2α− (n− (j + 1)))

ai,n−3−i =− i(i+ 3)!


and for 1 ≤ i ≤ n− 5

bi,n−3 =6(−1)i+3((i+ 3)α− (n− 2))

bi,n−3−i =(i+ 3)!(3α− (n− 2− i))

bi,n−5−i =− i(i+ 5)!

Finally,

0 0 0 · · · 0 0 0 a1,n−4 a1,n−3 a1,n−2

0 0 0 · · · 0 0 a2,n−5 0 a2,n−3 a′2,n−2

0 0 0 · · · 0 a3,n−6 0 0 a4,n−3 a′3,n−2...

......

......

......

......

0 an−4,1 0 · · · 0 0 0 0 an−4,n−3 a′n−4,n−2

an−3,0 0 0 · · · 0 0 0 0 an−3,n−3 a′n−3,n−2

0 0 0 · · · 0 0 0 0 b′1,n−3 b1,n−2

0 0 0 · · · 0 0 0 0 b′2,n−3 b2,n−2...

......

......

......

......

0 0 0 · · · 0 0 0 0 b′n−7,n−3 bn−7,n−2

0 0 0 · · · 0 0 0 0 b′n−6,n−3 bn−6,n−2

0 0 0 · · · 0 0 0 0 b′n−5,n−3 bn−5,n−2

where, for 2 ≤ i ≤ n− 3,

a1,n−2 =− 2(3α− (n− 1))

a1,n−3 =− 6(2α− (n− 2))

a1,n−4 =− 24


a′i,n−2 =2(−1)i

(i− 1)!

([i+1∑j=3

(j − 3)!(jα− (n− 1))i+1∏`=j

(2α− (n− `))

]

+ (i− 1)!((i+ 2)α− (n− 1))

)

ai,n−3 =6(−1)i

(i− 1)!

i∏j=1

(2α− (n− (j + 1)))

ai,n−3−i =− i(i+ 3)!

and for 1 ≤ i ≤ n− 5,

bi,n−2 =

((3α− (n− 2− i))

i

)a′i,n−2 −

(i

i+ 2

)a′i+2,n−2

b′i,n−3 =6(−1)i+3((i+ 3)α− (n− 2))−(

i

i+ 2

)ai+2,n−3

+

((3α− (n− 2− i))

i

)ai,n−3

which gives us the explicit formulae for 2 ≤ i ≤ n− 5

b1,n−2 =1

3(−n3 + 7αn2 + 3n2 − 16α2n− 9αn− 2n+ 12α3 + 6α)

b′1,n−3 = −n3 + 6αn2 + 3n2 − 12α2n− 6αn− 2n+ 8α3 + 4α

bi,n−2 =2(−1)i

i!

[i

(i+ 1)(i+ 2)

(i+3∑j=1

(j − 3)!(jα− n+ 1)i+3∏`=j

(2α− (n− `))

)

+ (3α− n+ 2 + i)

(i+1∑j=1

(j − 3)!(jα− (n− 1))i+1∏`=j

(2α− (n− `))

)

(i− 1)!((i+ 2)α− n+ 1)(3α− n+ 2 + i) +i(i+ 1)!

(i+ 2)(i+ 1)((i+ 5)α− n+ 1)

]

b′i,n−3 =6(−1)i+1

(i+ 2)!

[(i+ 2)!

((i+ 3)α− n+ 2) +

i∏l=j

(2α− n+ j + 1)

)

× (i(2α− n+ i+ 2)(2α− n+ i+ 3)− (i+ 1)(i+ 2)(3α− n+ 2 + i))

]


Remark 4.3.1. This last matrix shows the rank is always at least n − 3, which

means there is always at most a 1-dimensional non-trivial solution space.


When n = 5, we only get two formulae and thus our complete coefficient matrix

is only a 2×2 matrix, so the rank is automatically less than or equal to n−3 = 2.

Thus there exists a non-trivial solution, and we simply solve the matrix to find

a solution then solve for the trivial cocycle and find a satisfactory cocycle not

equivalent to the trivial cocycle, as done in the previous sections. This gives an

equivalence class of:

(α− 1)k5θ(0) − k4θ(1) − 12k3θ(2) + 24k2θ(3), for α− β = 4.


When n = 6, the rank of the coefficient matrix will only be (n− 3) when b′1,n−3 =

b1,n−2 = 0. This means there is only a non-trivial solution where both polynomial

coefficients (4.3.1), (4.3.2) are zero at the same time.

b1,n−2 =1

3(−n3 + 7αn2 + 3n2 − 16α2n− 9αn− 2n+ 12α3 + 6α) (4.3.1)

b′1,n−3 = −n3 + 6αn2 + 3n2 − 12α2n− 6αn− 2n+ 8α3 + 4α (4.3.2)

Solving for the roots of these polynomials, we get two shared roots of the form

α =n±√−2 + 3n

2which reduces to α = 5 and α = 1. Similar to the methods

used in the previous section, there is a 1-dimensional non-trivial solution space


given by the equivalence classes of:

2k5θ(1) + 10k4θ(2) + 60k3θ(3) − 120k2θ(4), for α = 5, β = 0.

12k6θ(0) − 22k5θ(1) + 10k4θ(2) + 60k3θ(3) − 120k2θ(4), for α = 1, β = −4.


When n = 7, the rank of the coefficient matrix will equal (n − 3) when b′1,n−3 =

b1,n−2 = b′2,n−3 = b2,n−2 = 0. This means, there is only a non-trivial solution space

when all four polynomial coefficients are zero at the same time.

b2,n−2 = −1

6(24α4 + 60α3 − 44α3n+ 12α2 − 98α2n+ 30α2n2

+ 30α− 55αn+ 50αn2 − 9αn3 − 10n+ 17n2

− 8n3 + n4)

(4.3.3)

b′2,n−3 = −1

2(− 10n+ 20α + 8α2 − 38αn+ 17n2 − 72α2n

+ 42αn2 + 40α3 − 8n3 − 32α3n− 8αn3 + 24α2n2

+ 16α4 + n4)

(4.3.4)

Solving for the roots, (4.3.1), (4.3.2), (4.3.3), and (4.3.4) all share common roots

of α =n±√−2 + 3n

2. This gives 2 cases of non-trivial cocycles

For α =7 +√

19

2:− 22 +

√19

4k7θ(0) +

31 + 7√

19

2k6θ(1)

− (25 + 7√

19)k5θ(2) + 30k4θ(3) + 120k3θ(4) − 240k2θ(5)

For α =7−√

19

2:− 22−

√19

4k7θ(0) +

31− 7√

19

2k6θ(1)

− (25− 7√

19)k5θ(2) + 30k4θ(3) + 120k3θ(4) − 240k2θ(5)


4.3.4 Cocycles of degree greater than or equal to 8

When n ≥ 8, the system will only admit non-trivial cocycles when

b′1,n−3 = b1,n−2 = b′2,n−3 = b2,n−2 = b′3,n−3 = b3,n−2 = · · · = b′n−5,n−3 = bn−5,n−2 = 0

If there exists a common solution to these formulae, then this solution must satisfy

all of the first four formulae, which means that the solution must necessarily be

of the form

α =n±√−2 + 3n

2

This is not a solution to the 5th or 6th formulae (4.3.5), (4.3.6) as shown in (4.3.7)

and (4.3.8). This means the rank of the matrix is always at least (n− 2) and thus

this case never admits a non-trivial solution space.

b3,n−2 =1

60

(− 172n+ 516α + 360α2 − 1080αn+ 320n2

− 2060α2n+ 1115αn2 + 1140α3 − 185n3 − 1440α3n

− 336α4n+ 312α3n2 − 144α2n3 + 33αn4 − 340αn3

+ 1060α2n2 + 144α5 − 3n5 + 720α4 + 40n4)

(4.3.5)

b3,n−3 =1

20

(− 172n+ 344α + 240α2 − 760αn+ 320n2

− 1500α2n+ 930αn2 + 760α3 − 185n3 − 1040α3n

− 240α4n+ 240α3n2 − 120α2n3 + 30αn4 − 300αn3

+ 840α2n2 + 96α5 − 3n5 + 480α4 + 40n4)

(4.3.6)


At α =n±√−2 + 3n

2,

b3,n−2 =1

40

(− 2n∓ 6

√3n− 2± 9n

√3n− 2

+ 3n2 ∓ 3n2√

3n− 2− n3) (4.3.7)

b3,n−3 =± 3√

3n− 2

20− (n− 1)! (4.3.8)

4.4 Dual Modules

The dual vector space of a vector space V is defined as

V ∗ = {ϕ : V → C;ϕ is linear}

or V ∗ is the set of all linear functionals on V .

If (V, ρ) is a module for a Lie algebra L, then its dual vector space V ∗ admits

the action of L in the following way:

(ρ∗(x)ϕ)v = −ϕ(ρ(x)v) (4.4.1)

where ρ∗ is the action on the dual vector space.

When V = T (α, γ), T (α, γ)∗ contains the elements

v∗m : T (α, γ)→ C, v∗m(vn) = δm,n

For finite dimensional vector spaces, the dual vector space is isomorphic to the

original space. In infinite dimensions though, the dual vector space is, in general,

a much larger space. In this case, the dual vector space is larger than we want,


so we introduce the restricted dual of T (α, γ), defined as T (α, γ)∗ = ⊕m∈γ+ZCv∗m.

From now on, when we refer to the dual space, we mean the restricted dual space.

First, we would like to shift the basis of T (α, γ)∗ to behave more like the

original space. Let vm = v∗−m. As m ∈ γ + Z, then the basis vm is indexed by

−γ + Z. The W1-action on this basis is given by

(ek.vm)vm =− vm(ekvs)

=− (s+ αk)v∗−m(vs+k)

=− (s+ αk)δ−m−s−k,0

=− (−m− k + αk)δ−m−s−k,0

=(m+ (1− α)k)vm+kvs

=⇒ ekvm =(m+ (1− α)k)vm+k

As W1-modules, T (α, γ)∗ ∼= T (1−α,−γ) by the isomorphism vm = v∗−m → vm.

For the short exact sequence

0→ T (α, γ)→M → T (β, γ)→ 0

with the cocycle τ(k,m), we can compare the dual spaces of these modules. Then

ekvm =(m+ (1− α)k)vm+k − τ(k,−m− k)wm+k

ekwm =(m+ (1− β)k)wm+k

As a consequence, we can relate the module extensions of the dual modules to

module extensions of the original modules. The short exact sequence

0→ T (1− β,−γ)→M∗ → T (1− α,−γ)→ 0


admits the cocycle τ ∗(k,m) = −τ(k,−m− k). Thus from finding one cocycle, we

can easily derive another one by looking at the dual extension.

It is of note that for our equivalence classes of polynomial cocycles, the cases

that give a range of possible values for α admit dual cocycles in the same equiv-

alence class. The cases of degree 2, degree 6 and degree 7 cocycles do not admit

a range of values for α, but only have non-trivial cocycles for exactly two values

of α. In these cases, the dual cocycle is not in the same equivalence class as the

original cocycle, but is of the same degree. Thus the two cases of degree 2 cocycles

are dual to each other, the two cases of degree 6 cocycles are dual to each other

and the two cases of degree 7 cocycles are also dual to each other.

Chapter 5

Delta Cocycles

As discussed in Chapter 3, our cocycles may have a delta function of δk+m,0 when

γ ∈ Z and α = 0. In this chapter, we will assume that α = 0 and γ ∈ Z and that

our τ -functions are delta functions. These cocycles will classify extensions of the

form

0→ D(0, 0)→M → T (β, 0)→ 0

where D(0, 0) is the 1-dimensional trivial submodule of T (0, 0) as defined in Chap-

ter 2. As D(0, 0) ⊂ T (0, 0), this extension will admit an extension

0→ T (0, 0)→M ′ → T (β, 0)→ 0

In this section, we will find these delta cocycles in the context M ′ rather than

MN , where δk(tm) = δk+m,0vm+k.

These delta cocycles will be of the form

τ(k,m) = δm+k,0f(k,m) = δm+k,0f(k,−k)

Thus, f is a function of k. In fact, by Theorem 3.4.1 for ek.w−k and β 6= 1, f(k)

65

CHAPTER 5. DELTA COCYCLES 66

must be a polynomial function in k. As we will deal with the special case of β = 1

in the next chapter, we will assume that f(k) is a polynomial in k for every β for

now.

The dual spaces also admit a delta function cocycle, given by the short exact

sequence

0→ T (1− β, 0)→M → T (1, 0)→ 0

where τ(k,m) = δm,0g(k).

We will first handle the case that α = 0, then see what results we can obtain

from the dual case.

5.1 Conditions for delta functions

As derived in Chapter 3, cocycles must satisfy the relation (3.1.2) so for τ(k,m) =

δk+m,0f(k),

(s− k)δk+m+s,0f(k + s) =(m+ s)δs+m,0f(s)− (m+ k)δk+m,0f(k)

+ (m+ βs)δk+m+s,0f(k)− (m+ βk)δk+m+s,0f(s)

This reduces to:

(s− k)f(k + s) = ((β − 1)s− k)f(k)− ((β − 1)k − s)f(s) (5.1.1)

Recall a trivial cocycle is a 1-coboundary which becomes the zero function

under some change of basis. Let um = wm+ cδm,0v0 be a new basis for the module

M . Then

ekum = (m+ βk)wm + cδm,0ekv0


= (m+ βk)wm

= (m+ βk)wm + (m+ βk)δm+k,0cv0 − (m+ βk)δm+k,0cv0

= (m+ βk)um + k(1− β)δm+k,0cv0

Thus τ(k,m) is trivial if there exists a constant c ∈ C such that

τ(k,m) = ck(1− β)δk+m,0 (5.1.2)

To solve for cocycles τ(k,m) = δk+m,0f(k), we know that the function f(k)

must be polynomial so that for some n ∈ N, f(k) is given by

f(k) = ankn + an−1k

n−1 + · · ·+ a1k + a0.

Equation (5.1.1) is homogeneous in s and k so each f(k) = km can be treated

separately.

When n = 0, we only need to solve a simple linear equation for a0 using (5.1.1):

(s− k)a0 = ((β − 1)s− k)a0 − ((β − 1)k − s)a0

= (βs− s− k − βk + k + s)a0

= β(s− k)a0

So f(k) = a0 is a valid function only when β = 1. Since

a0δk+m,0 = ck(1− β)δk+m,0 ⇐⇒ a0 = 0

then δk+m,0 is non-trivial for β = 1. This gives an equivalence class of δk+m,0 for

β = 1.


When n = 1, f(k) = k so that

(s− k)(k + s) = ((β − 1)s− k)k − ((β − 1)k − s)s

= (s− k)(s+ k)

which holds for all β, but is only non-trivial when β = 1. This gives an equivalence

class of δk+m,0k for β = 1.

When n = 2, f(k) = k2 so that

(s− k)(k + s)2 = ((β − 1)s− k)k2 − ((β − 1)k − s)s2

= (β)(sk2 − s2k) + (s3 + s2k − sk2 − k3)

which holds only if β = 0. This gives an equivalence class of δk+m,0k2 for β = 0.

When n = 3, f(k) = k3 so that

(s− k)(k + s)3 = ((β − 1)s− k)k3 − ((β − 1)k − s)s3

= (β + 1)(sk3 − s3k) + (s4 − 2sk3 + 2ks3 − k4)

which holds only if β = −1. This gives an equivalence class of δk+m,0k3 for β = −1.

When n ≥ 4, by (5.1.1) we can obtain the system

0 = (s− k)(k + s)n − ((β − 1)s− k)kn + ((β − 1)k − s)sn

The coefficient at sn−1k2 n ≥ 4 will be given by:

((n

2

)−(n

1

))=

1

2n(n− 1)− n 6= 0 when n 6= 3 (5.1.3)

Thus, this coefficient is non-zero for any value of β. Therefore this system will

only admit the trivial cocycle as a solution.


We obtain the equivalence classes of:

δk+m,0, when β = 1 (5.1.4)

δk+m,0k, when β = 1 (5.1.5)

δk+m,0k2, when β = 0 (5.1.6)

δk+m,0k3, when β = −1 (5.1.7)

This last cocycle δk+m,0k3 for β = −1 corresponds to the Virasoro cocycle, the

central extension of the Witt algebra described in Chapter 2. In this equivalence

class is δm+k112

(k3 − k), the more common form of the Virasoro cocycle.

Now, we can look at the dual extension.

0→ T (1− β, 0)→M → T (1, 0)→ 0

As was shown at the end of the previous section, τ ∗(k,m) = τ(k,−m− k) so that

if τ(k,m) = δk+m,0f(k), τ ∗(k,m) = δk−m−k,0f(k) = δm,0f(k) when β = 1. This

gives us the following equivalence classes:

δm,0, when α = 0, β = 1 (5.1.8)

δm,0k, when α = 0, β = 1 (5.1.9)

δm,0k2, when α = 1, β = 1 (5.1.10)

δm,0k3, when α = 2, β = 1 (5.1.11)

Notice here that when α = 0, β = 1, we obtain both cocycles δk+m,0 and δm,0.

As these two differ by the coboundary δk+m,0 − δm,0 (4.1.6), these cocycles are

contained in the same equivalence classes. Thus, (5.1.8) is not a new equivalence

class.

Chapter 6

Special case of β = 1

The approach used in Chapter 3 to find conditions on our cocycles depended on the

fact that β 6= 1. Recall that this method did not work because our homomorphism

from T (β, γ)→ T (β, γ) was not injective for β = 1, so that there was no obvious

way in which to define MN . In particular, εj /∈ T (1, γ) for any j ∈ γ+Z. We deal

with this case here, using a similar approach by defining a basis of M using the

A-cover M .

From Chapter 4, we found all possible polynomial cocycles including the case

of β = 1. Now, we will assume they are strictly not polynomial and see what

cocycles arise from this case.

6.1 Conditions on non-polynomial cocycles

We would like to find an appropriate basis of M so that we can determine condi-

tions on cocycles. First, introduce σ0 ∈ M as σ0 = φ(e0, wγ). Then

σ0(tm) = φ(e0, wγ)(t

m) = em.wγ = (m+ γ)wm+γ + τ(m, γ)vm+γ

70

CHAPTER 6. SPECIAL CASE OF β = 1 71

for all m ∈ Z, k ∈ Z.

Now, introduce a new basis of M by

wm+γ =1

m+ γσ0(t

m), for m+ γ 6= 0

Note that in the case of γ ∈ Z, we have a basis for the module extension given

by the short exact sequence

0→ T (α, 0)→M ′ → T ◦(1, 0)→ 0

By Theorem 3.4.1, M ∼= A ⊗ Mγ for some finite representation Mγ of L+.

Then σ0(tm) = tm ⊗ σ for some σ ∈ Mγ. For m+ γ 6= 0,

ek.wm+γ =1

m+ γek.(σ0(t

m))

=1

m+ γek.(t

m ⊗ σ)

=1

m+ γ

((m+ γ)tm+k ⊗ σ + ku1(t

m+k) + · · ·+ knun(tm+k))

=1

m+ γ

((m+ γ)σ0(t

m+k) + ku1(tm+k) + · · ·+ knun(tm+k)

)= (m+ k + γ)wm+k+γ +

1

m+ γ

(ku1(t

m+k) + · · ·+ knun(tm+k))

for all k ∈ Z, where ui =1

i!ρ

(zid

dz

)σ as in Theorem 3.4.1. Note that

for γ ∈ Z, σ0(1) = τ(0, 0)v0 ∈ M(α, 0). Then the image of σ0 under the map

π : M → T (1, γ) is simply zero since

(ekπ(σ0))(tm) = (ekψ(e0, wγ))(t

m)

= ek(ψ(e0, wγ)tm)− ψ(e0, wγ)(ekt

m)

= (m+ γ)(m+ γ + k)wm+γ+k − (m+ γ)(m+ γ + k)wm+γ+k

= 0, ∀m ∈ γ + Z


so that

ekφ(e0, wγ) = 0 =⇒ φ(e0, wγ) = 0

and so π(σ0) ∈ M(α, γ). This shows that ku1 + · · ·+ knun ∈ M(α, γ), so that for

m ∈ γ+Z, the cocycle τ(k,m) =1

m(ku1 + · · ·+knun), mτ(k,m) is contained in a

submodule of the form described in Proposition 3.3.2. In other words, this cocycle

will be a polynomial function or a δm+k,0 function when α = 0, with a factor of

m−1. Notice here that we may apply Theorem 3.4.1 to obtain that if a cocycle is

given by 1mδm+k,0f(k), f(k) is polynomial in k.

The function π(τ) = τ ′(k,m) is not defined when m = 0. This τ -function

completely determines the cocycle on the short exact sequence with T ◦(1, 0). To

extend this cocycle onto M , we introduce a possible δm,0 function to determine

the cocycle on the basis vector w0.

ekw0 = kwk + µ(k, 0)vk

Note that µ(k, 0) is simply a function on k, and so we will use the notation µ(k).

We can extend the cocycle to the whole space of M by setting τ(k,m) to be

the piecewise function:

τ(k,m) =

τ ′(k,m), m 6= 0

µ(k), m = 0

These two components are independent so that it admits two cases, when the

cocycle is of the form τ ′(k,m) or when τ(k,m) = δm,0µ(k). Hence, for β = 1,

cocycles in M can be polynomial functions in k with a possible factor of m−1,

δk+m,0 functions with a possible factor of m−1 or δm,0 functions.


6.2 Cocycles with a factor of m−1

Suppose that τ ′(k,m) = m−1µ(k,m) for k ∈ Z,m ∈ γ + Z. If this is a cocycle

then by (3.1.2),

(s− k)m−1µ(k + s,m) =(m+ s+ αk)m−1µ(s,m) + (m+ s)(m+ s)−1µ(k,m+ s)

− (m+ k + αs)m−1µ(k,m)− (m+ k)(m+ k)−1µ(s,m+ k)

(s− k)µ(k + s,m) =(m+ s+ αk)µ(s,m) +mµ(k,m+ s)

− (m+ k + αs)µ(k,m)−mµ(s,m+ k)

So this reduces to the case that µ(k,m) is a cocycle for β = 0.

In the case that α = 0 and µ(k,m) = δk+m,0f(k) there are two cocycles: the

trivial cocycle δk+m,0k and the nontrivial cocycle δk+m,0k2.

δk+m,0m−1k = δm+k

k

−k= −δk+m,0

This first case gives a cocycle of the form δk+m,0 which was found to be non-trivial

in Chapter 5.

δk+m,0m−1k2 = δk+m,0

k2

−k= −kδk+m,0

The second case is exactly the equivalence class of δk+m,0k found in Chapter 5.

In the case that µ(k,m) is a polynomial function, then we would like to see

when this cocycle is a coboundary. Then there exits some change of basis of M

such that the cocycle becomes trivial. In other words, there exists g : γ + Z→ C

such that

g(m)(m+ αk)− g(m+ k)(m+ k) =1

mµ(k,m)


But if kg(k) = f(k), then this reduces to

mg(m)(m+ αk)−mg(m+ k)(m+ k) = µ(k,m)

f(m)(m+ αk)− f(m+ k)m = µ(k,m)

which is the same condition for µ(k,m) being a non-trivial polynomial cocycle for

β = 0. In Chapter 4, we described all polynomial cocycles in MN . In Chapter 7,

we describe the resulting polynomial cocycles in M . Then using Table 7.2, µ(k,m)

can be any of the following:

α = 0, k

α = 1, km

α = 1, k2

α = 2, k3 + k2m

α = 3, k3m+ k2m2

α = 4, k4m− 6k3m2 − 4k2m3

α = 5, 2k5m− 5k4m2 + 10k3m3 + 5k2m4

Most of these cases will reduce to equivalence classes from Chapter 4. Four cases

admit new partial polynomial cocycles, giving the equivalence classes

α = 0, β = 1, m−1k

α = 1, β = 1, m−1k2

α = 2, β = 1, m−1k3 + k2

Remark 6.2.1. These functions are not polynomial but can be considered such

under some change of basis.

Consider the homomorphism from T (0, γ) → T (1, γ) by vm → mwm. For


γ /∈ Z, this map is bijective so that T (0, γ) ∼= T (1, γ). In the case that γ ∈ Z, the

image of this map is T ◦(1, 0) and the kernel is Cv0 = D(0, 0) so that

T (0, 0)/D(0, 0) ∼= T ◦(1, 0)

But as these τ ′-functions zero at m = 0, they can be considered cocycles defined

on T ◦(1, 0).

By applying this map to wm ∈ T (1, γ) to w′m ∈ T (0, γ),

ek.wm = (k +m)wm+k +1

mµ(k,m)vm+k

mek.wm = m(k +m)wm+k + µ(k,m)vm+k

ek.w′m = mw′m + µ(k,m)vm+k

and µ(k,m) is a polynomial cocycle. Although these module extensions are not

identical, they are equivalent under some change of basis so that these cocycles

can be thought of as polynomials in some sense.

6.3 Delta cocycles

As in Chapter 5, we will consider delta functions of the form δm,0f(k,m). This

reduces to f(k,m) = µ(k) where µ is polynomial in k. If we take the change of

basis w′0 = w0 + v0, then we find that δm,0αk is the trivial cocycle.

Now, we take τ(k,m) = δm,0µ(k) so that the W1-action on M is given by

ekw0 =kwk + µ(k)vk

ekwm =(m+ k)wm+k


By the cocycle condition (3.1.2),

(k − s)µ(k + s) + (k + αs)µ(k)− (s+ αk)µ(s) = 0 (6.3.1)

Using this condition, we may derive the following equations:

For k = 0, s = 1 : αµ(0) = 0

For k = 1, s = 2 : µ(3) = (2 + α)µ(2)− (1 + 2α)µ(1)

For k = 1, s = 3 : µ(4) = 12(3 + α)(2 + α)µ(2)− (2 + 5α + α2)µ(1)

For k = 1, s = 4 : µ(5) = 16(4 + α)(3 + α)(2 + α)µ(2)− 1

3(9 + 26α + 9α2 + α3)µ(1)

For k = 2, s = 3 : µ(5) = (4 + 4α + 2α2)µ(2)− (3 + 8α + 4α2)µ(1)

By equating the last two equations, we obtain the result that

(2α− 3α2 + α3)µ(2) = 2(2α− 3α2 + α3)µ(1)

As 2α− 3α2 + α3 = α(α− 1)(α− 2), then µ(2) = 2µ(1) as long as α 6= 0, 1, 2.

Lemma 6.3.1. If α 6= 0, 1, 2, then µ(n) = nµ(1) for n ≥ 3.

Proof. If n = 3, then µ(3) = (2(2 + α)− (1 + 2α))µ(1) = 3µ(1).

By induction, set s = n, k = 1, then

µ(n+ 1) =1

n− 1((n+ α)µ(n)− (1 + nα)µ(1))

=1

n− 1((n+ α)n− (1 + nα))µ(1)

=1

n− 1

(n2 − 1

)µ(1)

= (n+ 1)µ(1)


Thus by the above recurrence relation µ(k) = k except when α = 0, 1 or 2. In

the next three lemmas, we consider µ(k) in these three special cases.

Lemma 6.3.2. If α = 0, then µ(n) = (n− 1)µ(2)− (n− 2)µ(1) for n ≥ 3.

Proof. If n = 3, then µ(3) = 2µ(2)− µ(1).


µ(n+ 1) =1

n− 1((n)µ(n)− µ(1))

=1

n− 1((n(n− 1))µ(2)− (n(n− 2) + 1)µ(1))

= nµ(2)− 1

n− 1(n2 − 2n+ 1)µ(1)

= nµ(2)− (n− 1)µ(1)

In this case, µ(k) will be a polynomial of degree 1.

Lemma 6.3.3. If α = 1, then µ(n) =n(n− 1)

2µ(2)− (n2 − 2n)µ(1) for n ≥ 3.

Proof. If n = 3, then µ(3) = 3µ(2)− 3µ(1), so that

n(n− 1)

2µ(2)− (n2 − 2n)µ(1) = 3µ(2)− 3µ(1) = µ(3)


µ(n+ 1) =1

n− 1((n+ 1)µ(n)− (n+ 1)µ(1))

µ(n+ 1) =1

n− 1

(((n+ 1)

n(n− 1)

2

)µ(2)− ((n+ 1)(n2 − 2n) + (n+ 1))µ(1)

)µ(n+ 1) =

n(n+ 1)

2µ(2)− 1

n− 1(n3 − n2 − n+ 1)µ(1)

µ(n+ 1) =n(n+ 1)

2µ(2)− (n2 − 1)µ(1)


µ(n+ 1) =n(n+ 1)

2µ(2)− ((n+ 1)2 − 2(n+ 1))µ(1)

Hence, µ(k) is a polynomial of degree 2.

Lemma 6.3.4. If α = 2, then µ(n) =n3 − n

6µ(2)− n3 − 4n

3µ(1) for n ≥ 3.

Proof. If n = 3, then µ(3) = 4µ(2)− 5µ(3), so that

n3 − n6

µ(2)− n3 − 4n

3µ(1) = 4µ(2)− 5µ(1) = µ(3)


µ(n+ 1) =1

n− 1

((n+ 2)

n(n+ 1)(n− 1)

6µ(2)

−(

(n+ 2)(n3 − 4n)

3+ (2n+ 1)

)µ(1)

)µ(n+ 1) =

n(n+ 1)(n+ 2)

6µ(2)− 1

3

(n4 + 2n3 − 4n2 − 8n+ 6n+ 3)

)µ(1)

µ(n+ 1) =(n+ 1)2 − (n+ 1)

6µ(2)− 1

3

(n4 + 2n3 − 4n2 − 2 + 3

)µ(1)

µ(n+ 1) =(n+ 1)2 − (n+ 1)

6µ(2)− (n+ 1)3 − 4(n+ 1)

3µ(1)

This last case gives the result that µ(k) is a polynomial of degree 3.

Thus if µ(k) is non-trivial, then it has possible values of:

α = 0, µ(k) = k or 1

α = 1, µ(k) = k2

α = 2, µ(k) = k3

which are exactly the dual delta function cocycles we found in Chapter 5.

Chapter 7

Conclusion

In Chapter 4, we found all possible τ(k,m) cocycles contained in the span of{θ(0)k , . . . , θ

(N)k

}, which are summarized below. These functions yield polynomial

cocycles in M under the map π : MN → M , which are listed in Table 7.2. Re-

call that these cocycles determine the action of the Witt algebra on the module

M , which precisely determines the modules M so that these results give us a

classification of all length two extensions of tensor modules.

Most extensions admit classes of cocycles for a range of α and β. The few

exceptions occur in pairs; these pairs are modules extensions that are dual to

each other in the sense that for the extension M/T (α, γ) ∼= T (β, γ), the dual

extension is given by M ′/T (1− β,−γ) ∼= T (1 − α,−γ) with the dual cocycle

τ ∗(k,m) = τ(k,−m− k). In the case where a range of α and β are possible, these

extensions are dual to themselves.

These polynomial functions have at most degree 7. This comes from the equa-

tions we found that determine the coefficients ci in Lemma 4.1.3. When n gets

large, we quickly obtain systems that overdetermine the coefficients. The point

when n ≥ 8 is precisely where there is no non-trivial solution to the whole system

79

CHAPTER 7. CONCLUSION 80

of equations.

Table 7.1: Polynomial cocycles in MN

α− β = 0 n = 1 kθ(0)

α = 1, β = 0 n = 2 kθ(1)

α = 1, β = 0 n = 2 k2θ(0)

α− β = 2 n = 3 k3θ(0) − 2k2θ(1)

α− β = 3 n = 4 k3θ(1) − 2k2θ(2)

α− β = 4 n = 5 (α− 1)k5θ(0) − k4θ(1) − 12k3θ(2) + 24k2θ(3)

α = 1, β = −4 n = 6 2k5θ(1) + 10k4θ(2) + 60k3θ(3) − 120k2θ(4)

α = 5, β = 0 n = 6 12k6θ(0) − 22k5θ(1) + 10k4θ(2) + 60k3θ(3) − 120k2θ(4)

α =7 +√19

2, n = 7 −22 +

√19

4k7θ(0) +

31 + 7√19

2k6θ(1) − (25 + 7

√19)k5θ(2)

β = −5 +√19

2+30k4θ(3) + 120k3θ(4) − 240k2θ(5)

α =7−√19

2, n = 7 −22−

√19

4k7θ(0) +

31− 7√19

2k6θ(1) − (25− 7

√19)k5θ(2)

β = −5−√19

2+30k4θ(3) + 120k3θ(4) − 240k2θ(5)

There are a few remarks we can make about these functions in Table 7.2. By

directly applying the map π : MN → M , we do not obtain exactly the same

cocycle, only a cocycle that is in the same equivalence class. This will not change

the equivalence class though, so that we can shift the representative of the class

to a function of the same form. The choice of these equivalence classes are a bit

arbitrary; they are modelled after the results of Feigin and Fuks [8]. There is

a typo in [8] in the case of the degree 7 cocycles, otherwise we obtain the same

results in Table 7.2.

At the beginning of Chapter 4, we stated that it was enough to look at ho-

mogeneous polynomial cocycles. To obtain a general cocycle, we may have non-

homogeneous polynomials where each homogeneous component will be in one of

these equivalence classes. The values of α and β therefore put some limitations

on the general cocycles. For example, there will not exist a cocycle that has both


a non-trivial degree 6 component and a non-trivial degree 7 component.

Table 7.2: Polynomial cocycles in M

α− β = 0 n = 1 k

α = 1, β = 0 n = 2 km

α = 1, β = 0 n = 2 k2

α− β = 2 n = 3 k3 + 2k2m

α− β = 3 n = 4 k3m+ k2m2

α− β = 4 n = 5 (α− 4)k5 + k4m− 6k3m2 − 4k2m3

α = 1, β = −4 n = 6 12k6 + 22k5m+ 5k4m2 − 10k3m3 − 5k2m4

α = 5, β = 0 n = 6 2k5m− 5k4m2 + 10k3m3 + 5k2m4

α =7 +√19

2, n = 7 −22 +

√19

4k7 − 31 + 7

√19

2k6m

β = −5 +√19

2−25 + 7

√19

2k5m2 − 5k4m3 + 5k3m4 + 2k2m5

α =7−√19

2, n = 7 −22−

√19

4k7 − 31− 7

√19

2k6m

β = −5−√19

2−25− 7

√19

2k5m2 − 5k4m3 + 5k3m4 + 2k2m5

From the cocycle condition in given by (3.1.2), other extensions can be found.

For example, the function τ(k,m) = 1 will satisfy the cocycle condition when

α = β. The action of e0 on the basis vectors will be

e0vm = mvm

e0wm = mwm + vm

This cocycle does not produce a weight module as e0 will not act diagonally.

Equation (3.1.2) is not enough to obtain a weight extension, while the τ cocycles

are in one-to-one correspondence to τ cocycles that yield weight module extensions.

For β = 1, α = 0 and γ ∈ Z, we obtained a few non-polynomial cocycles.

Recall that for τ(k,m) = 1mµ(k,m), where τ(k, 0) is defined to be zero. As was

discussed in Section 6.2, these functions are almost polynomial in the sense that


they admit polynomial cocycles under the change of basis T (0, γ)→ T (1, γ).

Table 7.3: Non-polynomial cocycles

γ ∈ Z α = 0, β = −1 δk+m,0k3

γ ∈ Z α = 0, β = 0 δk+m,0k2

γ ∈ C α = 0, β = 1 m−1k

γ ∈ Z α = 0, β = 1 δk+m,0

γ ∈ Z α = 0, β = 1 δk+m,0k

γ ∈ Z α = 0, β = 1 δm,0k

γ ∈ C α = 1, β = 1 m−1k2

γ ∈ Z α = 1, β = 1 δm,0k2

γ ∈ C α = 2, β = 1 m−1k3 + k2

γ ∈ Z α = 2, β = 1 δm,0k3

These delta cocycles are of interest for another reason. Using the delta func-

tions that give an extra factor on v0 in the W1-action on T (β, γ), i.e. cocycles of

the form δk+m,0µ(k), we can construct new Lie algebras.

Let L be the Virasoro algebra which is spanned by {L(m), c1} where c1 is

the central extension, m ∈ N. Suppose that module V is a module for the

Virasoro algebra with basis {W (m), c3}. By viewing V as a (possibly abelian)

Lie algebra, we can look at the semi-direct product of L with V spanned by

{L(m),W (m), c1, c2, c3} with bracket:

[L(k), L(m)] =(m− k)L(k +m) + δk+m,0k3 − k

12c1

[L(k),W (m)] =(m− βk)W (k +m) + δk+m,0µ(k)c2

where V is a subalgebra and the bracket of any element with a central extension

c1, c2, c3 is simply zero.

In this way, we can construct what is called the W (2, 2) algebra [17]. Take V


to be the module corresponding to the adjoint representation of L and let V be

an abelian Lie algebra. Then the bracket of the new algebra is given by

[L(k), L(m)] =(m− k)L(m+ k) + δk+m,0k3 − k

12c1

[L(k),W (m)] =(m− k)W (m+ k) + δk+m,0k3 − k

12c1

[W (k),W (m)] =0

This corresponds to the cocycle δk+m,0k3, β = −1.

Similarly, we can construct the twisted Heisenberg-Virasoro algebra [4]. Here,

V is the Heisenberg Lie algebra which is itself an L-module.

[L(k), L(m)] =(m− k)L(m+ n) + δk+m,0k3 − k

12c1

[L(k),W (m)] =mW (k +m) + δk+m,0(k2 − k)c2

[W (k),W (m)] =δk+m,0nc3

This corresponds to the cocycle δk+m,0k2, β = 0.

We can construct two more algebras of this form, given by the cases of δk+m,0, β =

1, δk+m,0k, β = 1. As in the previous cases, V can be taken as an abelian alge-

bra or as a Heisenberg algebra. In the case that V is a Heisenberg algebra, this

construction is only a Lie algebra if β = 0.

For the cocycle δk+m,0, β = 1, we take V to be an abelian Lie algebra. The

resulting algebra is given below, where the bracket with any central element is

trivial.

[L(k), L(m)] =(m− k)L(m+ n) + δk+m,0k3 − k

12c1

[L(k),W (m)] =(m+ k)W (k +m) + δm+k,0c2

[W (k),W (m)] =0


The case of δk+m,0k, β = 1 will construct the algebra given by the following,

where the the bracket with any central element is trivial.

[L(k), L(m)] =(m− k)L(m+ n) + δk+m,0k3 − k

12c1

[L(k),W (m)] =(m+ k)W (k +m) + δm+k,0kc2

[W (k),W (m)] =0

Notice that if we set I(m) = mW (m),

[L(k), I(m)] = mI(m+ k)−m2δm+k,0

This action is very close to the twisted Heisenberg-Virasoro algebra. The difference

is the action on W (0) is given by [L(k),W (0)] = kW (k).

The corresponding Virasoro modules are given by the extensions

V/D(0, 0) ∼= T (0, 0), for δk+m,0

V/D(0, 0) ∼= T (1, 0), for δk+m,0k

The main goal of this work was to produce an explicit classification of length

two module extensions of the Witt algebra. However, this method is promising in

finding a similar classification of length two module extensions of the solenoidal

subalgebra (see Definition 2.2 of[6]) of Wn, and may even be used to find a classi-

fication of module extensions of this type for Wn.

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Length two extensions of modules for the Witt algebra

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