periodisations of contragredient lie superalgebras and their

Periodisations of ContragredientLie Superalgebras and Their

Presentations

Anna Larsson

Doctorial Dissertation 2003Department of MathematicsUniversity of StockholmS-106 91 Stockholm

Typeset by LATEXc©2003 Anna Larssone-mail: [email protected] 91-7265-725-1Akademitryck AB, Edsbruk, 2003

AbstractThe motivation for this thesis comes from the theory of commutativerings. Let R = k ⊕R1 ⊕R2 ⊕ · · · be a commutative algebra over a fieldk. It is a well known fact that there exists a Z+-graded Lie superalgebraLR = ⊕i≥1(LR)i such that

Ext∗R(k, k) = U(LR),

where U(LR) is the universal enveloping algebra of LR. The Lie superal-

gebra L(1)R generated by (LR)1 is a consistent Z+-graded Lie superalgebra,

that is, the Z2-grading is obtained from the Z+-grading by reducing mod-

ulo 2. Furthermore, Clas Lofwall has showed that L(1)R can be presented

by generators of degree 1 and relations of degree 2 (L(1)R is 1-2-presented).

Also, Clas Lofwall and Jan-Erik Roos have discovered a commutative

ring R such that L(1)R is a consistent Z+-graded ”periodisation” of a Lie

superalgebra. The aim of this thesis is to systematically investigate theconsistent Z+-graded periodisations g1⊕g0⊕g1⊕· · · of a certain type ofLie superalgebras g = g0⊕ g1. We are in particular interested in findingLie superalgebras for which these periodisations are 1-2-presented. Thisinvestigation is carried out in three steps. First we develop a method forLie algebras and prove that the periodisation g⊕g⊕g · · · of a semisimpleLie algebra over an algebraically closed field of characteristic 0 withoutsl(2)-component is 1-2-presented. This method is then generalized andwe show three statements to be equivalent about a contragredient Liesuperalgebra g over a field of characteristic 0 whose Cartan matrix hasnon-zero determinant:

The Cartan matrix of g contains no ”even unit-column”

g is not a semi-direct product of sl(2)

The periodisation g⊕ g⊕ g · · · of g is 1-2-presented

However, the periodisations above do not need to be consistent. In thelast part of the thesis the aim is to enable an investigation of the consis-tent Z+-graded periodisations of contragredient Lie superalgebras withthe same method as for semisimple Lie algebras. We show the sufficiencyof relations of degree at most 4 in presenting the consistent Z+-gradedperiodisation of a simple contragredient Lie superalgebra over a field ofcharacteristic 0 and of rank ≥ 2. The thesis is ended by a detailed studyof some of the classical contragredient Lie superalgebras in order to showwhether their consistent Z+-graded periodisations are 1-2-presented ornot.

Acknowledgements

I am most grateful to my adviser Clas Lofwall at Stockholm Univer-sity for his support during graduate studies and for introducing me toand teaching me about Lie algebras and Lie superalgebras. Also, I amgrateful for his guidance during my work and for his numerous sugges-tion that have improved this thesis. His enthusiasm and commitmenthave helped me to complete this work.

I would like to thank all the employees at the Department of Mathe-matics at Stockholm University. They have been most helpful wheneverneeded and have contributed to a pleasant working atmosphere. I alsowould like to take this opportunity to thank Christian Gottlieb for hisguidance during my teaching assignments as a graduate student.

To my friends outside the department, I thank you all for your sup-port and for your ability to get my mind off mathematics occasionally.

I am grateful to my family, to my parents who always are present, toHakan for his encouragement and to Tomas for all his English words.

Most of all, I would like to thank Mika for his endless patience andfor that he always believes in me.

Contents

Introduction ix0.1 Lie Algebras and Lie Superalgebras . . . . . . . . . . . . ix0.2 Research Problems and Summary of Results . . . . . . . xv0.3 The Structure of the Thesis . . . . . . . . . . . . . . . . xix

1 A Periodisation of Semisimple Lie Algebras 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Root System and Chevalley Basis . . . . . . . . . . . . . 21.3 The Periodisation and a Free Lie Algebra . . . . . . . . . 4

1.3.1 The Periodisation . . . . . . . . . . . . . . . . . . 51.3.2 The Free Lie Algebra . . . . . . . . . . . . . . . . 5

1.4 Proof of Theorem 1 . . . . . . . . . . . . . . . . . . . . . 10

2 Contragredient Lie Superalgebras 232.1 Definition of Contragredient Lie Superalgebras . . . . . . 232.2 The Root System and the Root-tuple System . . . . . . 262.3 The Structure Constants . . . . . . . . . . . . . . . . . . 31

3 Non-consistent Periodisations 353.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 The Periodisation and the Free Lie Superalgebra . . . . . 36

3.2.1 The Periodisation . . . . . . . . . . . . . . . . . . 373.2.2 The Free Lie Superalgebra . . . . . . . . . . . . . 37

3.3 Formulation of Theorem 2 . . . . . . . . . . . . . . . . . 423.4 The First Step: (ii) Implies (i) . . . . . . . . . . . . . . 433.5 The Second Step: (iii) Implies (ii) . . . . . . . . . . . . 553.6 The Third Step: (i) Implies (iii) . . . . . . . . . . . . . 573.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4 Consistent Periodisations 634.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 63

viii CONTENTS

4.2 Simple Lie Superalgebras . . . . . . . . . . . . . . . . . . 644.3 The Consistent Periodisation and the Free Lie Superalgebra 68

4.3.1 The Periodisation . . . . . . . . . . . . . . . . . . 684.3.2 The Free Lie Superalgebra . . . . . . . . . . . . . 69

4.4 Formulation and Proof of Theorem 3 . . . . . . . . . . . 724.5 An Investigation of Some of the Classical Lie Superalgebras 83

Introduction

0.1 Lie Algebras and Lie Superalgebras

Lie algebras was first introduced in the 1880 article [20] by Sophus Lie inthe context of Lie groups. A Lie group G is a group and a manifold suchthat the mappings (x, y) 7→ x ·y and x 7→ x−1 are mappings of manifolds.The so-called infinitesimal transformations of a Lie group generate a Liealgebra. In modern terms without referring to Lie groups, a Lie algebra gover a field k is a vector space endowed with a bilinear bracket operationsuch that

[x, y] = −[y, x]

[x, [y, z]] = [[x, y], z] + [y, [x, z]]

for all x, y ∈ g. The first condition is called the anticommutativity andthe second the Jacobi identity. A Lie algebra g is simple if [g, g] 6= 0and the only ideals are the zero ideal and g itself. For a Lie algebra g,define two sequences of ideals by g(0) = g0 = g, g(i) = [g(i−1), g(i−1)] andgi = [g, gi−1] for all i ≥ 1. Then, g is called nilpotent if gi = 0 for somei ≥ 0. Furthermore, g is called solvable if g(i) = 0 for some i ≥ 0 andsemisimple if the only solvable ideal of g is the zero ideal. A semisimpleLie algebra is a direct sum of simple ones.

Examples of Lie algebras

Let k be a field.

1. An associative k-algebra A is a Lie algebra with the bracket oper-ation defined as

[x, y] = xy − yx.

2. A derivation D : A → A of a not necessarily associative k-algebraA is a linear map such that D(xy) = D(x)y + xD(y). The set

x 0. Introduction

Der(A) of all derivations of A is a Lie algebra with the bracketoperation defined as

[D1, D2] = D1D2 −D2D1

for all D1, D2 ∈ Der(A).

3. Let V be a vector space over k of dimension l. The set End(V )of all endomorphisms of V is an associative k-algebra. The Liealgebra induced by End(V ) (as in 1.) is called the general linearalgebra gl(V ) or gl(l). The following Lie algebras are subalgebrasof the general linear algebra:

• The special linear algebra sl(l + 1), or Al, is the subalgebraof gl(l + 1) consisting of all endomorphisms of trace 0.

• The symplectic algebra sp(2l + 1) or Cl.

• The orthogonal algebras o(2l + 1) and o(2l), or Bl and Dl

respectively.

For more details about the symplectic and the orthogonal Lie algebras,see e.g. [5] and [10]. The Lie algebras Al, Bl, Cl and Dl are calledclassical.

After the introduction, the theory of Lie algebras was developed byWilhelm Killing and Elie Cartan. In his 1894 thesis [3], E. Cartan classi-fied all finite-dimensional simple Lie algebras over C. These Lie algebrasare either isomorphic to a classical Lie algebra or to one of the so-calledexceptional Lie algebras G2, F4, E6, E7 or E8 (for more information aboutthe exceptional Lie algebras see e.g. [5] and [10]). This classification waslater generalized to finite-dimensional Lie algebras over an arbitrary fieldof characteristic 0, mainly by Claude Chevalley and Harish-Chandra.

In his classification, E. Cartan introduced the concept of Cartan ma-trices, root spaces and root-space decomposition of a Lie algebra. Herewe present a brief up-to-date account of this theory. A semisimple Liealgebra can be defined in terms of a Cartan matrix and Chevalley gen-erators. Let g be a semisimple Lie algebra over a field k of characteristic0 and let h be a Cartan subalgebra of g, i.e., a nilpotent subalgebra suchthat h = {x ∈ g; [x, h] ⊂ h}. Then g decomposes as a direct sum

g = h⊕ (⊕

α∈∆

gα

), (∗)

0.1. Lie Algebras and Lie Superalgebras xi

where gα = {x ∈ g; [h, x] = α(h)x for all h ∈ h} for all α ∈ h∗ andwhere ∆ = {α ∈ h∗; gα 6= 0}. The set ∆ is called the root system, thesubspaces gα, for α ∈ ∆, are called the root spaces and the decomposition(∗) is called the root-space decomposition of g. The root spaces gα

are one-dimensional. Let {h1, . . . , hr} be a basis of h and fix a basis{α1, . . . , αr} of h∗. The matrix A = (αj(hi))

ri,j=1 is the Cartan matrix of

g. The information in the Cartan matrix can also be presented in theso-called Dynkin diagram.

The probably most studied Lie algebras are the Kac-Moody alge-bras, and in particular the affine Kac-Moody algebras, with applicationsin both mathematics and physics. The theory of Kac-Moody algebras isbasically a generalization of the concept of Lie algebras with Cartan ma-trices to the infinite-dimensional case and was developed independentlyby Victor Kac [16] and Robert Moody [25] in the 1960s. For a surveyof this development we refer to the article [2] by Stephen Berman andKaren Hunger Parshall from where some of the references are taken. Fora more detailed description of the Kac-Moody algebras, we refer to [8],[10] and [13]. An integral matrix A = (aij)

ri,j=1 is called a generalized

Cartan matrix if

aii = 2 for all 1 ≤ i ≤ r,

aij ≤ 0 for all 1 ≤ i, j ≤ r such that i 6= j,

aij = 0 =⇒ aji = 0 for all 1 ≤ i, j ≤ r.

The Kac-Moody algebra is the complex Lie algebra generated by the set{fi, hi, ei}r

i=1 and with the defining relations

[hi, ej] = aijej, [hi, fj] = −aijfj

[hi, hj] = 0, [ei, fj] = δijhi(i)

(ad ei)1−aijej = 0, (ad fi)

1−aijfj = 0 (ii)

The relations in (ii) are the so-called Serre relations (see Jean-PierreSerre [28]). In an alternative setting, the Kac-Moody algebra is definedas the quotient of the Lie algebra generated by {fi, hi, e,}r

i=1 and withthe relations (i) by the maximal Z-graded ideal intersecting the span ofhi, ei and fi, for 1 ≤ i ≤ r, trivially. This ideal is in fact generatedby the Serre relations. The simple finite-dimensional Lie algebras areKac-Moody algebras and the Cartan matrix of such a Lie algebra is saidto be of finite type. An affine Lie algebra is an infinite-dimensional Kac-Moody algebra of polynomial growth. The Cartan matrix of an affine

xii 0. Introduction

Lie algebra is said to be of affine type. An affine Lie superalgebra iseither ”non-twisted” or ”twisted”.

Let C[t, t−1] be the algebra of Laurent polynomials in t and let gbe a Lie algebra. The Lie algebra L(g) = g ⊗ C[t, t−1] is called the

Loop algebra of g. Let◦A be a generalized Cartan matrix of finite type.

The so-called extended Cartan matrix A is obtained from◦A by adding

a certain row and column. Then g(A) is an affine Kac-Moody algebra

and the Loop algebra L(g(◦A)) of g(

◦A) is isomorphic to [g(A), g(A)]/c (c

being the center of [g(A), g(A)]).The first introduction of Lie superalgebras in the 1970s was motivated

by the interest of supersymmetries in physics (see e.g. the notes aboutthis in the introduction of [14] and [12]). A Lie superalgebra is a Z2-graded vector space g = g0⊕g1 endowed with a bilinear bracket operationsuch that:

[x, y] = −(−1)uv[y, x] for all x ∈ gu, y ∈ gv

[x, [y, z]] = [[x, y], z] + (−1)uv[y, [x, z]] for all x ∈ gu, y ∈ gv, z ∈ g

The second condition is the Jacobi identity. The elements in g0 and g1are called even and odd respectively. From now on, the super-degree(sdeg) refers to the Z2-grading of a vector space, an algebra or a Liesuperalgebra. The following are examples of Lie superalgebras.

1. An associative Z2-graded algebra A = A0 ⊕A1, where the bracketoperation is defined by

[x, y] = xy − (−1)uvyx

for all x ∈ Au, y ∈ Av, is a Lie superalgebra. A Z2-graded algebrais called a superalgebra.

2. The Grassman algebra Λ(n), generated by the set {1, ξ1, . . . , ξn}such that ξiξj = −ξjξi for all i, j ∈ {1, . . . , n}, becomes an as-sociative super-commutative superalgebra by letting sdeg(ξi) ≡ 1(mod 2). The Lie superalgebra induced by Λ(n) is abelian.

3. Let A = A0⊕A1 be a superalgebra. A derivation D of super-degreed ∈ Z2 is an endomorphism of A such that

D(xy) = D(x)y + (−1)duxD(y)

0.1. Lie Algebras and Lie Superalgebras xiii

for all x ∈ Au, y ∈ A. Let Deri(A), for i = 0, 1, be the set of allderivations of A of super-degree i. The set Der(A) = Der0(A) ⊕Der1(A) of all derivations of A is a Lie superalgebra with thebracket operation defined as

[D1, D2] = D1D2 − (−1)uvD2D1

for all D1 ∈ Deru(A) and D2 ∈ Derv(A). The Lie superalgebraW (n) = Der(Λ(n)) is an example of a Lie superalgebra of Cartantype, that is a simple Lie superalgebra g = g0 ⊕ g1 such that therepresentation of g0 on g1 is not completely reducible. There arefour classes of Cartan type Lie superalgebras: W (n) for n ≥ 2,S(n) for n ≥ 3, S(n) for n ≥ 4 and the Hamiltonian H(n) forn ≥ 4 (for more details about the Cartan type Lie superalgebras,see ref. [5] and [14]).

4. Let L and g be two Lie superalgebras and let

ϕ : L −→ Der(g)

be a Lie superalgebra homomorphism. The vector space

Ln g = L ⊕ g

is a Lie superalgebra with the bracket operation defined by

[(l, g), (l′, g′)] = ( [l, l′], [g, g′] + ϕ(l)(g′)− (−1)uvϕ(l′)(g) )

for all l ∈ L, g′ ∈ g, l′ ∈ Lu and g ∈ gv. The Lie superalgebraLn g is the semi-direct product of L and g with respect to ϕ.

5. Let V = V0⊕V1 be a Z2-graded vector space and let End(V ) denotethe set of all endomorphisms of V . Then End(V ) = End0(V ) ⊕End1(V ), where

End0(V ) = {f ∈ End(V ); f(Vi) ⊆ Vi for i = 0, 1}End1(V ) = {f ∈ End(V ); f(Vi) ⊆ Vi+1 for i = 0, 1},

is an associative superalgebra and, hence, a Lie superalgebra withthe bracket operation defined as in 1. As a Lie superalgebra ,End(V ) is denoted as gl(V ) or gl(m,n) where m = dim(V0) andn = dim(V1).

xiv 0. Introduction

Let G be an Abelian group. A Lie superalgebra g is G-graded if g =⊕a∈Gga as a vector space and [ga, gb] ⊆ ga+b for all a, b ∈ G. A Z+-gradedLie superalgebra is a Z-graded Lie superalgebra g such that gl = 0 forall l ≤ 0. A Z-grading (Z+-grading) of a Lie superalgebra is calledconsistent if the Z2-grading is obtained by reducing modulo 2.

The concept of Kac-Moody algebras (sometimes referred to as con-tragredient Lie algebras) was first generalized to the superalgebra caseby V. Kac in [14]. Below is a brief description of the so-called contragre-dient Lie superalgebras. A more extensive account of the subject is givenin Chapter 2. Given a matrix A = (aij)

ri,j=1 over a field k and a sub-

set τ ⊆ {1, . . . , r}, let g(A, τ) = ⊕i∈Zgi be a Z-graded Lie superalgebragenerated by the set {fi, hi, ei}r

i=1 and with the relations

[ei, fj] = δijhi, [hi, hj] = 0,

[hi, ej] = aijej, [hi, fj] = −aijfj,

sdeg(hi) ≡ 0 (mod 2)

sdeg(ei) = sdeg(fi) ≡{

0 (mod 2) if i 6∈ τ

1 (mod 2) if i ∈ τ

for all i, j ∈ {1, . . . , r} and where δij is the Kronecker-delta. The con-tragredient Lie superalgebra g(A, τ) is defined as the Z-graded Lie su-peralgebra g(A, τ)/m, where m is the maximal Z-graded ideal in g(A, τ)intersecting the subspace g−1 ⊕ g0 ⊕ g1 trivially. This maximal ideal isnot as easy to describe as in the Lie algebra case with the Serre relations.In the article [9] by Pavel Grozman and Dimitry Leites, they derive thedefining relations for some Lie superalgebras with Cartan matrix.

Some examples of contragredient Lie superalgebras are the special lin-ear Lie superalgebras sl(m,n) for m,n ≥ 1 and the so called orthogonal-symplectic Lie superalgebras osp(m, 2n) for m ≥ 1 and n ≥ 0 (if m = 2,n ≥ 1). The contragredient Lie superalgebra sl(m,n) is the subalgebraof gl(m,n) consisting of all endomorphisms with supertrace 0. These canbe regarded as the set of matrices of the form M = ( a b

c d ) for which thesupertrace str(M) = tr(a)− tr(d) is zero. The orthogonal-symplectic Liesuperalgebras are also subalgebras of gl(m,n). For a description of theorthogonal-symplectic Lie superalgebras, see e.g. [5] and [14]. FollowingV. Kac notation in [14], which is analogues to the one for Lie algebras,

0.2. Research Problems and Summary of Results xv

we have that

A(m,n) = sl(m + 1, n + 1) for m,n ≥ 0 such that m 6= n

A(m,m) = sl(m + 1,m + 1)/c for n > 1 where c is the center

B(m,n) = osp(2m + 1, 2n) for m ≥ 0, n > 1

C(n) = osp(2, 2n− 2) for n ≥ 2

D(m,n) = osp(2m, 2n) for m ≥ 2, n > 0

The Lie superalgebra A(m,m) is not contragredient but almost sincesl(m + 1,m + 1) is contragredient. The series A(m,n), B(m,n), C(n)and D(m,n), together with the so-called ”exceptional” contragredientLie superalgebras F (4), G(3) and D(2, 1; α) and the ”strange” seriesP (n) and Q(n), are the ”classical” Lie superalgebras (The strange se-ries P (n) and Q(n) are not contragredient). These are the simple Liesuperalgebras g = g0 ⊕ g1 such that the representation of g0 on g1 iscompletely reducible.

The simple finite-dimensional Lie superalgebras was classified by V.Kac in [14]. A simple finite-dimensional Lie superalgebra is isomorphic toeither a classical Lie superalgebra or to a Cartan type Lie superalgebra.As a part of this classification V. Kac showed that a finite-dimensionalcontragredient Lie superalgebra g(A, τ) such that g(A, τ)/c is simple (cbeing the center of g(A, τ)) is isomorphic either to a simple Lie alge-bra or to one of the classical Lie superalgebras A(m,n), B(m,n), C(n),D(m,n), D(2, 1; α), F (4) or G(3).

A Kac-Moody Lie superalgebra is a contragredient Lie superalgebrag(A, τ) where A = (aij)

ri,j=1 is a so-called generalized Cartan matrix, i.e.:

aii ∈ {0, 2} for all i ∈ [r] and aii = 0 implies that i ∈ τ

aij ≤ 0 for all i, j ∈ [r] such that i 6= j and aii 6= 0

aij = 0 implies that aji = 0 for all i, j ∈ [r]

A Kac-Moody algebra is called affine if it is infinite-dimensional but ofpolynomial growth. In [31] John W. van de Leur classifies the contra-gredient affine Lie superalgebras with symmetrizable Cartan matrix.

0.2 Research Problems and Summary ofResults

The motivation for this thesis comes from the theory of k-algebras. Givena commutative, graded and connected k-algebra R = k ⊕ R1 ⊕ R2 · · · ,

xvi 0. Introduction

there exists a Z+-graded Lie superalgebra LR = ⊕i≥1 (LR)i such that

Ext∗R(k, k) = U(LR),

where U(LR) is the universal enveloping algebra of LR. Let L(1)R be

the Lie subalgebra of LR generated by (LR)1, first introduced in [23],

a part of the 1976 thesis of Clas Lofwall. The Lie subalgebra L(1)R is a

consistent Z+-graded Lie superalgebra. Furthermore, it was proved in

[23] that L(1)R is 1-2-presented, i.e., L(1)

R can be presented as a free Liesuperalgebra with generators of degree 1 and relations of degree 2. The

k-algebra R is Koszul if and only if LR = L(1)R , i.e., if and only if LR is

1-2-presented. The paper [24] by Clas Lofwall and Jan-Erik Roos gives

an example of a k-algebra R such that L(1)R is a consistent Z+-graded

periodisation of a 17-dimensional Lie superalgebra. By the consistentZ+-graded periodisation of a Lie superalgebra g, we mean the the Z+-graded Lie superalgebra ⊕i≥1g(i mod 2). This caused the question: WhichLie superalgebras emerge like this from a k-algebra? On the other hand,the Koszul dual R = U(L)! of a 1-2-presented consistent Z+-graded Lie

superalgebra L = ⊕i≥1Li is a commutative ring and L = L(1)R . The

Koszul dual R is the quotient of a polynomial ring by elements of degree2. For examples and more details on the subject, we refer the readerto e.g. Lofwall, [21] and [23], Froberg [6] and Backelin-Froberg [1]. Thetask to find Lie superalgebras whose consistent Z+-graded periodisationsare presented with generators of degree 1 and relations of degree 2 wasobvious. This is the issue and the motivation for this thesis.

Let g = ⊕gi be a Z+-graded Lie (super)algebra. Then, a minimal setof generators for g is in one-to-one correspondence with a basis of thehomology group H1(g, k). Also, given a minimal set of generators forg, a minimal set of the relations is in one-to-one correspondence with abasis of H2(g, k) (see Lemma 1.0.4 and Theorem 1.2.1 in [19]). In termsof Lie superalgebra homology the issue may be expressed as follows:Which Lie superalgebras g have a consistent Z+-graded periodisationgper such that H1(gper, k) is concentrated in degree 1 and H2(gper, k)is concentrated in degree 2. However, our approach to the problemwill not be homological but through the root system and root spacedecomposition of contragredient Lie (super)algebras.

As a first step to approach the problem, we looked at contragredientLie algebras (non super). This resulted in the following theorem:

0.2. Research Problems and Summary of Results xvii

Theorem 1. Let L be a finite-dimensional semisimple Lie algebra overan algebraically closed field of characteristic 0 without sl2-component.Let Lper be the Z+-graded periodisation

Lper = L⊕ L⊕ L⊕ · · ·

of L. Then there is a set of generators G of degree 1 and an ideal R ofF(M), generated by expressions of degree 2, such that Lper

∼= F(M)/R.

It is possible that Theorem 1 could be derived from a result due toHoward Garland and James Lepowski [8] in terms of Lie algebra homol-ogy and representation theory. However, the aim was partly to developa method that could be generalized to Lie superalgebras. We did notbelieve that this result was suitable for this purpose since the theory forrepresentations of Lie superalgebras is not as extensive as the one of Liealgebras.

The next step in the investigation of finding Lie superalgebras whoseconsistent Z+-graded periodisations are 1-2-presented was to generalizethe method for contragredient Lie algebras to contragredient Lie superal-gebras. For Lie algebras, the consistency of the Z+-graded periodisationwas not a problem and the Lie algebra was placed in degree 1. The mostnatural continuation was to do the same thing for Lie superalgebras.Thus, we started the investigation of Lie superalgebras by consideringthe Z+-graded periodisation

gper =⊕i≥1

(gper)i = g⊕ g⊕ g⊕ · · ·

of a Lie superalgebra g. Clearly, this periodisation is not necessarilyconsistent. The main result of this investigation is the following theorem:

Theorem 2. Let g = g(A, τ) be a contragredient Lie superalgebra over afield k of characteristic 0 and such that det(A) 6= 0. Then the followingare equivalent:

(i) gper is 1-2-presented

(ii) A does not contain an even column where the elements are zerooutside the diagonal entry

(iii) g is not a semidirect product of sl(2, k)

xviii 0. Introduction

For semisimple Lie algebras, a direct product is equivalent to a semi-direct product. Hence, Theorem 1 is a special case of Theorem 2.

Even though the Z+-graded periodisation in Theorem 2 need not beconsistent, there is a connection to the theory of k-algebras. Given a1-2-presented Z+-graded Lie superalgebra L = ⊕i≥1Li, not necessarilyconsistent, the Koszul dual R = U(L)! is a skew-commutative ring and

L = L(1)R where L(1)

R is defined as above. Furthermore, R is a quotientof the tensor product of a polynomial ring and an exterior algebra byelements of degree 2.

Let g be a Lie superalgebra and define the consistent Z+-gradedperiodisation gper of g as

gper =⊕

i≥1

(gper)i where (gper)i =

{g0 if i ≡ 0 (mod 2)

g1 if i ≡ 1 (mod 2)

for all i ≥ 1. To be able to use the same method to investigate Liesuperalgebras g and the consistent Z+-graded periodisation gper, we hadto confine ourselves to simple contragredient Lie superalgebras and showthe sufficiency of relations of degree at most 4.

Theorem 3. Let g = g(A, τ) be a simple contragredient Lie superalgebraover a field of characteristic 0. Assume that rank(g) ≥ 2. Then theconsistent Z+-graded periodisation gper of g is isomorphic to a free Liesuperalgebra with generators of degree 1 and relations of degree at most4.

Theorem 2 and Theorem 3 includes the classical Lie superalgebrasA(m,n) for m 6= n, B(m,n), C(n + 1), D(m,n), D(2, 1; α), F (4) andG(3) (classified by Kac in e.g. [14]). Note however that the contragre-dient Lie superalgebras included here do not have to be classical, theymight be of infinite dimension. We end the thesis with an investigation ofthe consistent Z+-graded periodisation of some classical contragredientLie superalgebras. According to Theorem 3, it suffices to use relations ofdegree at most 4 in presenting this periodisation of the classical contra-gredient Lie superalgebras. We use this result in order to investigate forwhich of the classical contragredient Lie superalgebras it suffices withrelations of degree 2. Our conclusion is that gper is 1-2-presented if gis one of A(m,n) for m 6= n and m,n ≥ 2, B(m,n) for m,n ≥ 2 orC(n + 1) for n ≥ 2. Furthermore, we show that the presentation of gper

needs relations of degree 3 or 4 if g is one of A(1, 4), B(0, n) for n > 1or C(1 + 1). The remaining classical Lie superalgebras are yet to beinvestigated.

0.3. The Structure of the Thesis xix

0.3 The Structure of the Thesis

Chapter 1: In this chapter we investigate semisimple Lie algebras andthe main result is Theorem 1 stated above. Proposition 1.3.6 presents alist of the quadratic relations. In Section 1.2 we describe the root systemand the Chevalley basis of a semisimple finite-dimensional Lie algebraand in Section 1.3 we construct the periodisation and the free Lie alge-bra in question. Table 1.1 in Section 1.3.2 shows the maximal roots androot vectors for the finite-dimensional simple Lie algebras. A version ofChapter 1 is published in Homology, Homotopy and Application [17].Consequently, the notation in this chapter does not completely followthe notation in the rest of the thesis.

Chapter 2: Chapter 2 is a preparation for Chapter 3 and 4. In thischapter we give a definition of contragredient Lie superalgebras. Also,we define the root system of contragredient Lie superalgebras with Car-tan matrices with non-zero determinant. Some well-known results aboutcontragredient Lie superalgebras are stated and proved. For semisimpleLie algebras and most classical Lie superalgebras, the root spaces areone-dimensional. This is not true for all contragredient Lie superalge-bras and to be able to work with one-dimensional spaces, we refine theroot space decomposition and introduce the root-tuple system.

Chapter 3: In this chapter we make the first generalization of themethod for semisimple Lie algebras to contragredient Lie superalgebras.At this point, we do not require consistency of the Z+-graded periodis-ation in question. The periodisation is simply generated by the the Liesuperalgebra itself and is described in Section 3.2.1. The main result ofthis chapter is Theorem 2 stated above (also, see Section 3.3) and thequadratic relations are listed in Proposition 3.2.4. We end the chap-ter with some examples. In the first (Example 3.7.1) we show that theperiodisation of the finite-dimensional Lie superalgebra sl(2, 2) is not 1-2-presented. This shows that the condition of non-zero determinant cannot be omitted in Theorem 2. In the second example (Example 3.7.2)we calculate the generators and the quadratic relations for the periodis-ation of the 5-dimensional contragredient Lie superalgebra with Cartanmatrix (1) and τ = {1}. We also construct the Koszul dual of this peri-odisation as the quotient of the tensor product of a polynomial ring andan exterior algebra by elements of degree 2 (c.f. Section 0.2).

xx 0. Introduction

Chapter 4: In this chapter we continue the generalization of the methodfor semisimple Lie algebras in order to investigate consistent Z+-gradedperiodisations of contragredient Lie superalgebras. This periodisation isdescribed in Section 4.3.1 and the main result of this chapter is Theo-rem 3. In this chapter we confine ourself to the study of simple contra-gredient Lie superalgebras and in Section 4.2, we state and prove somewell-known results about simple Lie superalgebras. In Theorem 3 weshow the sufficiency of relations of degree at most 4. In Section 4.5,we end this chapter and the thesis with an investigation of some of theclassical contragredient Lie superalgebras in order to show whether theirconsistent Z+-graded periodisations are 1-2-presented or not.

Remark. In this text, we consider several gradings of the Lie superalge-bras involved. Clearly, there is the Z2-grading coming from the divisioninto odd and even elements. But a contragredient Lie superalgebra isalso Z-graded and then there is a Z+-grading defined from the periodi-sation. When we talk about the super-degree, or sdeg, we always referto the Z2-grading. The degree, or deg, refers to the Z-grading or to theZ+-grading. When necessary, we will point out to which grading thedegree refers to.

Chapter 1

A Periodisation ofSemisimple Lie Algebras

A version of this chapter is published in Homology, Homotopy and Appli-cation ([17]). Consequently, the notation in this chapter does not com-pletely follow the notation in the rest of the thesis.

1.1 Introduction

The aim of this chapter is to study the periodisation of finite-dimen-sional semisimple Lie algebras over an algebraically closed field F of char-acteristic 0 without sl2-component. Every finite-dimensional semisimpleLie algebra can be written as the direct sum of its simple ideals. Thecondition that there is no sl2-component means that no simple ideal inthis sum is of sl2-type.

Let L be a finite-dimensional semisimple Lie algebra over a alge-braically closed field F of characteristic 0. In this chapter we assumethat L has no sl2-component. Let Lper denote the periodisation of L,i.e., let Lper be the graded Lie algebra with L in each positive degree (aformal definition of Lper is given in section 1.3). Moreover, let F(G) bethe free Lie algebra generated by G, which we consider as graded lettingthe elements of G have degree 1. We call the homogeneous elementsin F(G)2 quadratic. The main result in this chapter is the followingtheorem:

Theorem 1. Let L be a finite-dimensional semisimple Lie algebra overan algebraically closed field F of characteristic 0 without sl2-component.Then there is a set of generators G and an ideal R of F(G), generatedby quadratic expressions only, such that Lper

∼= F(G)/R.

2 1. A Periodisation of Semisimple Lie Algebras

Our approach to Theorem 1 will be through a Chevalley basis of Land we use only elementary methods. A short description of root systemsand a Chevalley basis will be given in section 1.2. We construct F(G) andan ideal R generated by quadratic expressions only in accordance withthis Chevalley basis (section 1.3.2). We use induction to show that thereis a set of generators for the vector space F(G)/R which corresponds toa basis of Lper (Proposition 1.4.11). Then one easily finds a graded Liealgebra isomorphism between Lper and F(G)/R (section 1.4).

1.2 Root System and Chevalley Basis

Let L be a finite-dimensional semisimple Lie algebra over a algebraicallyclosed field F of characteristic 0 without sl2-component. Then L has aChevalley basis and, as we mentioned in the Introduction, this basis isan important part of our proof of Theorem 1. In this section a descrip-tion of the Chevalley basis will be given and some useful properties willbe emphasized. These properties are well-known results (cf. e.g. [10]).However, Lemma 1.2.1 is a consequence of the assumption that L hasno sl2-component.

Let h be a Cartan subalgebra of L, Φ the set of roots of L relative toh, that is

Φ = {α ∈ h∗; α 6= 0 and Lα 6= 0}where

Lα = {x ∈ L; [H, x] = α(H)x for all H ∈ h}.Then Φ spans h∗. For every α ∈ Φ, there is a unique element Hα ∈ hsuch that α(Hα) = 2. In particular, α(Hα) 6= 0.

Observation 1. Let α, β ∈ Φ. Then, β(Hα) is an integer.

Observation 2. If α, β ∈ Φ, then β − β(Hα)α ∈ Φ.

Observation 3. If α ∈ Φ, then −α ∈ Φ. If α ∈ Φ and mα ∈ Φ for somescalar m, then m = ±1.

Notice that Observation 3 implies that α, β ∈ Φ are linearly depen-dent if and only if α = ±β.

Observation 4. If α, β ∈ Φ are linearly independent, then all roots ofthe form β + iα (i ∈ Z) form a string

β − rα, β − (r − 1)α, . . . , β + (q − 1)α, β + qα

where r, q ≥ 0. This string is called the α-string through β.

1.2. Root System and Chevalley Basis 3

Observation 5. Let α, β ∈ Φ such that α 6= ±β. Then

α(Hβ)β(Hα) 6= α(Hα)β(Hβ).

Let Π = {α1, . . . , αl} be the set of fundamental or simple roots, thatis, α1, . . . , αl is a basis of h∗ and each root α ∈ Φ can be written asα =

∑li=1 ciαi where ci are integral coefficients, either all positive (α is

a positive root) or all negative (α is a negative root). Let Φ+ and Φ− bethe sets of positive and negative roots respectively. Then Φ = Φ+ ∪Φ−.

Observation 6. If α ∈ Φ+ such that α /∈ Π, then α− β ∈ Φ+ for someβ ∈ Π (see [10], Lemma A in section 10.2).

To abbreviate, let Hi = Hαifor all 1 ≤ i ≤ l. Then {Hi; 1 ≤ i ≤ l}

is a basis of h. From now on, let {Xα; α ∈ Φ} ∪ {Hi; 1 ≤ i ≤ l} be aChevalley basis of L. This means that:

1. [Hα, Hβ] = 0 for all α, β ∈ Φ

2. [Hβ, Xα] = α(Hβ)Xα for all α, β ∈ Φ

3. [Xα, X−α] = Hα for all α ∈ Φ

4. Let α, β ∈ Φ such that α 6= −β. Then

[Xα, Xβ] =

{Nα,βXα+β if α + β ∈ Φ0 if α + β /∈ Φ

where Nα,β is a non-zero integer.

5. If α, β, α + β ∈ Φ then Nα,β = −N−α,−β

To simplify calculations, let Nα,β = 0 if α, β or α+β /∈ Φ. The followinglemma is a consequence of the fact that L has no sl2-component. Letrank(L) denote the dimension of h∗. This is well defined when L issemisimple.

Lemma 1.2.1. If α ∈ Π, then α + β ∈ Φ for some β ∈ Φ.

Proof. L is a direct sum of simple Lie algebras, none of sl2-type, i.e.,all of rank ≥ 2. The union of the root systems of these simple Liealgebras gives the decomposition of Φ into its irreducible components.If α ∈ Π, let Φ′ be the component such that α ∈ Φ′ and let L′ be thecorresponding simple Lie algebra. Since α is a fundamental root of L, α


is a fundamental root of L′. Now, Φ′ is irreducible and hence, the Dynkindiagram of Φ′ is connected. Since dim(Φ′) ≥ 2, there is a fundamentalroot β ∈ Φ′ such that α 6= β. Choose β to be a neighbor of α in thediagram, i.e., α(Hβ) 6= 0. In fact, then α(Hβ) < 0 since α and β arefundamental roots. Furthermore, α − α(Hβ)β ∈ Φ′. In view of this andof the β-string through α, we have that α + β ∈ Φ.

We end this section with a lemma about the structure constants.

Lemma 1.2.2. Let α, β ∈ Φ such that α 6= ±β. Then

−α(Hβ) = Nα,−βNβ,α−β + Nβ,αN−β,β+α.

Proof. We have that:

[Xα, Hβ] = [Xα, [Xβ, X−β]] = [Xβ, [Xα, X−β]] + [X−β, [Xβ, Xα]]

= Nα,−βNβ,α−βXα + Nβ,αN−β,β+αXα

= (Nα,−βNβ,α−β + Nβ,αN−β,β+α)Xα

On the other hand, [Xα, Hβ] = −α(Hβ)Xα and hence,

−α(Hβ) = Nα,−βNβ,α−β + Nβ,αN−β,β+α.

1.3 The Periodisation and a Free Lie Al-gebra

The notation follows the previous sections, L is a finite-dimensionalsemisimple Lie algebra over a algebraically closed field F of characteris-tic 0 without sl2-component. Furthermore, h is a Cartan subalgebra ofL, Φ the root system relative to h, Π = {α1, . . . αl} a set of fundamentalroots of Φ and {Xα ; α ∈ Φ } ∪ {Hi ; 1 ≤ i ≤ l } a Chevalley basisof L. In this section, we construct the objects appearing in Theorem 1:The periodisation of L and a free Lie algebra.

1.3. The Periodisation and a Free Lie Algebra 5

1.3.1 The Periodisation

Define the periodisation Lper of L as follows:

Definition 1.3.1. Let Lper be the positive graded Lie algebra definedby

Lper = L⊗ tF [t] =⊕i≥1

Lti

with [x⊗ a, y⊗ b] = [x, y]⊗ ab for all x, y ∈ L and a, b ∈ tF [t]. If x ∈ L,let x(i) denote the homogeneous element x⊗ ti.

The Lie algebra Lper can be considered as the graded Lie algebrawith L in each degree. We write (Lper)i for the homogeneous part inLper of degree i (i.e., (Lper)i = Lti). Then, if x ∈ L, x(i) is the corre-

sponding element in (Lper)i. Observe that the set {X(i)α ; α ∈ Φ } ∪

{H(i)i ; 1 ≤ i ≤ l } is a basis of the vector space (Lper)i for all i ≥ 1.

Furthermore, [x(i), y(j)] = [x, y](i+j) for all x, y ∈ L and i, j ≥ 1.

Remark 1.3.2. Let C[t, t−1] be the algebra of Laurent polynomials int and let g be a Lie algebra. The well known loop algebra, L(g) =g⊗C[t, t−1] =

⊕∞−∞ gti, is constructed in the same manner as Lper, Lper

is the positive part of the loop algebra of L. Given a matrix A, let g(A)

be the Kac-Moody algebra. Let◦A be a Cartan matrix of finite type

and let A be the affine extended Cartan matrix of◦A (obtained from

◦A

by adding a certain row and column). Then L(g(◦A)) is isomorphic to

[g(A), g(A)]/c where c is the center of [g(A), g(A)] (see Theorem 7.4 in[13]). For further information about this subject, we refer to [13]. It isprobably possible to deduce Theorem 1 from this isomorphism and theSerre relations.

1.3.2 The Free Lie Algebra

Recall that Lper is the graded Lie algebra with L in each degree and ifx ∈ L, we denote x ∈ (Lper)i by x(i). Also, to abbreviate, let x(1) = x. Inthis section, the object is to construct a set G, the free Lie algebra F(G)and an ideal R in F(G) generated by quadratic elements. In doing this,we keep the Chevalley basis in mind.

Let G = {xα; α ∈ Φ} ∪ {hi; 1 ≤ i ≤ l} and deg(x) = 1 for all x ∈ G.Let F = F(G) be the free Lie algebra generated by G.


Definition 1.3.3. Define a Lie algebra homomorphism φ : F −→ Lper

by induction and linear extension:

1. φ : xα 7→ Xα, φ : hi 7→ Hi for all α ∈ Φ and 1 ≤ i ≤ l

2. φ : [x, y] 7→ [φ(x), φ(y)] for all x, y ∈ FRemark 1.3.4. An easy induction on i shows that φ : Fi −→ (Lper)i

for all i ≥ 1. Hence, φ is in fact a graded Lie algebra homomorphism.Since [L,L] = L, induction on i shows that φ maps F onto Lper.

Recall that, for all α ∈ Φ, there is Hα ∈ h such that α(Hα) = 2. If

Hα =∑l

i=1 ciHi, let hα =∑l

i=1 cihi. Then φ(hα) = Hα for all α ∈ Φ. Toconstruct R, consider the restriction of φ to F2:

φ|F2: F2 −→ Lper :

[xα, xβ] 7→ [Xα, Xβ], α, β ∈ Φ

[hi, xα] 7→ [Hi, Xα], α ∈ Φ, 1 ≤ i ≤ l

[hi, hj] 7→ [Hi, Hj], 1 ≤ i, j ≤ l

Let R be the ideal in F generated by ker(φ|F2). Since ker(φ|F2

) ⊆ F2, R isgenerated by quadratic expressions. Next is an example of an expressionin this kernel.

Example 1.3.5. If α, β, α + β ∈ Φ, then we have the following relationin L

Nα,β[Xα+β, X−α−β] = −N−α,−β[Xα+β, X−α−β] = −[Xα+β, [X−α, X−β]]

= −[X−α, [Xα+β, X−β]]− [X−β, [X−α, Xα+β]]

= −Nα+β,−β[X−α, Xα] + N−α,α+β[Xβ, X−β].

Using this relation in L, we get the following expression in ker(φ|F2) ⊂ F

Nα,β[xα+β, x−α−β] + Nα+β,−β[x−α, xα]−N−α,α+β[xβ, x−β].

Even though we do not need it, the following proposition gives anexplicit list of quadratic expressions generating R.

Proposition 1.3.6. Let L be a finite-dimensional semisimple Lie algebraover a algebraically closed field F of characteristic 0, h a Cartan subalge-bra of L, Φ the root system of L relative to h with a set Π = {α1, . . . , αl}of fundamental roots. Let G be the set given above, φ the graded Liealgebra epimorphism given by Definition 1.3.3 and let R be the ideal inF = F(G) generated by ker(φ|F2

). Then R is generated by the followingquadratic expressions:


1. α(Hα)[hi, xα]− α(Hi)[hα, xα] for all α ∈ Φ, 1 ≤ i ≤ l

2. (α + β)(Hα+β)[xα, xβ]−Nα,β[hα+β, xα+β] for all α, β ∈ Φ such thatα + β ∈ Φ

3. [xα, xβ] for all α, β ∈ Φ such that α + β /∈ Φ, α 6= −β

4. [hi, hj] for all 1 ≤ i, j ≤ l

5. [xα, x−α]−∑li=1 ci[xαi

, x−αi] for all α ∈ Φ where Hα =

∑li=1 ciHi

Proof. First to abbreviate, let φ2 = φ|F2. Let R′ be the ideal in F

generated by the quadratic expressions in 1-5. We must show that R′ =R. Let R′

2 = F2∩R′. Clearly, R′ is generated by R′2. Since R is generated

by ker(φ2) by definition, it suffices to show that R′2 = ker(φ2).

To take φ2 on an expression in F2 is to change small letters in the ex-pression to capital ones. In view of the list of properties of the Chevalleybasis given in section 1.2, it is easy to see that the quadratic expressionsgiven in Proposition 1.3.6 are in ker(φ2). Hence, R′

2 ⊂ ker(φ2). Considerthe vector space epimorphism:

ϕ : F2/R′2 −→ F2/ ker(φ2) : x + R′

2 7→ x + ker(φ2)

If we can show that ϕ is an isomorphism, then we know that R′ = ker(φ2)and we are done.

Now, F2/R′2 and F2/ ker(φ2) are finite-dimensional vector spaces.

Thus, to show that ϕ is an isomorphism, it suffices to show thatdimF2/R

′2 = dimF2/ ker(φ2). Since ϕ is an epimorphism, we have that

dimF2/R′2 ≥ dimF2/ ker(φ2). Furthermore, Im(φ2) = (Lper)2

∼= L.Hence, F2/ ker(φ2) ∼= L, and we must show that dimF2/R

′2 ≤ dim L.

If x ∈ F2, let x denote the image of x in F2/R′2. F2 is generated by

{[xα, xβ]; α, β ∈ Φ} ∪ {[xα, hi]; α ∈ Φ, 1 ≤ i ≤ l} ∪ {[hi, hj; 1 ≤ i, j ≤ l]}

and then, F2/R′2 is generated by

{[xα, xβ]; α, β ∈ Φ} ∪ {[hi, xα]; α ∈ Φ, 1 ≤ i ≤ l} ∪ {[hi, hj; 1 ≤ i, j ≤ l]}.


From 1-5 in Proposition 1.3.6 we have that:

[hi, xα] =α(Hi)

α(Hα)[hα, xα] for all α ∈ Φ, 1 ≤ i ≤ l

[xα, xβ] =Nα,β

(α + β)(Hα+β)[hα+β, xα+β] for all α, β ∈ Φ such

that α + β ∈ Φ

[xα, xβ] = 0 for all α, β ∈ Φ such that α + β /∈ Φ, α 6= −β

[hi, hj] = 0 for all 1 ≤ i 6= j ≤ l

[xα, x−α] =l∑

i=1

ci[xαi, x−αi

] for all α ∈ Φ where Hα =l∑

i=1

ciHi

Hence, F2/R′2 is generated by

{[hα, xα]; α ∈ Φ} ∪ {[xαi, x−αi

]; 1 ≤ i ≤ l}.

Since {Xα; α ∈ Φ} ∪ {Hi; 1 ≤ i ≤ l} is a basis of L, we have thatdimF2/R

′2 ≤ dim L.

Remark 1.3.7. If L is a simple Lie algebra, there is a unique maximalroot γ ∈ Φ (the root such that γ +α /∈ Φ for any α ∈ Φ+), and Xγ is theunique maximal vector up to non-zero scalar multiples in L viewed as anL-module (i.e., the vector such that Xα.Xγ = 0 for all α ∈ Φ+). Let Vbe the subspace in F2 generated by the expressions in Proposition 1.3.6.Then V is an L-module and the vectors

[xγ−αi, xγ] where αi ∈ Π such that γ − αi ∈ Φ (1.1)

are the maximal vectors up to non-zero scalar multiples. Hence, V isgenerated by the vectors (1.1) as an L-module. Table 1.1 shows themaximal root and the maximal vectors, or generators, for the differenttypes of L. We see that if L is of type Al (l ≥ 2), V has two generatorsand otherwise, V is generated by one vector only.

From now on, we mainly do calculations in F/R. To simplify thenotations, we therefore let {xα, α ∈ Φ} ∪ {hi, 1 ≤ i ≤ l} denote theimages of the generators for F in F/R. Hopefully, no confusion will arise.To make later calculations more understandable, we do the followingabbreviations. Note that α(Hα) 6= 0.


Table 1.1: Generators for V

Type Maximal root γ GeneratorsAl α1 + . . . + αl [xγ−α1

, xγ], [xγ−αl, xγ]

Bl α1 + 2α2 + . . . + 2αl [xγ−α2, xγ]

Cl 2α1 + . . . + 2αl−1 + αl [xγ−α1, xγ]

Dl α1 +2α2 + . . .+2αl−2 +αl−1 +αl [xγ−α2, xγ]

E6 α1 + 2α2 + . . . + 2α5 + α6 [xγ−α2, xγ]

E7 2α1 + 2α2 + 3α3 + 4α4 + 3α5 +2α6 + α7

[xγ−α1, xγ]

E8 2α1 + 3α2 + 4α3 + 6α4 + 5α5 +4α6 + 3α7 + 2α8

[xγ−α8, xγ]

F4 2α1 + 3α2 + 4α3 + 2α4 [xγ−α1, xγ]

G2 3α1 + 2α2 [xγ−α2, xγ]

Definition 1.3.8. Let {xα, α ∈ Φ} ∪ {hi, 1 ≤ i ≤ l} be the images ofthe generators for F in F/R. Define by induction:

1. x(2)α = 1

α(Hα) [hα, xα], x(n)α = 1

α(Hα) [hα, x(n−1)α ]

2. h(2)α = [xα, x−α], h

(n)α = [xα, x

(n−1)−α ]

Furthermore, let x(1)α = xα, h

(1)i = hi

Example 1.3.9. In this example, we derive two relations in F(G)/Rthat will be used many times later on. Take any α, β ∈ Φ such thatα + β ∈ Φ.

1. Using the properties of the Chevalley basis listed in section 1.2,the relation in L given in Example 1.3.5 can be written as

Nα,βHα+β + Nα+β,−βH−α −N−α,α+βHβ = 0.

By the definition of hδ when δ ∈ Φ, we have that

Nα,βhα+β + Nα+β,−βh−α −N−α,α+βhβ = 0

in F(G)/R.

2. Take the expression in ker(φ|F2) given in Example 1.3.5 and use

Definition 1.3.8. Then we have that

Nα,βh(2)α+β + Nα+β,−βh

(2)−α −N−α,α+βh

(2)β = 0

in F(G)/R.


1.4 Proof of Theorem 1

In this section we prove Theorem 1. However, first we state and prove apreparatory proposition, Proposition 1.4.1. This is an important result,since these relations reduce the dimension in each homogeneous part ofF/R (c.f. Proposition 1.4.11). In fact, this proposition does all the hardwork in proving Theorem 1.

Proposition 1.4.1. Let L be a finite-dimensional semisimple Lie alge-bra over a algebraically closed field F of characteristic 0 without sl2-component, h a Cartan subalgebra of L and Φ the root system of Lrelative to h with a set Π = {α1, . . . αl} of fundamental roots. Let{Xα ; α ∈ Φ } ∪ {Hi ; 1 ≤ i ≤ l } be a Chevalley basis ofL. Moreover, let G be the set of generators and R the ideal of F(G)given in section 1.3.2. Take any α, β ∈ Φ and 1 ≤ i, j ≤ l. Then, forevery k ≥ 1 we have the following relations in F(G)/R:

1. [hi, x(k)α ] = α(Hi)x

(k+1)α

2. If α 6= −β, then [xα, x(k)β ] = Nα,βx

(k+1)α+β

3. [xα, h(k)i ] = −α(Hi)x

(k+1)α

4. [hi, h(k)j ] = 0

5. [xα, x(k)−α] is a linear combination of h

(k+1)1 , . . . , h

(k+1)l

Notice that the statement in 2. is a bit unclear since xα+β is notdefined when α + β /∈ Φ. However, Nα,β = 0 if α + β /∈ Φ, so what weactually mean is that

[xα, x(k)β ] =

{Nα,βx

(k+1)α+β if α + β ∈ Φ

0 if α + β /∈ Φ

Hence, let x(k)δ = 0 for all δ /∈ Φ and all k ≥ 1 and we get the state-

ment in 2. The advantage in writing like this is the simplicity in thecalculations.

The proof of Proposition 1.4.1 is done by induction and depends onseveral lemmas. Since these lemmas depend on the induction hypothesis,they are included in the proof.

1.4. Proof of Theorem 1 11

Proof of Proposition 1.4.1. We use induction on k. For k = 1, 1-5 ofProposition 1.4.1 are just consequences of the quadratic expressions inR and Definition 1.3.8 (c.f. Proposition 1.3.6). Assume that Proposition1.4.1 is true for all k < n for some n ≥ 2. The induction step is given inthe lemmas below. Observe that the induction hypothesis is assumed tobe valid in the lemmas.

Lemma 1.4.2. Let i ∈ {1, . . . , l}. Then for any α ∈ Φ:

[hi, x(n)α ] = α(Hi)x

(n+1)α

Proof of Lemma 1.4.2.

[hi, x(n)α ] =

1

α(Hα)[hi, [hα, x(n−1)

α ]]

=1

α(Hα)[hα, [hi, x

(n−1)α ]] +

1

α(Hα)[x(n−1)

α , [hα, hi]]

=α(Hi)

α(Hα)[hα, x(n)

α ] = α(Hi)x(n+1)α

The first and the last equality follows from Definition 1.3.8, the secondfrom the Jacobi identity and the third equality follows from the fact hα isa linear combination of h1, . . . , hl and the quadratic relations [hj, hi] = 0for all 1 ≤ i , j ≤ l.

Lemma 1.4.3. Let H =∑l

i=1 ciHi and let h =∑l

i=1 cihi be the corre-sponding element in F/R. Then

[h, x(k)α ] = α(H)x(k+1)

α

for all k ≤ n.

Proof of Lemma 1.4.3. According to the induction hypothesis and toLemma 1.4.2 we have that

[hi, x(k)α ] = α(Hi)x

(k+1)α

for all 1 ≤ i ≤ l and k ≤ n. Then, by linearity of the bracket-operatorand of α, we have the desired equality.

Lemma 1.4.4. Let α, β ∈ Φ such that α 6= −β. Then

[xα, x(n)β ] = Nα,βx

(n+1)α+β .


Proof of Lemma 1.4.4. We proceed in steps:(i) If α, β are linearly independent (α, β ∈ h∗), there is H ∈ h such thatα(H) = 0 and β(H) 6= 0. Let h denote the corresponding element inF/R. Then:

[xα, x(n)β ] =

1

β(H)[xα, [h, x

(n−1)β ]]

=1

β(H)[h, [xα, x

(n−1)β ]] +

1

β(H)[x

(n−1)β , [h, xα]]

=Nα,β

β(H)[h, x

(n)α+β] +

α(H)

β(H)[x

(n−1)β , x(2)

α ] = Nα,βx(n+1)α+β

The first equation follows from Lemma 1.4.3, the third from the in-duction hypothesis and Lemma 1.4.3 and the last equality follows fromLemma 1.4.3 and the fact that α(H) = 0.(ii) If α and β are linearly dependent then α = β according to Obser-vation 3 and the assumption that α 6= −β. Hence, we must show that

[xα, x(n)α ] = 0. By combining Observation 6 and Lemma 1.2.1 and con-

sidering the different cases when ±α ∈ Π and ±α ∈ Φ+ but ±α /∈ Πrespectively, we have that there is a γ ∈ Φ such that α 6= ±γ andα + γ ∈ Φ. Then Nα+γ,−γ 6= 0 and we have that:

[xα, x(n)α ] =

1

Nα+γ,−γ[xα, [xα+γ, x

(n−1)−γ ]]

=1

Nα+γ,−γ[xα+γ, [xα, x

(n−1)−γ ]] +

1

Nα+γ,−γ[x

(n−1)−γ , [xα+γ, xα]]

=Nα,−γ

Nα+γ,−γ[xα+γ, x

(n)α−γ] +

Nα+γ,α

Nα+γ,−γ[x

(n−1)−γ , x

(2)2α+γ]

The first and the last equality follows from the induction hypothesis.We will show that both terms in this sum are zero. If α − γ /∈ Φ thenNα,−γ = 0. If α−γ ∈ Φ then α+γ and α−γ must be linear independent.

By the first part of the proof we have that [xα+γ, x(n)α−γ] = 0 since 2α /∈ Φ

according to Observation 3. Let δ = 2α + γ. If δ /∈ Φ then Nα+γ,α = 0.If δ ∈ Φ then

[x(n−1)−γ , x

(2)2α+γ] =

1

δ(Hδ)[x

(n−1)−γ , [hδ, x2α+γ]]

=1

δ(Hδ)(−[hδ, [x2α+γ, x

(n−1)−γ ]] + [x2α+γ, [hδ, x

(n−1)−γ ]])

=−γ(Hδ)

δ(Hδ)[x2α+γ, x

(n)−γ ]


The first equation follows from Lemma 1.4.3 and the last equality followsfrom Lemma 1.4.3, the induction hypothesis and the fact that 2α + γ −γ = 2α /∈ Φ. Now, −γ and 2α + γ are linearly independent. Hence,

according to the first part of the proof [x2α+γ, x(n)−γ ] = 0.

Lemma 1.4.5. Let α, β ∈ Φ such that α 6= −β. Then

[x(2)α , x

(k−1)β ] = Nα,βx

(k+1)α+β for all k ≤ n.

Proof of Lemma 1.4.5. According to the induction hypothesis and

Lemma 1.4.4, we have that [xα, x(k)β ] = Nα,βx

(k+1)α+β for all k ≤ n. Hence:

[x(2)α , x

(k−1)β ] = − 1

α(Hα)[x

(k−1)β , [hα, xα]]

= − 1

α(Hα)[hα, [x

(k−1)β , xα]]− 1

α(Hα)[xα, [hα, x

(k−1)β ]]

=(α + β)(Hα)

α(Hα)Nα,βx

(k+1)α+β − β(Hα)

α(Hα)Nα,βx

(k+1)α+β

= Nα,βx(k+1)α+β

We have now done the induction step for statement 1 - 2 in Propo-sition 1.4.1. Before proceeding, we give a useful lemma that generalizesthe relations given in Example 1.3.9.

Lemma 1.4.6. Let α, β, α + β ∈ Φ. Then

Nα,βh(n)α+β + Nα+β,−βh

(n)−α −N−α,α+βh

(n)β = 0

Proof of Lemma 1.4.6. If n = 2, Lemma 1.4.6 follows from Example1.3.9. Assume that n > 2. Since α + β ∈ Φ, it follows that −α− β ∈ Φ


and N−α,−β 6= 0. Then:

Nα,βh(n)α+β = −N−α,−β[xα+β, x

(n−1)−α−β] = −[xα+β, [x−α, x

(n−2)−β ]]

= −[x−α, [xα+β, x(n−2)−β ]]− [x

(n−2)−β , [x−α, xα+β]]

= −Nα+β,−β[x−α, x(n−1)α ]−N−α,α+β[x

(n−2)−β , x

(2)β ]

= −Nα+β,−βh(n)−α]−N−α,α+β[x

(n−2)−β , x

(2)β ]

[x(n−2)−β , x

(2)β ] =

1

β(Hβ)[x

(n−2)−β , [hβ, xβ]]

=1

β(Hβ)[hβ, [x

(n−2)−β , xβ]] +

1

β(Hβ)[xβ, [hβ, x

(n−2)−β ]]

= − 1

β(Hβ)[hβ, h

(n−1)β ]− [xβ, x

(n−1)−β ]] = −h

(n)β

Hence, Nα,βh(n)α+β = −Nα+β,−βh

(n)−α + N−α,α+βh

(n)β

Lemma 1.4.7. Let i ∈ {1, . . . , l} and let α ∈ Φ. Then

[xα, h(n)i ] = −α(Hi)x

(n+1)α

Proof of Lemma 1.4.7. First, take any δ ∈ Φ such that α 6= ±δ. Then:

[xα, h(n)δ ] = [xα, [xδ, x

(n−1)−δ ]] = [xδ, [xα, x

(n−1)−δ ]] + [x

(n−1)−δ , [xδ, xα]]

= Nα,−δ[xδ, x(n)α−δ] + Nδ,α[x

(n−1)−δ , x

(2)δ+α]

= Nα,−δNδ,α−δx(n+1)α + Nδ,αN−δ,δ+αx(n+1)

α

= (Nα,−δNδ,α−δ + Nδ,αN−δ,δ+α)x(n+1)α = −α(Hδ)x

(n+1)α

The third equality follows from Lemma 1.4.4 and Lemma 1.4.5 and thelast equality follows from Lemma 1.2.2. Hence:

[xα, h(n)δ ] = −α(Hδ)x

(n+1)α for all δ ∈ Φ such that α 6= ±δ (1.2)

If α 6= ±αi, take δ = αi in equation (1.2) and we are done. Assume thatα = ±αi. According to Lemma 1.2.1 there is a β ∈ Φ such that αi 6= ±βand αi + β ∈ Φ. Now, use the following relation given in Lemma 1.4.6:

Nβ,αih

(n)β+αi

+ Nβ+αi,−αih

(n)−β −N−β,β+αi

h(n)αi

= 0


Then N−β,β+αi6= 0 and we have that:

[xα, h(n)i ] =

Nβ,αi

N−β,β+αi

[xα, h(n)β+αi

] +Nβ+αi,−αi

N−β,β+αi

[xα, h(n)−β]

=Nβ,αi

N−β,β+αi

[x(n)α , hβ+αi

] +Nβ+αi,−αi

N−β,β+αi

[x(n)α , h−β]

= [x(n)α , hi] = −α(Hi)x

(n+1)α

The second equality follows from the fact that α 6= ±(β + αi) and α 6=±β, equation (1.2) and Lemma 1.4.3. The third equality follows fromExample 1.3.9 and the last equality follows from Lemma 1.4.2.

Before doing the induction step for the last two statements in Propo-sition 1.4.1, we need the following lemma.

Lemma 1.4.8. Let α ∈ Π. Then:

[xα, x(n)−α] + [x

(n−1)−α , x(2)

α ] = 0

[x−α, x(n)α ] + [x(n−1)

α , x(2)−α] = 0

[xα, x(n)−α] + [x−α, x(n)

α ] = 0

Proof of Lemma 1.4.8. By Lemma 1.2.1, there is β ∈ Φ such that α+β ∈Φ. Furthermore, according to Observation 3 and the fact that 0 /∈ Φ, wehave that α 6= ±β. Consider the relation given in Lemma 1.4.6:

Nα,βh(n)α+β + Nα+β,−βh

(n)−α −N−α,α+βh

(n)β = 0

Taking the Lie bracket with this relation and, on one hand hα and, onthe other hand hβ, we get the following system of equations:

{Nα,β[hα, h

(n)α+β] + Nα+β,−β[hα, h

(n)−α]−N−α,α+β[hα, h

(n)β ] = 0

Nα,β[hβ, h(n)α+β] + Nα+β,−β[hβ, h

(n)−α]−N−α,α+β[hβ, h

(n)β ] = 0

(1.3)

Now, take any δ, γ ∈ Φ. Then we have that:

[hγ, h(n)δ ] = [hγ, [xδ, x

(n−1)−δ ]]

= [xδ, [hγ, x(n−1)−δ ]] + [x

(n−1)−δ , [xδ, hγ]]

= −δ(Hγ)[xδ, x(n)−δ ]− δ(Hγ)[x

(n−1)−δ , x

(2)δ ] (1.4)

The first equality follows from Definition 1.3.8, the second from the Ja-cobi identity. Recall that if Hγ =

∑li=1 ciHi, we defined hγ =

∑li=1 cihi


and the last equality follows from Lemma 1.4.3. Apply equation (1.4) toeach bracket in system (1.3). To simplify, put:

x = Nα,β([xα+β, x(n)−α−β] + [x

(n−1)−α−β, x

(2)α+β])

y = Nα+β,−β([x−α, x(n)α ] + [x

(n−1)α , x

(2)−α])

z = N−α,α+β([xβ, x(n)−β] + [x

(n−1)−β , x

(2)β ])

System (1.3) then becomes:

{ −(α + β)(Hα)x + α(Hα)y + β(Hα)z = 0−(α + β)(Hβ)x + α(Hβ)y + β(Hβ)z = 0

Adding the multiple −β(Hα) of the second row to the multiple β(Hβ) ofthe first row, and adding the multiple −α(Hα) of the second row to themultiple α(Hβ) of the first row we obtain:

{(α(Hβ)β(Hα)− α(Hα)β(Hβ)

)(x− y) = 0(

α(Hα)β(Hβ)− α(Hβ)β(Hα))(x− z) = 0

(1.5)

According to Observation 5 (and the fact that α 6= ±β), we have thatα(Hβ)β(Hα) 6= α(Hα)β(Hβ). Hence, dividing the equations in sys-tem (1.5) by α(Hβ)β(Hα)−α(Hα)β(Hβ) and α(Hα)β(Hβ)−α(Hβ)β(Hα)respectively, we get {

x− y = 0x− z = 0.

(1.6)

The idea is to find a third relation between x, y and z which is linearlyindependent of the first two.

Consider the following relations:

Nα,β[xα+β, x(n)−α−β] = −N−α,−β[xα+β, x

(n)−α−β] = −[xα+β, [x−α, x

(n−1)−β ]]

= −[x−α, [xα+β, x(n−1)−β ]]− [x

(n−1)−β , [x−α, xα+β]]

= −Nα+β,−β[x−α, x(n)α ]−N−α,α+β[x

(n−1)−β , x

(2)β ]

Nα,β[x(n−1)−α−β, x

(2)α+β] = [x

(n−1)−α−β, [xα, xβ]]

= [xα, [x(n−1)−α−β, xβ]] + [xβ, [xα, x

(n−1)−α−β]]

= N−α−β,β[xα, x(n)−α] + Nα,−α−β[xβ, x

(n)−β]

= −Nα+β,−β[xα, x(n)−α]−N−α,α+β[xβ, x

(n)−β]


Adding these two equalities we get

x + Nα+β,−β([xα, x(n)−α] + [x−α, x(n)

α ]) + z = 0 (1.7)

We proceed in steps:

(i) When n = 2, we must show that [xα, x(2)−α] + [x−α, x

(2)α ] = 0 for all

α ∈ Π. If n = 2, then

y = Nα+β,−β([xα, x(2)−α] + [x−α, x(2)

α ])

and equation (1.7) becomes x + y + z = 0. This equation is linearlyindependent with the two in system (1.6). Hence, x = y = z = 0 and

particularly, [xα, x(2)−α] + [x−α, x

(2)α ] = 0.

(ii) For n > 2, we first show that

[xα, x(k)−α] = [x(m1)

α , x(m2)−α ] (1.8)

for all 1 ≤ k ≤ n − 1 and all m1,m2 ≥ 1 such that m1 + m2 = k + 1.

Observe that the induction hypothesis gives that [hα, h(k)α ] = 0 for all k <

n. If k = 2 equation (1.8) follows from (i). Assume that equation (1.8)is true for all k < l where 2 < l ≤ n− 1. Take any 1 ≤ m ≤ l− 1. Then:

[x(m)α , x

(l−m+1)−α ] = − 1

α(Hα)[x(m)

α , [hα, x(l−m)−α ]]

= − 1

α(Hα)[hα, [x(m)

α , x(l−m)−α ]]− 1

α(Hα)[x

(l−m)−α , [hα, x(m)

α ]]

= − 1

α(Hα)[hα, [xα, x

(l−1)−α ]]− [x

(l−m)−α , x(m+1)

α ]

= − 1

α(Hα)[hα, h(l)

α ] + [x(m+1)α , x

(l−m)−α ] = [x(m+1)

α , x(l−m)−α ]

The third equality follows from the fact that l−1 < l and the assumptionthat equation (1.8) is true for all k < l. The last equality follows fromthe fact that l ≤ n − 1 and the induction hypothesis. Now, let m runfrom 1 to l − 1 and we are done. Hence (1.8) is proved.

Consider the following relation for all 1 ≤ m ≤ n− 1:

[hα, h(n)α ] = [hα, [xα, x

(n−1)−α ]] = [hα, [x(m)

α , x(n−m)−α ]]

= [x(m)α , [hα, x

(n−m)−α ]] + [x

(n−m)−α , [x(m)

α , hα]]

= −α(hα)[x(m)α , x

(n−m+1)−α ]− α(hα)[x

(n−m)−α , x(m+1)

α ]

= −α(hα)([x(m)α , x

(n−m+1)−α ] + [x

(n−m)−α , x(m+1)

α ])


The second equality follows from equation (1.8) for k = n−1. By lettingm run from 1 to n− 1 we get:

[xα,x(n)−α] + [x

(n−1)−α , x(2)

α ] = −[x(n−1)−α , x(2)

α ] + [x(n−2)−α , x(3)

α ]

= −[x(n−2)−α , x(3)

α ] + [x(n−3)−α , x(4)

α ] = · · · = −[x(2)−α, x(n−1)

α ] + [x−α, x(n)α ](1.9)

Adding all the equations in (1.9) gives:

(n− 1)([x−α, x(n)α ] + [x(n−1)

α , x(2)−α]) = [xα, x

(n)−α] + [x−α, x(n)

α ] (1.10)

This inserted in equation (1.7) gives x + (n − 1)y + z = 0. Since thecharacteristic is 0, this equation is linearly independent with the two

in system (1.6) and hence, x = y = z = 0. Particularly, [x−α, x(n)α ] +

[x(n−1)α , x

(2)−α] = 0. By equation (1.9), we also have that

[xα, x(n)−α] + [x

(n−1)−α , x(2)

α ] = [x−α, x(n)α ] + [x(n−1)

α , x(2)−α] = 0.

Finally, by equation (1.10) we have that [xα, x(n)−α] + [x−α, x

(n)α ] = 0.

Lemma 1.4.9. Let i, j ∈ {1, . . . , l}. Then for all n ≥ 1:

[hi, h(n)j ] = 0

Proof of Lemma 1.4.9. Take any 1 ≤ i, j ≤ l. Then:

[hi, h(n)j ] = [hi, [xαj

, x(n−1)−αj

]]

= [xαj, [hi, x

(n−1)−αj

]] + [x(n−1)−αj

, [xαj, hi]]

= −αj(hi)[xαj, x

(n)−αj

]− αj(hi)[x(n−1)−αj

, x(2)αj

] = 0

The last equality follows from Lemma 1.4.8.

Lemma 1.4.10. Let α ∈ Φ. Then [xα, x(n)−α] is a linear combination of

h(n+1)1 , . . . , h

(n+1)l .

Proof of Lemma 1.4.10. Recall from section 1.2 that every positive rootcan be written uniquely as a sum of fundamental roots. Hence, we cantalk about the length of a root. If α ∈ Φ, let l(α) denote the length ofα. Clearly, l(α) = l(−α). We use induction over this length to proveLemma 1.4.10.


If l(α) = 1, then α = ±αi for some i ∈ {1, . . . l}. Clearly, [xαi, x

(n)−αi

] =

h(n+1)i by Definition 1.3.8. According to Lemma 1.4.8, we have that

[x−αi, x

(n)αi ] = −[xαi

, x(n)−αi

] = h(n+1)i . Now, assume that Lemma 1.4.10 is

true for all roots of length < m for some m > 1, and let α ∈ Φ suchthat l(α) = m. Since m > 1, we have that ±α is not fundamental. Weproceed in steps:(i) If α ∈ Φ+ there is, according to Observation 6, αi ∈ Π such thatα− αi ∈ Φ+. Clearly, l(α− αi) < m. Furthermore, N−α+αi,−αi

6= 0 and

[xα, x(n)−α] =

1

N−α+αi,−αi

[xα, [x−α+αi, x

(n−1)−αi

]]

=1

N−α+αi,−αi

[x−α+αi, [xα, x

(n−1)−αi

]]

+1

N−α+αi,−αi

[x(n−1)−αi

, [x−α+αi, xα]]

=Nα,−αi

N−α+αi,−αi

[x−α+αi, x

(n)α−αi

] +N−α+αi,α

N−α+αi,−αi

[x(n−1)−αi

, x(2)αi

]

=Nα,−αi

N−α+αi,−αi

[x−α+αi, x

(n)α−αi

]− N−α+αi,α

N−α+αi,−αi

[xαi, x

(n)−αi

]

The last equality follows from Lemma 1.4.8. Now, since l(αi) < m, and

l(−α + αi) = l(α − αi) < m we have that [xαi, x

(n)−αi

] and [x−α+αi, x

(n)α−αi

]

are linear combinations of h(n+1)1 , . . . , h

(n+1)l . Hence, [xα, x

(n)−α] is a linear

combination of h(n+1)1 , . . . , h

(n+1)l .

(ii) If α ∈ Φ− then −α ∈ Φ+. Since −α /∈ Π, there is, according toObservation 6, αi ∈ Π such that −α − αi ∈ Φ+. Hence, there is αi ∈ Πsuch that α+αi ∈ Φ. Clearly, l(α+αi) < m. Furthermore N−α−αi,αi

6= 0and

[xα, x(n)−α] =

1

N−α−αi,αi

[xα, [x−α−αi, x(n−1)

αi]]

=1

N−α−αi,αi

[x−α−αi, [xα, x(n−1)

αi]]

+1

N−α−αi,αi

[x(n−1)αi

, [x−α−αi, xα]]

=Nα,αi

N−α−αi,αi

[x−α−αi, x

(n)α+αi

] +N−α−αi,α

N−α−αi,αi

[x(n−1)αi

, x(2)−αi

]

=Nα,αi

N−α+αi,αi

[x−α−αi, x

(n)α+αi

] +N−α−αi,α

N−α−αi,αi

[xαi, x

(n)−αi

]


The last equality follows from Lemma 1.4.8. Since l(αi) < m and

l(−α− αi) = l(α + αi) < m, we have that [xαi, x

(n)−αi

] and [x−α−αi, x

(n)α+αi

]

are linear combinations of h(n+1)1 , . . . , h

(n+1)l . Hence, [xα, x

(n)−α] is a linear

combination of h(n+1)1 , . . . , h

(n+1)l .

In Lemma 1.4.2, 1.4.4, 1.4.7, 1.4.9 and 1.4.10 we have done the induc-tion step for statement 1-5 in Proposition 1.4.1. Hence Proposition 1.4.1is true for all k ≥ 1.

The next proposition is a consequence of Proposition 1.4.1.

Proposition 1.4.11. Let L be a finite-dimensional semisimple Lie al-gebra over a algebraically closed field F of characteristic 0 without sl2-component. Let {Xα ; α ∈ Φ } ∪ {Hi ; 1 ≤ i ≤ l } be a Chevalleybasis of L, where Φ is the root system of L relative to a Cartan subalgebrah and Π = {α1, . . . αl} is a set of fundamental roots. Take G to be the setof generators and R to be the ideal of F = F(G) given in section 1.3.2.Then the set

{x(k)α ; α ∈ Φ} ∪ {h(k)

i ; 1 ≤ i ≤ l}(see Definition 1.3.8) generates (F/R)k for all k ≥ 1.

Proof. This is clear for k = 1. Assume that { x(k)α ; α ∈ Φ } ∪

{h(k)i ; 1 ≤ i ≤ l} generates (F/R)k for all k ≤ n, for some n ≥ 1.

Take any y ∈ (F/R)1, z ∈ (F/R)n. Then

y =l∑

i=1

aihi +∑

α∈Φ

aαxα, z =l∑

i=1

bih(n)i +

∑

α∈Φ

bαx(n)α

for some coefficients ai, aα, bi, bα. Hence,


[y, z] =

[l∑

i=1

aihi +∑

α∈Φ

aαxα,

l∑i=1

bih(n)i +

∑

α∈Φ

bαx(n)α

]

=l∑

i=1

l∑j=1

aibj[hi, h(n)j ] +

∑

α∈Φ

l∑i=1

aibα[hi, x(n)α ] +

∑

α∈Φ

l∑i=1

aαbi[xα, h(n)i ]

+∑

α∈Φ

∑

β∈Φ

aαbβ[xα, x(n)β ]

=∑

α∈Φ

l∑i=1

(aibα − aαbi)α(Hi)x(n+1)α +

∑

α∈Φ

∑

β∈Φβ 6=−α

aαbβNα,βx(n+1)α+β

+∑

α∈Φ

aαb−αh(n+1)α .

The third equality follows from Proposition 1.4.1. Moreover, accord-

ing to Proposition 1.4.1 we have that h(n+1)α is a linear combination of

h(n+1)1 , . . . , h

(n+1)l . Since every element in (F/R)n+1 can be written as a

finite sum of terms of the form [y, z] where y ∈ (F/R)1 and z ∈ (F/R)n

we are done.

We are now in a position to restate and prove Theorem 1.

Theorem 1. Let L be a finite-dimensional semisimple Lie algebra overan algebraically closed field F of characteristic 0 without sl2-component.Then there is a set of generators G and an ideal R of F(G), generatedby quadratic relations only, such that Lper

∼= F(G)/R.

Proof. Let G be the set of generators and R the ideal of F(G) given insection 1.3.2. To abbreviate, let F = F(G). Consider the graded Liealgebra epimorphism φ : F → Lper also given in Section 1.3.2. Now,R ⊂ ker(φ) and let ϕ be the graded Lie algebra epimorphism such that

ϕ : F/R −→ Lper : ϕ(x + R) = φ(x) for all x ∈ F .

To show that Lper∼= F/R it suffices to show that ϕ is one-to-one.

The set ⋃

k≥1

({X(k)

α ; α ∈ Φ} ∪ {H(k)i ; 1 ≤ i ≤ l}

)


is a basis of Lper (see the paragraph after Definition 1.3.1) and, accordingto Proposition 1.4.11, the set

⋃

k≥1

({x(k)

α ; α ∈ Φ} ∪ {h(k)i ; 1 ≤ i ≤ l}

)

generates F/R. Furthermore, by induction over k, we see that

ϕ(x(k)α ) = X(k)

α , ϕ(h(k)i ) = H

(k)i

for all α ∈ Φ, 1 ≤ i ≤ l and all k ≥ 1. Hence, ϕ maps a set of generatorsto a basis and thus, ϕ is one-to-one.

We end with an example of a Lie algebra with a sl2-component,namely sl2 itself, and for which Theorem 1 not is fulfilled.

Example 1.4.12. Consider the Lie algebra sl2 over an algebraicallyclosed field F of characteristic 0. For a Chevalley basis, take

X =

(0 10 0

)Y =

(0 01 0

)H =

(1 00 −1

)

where [X, Y ] = H, [H,X] = 2X and [H, Y ] = −2Y . Since char(F ) = 0,there are no quadratic relations in (sl2)per. Hence, no presentation of(sl2)per with generators and relations can have any quadratic relations.On the other hand, there are cubic relations in (sl2)per, e.g. [Y, [H, Y ]].In fact, (sl2)per can be presented by cubic relations. Let G = {x, y, h}and let R be generated by [x, [h, x]], [h, [h, x]] + 2[x, [x, y]], [h, [x, y]],[h, [h, y]] + 2[y, [y, x]] and [y, [h, y]]. Then (sl2)per

∼= F(G)/R. Fur-thermore, the vector space R ∩ F(G)3 is generated by [x, [h, x]] as ansl2-module (c.f. Remark 1.3.7).

Chapter 2

Contragredient LieSuperalgebras

In this chapter we define contragredient Lie superalgebras, the root sys-tem and the structure constants (for more detailed information we referto e.g. [14], [27], [10], [13] and [5]). For semisimple Lie algebras and formost classical Lie superalgebras, all root spaces are one-dimensional. Toobtain a decomposition into one-dimensional components for more gen-eral contragredient Lie superalgebras, we introduce the so called root-tuples. In Section 2.1, we work with a general Cartan matrix. In therest of the chapter we assume that the Cartan matrices in question havenon-zero determinant.

2.1 Definition of Contragredient Lie Su-peralgebras

Given a matrix A = (aij)ri,j=1 over a field k and a subset τ of [r] =

{1, . . . , r}, let g(A, τ) =⊕

l∈Z gl be the free Z-graded Lie superalgebragenerated by {fi, hi, ei}i∈[r] with defining relations

[ei, fj] = δijhi, [hi, hj] = 0,

[hi, ej] = aijej, [hi, fj] = −aijfj

sdeg(hi) ≡ 0 (mod 2)

sdeg(ei) ≡ sdeg(fi) ≡{

0 (mod 2) if i 6∈ τ

1 (mod 2) if i ∈ τ

(2.1)

24 2. Contragredient Lie Superalgebras

for all i, j ∈ [r], where δij is the Kronecker-delta. The subspaces g−1, g0and g1 of g(A, τ) are vector spaces over k with bases {fi}i∈[r], {hi}i∈[r]and {ei}i∈[r] respectively. Furthermore, for all l ≥ 2, gl and g−l are thesubspaces generated by

{[· · · [ei1, ei2] · · · , eil]}i1,...il∈[r] and {[· · · [fi1, fi2] · · · , fil]}i1,...il∈[r]

respectively (see e.g. [11] and [31]). The subspace g−1⊕ g0⊕ g1 is calledthe local part of g(A, τ).

The sum of all Z-graded ideals in g(A, τ) intersecting the local parttrivially is maximal with this property. Denote this maximal Z-gradedideal by m. The next proposition gives an explicit description of thegraded components of m (see also [13] and [31]).

Proposition 2.1.1. Let A be an r× r-matrix and let m = ⊕l∈Z (m ∩ gl)be the maximal Z-graded ideal in g(A, τ) intersecting the local part triv-ially. Then m ∩ gl = 0 for l = −1, 0, 1 and

m ∩ gl = {x ∈ gl; [x, fi] ∈ m for all i ∈ [r]}m ∩ g−l = {x ∈ g−l; [x, ei] ∈ m for all i ∈ [r]}

for all l ≥ 2.

Proof. Since m intersects the local part trivially, it is clear that m∩gl = 0for l = −1, 0, 1. Before continuing, we make an observation. Take y =[· · · [ei1, ei2] · · · , eil] ∈ gl for some |l| ≥ 1 and let α =

∑lj=1 αij . According

to the Jacobi identity and the defining relations (2.1), [y, hi] = −α(hi)yfor all i ∈ [r]. Hence, for each y ∈ gl and i ∈ [r] there exist a constantC such that [y, hi] = Cy. Take i, j ∈ [r] and let si = sdeg(ei) andsj = sdeg(fj). Then

[[y, ei], fj] = δij[y, hi] + (−1)si sj [[y, fj], ei]

= C δijy + (−1)si sj [[y, fj], ei].

for some constant C.Take l ≥ 2. The inclusions ⊆ are clear. Let x ∈ gl such that [x, fi] ∈

m for all i ∈ [r]. To show that x ∈ m, it suffices to show that 〈x〉∩gn ⊆ mfor n = −1, 0, 1. Then, since m intersects the local part of g(A, τ)trivially, so does 〈x〉. Since m is maximal with this property, x ∈ m.

Consider the spaces 〈x〉∩gn, n = −1, 0, 1. These spaces are generatedby elements of the form

[· · · [x, x1] · · · , xm],

2.1. Definition of Contragredient Lie Superalgebras 25

where x1, . . . , xm ∈ g1 ∪ g−1. Take n ∈ {−1, 0, 1} and let

y = [· · · [x, x1] · · · , xm] ∈ 〈x〉 ∩ gn.

Recall that x ∈ gl for l ≥ 2. If n = −1, at least l + 1 of the elementsx1, . . . xm lie in g−1. If n = 0, at least l of the elements x1, . . . xm lie ing−1 and if n = 1, at least l − 1 of the elements x1, . . . xm lie in g−1. Thereminding elements are equally divided between g−1 and g1. Accordingto the observation above, we can assume that x1 ∈ g−1. Then, the factthat [x, fi] ∈ m for all i ∈ [r], implies that y ∈ m. Hence, 〈x〉 ∩ gn ⊆ mfor n = −1, 0, 1. Thus, the ideal 〈x〉 intersects the local part of g(A, τ)trivially, and since m is maximal with this property, x ∈ m. The inclusion⊇ for g−l is similar.

We are now in a position to define contragredient Lie superalgebras.Given an r× r-matrix A over a field k and a subset τ of [r] = {1, . . . , r},let g(A, τ) = g(A, τ)/m. Since g(A, τ) and m are Z-graded, g(A, τ) isZ-graded and

g(A, τ) =⊕

l∈Zgl =

⊕

l∈Zgl/m ∩ gl.

The Lie superalgebra g(A, τ) is called the contragredient Lie superalge-bra with respect to A and τ . A is called the Cartan matrix of g(A, τ)and r is the rank of g(A, τ). We usually write g = g(A, τ) when it is clearwhat A and τ is. If a pair (A′, τ ′) is obtained from (A, τ) by multiplyinga row with a non-zero field element or by a permutation of the index set{1, . . . r}, then g(A, τ) ∼= g(A′, τ ′) as Z-graded Lie superalgebras and thepairs (A, τ) and (A′, τ ′) are called equivalent. The following Lemma willbe useful later on.

Lemma 2.1.2. Let g = g(A, τ) be a contragredient Lie superalgebra suchthat A does not contain a column where all the elements are zero. Theng = [g, g].

Proof. Let r be the rank of g and let g = ⊕l∈Zgl be the Z-decompositionof g. Clearly, gl ⊆ [g, g] for all |l| ≥ 2. Furthermore, g0 ⊆ [g, g] sincehi = [ei, fi] for all i ∈ [r]. Take i ∈ [r] and consider the elements ei andfi in g1 and g−1 respectively. According to the assumption in the lemma,there exists j ∈ [r] such that aji 6= 0. Then

ei =1

aji[hj, ei] ∈ [g, g], fi = − 1

aji[hj, fi] ∈ [g, g].

Hence, g1, g−1 ⊆ [g, g] and g = [g, g].


Let h = 〈h1, . . . , hr〉 be the vector space over k with basis {hi}i∈[r]and define functionals αi ∈ h∗ by αi(hj) = aji for all i, j ∈ [r]. Let G bethe free Abelian group generated by α1, . . . αr. For each α ∈ G, definethe space gα as follows:

gα =

〈[· · · [ei1, ei2] · · · , eil]; α = αi1 + · · ·+ αil〉〈[· · · [fi1, fi2] · · · , fil]; α = −αi1 − · · · − αil〉h if α = 0

0 otherwise

(2.2)

Define the height of α =∑r

i=1 niαi as ht(α) =∑r

i=1 ni. Then we havethat gl =

⊕ht(α)=l gα and

g(A, τ) =⊕

α∈G

gα.

We end this section by introducing an automorphism of a contragredientLie superalgebra.

Automorphism. Let L be a Lie superalgebra and assume that

f : L → L

is an automorphism of L. If I is an ideal of L such that f(I) = I thenf induces an automorphism of L/I. The mapping

eif7→ fi, hi

f7→ −hi, fif7→ (−1)siei,

where si = sdeg ei, induces an automorphism on g(A, τ) of order 4. Theimage of m is a Z-graded ideal in g(A, τ) intersecting the local parttrivially. Since m is maximal in sense, f(m) ⊆ m, and since f is offinite order, f(m) = m. Hence f induces an Z-graded automorphism ofg(A, τ). This implies that gl

∼= g−l for all l ≥ 1.

2.2 The Root System and the Root-tupleSystem

In this section we continue the outline of the previous section and definethe root system and the root spaces of a contragredient Lie superalgebra

2.2. The Root System and the Root-tuple System 27

g(A, τ) such that det(A) 6= 0. We will also define the root-tuple sys-tem and the one-dimensional root-tuple spaces and introduce some newnotation.

Given a matrix A = (aij)ri,j=1 over a field k such that det(A) 6=

0 and a subset τ of [r] = {1, . . . r}, the Cartan subalgebra h of g =g(A, τ) is given by h = 〈h1, . . . hr〉. The fact that det(A) 6= 0 impliesthat the functionals α1, . . . , αr ∈ h∗, given by αi(hj) = aji for all i, j ∈[r], are linearly independent. Hence, the formal linear combinationswith integer coefficients introduced in the previous section coincides withthe linear combinations of α1, . . . , αr with integer coefficients, viewed asfunctionals. In other words, G ⊂ h∗. We extend the definition in (2.2)to include all functionals in h∗. Hence, for each α ∈ h∗ define the vectorspaces gα by:

gα =

〈[· · · [ei1, ei2] · · · , eil]; α = αi1 + · · ·+ αil〉〈[· · · [fi1, fi2] · · · , fil]; α = −αi1 − · · · − αil〉h if α = 0

0 otherwise

Since α1, . . . αr are linearly independent, the generators of gα in the firsttwo cases contain the same elements ei or fi, but in different order. Also,note that if α = 0, then gα = h = g0, where g0 is the part of degree 0 inthe Z-grading of g.

The set∆ = {α ∈ h∗; α 6= 0 and gα 6= 0}

is called the root system of g. A functional α ∈ ∆ is called a root of gand the spaces gα for α ∈ ∆ are called root spaces. The roots α1, . . . , αr

are called fundamental.

Lemma 2.2.1. Take i ∈ [r]. Then n ·αi 6∈ ∆ for all integers n such that|n| ≥ 3. Furthermore, ±2αi ∈ ∆ if and only if i ∈ τ and aii 6= 0.

Proof. This is a consequence of the definition of the root system and thefact that [[ei, ei], ei] = [[fi, fi], fi] = 0 for all i ∈ [r].

If α ∈ ∆, then α is a linear combination of α1, . . . , αr with integercoefficients either all positive (α is positive) or all negative (α is nega-tive). The set of positive roots is denoted by ∆+ and the set of negativeroots by ∆−. Then ∆ = ∆+ ∪∆− and, since det(A) 6= 0, this union isdisjoint.


There is a Z2-grading g = g0 ⊕ g1 of the Lie superalgebra g. Thefact that the generators of gα contain the same elements but in differentorder, implies that the root spaces are homogeneous with respect to thisZ2-grading. A root α ∈ ∆ is called even (sdegα = 0) if gα ⊆ g0 andodd (sdegα = 1) if gα ⊆ g1. The set of even and the set of odd rootsis denoted by ∆0 and ∆1 respectively. Then, ∆ = ∆0 ∪∆1 is a disjointunion and:

sdeg(x) ≡{

0 (mod 2) if x ∈ gα and α ∈ ∆0

1 (mod 2) if x ∈ gα and α ∈ ∆1

sdeg(h) ≡ 0 (mod 2) for all h ∈ h

If α ∈ ∆0 (α ∈ ∆1) then −α ∈ ∆0 (−α ∈ ∆1).As in the previous section, we have the following decomposition of g,

the so-called root space decomposition:

g = h⊕ (⊕

α∈∆

gα

)

The fact that det(A) 6= 0 implies that we may define the height of aroot as before and we have that gl =

⊕ht(α)=l gα for all l ∈ Z. The next

proposition gives an alternative description of the root spaces.

Proposition 2.2.2. Let α ∈ h∗. Then,

gα = {x ∈ g(A, τ); [h, x] = α(h)x, for all h ∈ h}Proof. The inclusion ⊆ is clear. Take α ∈ h∗ and x ∈ g(A, τ) such that[h, x] = α(h)x for all h ∈ h. Let x =

∑β∈h∗ xβ be the decomposition of

x. Then,∑

β∈h∗α(h)xβ = α(h)x = [h, x] =

∑

β∈h∗[h, xβ] =

∑

β∈h∗β(h)xβ

for all h ∈ h∗. This shows that, for all β ∈ h∗ such that xβ 6= 0,α(h) = β(h) for all h ∈ h. That is, xβ 6= 0 implies that α = β and hence,x ∈ gα.

By presenting gα as in Proposition 2.2.2, it is evident that [gα, gβ] ⊆gα+β for all α, β ∈ ∆ and thereby [gα, g−α] ⊆ g0 = h.

For most classical Lie superalgebras the root spaces are of dimensionone. However, this is not necessarily the case for the contragredient

2.2. The Root System and the Root-tuple System 29

Lie superalgebras given here. To be able to work with one-dimensionalspaces in a similar way as in the classical case, we introduce the root-tuples. For each l ∈ Z, define the set of tuples Γl inductively by:

1. Γ0 = ∅ and Γ1 = {(i); i ∈ [r]}2. For all l ≥ 2, choose Γl ⊆ {(a, i); i ∈ [r], (a) ∈ Γl−1} such that the

set {[· · · [ei1, ei2] · · · , eil]; (i1, . . . il) ∈ Γl} is a basis of gl.

3. For each l ≥ 1, let Γ−l = {(−i1, . . . ,−il); (i1, . . . , il) ∈ Γl)}The set Γ =

⋃l∈Z Γl is called a root-tuple system of g. A tuple (a) ∈ Γ

is called a root-tuple of g and the root-tuples (1), . . . , (r) are called fun-damental. If (a) = (i1, . . . , il) ∈ Γ, denote the root-tuple (−i1, . . . ,−il)by (−a). Define the set of positive and negative root-tuples as follows:

Γ+ =⋃

l≥1

Γl, Γ− =⋃

l≤−1

Γl

Then Γ = Γ+ ∪ Γ− and this union is disjoint.For each (a) = (i1, . . . , il) ∈ Γ+, define the root-tuple spaces g(a) and

g(−a) as the one-dimensional spaces generated by

x(a) = [· · · [ei1, ei2] · · · , eil] and x(−a) = [· · · [fi1, fi2], · · · fil]

respectively. The elements x(a) are called the root-tuple vectors. Also,we make the following definition:

Definition 2.2.3. For each (a) ∈ Γ define h(a) ∈ h by h(a) = [x(a), x(−a)].Note that h(i) = hi for all i ∈ [r]. We will use the notation hi instead ofh(i). Furthermore, for each α ∈ ∆+ let hα be an element in h such thatα(hα) = 1 and let h−α = −hα. Then −α(h−α) = 1.

Note that for each (a) ∈ Γ we have defined h(a) ∈ h and h(a)∗ ∈ h.The elements h(a) and h(a)∗ are not necessarily equal. For example, takei ∈ [r] such that aii = 0. Then αi(hi) = 0 and hence, hαi

6= hi.The root-tuple spaces are homogeneous with respect to the Z2-grad-

ing of g. A root-tuple (a) is called even if g(a) ⊆ g0 and odd if g(a) ⊆ g1.Let Γ0 denote the set of even root-tuples and Γ1 the set of odd root-tuples. Then the union Γ = Γ0 ∪ Γ1 is disjoint and

sdeg(x(a)) =

{0 if (a) ∈ Γ0

1 if (a) ∈ Γ1.


We will use the notation sdeg(a) = sdegx(a) for a root-tuple (a) ∈ Γ.From the definition of Γ, it follows that gl =

⊕(a)∈Γl

g(a) for all integersl 6= 0. Hence, we have the following root-tuple space decomposition ofg:

g = h⊕

⊕

(a)∈Γ

g(a)

.

There is a connection between the roots and the root-tuples. Foreach (a) = (i1, . . . , il) ∈ Γ+ define the functionals (a)∗, (−a)∗ ∈ h∗ by(a)∗ = αi1 + · · · + αil and (−a)∗ = −(a)∗. Then (a) ∈ Γ implies that(a)∗ ∈ ∆ and

gα =⊕

(a)∈Γ,(a)∗=α

g(a).

In other words, for α ∈ ∆, the set {x(a); (a) ∈ Γ such that (a)∗ = α} is abasis of gα. For each α ∈ ∆+, fix an order of the tuples (a1), . . . , (al) ∈ Γsuch that (ai)

∗ = α, i = 1, . . . , l, and let

xα =

x(a1)...

x(al)

, x−α =

x(−a1)...

x(−al)

.

Thus, if α ∈ ∆, xα is a column vector containing a basis of gα. If α 6∈ ∆,put xα = 0. Note that sdeg(α) ≡ sdeg(a) (mod 2) for all α ∈ ∆ and(a) ∈ Γ such that (a)∗ = α. Let

sdeg(xα) =

{0 (mod 2) if α ∈ ∆0

1 (mod 2) if α ∈ ∆1.

Finally, introduce the following notations:

[xα, y] =

[x(a1), y]...

[x(al), y]

and [y,xα] =

[y, x(a1)]...

[y, x(al)]

for all α ∈ ∆ and y ∈ g, where {x(a1), . . . , x(al)} is the ordered basis

of gα chosen above. Clearly, [y,xα] = −(−1)sdeg(y) sdeg(xα)[xα, y] if y issuper-homogeneous.

2.3. The Structure Constants 31

2.3 The Structure Constants

Let g = g(A, τ) be a contragredient Lie superalgebra of rank r such thatdet(A) 6= 0. Recall that [gα, gβ] ⊆ gα+β for all α, β ∈ ∆. This impliesthat [g(a), g(b)] ⊆ g(a)∗+(b)∗ for all (a), (b) ∈ Γ. Hence, for each (a), (b) ∈ Γsuch that (a)∗ 6= −(b)∗, there is a row vector N(a),(b) such that

[x(a), x(b)] = N(a),(b)x(a)∗+(b)∗.

If the space g(a)∗+(b)∗ is one-dimensional, we write n(a),(b) to denote thefield element such that N(a),(b) = (n(a),(b)). Note for example that gα isone-dimensional if ht(α) = 1 or if ht(α) = 2.

Let (a) ∈ Γ and β ∈ ∆ such that (a)∗ 6= −β and let {x(b1), . . . , x(bl)}be the ordered basis of gβ chosen in the previous section. Then

[x(a), x(bi)] ∈ g(a)∗+β

for all i = 1, . . . , l and

[x(a),xβ] =

[x(a), x(b1)]...

[x(a), x(bl)]

=

N(a),(b1)x(a)∗+β...

N(a),(bl)x(a)∗+β

= N(a),βx(a)∗+β.

for a matrix N(a),β. A matrix Nβ,(a) is defined similarly and

Nβ,(a) = −(−1)sdeg(a)·sdeg(β)N(a),β.

We call N(a),(b) and N(a),β, for (a), (b) ∈ Γ and β ∈ ∆, the structureconstants for g. To simplify succeeding calculations, we put N(a),(b) = 0if (a) 6∈ Γ or if (b) 6∈ Γ and N(a),β = Nβ,(a) = 0 if (a) 6∈ Γ or if β 6∈ Γ.The propositions below state some simple properties about the structureconstants. The first proposition follows immediately from the Jacobiidentity.

Proposition 2.3.1. Let (a), (b) ∈ Γ such that (a)∗ 6= −(b)∗. Then

N(a),(b) = −(−1)sa sbN(b),(a)

N(a),(b)∗ = −(−1)sa sbN(b)∗,(a),

where sa = sdeg(x(a)) and sb = sdeg(x(b)).


Proposition 2.3.2. Let i, j ∈ [r] such that (i, j) ∈ Γ. Then

n(i),(j) = n(−i),(−j) = 1

n(i,j),(−i) = (−1)sin(−i,−j),(i)

n(i,j),(−j) = (−1)sjn(−i,−j),(j),

where si = sdeg(x(i)) and sj = sdeg(x(j)).

Proof of Proposition 2.3.2. The first equality is clear and follows fromthe definition of Γ and from the definition of the root-tuple vectors.Recall from the end of Section 2.1 that the mapping

eif7→ fi, hi

f7→ −hi, fif7→ (−1)sdeg(ei)ei

induces an automorphism on g. To show the second and third equality,we apply this automorphism on both sides of [x(i,j), x(−i)] = n(i,j),(−i)x(j)and [x(i,j), x(−j)] = n(i,j),(−j)x(i) respectively. Note that x(i)∗ = (x(i)) andx(j)∗ = (x(j)).

We end this section with two lemmas that specify a way to decomposecertain root-tuples. In the last lemma we make an assumption that weare going to use in Chapter 3.

Lemma 2.3.3. Let (a) be a positive root-tuple, but not simple. Thenthere exists i ∈ [r] and (b) ∈ Γ+ such that (a) = (b, i), (−a) = (−b,−i)and such that

x(a) = [x(b), x(i)]

x(−a) = [x(−b), x(−i)].

Furthermore, if (a) 6= (i, i) for all i ∈ [r], then (b)∗ and (i)∗ are linearlyindependent.

Proof. Let (a) = (i1, . . . , il) where i1, . . . , il ∈ [r] and l ≥ 2. Thenx(a) = [. . . [ei1, ei2], . . . eil] and x(−a) = [. . . [fi1, fi2], . . . fil]. Let i = il and(b) = (i1, . . . , il−1). Clearly, (a) = (b, i) and (−a) = (−b,−i). Since(a) ∈ Γ+, we have that (b) ∈ Γ+ and (−b) ∈ Γ−. Furthermore,

[x(b), x(i)] = [. . . [ei1, ei2] . . . , eil−1], eil] = x(a)

[x(−b), x(−i)] = [. . . [fi1, fi2], . . . , fil−1], fil] = x(−a).

Finally, from Lemma 2.2.1 and the choice of (b) and (i), we see that (b)∗

and (i)∗ are linearly independent if (a) 6= (i, i) for all i ∈ [r].

2.3. The Structure Constants 33

Lemma 2.3.4. Assume that A does not contain an even unit-column.Then for each i ∈ [r], there exists j ∈ [r] such that (i, j) ∈ Γ andn(i,j),(−j) 6= 0 or such that (j, i) ∈ Γ and n(j,i),(−j) 6= 0. Furthermore, ifi ∈ [r] \ τ , or if i ∈ [r] and aii = 0, we can chose j 6= i.

Proof. Given i ∈ [r], choose j ∈ [r] such that aji = αi(hj) 6= 0. Letsi and sj denote the super-degree of ei and ej respectively. If the onlypossible choice of j is j = i, then the i’th column in A is a unit-columnand, therefore, (i) ∈ Γ1. We now have that [[ei, ei], fi] = −2αi(hi)ei 6= 0and, hence (i, i) ∈ Γ and n(i,i),(−i) 6= 0. Assume that i 6= j. Then

αi(hj)x(i) = αi(hj)ei = [hj, ei] = [[ej, fj], ei]

= [ej, [fj, ei]]− (−1)sj [fj, [ej, ei]]

= −(−1)sj [fj, [ej, ei]].

(2.3)

The first equality follows from the definition of x(i) and the fourth equal-ity is the Jacobi identity. The remaining equalities follow from the rela-tions defining g given in Section 2.1. Since αi(hj) 6=, [fj, [ej, ei]] 6= 0 andin particular, [ej, ei] 6= 0. This implies that (i, j) ∈ Γ or (j, i) ∈ Γ. If(j, i) ∈ Γ, we have

[fj, [ej, ei]] = [x(−j), x(j,i)] = n(−j),(j,i)x(i).

Hence, n(j,i),(−j) = −(−1)(si+sj)sjn(−j),(j,i) 6= 0. Similarly, if (j, i) ∈ Γ,n(j,i),(−j) 6= 0.

In the proof above we conclude that

n(i,j),(−j) = −aji, n(i,j),(−i) = (−1)sdeg(ei)sdeg(ej)aij

for all i, j ∈ [r] such that i 6= j and (i, j) ∈ Γ, and that n(i,i),(−i) = −2aii

for all i ∈ [r]. Furthermore, Proposition 2.1.1 implies that [ei, ej] = 0for some i, j ∈ [r] if and only if [fk, [ei, ej]] = 0 for all k ∈ [r]. This andthe calculations in the proof above implies that [ei, ej] = 0 if and only ifaij = aji = 0 for all i, j ∈ [r] such that i 6= j, and that [ei, ei] 6= 0 if andonly if i ∈ τ and aii 6= 0.

Chapter 3

Non-consistentPeriodisations

3.1 Introduction

The aim of this chapter is to study the non-consistent Z+-graded pe-riodisation gper of a contragredient Lie superalgebra g over a field ofcharacteristic 0 (for a formal definition see section 3.2.1). We will in-vestigate for which Lie superalgebras this periodisation is 1-2-presented,i.e., for which Lie superalgebras this periodisation is isomorphic to a freeLie superalgebra with generators of degree 1 and relations of degree 2.The main result is the following theorem:

Theorem 2. Let g = g(A, τ) be a contragredient Lie superalgebra overa field k of characteristic 0, such that det(A) 6= 0. Then the followingare equivalent:




For semisimple Lie algebras, a semidirect product is equivalent to a directproduct. Hence, Theorem 1 is a special case of Theorem 2.

When gper is 1-2-presented, Proposition 3.2.4 gives an explicit listof the quadratic relations. Theorem 2 shows that the periodisations ofthe finite-dimensional classical Lie superalgebras A(m,n) when m 6= n,B(m,n), C(n), D(m,n), D(2, 1; α), F (4) and G(3) (classified by Kac

36 3. Non-consistent Periodisations

in e.g. [14]) are 1-2-presented. Note however that the contragredientLie superalgebras considered in Theorem 2 do not have to be classical.They might for example be infinite-dimensional. The contragredient Liesuperalgebra sl(m,m) is not included in Theorem 2 since its Cartanmatrix has determinant 0. Example 3.7.1 shows that sl(2, 2)per is not1-2-presented.

The proof of Theorem 2 is carried out in three steps. In the firststep (Section 3.4) we show that (ii) implies (i). This is shown witha generalization of the method for proving Theorem 1 in Chapter 1.Consequently, some parts are similar to the proof of Theorem 1. In thesecond step we prove that (iii) implies (ii) (Proposition 3.5.1). In the laststep we show that (i) implies (iii). This is proved by homological methodsin Section 3.6. However, we start with the setting: The definition of theperiodisation in question, not necessarily consistent, and the free Liesuperalgebra with generators of degree 1 and relations of degree 2. Weend this chapter with two examples. In the first example we show thatsl(2, 2)per is not 1-2-presented. In Example 3.7.2 we demonstrate theconnection with the theory of k-algebras.

Remark. We will consider several gradings of the Lie superalgebrasinvolved. Clearly, there is the Z2-grading coming from the division intoodd and even elements. But a contragredient Lie superalgebra is alsoZ-graded and then there is a Z+-grading defined from the periodisation.When we talk about the super-degree, or sdeg, we always refer to theZ2-grading. The degree, or deg, refers to the Z-grading or to the Z+-grading. When necessary, we will point out which grading the degreerefers to.

3.2 The Periodisation and the Free Lie Su-peralgebra

In this chapter we construct the objects appearing in Theorem 2: Theperiodisation of a Lie superalgebra and a free Lie superalgebra. Theseconstructions are Z+-graded Lie superalgebras. Hence, there are twogradings to consider, the Z+-grading and the Z2-grading. The degree, ordeg, of an element refers to the Z+-grading and the super-degree, or sdeg,refers to the Z2-grading. In the same way we use the terms homogeneousand super-homogeneous.

3.2. The Periodisation and the Free Lie Superalgebra 37


Definition 3.2.1. Given a Lie superalgebra g over a field k, let gper bethe Z+-graded Lie superalgebra defined by

gper = g⊗ tk[t] =⊕i≥1

(g⊗ kti

),

where [x ⊗ f(t), y ⊗ g(t)] = [x, y] ⊗ f(t)g(t). The Z2-grading of gper isgiven by

(g⊗ tk[t])α = gα ⊗ tk[t] for all α ∈ Z2.

Hence, an element x⊗ ti ∈ gper is super-homogeneous if and only if x issuper-homogeneous in g, and sdeg(x ⊗ ti) = sdeg(x). In particular, forour contragredient Lie superalgebras, we have that:

sdeg(hi ⊗ tj) ≡ 0 (mod 2) for all i ∈ [r] and all j ≥ 1,

sdeg(x(a) ⊗ tj) ≡{

0 (mod 2) if (a) ∈ Γ0

1 (mod 2) if (a) ∈ Γ1for all j ≥ 1.

Remark. Let C[t, t−1] be the algebra of Laurent polynomials in t andlet g be a Lie superalgebra. Now, gper is the positive part of the Loop

superalgebra L(g) = g⊗ C[t, t−1]. If g = g(◦A,

◦τ) is a Kac-Moody super-

algebra with symmetrizable and indecomposable Cartan matrix◦A, then

an extension of L(g) is isomorphic to g(A, τ), where A is the extended

Cartan matrix of◦A (see van de Leur [31]).

3.2.2 The Free Lie Superalgebra

In this section, let g = g(A, τ) be a contragredient Lie superalgebra ofrank r such that det(A) 6= 0 and let Γ be a set of root-tuples defined inSection 2.2. We will construct a free Lie superalgebra with generatorsof degree 1, and an ideal of this superalgebra with generators of degree2. Note that the Z-grading of g is not considered here. The free Liesuperalgebra is Z+-graded and the degree refers to this grading.

To construct a free Lie superalgebra, we have to define the generatorsand specify the degree and the super-degree of each generator. Set

M = {x(1)(a); (a) ∈ Γ} ∪ {h(1)

i ; 1 ≤ i ≤ r}


and let

deg(x) = 1 for all x ∈ M

sdeg(h(1)i ) ≡ 0 (mod 2) for all i ∈ [r]

sdeg(x(1)(a)) ≡

{0 (mod 2) if (a) ∈ Γ0

1 (mod 2) if (a) ∈ Γ1.

Let F(M) denote the free Lie superalgebra generated by M and letFi(M) denote the homogeneous part of degree i in F(M). We call theelements in F2(M) quadratic. Note that the Z+-grading need not to beconsistent, i.e., there might exist x ∈ F(M) such that deg(x) 6≡ sdeg(x)(mod 2). For example, there may be elements of both super degrees inF1(M). For each α ∈ ∆, let

x(1)α =

x(1)(a1)...

x(1)(an)

and sdeg(x(1)

α ) ≡{

0 (mod 2) if α ∈ ∆0

1 (mod 2) if α ∈ ∆1,

where {x(a1), . . . , x(an)} is the ordered basis of gα chosen in Section 2.2.Also, let the Lie-bracket of a column vector and an element in F(M) bedefined as in Section 2.2.

To define the ideal R generated by quadratic expression, we constructa graded Lie superalgebra homomorphism ϕ from F(M) to gper, and letR be the ideal in F(M) generated by the kernel of the restriction of ϕto F2(M).

Definition 3.2.2. Let ϕ : F(M) → gper be the unique graded Lie su-peralgebra homomorphism defined by:

ϕ : x(1)(a) 7→ x(a) ⊗ t for all (a) ∈ Γ

ϕ : h(1)i 7→ hi ⊗ t for all 1 ≤ i ≤ r

This is a graded Lie superalgebra homomorphism since ϕ preserves thedegree and the super-degree of each generator of F(M). Denote by ϕi,i = 1, 2, . . ., the graded components of ϕ.

Lemma 3.2.3. The graded Lie superalgebra homomorphism ϕ is an epi-morphism and the graded component ϕ1 is an isomorphism.


Proof. The first statement follows by using induction over the degreeof gper and the fact that g = [g, g] (see Lemma 2.1.2). That ϕ1 is anisomorphism follows since {x(a); (a) ∈ Γ} ∪ {hi; 1 ≤ i ≤ r} is a basis ofg.

Let R be the ideal in F(M) generated by ker ϕ2. The next propositiongives an explicit list of the generators of R.

Proposition 3.2.4. Let g = g(A, τ) be a contragredient Lie superalgebraof rank r and such that det(A) 6= 0. Let ∆ and Γ be the root system andthe root-tuple system of g respectively. Recall the definition of hα ∈ h,

α ∈ ∆, given in Definition 2.2.3 and let h(1)α = ϕ−1

1 (hα). Then, the idealR is generated by the following elements:

1. [h(1)i , x

(1)(a)]− (a)∗(hi)[h

(1)(a)∗, x

(1)(a)] for all (a) ∈ Γ and i ∈ [r].

2. [x(1)(a), x

(1)(b) ]−N(a),(b)[h

(1)(a)∗+(b)∗,x

(1)(a)∗+(b)∗] for all pairs (a), (b) ∈ Γ such

that (a)∗ 6= −(b)∗.

3. [h(1)i , h

(1)j ] for all 1 ≤ i < j ≤ r.

4. [x(1)(a), x

(1)(b) ]−

∑ri=i ci[x

(1)(i) , x

(1)(−i)] for all pairs (a), (b) ∈ Γ such that

(a)∗ = −(b)∗, where [x(a), x(b)] =∑

i∈[r] cihi.

Proof. Denote by I2 the vector subspace of F(M) generated by the ex-pressions above. To show that R is generated as an ideal by I2, it sufficesto show that I2 = ker(ϕ2). Clearly, I2 ⊆ ker(ϕ2).

Consider the vector space epimorphism

ϕ2 : F2(M) → (gper)2 .

Since I2 ⊆ ker(ϕ2), there exists an epimorphism

φ2 : F2(M)/I2 → (gper)2 ; φ2(x + I2) = ϕ2(x).

To show that I2 = ker(ϕ2), it suffices to show that φ2 is injective. Then

F2(M)/I2∼= (gper)2

∼= F2(M)/ ker(ϕ2)

and, thus, I2 = ker(ϕ2).


The following set of elements generates F2(M):

{[h(1)i , x

(1)(a)]; i ∈ [r], (a) ∈ Γ}∪

{[x(1)(a), x

(1)(b) ]; (a), (b) ∈ Γ} ∪ {[h(1)

i , h(1)j ]; i, j ∈ [r]}

By applying the expressions in I2, we see that F2(M)/I2 is generatedby

{[h(1)(a)∗, x

(1)(a)] + I2; (a) ∈ Γ} ∪ {[x(1)

(i) , x(1)(−i)] + I2; i ∈ [r]}.

Furthermore,

[h(1)(a)∗, x

(1)(a)] + I2

φ27−→ [h(a)∗, x(a)]⊗ t2 = x(a) ⊗ t2

[x(1)(i) , x

(1)(−i)] + I2

φ27−→ [x(i), x(−i)]⊗ t2 = hi ⊗ t2,

for all (a) ∈ Γ and i ∈ [r]. Since the set {x(a)⊗t2; (a) ∈ Γ}∪{hi⊗t2; i ∈[r]} is a basis for (gper)2 = g⊗ t2, φ2 maps a set of generators injectivelyto a basis. Hence, φ2 is injective and we are done.

Since R ⊆ ker(ϕ), there exists a graded Lie superalgebra epimorphism

φ : F(M)/R −→ gper; φ(x + R) = ϕ(x).

Denote by φi, i ≥ 1, the graded components of φ. Clearly, φi are vectorspace epimorphisms, and for i = 1, 2, φi is an isomorphism.

In what follows, our main interest will be in the Lie superalgebra

F(M)/R. To save notations, from now on we let {x(1)(a); (a) ∈ Γ} ∪

{h(1)i ; 1 ≤ i ≤ r} stand for the image of the generators in F(M)/R.

Hopefully, no confusion will arise. To simplify notations even further,we make the following definition:

Definition 3.2.5. Let {x(1)(a); (a) ∈ Γ} ∪ {h(1)

i ; 1 ≤ i ≤ r} be the images

of the generators for F(M) in F(M)/R. Also, let the Lie-bracket of acolumn-vector and an element be defined as in Section 2.2. Recall thedefinition of hα ∈ h given in Definition 2.2.3. For all h ∈ h, (a) ∈ Γ,α ∈ ∆ and k ≥ 2, define:

h(1) = φ−11 (h⊗ t) , h

(k)(a) = [x

(1)(a), x

(k−1)(−a) ],

x(k)(a) = [h

(1)(a)∗, x

(k−1)(a) ], x(k)

α = [h(1)α ,x(k−1)

α ]

We write h(k)i instead of h

(k)(i) .


Remark. As noted after Definition 2.2.3, for each (a) ∈ Γ we havedefined h(a) ∈ h and h(a)∗ ∈ h. The element h(a) ∈ h is defined ash(a) = [x(a), x(−a)] and h(a)∗ is defined to be an element in h such that

(a)∗(h(a)∗) = 1. Also, note that if h =∑

cihi ∈ h, then h(1) =∑

cih(1)i .

The definition of x(k)α above implies that

x(k)α =

x(k)(a1)...

x(k)(al)

for all k ≥ 1, where {x(a1), . . . , x(al)} is the ordered basis of gα chosen in

Section 2.2. If α 6∈ ∆, let x(k)α = 0 for all k ≥ 1. By induction over k, it

is easy to see that

sdeg(h(k)i ) ≡ 0 (mod 2) for all i ∈ Γ

sdeg(x(k)(a)) ≡

{0 (mod 2) for all (a) ∈ Γ0

1 (mod 2) for all (a) ∈ Γ1

sdeg(x(k)α ) ≡

{0 (mod 2) for all α ∈ ∆0

1 (mod 2) for all α ∈ ∆1

for all k ≥ 1. As a corollary of Proposition 3.2.4, we rewrite some of thegenerators of R as quadratic relations in F(M)/R using the definitionabove.

Corollary 3.2.6. Let {x(1)(a), (a) ∈ Γ} ∪ {h(1)

i , 1 ≤ i ≤ r} be the images

of the generators for F(M) in F(M)/R. Then we have the followingquadratic relations in F(M)/R:

1. [h(1)i , x

(1)(a)] = (a)∗(hi)x

(2)(a) for all (a) ∈ Γ and all 1 ≤ i ≤ r.

2. [x(1)(a), x

(1)(b) ] = N(a),(b)x

(2)(a)∗+(b)∗ for all (a), (b) ∈ Γ such that (a)∗ 6=

−(b)∗.

3. [h(1)i , h

(1)j ] = 0 for all 1 ≤ i, j ≤ r.

4. [x(1)(a), x

(1)(b) ] =

∑ri=1 cih

(2)i for all (a), (b) ∈ Γ such that (a)∗ = −(b)∗,

where [x(a), x(b)] =∑r

i=1 cihi.


We end this section by introducing an automorphism of F(M)/Rthat will simplify succeeding calculations. Recall the notations aboutautomorphisms of Lie superalgebras given at the end of Section 2.1.The mapping

x(1)(a) 7→ x

(1)(−a), h

(1)i 7→ −h

(1)i , x

(1)(−a) 7→ (−1)sax

(1)(a)

for all (a) ∈ Γ+ and i ∈ [r], where sa ≡ sdeg(x(a)) (mod 2), inducesan automorphisms of F(M) of order 4. This mapping preserves thegenerators of R and, hence, R is mapped to itself under this map. Thus,this map induces an automorphism of F(M)/R. It is easy to see that

x(k)(a) 7→ x

(k)(−a), x

(k)(−a) 7→ (−1)sax

(k)(a)

for all (a) ∈ Γ+ and all k ≥ 1.

3.3 Formulation of Theorem 2

We state our main theorem.

Theorem 2. Let g = g(A, τ) be a contragredient Lie superalgebra overa field k of characteristic 0, such that det(A) 6= 0. Then the followingare equivalent:




From now on we say that the matrix A contains an even unit-columnif A contains an even column where the elements are zero outside thediagonal entry. We will prove Theorem 2 by showing that (ii) implies(i) (Section 3.4), (iii) implies (ii) (Section 3.5) and that (i) implies (iii)(Section 3.6). The proof of the first implication is a generalization ofthe proof of Theorem 1. The second implication is essentially Proposi-tion 3.5.1 and the last implication is shown with homological methods.

3.4. The First Step: (ii) Implies (i) 43

3.4 The First Step: (ii) Implies (i)

In this section let g = g(A, τ) be a contragredient Lie superalgebra, overa field of characteristic 0, such that det(A) 6= 0 and such that A does notcontain an even unit-column. Let M be the set of generators of degree 1and let R be the ideal in F(M) generated by quadratic elements, bothconstructed in Section 3.2. The aim of this section is to prove that (ii)implies (i) in Theorem 2, i.e., to show that gper is isomorphic to F(M)/R.In proving this, we will use induction over the Z+-grading of gper andF(M)/R. We will show that each graded component of

φ : F(M)/R 7−→ gper,

given in Section 3.2.2, is an isomorphism. The following remark will beuseful.

Remark 3.4.1. Let φ be the graded Lie superalgebra epimorphism in-troduced in Section 3.2.2. Then by induction over i, it is easy to seethat

x(i)(a)

φ7−→ x(a) ⊗ ti, h(i)(a)

φ7−→ h(a) ⊗ ti

for all (a) ∈ Γ and all i ≥ 1.Take k ≥ 2 and assume that φ is an isomorphism up to and including

degree k. If a relation holds in g, then the corresponding relation holdsin (F(M)/R)i for all 1 ≤ i ≤ k. For example, take (a), (b) ∈ Γ andconsider the relation

[x(a), x(b)] = N(a),(b)x(a)∗+(b)∗

in g. Take any m,n ≥ 1 such that m + n = k. Then,

φm+n

([x

(m)(a) , x

(n)(b) ]

)= φ

([x

(m)(a) , x

(n)(b) ]

)= [φ(x

(m)(a) ), φ(x

(n)(b) )]

= [x(a) ⊗ tm, x(b) ⊗ tn] = [x(a), x(b)]⊗ tm+n

=(N(a),(b)x(a)∗+(b)∗

)⊗ tm+n

= φm+n

(N(a),(b)x

(m+n)(a)∗+(b)∗

),

where φm+n is graded component of φ of degree m+n. Since ker(φm+n) =

0, the relation [x(m)(a) , x

(n)(b) ] = N(a),(b)x

(m+n)(a)∗+(b)∗ holds in F(M)/R.

Before proving that gper and F(M)/R are isomorphic, we state and provesome lemmas.


Lemma 3.4.2. Take k ≥ 2 and assume that φ is a graded Lie superal-gebra isomorphism up to and including degree k. Then

[h(1), x(k)(a)] = (a)∗(h)x

(k+1)(a)

[h(1),x(k)(a)∗] = (a)∗(h)x

(k+1)(a)∗

holds for all (a) ∈ Γ and h ∈ h.

Proof of Lemma 3.4.2. Take (a) ∈ Γ, h ∈ h and let α = (a)∗. Then,

[h(1), x(k)(a)] = [h(1), [h(1)

α , x(k−1)(a) ]]

= [[h(1), h(1)α ], x

(k−1)(a) ] + [h(1)

α , [h(1), x(k−1)(a) ]]

= α(h)[h(1)α , x

(k)(a)] = α(h)x

(k+1)(a) .

The first and the third equality follows from the assumption that φ isan isomorphism up to and including degree k (c.f. Remark 3.4.1). Thesecond equality follows from the Jacobi identity and the last equalityfollows from Definition 3.2.5.

The second statement follows from the first.

Lemma 3.4.3. Take k ≥ 2 and assume that φ is an isomorphism up toand including degree k. Then

h(k)(i,j) = n(i,j),(−i)h

(k)j − (−1)sisjn(i,j),(−j)h

(k)i

for all i, j ∈ Γ such that (i, j) ∈ Γ, where si = sdeg(x(i)) and sj =sdeg(x(j)).

Proof of Lemma 3.4.3. In view of Remark 3.4.1, it suffices to show thatthe corresponding relation holds in g. Observe that dim g(i) = dim g(j) =dim g(i,j) = 1 and that n(i),(j) = 1. Then,

h(i,j) = [x(i,j), x(−i,−j)] = [x(i,j), [x(−i), x(−j)]]

= [[x(i,j), x(−i)], x(−j)] + (−1)(si+sj)si[x(−i), [x(i,j), x(−j)]]

= n(i,j),(−i)[x(j), x(−j)] + (−1)(si+sj)sin(i,j),(−j)[x(−i), x(i)]

= n(i,j),(−i)hj − (−1)sisjn(i,j),(−j)hi,

and we are done.



[x(m)(a) , x

(n)(b) ] = N(a),(b)x

(k+1)(a)∗+(b)∗

[x(m)(a) ,x

(n)(b)∗] = N(a),(b)∗x

(k+1)(a)∗+(b)∗

holds for all m,n ≥ 1 such that m + n = k + 1 and for all (a), (b) ∈ Γsuch that (a)∗ 6= −(b)∗.

Proof of Lemma 3.4.4. Take (a), (b) ∈ Γ such that (a)∗ 6= −(b)∗ and letα = (a)∗ and β = (b)∗. Take h ∈ h, such that β(h) 6= 0. Suppose thatm,n ≥ 1 such that m + n = k + 1 and n ≥ 2. Then we have that

[x(m)(a) , x

(n)(b) ] =

1

β(h)[x

(m)(a) , [h(1), x

(n−1)(b) ]]

=1

β(h)[[x

(m)(a) , h(1)], x

(n−1)(b) ] +

1

β(h)[h(1), [x

(m)(a) , x

(n−1)(b) ]]

= −α(h)

β(h)[x

(m+1)(a) , x

(n−1)(b) ] +

N(a),(b)

β(h)[h(1),x

(m+n−1)α+β ]

= −α(h)

β(h)[x

(m+1)(a) , x

(n−1)(b) ] +

N(a),(b)

β(h)(α + β)(h)x

(m+n)α+β

=α(h)

β(h)

(−[x

(m+1)(a) , x

(n−1)(b) ] + N(a),(b)x

(m+n)α+β

)+ N(a),(b)x

(m+n)α+β .

(3.1)

The first and the third equality follows from the assumption that φis an isomorphism up to and including degree k (c.f. Remark 3.4.1).The last equality follows from Lemma 3.4.2. In particular, this showsthat it suffices to show the lemma for a fixed pair m,n ≥ 1 such thatm + n = k + 1. We proceed in steps:

1. If α and β are linearly independent there exists h ∈ h such thatα(h) = 0 and β(h) 6= 0. Using this h in equation (3.1) gives that

[x(m)(a) , x

(n)(b) ] = N(a),(b)x

(m+n)α+β for n ≥ 2 and, then, for all m,n ≥ 1

such that m+n = k +1. Moreover, this implies that [x(m)(a) ,x

(n)(b)∗] =

N(a),(b)∗x(m+n)(a)∗+(b)∗ for all (a), (b) ∈ Γ such that (a)∗ and (b)∗ are

linearly independent and all m,n ≥ 1 such that m + n = k + 1.

2. Let α and β be linearly dependent. Suppose that there exist(c), (d) ∈ Γ such that (c)∗ and (d)∗ are linearly independent and

x(b) = K[x(c), x(d)]


for some non-zero constant K. The fact that α and β are linearlydependent roots such that α 6= −β, that (c)∗ and (d)∗ are linearlyindependent and that β = (c)∗ + (d)∗ implies that α 6= −(c)∗ andα 6= −(d)∗. Furthermore, α+(c)∗ and (d)∗ are linearly independent,and so are α + (d)∗ and (c)∗. Let sa = sdegx(a) and sc = sdegx(c).Take n ≥ 2. Then:

[x(m)(a) , x

(n)(b) ] = K[x

(m)(a) , [x

(1)(c), x

(n−1)(d) ]]

= K([[x

(m)(a) , x

(1)(c)], x

(n−1)(d) ] + (−1)sasc[x

(1)(c), [x

(m)(a) , x

(n−1)(d) ]]

)

= K(N(a),(c)[x

(m+1)α+(c)∗, x

(n−1)(d) ] + (−1)sascN(a),(d)[x

(1)(c),x

(m+n−1)α+(d)∗ ]

)

= K(N(a),(c)Nα+(c)∗,(d) + (−1)sascN(a),(d)N(c),α+(d)∗

)x

(m+n)α+β

(3.2)

The first and the third equality follows from the assumption that φis an isomorphism up to and including degree k (c.f. Remark 3.4.1).The fourth equality follows from step 1. By doing the same calcu-lations in g without the superscript, we see that

N(a),(b) = K(N(a),(c)Nα+(c)∗,(d) + (−1)sascN(a),(d)N(c),α+(d)∗

).

Hence,

[x(m)(a) , x

(n)(b) ] = N(a),(b)x

(m+n)α+β

holds for n ≥ 2 and, thereby, for all m,n ≥ 1 such that m + n =k + 1.

These calculations also shows that

[x(m)(i,i), x

(n)(i,i)] = [x

(m)(−i,−i), x

(n)(−i,−i)] = 0

for all m,n ≥ 1 such that m + n = k + 1. Take (c) = (d) = (i) and(c) = (d) = (−i) respectively. Then we can use the three first linesin (3.2). According to Lemma 2.2.1, N(i,i),(i,i) = N(−i,−i),(−i,−i) = 0

and hence, [x(m)(i,i), x

(n)(i,i)] = [x

(m)(−i,−i), x

(n)(−i,−i)] = 0.


3. According to Lemma 2.2.1, 2.3.3 and 2.3.4, the following remainsto be shown:

[x(m)(i) , x

(n)(i) ] = N(i),(i)x

(m+n)(i,i)∗ , [x

(m)(−i), x

(n)(−i)] = N(−i),(−i)x

(m+n)(−i,−i)∗,

[x(m)(−i), x

(n)(i,i)] = N(−i),(i,i)x

(m+n)(i)∗ , [x

(m)(i) , x

(n)(−i,−i)] = N(i),(−i,−i)x

(m+n)(−i)∗ ,

[x(m)(i) , x

(n)(i,i)] = 0, [x

(m)(−i), x

(n)(−i,−i)] = 0

for all m,n ≥ 1 such that m + n = k + 1 and all i ∈ τ such thataii 6= 0 and aji = 0 for all 1 ≤ j 6= i ≤ r. (That is, column i isan odd column where the elements are zero outside the diagonalentry. Hence, in this step we show the lemma for the case whenA = (1) and τ = {1}.) Take i ∈ τ satisfying this. Note thatdim gαi

= dim g2αi= 1 and that n(i),(i) = 1. Also, note that

equation (3.1) implies that it suffices to show the equalities fora fixed pair m,n ≥ 1 such that m + n = k + 1. In view of this andof the automorphism of F(M)/R given in Section 3.2.2, it sufficesto show that

[x(k)(i) , x

(1)(i) ] = x

(k+1)(i,i) , [x

(k)(−i), x

(1)(i,i)] = n(−i),(i,i)x

(k+1)(i) ,

[x(k)(i) , x

(1)(i,i)] = 0

(3.3)

Before we start, observe that (i)∗(hi) = αi(hi) = aii 6= 0 and that(i, i)∗(hi) = 2αi(hi) = 2aii.

We start with the first equality.

[x(k)(i) , x

(1)(i) ] =

1

aii[[h

(k−1)i , x

(1)(i) ], x

(1)(i) ] =

1

2aii[h

(k−1)i [x

(1)(i) , x

(1)(i) ]]

=1

4a2ii

[h(k−1)i , [h

(1)i , x

(1)(i,i)]] =

1

4a2ii

[h(1)i , [h

(k−1)i , x

(1)(i,i)]]

=1

2aii[h

(1)i , x

(k)(i,i)]] = x

(k+1)(i,i)

The first, third, fourth and the fifth equality follows from the as-sumption that φ is an isomorphism up to and including degreek (c.f. Remark 3.4.1) (the fourth equality also follows from theJacobi identity). The last equality follows from Lemma 3.4.2.

To show the second equality in (3.3), we use the relation

h(k)(i,i) = 2n(i,i),(−i)h

(k)(i)


following from Lemma 3.4.3. Note that n(i,i),(−i) = −2a(ii) 6= 0.

[x(k−1)(−i) ,x

(2)(i,i)] = [x

(k−1)(−i) , [x

(1)(i) , x

(1)(i) ]] = 2[[x

(k−1)(−i) , x

(1)(i) ], x

(1)(i) ]

= 2[h(k)(i) , x

(1)(i) ] =

1

n(i,i),(−i)[h

(k)(i,i), x

(1)(i) ]

=1

n(i,i),(−i)[[x

(1)(i,i), x

(k−1)(−i,−i)], x

(1)(i) ]

=1

n(i,i),(−i)

([x

(1)(i,i), [x

(k−1)(−i,−i), x

(1)(i) ]]− [x

(k−1)(−i,−i), [x

(1)(i,i), x

(1)(i) ]]

)

=n(−i,−i),(i)

n(i,i),(−i)[x

(1)(i,i), x

(k)(−i)] = [x

(k)(−i), x

(1)(i,i)]

The first and the second last equality follows from the assump-tion that φ is an isomorphism up to and including degree k (c.f.Remark 3.4.1). The third and fifth equality follows from Defini-tion 3.2.5. The last equality follows from the anticommutativity

and Proposition 2.3.2. Hence, [x(k−1)(−i) , x

(2)(i,i)] = [x

(k)(−i), x

(1)(i,i)]. This

inserted in equation (3.1) yields the desired equality.

To show the last equation in (3.3), we start with the case whenk = 2. We have that

[x(1)(i) , x

(2)(i,i)] = [x

(1)(i) , [x

(1)(i) , x

(1)(i) ]] = 0.

Hence, [x(2)(i) , x

(1)(i,i)] = 0 by (3.1). Take k ≥ 3. By applying equa-

tion (3.1) k − 1 times, we have that

[x(1)(i) , x

(k)(i,i)] =

(−1

2

)k−1

[x(k)(i) , x

(1)(i,i)].

On the other hand, we have that

[x(1)(i) , x

(k)(i,i)] =

1

2aii[x

(1)(i) , [h

(k−1)i , x

(1)(i,i)]]

=1

2aii

([[x

(1)(i) , h

(k−1)i ], x

(1)(i,i)] + [h

(k−1)i , [x

(1)(i) , x

(1)(i,i)]]

)

= −1

2[x

(k)(i) , x

(1)(i,i)]

The first and the last equality follows from the assumption that φ isan isomorphism up to and including degree k (c.f. Remark 3.4.1).


Hence,

[x(k)(i) , x

(1)(i,i)] =

(−1

2

)k−2

[x(k)(i) , x

(1)(i,i)]

and since the characteristic is zero, we have that [x(k)(i) , x

(1)(i,i)] = 0.

Finally, the first statement in the lemma implies that second one and weare done.


[x(1)(i) , x

(k)(−i)]− [x

(m)(i) , x

(n)(−i)] = 0

for all integers m,n ≥ 1 such that m + n = k + 1 and for all i ∈ [r].

Proof of Lemma 3.4.5. Take any i ∈ [r]. We begin with an observation.Take any h ∈ h and let s, t be any integers such that s, t ≥ 1 ands + t = k. Then:

[h(1), h(k)i ] = [h(1), [x

(s)(i) , x

(t)(−i)]]

= [[h(1), x(s)(i) ], x

(t)(−i)] + [x

(s)(i) , [h

(1), x(t)(−i)]]

= −(i)∗(h)([x

(s)(i) , x

(t+1)(−i) ]− [x

(s+1)(i) , x

(t)(−i)]

)

The first and the last equality follows from the assumption that φ isan isomorphism up to and including degree k (c.f. Remark 3.4.1). Thesecond equality follows from the Jacobi identity. Choosing h ∈ h suchthat (i)∗(h) 6= 0 and letting s run from 1 to k − 1 yields the followingequalities:

[x(1)(i) , x

(k)(−i)]− [x

(2)(i) , x

(k−1)(−i) ] = [x

(2)(i) , x

(k−1)(−i) ]− [x

(3)(i) , x

(k−1)(−i) ] = . . .

. . . = [x(k−2)(i) , x

(3)(−i)]− [x

(k−1)(i) , x

(2)(−i)] = [x

(k−1)(i) , x

(2)(−i)]− [x

(k)(i) , x

(1)(−i)]

Adding an appropriate number of these equalities, if follows that

[x(1)(i) , x

(k)(−i)]− [x

(m)(i) , x

(n)(−i)] = (m− 1)

([x

(1)(i) , x

(k)(−i)]− [x

(2)(i) , x

(k−1)(−i) ]

)(3.4)

for all m,n ≥ 1 such that m + n = k + 1. Hence, to show the lemma, if

suffices to show that [x(1)(i) , x

(k)(−i)]− [x

(2)(i) , x

(k−1)(−i) ] = 0.

According to Lemma 2.3.4, there exists j ∈ [r] such that (i, j) ∈ Γ andn(i,j),(−j) 6= 0, or such that (j, i) ∈ Γ and n(j,i),(−j) 6= 0. We treat the case


when (i, j) ∈ Γ here. The case when (j, i) ∈ Γ is similar. Furthermore,if it is possible , choose j 6= i. If not, note that sdegx(i) = 1 according tothe proof of Lemma 2.3.4. Let si = sdegx(i) and sj = sdegx(j) and define

A = n(i,j),(−j)

([x

(1)(i) , x

(k)(−i)]− [x

(2)(i) , x

(k−1)(−i) ]

)

B = n(i,j),(−i)

([x

(1)(j), x

(k)(−j)]− [x

(2)(j), x

(k−1)(−j) ]

)

C = [x(1)(i,j), x

(k)(−i,−j)]− [x

(2)(i,j), x

(k−1)(−i,−j)].

The idea is to find three independent linear relations of A,B and C.

Then A = 0, and since n(i,j),(−j) 6= 0, [x(1)(i) , x

(k)(−i)] − [x

(2)(i) , x

(k−1)(−i) ] = 0. We

continue in steps:

1. For the first linear relation, we expand [x(1)(i,j), x

(k)(−i,−j)] and

[x(2)(i,j), x

(k−1)(−i,−j)]:

[x(1)(i,j), x

(k)(−i,−j)] = [x

(1)(i,j), [x

(1)(−i), x

(k−1)(−j) ]]

= [[x(1)(i,j), x

(1)(−i)], x

(k−1)(−j) ] + (−1)si(si+sj)[x

(1)(−i), [x

(1)(i,j), x

(k−1)(−j) ]]

= n(i,j),(−i)[x(2)(j), x

(k−1)(−j) ] + (−1)si(si+sj)n(i,j),(−j)[x

(1)(−i), x

(k)(i) ]

= n(i,j),(−i)[x(2)(j), x

(k−1)(−j) ]− (−1)sisjn(i,j),(−j)[x

(k)(i) , x

(1)(−i)]

(3.5)

The first and the third equality follows from the assumption that φis an isomorphism up to and including degree k (c.f. Remark 3.4.1).The last equality follows from the anticommutativity. Similarly:

[x(2)(i,j), x

(k−1)(−i,−j)] = [[x

(1)(i) , x

(1)(j)], x

(k−1)(−i,−j)]

= [x(1)(i) , [x

(1)(j), x

(k−1)(−i,−j)]]− (−1)sisj [x

(1)(j), [x

(1)(i) , x

(k−1)(−i,−j)]]

= n(j),(−i,−j)[x(1)(i) , x

(k)(−i)]− (−1)sisjn(i),(−i,−j)[x

(1)(j), x

(k)(−j)]

= −(−1)sj(si+sj)n(−i,−j),(j)[x(1)(i) , x

(k)(−i)] + (−1)sin(−i,−j),(i)[x

(1)(j), x

(k)(−j)]

The last equality follows from Proposition 2.3.1. By using Propo-sition 2.3.2, we have that:

[x(2)(i,j), x

(k−1)(−i,−j)] =

n(i,j),(−i)[x(1)(j), x

(k)(−j)]− (−1)sisjn(i,j),(−j)[x

(1)(i) , x

(k)(−i)] (3.6)


By subtracting equation (3.6) from equation (3.5) and using thenotation above, we have that

C = −B + (−1)sisjn(i,j),(−j)

([x

(1)(i) , x

(k)(−i)]− [x

(k)(i) , x

(1)(−i)]

). (3.7)

Applying equation (3.4), for the case when m = k, on this, we havethat

C = −B + (−1)sisj(k − 1)A.

2. Take any h ∈ h. Then,

[h(1), h(k)(c) ] = [h(1), [x

(1)(c), x

(k−1)(−c) ]]

= [[h(1), x(1)(c)], x

(k−1)(−c) ] + [x

(1)(c), [h

(1), x(k−1)(−c) ]]

= −(c)∗(h)([x

(1)(c) , x

(k)(−c)]− [x

(2)(c), x

(k−1)(−c) ]

) (3.8)

for any (c) ∈ Γ. The first and the last equality follows from theassumption that φ is an isomorphism up to and including degreek (c.f. Remark 3.4.1). Using Lemma 3.4.3, we have that

[h(1), h(k)(i,j)] = n(i,j),(−i)[h

(1), h(k)j ]− (−1)sisjn(i,j),(−j)[h

(1), h(k)i ].

Applying (3.8) on this yields that:

(i, j)∗(h)C = (j)∗(h)B − (−1)sisj(i)∗(h)A

From step 1 and 2, we have that

C = −B + (−1)sisj(k − 1)A

(i, j)∗(h)C = (j)∗(h)B − (−1)sisj(i)∗(h)A(3.9)

If i = j, then si = sj = 1 and A = B. Choose any h ∈ h such that(i)∗(h) 6= 0. Then equations in (3.9) become C = −kA and C = A. Thisgives that (k + 1)A = 0 and since the characteristic of the underlyingfield is zero, A = 0. If i 6= j, then (i)∗ and (j)∗ are linearly independent.Choose h, h′ ∈ h such that (i)∗(h) = 0, (j)∗(h) 6= 0 and (i)∗(h′) 6= 0,(j)∗(h′) = 0. Then the equations in (3.9) become C = −B+(−1)sisj(k−1)A, C = B and C = −(−1)sisjA. This gives that (k + 1)A = 0 andsince the characteristic of the underlying field is zero, A = 0 and we aredone.



[h(1)i , h

(k)j ] = 0

for all i, j ∈ [r].

Proof of Lemma 3.4.6. Take any i, j ∈ [r]. Then:

[h(1)i , h

(k)j ] = [h

(1)i , [x

(1)(j), x

(k−1)(−j) ]]

= [[h(1)i , x

(1)(j)], x

(k−1)(−j) ] + [x

(1)(j), [h

(1)i , x

(k−1)(−j) ]]

= (j)∗(hi)([x

(2)(j), x

(k−1)(−j) ]− [x

(1)(j), x

(k)(−j)]

)= 0

The first and the third equality follows from the assumption that φ is anisomorphism up to and including degree k (c.f. Remark 3.4.1) and thelast equality follows from Lemma 3.4.5.


[h(k)(i) , x

(1)(b) ] = (b)∗(hi)x

(k+1)(b)

for all i ∈ [r] and all (b) ∈ Γ.

Proof of Lemma 3.4.7. Take (b) ∈ Γ. We continue in steps:

1. Let (a) ∈ Γ such that (a)∗ 6= ±(b)∗ and let sa = sdeg(x(a)). Then:

[h(k)(a), x

(1)(b) ] = [[x

(1)(a), x

(k−1)(−a) ], x

(1)(b) ]

= [x(1)(a), [x

(k−1)(−a) , x

(1)(b) ]]− (−1)sa[x

(k−1)(−a) , [x

(1)(a), x

(1)(b) ]]

= N(−a),(b)[x(1)(a),x

(k)(b)∗−(a)∗]− (−1)saN(a),(b)[x

(k−1)(−a) ,x

(2)(a)∗+(b)∗]

=(N(−a),(b)N(a),(b)∗−(a)∗ − (−1)saN(a),(b)N(−a),(a)∗+(b)∗

)x

(k+1)(b)∗

The first equality follows from Definition 3.2.5, the third from theassumption that φ is an isomorphism up to and including degreek (c.f. Remark 3.4.1) and last equality follows from Lemma 3.4.4.By doing the same calculations in g without the superscript, wesee that

(b)∗(h(a))x(b) =(N(−a),(b)N(a),(b)∗−(a)∗ − (−1)saN(a),(b)N(−a),(a)∗+(b)∗

)x(b)∗

and, hence, [h(k)(a), x

(1)(b) ] = (b)∗(h(a))x

(k+1)(b) .


2. Assume that (b) = (±i). According to Lemma 2.3.4 there existsj ∈ [r] such that (i, j) ∈ Γ and n(i,j),(−j) 6= 0, or such that (j, i) ∈ Γand n(j,i),(−j) 6= 0. We treat the case when (i, j) ∈ Γ here. The casewhen (j, i) ∈ Γ is similar. Let si = sdeg(x(i)) and sj = sdeg(x(j)).From Lemma 3.4.3 we have

h(k)(i) =

(−1)sisj

n(i,j),(−j)

(n(i,j),(−i)h

(k)j − h

(k)(i,j)

)

and this implies that

[h(k)(i) , x

(1)(b) ] =

(−1)sisj

n(i,j),(−j)

(n(i,j),(−i)[h

(k)j , x

(1)(b) ]− [h

(k)(i,j), x

(1)(b) ]

). (3.10)

If we can choose j ∈ [r] such that j 6= i, then the first step impliesthat

[h(k)(i) , x

(1)(b) ] =

(−1)sisj

n(i,j),(−j)

(n(i,j),(−i)(b)

∗(hj)− (b)∗(h(i,j)))x

(k+1)(b) .

By doing the same calculations in g without the superscript, wesee that

(b)∗(hi) =(−1)sisj

n(i,j),(−j)

(n(i,j),(−i)(b)

∗(hj)− (b)∗(h(i,j)))

and, hence, [h(k)(i) , x

(1)(b) ] = (b)∗(hi)x

(k+1)(b) .

If the only possible choice is j = i, then i ∈ τ and equation (3.10)becomes

2[h(k)(i) , x

(1)(b) ] =

1

n(i,i),(−i)[h

(k)(i,i), x

(1)(b) ].

Again, by using the first step and by doing the same calculations in

g without the superscript, we have that [h(k)(i) , x

(1)(b) ] = (b)∗(hi)x

(k+1)(b) .

Lemma 3.4.8. Take k ≥ 2 and assume that φ is a graded Lie super-

algebra isomorphism up to and including degree k. Then, [x(m)(a) , x

(n)(b) ] is

a linear combination of h(k+1)1 , . . . , h

(k+1)r for all (a), (b) ∈ Γ such that

(a)∗ = −(b)∗ and all m,n ≥ 1 such that m + n = k + 1.


Proof of Lemma 3.4.8. We will prove Lemma 3.4.8 with induction overthe height of (b)∗. If |ht(b)∗| = 1 the lemma follows from Lemma 3.4.5and Definition 3.2.5. Take l ≥ 1 and assume that the lemma is truefor all (b) ∈ Γ such that |ht(b∗)| ≤ l. Take (a), (b) ∈ Γ such that(a)∗ = −(b)∗ and |ht(b)∗| = l + 1. Then |ht(a)∗| = l + 1 and in view ofthe automorphism of F(M)/R introduced at the end of Section 3.2, wecan assume that (b) ∈ Γ+ and n ≥ 2. According to Lemma 2.3.4, thereexists i ∈ [r] and (c) ∈ Γ+ such that x(b) = [x(c), x(i)]. According to theassumption that φ is an isomorphism up to and including degree k, wehave that

x(n)(b) = [x

(1)(c) , x

(n−1)(i) ]

(c.f. Remark 3.4.1). Let sa = sdeg(x(a)) and sc = sdeg(x(c)). Then:

[x(m)(a) , x

(n)(b) ] = [x

(m)(a) , [x

(1)(c), x

(n−1)(i) ]]

= [[x(m)(a) , x

(1)(c)], x

(n−1)(i) ] + (−1)sasc[x

(1)(c), [x

(m)(a) , x

(n−1)(i) ]]

= N(a),(c)[x(m+1)(a)∗+(c)∗, x

(n−1)(i) ] + (−1)sascN(a),(i)[x

(1)(c),x

(m+n−1)(a)∗+(i)∗ ]

The last equality follows from the assumption that φ is an isomorphismup to and including degree k (c.f. Remark 3.4.1). Now, ht(c)∗ < ht(b)∗

and (a)∗ + (i)∗ = −(c)∗. Then, according to the induction hypothesis,

we have that N(a),(c)[x(m+1)(a)∗+(c)∗, x

(n−1)(i) ] and N(a),(i)[x

(1)(c),x

(m+n−1)(a)∗+(i)∗ ] are linear

combinations of h(k+1)1 , . . . , h

(k+1)r . Hence, so is [x

(m)(a) , x

(n)(b) ] and we are

done.

Final argument. We are now in a position to show that gper andF(M)/R are isomorphic. We will use the graded Lie superalgebra ho-momorphism φ defined in Section 3.2.2. Recall that

x(i)(a)

φ7→ x(a) ⊗ ti, h(i)(a)

φ7→ h(a) ⊗ ti (3.11)

for all (a) ∈ Γ and all i ≥ 1. Since φ is an epimorphism, if suffices toshow that φ is injective, or that each graded component of φ is injective.Clearly, φ1 and φ2 are injective. Take k ≥ 2 and assume that φk isinjective. Then, since

{x(a) ⊗ tk; (a) ∈ Γ} ∪ {hi ⊗ tk; i ∈ [r]}is a basis of (gper)k, the set

{x(k)(a); (a) ∈ Γ} ∪ {h(k)

i ; i ∈ [r]}

3.5. The Second Step: (iii) Implies (ii) 55

is a basis of (F(M)/R)k. This implies that (F(M)/R)k+1 is generatedby

{[x(1)(a), x

(k)(b) ]; (a), (b) ∈ Γ} ∪ {[x(1)

(a), h(k)i ]; (a) ∈ Γ, i ∈ [r]}

∪ {[h(1)i , x

(k)(a)]; (a) ∈ Γ, i ∈ [r]} ∪ {[h(1)

i , h(k)j ]; i, j ∈ [r]}.

By applying Lemma 3.4.2-3.4.8, we see that (F(M)/R)k+1 is generatedby

{x(k+1)(a) ; (a) ∈ Γ} ∪ {h(k+1)

i ; i ∈ [r]}In view of (3.11), φk+1 brings a set of generators injectively into a basis,and hence, is injective.

By induction, we have shown that each graded component of φ is anisomorphism and, hence, φ is an isomorphism. This implies that gper

and F(M)/R are isomorphic, and we are done.

3.5 The Second Step: (iii) Implies (ii)

Let g = g(A, τ) be a contragredient Lie superalgebra over a field k ofcharacteristic 0 such that det(A) 6= 0. The aim of this section is to provethat (iii) implies (ii) in Theorem 2. Therefore, assume that A containsan even unit-column. Without loss of generality, we may assume thatthis column is the first column in A. Then kf1⊕kh1⊕ke1 is a subalgebraof g. Recall the definition of a semi-direct product of Lie superalgebrasgiven in the introduction. In the next proposition we will show thatthere exists a complement I of sl(2, k) = kf1 ⊕ kh1 ⊕ ke1 in g that isan ideal in g. Then adx(I) ⊆ I for all x ∈ g, and in particular, for allx ∈ sl(2, k). Hence,

ad : sl(2, k) −→ Der(I)

is a Lie superalgebra homomorphism and g is isomorphic to the semi-direct product of sl(2, k) and I with respect to ad.

Proposition 3.5.1. Let g = g(A, τ) be a contragredient Lie superalgebraover a field k of characteristic 0 such that det(A) 6= 0. Assume thatthe first column in A is an even column where the elements outside thediagonal entry is zero. Then there exists a complement of sl(2, k) =ke1 ⊕ kh1 ⊕ kf1 in g that is an ideal in g.

Proof. Let g = ⊕l∈Zgl be the Z-decomposition of g and let r be the rankof g. Recall the relations defining g given in Section 2.1. The fact that


the first column in A is a even unit-column, implies that sdeg(e1) ≡ 0(mod 2) and that α1(hi) = 0 for all i ∈ [r] \ {1}. Let I be the followingcomplement of sl(2, k) in g:

I =⊕

i∈[r]i6=1

(kfi ⊕ khi ⊕ kei)⊕

⊕

|n|>1

gn

We start by showing the following inclusions:

[g2, g−1] ⊆ I, [g−2, g1] ⊆ I, [g2, g−2] ⊆ I (3.12)

Take i1, i2, j ∈ [r]. Clearly, if i1 6= j and i2 6= j, then

[[ei1, ei2], fj] = [[fi1, fi2], ej] = 0 ∈ I,

and if i1 6= 1 and i2 6= 1, then [[ei1, ei2], fj], [[fi1, fi2], ej] ∈ I. Also,[[e1, ei], f1], [[f1, fi], e1] ∈ I for all i ∈ [r]. Hence, to show the first twoinclusions in (3.12), we must show that [[e1, ei], fi], [[f1, fi], ei] ∈ I for alli ∈ [r] \ {1}. Take i ∈ [r] \ {1} and let si = sdeg(ei). Then we have:

[[e1, ei], fi] = [e1, [ei, fi]]− [ei, [e1, fi]] = [e1, hi] = −α1(hi)e1 = 0 ∈ I[[f1, fi], ei] = [f1, [fi, ei]]− [fi, [f1, ei]] = −(−1)si[f1, hi]

= −(−1)siα1(hi)f1 = 0 ∈ ITo show the last inclusion, observe that [g1 ∩ I, g−1] ⊆ I. This fol-lows since [ei, fj] = δijhi and i 6= 1 for ei ∈ I. Let x ∈ g2 be super-homogeneous and take i, j ∈ [r]. Let sx = sdeg(x) and si = sdeg(ei).Then:

[x, [fi, fj]] = [[x, fi], fj] + (−1)sx si[fi, [x, fj]]

The first inclusion in (3.12) and the observation above implies that thisis an element in I. Hence, [g2, g−2] ⊆ I.

To show that I is an ideal in g, we must show that [sl(2, k), I] ⊆ Iand that [I, I] ⊆ I. The inclusion [sl(2, k), I] ⊆ I is clear from the firsttwo inclusions in (3.12) and from the fact [f1, ei] = [e1, fi] = [h1, hi] = 0,[f1, hi] = α1(hi)f1 = 0 and [e1, hi] = −α1(hi)e1 = 0 for all i ∈ [r] \ {1}.To show that [I, I] ⊆ I, it suffices to show that:

[I ∩ gl, I ∩ g−l+1] ⊆ I for all l ≥ 0 (i)

[I ∩ gl, I ∩ g−l] ⊆ I for all l ≥ 0 (ii)

[I ∩ g−l, I ∩ gl−1] ⊆ I for all l ≥ 0 (iii)

3.6. The Third Step: (i) Implies (iii) 57

We will show the first two inclusions. The last inclusion is similar to thefirst. From now on let [x1, . . . , xn] denote the element [· · · [x1, x2] · · · , xn].

The inclusion (i) is clear for l = 0, 1. We will show that [gl, g−l+1] ⊆ Ifor all l ≤ 2 by induction over l. For l = 2 this follows from (3.12). Taken ≥ 3 and assume that [gl, g−l+1] ⊆ I for all 2 ≤ l < n. Let x ∈ gn besuper-homogeneous and take y = [fi1, . . . , fin−1

] ∈ g−n+1. Then:

[x, y] = [x, [fi1, . . . , fin−1]]

= [[x, [fi1, . . . , fin−2]], fin−1

]± [[x, fin−1], [fi1, . . . , fin−2

]]

We have that:

[[x, [fi1, . . . , fin−2]], fin−1

] ∈ [g2, g−1] ⊆ I[[x, fin−1

], [fi1, . . . , fin−2]] ∈ [gn−1, g−n+2]

Hence, [x, y] ∈ I and, by the induction hypothesis, [gn, g−n+1] ⊆ I.The inclusion (ii) is clear for l = 0, 1 and for l = 2, it follows from

(3.12). Take n ≥ 3 and assume that [I∩gl, I∩g−l] ⊆ I for all 0 ≤ l < n.Let x ∈ I∩gn be super-homogeneous and take y = [fi1, . . . , fin] ∈ I∩g−n.Then:

[x, y] = [x, [fi1, . . . , fin]]

= [[x, [fi1, . . . , fin−1]], fin]± [[x, fin], [fi1, . . . , fin−1

]]

Note that I ∩ gl = gl for all |l| ≥ 2. Then we have that:

[[x, [fi1, . . . , fin−1]], fin] ∈ [I ∩ g1, g−1] ⊆ I

[[x, fin], [fi1, . . . , fin−1]] ∈ [gn−1, g−n+1] = [I ∩ gn−1, I ∩ g−n+1] ⊆ I

Hence, [x, y] ∈ I and we are done.

3.6 The Third Step: (i) Implies (iii)

We are now in a position to show the last step of the proof of Theorem 2,i.e., to show that (i) implies (iii). Let g = g(A, τ) be a contragredient Liesuperalgebra over a field k of characteristic 0 and such that det(A) 6= 0.Furthermore, assume that g is a semi-direct product of sl(2, k). As wementioned in the introduction, given a minimal set of generators for gper,then a minimal set of relations for gper is in one-to-one correspondenceto a basis of H2(g, k). Since {x(a); (a) ∈ Γ} ∪ {hi; i ∈ [r]} is a basis


of g, the set {x(a) ⊗ t; (a) ∈ Γ} ∪ {hi ⊗ t; i ∈ [r]} is a minimal set ofgenerators for gper. Hence, if we can show that H2,3(gper, k) 6= 0, gper

must have cubic relations and, therefore, gper is not 1-2-presented.The fact that g is a semi-direct product of sl(2, k) implies that there

are Lie superalgebra homomorphisms ϕ, ψ such that

g

ψ33sl(2, k)Y9

ϕ

ww and ψ ◦ ϕ = Idsl(2,k).

This implies that there are Z+-graded Lie superalgebra homomorphismsϕper and ψper such that

gper

ψper

22sl(2, k)per

[;

ϕper

uuand ψper ◦ ϕper = Idsl(2,k)per

.

By applying the functor H2(−, k) on this diagram, we obtain vector spacehomomorphisms f, g such that

H2(gper, k)f

00H2(sl(2, k)per, k)]=

gqq

, f ◦ g = IdH2(sl(2,k)per,k).

Since sl(2, k)per contains cubic relations (see Example 1.4.12 in Chap-ter 1) H2,3(sl(2, k)per, k) 6= 0 and, hence, H2,3(gper, k) 6= 0. Thus, gper

contains cubic relations and is not 1-2-presented.

3.7 Examples

We end this chapter with two examples. In the first example, we in-vestigate the generators and the relations for sl(2, 2)per. According toLemma 2.1.2, sl(2, 2)per is 1-generated (c.f. Lemma 3.2.3). The Lie su-peralgebra sl(2, 2) is not included in the Theorem 2 since the determinantof the Cartan matrix is zero. This example shows that the condition of anon-zero determinant can not be omitted in Theorem 2 since sl(2, 2)per

contains cubic relations. In this example, we use “Liedim”, a softwarebased on Mathematica for Lie-calculations, created by C.Lofwall (see[22]).

3.7. Examples 59

Example 3.7.1. The Lie superalgebra sl(2, 2) is contragredient withCartan matrix

A =

2 −1 0−1 0 10 −1 2

and index set τ = {2}. Let h = 〈h1, h2, h3〉 be the Cartan subalgebraand let Π = {α1, α2, α3} be the set of fundamental roots. Then:

∆+ = {α1, α2, α3, α1 + α2, α2 + α3, α1 + α2 + α3}Γ+ = {(1), (2), (3), (1, 2), (2, 3), (1, 2, 3)}

Following the constructions in Section 3.2.2, we create F(M)/R, where

M = {x(1)(a); (a) ∈ Γ} ∪ {h1, h2, h3}

and R is the ideal generated by the 98 quadratic expressions listed inProposition 3.2.4. Using Liedim, we see that dim(F(M)/R)3 = 17. Sincedim sl(2, 2)3 = dim sl(2, 2) = 15, we have that

F(M)/R 6∼= sl(2, 2)per.

Also, by using Liedim, we see that the ideal generated by ker(ϕ) (seeSection 3.2.2) contains the two cubic expressions

[x(1)(1,2,3), [h

(1)3 , x

(1)(2)]] and [x

(1)(−1,−2,−3), [h

(1)3 , x

(1)(−2)]].

These expressions do not follow from the quadratic expressions. There-fore, the ideal appearing in any presentation of sl(2, 2)per, with genera-tors of degree 1, must contain cubic expressions. Hence, sl(2, 2)per is not1-2-presented.

A skew-commutative k-algebra S is a k-algebra endowed with a Z2-grading such that

xy = (−1)|x||y|yx for all x, y ∈ S

x2 = 0 for all x ∈ S such that |x| = 1,

where |x| and |y| denote the Z2-degree of x and y respectively. If g is aZ+-graded Lie superalgebra over a field k, the Koszul dual S = U(g)! isan N-graded skew-commutative k-algebra. Moreover,

ExtS(k, k) = U(gS)


for a Z+-graded Lie superalgebra gS, and g = g(1)S , where g

(1)S is the

subalgebra of gS generated by the elements of degree 1. In the examplebelow, we will calculate the generators and relations of the 1-2-presentedLie superalgebra gper = g(A, τ)per, where A = (1) and τ = {1}, andconstruct the Koszul dual of U(gper). For more details about the Koszuldual see e.g. [7].

Example 3.7.2. In this example, we write |x| to denote the Z2-degree ofan element x, when x belongs to a Lie superalgebra or to a k-algebra en-dowed with a Z2-structure. Let g be the 5-dimensional Lie superalgebrag(A, τ), where A = (1) and τ = {1}. Then ∆ = {α1,−α1, 2α1,−2α1}and Γ = {(1), (−1), (1, 1), (−1,−1)}. According to the Theorem 2,gper

∼= F(M)/R, where

M = {x(1)(1), x

(1)(−1), x

(1)(1,1)x

(1)(−1,−1), h

(1)(1)}

and R is generated by

[x(1)(1), x

(1)(1,1)], [x

(1)(−1), x

(1)(−1,−1)], [h

(1)(1), x

(1)(1,1)]− 2[x

(1)(1), x

(1)(1)],

[h(1)(1), x

(1)(−1,−1)] + 2[x

(1)(−1), x

(1)(−1)], 2[h

(1)(1), x

(1)(1)]− [x

(1)(−1), x

(1)(1,1)],

2[h(1)(1), x

(1)(−1)]− [x

(1)(1), x

(1)(−1,−1)], [x

(1)(1,1), x

(1)(−1,−1)] + 4[x

(1)(1), x

(1)(−1)].

To simplify notations, let x1 = x(1)(1), x2 = x

(1)(−1), x3 = x

(1)(1,1), x4 = x

(1)(−1,−1)

and x5 = h(1)(1). Note that |x1| = |x2| = 1 and |x3| = |x4| = |x5| = 0. Let

V be the vector space spanned by x1, x2, x3, x4 and x5. Let U be the7-dimensional subspace of V ⊗ V obtained from the generators of R byreplacing [x, y] by x⊗ y − (−1)|x||y|y ⊗ x. Hence, U is generated by

U =< x1 ⊗ x3 − x3 ⊗ x1, x2 ⊗ x4 − x4 ⊗ x2,

x5 ⊗ x3 − x3 ⊗ x5 − 4x1 ⊗ x1, x5 ⊗ x4 − x4 ⊗ x5 + 4x2 ⊗ x2,

2x5 ⊗ x1 − 2x1 ⊗ x5 − x2 ⊗ x3 + x3 ⊗ x2,

2x5 ⊗ x2 − 2x2 ⊗ x5 − x1 ⊗ x4 + x4 ⊗ x1, x1 ⊗ x3 − x3 ⊗ x1

x3 ⊗ x4 − x4 ⊗ x3 + 4x1 ⊗ x2 + 4x2 ⊗ x1 > .

Then U(gper) can be considered as T (V )/ 〈U〉.Take the dual space V ∗ = 〈y1, y2, y3, y4, y5〉, where yi(xj) = δij (δij is

the Kronecker-delta), and let |y1| = |y2| = 0 and |y3| = |y4| = |y5| = 1.Then U(gper)

! = T (V ∗)/ < U ◦ >, where U◦ = {f ∈ V ∗⊗V ∗; f(U) = 0}.

3.7. Examples 61

Hence,

U◦ =<y3 ⊗ y3, y4 ⊗ y4, y5 ⊗ y5, y1 ⊗ y3 − y3 ⊗ y1,

y2 ⊗ y4 − y4 ⊗ y2, y5 ⊗ y3 + y3 ⊗ y5, −4y3 ⊗ y5 + y1 ⊗ y1,

y5 ⊗ y4 + y4 ⊗ y5, 4y4 ⊗ y5 + y2 ⊗ y2, y5 ⊗ y1 − y1 ⊗ y5,

y1 ⊗ y5 − 2y2 ⊗ y3, y2 ⊗ y3 − y3 ⊗ y2, y5 ⊗ y2 − y2 ⊗ y5,

y2 ⊗ y5 − 2y1 ⊗ y4, y1 ⊗ y4 − y4 ⊗ y1, y3 ⊗ y4 + y4 ⊗ y3,

4y4 ⊗ y3 + y1 ⊗ y2, y1 ⊗ y2 − y2 ⊗ y1 > .

By excluding the elements in U◦ that express the fact that U(gper)! is a

skew-commutative k-algebra, we have that

U(gper)! =

k[y1, y2]⊗ Λ(y3, y4, y5)⟨y2

1 − 4y3y5, y22 + 4y4y5, y1y5 − 2y2y3,

y2y5 − 2y1y4, y1y2 + 4y4y3

⟩ ,

where Λ denotes the exterior algebra.

Chapter 4

ConsistentPeriodisations

4.1 Introduction

Recall from the introduction that the motivation for this thesis was thetask to find Lie superalgebras whose consistent Z+-graded periodisationsare 1-2-presented, that is, that could be presented with generators of de-gree 1 and relations of degree 2. In Chapter 1, we developed a method toinvestigate this question for Lie algebras. This method was generalizedin Chapter 3 to Lie superalgebras. However, the Z+-graded periodisa-tion considered in Chapter 3 was not necessarily consistent. Consistentmeans that the Z2-grading is obtained from the Z+-grading by reducingmodulo 2. The aim of this chapter is to study the consistent Z+-gradedperiodisation of simple contragredient Lie superalgebras with respect tothe generators and relations with the same method. As a consequence,we restrict ourself to show the sufficiency of relations of degree at most4. We then get the following theorem:

Theorem 3. Let g = g(A, τ) be a simple contragredient Lie superalgebraover a field of characteristic 0. Assume that rank(g) ≥ 2. Then theconsistent Z+-graded periodisation gper of g is isomorphic to a free Liesuperalgebra with generators of degree 1 and relations of degree at most4.

As the object is to prove Theorem 3 in the same way as Theorem 2, thischapter is in many ways similar to Chapter 3. However, the fact thatthe Z+-grading here is consistent generates a great deal of complicationsand we must always consider the super-degree of the elements.

64 4. Consistent Periodisations

The contragredient Lie superalgebras included in Theorem 3 do notneed to be classical, they might be of infinite dimension. However, The-orem 3 shows that the periodisations of the finite-dimensional classicalcontragredient Lie superalgebras A(m,n) when m 6= n, B(m,n), C(n),D(m,n), D(2, 1; α), F (4) and G(3) are presented with generators of de-gree 1 and relations of degree 4. In Section 4.5 we will use this fact inorder to investigate whether their consistent Z+-graded periodisationsare 1-2-presented or not.

Notation. We consider several gradings of the Lie superalgebras in-volved here. First there is the Z2-grading coming from the division intoodd and even elements. However, a contragredient Lie superalgebra isalso Z-graded and finally there is a Z+-grading defined from the periodi-sation. When we talk about the super-degree, or sdeg, we always referto the Z2-grading. The degree, or deg, refers to the Z-grading or to theZ+-grading. When necessary, we will point out which grading the degreerefers to.

4.2 Simple Lie Superalgebras

In Theorem 3 we restrict ourselves to simple contragredient Lie superal-gebras. Therefore, we start this chapter by deriving some results aboutsimple Lie superalgebras.

Lemma 4.2.1. Let g = g0⊕g1 be a simple Lie superalgebra, not necessar-ily contragredient, such that g1 6= 0. Then [g1, g1] = g0 and [g1, g0] = g1.

Proof. Let I = [g1, g]. We will show that I is a non-zero ideal in g. Sinceg is simple this implies that g = [g1, g] and, hence, that [g1, g1] = g0 and[g1, g0] = g1.

To show that I is an ideal in g, it suffices to show that [g0, I], [g1, I] ⊆I. Clearly, [g1, I] ⊆ I and

[g0, I] = [g0, [g1, g]] ⊆ [[g0, g1], g] + [g1, [g0, g]] ⊆ [g1, g] + [g1, g] ⊆ I

Hence, I is an ideal in g.If I=0 then g1 is an ideal in g and, since g1 6= 0, g1 = g. This

implies that [g, g] = 0 which contradicts the fact that g is simple. Hence,I 6= 0.

4.2. Simple Lie Superalgebras 65

Proposition 4.2.2. Let g = g(A, τ) = ⊕l∈Zgl be a contragredient Liesuperalgebra generated by the set {fi, hi, ei}r

i=1 and let h = 〈h1, . . . , hr〉.Recall that αi(hj) = aji for all i, j ∈ {1, . . . , r}, where A = (aij)

ri,j=1.

Thenc = {h ∈ h; αi(h) = 0 for all i ∈ [r]},

where c is the center of g.

Proof. Take x ∈ c and let x =∑

l∈Z xl be the Z-decomposition of x.According to Proposition 2.1.1, for each non-zero element xl ∈ gl, |l| ≥ 1,we can choose y ∈ g−1 ∪ g1 such that [xl, y] 6= 0. Since [x, g] = 0, thisimplies that xl = 0 for all |l| ≥ 1. Hence, x = h for some h ∈ g0 = h.Furthermore, [h, ei] = αi(h)ei for all i ∈ [r]. Thus, αi(h) = 0 for alli ∈ [r].

Take h ∈ h such that αi(h) = 0 for all i ∈ [r]. Let l ∈ Z. Toshow that h ∈ c it suffices to show that [h, xl] = 0 for all xl ∈ gl. Ifl = 0, xl ∈ g0 = h and this is clear. If l ≥ 1, we can assume thatxl = [· · · [ei1, ei2] · · · , eil]. Then

[h, xl] =l∑

j=1

αij(h)xl = 0.

The case when l ≤ 1 is similar.

According to Proposition 4.2.2, c = 0 if and only if det(A) 6= 0. Ifdet(A) 6= 0, c = 0 since α1, . . . , αr are linearly independent. If det(A) =0, there exist b1, . . . , br such that

∑ri=1 biaij = 0 for all j ∈ [r]. Then

h =∑r

i=1 bihi ∈ c.The next proposition states a necessary and sufficient condition for

a contragredient Lie superalgebra to be simple. The same propositionis stated in [14] for finite-dimensional contragredient Lie superalgebrasand is stated and proved in [13] for contragredient Lie algebras.

Proposition 4.2.3. Let g = g(A, τ) be a contragredient Lie superalgebraof rank r. Then g is simple if and only if det(A) 6= 0 and, for each pairi, j ∈ [r], there exist indices i1, . . . , is, s ≥ 0, such that aii1 · . . . · aisj 6= 0.

Proof. In this proof we will use the notation [x1, . . . , xn] to denote theelement [. . . [[x1, x2], x3], . . . xn].

Assume that g is simple. According to Proposition 4.2.2, det(A) 6= 0.Let ∆ be the root system of g. Take i, j ∈ [r] such that i 6= j. Con-sider the ideal I consisting of all linear combinations of elements of the


form [ei, x1, . . . , xn], where n = 0, 1, 2, . . . and xl ∈ {f1, . . . , fr, e1, . . . , er}for all 1 ≤ l ≤ n. Since g is simple, ej ∈ I. Furthermore, since gis a direct sum of the root spaces and each [ei, x1, . . . , xn] is homo-geneous with respect to this decomposition, there exist x1, . . . , xn ∈{f1, . . . , fr, e1, . . . , er} such that

0 6= [ei, x1, . . . , xn] ∈ gαj.

Note that [ei, x1, . . . , xn] ∈ g1. Hence, n is an even number and n/2of the elements x1, . . . , xn lie in g−1 and n/2 lie in g1. By rearrangingthe order with the Jacobi identity, we see that [ei, x1, . . . , xn] is a linearcombination of elements of the form

[ei, fj0, ei1, fj1, . . . eis, fjs, eis+1

],

not all zero. The fact that [ei, x1, . . . , xn] ∈ gαjand that [ek, fl] = 0 for

k 6= l implies that there exist indices i1, . . . , is such that

0 6= [ei, fi, ei1, fi1, . . . eis, fis, ej] = aii1 · . . . · aisjej 6= 0.

Thus, aii1 · . . . · aisj 6= 0 and we are done with the case when i 6= j.Let i = j. If aii 6= 0, there is nothing to prove. Otherwise, since

det(A) 6= 0, there exists l ∈ [r] such that l 6= i and ail 6= 0. Fromabove, there exist indices i1, . . . , is such that ali1 · . . . · aisi 6= 0. Hence,ail · ali1 · . . . · aisi 6= 0.

Assume that det(A) 6= 0 and that for all pairs i, j ∈ [r] there existindices i1, . . . , is, for some s ≥ 0, such that

aii1 · . . . · aisj 6= 0.

Let I be a non-zero ideal in g. To show that g is simple, it suffices toshow that ej ∈ I for all j ∈ [r]. Then, I = g.

Since g is a contragredient Lie superalgebra, I intersects the localpart of g nontrivially. Hence, I ∩ (g−1 ⊕ g0 ⊕ g1) 6= 0 and there exists anon-zero element x ∈ I such that x =

∑aifi +

∑bihi +

∑ciei. By lie-

multiply x with suitable elements in g, we see that there exists ei ∈ I forsome i ∈ [r]. Here we use the defining relations (2.1) and the fact thatdet(A) 6= 0. Now, take any j ∈ [r]. Then there are indices i1, . . . , is ∈ [r]such that aii1 · . . . · aisj 6= 0. Clearly, [ei, fi, ei1, fi1, . . . , eis, fis, ej] ∈ I and

[ei, fi, ei1, fi1, . . . , eis, fis, ej] = aii1 · . . . · aisjej 6= 0.

Hence, ej ∈ I for all j ∈ [r] and we are done.

4.2. Simple Lie Superalgebras 67

In view of Proposition 4.2.3, the condition that g(A, τ) is simple isstronger then the condition that det(A) 6= 0 and A does not contain aneven unit-column used in Chapter 3. In fact, the simplicity of g(A, τ)also implies that A does not contain an odd column where the elementsare zero outside the diagonal entry. The next corollary is a direct con-sequence of Proposition 4.2.3 (c.f. Lemma 2.3.4).

Corollary 4.2.4. Let g(A, τ) be a simple contragredient Lie superalgebraof rank r ≥ 2. Then, for each i ∈ [r], there exists j ∈ [r] such that j 6= iand aji 6= 0.

We end this section with a proposition that specifies a way to decom-pose root-tuple vectors.

Proposition 4.2.5. Let g(A, τ) be a simple contragredient Lie superal-gebra of rank r ≥ 2. Then, for each (a) ∈ Γ there exist (c), (d) ∈ Γ suchthat (c)∗ and (d)∗ are linearly independent and

x(a) = C[x(c), x(d)],

for some non-zero field element C.

Proof. If (a) = (i, j), (−i,−j) where i 6= j, or if |ht (a)| ≥ 3, this followsfrom the definition of x(a), the definition of Γ and from Lemma 2.2.1 byletting (d) be the last index in (a).

Let (a) = (i) for some i ∈ [r]. According to Corollary 4.2.4, thereexists j ∈ [r] such that j 6= i and aji 6= 0. Let si = sdeg(ei) and sj =sdeg(ej). Then, by the Jacobi identity and the defining relations (2.1)in Section 2.1,

[[ei, ej], fj] = [ei, [ej, fj]]− (−1)sisj [ej, [ei, fj]]

= [ei, hj] = −ajiei 6= 0.

Let (c) = (i, j) if (i, j) ∈ Γ or let (c) = (j, i) if (j, i) ∈ Γ. Furthermore,let (d) = (−j). Then, x(a) = C [x(c), x(d)] for some field element C 6= 0.Also, (c)∗ and (d)∗ are linearly independent. The case when (a) = (−i)is similar.

Let (a) = (i, i) ∈ Γ for some i ∈ [r]. Note that (a) ∈ Γ implies thatsdeg(ei) = 1. According to Corollary 4.2.4, there exists j ∈ [r] suchthat j 6= i and aji 6= 0. Then, by the Jacobi identity and the definingrelations (2.1) in Section 2.1,

[[[ei, ei], ej], fj] = [[ei, ei], [ej, fj]]− [ej, [[ei, ei], fj]]

= [[ei, ei], hj] = −2[ei, [hj, ei]] = −2aji[ei, ei] 6= 0.


Let (c) = (i, i, j) if (i, i, j) ∈ Γ, (c) = (i, j, i) if (i, j, i) ∈ Γ or let(c) = (j, i, i) if (j, i, i) ∈ Γ. Furthermore, let (d) = (−j). Then,x(a) = C [x(c), x(d)] for some field element C 6= 0. Also, (c)∗ and (d)∗

are linearly independent. The case when (a) = (−i,−i) ∈ Γ for somei ∈ [r] is similar.

Remark 4.2.6. Let (a) ∈ Γ such that g(a)∗ is one-dimensional. ThenProposition 4.2.5 states that there exist (c), (d) ∈ Γ such that n(c),(d) 6= 0and

[x(c), x(d)] = n(c),(d) x(a).

In particular, if (a) = (i) for some i ∈ [r], the proof above implies that forall i ∈ [r] there exists j ∈ [r] such that i 6= j and such that n(i,j),(−j) 6= 0or n(j,i),(−j) 6= 0. (C.f. Lemma 2.3.4).)

4.3 The Consistent Periodisation and theFree Lie Superalgebra

In this section we introduce two Z+-graded Lie superalgebras, the pe-riodisation of a Lie superalgebra and the free Lie superalgebra. Hence,there are two gradings to consider here, the Z+-grading and the Z2-grading. The degree, or deg, of an element refers to the Z+-grading andthe super-degree, or sdeg, refers to the Z2-grading.


Definition 4.3.1. Given a Lie superalgebra g = g0 ⊕ g1 over a field k,let gper be the Z+-graded Lie superalgebra defined by

gper =⊕i≥1

(gi mod 2 ⊗ kti

),

where [x ⊗ f(t), y ⊗ g(t)] = [x, y] ⊗ f(t)g(t). The Z2-grading of gper isdetermined by:

(gper)l mod 2 =

{⊕i≥1

(g0 ⊗ kt2i

)if l ≡ 0 (mod 2)⊕

i≥1

(g1 ⊗ kt2i−1

)if l ≡ 1 (mod 2)

Note that the Z+-grading is consistent, that is deg(x) ≡ sdeg(x) (mod 2)for all homogeneous elements x ∈ gper.

4.3. The Consistent Periodisation and the Free Lie Superalgebra 69

Remark. Given a Lie superalgebra g, let µ be the automorphism of g byletting µ(x) = (−1)sdeg(x)x. Then gper is the positive part of the twistedLoop superalgebra

L(g, µ) =⊕

i∈Z

(g(µ)i mod 2 ⊗ ti

),

where g(µ)i mod 2 are the eigenspaces of µ.

4.3.2 The Free Lie Superalgebra

Let g = g(A, τ) be a simple contragredient Lie superalgebra of rankr ≥ 2. According to Proposition 4.2.3, det(A) 6= 0. Let h, ∆ andΓ be the Cartan subalgebra, the root system and a set of root-tuplesconstructed in Section 2.2. In this section, we will construct a free Liesuperalgebra F(M) with generators of degree 1, and an ideal R of thissuperalgebra with generators of degree at most 4. The Z-grading of g willnot be considered here. The free Lie superalgebra is Z+-graded and thedegree refers to this grading. The idea is to construct F(M) and R suchthat F(M)/R ∼= gper as Z+-graded Lie superalgebras (see Section 4.4).Since gper is consistent, (gper)1 ⊆ g1. Hence, unlike in Chapter 3, thegenerators of F(M) must have super-degree 1.

Let F(M) be the free Lie superalgebra generated by

M = {x(1)(a); (a) ∈ Γ1},

where the degree and the super-degree is determined by

deg(x) = 1

sdeg(x) ≡ 1 (mod 2)

for all x ∈ M . Let Fi(M) denote the Z+-homogeneous part of degree iin F(M). Note that the Z+-grading is consistent, that is, the Z2-gradingof F(M) is obtained from the Z+-grading by reducing modulo 2.

The ideal R generated by expression of degree ≤ 4 is defined in thesame way as in Chapter 3. We construct a graded Lie superalgebrahomomorphism ϕ from F(M) to gper and we let R be the ideal in F(M)generated by the kernels of the restrictions of ϕ to F2(M), F3(M) andF4(M).


Definition 4.3.2. Let ϕ : F(M) → gper be the unique graded Lie su-peralgebra homomorphism defined by

ϕ : x(1)(a) 7→ x(a) ⊗ t

for all (a) ∈ Γ1.

This is in fact a graded Lie superalgebra homomorphism since ϕ pre-serves the degree and the super-degree of each generator of F(M). Letx⊗ ti ∈ (gper)i for some i ≥ 1. According to Lemma 4.2.1, x = [y, z] forsome y ∈ g1 and z ∈ g. Then, x ⊗ ti = [y ⊗ t1, z ⊗ ti−1]. By using thisand induction over the degree of gper it follows that ϕ is an epimorphism.Let ϕi, i = 1, 2, . . ., denote the graded components of ϕ and let R be theideal in F(M) generated by ker ϕ2 ∪ ker ϕ3 ∪ ker ϕ4. The ideal R is ho-mogeneous in F(M). Hence, F(M)/R is a Z+-graded Lie superalgebraand

(F(M)/R)i = {x + R ∈ F(M)/R; x ∈ Fi(M)}.Furthermore, since R ⊆ ker ϕ, there exist a graded Lie superalgebrahomomorphism

φ : F(M)/R −→ gper; φ(x + R) = ϕ(x). (4.1)

Denote by φi, i ≥ 1, the graded components of φ. The following lemmafollows from the definition of R and the fact that ϕ is surjective.

Lemma 4.3.3. The graded Lie superalgebra homomorphism φ is surjec-tive and for i = 1, 2, 3, 4, φi is an isomorphism.

In what follows, our main interest will be in the Lie superalgebra

F(M)/R. To avoid too many symbols, from now on we let {x(1)(a); (a) ∈

Γ1} stand for the image of the generators in F(M)/R. Hopefully, noconfusion will arise. To simplify notations even further, we make thefollowing definition. This definition corresponds to Definition 3.2.5 inChapter 3. The complication here is that we have access to the Cartan

subalgebra h only in even degrees. for example, we can not define x(2)(a)

as before. Also, in defining h(2k)(a) , for k ≥ 2 and (a) ∈ Γ, we must take

into consideration if (a) ∈ Γ0 or if (a) ∈ Γ1.

4.3. The Consistent Periodisation and the Free Lie Superalgebra 71

Definition 4.3.4. Let {x(1)(a); (a) ∈ Γ1} be the images of the generators

of F(M) in F(M)/R.

1. For all (a) ∈ Γ0 and h ∈ h, define

x(2)(a) = φ−1

2 (x(a) ⊗ t2) and h(2) = φ−12 (h⊗ t2).

2. Take α ∈ ∆1 and β ∈ ∆0 and let the set {x(a1), . . . , x(as)} and{x(b1), . . . , x(bt)} be the ordered basis of gα and gβ respectively chosen inSection 2.2. Define

x(1)α =

x(1)(a1)...

x(1)(as)

and x

(2)β =

x(2)(b1)...

x(2)(bt)

.

3. For each α ∈ ∆, fix hα ∈ h such that α(hα) = 1. Let

x(k)(a) = [h

(2)(a)∗, x

(k−2)(a) ] and x(k)

α = [h(2)α ,x(k−2)

α ]

for all k ≥ 3, (a) ∈ Γk and α ∈ ∆k.

4. For all k ≥ 2, define

h(2k)(a) =

{[x

(1)(a), x

(2k−1)(−a) ] for all (a) ∈ Γ1

[x(2)(a), x

(2k−2)(−a) ] for all (a) ∈ Γ0

.

We write h(2k)i instead of h

(2k)(i) for all k ≥ 1. Note that the definition of

x(k)α in 3. above implies that

x(k)α =

x(k)(a1)...

x(k)(al)

,

where {x(a1), . . . , x(al)} is the ordered basis for gα chosen in Section 2.2.

If α 6= ∆, let x(k)α = 0 for all k ≥ 1.

The Z+-grading of F(M)/R is consistent, that is, the Z2-grading isobtained from the Z+-grading by reducing modulo 2. This implies that

sdeg(x

(k)(a)

)≡ k (mod 2)

sdeg(h

(2k)(b)

)≡ 2k ≡ 0 (mod 2)


for all k ≥ 1, (a) ∈ Γk and all (b) ∈ Γ. Observe that sdeg(h

(2k)(b)

)≡ 0

(mod 2) also when (b) ∈ Γ1. Let

sdeg(x(k)

α

)≡ k (mod 2)

for all k ≥ 1 and all α ∈ Γk.

Remark. As in Chapter 3, for each (a) ∈ Γ we have defined h(a) ∈ hand h(a)∗ ∈ h. The element h(a) is defined as h(a) = [x(a), x(−a)] and h(a)∗

is defined such that (a)∗(h(a)∗) = 1. The elements h(a) and h(a)∗ are notnecessarily equal. For example, take i ∈ [r] such that aii = 0. Thenαi(hi) = 0 and hence, hαi

6= hi.

We end this section by introducing an automorphism of F(M)/R.The mapping

x(1)(a) 7→ x

(1)(−a), x

(1)(−a) 7→ −x

(1)(a),

for all (a) ∈ Γ+ ∩ Γ1, induces an automorphism of F(M) of order 4.Since R is mapped into R, this gives an automorphism of F(M)/R (c.fthe notation at the end of Section 2.1). Then

x(k)(a) 7→ x

(k)(−a), h

(k)i 7→ (−1)sih

(k)(−i), x

(k)(−a) 7→ (−1)kx

(k)(−a)

for all k ≥ 1, h ∈ h and all (a) ∈ Γ+ ∩ Γk, where si ≡ sdeg(i) (mod 2).For k = 2, this is realized by the connection between the automorphismabove and the automorphism of g introduced at the end of Section 2.1.

4.4 Formulation and Proof of Theorem 3

In this section let g = g(A, τ) be a simple contragredient Lie superalgebraover a field of characteristic 0 and let r ≥ 2 be the rank of g. We willformulate and and prove the main theorem of this chapter.

Theorem 3. Let g = g(A, τ) be a simple contragredient Lie superalgebraover a field of characteristic 0 and let r ≥ 2 be the rank of g. Then gper

can be presented with generators of degree 1 and with relations of degreeat most 4.

As mentioned before, the proof of Theorem 3 is a generalization of theproof of Theorem 2. This section is similar to Section 3.4 in Chapter 3.

4.4. Formulation and Proof of Theorem 3 73

Recall the graded Lie superalgebra epimorphism φ given by (4.1)in Section 4.3.2. To show Theorem 3, it suffices to show that φ is anisomorphism. We will show, by induction over the degree, that eachgraded component of φ is injective. Before proving Theorem 3, we willstate and prove some lemmas that will be the foundation for this in-duction (c.f. Lemma 3.4.2-4.4.6). However, we start with a remark (c.f.Remark 3.4.1).

Remark 4.4.1. Consider the graded Lie superalgebra epimorphism

φ : F(M)/R −→ gper

given by (4.1) in Section 4.3.2. By induction over k, it is easy to see that

x(k)(a)

φ7−→ x(a) ⊗ tk, h(2k)(b)

φ7−→ h(b) ⊗ t2k

for all k ≥ 1, (a) ∈ Γk and all (b) ∈ Γ. Recall that we denote by φi,i ≥ 1, the graded components of φ. Take k ≥ 1 and assume that theφi are isomorphisms for all 1 ≤ i ≤ k. If a relation holds in g, thenthe corresponding relation holds in F(M)/R up to and including thisdegree. For example, let (a), (b) ∈ Γ and consider the following relationin g:

[x(a), x(b)] = N(a),(b)x(a)∗+(b)∗

Take m,n ≥ 1 such that m+n ≤ k and such that (a) ∈ Γm and (b) ∈ Γn.Then,

φm+n

([x

(m)(a) , x

(n)(b) ]

)=

[φ

(x

(m)(a)

), φ

(x

(n)(b)

)]=

[x(a) ⊗ tm, x(b) ⊗ tn

]

= [x(a), x(b)]⊗ tm+n =(N(a),(b) x(a)∗+(b)∗

)⊗ tm+n

= φm+n

(N(a),(b)x

(m+n)(a)∗+(b)∗

).

Since ker(φm+n) = 0,

[x(m)(a) , x

(n)(b) ] = N(a),(b)x

(m+n)(a)∗+(b)∗

holds in (F(M)/R)m+n.


[h(2), x(k−1)(a) ] = (a)∗(h)x

(k+1)(a) (i)

[h(2),x(k−1)(a)∗ ] = (a)∗(h)x

(k+1)(a)∗ (ii)

holds for all (a) ∈ Γk−1 and all h ∈ h.


Proof of Lemma 4.4.2. Clearly, (i) implies (ii). Take h ∈ h, (a) ∈ Γk−1and let α = (a)∗. Then,

[h(2), x(k−1)(a) ] = [h(2), [h(2)

α , x(k−3)(a) ]]

= [[h(2), h(2)α ], x

(k−3)(a) ] + [h(2)

α , [h(2), x(k−3)(a) ]]

= α(h)[h(2)α , x

(k−1)(a) ] = α(h)x

(k+1)(a) .

The first and the last equality follows from Definition 4.3.4, the sec-ond from the Jacobi identity and the third equality follows from theassumption that φ is an isomorphism up to and including degree k (c.f.Remark 4.4.1).

Lemma 4.4.3. Take k ≥ 4. Assume that φ is an isomorphism up toand including degree k. Then

[x(m)(a) , x

(n)(b) ] = N(a),(b)x

(k+1)(a)∗+(b)∗ (i)

[x(m)(a) ,x

(n)(b)∗] = N(a),(b)∗x

(k+1)(a)∗+(b)∗ (ii)

holds for all integers m,n ≥ 1 such that m + n = k + 1 and for all(a) ∈ Γm, (b) ∈ Γn such that (a)∗ 6= −(b)∗.

Proof of Lemma 4.4.3. Take m,n ≥ 1 such that m + n = k + 1 and let(a) ∈ Γm, (b) ∈ Γn such that (a)∗ 6= −(b)∗. Since k ≥ 4, we can assumethat n ≥ 3.

We start by showing (i) when (a)∗ and (b)∗ are linearly independent.Choose h ∈ h such that (a)∗(h) = 0 and (b)∗(h) = 1. Then:

[x(m)(a) , x

(n)(b) ] = [x

(m)(a) , [h(2), x

(n−2)(b) ]]

= [[x(m)(a) , h(2)], x

(n−2)(b) ] + [h(2), [x

(m)(a) , x

(n−2)(b) ]]

= −(a)∗(h)[x(m+2)(a) , x

(n−2)(b) ] + N(a),(b)[h

(2),x(m+n−2)(a)∗+(b)∗ ]

= N(a),(b)x(k+1)(a)∗+(b)∗

The first and the third equality follows from the assumption that φ isan isomorphism up to and including degree k (c.f. Remark 4.4.1). Thesecond equality follows from the Jacobi identity and the last equalityfollows from Lemma 4.4.2 and the fact that (a)∗(h) = 0. Hence, (i)holds for all (a) ∈ Γm and (b) ∈ Γn such that (a)∗ and (b)∗ are linearlyindependent. For (ii) when (a)∗ and (b)∗ are linearly independent, let


{x(b1), . . . , x(bl)} be the ordered basis of g(b)∗ chosen in Section 2.2. Since(bi)

∗ = (b)∗ for all 1 ≤ i ≤ l, (a)∗ and (bi)∗ are linearly independent and

it follows from above that

[x(m)(a) , x

(n)(bi)

] = N(a),(bi)x(k+1)(a)∗+(b)∗

for all 1 ≤ i ≤ l. This, together with the definition of [x(m)(a) ,x

(n)(b) ] and

N(a),(b)∗, implies (ii).Assume that (a)∗ and (b)∗ are linearly dependent roots such that

(a)∗ 6= −(b)∗. According to Proposition 4.2.5, there exist (c), (d) ∈ Γsuch that (c)∗ and (d)∗ are linearly independent and

x(b) = C[x(c), x(d)]

for some non-zero field element C. Note that this implies that (a)∗ 6=−(c)∗, (a)∗ 6= −(d)∗, (a)∗ + (c)∗ and (d)∗ are linearly independent andthat (a)∗ + (d)∗ and (c)∗ are linearly independent. Furthermore, thefact that (a)∗ 6= −(b)∗implies that (a)∗ + (c)∗ + (d)∗ 6= 0. Since n ≤ k,the assumption that φ is an isomorphism up to and including degree kimplies that

x(n)(b) = C[x

(s)(c), x

(t)(d)]

for all s, t ≥ 1 such that s + t = n and such that (c) ∈ Γs and (d) ∈ Γt

(c.f. Remark 4.4.1). Then,

[x(m)(a) , x

(n)(b) ] = C[x

(m)(a) , [x

(s)(c), x

(t)(d)]]

= C([[x

(m)(a) , x

(s)(c)], x

(t)(d)] + (−1)m·s[x(s)

(c), [x(m)(a) , x

(t)(d)]]

)

= C(N(a),(c)[x

(m+s)(a)∗+(c)∗, x

(t)(d)] + (−1)m·sN(a),(d)[x

(s)(c),x

(m+t)(a)∗+(d)∗]

)

= C(N(a),(c)N(a)∗+(c)∗,(d) + (−1)m·sN(a),(d)N(c),(a)∗+(d)∗

)x

(k+1)(a)∗+(b)∗.

The second equality follows from the Jacobi identity and the third equal-ity follows from the assumption that φ is an isomorphism up to andincluding degree k (c.f. Remark 4.4.1). The last equality follows fromabove. Moreover, [x(a), x(b)] = N(a),(b)x(a)∗+(b)∗ in g, and by doing thesame calculations in g without the superscript, we see that

N(a),(b) = C(N(a),(c)N(a)∗+(c)∗,(d) + (−1)m·sN(a),(d)N(c),(a)∗+(d)∗

).

Hence, [x(m)(a) , x

(n)(b) ] = N(a),(b)x

(k+1)(a)∗+(b)∗ for (a)∗ 6= −(b)∗ such that (a)∗ and

(b)∗ are linearly dependent. Finally, (ii) follows from (i).


Lemma 4.4.4. Take k ≥ 2 and assume that φ is an isomorphism up toand including degree 2k. Then

[x(1)(a), h

(2k)i ] = −(a)∗(hi)x

(2k+1)(a)

holds for all i ∈ [r] and all (a) ∈ Γ1.

Proof of Lemma 4.4.4. First, assume that (a) 6= (±i). Take any integerss, t ≥ 1 such that s + t = 2k and s ≡ t ≡ sdeg(i) (mod 2). Then,

[x(1)(a), h

(2k)(i) ] = [x

(1)(a), [x

(s)(i) , x

(t)(−i)]]

= [[x(1)(a), x

(s)(i) ], x

(t)(−i)]] + (−1)s[x

(s)(i) , [x

(1)(a), x

(t)(−i)]]

= N(a),(i)[x(s+1)(a)∗+(i)∗, x

(t)(−i), ] + (−1)sN(a),(−i)[x

(s)(i) ,x

(t+1)(a)∗−(i)∗]

=(N(a),(i)N(a)∗+(i)∗,(−i) + (−1)sN(a),(−i)N(i),(a)∗−(i)∗

)x

(2k+1)(a)∗

The first and the third equality follows from the assumption that φ is anisomorphism up to and including degree 2k (c.f. Remark 4.4.1). The lastequality follows from Lemma 4.4.3. Moreover, [x(a), h(i)] = −(a)∗(hi)x(a)holds in g, and by preforming the same calculation as above in g withoutthe superscript, we see that

−(a)∗(hi)x(a) =(N(a),(i)N(a)∗+(i)∗,(−i) + (−1)sN(a),(−i)N(i),(a)∗−(i)∗

)x(a)∗.

Hence, [x(1)(a), h

(2k)i ] = −(a)∗(hi)x

(2k+1)(a) .

Assume that (a) = (i). Then sdeg(i) = 1 and

2[x(1)(i) , h

(2k)(i) ] = 2[x

(1)(i) , [x

(1)(i) , x

(2k−1)(−i) ]] = [[x

(1)(i) , x

(1)(i) ], x

(2k−1)(−i) ]]

= n(i),(i)[x(2)(i,i), x

(2k−1)(−i) ] = n(i),(i)n(i,i),(−i)x

(2k+1)(i) .

The first and the third equality follows from Definition 4.3.4 and thesecond follows from the Jacobi identity. The last equality follows fromLemma 4.4.3. Again, by preforming the same calculations in g we seethat

−2αi(hi) = n(i),(i)n(i,i),(−i)

and, hence,

[x(1)(i) , h

(2k)(i) ] = −αi(hi)x

(2k+1)(i) .

The case when (a) = (−i) follows from the automorphism of F(M)/Rgiven at the end of Section 4.3.2. Observe that hi = −h(−i) in g. Thisand the assumption that φ is an isomorphism up to and including degree

2k implies that h(2k)i = −h

(2k)(−i).


The next lemma states a relation that appears only in even degrees.Therefore, we assume that k is an odd integer.

Lemma 4.4.5. Let k > 4 be an odd integer and assume that φ is anisomorphism up to and including degree k. Let i ∈ [r]. Then,

[x(1)(i) , x

(k)(−i)]− [x

(m)(i) , x

(n)(−i)] = 0 if i ∈ τ

[x(2)(i) , x

(k−1)(−i) ]− [x

(m)(i) , x

(n)(−i)] = 0 if i ∈ [r] \ τ,

for all integers m,n ≥ 1 such that m + n = k + 1 and m ≡ n ≡ sdeg(i).

Proof. We begin with an observation. Take (c) ∈ Γ, h ∈ h and let s, t beany integers such that s, t ≥ 1, s + t = k− 1 and s ≡ t ≡ sdeg(c). Then:

[h(2), h(k−1)(c) ] = [h(2), [x

(s)(c), x

(t)(−c)]]

= [[h(2), x(s)(c)], x

(t)(−c)] + [x

(s)(c), [h

(2), x(t)(−c)]]

= −(c)∗(h)([x

(s)(c), x

(t+2)(−c) ]− [x

(s+2)(c) , x

(t)(−c)]

) (4.2)

The first and the last equality follows from the assumption that φ is anisomorphism up to and including degree k (c.f. Remark 4.4.1). Takei ∈ [r] and choose h ∈ h such that (i)∗(h) 6= 0. If i ∈ τ , let s run throughodd integers from 1 to k − 2 and if i ∈ [r] \ τ , let s run through evenintegers from 2 to k − 3. This yields the equalities

[x(1)(i) , x

(k)(−i)]− [x

(3)(i) , x

(k−2)(−i) ] = [x

(3)(i) , x

(k−2)(−i) ]− [x

(5)(i) , x

(k−4)(−i) ] = . . .

. . . =[x(k−4)(i) , x

(5)(−i)]− [x

(k−2)(i) , x

(3)(−i)] = [x

(k−2)(i) , x

(3)(−i)]− [x

(k)(i) , x

(1)(−i)]

if i ∈ τ and the equalities

[x(2)(i) , x

(k−1)(−i) ]− [x

(4)(i) , x

(k−3)(−i) ] = [x

(4)(i) , x

(k−3)(−i) ]− [x

(6)(i) , x

(k−5)(−i) ] = . . .

. . . =[x(k−5)(i) , x

(6)(−i)]− [x

(k−3)(i) , x

(4)(−i)] = [x

(k−3)(i) , x

(4)(−i)]− [x

(k−1)(i) , x

(2)(−i)]

if i ∈ [r]\τ . Adding an appropriate number of these equalities, it followsthat

[x(1)(i) , x

(k)(−i)]− [x

(m)(i) , x

(n)(−i)] =

m− 1

2

([x

(1)(i) , x

(k)(−i)]− [x

(3)(i) , x

(k−2)(−i) ]

)(4.3)

if i ∈ τ and that

[x(2)(i) , x

(k−1)(−i) ]− [x

(m)(i) , x

(n)(−i)] =

m− 2

2

([x

(2)(i) , x

(k−1)(−i) ]− [x

(4)(i) , x

(k−3)(−i) ]

)(4.3′)


if i ∈ [r] \ τ for all m,n ≥ 1 such that m + n = k + 1 and m ≡ n ≡sdeg(i). Hence, to show the lemma, if suffices to show that [x

(1)(i) , x

(k)(−i)]−

[x(3)(i) , x

(k−2)(−i) ] = 0 if i ∈ τ , and that [x

(2)(i) , x

(k−1)(−i) ] − [x

(4)(i) , x

(k−3)(−i) ] = 0 if

i ∈ [r] \ τ .Choose any i ∈ [r]. According to Proposition 4.2.5 and Remark 4.2.6,

there exists j ∈ [r] such that j 6= i, (i, j) ∈ Γ and n(i,j),(−j) 6= 0, orsuch that j 6= i, (j, i) ∈ Γ and n(j,i),(−j) 6= 0. Let si = sdeg(i) andsj = sdeg(j). We treat the case when (i, j) ∈ Γ here. The case when(j, i) ∈ Γ is similar. Define

A = n(i,j),(−j)

([x

(1)(i) , x

(k)(−i)]− [x

(3)(i) , x

(k−2)(−i) ]

)

B = n(i,j),(−i)

([x

(1)(j), x

(k)(−j)]− [x

(3)(j), x

(k−2)(−j) ]

)if i, j ∈ τ ,

C = [x(2)(i,j), x

(k−1)(−i,−j)]− [x

(4)(i,j), x

(k−3)(−i,−j)]

(i)

A = n(i,j),(−j)

([x

(1)(i) , x

(k)(−i)]− [x

(3)(i) , x

(k−2)(−i) ]

)

B = n(i,j),(−i)

([x

(2)(j), x

(k−1)(−j) ]− [x

(4)(j), x

(k−3)(−j) ]

)if i ∈ τ and j ∈ [r] \ τ ,

C = [x(1)(i,j), x

(k)(−i,−j)]− [x

(3)(i,j), x

(k−2)(−i,−j)]

(ii)

A = n(i,j),(−j)

([x

(2)(i) , x

(k−1)(−i) ]− [x

(4)(i) , x

(k−3)(−i) ]

)

B = n(i,j),(−i)

([x

(1)(j), x

(k)(−j)]− [x

(3)(j), x

(k−2)(−j) ]

)if i ∈ [r] \ τ and j ∈ τ ,

C = [x(1)(i,j), x

(k)(−i,−j)]− [x

(3)(i,j), x

(k−2)(−i,−j)],

(iii)

A = n(i,j),(−j)

([x

(2)(i) , x

(k−1)(−i) ]− [x

(4)(i) , x

(k−3)(−i) ]

)

B = n(i,j),(−i)

([x

(2)(j), x

(k−1)(−j) ]− [x

(4)(j), x

(k−3)(−j) ]

)if i, j ∈ [r] \ τ .

C = [x(2)(i,j), x

(k−1)(−i,−j)]− [x

(4)(i,j), x

(k−3)(−i,−j)]

(iv)

The idea is to, in each case, find three independent linear relations ofA,B and C. Then A = 0 and, since n(i,j),(−j) 6= 0, we are done. Wecontinue in steps:


1. For the first linear relation in case (i) above, we expand the two

Lie brackets [x(2)(i,j), x

(k−1)(−i,−j)] and [x

(4)(i,j), x

(k−3)(−i,−j)].

[x(2)(i,j), x

(k−1)(−i,−j)] = [x

(2)(i,j), [x

(1)(−i), x

(k−2)(−j) ]]

= [[x(2)(i,j), x

(1)(−i)], x

(k−2)(−j) ] + [x

(1)(−i), [x

(2)(i,j), x

(k−2)(−j) ]]

= n(i,j),(−i)[x(3)(j), x

(k−2)(−j) ] + n(i,j),(−j)[x

(1)(−i), x

(k)(i) ]

= n(i,j),(−i)[x(3)(j), x

(k−2)(−j) ] + n(i,j),(−j)[x

(k)(i) , x

(1)(−i)]

(4.4)

The first and the third equality follows the assumption that φ isan isomorphism up to and including degree k (c.f. Remark 4.4.1).The second equality follows from the Jacobi identity and the lastequality follows from the anticommutativity. Similarly:

[x(4)(i,j), x

(k−3)(−i,−j)] = [[x

(1)(i) , x

(3)(j)], x

(k−3)(−i,−j)]

= [x(1)(i) , [x

(3)(j), x

(k−3)(−i,−j)]] + [x

(3)(j), [x

(1)(i) , x

(k−3)(−i,−j)]]

= n(j),(−i,−j)[x(1)(i) , x

(k)(−i)] + n(i),(−i,−j)[x

(3)(j), x

(k−2)(−j) ]

= n(i,j),(−i)[x(3)(j), x

(k−2)(−j) ] + n(i,j),(−j)[x

(1)(i) , x

(k)(−i)]

(4.5)

The last equality follows from Proposition 2.3.1 and 2.3.2.

By subtracting equation (4.5) from equation (4.4) and using thenotation for A, B and C given above, we have that:

C = −n(i,j),(−j)

([x

(1)(i) , x

(k)(−i)]− [x

(k)(i) , x

(1)(−i)]

)

Similarly, by expanding [x(1)(i,j), x

(k)(−i,−j)] and [x

(3)(i,j), x

(k−2)(−i,−j)] in case

(ii) and (iii), and by expanding [x(2)(i,j), x

(k−1)(−i,−j)] and [x

(4)(i,j), x

(k−3)(−i,−j)]

in case (iv), we have that

C = (−1)sjn(i,j),(−j)

([x

(1)(i) , x

(k)(−i)]− [x

(k)(i) , x

(1)(−i)]

)if i ∈ τ

C = −B + n(i,j),(−j)

([x

(2)(i) , x

(k−1)(−i) ]− [x

(k−1)(i) , x

(2)(−i)]

)if i ∈ [r] \ τ

where sj = sdeg(j). By applying equation (4.3) and equation (4.3′)


on this, we have that:

C = (−1)sjk − 1

2A if i ∈ τ

C = −B +k − 3

2A if i ∈ [r] \ τ

2. For the second linear relation of A, B and C, we start with a cal-culation in g. Let si ≡ sdeg(i) (mod 2) and sj ≡ sdeg(j) (mod 2).

h(i,j) = [x(i,j), x(−i,−j)] = [x(i,j), [x(−i), x(−j)]]

= [[x(i,j), x(−i)], x(−j)] + (−1)(si+sj)si[x(−i), [x(i,j), x(−j)]]

= n(i,j),(−i)[x(j), x(−j)]− (−1)sisjn(i,j),(−j)[x(i), x(−i)]

= n(i,j),(−i)hj − (−1)sisjn(i,j),(−j)hi

According to the assumption that φ is an isomorphism up to degree

k, this implies that h(k−1)(i,j) = n(i,j),(−i)h

(k−1)j − (−1)sisjn(i,j),(−j)h

(k−1)i

(c.f. Remark 4.4.1). Take any h ∈ h. Then

[h(2), h(k−1)(i,j) ] = n(i,j),(−i)[h

(2), h(k−1)j ]− (−1)sisjn(i,j),(−j)[h

(2), h(k−1)i ].

(4.6)Consider equation (4.2) for (c) = (i, j), (c) = (j) and (c) = (i)respectively. If sdeg(c) ≡ 1 (mod 2), let s = 1 and if sdeg(c) ≡ 0(mod 2), let s = 2. By applying equation (4.2), equation (4.6)becomes

(i, j)∗(h)C = (j)∗(h)B − (−1)sisj(i)∗(h)A. (4.7)

Since i 6= j, (i)∗ and (j)∗ are linearly independent. Choose h, h′ ∈ hsuch that (i)∗(h) = 0, (j)∗(h) 6= 0 and (i)∗(h′) 6= 0, (j)∗(h′) = 0.Then, from (4.7), we get the two linear combinations

C = B and C = −(−1)sisjA

Summing up, from step 1 and 2, we have the following linear relationsof A, B and C:

C = (−1)sjk − 1

2A

C = B

C = −(−1)sjA

if i ∈ τ

C = −B +k − 3

2A

C = B

C = −A

if i ∈ [r] \ τ


This gives that (k + 1)A = 0 for all i ∈ [r]. Since the characteristic ofthe underlying field is zero, A = 0, and since n(i,j),(−j) 6= 0 we are done.

Lemma 4.4.6. Let k > 4 be an odd integer and assume that φ is anisomorphism up to and including degree k. Take integers m,n ≥ 1 such

that m ≡ n (mod 2) and m + n = k + 1. Then [x(m)(a) , x

(n)(b) ] is a linear

combination of

h(k+1)1 , . . . , h(k+1)

r

for all (a), (b) ∈ Γm = Γn such that (a)∗ = −(b)∗.

Proof of Lemma 4.4.6. Recall the height of a root defined in Section 2.2.Note that this height is well defined here since the fact that g is simpleimplies that det(A) 6= 0. We will use induction over the height of (b)∗

to prove Lemma 4.4.6. If |ht(b)∗| = 1, this follows from Lemma 4.4.5and Definition 4.3.4. Take l ≥ 1 and assume that the lemma is truewhen |ht(b)∗| ≤ l. Take integers m, n ≥ 1 such that m ≡ n (mod 2)and m + n = k + 1. Let (a), (b) ∈ Γm = Γn be root-tuples such that(a)∗ = −(b)∗ and |ht(b)∗| = l + 1. In view of the automorphism ofF(M)/R given in the end of Section 4.3.2 and of Lemma 4.4.5, we canassume that (b) ∈ Γ− and that n ≥ 3. According to the definition of Γ,there exists j ∈ [r] and (c) ∈ Γ− such that

x(b) = [x(c), x(−j)].

Clearly, |ht(c)∗| = l. According to the assumption that φ is an isomor-phism up to and including degree k,

x(n)(b) = [x

(s)(c), x

(t)(−j)]

holds for all integers s, t ≥ 1 such that s + t = n, s ≡ sdeg(c) (mod 2)and t ≡ sdeg(−j) (mod 2) (c.f. Remark 4.4.1). Hence,

[x(m)(a) , x

(n)(b) ] = [x

(m)(a) , [x

(s)(c), x

(t)(−j)]]

= [[x(m)(a) , x

(s)(c)], x

(t)(−j)] + (−1)ms[x

(s)(c), [x

(m)(a) , x

(t)(−j)]]

= N(a),(c)[x(m+s)(a)∗+(c)∗, x

(t)(−j)] + (−1)msN(a),(−j)[x

(s)(c),x

(m+t)(a)∗−(j)∗]

The last equation follows from Lemma 4.4.3. Note that

(a)∗ + (c)∗ = (j)∗ and (a)∗ − (j)∗ = −(c)∗.


According to the induction hypothesis, the expressions [x(m+s)(a)∗+(c)∗, x

(t)(−j)]

and [x(s)(c),x

(m+t)(a)∗−(j)∗] are column-vectors consisting of linear combinations

of h(k+1)1 , . . . , h

(k+1)r . Hence,

N(a),(c)[x(m+s)(a)∗+(c)∗, x

(t)(−j)] and N(a),(−j)[x

(s)(c),x

(m+t)(a)∗−(j)∗]

are linear combinations of h(k+1)1 , . . . , h

(k+1)r , and so is [x

(m)(a) , x

(n)(b) ].

We are now in a position to prove Theorem 3.

Proof of Theorem 3. We will show that F(M)/R and gper are isomor-phic, where M and R is the set and ideal constructed in Section 4.3.Consider the graded Lie superalgebra epimorphism φ defined in Sec-tion 4.3. Note that

φ(x(k)(a)) = x(a) ⊗ tk for all k ≥ 1 and all (a) ∈ Γk

φ(h(2k)i ) = hi ⊗ t2k for all k ≥ 1 and all i ∈ [r].

(4.8)

To show that F(M)/R and gper are isomorphic, it suffices to show that φis injective and, hence, it suffices to prove that each graded component ofφ is injective. Let φi, i ≥ 1, denote the graded components of φ. Clearly,φ1, φ2, φ3 and φ4 are injective. Assume that φk is injective for some

k ≥ 4. Then, {x(k)(a); (a) ∈ Γ0}∪ {h(k)

i ; i ∈ [r]} is a basis of (F(M)/R)k if

k is even, and {x(k)(a); (a) ∈ Γ1} is a basis of (F(M)/R)k if k is odd. This

follows by (4.8) and the fact that {x(a)⊗ tk; (a) ∈ Γ0}∪{hi⊗ tk; i ∈ [r]}is a basis of (gper)k if k is even, and {x(a) ⊗ tk; (a) ∈ Γ1} is a basis of(gper)k if k is odd. Hence, (F(M)/R)k+1 is generated by

{[x(1)(a), x

(k)(b) ]; (a) ∈ Γ1, (b) ∈ Γ0} ∪ {[x(1)

(a), h(k)i ]; (a) ∈ Γ1, i ∈ [r]}

if k is even and by

{[x(1)(a), x

(k)(b) ]; (a), (b) ∈ Γ1}

if k is odd. By applying Lemma 4.4.2-4.4.6, we see that the vector space(F(M)/R)k+1 is generated by

{x(k+1)(a) ; (a) ∈ Γ1}

if k is even and by

{x(k+1)(a) ; (a) ∈ Γ0} ∪ {h(k+1)

i ; i ∈ [r]}

4.5. An Investigation of Some of the Classical Lie Superalgebras 83

if k is odd. In view of (4.8), φk+1 brings a set of generators of the vectorspace (F(M)/R)k+1 injectively into a basis of (gper)k+1. Thus, φk+1 isinjective and, by induction over k, φ is an isomorphism.

4.5 An Investigation of Some of the Clas-sical Lie Superalgebras

Theorem 3 shows the sufficiency of relations of degree at most 4 in pre-senting the consistent Z+-graded periodisation of simple contragredientLie superalgebras. The aim of this last section is to use Theorem 3to find Lie superalgebras for which this Z+-graded periodisation is 1-2-presented among the classical contragredient Lie superalgebras. We alsogive a description of some of the classical Lie superalgebras as contra-gredient. The following is a summary of this investigation:

Summary of investigation:Let g = g(A, τ) be a classical contragredient Lie superalgebra. The con-sistent Z+-graded periodisation gper of g is 1-2-presented if g is one ofthe following:

A(m− 1, n− 1) = sl(m,n) for m 6= n such that m,n ≥ 2

B(m,n) = osp(2m + 1, 2n) for m,n ≥ 2

C(n + 1) = osp(2, 2n− 2) for n ≥ 2

G(3) or F (4)

Furthermore, it is necessary with relations of degree 3 or 4 in presentinggper when g is one of D(2, 1; α), A(1, 4), B(0, n) for n > 1 or C(2). Theremaining classical Lie superalgebras are yet to be investigated.

Let g = g(A, τ) be a simple contragredient Lie superalgebra. Recallthe definition of M , the definition of the Z+-graded Lie superalgebraepimorphism

ϕ : F(M) −→ gper

and the definition of the ideal R generated by ker ϕ2∪ker ϕ3∪ker ϕ4 givenin Section 4.3.2. This induces a Z+-graded Lie superalgebra epimorphism

φ : F(M)/R −→ gper


such that φi is injective for i = 1, 2, 3, 4. Clearly, we are not obliged todefine R in this way. We may as well let R be the ideal generatedby ker ϕ2. Then φi is an isomorphism for i = 1, 2. However, if φi

still is an isomorphism for i = 3, 4 as well, Definition 4.3.4 and theproof of Theorem 3 still holds. Hence, then gper is 1-2-presented. Sincethe classical Lie superalgebras are finite-dimensional, to show that φi isinjective for i = 3, 4, it suffices to show that

dim (F(M)/R)3 = dim (gper)3 = dim g1

dim (F(M)/R)4 = dim (gper)4 = dim g0(∗)

To show the equalities in (∗) we use two different methods. For G(3),F (4) and for single Lie superalgebras from the series A(m,n) whenm 6= n, B(m,n), C(n), D(m,n) or D(2, 1; α), we use ”Liedim”, a Math-ematica based software for Lie-calculations created by C. Lofwall [22].We construct the quadratic relations, that is the generators for ker ϕ2,and use Liedim to calculate the left hand side dimensions in (∗). Theresults of these investigations are presented below. However, it is unfea-sible to compute the quadratic relations for series of Lie superalgebras.Instead we use a method similar to the proof of Theorem 3. This methodis described in more details later on in this section.

Before continuing, we settle some notations. The root spaces ofthe contragredient classical Lie superalgebras g considered here are one-dimensional. Therefore, the root system ∆ and the root-tuple system Γintroduced in Chapter 2 coincide, and we prefer to use the root system.We use xα to denote the root vector of the root space gα for α ∈ ∆.Note however that there is a choice in assigning the root vectors xα forα ∈ ∆. In calculating these examples, we always use the most naturalchoice of the root vectors. Let r be the rank of g. If α = αi for somei ∈ [r], we let xα = ei and x−α = fi. Furthermore, if α = α1+ · · ·+αk forsome k ∈ [r] \ {1} and xα1+···+αk−1

∈ ∆, we let xα = [xα1+···+αk−1, xαk

] andx−α = [x−α1−···−αk−1

, x−αk]. In Chapter 2 we also defined the elements

h(a) and h(a)∗ in h for each (a) ∈ Γ. As before, let hα be a fixed elementin h such that α(hα) = 1 for all α ∈ ∆. However, since the root-tuplesare not used here, we introduce the notation

h(α) = [xα, x−α]

for all α ∈ ∆. Finally, for each α, β ∈ ∆ such that α 6= −β, let nα,β bethe number such that

[xα, xβ] = nα,β xα+β.


Let nα,β = 0 if α 6∈ ∆ or if β 6∈ ∆. For the contragredient Lie superalge-bras considered here, nα,β 6= 0 if and only if α, β, α + β ∈ ∆.

The exceptional Lie superalgebra G(3):

The exceptional Lie superalgebra G(3) is a 31-dimensional contragredi-ent Lie superalgebra g = g(A, τ) with

A =

0 1 0−1 2 −30 −1 2

and τ = {1}.

The even part g0 of g is 17-dimensional and the odd part g1 is 14-dimensional. Hence,

dim(gper)i =

{14 if i ≡ 1 (mod 2)

17 if i ≡ 0 (mod 2).

Let α1, α2 and α3 be the fundamental roots of G(3). Then:

∆+0 = {α2, α3, α2 + α3, 2α2 + α3, 3α2 + α3, 3α2 + 2α3,

2α1 + 4α2 + 2α3}∆+

1 = {α1, α1 + α2, α1 + α2 + α3, α1 + 2α2 + α3,

α1 + 3α2 + α3, α1 + 3α2 + 2α3, α1 + 4α2 + 2α3}Let F(M) be the free Lie superalgebra generated by the set M ={xα; α ∈ g1}. The number of quadratic expressions generating ker ϕ2, is88. Below is a list of the quadratic relations:

1. The 48 expressions [xα, xβ] and [x−α, x−β] for all α, β ∈ ∆+1 such

that α + β 6= 2α1 + 4α2 + 2α3.

2. The 6 expressions [xα, x−β] for all α, β ∈ ∆+1 such that α 6= β and

α + β = 2α1 + 4α2 + 2α3.

3. The 15 expressions of the form

nγ,δ[xα, xβ]− nα,β[xγ, xδ]

where α + β = γ + δ ∈ ∆1 and α, γ ∈ ∆+1 , and the corresponding

15 expressions

nγ,δ[x−α, x−β]− nα,β[x−γ, x−δ].


4. The four expressions:

h(α1+2α2+α3) − 2h(α1) − 2h(α1+α2) + 2h(α1+α2+α3)

h(α1+3α2+α3) + 8h(α1) + 8h(α1+α2) + 4h(α1+α2+α3)

h(α1+3α2+2α3) − 8h(α1) − 4h(α1+α2) + 8h(α1+α2+α3)

h(α1+4α2+2α3) + 12h(α1) + 8h(α1+α2) − 8h(α1+α2+α3)

Let R be the ideal in F(M) generated by the expressions above. Byusing ”Liedim” we see that

dim(F(M)/R)1 = 14, dim(F(M)/R)2 = 17,

dim(F(M)/R)3 = 14, dim(F(M)/R)4 = 17.

Hence, the consistent Z+-graded periodisation of G(3) is 1-2-presented.

The exceptional Lie superalgebra F(4):

F (3) is a 40-dimensional contragredient Lie superalgebra g = g(A, τ)with

A =

0 1 0 0−1 2 −2 00 −1 2 −10 0 −1 2

and τ = {1}.

Furthermore, dim g0 = 24 and dim g1 = 16. The number of quadraticexpressions generating the kernel of ϕ2 is

(16

2

)+ 16− 24 = 112.

Let α1, α2, α3 and α4 be the fundamental roots of F (4). Then:

∆+0 = {α2, α3, α4, α2 + α3, α3 + α4, 2α2 + α3, α2 + α3 + α4,

2α2 + α3 + α4, 2α2 + 2α3 + α4, 2α1 + 3α2 + 2α3 + α4}∆+

1 = {α1, α1 + α2, α1 + α2 + α3, α1 + α2 + α3 + α4, α1 + 2α2 + α3,

α1 + 2α2 + α3 + α4, α1 + 2α2 + 2α3 + α4, α1 + 3α2 + 2α3 + α4}.Let F(M) be the free Lie superalgebra generated by M = {xα; α ∈ g1}.Below is a list of the 112 quadratic expressions generating ker ϕ2.


that α + β 6= 2α1 + 3α2 + 2α3 + α4.


2. The 8 expressions [xα, x−β] for all α, β ∈ ∆+1 such that α + β =

2α1 + 3α2 + 2α3 + α4.

3. The 18 expressions of the form nγ,δ[xα, xβ] − nα,β[xγ, xδ] whereα + β = γ + δ ∈ ∆1 and α, γ ∈ ∆+

1 , and the corresponding 18expressions nγ,δ[x−α, x−β]− nα,β[x−γ, x−δ].

4. The four expressions:

h(α1) + h(α1+α2) − h(α1+α2+α3) − h(α1+2α2+α3)

h(α1) + h(α1+α2) + h(α1+α2+α3+α4) + h(α1+2α2+α3+α4)

h(α1) − h(α1+α2+α3) + h(α1+α2+α3+α4) − h(α1+2α2+2α3+α4)

h(α1+α2) + h(α1+α2+α3) − h(α1+α2+α3+α4) − h(α1+3α2+2α3+α4)

Let R be the ideal in F(M) generated by the expressions above. Byusing ”Liedim” we see that

dim(F(M)/R)1 = 16, dim(F(M)/R)2 = 24,

dim(F(M)/R)3 = 16, dim(F(M)/R)4 = 24.

Hence, F (4)per is 1-2-presented.

The exceptional Lie superalgebra D(2,1; α):

The exceptional Lie superalgebras D(2, 1; α) for α 6= {0,−1} are a one-parameter family of 17-dimensional Lie superalgebras. D(2, 1; α) is adeformation of D(2, 1) and D(2, 1; 1) ∼= D(2, 1). Moreover, the Lie su-peralgebras D(2, 1; β) such that

β ∈{

α±1, −(1 + α)±1,

( −α

1 + α

)±1}

for some α 6∈ {0,−1} are isomorphic.For α 6∈ {0,−1}, D(2, 1; α) is a contragredient Lie superalgebra g =

g(A, τ) where

A =

0 1 α−1 2 0−1 0 2

and τ = {1}.

Furthermore, dim g0 = 9 and dim g1 = 8. Hence,

dim (gper)i =

{9 if i ≡ 0 (mod 2)

8 if i ≡ 1 (mod 2).


Let α1, α2 and α3 be the fundamental roots of D(2, 1; α). Then

∆+0 = {α2, α3, 2α1 + α2 + α3}

∆+1 = {α1, α1 + α2, α1 + α3, α1 + α2 + α3}

The dimension of ker ϕ2 is 27. The following expressions form a basis ofker ϕ2. These expressions are independent of the parameter α.


that α + β 6= 2α1 + α2 + α3.

2. The 4 expressions [xα, x−β] for all α, β ∈ ∆+1 such that α + β =

2α1 + α2 + α3.

3. Three expressions of the form nγ,δ[xα, xβ]− nα,β[xγ, xδ] where α +β = γ + δ ∈ ∆1 and α, γ ∈ ∆+

1 , and the corresponding threeexpressions nγ,δ[x−α, x−β]− nα,β[x−γ, x−δ].

4. The quadratic expression h(α1) + h(α1+α2) + h(α1+α3) + h(α1+α2+α3).

By using Liedim to calculate the dimensions of the graded componentsof the free Lie superalgebra generated by {xα; α ∈ ∆1} and with therelations above, we see that

dim(F(M)/R)1 = 8, dim(F(M)/R)2 = 9,

dim(F(M)/R)3 = 16, dim(F(M)/R)4 = 45.

Hence, D(2, 1; α)per is not 1-2-presented. Furthermore, the presentationof D(2, 1; α)per needs relations of degree 3.

The series A, B and C:

The method of using Liedim as for G(3), F (4) and D(2, 1; α) above canbe applied to show whether the periodisation of a single Lie superalgebrais 1-2-presented or not. However, it is unfeasible to calculate the gener-ators of ker ϕ2 for a series of Lie superalgebras. Instead, to investigateA(m,n) for m 6= n, B(m,n) and C(n + 1), we use a similar methodas in the proof of Theorem 3. Let g be a finite-dimensional contragre-dient Lie superalgebra, M = {xα; α ∈ ∆1} and let R be the ideal inF(M) generated by ker ϕ2. According to Theorem 3, to show that gper

is 1-2-presented, it suffices to show that dim (F(M)/R)3 = dim g1 anddim (F(M)/R)4 = dim g0. Actually, it suffices to show that

dim (F(M)/R)3 ≤ dim g1

dim (F(M)/R)4 ≤ dim g0.


Hence, analogous to the proof of Theorem 3 and in view of the auto-morphism of F(M)/R, to show that the periodisation of a classical Liesuperalgebras is 1-2-presented, we need to show that:

[h(2)(α), x

(1)β ] = β(h(α)) x

(3)β for all α, β ∈ ∆+

1 (1)

[x(1)α , x

(2)β ] = nα,β x

(3)α+β for all α ∈ ∆+

1 , β ∈ ∆0 such that

α 6= −β(2)

[h(2)(α), x

(2)β ] = β(h(α)) x

(4)β for all α ∈ ∆+

1 , β ∈ ∆+0 (3)

[x(m)α , x

(n)β ] = nα,β x

(4)α+β for all m,n ≥ 1 such that m + n = 4

and all α ∈ ∆+m ,β ∈ ∆n such that α 6= −β.

(4)

[x(1)αi

, x(3)−αi

]− [x(3)αi

, x(1)−αi

] = 0 for all i ∈ τ (5)

[x(m)α , x

(n)−α] is a linear combination of h

(4)(α1)

, . . . , h(4)(αr)

for all

m,n ≥ 1 such that m + n = 4 and all α ∈ ∆m = ∆n

(6)

The equalities (1)-(6) above are analogous to Lemma 4.4.2-4.4.6 for k =2, 3. The difficulty in showing these equalities in small degrees originatesin the fact that the Z+-graded periodisation in question is consistent.Consequently, we do not have access to g0 in degree 1. This was never aproblem in Chapter 3 where the periodisation gper of g was not consistentand where (gper)1 = g. For example, let α ∈ ∆0. In Chapter 3 we

used the definition xα = [hα, xα] in order to split up the element x(2)α as

[h(1)α , x

(1)α ]. This is not possible here since g0 6⊆ (gper)1. Instead, we have

to use the fact that g0 = [g1, g1] and spit up x(2)α as a sum

∑ci[x

(1)γi , x

(1)δi

]where γi, δi ∈ ∆1. As a consequence, we become more dependent ofthe structure of the root system than before. Another example of this

problem is how to deal with h(2)(α) for α ∈ ∆0. We can not use the fact

that h(α) = [xα, x−α]. However, this problem is already solved in (1)-(6)where we only consider h(α) for α ∈ ∆1. This can be done since h isgenerated by the set {h(α); α ∈ ∆1}.

As mentioned before, in showing the equalities (1)-(6), we are moredependent of the structure of the root system in question than in theproofs of the corresponding lemmas. As a consequence the proofs of(1)-(6) are quite extensive and we will not show all the technicalitieshere. However, given the equalities (1), (2) and (3) for A(m,n) where2 ≤ m < n, B(m,n) where m,n ≥ 2 and for C(n + 1) where n ≥ 3,then (4), (5) and (6) can be shown in a more general way. For (4), take


α, β ∈ ∆1 such that α 6= −β. We can show that

[x(1)α , x

(3)β ] = nα,β x

(4)α+β

by a small modification in the proof of Lemma 4.4.4. The only differenceis when α and β are linearly dependent. Since the relation is of degree4, the dependency of α and β implies that α = β. This case is treatedseparately. Then, (4) follows for α, β ∈ ∆1 and m = n = 2 by splitting

x(2)β into root vectors of degree 1 and using the Jacobi identity.

The root system for the classical Lie superalgebras may be chosensuch that |τ | = 1. Let τ = {i}. For A(m,n) where 2 ≤ m < n, B(m,n)where m,n ≥ 2 and for C(n + 1) where n ≥ 3, it is easy to see thatαi+1, αi+2 ∈ ∆0 and that αi + αi+1, αi + αi+1 + αi+2 ∈ ∆1. To show (5),much can be done in the same way as in the proof of Lemma 4.4.5. Let

A = nαi+αi+1,−αi+1

([x(1)

αi, x

(3)−αi

]− [x(3)αi

, x(1)−αi

])

C = [x(1)αi+αi+1

, x(3)−αi−αi+1

]− [x(3)αi+αi+1

, x(1)−αi−αi+1

].

Then, as in the proof of Lemma 4.4.5, A = C. Also, it can be shownthat

[h(2), h(2)(αi+αi+1)

] = nαi+αi+1,−αi[h(2), h

(2)(αi+1)

]− nαi+αi+1,−αi+1[h(2), h

(2)(αi)

]

for all h ∈ h. Since αi, αi +αi+1 ∈ ∆1, h(2)(αi)

= [x(1)αi , x

(1)−αi

] and h(2)(αi+αi+1)

=

[x(1)αi+αi+1

, x(1)−αi−αi+1

]. However, we can not expand h(2)(αi+1)

in this way since

αi+1 ∈ ∆0. Still, this yields that

(αi + αi+1)(h)C = nαi+αi+1,−αi[h(2), h

(2)(αi+1)

]− αi(h)A

and hence,

(2αi + αi+1)(h)A = nαi+αi+1,−αi[h(2), h

(2)(αi+1)

]

for all h ∈ h. Let β = αi +αi+1 +αi+2 ∈ ∆1. It can be shown, for our Liesuperalgebras in question, that β(h(αi+1)) = 0 and (2αi +αi+1)(h(β)) 6= 0.Hence, A = 0.

The equality (6) for α ∈ ∆1, m = 1 and n = 3 is shown in the sameway as Lemma 4.4.6. Although, observe that for each β ∈ ∆+

1 , there

exist j ∈ [r] \ τ and γ ∈ ∆+1 such that β = αj + γ. The case when


α ∈ ∆0 and m = n = 2 then follows by splitting x(2)−α into root vectors of

degree 1.Now follows a description of the classical Lie superalgebras A(m,n)

for m 6= n, B(m,n) and C(n + 1) as contragredient Lie superalgebras.We give an account of our investigation of whether the periodisationsare 1-2-presented or not. As mentioned before, if it is 1-2-presented theproof (1), (2) and (3) above is quite extensive. We will omit to present allthe technicalities here and just show some examples of the calculationsinvolved.

The classical Lie superalgebra A(m,n):

The classical Lie superalgebra A(m,n) = sl(m + 1, n + 1) for m 6= n isa (m + n)2 − 1-dimensional contragredient Lie superalgebra g = g(A, τ)of rank m + n− 1, where A is the matrix

2 −1 0 · · · 0 · · · · · · · · · 0

−1 . . . . . . . . . ...

0 . . . . . . 0... . . . . . . . . . −1 . . .

0 0 −1 2 −1 . . . ...... . . . −1 0 1 . . . ...... . . . −1 2 −1 0 0

. . . −1 . . . . . . . . . ...

0 . . . . . . 0... . . . . . . . . . −10 · · · · · · · · · 0 · · · 0 −1 2

and τ = {n}. Let α1, . . . , αm+n−1 be the simple roots of g. Then

∆+0 = {αk, · · ·αl; 1 ≤ k ≤ l < n or n < k ≤ l ≤ m + n− 1}

∆+1 = {αk, · · ·αl; 1 ≤ k ≤ n ≤ l ≤ m + n− 1}.

Furthermore, dim g0 = m2+n2−m−n and dim g1 = 2mn. We have thatA(m,n) ∼= A(n,m) (see [14]) and, hence, we can assume that m < n.

If m is to small, the presentation of gper probably needs relationsof degree 3. For example, let m = 1 and n = 4. By calculating thegenerators for ker ϕ2 and using Liedim, we see that dim (F(M)/R)3 = 48and dim (F(M)/R)4 = 188. Hence, A(1, 4)per is not 1-2-presented and


the presentation needs relations of degree 3. However, A(m,n)per is 1-2-presented for m ≥ 2 and n ≥ 3. To show this it remains to showthat:

[h(2)(α), x

(1)β ] = β(h(α)) x


1 (1)

[x(1)α , x

(2)β ] = nα,β x


1 , β ∈ ∆0 (2)

[h(2)(α), x

(2)β ] = β(h(α)) x


0 , β ∈ ∆+1 (3)

We will outline the proof of (1). First, for each β ∈ ∆ choose δβ ∈ ∆+1

such that β(h(δβ)) 6= 0 and let

hβ =1

β(h(δβ))h(δβ).

Also, note that γ+δ 6∈ ∆ for all γ, δ ∈ ∆+1 . This implies that [h

(2)(γ), x

(1)δ ] =

0 if δ − γ 6∈ ∆ and that δ(h(γ)) 6= 0 if and only if γ − δ ∈ ∆.Take α, β ∈ ∆+

1 such that β − α ∈ ∆. First we will show that if

β − α− δ ∈ ∆ for some δ ∈ ∆+1 , then

[h(2)(α), x

(2)β ] =

β(h(α))

β(h(δ))[h

(2)(δ), x

(2)β ].

We have that:

[h(2)(α), x

(1)β ] = [[x(1)

α , x(1)−α], x

(1)β ] = [x(1)

α , [x(1)−α, x

(1)β ]] + [x

(1)−α, [x(1)

α , x(1)β ]]

= n−α,β[x(1)α , x

(2)β−α] =

n−α,β

nδ,β−α−δ[x(1)

α , [x(1)δ , x

(1)β−α−δ]]

=n−α,β

nδ,β−α−δ

([[x(1)

α , x(1)δ ], x

(1)β−α−δ]− [x

(1)δ , [x(1)

α , x(1)β−α−δ]]

)

=n−α,β nα,β−α−δ

nδ,β−α−δ[x

(1)δ , x

(2)β−δ]]


nδ,β−α−δ nβ,−δ[x

(1)δ , [x

(1)β , x

(1)−δ]]


nδ,β−α−δ nβ,−δ

([[x

(1)δ , x

(1)β ], x

(1)−δ]− [x

(1)β , [x

(1)δ , x

(1)−δ]]

)


nδ,β−α−δ nβ,−δ[h

(2)(δ), x

(1)β ] =

β(h(α))

β(h(δ))[h

(2)(δ), x

(2)β ]

The last equality is just an equality between two constants and is easyto derive by doing the same calculations as above in g without the su-perscript.


The proof of (1) splits into two cases. In the first case, β−α−δβ ∈ ∆.Hence,

[h(2)(α), x

(2)β ] =

β(h(α))

β(h(δβ))[h

(2)(δβ), x

(2)β ] = β(h(α))[h

(2)β , x

(2)β ] = β(h(α))x

(4)β .

In the second case β − α − δβ 6∈ ∆, but there exists δ ∈ ∆+1 such that

β − δ, α− β − δ, β − δ − δβ ∈ ∆. Hence,

[h(2)(α), x

(2)β ] =

β(h(α))

β(h(δ))[h

(2)(δ), x

(2)β ] =

β(h(α))β(h(δ))

β(h(δ))β(h(δβ))[h

(2)(δβ), x

(2)β ]

= β(h(α))[h(2)β , x

(2)β ] = β(h(α))x

(4)β

and we are done.

The Lie superalgebra B(m,n)

The simple contragredient Lie superalgebra B(m,n) = osp(2m+1, 2n) isof rank m+n and has dimension 2(m+n)2 +m+3n. Let g = g(A, τ) =B(m,n). The dimension of g0 is 2m2 + 2n2 + m + n and the dimensionof g1 is 4mn + 2n.

For m = 0,

A =

2 −1 0 · · · · · · 0

−1 . . . . . . . . . ...

0 . . .... . . . . . . . . . ...

. . . . . . −1 0... . . . −1 2 −10 · · · · · · 0 −2 2

and τ = {n}.

Let α1, . . . , αn be the fundamental roots of B(0, n). Then

∆+0 = {αk + · · ·αl, αk + · · ·αl + 2αl+1 + · · · 2αn; 1 ≤ k ≤ l < n} ∪

∪ {2αk + · · ·+ 2αn; 1 ≤ k ≤ n}∆+

1 = {αk + · · ·+ αn; 1 ≤ k ≤ n}

The dimension of ker(ϕ2) is(2n

2

)+2n− 2n2−n = 0. On the other hand,

[xαk+···+αn, [xαn

, xαn]] = 0


for all 1 ≤ k < n. This implies that B(0, n) is not 1-2-presented forn > 1.

For m ≥ 1, A is the following (m + n)× (m + n)-matrix

2 −1 0 · · · 0 · · · · · · · · · 0

−1 . . . . . . . . . ...

0 . . . . . . 0... . . . . . . . . . −1 . . .

0 0 −1 2 −1 . . . ...... . . . −1 0 1 . . . ...... . . . −1 2 −1 0 0

. . . −1 . . . . . . . . . ...

0 . . . . . . −1 0. . . −1 2 −1

0 · · · · · · · · · 0 · · · 0 −2 2

and τ = {n}. Let α1, . . . , αm+n be the fundamental roots of g. Then

∆+0 = {αk + · · ·αl, αk + · · ·αl + 2αl+1 + · · · 2αn+m; 1 ≤ k ≤ l < n}

∪ {αn+i + · · ·αn+j; 1 ≤ i ≤ j ≤ m}∪ {αn+i + · · ·αn+j + 2αn+j+1 + · · ·+ 2αn+m; 1 ≤ i ≤ j < m}

∪ {2αk + · · ·+ 2αn+m; 1 ≤ k ≤ n}∆+

1 = {αk + · · ·αn+i; 1 ≤ k ≤ n, 0 ≤ i ≤ m}∪ {αk + · · ·αn+i + 2αn+i+1 + · · · 2αn+m; 1 ≤ k ≤ n, 0 ≤ i < m}.

For m = 2 and n = 1, B(m,n)per is not 1-2-presented. By calculatingthe generators for ker ϕ2 and using Liedim, we see that

dim (F(M)/R)3 = 30 6= dim g1.

However, B(m, n)per is 1-2-presented for all m,n ≥ 2. To show this, itremains to show that:

[h(2)(α), x

(1)β ] = β(h(α)) x


1 (1)

[x(1)α , x

(2)β ] = nα,β x


1 , β ∈ ∆0 (2)

[h(2)(α), x

(2)β ] = β(h(α)) x


1 , β ∈ ∆+0 (3)

We will give an example of how to show (2). Take α ∈ ∆+1 and β ∈ ∆+

0 .Then there exist δ1, δ2 ∈ ∆1 such that β = δ1 +δ2. Furthermore, we have


that α + δi 6∈ ∆ for i = 1, 2 if α + β 6∈ ∆ and α − δi 6∈ ∆ for i = 1, 2 ifα− β 6∈ ∆. For example, take

α = α1 + · · ·+ αn, β = β1 + · · ·+ αn−1,

δ1 = α1 + · · ·+ αn+1 δ2 = −αn − αn+1.

Clearly α + β 6∈ ∆. Furthermore, β = δ1 + δ2 and α + δi 6∈ ∆ for

i = 1, 2. Hence, by splitting x(2)β and using the Jacobi identity, we see

that [x(1)α , x

(2)β ] = 0 for all α ∈ ∆+

1 and β ∈ ∆0.

The Lie superalgebra C(n + 1):

The classical Lie superalgebra C(n + 1) = osp(2, 2n) is a (2n2 + 5n + 1)-dimensional contragredient Lie superalgebra g = g(A, τ). The Cartanmatrix A is the following (n + 1)× (n + 1)-matrix

0 1 0 · · · · · · 0

−1 2 −1 0...

0 −1 . . . . . . . . . ...

0 0 . . . . . . −1 0... . . . −1 2 −20 · · · · · · 0 −1 2

and τ = {1}. The dimension of g0 is 2n2 + n + 1 and the dimension ofg1 is 4n. Let α1, . . . , αn+1 be the fundamental roots of g. Then:

∆+0 = {αk+1 + · · ·+ αl; 1 ≤ k < l ≤ n + 1}

∪ {αk+1 + · · ·+ αl + 2αl+1 + · · ·+ 2αn + αn+1; 1 ≤ k < l < n}∪ {2αk+1 + · · ·+ 2αn + αn+1; 1 ≤ k < n}

∆+1 = {α1 + · · ·+ αk; 1 ≤ k ≤ n + 1}

∪ {α1 + · · ·+ αk + 2αk+1 + · · ·+ 2αn + αn+1; 1 ≤ k < n}

By calculating the quadratic expressions generating ker ϕ2 and usingLiedim, we see that C(1+1)per is not 1-2-presented and that C(2+1)per

is 1-2-presented. In fact, C(n + 1)per is 1-2-presented for all n ≥ 2. Inview of the discussion on page 89, to show this for n ≥ 3 it suffices to


show that:

[h(2)(α), x

(1)β ] = β(h(α)) x


1 (1)

[x(1)α , x

(2)β ] = nα,β x


1 , β ∈ ∆0 such that

α 6= −β(2)

[h(2)(α), x

(2)β ] = β(h(α)) x


1 , β ∈ ∆+0 (3)

As mentioned before, the proofs of these equalities are quite extensive.Here we outline the proof of (3). The following properties of the rootsystem will be needed:

• For each α ∈ ∆0 there exists β ∈ ∆1 such that α ± β 6∈ ∆. Thisimplies that α(h(β)) = 0.

• For each α, β ∈ ∆+1 , α+β 6∈ ∆. Furthermore, if α 6= β, α−β ∈ ∆.

This implies that α(h(β)) 6= 0 for all α, β ∈ ∆+1 such that α 6= β.

• For each α ∈ ∆0 there exists β ∈ ∆1 such that α − β ∈ ∆1. Thisimplies that nβ,α−β 6= 0.

To show (3), first observe that [h(2)(α), x

(2)β ] = 0 for all α ∈ ∆1 and β ∈ ∆0

such that β ± α 6∈ ∆. Before continuing, we will show that

[x(1)γ , x

(3)δ ] = [x(3)

γ , x(1)δ ]

for all γ, δ ∈ ∆1 such that γ 6= −δ and γ + δ ∈ ∆. Since γ + δ ∈ ∆0,there exists τ ∈ ∆1 such that γ + δ ± τ 6∈ ∆. Furthermore, this impliesthat δ 6= ±τ and δ(h(τ)) 6= 0. Then:

[x(1)γ , x

(3)δ ] =

1

δ(h(τ))[x(1)

γ , [h(2)(τ), x

(1)δ ]]

=1

δ(h(τ))

([[x(1)

γ , h(2)(τ)], x

(1)δ ] + [h

(2)(τ), [x

(1)γ , x

(1)δ ]]

)

=1

δ(h(τ))

(−γ(h(τ))[x

(3)γ , x

(1)δ ] + nγ,δ[h

(2)(τ), x

(2)γ+δ]

)

= −γ(h(τ))

δ(h(τ))[x(3)

γ , x(1)δ ] = [x(3)

γ , x(1)δ ]

The last equality follows since (γ + δ)(h(τ)) = 0 and, hence, γ(h(τ)) =−δ(h(τ)). Take α ∈ ∆+

1 and β ∈ ∆+0 . Then there exists γ ∈ ∆1 such that


β − γ ∈ ∆1 and

[h(2)(α), x

(2)β ] =

1

nγ,β−γ[h

(2)(α), [x

(1)γ , x

(1)β−γ]]

=1

nγ,β−γ

([[h

(2)(α), x

(1)γ ], x

(1)β−γ] + [x(1)

γ , [h(2)(α), x

(1)β−γ]]

)

=1

nγ,β−γ

(γ(h(α))[x

(3)γ , x

(1)β−γ] + (β − γ)(h(α))[x

(1)γ , x

(3)β−γ]

)

=β(h(α))

nγ,β−γ[x(1)

γ , x(3)β−γ].

Hence, for β ∈ ∆0, we have that

β(h(α′))[h(2)(α), x

(2)β ] = β(h(α))[h

(2)(α′), x

(2)β ]

for all α, α′ ∈ ∆1. Now, we may define hβ as

hβ =1

β(h(α′))h(α′)

for some α′ ∈ ∆1 such that β(h(α′)) 6= 0. Then

[h(2)(α), x

(2)β ] =

β(h(α))

β(h(α′))[h

(2)(α′), x

(2)β ] = β(h(α))[h

(2)β , x

(2)β ] = β(h(α))x

(4)β

for all α ∈ ∆1 and we are done.This ends the investigation of the classical contragredient Lie su-

peralgebras A(m,n) for m 6= n, B(m,n), C(n + 1), G(3), F (4) andD(2, 1; α).

Additional Questions

During the work on this thesis, some new questions that we found inter-esting and worth mentioning have arisen. The object here is to give abrief account of these questions.

• For each consistent Z+-graded periodisation Lper of a Lie superal-gebra L, such that Lper is 1-2-presented, there exists a commutativering R = U(Lper)

!. Is it possible to describe this class of rings inanother way?

• Through this thesis we have worked over a field of characteristic 0.Is it possible to get some similar results in positive characteristic?

• Let g be a Lie superalgebra. In this work we have consideredtwo Z+-graded periodisations of g. Other ”twisted” periodisationsmight be of interest (c.f. [18] and [24]).

• Let g be a Lie (super)algebra over a field k. In terms of Lie(super)algebra homology, we have studied the homology groupsH1(gper, k) and H2(gper, k) for the different periodisations of g. Apossible continuation of this work is to study the higher order ho-mology groups. Furthermore, we have studied the homology groupsas vector spaces. They might be studied as g-modules as well (c.f.Remark 1.3.7).

• The investigation in this thesis includes the classical contragredi-ent Lie superalgebras. What can be said about the strange seriesP (n) and Q(n) and what about the Cartan type Lie superalgebrasW (n), S(n), S(n) and H(n)? In his degree project [30], MagnusThedsen showed that H(5)per = H(5) ⊕ H(5) ⊕ H(5) · · · is al-most 1-2-presented in the sense that H(5)per is the quotient of a1-2-presented Lie superalgebra by central elements.

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periodisations of contragredient lie superalgebras and their

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