Leibniz Algebras: Structure and Classification [1 ed.] 0367354810, 9780367354817

Table of contents :
Contents
Preface
Acknowledgments
Authors
Notations And Conventions
1 Introduction
1.1 Algebras
1.2 Associative Algebras
1.3 Lie Algebras
1.4 Loday Algebras
2 Structure Of Leibniz Algebras
2.1 Some Properties Of Leibniz Algebras
2.2 Nilpotent And Solvable Leibniz Algebras
2.3 On Levi’s Theorem For Leibniz Algebras
2.4 Semisimple Leibniz Algebras
2.5 On Cartan Subalgebras Of Finite-dimensional Leibniz Algebras
2.6 Some Properties Of Weight Spaces Of Leibniz Algebras And Cartan’s Criterion Of Solvability
3 Classification Problem In Low Dimensions
3.1 Algebraic Classification Of Low-dimensional Leibniz Algebras
3.2 Application
3.3 Low-dimensional Nilpotent Leibniz Algebras
3.4 Four-dimensional Solvable Leibniz Algebras
3.5 Rigidity Of Lie And Leibniz Algebras
3.6 Leibniz Cohomology Computations
3.7 Rigid Leibniz Algebra With Non-trivial
3.8 Lie-rigidity Versus Leibniz-rigidity
4 On Some Classes Of Leibniz Algebras
4.1 Irreducible Components Of A Subclass Of Nilpotent Leibniz Algebras
4.2 Classification Of Naturally Graded Complex Filiform Leibniz Algebras
4.3 Classification Of Some Solvable Leibniz Algebras
5 Isomorphism Criteria For Filiform Leibniz Algebras
5.1 On Base Changes In Complex Filiform Leibniz Algebras
5.2 Criteria Of Isomorphisms Of Complex Filiform Non-lie Leibniz Algebras
6 Classification Of Filiform Leibniz Algebras In Low Dimensions
6.1 Isomorphism Criteria For First Class
6.2 Classification Of First Class In Low Dimensions
6.3 Isomorphism Criteria For Second Class
6.4 Classification Of Second Class In Low Dimensions
6.5 Simplifications And Notations In Third Class
6.6 Classification In Dimension Five
6.7 Classification In Dimension Six
Appendix A  Linear Algebra
A.1 Vector Spaces And Subspaces
A.2 Linear Transformations
A.3 Dual Vector Space
A.4 Tensor Products
A.5 Tensor Algebra
A.6 Matrix Of A Linear Transformation
A.7 Jordan Normal Form
Appendix B  Elements of Representation Theory
Appendix C  Zariski Topology
C.1 Action Of A Group
C.2 Algebraic Groups
Bibliography
Index

Citation preview

Leibniz Algebras Structure and Classification

Leibniz Algebras Structure and Classification

Shavkat Ayupov Institute of Mathematics Uzbekistan Academy of Sciences

Bakhrom Omirov National University of Uzbekistan

Isamiddin Rakhimov University Technology MARA (UiTM)

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742

c 2020 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper International Standard Book Number-13: 978-0-367-35481-7 (Hardback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Control Number: 2019953041 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Dedicated to the memory of Professor J.-L. Loday (1946–2012)

Contents

Preface

xiii

Acknowledgments

xv

Authors

xvii

Notations and Conventions

xix

Chapter

1  INTRODUCTION

1

1.1

ALGEBRAS

1.2

ASSOCIATIVE ALGEBRAS

10

1.3

LIE ALGEBRAS

12

1.3.1

Simple Lie algebras

14

1.3.2

Solvable and nilpotent Lie algebras

17

1.3.3

Semisimple Lie algebras

20

1.3.4

More on finite-dimensional Lie algebras

25

1.4

1

LODAY ALGEBRAS

26

1.4.1

Leibniz algebras

27

1.4.2

Associative dialgebras

34

1.4.3

Universal enveloping algebra (Poincar´e-BirkhoffWitt Theorem)

36

1.4.4

Zinbiel algebras

37

1.4.5

Dendriform algebras: Koszul dual of associative dialgebras

38

Loday diagram and Koszul duality

40

1.4.6

vii

viii  Contents

Chapter

2  STRUCTURE OF LEIBNIZ ALGEBRAS

41

2.1

SOME PROPERTIES OF LEIBNIZ ALGEBRAS

41

2.2

NILPOTENT AND SOLVABLE LEIBNIZ ALGEBRAS

42

2.3

ON LEVI’S THEOREM FOR LEIBNIZ ALGEBRAS

46

2.3.1 2.3.2

On conjugacy of Levi subalgebras of Leibniz algebras

48

On conjugacy of Levi subalgebras of complex Leibniz algebras

55

2.4

SEMISIMPLE LEIBNIZ ALGEBRAS

60

2.5

ON CARTAN SUBALGEBRAS OF FINITE-DIMENSIONAL LEIBNIZ ALGEBRAS

67

SOME PROPERTIES OF WEIGHT SPACES OF LEIBNIZ ALGEBRAS AND CARTAN’S CRITERION OF SOLVABILITY

74

3  CLASSIFICATION PROBLEM IN LOW DIMENSIONS

79

2.6

Chapter

3.1

ALGEBRAIC CLASSIFICATION OF LOW-DIMENSIONAL LEIBNIZ ALGEBRAS

79

3.1.1

Classification of Lie algebras

79

3.1.2

Low-dimensional complex Leibniz algebras

80

3.1.2.1

Two-dimensional Leibniz algebras

80

3.1.2.2

Three-dimensional Leibniz algebras

81

3.2

APPLICATION

88

3.3

LOW-DIMENSIONAL NILPOTENT LEIBNIZ ALGEBRAS

91

3.3.1 3.4

Four-dimensional nilpotent complex Leibniz algebras

94

FOUR-DIMENSIONAL SOLVABLE LEIBNIZ ALGEBRAS 104

3.4.1 3.4.2

Four-dimensional solvable Leibniz algebras with two-dimensional nilradical

108

Four-dimensional solvable Leibniz algebras with three-dimensional nilradical

110

Contents  ix

3.5

RIGIDITY OF LIE AND LEIBNIZ ALGEBRAS

124

3.6

LEIBNIZ COHOMOLOGY COMPUTATIONS

129

3.6.1

Leibniz cohomology

129

3.6.2

Leibniz cohomology of rigid Lie algebras

130

3.6.3

Non-triviality of Leibniz cohomology

3.7

RIGID LEIBNIZ ALGEBRA WITH NON-TRIVIAL HL

3.8

LIE-RIGIDITY VERSUS LEIBNIZ-RIGIDITY

3.8.1

3.8.2

3.8.3

Chapter

4.1 4.2 4.3

132 2

134 138

Necessary criteria for rigidity of Leibniz algebras

147

3.8.1.1

Invariance Argument 1

147

3.8.1.2

Invariance Argument 2

148

3.8.1.3

Invariance Argument 3

148

3.8.1.4

Invariance Argument 4

149

3.8.1.5

Invariance Argument 5

149

Applications of invariance arguments to varieties of low-dimensional Leibniz algebras

150

3.8.2.1

150

Two-dimensional Leibniz algebras

Variety of three-dimensional nilpotent Leibniz algebras

150

3.8.4

Three-dimensional rigid Leibniz algebras

152

3.8.5

Rigid nilpotent Leibniz algebras in dimension four 153

4  ON SOME CLASSES OF LEIBNIZ ALGEBRAS

159

IRREDUCIBLE COMPONENTS OF A SUBCLASS OF NILPOTENT LEIBNIZ ALGEBRAS

159

CLASSIFICATION OF NATURALLY GRADED COMPLEX FILIFORM LEIBNIZ ALGEBRAS

164

CLASSIFICATION OF SOME SOLVABLE LEIBNIZ ALGEBRAS

176

x  Contents

Chapter

5.1 5.2

5  ISOMORPHISM CRITERIA FOR FILIFORM LEIBNIZ ALGEBRAS ON BASE CHANGES IN COMPLEX FILIFORM LEIBNIZ ALGEBRAS

195

CRITERIA OF ISOMORPHISMS OF COMPLEX FILIFORM NON-LIE LEIBNIZ ALGEBRAS

200

5.2.1

Isomorphism criteria

210

5.2.2

Classification procedure

212

5.2.2.1 5.2.2.2 5.2.2.3

Chapter

195

Classification algorithm and invariants for FLbn+1

213

Classification algorithm and invariants for SLbn+1

216

Classification algorithm and invariants for TLbn+1

219

6  CLASSIFICATION OF FILIFORM LEIBNIZ ALGEBRAS IN LOW DIMENSIONS

223

6.1

ISOMORPHISM CRITERIA FOR FIRST CLASS

223

6.2

CLASSIFICATION OF FIRST CLASS IN LOW DIMENSIONS

229

6.2.1

Classification in dimension five

229

6.2.2

Classification in dimension six

231

6.3

ISOMORPHISM CRITERIA FOR SECOND CLASS

234

6.4

CLASSIFICATION OF SECOND CLASS IN LOW DIMENSIONS

234

6.4.1

Classification in dimension five

235

6.4.2

Classification in dimension six

237

6.5

SIMPLIFICATIONS AND NOTATIONS IN THIRD CLASS

239

6.6

CLASSIFICATION IN DIMENSION FIVE

240

6.6.1

242

Isomorphism classes in TLb5

Contents  xi

6.7

Appendix

CLASSIFICATION IN DIMENSION SIX

245

6.7.1

246

Isomorphism classes in TLb6

A  Linear Algebra

251

A.1

VECTOR SPACES AND SUBSPACES

251

A.2

LINEAR TRANSFORMATIONS

254

A.3

DUAL VECTOR SPACE

254

A.4

TENSOR PRODUCTS

255

A.5

TENSOR ALGEBRA

256

A.6

MATRIX OF A LINEAR TRANSFORMATION

257

A.7

JORDAN NORMAL FORM

261

Appendix

B  Elements of Representation Theory

Appendix C  Zariski Topology

269

273

C.1

ACTION OF A GROUP

279

C.2

ALGEBRAIC GROUPS

279

Bibliography

283

Index

301

Preface The book is designed to introduce the reader to the theory of Leibniz algebras. Knowledge of a standard course of linear algebra is presupposed, as well as some acquaintance with the methods of Lie algebras is required. Due to this we omitted the proofs of the facts from the theory of Lie algebras. Leibniz algebra is a generalization of the Lie algebras and these algebras preserve a unique property of Lie algebras that the right multiplication operators are derivations. They were first called D-algebras in papers by A.M. Blokh published in the 1960s to indicate their close relations with derivations. The theory of D-algebras did not receive great attention immediately after their introduction. Later the same algebras were introduced by J.-L. Loday in 1993, who called them Leibniz algebras due to the identity they satisfy. The main motivation to the introduction of Leibniz algebras was to study periodicity phenomena in algebraic K-theory. Nowadays the theory of Leibniz algebras is one of actively developing areas of modern algebra. Along with cohomological, structural and classification results on Leibniz algebras some papers with their various applications also appear. But the main goal of the book is specific. We focus mainly on description problems of Leibniz algebras. Particularly, in the book we propose a method of classification of a subclass of Leibniz algebras based on algebraic invariants. The method is applicable in the Lie algebras case as well. Our motivation to write this book came from a 2001 book titled Dialgebras and Related Operands by Jean-Louis Loday et al. (volume 1763 in Springer’s Lecture Notes in Mathematics Series). Many readers are unaware of recent results obtained in the theory of Leibniz algebras. We hope that the review and citations given in Chapter 1 will provide such awareness. Nevertheless, we had to omit many topics which are out of the description problem of algebras, i.e., (co)homology, analogues of some results from Lie algebras, classification of graded Leibniz algebras (p-filiform, quasi-filiform, algebras of maximum length, etc.) and over non-algebraically closed fields, Leibniz algebras in prime characteristic, descriptions of the derivations and automorphisms of algebras, recent results on geometric classifications and

xiii

xiv  Preface

much more. We also hope the reader will be stimulated to pursue these topics in the references listed in the Bibliography at the end of this book. We start Chapter 2 with some basic properties of Leibniz algebras followed by introducing the notion of semisimplicity of Leibniz algebras and presenting the structure of semisimple Leibniz algebras. Here we discuss the properties of Lie algebras that can be extended to the Leibniz algebras case and those which fail to extend, and we provide examples for each case. Chapter 3 deals with the problem of algebraic and geometric classification of low-dimensional complex Leibniz algebras and their subclasses. Chapter 4 of the book is devoted to the classification problem of some classes of complex Leibniz algebras in arbitrary finite dimension. Starting with Chapter 5, we focus on a subclass of nilpotent Leibniz algebras called filiform Leibniz algebras. We propose a classification technique based on algebraic invariants under action of base change (“transport of structure”). In Chapter 6 we apply the classification method proposed to low-dimensional cases. In the appendices, we provide summaries of basic linear algebra, representation theory and the Zariski topology.

Acknowledgments Several people have played important roles at various stages of the work that culminated in this book. They are S. Albeverio, U.D. Bekbaev, L.M. Camacho, J.M. Casas, J.R. G´omez, A.Kh. Khudoyberdiyev, K.K. Kudaybergenov, M. Ladra, K.A. Mohd Atan, Sh.K. Said Husain, I. Rikhsiboev, and others. The authors are grateful to them for fruitful discussions and collaborations. Some of the facts given in Section 2.1 are results of private discussions of one of the authors (B.O.) with V. Gorbatsevich to whom we are grateful. We also are indebted to Mrs. Elena E. Rakhimova for typing of an earlier version of the book. The authors’ special thanks go to Professor M. Ladra for the expert design, reading and in bringing the manuscript to its final form. Should errors appear in this book, the authors accept complete responsibility for their occurrence, and they will be rectified subsequently once they are brought to the authors’ attention. Any feedback from the readers in this context will be gratefully acknowledged.

Shavkat Ayupov Bakhrom Omirov Isamiddin Rakhimov

xv

Authors Shavkat Ayupov is the Director of the Institute of Mathematics at Uzbekistan Academy of Sciences, Professor at the National University of Uzbekistan, and Doctor of Sciences in Physics and Mathematics and Fellow of TWAS (The World Academy of Sciences). His research interests include functional analysis, algebra and topology. Professor Ayupov is the recipient of several international titles and awards, and his main research deals with the study of operator algebras, Jordan and Lie structures on von Neumann algebras, derivations and automorphisms on operator algebras, structure theory of Leibniz algebras and superalgebras and other non-associative algebras. Professor Ayupov is an organizer of CIMPA research school workshops, international conferences on nonassociative algebras and applications and on operator algebras and quantum probability. He has spoken at numerous plenary sessions and has been invited to numerous talks in various international conferences and workshops. Professor Ayupov is also the chief editor of the Uzbek Mathematical Journal and has authored 4 textbooks, 5 monographs, and more than 150 research papers which have been published in several international journals. Bakhrom Omirov is Professor at the National University of Uzbekistan, Doctor of Sciences in Physics and Mathematics, and Research Fellow at the Institute of Mathematics of the Uzbekistan Academy of Sciences. His research interests include non-associative algebras, Lie (super)algebras, Leibniz (super)algebras, n-Leibniz algebras, structure theory of algebras, p-adic analysis, evolution algebras and their applications. Professor Omirov has received several local and international awards. He is currently leading several international research projects and collaborations and has been invited as a speaker to many workshops and universities abroad. He has also authored more than 100 research papers in high impact international journals.

xvii

xviii  Authors

Isamiddin Rakhimov is Professor at the University Technology MARA (UiTM), Malaysia, Doctor of Sciences in Physics and Mathematics, and Research Fellow at the Institute for Mathematical Research (INSPEM), Universiti Putra Malaysia. His research interests focus on the theory of finitedimensional algebras and its applications. He has been invited to speak at a number of international conferences and workshops. He obtained a PhD in algebra from the Sankt Petersburg University of Russia. He has organized several international mathematical events and is on the editorial boards of a few international journals. He has also authored 1 textbook, 2 monographs and more than 70 scientific papers published in international cited journals.

Notations and Conventions • N, Z, Q, R, C - natural numbers, integers, the fields of rational, real and complex numbers, respectively. • LBn (F) - class of Leibniz algebras over F in dimension n. • LNn (F) - class of nilpotent Leibniz algebras over F in dimension n. • Lbn (F) - class of filiform Leibniz algebras over F in dimension n. • FLbn , SLbn and TLbn - subclasses of Lbn . • NGFi , i = 1, 2, 3 - classes of naturally graded filiform Leibniz algebras. • Cent(L) - center of the Leibniz algebra L. • Annl (L) - left annihilator of L. • Annr (L) - right annihilator of L. • Rad(L) - radical of L. • I - liezator of a Liebniz algebra L. • Aut(L) - group automorphisms of L. • Der(L) - algebra of derivations of L. • Norl (H) and Norl (H) - left and right normalizer of a subalgebra H in a Leibniz algebra L, respectively. • K(x, y) - Killing form. • Orb(µ) - orbit of an algebra µ. xix

xx  Notations and Conventions

• Z n (L, M) - the space of n-cocyles. • Bn (L, M) - the space of n-coboundaries. • H n (L, M) - nth -(co)homology group of an algebra L with coefficients in a L-module M. • ZLn (L, M), BLn (L, M) and HLn (L, M) - the space of n-cocyles, the space of n-coboundaries and nth -(co)homology group of a Leibniz algebra L with coefficients in a L-module M, respectively. • Ck =

(2k)! (k+1)!k! ,

k = 0, 1, 2, . . . . - Catalan numbers.

CHAPTER

1

INTRODUCTION

1.1

ALGEBRAS

An algebra A is a vector space V over a field F equipped by a bilinear binary operation λ : V × V −→ V on it. The vector space V is called the underlying vector space of A. In the book V is assumed to be finitedimensional over the field of complex numbers, although many of the statements can be proved over the fields of the characteristic different from two. Example 1.1. • Any field is an algebra over itself and over its subfield. Particularly, the number fields Q, R, C are examples of algebras. • The set of polynomials F[x1 , x2 , . . . , xn ] at variables x1 , x2 , . . . , xn with coefficients from a field F is an algebra over F. • The free algebra A = F < x1 , x2 , . . . , xn > at non-commuting variables x1 , x2 , . . . , xn with coefficients from a field F. A basis of this algebra consists of words in letters x1 , x2 , . . . , xn and multiplication in this basis is simply concatenation of words. • The set of n × n matrices with entries in a field F is an algebra over F. • The set of endomorphisms EndF (V) of a vector space V over a field F is an algebra over F.

1

2  Leibniz Algebras

• Let us consider α v u β

( O(F) =

) ! 3 α, β ∈ F and u, v ∈ F .

Define the binary operation λ on O(F) as follows: ! !! ! α v γ z αγ + (u, v) αz + δu − v × w λ , = , u β w δ γv + βw − u × z βδ + (v, z) where (x1 , x2 ) and x1 × x2 are the inner and cross products of the vectors x1 , x2 ∈ F, respectively. Then (O(F), λ) is an algebra over F called Cayley-Dickson matrix algebra.

γikj ,

If {e1 , e2 , . . . , en } is a basis of the vector space V then the coefficients where i, j, k = 1, 2, . . . , n of the linear combinations λ(ei , e j ) =

n X

γikj ek for i, j = 1, 2, . . . , n

k=1

are said to be the structure constants of the algebra A on the basis {e1 , e2 , . . . , en }. Therefore, if we fix a basis of the underlying vector space V over a field F then all possible algebra structures over V can be 3 identified by points {γikj } of n3 -dimensional affine space Fn . Note that the structure constants completely determine the product of any two elements x = α1 e1 + α2 e2 + · · · + αn en and y = β1 e1 + β2 e2 + · · · + βn en as follows: n n X X λ(x, y) = αi β j λ(ei , e j ) = αi β j γikj ek . (1.1) i, j=1

i, j,k=1

Thus the structure of an algebra can be completely determined by providing its structure constants. Such a definition is called the determination by the structure constants. The product λ(x, y) of elements x and y of an algebra is denoted by x · y, x ∗ y, x ? y, [x, y] or just xy etc., in each the cases above it is clear from the content of discussion what the specific notation means. Let (A1 , λ1 ) and (A2 , λ2 ) be two algebras with binary operations λ1 and λ2 , respectively. A function f : A1 −→ A2 is a homomorphism if f (λ1 (x, y)) = λ2 ( f (x), f (y)) for all x, y ∈ A1 .

Introduction  3

A bijective homomorphism is called isomorphism, this relation is denoted by “”. An automorphism of an algebra A is an isomorphism onto itself. The set of all automorphisms of an algebra A form a group with respect to the composition, the group is denoted by AutA. A linear transformation ı : A −→ A is called involution if ı(ı(x)) = x and ı(xy) = ı(x)ı(y) for all x, y ∈ A. Example 1.2. The algebra (O(F), λ) given in Example 1.1 has an involution defined as follows:  ! !  α v  α −v   ı  . = u β  −u β Definition 1.1. A subspace S of an algebra (A, λ) is said to be a subalgebra, if λ(S , S ) ⊂ S , i.e., λ(x, y) ∈ S whenever x, y ∈ S . Example 1.3. Let f : A1 −→ A2 be a homomorphism of algebras. Then the image Im f = {y ∈ A2 | ∃x ∈ A1 : y = f (x)} is a subalgebra of A2 . Definition 1.2. A subspace I of an algebra (A, λ) is called a left (respectively right) ideal of A if λ(A, I) ⊂ I (respectively λ(I, A) ⊂ I). Definition 1.3. A subspace I of an algebra (A, λ) is said to be a twosided ideal (or just ideal) of A if it is both left and right, i.e., λ(I, A) ⊂ I and λ(A, I) ⊂ I. Examples. 1. Any algebra has two trivial ideals, the subspace consisting only of the zero vector and the algebra itself. 2. A subset of (A, λ) defined by K(A) = {x ∈ A| λ(x, y) = λ(y, x) = 0 for all y ∈ A} is called the annihilator of A. It is an ideal of A.

4  Leibniz Algebras

3. The kernel

Ker( f ) = {x ∈ A1 | f (x) = 0}

of a homomorphism f : A1 −→ A2 is an ideal of A1 . 4. Let I1 and I2 be ideals of A and I1 + I2 = {x ∈ A| there exist xi ∈ Ii , i = 1, 2 such that x = x1 + x2 }. Then a. I1 + I2 is an ideal of A; b. I1 and I2 are ideals of I1 + I2 ; c. I1 ∩ I2 is an ideal of I1 and I2 . 5. Let f : A1 −→ A2 be a homomorphism of algebras. a. If J is an ideal of A2 then f −1 (J) is an ideal of A1 ; b. If f is an epimorphism and I is an ideal of A1 then f (I) is an ideal of A2 . Let A be an algebra and let I be its ideal. The quotient vector space A/I is an algebra with respect to the multiplication (product) λ : A/I × A/I −→ A/I defined by λ(x + I, y + I) = λ(x, y) + I. The operation λ is well-defined since if x+ I = x0 + I and y+ I = y0 + I then λ(x+ I, y+ I) = λ(x0 + I, y0 + I). Let I be an ideal of an algebra A. The canonical map π : A −→ A/I defined by x 7−→ x + I is a homomorphism from A onto A/I. The kernel Ker(π) = {x ∈ A| π(x) = 0} of the homomorphism π equals I. Theorem 1.1. Let A be an algebra and I be its ideal. Then there is one to one order-preserving correspondence between the ideals J of A which contain I and the ideals J of A/I given by J = π−1 J. Theorem 1.2. Let f : A1 −→ A2 be a homomorphism of algebras.

Introduction  5

a)

i) For any ideal J of A1 contained in Ker( f ), there exists an unique homomorphism ψ : A1 /J −→ A2 such that the following diagram f

A1 @ π@ R @

- A 2  ψ

A1 /J

commutes, where π is the canonical homomorphism. ii) If J = Ker f then ψ produces an isomorphism between A1 /Ker f and Im f . b) If I1 and I2 are ideals of an algebra A then (I1 + I2 )/I2 is isomorphic to I1 /I1 ∩ I2 . c) Suppose that I1 and I2 are ideals of an algebra A such that I1 ⊂ I2 . Then I2 /I1 is an ideal of A/I1 and (A/I1 )/(I2 /I1 )  (A/I2 ). Definition 1.4. Let (A1 , λ1 ) and (A2 , λ2 ) be two algebras over a field F; then an algebra with binary product λ, defined on vector space A = A1 ⊕ A2 = {(x1 , x2 )| x1 ∈ A1 and x2 ∈ A2 } as follows λ((x1 , x2 ), (y1 , y2 )) = (λ1 (x1 , y1 ), λ2 (x2 , y2 )) is said to be the direct sum of the algebras A1 and A2 . Note that A1 and A2 are ideals of A. The definition above can be extended to the case of the direct sum A = A1 ⊕ A2 ⊕ · · · ⊕ An of n algebras (A1 , λ1 ), (A2 , λ2 ), . . . (An , λn ) defining the product as follows λ((x1 , x2 , . . . , xn ), (y1 , y2 , . . . , yn )) = (λ1 (x1 , y1 ), λ2 (x2 , y2 ), . . . , λn (xn , yn )). Definition 1.5. A linear transformation d : A −→ A is said to be a derivation of an algebra (A, λ) if d(λ(x, y)) = λ(d(x), y) + λ(x, d(y)) whenever x, y ∈ A.

6  Leibniz Algebras

Proposition 1.1. The set of all derivations of an algebra A denoted by Der(A) is a Lie algebra with respect to the bracket operation [d1 , d2 ] = d1 ◦ d2 − d2 ◦ d1 . Proof. It is obvious that [d1 , d2 ] is a linear transformation, i.e., [d1 , d2 ](αx+βy) = α[d1 , d2 ](x)+β[d1 , d2 ](y), for all x, y ∈ A and α, β ∈ F. Let us show that [d1 , d2 ] is a derivation of A. Indeed, [d1 , d2 ](λ(x, y)) = (d1 ◦ d2 − d2 ◦ d1 )(λ(x, y)) = (d1 ◦ d2 )(λ(x, y)) − (d2 ◦ d1 (λ(x, y)) = d1 (λ(d2 (x), y) + λ(x, d2 (y)) −d2 (λ(d1 (x), y) + λ(x, d1 (y)) = λ((d1 ◦ d2 )(x), y) + λ(d 2 (x), d1 (y)) ::::::::::::: +λ(d1 (x), d2 (y)) + λ(x, (d1 ◦ d2 )(y)) −λ((d2 ◦ d1 )(x), y) − λ(d1 (x), d2 (y)) −λ(d 2 (x), d1 (y)) − λ(x, (d2 ◦ d1 )(y)) ::::::::::::: = λ((d1 ◦ d2 − d2 ◦ d1 )(x), y) +λ(x, (d1 ◦ d2 − d2 ◦ d1 )(y)) = λ([d1 , d2 ](x), y) + λ(x, [d1 , d2 ](y)).



Definition 1.6. It is called that an algebra (A, λ) is represented as a semidirect sum if there are subspaces A1 and A2 such that A = A1 ⊕ A2 (the direct sum of vector spaces), λ(A1 , A1 ) ⊂ A1 , λ(A1 , A2 ) ⊂ A1 , λ(A2 , A1 ) ⊂ A1 , and λ(A2 , A2 ) ⊂ A2 . Further the bilinear function λ(x, y) we denote just by xy. The associator in an algebra is the trilinear function (x, y, z) = (xy)z − x(yz). The Jacobian in an algebra is defined by J(x, y, z) = (xy)z + (yz)x + (zx)y. The left (right) Leibniz identity is defined by x(yz) = (xy)z + y(xz) ((xy)z = (xz)y + x(yz)) .

Introduction  7

Definition 1.7. • An algebra A – is said to be unital if there exists an element e ∈ A such that xe = ex = x for all x ∈ A. – is said to be a division algebra if any nonzero element of A is invertible, i.e., for any x , 0 ∈ A there exists y ∈ A such that xy = yx = e. – is commutative if xy = yx for all x, y ∈ A. – is anticommutative if x2 = 0 for all x ∈ A. – is associative if (x, y, z) = 0 for all x, y, z ∈ A. • A Jordan algebra is a commutative algebra satisfying the Jordan identity (x2 , y, x) = 0 for all x, y ∈ A. • An algebra A is left (right) alternative if (y, x, x) = 0 ((x, x, y) = 0) for all x, y ∈ A. • A Lie algebra is an anticommutative algebra A satisfying the Jacobi identity J(x, y, z) = 0 for all x, y, z ∈ A. • A Malcev algebra is an anticommutative algebra A satisfying the identity J(x, y, xz) = J(x, y, z)x for all x, y, z ∈ A. • A left (right) Leibniz algebra is an algebra A satisfying the left (right) Leibniz identity x(yz) = (xy)z + y(xz) ((xy)z = (xz)y + x(yz)) for all x, y, z ∈ A. FACTS: 1. Every associative algebra is alternative.

8  Leibniz Algebras

2. Every Lie algebra is a Malcev algebra. 3. Every Lie algebra is a Leibniz algebra. 4. A Leibniz algebra A is Lie if x2 = 0 for all x ∈ A. Therefore, the intersection of classes of Malcev and Leibniz algebras is the class of Lie algebras. 5. The commutator in an algebra A is the bilinear function [x, y] = xy − yx. The minus algebra A− of an algebra A is the algebra with the same underlying vector space as A but with [x, y] as the multiplication. Theorem 1.3. (Poincar´e-Birkhoff-Witt Theorem for Lie algebras) Every Lie algebra is isomorphic to a subalgebra of A− for some associative algebra A. 6. If A is an alternative algebra then A− is a Malcev algebra. 7. The plus algebra A+ of an algebra A is the algebra with the same underlying vector space as A but with x ◦ y = 21 (xy + yx) as the multiplication. The algebra A+ is always commutative. If A is alternative then A+ is a Jordan algebra. A Jordan algebra is called special if it is isomorphic to a subalgebra of A+ for some associative algebra A; otherwise it is called exceptional. The analogue of the PBW theorem for Jordan algebras is false: not every Jordan algebra is special. Example 1.4. Let A be an algebra with an involution ı. The set of all symmetric elements H(A, ı) = {a ∈ A | ı(a) = a} is closed under the Jordan product ◦, therefore, it is a Jordan algebra. • If A is an associative algebra then H(A, ı) is a special Jordan algebra.

Introduction  9

• If A = O3 is an algebra of 3 × 3 hermitian matrices, i.e., ı : O3 −→ O3 defined by ı(ai j ) = ai j , then H(O3 ) is an example of exceptional simple Jordan algebra. A Jordan algebra J is said to be an Albert algebra if by extending the field of scalars F of J to K one gets an isomorphic toN H(O3 ) algebra, i.e., there exists an extension K of F such that J K  H(O3 ). F

Definition 1.8. • An algebra A is simple if AA , {0} and A has no ideals apart from {0} and A. • An algebra is semisimple if it is the direct sum of simple algebras. Set A1 = A(1) = A and by induction we define X An+1 = Ai A j and A(n+1) = A(n) A(n) for n ≥ 1. i+ j=n+1

Definition 1.9. • An algebra A is called nilpotent if there exists k such that Ak = {0}. • An algebra A is called solvable if there exists s such that A(s) = {0}. The smallest natural number k (respectively, s) with the properties above is called the nilpotency index (respectively solvability index) of A. Remarks. • If algebra A is simple then AA = A. • An algebra A is nilpotent of index k if and only if any product of k elements (with any arrangement of parentheses) equals zero, and if there exists a nonzero product of k − 1 elements. • Every nilpotent algebra is solvable but the converse is not true in general. • Every solvable associative algebra is nilpotent. Now we briefly recall some classical results concerning the theory of associative and Lie algebras. Then we review a few classes of algebras, which are closely related to the algebras considered in this book.

10  Leibniz Algebras

1.2

ASSOCIATIVE ALGEBRAS

An algebra A is said to be associative if its product “ · ” has the associativity property (a · b) · c = a · (b · c) for all a, b, c ∈ A.

(1.2)

Lemma 1.1. An algebra A is associative if and only if (ei · e j ) · ek = ei · (e j · ek ), where i, j, k = 1, 2, . . . , n

(1.3)

for any basis {e1 , e2 , . . . , en } of A. Proof. This is seen immediately combining (1.2) and (1.1).



Note that the condition (1.3) gives polynomial identities for the structure constants {γikj } i, j, k = 1, 2, . . . , n as follows n X r=1

s γirj γrk =

n X

γrjk γirs for 1 ≤ i, j, k, s ≤ n.

r=1

Therefore, the points corresponding to the associative algebra struc3 3 tures in Fn form a Zariski closed subset of Fn . Example 1.5. 1. The field of complex numbers C is a two-dimensional associative algebra over R, but it is an infinite-dimensional algebra over Q. 2. The space of square n × n matrices with the entries from a field F is an associative algebra of dimension n2 . 3. The space of endomorphisms EndF (V) of an n-dimensional vector space V over a field F also is an n2 -dimensional associative algebra. 4. The space of polynomials with the coefficients from a field F is an infinite-dimensional associative algebra. The theory of finite-dimensional associative algebras is one of the ancient areas of the modern algebra. It originates primarily from works of Hamilton, who discovered the famous quaternions, and Cayley, who developed the theory of matrices. Later the structural theory of finite

Introduction  11

dimensional associative algebras have been treated by a number of mathematicians, notably B. Pierce, C.S. Pierce, Clifford, Weierstrass, Dedekind, Jordan, Frobenius. At the end of the 19th century, T. Molien and E. Cartan described semisimple algebras over the fields of complex and real numbers, and they made first attempts in the study of non semisimple algebras. A new era in the development of the theory of finite-dimensional associative algebras began due to works of Wedderburn, who obtained the fundamental results of this theory: description of the structure of semisimple algebras over a field, a theorem on the lifting of the quotient by the radical, the theorem on the commutativity of finite division rings, and others. Thanks to German school of algebraists, headed by Emmy Noether, E. Artin, R. Brouwer, the theory of semisimple algebras has found its modern form, and most of the results have been extended to some types of rings. Further development of the theory of associative algebras occurred in the 1980s when many open problems, remaining unsolved since the 1930s were solved. The classification problem of associative algebras can be reduced to the classification problem of semisimple and nilpotent associative algebras. Recall that an associative algebra (without unit) A is said to be nilpotent if An = 0 for some n (i.e., a1 ·a2 ·· · ··an = 0 for any ai ∈ A). The representatives of low-dimensional nilpotent associative algebras over a field F are given by their structure constants on a basis {e1 , e2 , . . . , en } as follows (we give only non zero products) n = 1. Trivial algebra. n = 2.

• Trivial algebra. • A1 := {SpanF {e1 , e2 } | e1 e1 = e2 }.

n = 3. Trivial algebra. A1 := {SpanF {e1 , e2 , e3 } | e1 e1 A2 (α) := {SpanF {e1 , e2 , e3 } | e1 e1 where α ∈ F∗ /(F∗ )2 ; A3 (α) := {SpanF {e1 , e2 , e3 } | e1 e1 where α ∈ F∗ ; A4 := {SpanF {e1 , e2 , e3 } | e1 e2 A5 := {SpanF {e1 , e2 , e3 } | e1 e1 A6 := {SpanF {e1 , e2 , e3 } | e1 e2

= e2 }; = αe3 , e2 e2 = e3 }, = αe3 , e2 e2 = e3 , e2 e1 = e3 }, = e3 , e2 e1 = −e3 }; = e2 , e1 e2 = e3 , e2 e1 = e3 }; = e3 , e2 e1 = −e3 , e2 e2 = e3 }.

12  Leibniz Algebras

The lists above are taken from the paper [98], to which we refer the reader for the list in dimension four. Complete lists of low-dimensional complex associative algebras have been given in [157]. Radical Rad(A) of an algebra A is the largest nilpotent ideal of A (containing all nilpotent ideals of the algebra). The quotient A/Rad(A) is semisimple, i.e., it has a zero radical. The following two theorems form a foundation of the structural theory of semisimple associative algebras ([145]). Theorem 1.4. (Wedderburn − Artin) Any finite-dimensional semisimple associative algebra A is uniquely decomposed into a direct sum of simple algebras: A = B1 ⊕ B2 ⊕ · · · ⊕ Bk . Recall that an algebra is simple if it has no nontrivial two-sided ideal. Theorem 1.5. Any finite-dimensional simple associative algebra A is isomorphic to the algebra of matrices Mn (D) over a division ring D, the number n and the division ring D are uniquely determined by the algebra A. These theorems give a complete description of semisimple associative algebras. At the same time the structure of non-semisimple algebras (nilpotent part), is not sufficiently investigated, even over an algebraically closed field. So far there is no systematic way to study this case. Complex associative algebras in dimensions up to 5 were first classified by B. Pierce in 1870, initially in the form of manuscripts, which appeared later in [145]. There are classifications of unital 3, 4 and 5-dimensional associative algebras by Scorza [165], Gabriel [81], and Mazzola [127], respectively. 1.3 LIE ALGEBRAS

The notion of a Lie algebra (the old name is Infinitesimal Lie group) arose in the study of Lie groups, but later became an object of selftheory. Lie algebra is a vector space L over a field F, with a bilinear binary operation denoted by [·, ·], and the operation satisfies the following conditions:

Introduction  13

• antisymmetry: [x, x] = 0 for all x ∈ L; • Jacobi identity: [[x, y], z] + [[y, z], x] + [[z, x], y] = 0 for all x, y, z ∈ L. Examples. 1. Let V = R3 and x = (x1 , x2 , x3 ), y = (y1 , y2 , y3 ) ∈ R3 . Define [x, y] = (x2 y3 − x3 y2 , x3 y1 − x1 y3 , x1 y2 − x2 y1 ). Then L = (R3 , [·, ·]) is a Lie algebra. 2. Let A be an associative algebra and x, y ∈ A. Define [x, y] = xy − yx. Then L = (A, [·, ·]) is a Lie algebra. Particularly, if A = EndF (V) for the Lie algebra L there is a special notation gl(V, F) (or gl(V) if the field is clearly specified). If A = Mn (F) the algebra of n × n matrices with the entries from F then the Lie algebra L is denoted by gl(n, F) (sometimes the notation gln (F) also is used). Definition 1.10. A subspace I of a Lie algebra L is said to be ideal if for x ∈ I and y ∈ L imply [x, y] ∈ I (Since [x, y] = −[y, x], in a Lie algebra all ideals are two-sided). Examples. 1. The center Cent(L) = {x ∈ L | [x, y] = 0, for all y ∈ L} of a Lie algebra L is an ideal of L. 2. The derived algebra [L, L] = SpanF {[x, y] ∈ L for all x, y ∈ L} is an ideal of L. Definition 1.11. A Lie algebra L is said to be simple if [L, L] , 0 and L has no ideals except itself and zero.

14  Leibniz Algebras

1.3.1

Simple Lie algebras

Every finite-dimensional complex simple Lie algebra is isomorphic to one of the following Lie algebras (the proof is based on works by Killing, Engel and Cartan): 1. Special Lie algebras sl(l + 1, C) = {A = (ai j ) ∈ gl(l + 1, C) | tr(A) =

l+1 X

aii = 0}.

i=1

Since Tr(AB) = Tr(BA) and Tr(A + B) = Tr(A) + Tr(B) the subspace sl(l + 1, C) is a subalgebra of gl(l + 1, C). The dimension of sl(l + 1, C) is l2 + 2l. Indeed, on the one hand sl(l + 1, C) is a proper subalgebra of gl(l + 1, C), hence its dimension at most (l + 1)2 − 1 = l2 + 2l. On the other hand we can exhibit this number of linearly independent matrices of trace zero: take all Ei j (i , j), along with all Hi = Eii − Ei+1,i+1 (1 ≤ i ≤ l), for total of l + (l + 1)2 − (l + 1) = l2 + 2l. 2. Orthogonal Lie algebras n o • o(2l + 1, C) = A ∈ gl(2l + 1, C) | AT J = −JA, ,    1 0 0  where J =  0 0 Il  and Il is l × l identity matrix. Let us 0 Il 0 write elements of o(2l + 1, C) as blocks, of shapes adapted to the blocks of J. Calculations show that    0 cT −bT  P  with P = −PT and Q = −QT . A =  b M  −c Q −M T For a basis, – take first the l diagonal matrices Eii − El+i,l+i (2 ≤ i ≤ l + 1). – add the 2l matrices involving only row one and column one: E1,l+i+1 − Ei+1,1 and Ei+1,1 − E1,l+i+1 (1 ≤ i ≤ l).

Introduction  15

– for the block M take Ei+1, j+1 − El+ j+1,l+i+1 (1 ≤ i , j ≤ l). – for the block P take Ei+1,l+ j+1 − E j+1,l+i+1 (1 ≤ i < j ≤ l). – for the block Q take Ei+l+1, j+1 − E j+l+1,i+1 (1 ≤ j , i ≤ l). The total number of basis vectors is 2l2 + l, i.e., dim o(2l + 1, C) = 2l2 + l. n o • o(2l, C) = A ∈ gl(2l, C) | AT J = −JA, ! 0 Il where J = and Il is l × l identity matrix. Il 0 Write elements of o(2l, C) as blocks, of shapes adapted to the blocks of J. Again calculations show that ! M P A= with P = −PT and Q = −QT . Q −M T For a basis, • take first the l diagonal matrices Eii − El+i,l+i (1 ≤ i ≤ l). • for the block M take Ei, j − El+ j,l+i (1 ≤ i , j ≤ l). • for the block P take Ei,l+ j − E j,l+i (1 ≤ i < j ≤ l). • for the block Q take Ei+l, j − E j+l,i (1 ≤ j < i ≤ l). The total number of basis vectors is 2l2 − l, i.e., dim o(2l, C) = 2l2 − l.

16  Leibniz Algebras

3. Symplectic Lie algebras n o sp(2l, C) = A ∈ gl(2l, C) | AT J = −JA, ! 0 Il where J = and Il is the identity matrix. Write ele−Il 0 ments of sp(2l, C) in a block form as follows ! M N where M, N, P, Q ∈ gl(l, C). P Q Then the condition of symplecticity AT J = −JA implies the following constraints N T = N, PT = P and M T = −Q for M, N, P, Q ∈ gl(l, C). As a basis of sp(2l, C) we take • l the diagonal matrices Eii − El+i,l+ j (1 ≤ i ≤ l); • add to these l2 − l vectors Ei j − El+ j,l+i (1 ≤ i , j ≤ l); • use l + 12 l(l − 1) matrices Ei,l+i (1 ≤ i ≤ l) and Ei,l+ j + E j,l+i (1 ≤ i , j ≤ l) for the positions in N; • and take l + 12 l(l − 1) matrices El+i,i (1 ≤ i ≤ l) and El+i, j + El+ j,i (1 ≤ i , j ≤ l) for the positions in Q. Summing up, we find dim sp(2l, C) = 2l2 + l. 4. Exceptional simple Lie algebras. The constructions of the next simple Lie algebras are more delicate and we give them without details. • G2 = Der(O), where O is the Cayley-Dickson matrix algebra (see Example 1.1). Note that dim G2 = 14. Here are Lie algebras given by some constructions over the Albert algebra J (see Example 1.4). • F4 = Der J and dim F4 = 52. • E6 = Strl0 J, where Strl0 J is the reduced structure of the Jordan algebra J. It is a subalgebra of codim 1 of the structure algebra Strl J of J. dim E6 = 78. • E7 = K(J) (The Tits-Kantor-Koecher algebra of J) and dim E7 = 133.

Introduction  17

The simple Lie algebra E8 is given as follows (Tits construction): • E8 = Der(O)⊕O0 ⊗J0 ⊕Der(J), where O0 and J0 are matrices with zero traces in O and J, respectively. The binary operation on the vector space E8 = Der(O) ⊕ O0 ⊗ J0 ⊕ Der(J) is given as follows: – The part Der(O) ⊕ Der(J) is the direct sum of the Lie algebras; – [a ⊗ x, D] = aD ⊗ x, [a ⊗ x, E] = a ⊗ xE for D ∈ Der (O), E ∈ Der (J), a ∈ O0 and x ∈ J0 ; – [a⊗x, b⊗y] = 121 tr(xy)Da,b +(a∗b)⊗(x∗y)+ 12 t(ab)[R x , Ry ] for a, b ∈ O0 , x, y ∈ J0 and Da,b = R[a,b] − L[a,b] − 3[La , Rb ] ∈ Der O, where t(a) is the trace of a ∈ O, tr(x) is the trace x ∈ J, a ∗ b = ab − 12 t(ab) and x ∗ y = xy − 13 tr(xy). The dimension of E8 is 248. The list above exhausts all finite-dimensional simple Lie algebras over an algebraic closed field F of characteristic zero. They are pairwise not isomorphic except for the following cases: • o(3, F)  sp(2, F)  sl(2, F); • o(4, F)  sl(2, F) ⊕ sl(2, F); • o(5, F)  sp(4, F); • o(6, F)  sl(4, F). 1.3.2

Solvable and nilpotent Lie algebras

Let (L, [·, ·]) be a Lie algebra. The lower central series and derived series of L are defined as follows: L1 = L, Lk+1 = [Lk , L], k ≥ 1, L[1] = L, L[s+1] = [L[s] , L[s] ], s ≥ 1. Obviously, L1 ⊇ L2 ⊇ · · · ⊇ Li ⊇ · · ·, and L[1] ⊇ L[2] ⊇ · · · ⊇ L[i] ⊇ · · ·. Note that L[i] ⊆ Li .

18  Leibniz Algebras

Definition 1.12. A Lie algebra L is said to be nilpotent (respectively, solvable), if there exists n ∈ N (m ∈ N) such that Ln = 0 (respectively, L[m] = 0). The proofs of the following two lemmas are straightforward and can be found in any standard book on Lie algebras. Lemma 1.2. Let L be a Lie algebra, I and J be its ideals. Then • If L is solvable, then so are all subalgebras and homomorphic images of L. • If I and L/I are solvable, then L itself also is solvable. • If I and J are solvable, so is I + J. As an application of the lemma for any Lie algebra L one can prove that there exists a unique maximal solvable ideal, called the radical of L and denoted Rad(L). If L is not trivial and Rad(L) = 0, then L is said to be semisimple. The problem of classification of finite dimensional Lie algebras can be reduced to the following three separate tasks: • Classification of nilpotent Lie algebras; • Description of solvable algebras with given nilradical; • Description Lie algebras with a given radical. The latter two problems are the most studied part of the classification problem of Lie algebras and brought to fruition for the complex Lie algebras in the mid-20th century. The third problem is reduced to a description of semisimple subalgebras of the derivation algebras of a solvable algebra [124]. The problem of how to construct, by a given solvable algebra R and a semisimple algebra S , all the algebras L with the radical R and the quotient algebra L/R isomorphic to S , also has been solved. It turned out that such algebras L are finite in number, and they correspond one-to-one to semisimple subalgebras of the algebra of derivations of R. Since semisimple algebras are completely described by the well-known theory of Cartan-Killing this problem reduces to the study of solvable algebras. The second problem is reduced to the description of the orbits of some nilpotent linear groups [125]. Thus the classification problem of Lie algebras is reduced to the study

Introduction  19

of nilpotent algebras. This problem is the most complicated and, unfortunately, still there is no any standard way to solve it. Recall that complete classification of complex Lie algebras is obtained only up to dimension six. As for nilpotent Lie algebras, then various methods of their classification have been implemented, and, accordingly, several lists of isomorphism classes have been presented. Recall just a few of them: the earliest results of the classification of complex nilpotent Lie algebras of dimension no greater than 6 were obtained in year 1891 by Umlauf in his thesis [177]. Umlauf’s list was inaccurate, it contained many mutually isomorphic algebras. In 1958, Dixmier refined this list and published the list of nilpotent Lie algebras of dimensions at most 5 over an arbitrary field [61]. We mention as well results of Mubarakzjanov (1963, over the field of real numbers) [130], [131]. In dimension 6 there are several classification results: Morozov (1958, over a field of characteristic zero) [129], Shedler (1964, over an arbitrary field) [171]. Vergne (1966, over C) [179], Skjelbred and Sund (1978, over R) [172], Beck and Kolman (1981, over R) [34]. In dimension 7, also there are several lists: Safiullina (1964, over C) [163]), Romdani (1985, over R and C) [162], Seeley (1988, over C) [166], and Ancochea and Goze (1989, over C) [20], [21]. All these results were obtained by using a variety of isomorphism invariants. In 1989, Carles, introducing a new type of invariants (the weights system), compared the results of Safiullina, Romdani and Seeley and in 1993 published his result in dimension 7 over C in [52]. In his PhD thesis Ming-Peng Gong [89] has applied the central extensions method to classify 7-dimensional nilpotent Lie algebras over algebraically closed fields and over R. Moreover, M.-P. Gong compared the results obtained earlier by the authors above with results he obtained and purified the lists of isomorphism classes of 7-dimensional Lie algebras. The classification of nilpotent Lie algebras in higher dimensions remains a vast open area. Again, as it was the case after Umlauf’s pioneering work, the most efficient way of sorting through the mess of facts in higher dimensions is not obvious. These days computers offer some hope of being able to deal with the kind of messy details which haunt this subject. There are results on classification of subclasses of nilpotent Lie algebras, such as Favre (1973, nilpotent Lie algebra of maximal rank) [67], Scheuneman (1967) [164], Gauger (1973) [83], Revoy (1980)

20  Leibniz Algebras

[158], and Galitssky and Timashev (1998) [82] (metabelian Lie algebra), G´omez, Jim´enez-Merch´an, Khakimdjanov (1998) [87] (filiform Lie algebras) and others. 1.3.3

Semisimple Lie algebras

In this section we shall give a short survey of the theory of complex semisimple Lie algebras (the reader interested in this subject is referred to the classical books on Lie algebras [105], [106], etc). Definition 1.13. A Lie algebra L is said to be semisimple if Rad(L) = {0}. It is easy to show that for a Lie algebra L the quotient L/Rad(L) is semisimple. As a consequence of this we get that a semisimple Lie algebra does not have a nonzero abelian ideals. Definition 1.14. A bilinear form B : L × L −→ C for a Lie algebra L is said to be invariant if B([x, y], z) + B(x, [y, z]) = 0 for all x, y, z ∈ L. The bilinear form defined by B(x, y) = Tr(ad x ◦ady ) on a Lie algebra L is an example of the invariant form. It is called the Killing form of L denoted by K(x, y). The following proposition shows the importance of the Killing form. Proposition 1.2. (Cartan-Killing Criterion) A Lie algebra L is semisimple if and only if its Killing form K(x, y) is non degenerate. Corollary 1.1. Let L be a semisimple Lie algebra and a its ideal. Then the set a⊥ = {x ∈ L | K(x, y) = 0, for all y ∈ a} is an ideal of L (called the kernel of the Killing form with respect to a) and L = a ⊕ a⊥ (direct sum of ideals). Corollary 1.2. A semisimple Lie algebra is represented as a direct sum of simple Lie algebras. Corollary 1.3. If a Lie algebra L is semisimple, then L = [L, L]. Definition 1.15. Let L be a Lie algebra over a field F.

Introduction  21

i) A representation of a Lie algebra L in a vector space V is a homomorphism of Lie algebras ρ : L −→ gl(V). ii) A vector space V over F endowed with an action L × V −→ V (denoted by x · v for x ∈ L and v ∈ V) is said to be L-module if • (αx + βy) · v = α(x · v) + β(y · v) for all α, β ∈ F, x, y ∈ L and v ∈ V; • x · (αv + βw) = α(x · v) + β(x · w) for all α, β ∈ F, x, y ∈ L and v ∈ V; • [x, y] · v = x · (y · v) − y · (x · v) for all x, y ∈ L and v ∈ V. Definition 1.16. A homomorphism of L-modules V1 and V2 is a linear function ϕ : V1 −→ V2 such that ϕ(x · v) = x · ϕ(v) for all x ∈ L and v ∈ V1 . The notions of the representation of a Lie algebra L in V and for V to be a L-module are equivalent. Indeed, if ρ is a representation of L in V then V is a L-module with the action x · v := ρ(x)(v). Conversely, if V is a L-module then the map ρ : L −→ gl(V) given by ρ(x)(v) := x · v is a representation of L in V. Examples. Every Lie algebra L can be regarded as a L-module by setting x · y := [x, y]. The corresponding representation is said to be adjoint representation of L. It is denoted by ad x , for x ∈ L, i.e., ad x (y) = [x, y]. Definition 1.17. A L-module V is called simple if V , {0} and V has no submodules other than {0} and V. A L-module V is called semisimple if it is a direct sum of simple submodules. In terms of representations these notions correspond to the irreducibility and complete reducibility of representations. There is a fundamental result on finite-dimensional (i.e., V is a finite-dimensional) representations known as Weyl’s Theorem. For the proof the reader is referred to [66] or [105]. Theorem 1.6. Let L be a semisimple Lie algebra and let V be a finitedimensional L-module. Then V is semisimple . Definition 1.18. A Lie algebra is said to be reductive if its adjoint representation is completely reducible.

22  Leibniz Algebras

Proposition 1.3. Let L be a Lie algebra. The following conditions are equivalent: i) L is reductive; ii) [L, L] is semisimple; iii) L = [L, L] ⊕ Cent(L); iv) Rad(L) = Cent(L). Here we give the following adapted version of well-known Schur’s Lemma from [185, p.57, Corollary 3] which is needed to prove some results on conjugacy of Leibniz algebras. Lemma 1.3. Let L be a complex Lie algebra and let V = V1 ⊕ · · · ⊕ Vm and W = W1 ⊕ · · · ⊕ Wn be completely reducible L-modules, where V1 , . . . , Vn , W1 , . . . , Wn are irreducible modules. Then any L-module homomorphism ϕ : V → W can be represented as ϕ=

m X n X

λi j ϕi j ,

i=1 j=1

where the operators ϕi j : Vi → W j are fixed L-module homomorphisms and λi j are complex numbers. Furthermore, ϕi j , 0 if and only if ϕi j is an isomorphism. Let us now discuss the following fundamental theorem on the structure of finite-dimensional Lie algebras. Theorem 1.7. (Levi’s Theorem) Every Lie algebra L can be written as a semi-direct sum of its radical Rad(L) and a semisimple subalgebra ˙ S such that L = Rad(L)⊕S. The semisimple subalgebra S in the theorem above is called Levi’s subalgebra (Levi’s Factor) of L. The Levi’s Factor is not unique. Note that if S1 and S2 are two Levi’s Factors such that ˙ 1 = Rad(L)⊕S ˙ 2, L = Rad(L)⊕S then there exists an automorphism σ of L such that σ(S1 ) = S2 . Moreover, the automorphism σ can be chosen of the form σ = exp(ad x ), where x is in the nilradical of L. The result is due to A.I. Malcev.

Introduction  23

Each element x of a complex semisimple Lie algebra is written uniquely as x = d+n, where d, n ∈ L are such that add is diagonalizable, adn is nilpotent and [d, n] = 0. Furthermore, if y ∈ L commutes with x, then [d, y] = 0 and [d, y] = 0. If in x = d + n the term d = 0, then x is said to be semisimple. Let L be a semisimple Lie algebra. There exists a subalgebra H of L consisting entirely of semisimple elements of L such that L is decomposed into weight spaces for the action of adH. Indeed, the subalgebra H consists of all semisimple elements of L. The algebra L has a basis of common eigenvalues for the elements of adH. Given a common eigenvector x ∈ L, the eigenvalues are given by the associated weight, α : H −→ C, defined by adh x = α(h)x, for all h ∈ H. Weights are elements of the dual space H ∗ . For each α ∈ H ∗ , let Lα = {x ∈ L | [h, x] = α(h)x for all h ∈ H} denote the corresponding weight space. One of these subspaces is the zero weight space: L0 = {x ∈ L | [h, x] = 0 for all h ∈ H}. The space L0 is the same as the centralizer C L (H) of H in L. As H is abelian, one has H ⊆ L0 . Let Φ denote the set of non-zero α ∈ L∗ for which Lα is non-zero. The decomposition of L into the direct sum of weight spaces for H is written as follows: M L = L0 ⊕ Lα . α∈Φ

Since L is finite-dimensional Φ is finite. The elements of the set Φ are said to be roots of L with respect to the Cartan subalgebra H. The following lemma expresses properties of weight spaces. Lemma 1.4. Suppose that α, β ∈ H ∗ . Then i) [Lα , Lβ ] ⊆ Lα+β . ii) If α + β , 0, then K(Lα , Lβ ) = 0.

24  Leibniz Algebras

iii) The restriction of K to L0 is non-degenerate. This property implies that for every α ∈ Φ there is a vector h ∈ H such that K(h, hα ) = α(h) for all h ∈ H. iv) If α ∈ Φ then −α ∈ Φ and we have [Lα , L−α ] = Chα , α(hα ) , 0. Definition 1.19. A subalgebra H of a Lie algebra L is said to be Cartan subalgebra if H is abelian and any element of H is semisimple and H is maximal with respect to these properties. Note that a Cartan subalgebra of a semisimple Lie algebra is not unique, all the Cartan subalgebras of a semisimple Lie algebra L are isomorphic. They also conjugate in the following sense: for any two subalgebras H1 and H2 of a semisimple Lie algebra L there is an automorphism σ of L of the form exp(ady ) with y ∈ L such that H2 = σ(H1 ). The dimension of the Cartan subalgebra is called the rank of L. Let r be the rank of the semisimple Lie algebra L with a Cartan subalgebra H. There is a basis {α1 , α2 , . . . , αr } of H ∗ such that 1. αi ∈ Φ, i = 1, 2, . . . , r. 2. Every root α is decomposed as a linear combination of {α1 , α2 , . . . , αr } with integer coefficients k1 , k2 , . . . , kr : α = k1 α1 + k2 α2 + · · · + kr αr such that the coefficients k1 , k2 , . . . , kr are all simultaneously positive or negative. Definition 1.20. • The root system S = {α1 , α2 , . . . , αr } satisfying the conditions above is said to be a system of simple roots. • A root α ∈ Φ is called positive (respectively, negative) if the coefficients k1 , k2 , . . . , kr in α = k1 α1 + k2 α2 + · · · + kr αr are positive (respectively, negative).

Introduction  25

For a given root system Φ its subsets consisting of the positive and negative roots are denoted by Φ+ and Φ− , respectively. Thus one has Φ = Φ+ ∪ Φ− , where Φ− = −Φ+ . Note also that some texts use a “maximal toral subalgebra” instead of Cartan subalgebra. Definition 1.21. A Lie algebra L is called toral if it consists entirely of semisimple elements of L. Definition 1.22. A maximal toral subalgebra of L is a subalgebra of L which is toral and is not contained in any larger toral subalgebra. The fact that the maximal toral subalgebras of a Lie algebra L are precisely the Cartan subalgebras is obtained from the statements below. Lemma 1.5. Any toral subalgebra of a semisimple Lie algebra is abelian. One immediately gets the following Corollary 1.4. A subalgebra of a semisimple Lie algebra L is maximal toral if and only if it is a Cartan subalgebra. 1.3.4

More on finite-dimensional Lie algebras

In this section we review a few more results from [106] for further usage. Lemma 1.6. (Fitting’s lemma) Let V be a vector space and A : V → V be a linear transformation. Then V = V0A ⊕ V1A , where A(V0A ) ⊆ V0A , A(V1A ) ⊆ V1A and V0A = {v ∈ V | Ai (v) = ∞ T 0 for some i}, V1A = Ai (V). Moreover, A|V0A is a nilpotent transfori=1

mation and A|V1A is an automorphism. The spaces V0A and V1A are called respectively Fitting null component and Fitting one component of the space V with respect to the transformation A.

26  Leibniz Algebras

Lemma 1.7. Let V be a vector space. And let A, B be linear transformations on V such that [. . . [[B, | A], A], A] = 0. Then the Fitting {z. . . } k times

components V0A , V1A of V with respect to A are invariant with respect to the transformation B. Further we make use the following results: Theorem 1.8. Let G be a nilpotent Lie algebra of linear transfor∞ T T mations of a vector space V and V0 = V0A , V1 = Gi (V). Then A∈G

i=1

the subspaces V0 and V1 are invariant with respect to G (i.e., V0 and V1 are invariant with respect P to every transformation B from G) and V = V0 ⊕ V1 . Moreover, V1 = V1A . A∈G

Theorem 1.9. Let G be a split nilpotent Lie algebra of linear transformations of a vector space M. Then G has a finite number of different weights, weight subspaces are submodules of M, and M is decomposed into the direct sum of these modules. Moreover, if M = M1 ⊕ M2 ⊕ . . . ⊕ Mr is an arbitrary decomposition of M into the sum of subspaces Mi (, 0), which are invariant with respect to G such that the following conditions hold: 1) for each i the restriction of A ∈ G on Mi has only one characteristic root αi (A) (of some multiplicity); 2) if i , j, then there exists A ∈ G such that αi (A) , α j (A); then the maps A → αi (A) are weights and Mi are weight subspaces. Proposition 1.4. Let L be a Lie algebra of linear transformations of a vector space over a field of zero characteristic, R be the radical of the algebra L, and let N be the radical of the associative algebra L∗ , generated by L. Then L ∩ N consists of all nilpotent elements of the radical R and [R, L] ⊆ N (N is considered as a nilpotent radical of the associative algebra L∗ ). 1.4 LODAY ALGEBRAS

In this section we briefly summarize basic facts about some classes of algebras which are closely related to Leibniz algebras. Any associative algebra gives rise to a Lie algebra by [x, y] = xy − yx. Leibniz algebras

Introduction  27

are non-commutative version of Lie algebras while associative dialgebras are generalization of associative algebras. One way to generalize the idea above, so as to obtain Leibniz algebra brackets that are not skew-symmetric, is to use two different associative algebra products. The notion of associative dialgebra was discovered by Loday while studying periodicity phenomena in algebraic K-theory. A few more classes of algebras and their relation with classical algebras are discussed. 1.4.1

Leibniz algebras

The concept of Leibniz algebra was introduced by Loday [117] in the study of Leibniz (co)homology as a noncommutative (to be more precise, as a non-anticommutative) analogue of Lie algebras (co)homology. A Leibniz algebra over a field F is a vector space over F, equipped with a F-bilinear map [·, ·] : L × L −→ L satisfying the Leibniz identity [[x, y], z] = [[x, z], y] + [x, [y, z]],

(1.4)

for all x, y, z ∈ L. In fact, the definition above is a definition of the right Leibniz algebra, whereas the identity for the left Leibniz algebra is as follows [x, [z, y]] = [[x, z], y] + [z, [x, y]], (1.5) for all x, y, z ∈ L. The passage from the right to the left Leibniz algebra can be easily done by considering a new product “[·, ·]opp ” on the same vector space defined by “[x, y]opp = [y, x]”. Therefore, the results proved for the right Leibniz algebras (L, [·.·]) can be easily reformulated for the left Leibniz algebras (L, [·, ·]opp ) and vice versa. Note that the versions [x, [y, z]] = [[x, y], z] − [[x, z], y], for all x, y, z ∈ L

(1.6)

[[x, z], y] = [x, [z, y]] − [z, [x, y]], for all x, y, z ∈ L. of the identities (1.4) and (1.5) are also often used.

(1.7)

and

Examples. 1. Let g be a Lie algebra and M be a g-module. Let f : M −→ g be a g-equivariant linear map, i.e., f (m · x) = [ f (m), x], for all m ∈ M and x ∈ g.

28  Leibniz Algebras

The bracket [·, ·] M defined by [m, n] M := m · f (n) provides a Leibniz algebra structure on M. Indeed, for m, n, h ∈ M we have [[m, n] M , h] M = [m · f (n), h] M = (m · f (n)) · f (h) = (m · f (h)) · f (n) − m · [ f (h), f (n)] = (m · f (h)) · f (n) + m · [ f (n), f (h)] = [m · f (h), n] M + m · f (n · f (h)) = [[m, h] M , n] M + [m, [n, h] M ] M . 2. Let g be a Lie algebra and M be a g-module. Then the vector space Q = g⊕ M equipped with the multiplication [x+m, y+n] = [x, y] + m · y, where m, n ∈ M, x, y ∈ g is a Leibniz algebra. Let us verify the Leibniz identity for the product. One has [[x + m, z + h]Q , y + n]Q + [x + m, [y + n, z + h]Q ]Q = [[x, z] + m · z, y + n]Q + [[x + m, [y, z] + n · z]Q = [[x, z], y] + (m · z) · y + [x, [z, y]] + (m · [y, z]) = [[x, z], y] + [x, [z, y]] + (m · z) · y + (m · [y, z]) = [[x, y], z] + (m · y) · z = [[x, y] + m · y, z + h]Q = [[x + m, y + n]Q , z + h]Q . 3. A vector space L = g ⊗ g, equipped with the product [x ⊗ y, a ⊗ b]L = [x, [a, b]] ⊗ y + x ⊗ [y, [a, b]] is a Leibniz algebra, where g is a Lie algebra, i.e.,     [x ⊗ y, a ⊗ b]L , s ⊗ t L = [x ⊗ y, s ⊗ t]L , a ⊗ b L   + x ⊗ y, [a ⊗ b, s ⊗ t]L L . Indeed,   [x ⊗ y, a ⊗ b]L , s ⊗ t L = [[x, [a, b]], [s, t]] ⊗ y +[x, [a, b]] ⊗ [y, [s, t]] | {z } :::::::::::::::::::: z }| { +[x, [s, t]] ⊗ [y, [a, b]] +x ⊗ [[y, [a, b]], [s, t]] and     [x ⊗ y, s ⊗ t]L , a ⊗ b L + x ⊗ y, [a ⊗ b, s ⊗ t]L L = ([[x, [a, b]], [s, t]] + [x, [[a, [s, t]], b]] + [x, [a, [b, [s, t]]]]) ⊗ y | {z }

Introduction  29

+[x, [a, b]] ⊗ [y, [s, t]]+[x, [s, t]] ⊗ [y, [a, b]] :::::::::::::::::::: z }| { +x ⊗ ([[y, [s, t]], [a, b]] + [y, [[a, [s, t]], b]] + [y, [a, [b, [s, t]]]]) . By using anti-symmetricity and the Jacobi identity of the Lie bracket. It it is easy to see that ([[x, [a, b]], [s, t]] + [x, [[a, [s, t]], b]] + [x, [a, [b, [s, t]]]]) and ([[y, [s, t]], [a, b]] + [y, [[a, [s, t]], b]] + [y, [a, [b, [s, t]]]]) are reduced to [[x, [a, b]], [s, t]] and [[y, [a, b]], [s, t]], respectively. 4. Let A be an associative algebra and D be a linear map D : A → A satisfying the condition D(aD(b)) = D(a)D(b) = D(D(a)b), ∀a, b ∈ A. Then A equipped with the multiplication [a, b]D = aD(b) − D(b)a is a Leibniz algebra (denoted by AD ). Remark. (Examples of the map D) • Let A = A+ ⊕ A− be a Z2 -graded associative algebra and D(x) := x+ ; • A derivation D of an associative algebra with D2 = 0; • An endomorphism D such that D2 = D. Clearly, a Lie algebra is a Leibniz algebra, and conversely, a Leibniz algebra L with property [x, y] = −[y, x], for all x, y ∈ L, is a Lie algebra. The inherent properties of non-Lie Leibniz algebras imply that the subspace spanned by squares of elements of the algebra L is a nontrivial ideal (further denoted by I.) Moreover, ideal I is abelian. There is an inclusion functor i : Lie ,→ Leib. This functor has a left adjoint p : Leib −→ Lie, which is defined on the objects by

30  Leibniz Algebras

LLie = L/I. That is any Leibniz algebra L gives rise to the Lie algebra LLie , obtained as the quotient of L by relation [x, x] = 0. The ideal I is the minimal ideal with respect to the property that g := L/I is a Lie algebra. The quotient mapping π : L −→ g is a homomorphism of Leibniz algebras. One has an exact sequence of Leibniz algebras: 0 −→ I −→ L −→ LLie −→ 0. We consider finite-dimensional algebras L over a field F of characteristic zero (in fact, it is sufficient that this characteristic is not equal to 2). A linear transformation d of a Leibniz algebra L is said to be a derivation if d([x, y]) = [d(x), y] + [x, d(y)], for all x, y ∈ L. Note that for x ∈ L the operator R x of right multiplication, R x : L → L, R x (y) = [y, x], y ∈ L, is a derivation (for a left Leibniz algebra L, the operator L x left multiplication, L x : L → L, L x (y) = [x, y], y ∈ L, also is a derivation). Such kind of derivations are called inner derivations on L. The set of all the inner derivations of a Leibniz algebra L is denoted by Inn(L). In other words, the right Leibniz algebra is characterized by this property, i.e., any right multiplication operator is a derivation of L. For the left Leibniz algebras, all left multiplication operators are derivations. The set of all derivations of L (further denoted by Der(L)) equipped with the commutator operation forms a Lie algebra. The set Inn(L) is an ideal of Der(L). The automorphism group Aut(L) of an algebra L also is naturally defined. If the field F is R or C, then the automorphism group is a Lie group and the Der(L) is its Lie algebra. One can consider Aut(L) as an algebraic group (or as a group of F-points of an algebraic group defined over the field F). The motivation to study the structural properties of Loday algebras (in fact, in [117] Loday introduced a few classes of algebras: diassociative, dendriform and Zinbiel algebras, and gave functorial diagram connecting these classes emphasizing their Koszulean duality, see Section 1.4.6) was provoked due to private discussions between J.L. Loday and Sh.A. Ayupov held in 1994 (Strasbourg).

Introduction  31

The appearance of Leibniz algebras was motivated by Loday as follows. The Chevalley-Eilenberg chain complex of a Lie algebra g is the sequence of chain modules given by the exterior powers of g as follows ^ dn+1 dn dn−1 ∗ g : · · · −→ ∧n+1 g −→ ∧n g −→ ∧n−1 g −→ . . . and boundary operators dn : ∧n g −→ ∧n−1 g classically defined by X dn (x1 ∧x2 ∧· · ·∧xn ) := (−1)i+ j+1 [xi , x j ]∧x1 ∧· · ·∧ xˆi ∧· · ·∧ xˆj ∧· · ·∧xn . i< j

The property dn ◦ dn+1 = 0, which makes this sequence a chain complex, is proved by using the antisymmetry x ∧ y = −y ∧ x of the exterior product, the Jacobi identity [[x, y], z] + [[y, z], x] + [z, x], y] = 0 and the antisymmetry [x, y] = −[y, x] of the Lie bracket given on g. Loday noticed that, if one replaces the exterior product with the tensor product and rewrites the Chevalley-Eilenberg boundary operator as X dn (x1 ⊗ x2 ⊗ · · · ⊗ xn ) = (−1)n− j x1 ⊗ · · · ⊗ [xi , x j ] ⊗ · · · ⊗ xˆj ⊗ · · · ⊗ xn , i< j

for the chain complex dn+1

dn

dn−1

· · · −→ L⊗(n+1) −→ L⊗n −→ L⊗(n−1) −→ . . . of an algebra L, then the property dn ◦ dn+1 = 0 is proved without making use of the antisymmetry properties of both the exterior product and the bracket on the algebra L. It suffices only that the bracket [·, ·] on L satisfies the so-called Leibniz identity [[x, y], z] = [[x, z], y] + [x, [y, z]]. Later on, besides this purely algebraic motivation to introduce the Leibniz algebras, some of their relationships with classical geometry, non-commutative geometry and as well as physics have been discovered. During the last 25 years the theory of Leibniz algebras has been a topic of active research. Some (co)homology and deformations properties; relations with R-matrices and Yang-Baxter equations; results on

32  Leibniz Algebras

various types of decompositions; structure of semisimple, solvable and nilpotent Leibniz algebras; classifications of some classes of graded nilpotent Leibniz algebras were obtained in numerous papers devoted to Leibniz algebras. In fact, many results of theory of Lie algebras have been extended to the Leibniz algebras case. For instance, the classical results on Cartan subalgebras, Levi’s decomposition, properties of solvable algebras with a given nilradical and others from the theory of Lie algebras are also true for Leibniz algebras. Recently, D. Barnes [32] has proved an analogue of Levi-Malcev Theorem for Leibniz algebras. He proved that a Leibniz algebra is decomposed into a semidirect sum of its solvable radical and a semisimple Lie algebra. Therefore, the main problem of the description of finite-dimensional Leibniz algebras consists of the study of solvable Leibniz algebras. D. Barnes also noted that the non-uniqueness of the semisimple subalgebra S may appear in Levi-Malcev theorem (the minimum dimension of Leibniz algebra where this phenomena appears is six). It is known that in the case of Lie algebras the semisimple Levi quotient is unique up to conjugation via an inner automorphism. The conjugation in the Leibniz algebras case is not true, in general. Recently, in [115] the authors studied the conditions for the semisimple parts of Leibniz algebras in the decomposition to be conjugated. With a given Leibniz algebra we can associate a few Lie algebras, namely: the quotient algebra by the ideal I; the algebra of all right multiplication operators with the commutator operation; the quotient algebra by the ideal of right annihilators etc. Some of properties of the Leibniz algebra can be stated in terms of these algebras. For instance, the solvability of the Leibniz algebra is equivalent to the solvability of each of the Lie algebras mentioned. Note that the nilradical of a Leibniz algebra (the maximal nilpotent ideal) is not radical in the sense of Kurosh, because the quotient algebra by the nilradical may contain a nilpotent ideal. Nevertheless, similarly to the case of Lie algebras, it plays a crucial role in the description of solvable Leibniz algebras. The fact that the square of a solvable Leibniz algebra is contained in the nilradical also motivates the study of nilpotent algebras. In the case of Lie algebras to classify the solvable part a result of the paper [131] has been used (see [17], [49], [132], [173] and [181] , etc). In the case of Leibniz algebras the analogue of Mubarakzjanov’s

Introduction  33

[131] results has been also successfully implemented in [50], [54], [55], [111] and [169]. A linear representation (sometimes referred as module) of a Leibniz algebra L is a vector space M, equipped with two actions (left and right) on L : [·, ·] : L × M −→ M and [·, ·] : M × L −→ M, such that the identity [[x, y], z] = [[x, z], y] + [x, [y, z]] is true whenever one (any) of the variables is in V, and the other two are in L, i.e., [[x, y], m] = [[x, m], y] + [x, [y, m]]; [[y, m], x] = [[y, x], m] + [y, [m, x]]; [[m, x], y] = [[m, y], x] + [m, [x, y]], for x, y ∈ L and m ∈ M. Note that to take the conditions above is predicted by the fact that on the direct sum L ⊕ M of vectors spaces the bracket [[l+m, l0 +m0 ]] := [l, l0 ]+[m, l0 ]+[l, m0 ], where l, l0 ∈ L and m, m0 ∈ M. defines a Leibniz algebra structure. Note that the concepts of representations for Lie and Leibniz algebras are different. Therefore, such an important theorem in the theory of Lie algebras, as the Ado theorem on the existence of faithful representation in the case of Leibniz algebras is proved much easier and gives a stronger result. Because the kernel of the Leibniz algebra representation is the intersection of the kernels (in general, different ones) of right and left actions, in contrast to the representations of Lie algebras, where these kernels are the same. Therefore, a faithful representation of Leibniz algebras can be obtained easier than a faithful representation in the case of Lie algebras (see [31]). The (co)homology theory, the representations and related problems for Leibniz algebras were studied by Frabetti, [80], Loday, and Pirashvili, [119] and others. A good survey of all these and related problems is the book [118].

34  Leibniz Algebras

The problems related to the group theoretical realizations and integrability problems of Leibniz algebras are studied by Kinyon, Weinstein, [112] and others. Different generalizations of Leibniz algebras as Leibniz superalgebras and n-ary Leibniz algebras also have been introduced and studied (see [9], [26], [147]). Deformation theory of Leibniz algebras and related physical applications are initiated by Fialowski, Mandal, Mukherjee [72]. The notion of simple Leibniz algebra has been introduced by Dzhumadil’daev in [63]. 1.4.2

Associative dialgebras

Here we present a generalization of associative algebras named by Loday as an associative dialgebra. A diassociative algebra D is a vector space equipped with two bilinear binary associative operations: a: D × D → D and `: D × D → D satisfying the axioms (x ` y) ` z = (x a y) ` z, (x ` y) a z = x ` (y a z), x a (y ` z) = x a (y a z). for all x, y, z ∈ D (the binary operations ` and a are said to be left and right products in D, respectively). Example 1.6. Let (A, d) be a differential associative algebra with d2 = 0. Define the left and right products on A as follows x a y := xd(y) and x ` y := d(x)y. Then (A, a, `) is a diassociative algebra. Example 1.7. Let A be an associative algebra, M be an A-bimodule and let f : M −→ A be an A-bimodule map. Define the left and right products on M as follows m a m0 := m f (m0 ) and m ` m0 := f (m)m0 . Then (M, a, `) is a diassociative algebra.

Introduction  35

Example 1.8. Let A be an associative algebra and n be a positive integer. The left and right products defined by  n  X  (x a y)i := xi  y j  for 1 ≤ i ≤ n j=1

and

 n  X  (x ` y)i :=  x j  yi for 1 ≤ i ≤ n j=1

on the A-module D = A make it an associative dialgebra. n

Example 1.9. Let D = F[x, y] be a polynomial algebra at two indeterminates x, y over a field F of characteristic zero. Define the left and right products a, ` on D by f (x, y) a g(x, y) := f (x, y)g(y, y) and f (x, y) ` g(x, y) := f (x, x)g(x, y). Then (D, a, `) is an associative dialgebra. Proposition 1.5. Let D be an associative dialgebra. Then the bracket [x, y] = x a y − y ` x turns D into a Leibniz algebra. Proof. It is a straightforward checking since [[x, y].z] = (x a y) a z − (y ` x) a z − z ` (x a y) + z ` (y ` x), [[x, z].y] = (x a z) a y − (z ` x) a y − y ` (x a z) + y ` (z ` x), [x, [y.z]] = x a (y a z) − x a (z ` y) − (y a z) ` x + (z ` y) ` x.  This construction defines a functor −

Dias → − Leib from category Dias of associative dialgebras to the category Leib of Leibniz algebras.

36  Leibniz Algebras

´ Universal enveloping algebra (Poincare-Birkhoff-Witt Theorem)

1.4.3

The universal enveloping algebra for a Leibniz algebra has been constructed by Loday and Pirashvili in [119] (also see [118]). Loday showed that such an algebra comes in as an algebra with two bilinear binary associative operations satisfying three axioms (the algebra has been called associative dialgebra by Loday). The enveloping algebra of a Leibniz algebra L was constructed by using the concept of free diassociative algebra. In algebras over fields this construction is interpreted as follows. Let V be a vector space over a field F. By definition the free dialgebra structure on V is the dialgebra D(V) equipped with a F-linear map i : V −→ D such that for any Flinear map f : V −→ D0 , where D0 is a dialgebra over F, there exists a unique factorization i

h

f : V→ − D(V) → − D0 , where h is a dialgebra morphism. The authors proved the existence of D(V) giving it as the tensor module D(V) = T (V) ⊗ V ⊗ T (V), where T (V) := F ⊕ V ⊕ V ⊗2 ⊕ · · · ⊕ V ⊗n ⊕ · · · with the following two products inductively defined by (v−n · · · v−1 ⊗ v0 ⊗ v1 · · · vm ) a (w−p · · · w−1 ⊗ w0 ⊗ w1 · · · wq ) = v−n · · · v−1 ⊗ v0 ⊗ v1 · · · vm w−p · · · wq , (v−n · · · v−1 ⊗ v0 ⊗ v1 · · · vm ) ` (w−p · · · w−1 ⊗ w0 ⊗ w1 · · · wq ) = v−n · · · · · · vm w−p · · · w−1 ⊗ w0 ⊗ w1 · · · wq ,

where vi , w j ∈ V. Let Ll and Lr be two copies of the Leibniz algebra L which is supposed to be free as F-module. We denote by l x and r x the elements of Ll and Lr corresponding to x ∈ L. Consider the tensor F-algebra T (Ll ⊕ Lr ), which is associative and unital. Let J be the two-sided ideal corresponding to the relations r[x,y] = r x ry −ry r x ,

l[x,y] = l x ry −ry l x ,

(ry +ly )l x = 0, for any x, y ∈ L.

Introduction  37

Definition 1.23. The universal enveloping algebra of the Leibniz algebra L is the associative and unital algebra UL(L) := T (Ll ⊕ Lr )/J. Let τ : V → W be an epimorphism of F-modules. Define the associative algebra S L(τ) as the quotient of S (W) ⊗ T (V) by the 2-sided ideal generated by 1 ⊗ xy + τ(x) ⊗ y, for all x, y ∈ V. Note that UL(L) is a filtered algebra, the filtration being induced by the filtration of T (Ll ⊕ Lr ), that is, Fn UL(L) = {image of F ⊕ E ⊕ · · · ⊕ E ⊗n in UL(L)}, where E = Ll ⊕ Lr . L The associated graded algebra is denoted grUL(L) := grn UL(L). n≥0

Then Poincar´e-Birkhoff-Witt Theorem for Leibniz algebras is given as follows. Theorem 1.10. For any Leibniz F-algebra L such that L and LLie are free as F-modules, there is an isomorphism of graded associative Falgebras grUL(L)  S L(L → LLie ). The classification of the associative dialgebras in low-dimensions has been given by using the structure constants and a computer program in Maple [33], [152], [153].

Classification Results.

1.4.4

Zinbiel algebras

The result intertwining diassociative algebras and dendriform algebras are best expressed in the framework of algebraic operads. In order to illustrate this point, Loday defined a class of Zinbiel algebras, which is Koszul dual to the category of Leibniz algebras. Definition 1.24. A Zinbiel algebra Z is an algebra with a binary operation · : Z × Z → Z, satisfying the condition: (a · b) · c = a · (b · c) + a · (c · b) for any a, b, c ∈ Z.

38  Leibniz Algebras

Example 1.10. Let V be the vector space of integrable functions. Let a product operation · on V be defined as follows ( f · g)(x) = f (x)

Zx g(t)dt for f, g ∈ V. 0

It is immediate to verify that (V, ·) is a Zinbiel algebra. Proposition 1.6. Let Z be a Zinbiel algebra. Then the product xy = x · y + y · x makes Z into a commutative algebra. 

Proof. Obvious.

Note that according to V.V. Gorbatsevich (see [93]) J.-L. Loday wrote as an author under a pseudonym “Guillaume William Zinbiel”, here Zinbiel is reading of Leibniz in the reverse order. Due to Proposition 1.6 we have a functor +

Zinb → − Com from category Zinb of Zinbiel algebras to the category Com of commutative algebras. A.S. Dzhumadildaev et al. proved that any finite-dimensional Zinbiel algebra is nilpotent and gave the list of isomorphism classes of Zinbiel algebras in dimension 3 [64]. By the way, Example 1.10 above is taken from [64] and it is an example of nonnilpotent infinite-dimensional Zinbiel algebra. Earlier 2-dimensional case has been classified by B. Omirov [137]. Further classifications have been carried out by using the isomorphism invariant called the characteristic sequence which was used successfully in the case of Lie algebras [2], [4], [5], [43]. Classification Results.

1.4.5

Dendriform algebras: Koszul dual of associative dialgebras

Definition 1.25. Dendriform algebra E is an algebra with two binary operations

Introduction  39

: E × E → E, ≺: E × E → E satisfying the following identities: (a ≺ b) ≺ c = (a ≺ c) ≺ b + a ≺ (b  c), (a  b) ≺ c = a  (b ≺ c),

(1.8)

(a ≺ b)  c + (a  b)  c = a  (b  c). Recall that an associative Rota-Baxter algebra (over a field F) is an associative algebra (A, ·) endowed with a linear operator R : A −→ A subject to the following relation: R(x) · R(y) = R(R(x) · y + x · R(y) + αx · y), α ∈ K. The operator R˜ := −αid − R is called a Rota-Baxter operator of weight α. The Rota-Baxter operator R˜ generates the following dendriform algebra structure on A. Example 1.11. Let (A, R, α) be a Rota-Baxter algebra of weight α and define ≺ and  as follows: ˜ ˜ · y. x ≺ y := −x · R(y) and x  y := R(x) It is immediate to check that (A, ≺, ) is a dendriform algebra. Proposition 1.7. Let E be a dendriform algebra. Then with the product defined by x∗y= x≺y−xy E becomes an associative algebra. Proof. It is an immediate checking if add up both sides of the equalities in (1.8).  Proposition 1.7 provides a functor +

Dend → − As from the category Dend of dendriform algebras to the category As of associative algebras.

40  Leibniz Algebras

The essential results on Dendriform algebras have been obtained by Aguiar, Guo and Ebrahimi-Fard [8], [65], [101]. There is a classification of two-dimensional dendriform algebras [161]. Classification Results.

1.4.6

Loday diagram and Koszul duality

The results intertwining Loday algebras are best expressed in the framework of algebraic operads. The notion of diassociative algebra defines an algebraic operad Dias, which is binary and quadratic. According to the theory of Ginzburg and Kapranov there is a well-defined “dual operad” Dias! . Loday has showed that this is exactly the operad Dend of the dendriform algebras, in other words a dual diassociative algebra is nothing but a dendriform algebra. A similar duality can be established between the algebraic operads Leib defined by the notion of Leibniz algebra and the algebraic operads Zinb defined by the notion of Zinbiel algebra. Operadic dualities Com! = Lie, As! = As have been proved in [86], Dias! = Dend is in [118] and Leib! = Zinb is in [120]. The categories of algebras over these operads assemble into the following commutative diagram of functors (see Figure 1.1) which reflects the Koszul duality (see [118]).

Figure 1.1

Loday diagram and Koszul duality

CHAPTER

2

STRUCTURE OF LEIBNIZ ALGEBRAS

2.1

SOME PROPERTIES OF LEIBNIZ ALGEBRAS

In this section we review extensions of main fundamental theorems of Lie algebras to Leibniz algebras. For those results which can not be extended counterexamples are provided. Let L be a (right) Leibniz algebra. The right and left multiplication operators for a fixed element a of L are denoted by Ra and La , respectively, i.e., Ra (x) = [x, a] and La (x) = [a, x] for x ∈ L. The set of right multiplication operators R(L) = {Ra | a ∈ L} forms an ideal of the Lie algebra Der(L). In a right Leibniz algebra the following identities R[b,a] = Ra ◦Rb −Rb ◦Ra , L[a,b] = Rb ◦La +La ◦Lb , L[a,b] = Rb ◦La −La ◦Rb hold true. Recall that I = SpanF {[x, x] | x ∈ L} (Loday has denoted it by Lann and another term “liezator” has been used by Gorbatsevich [93]). In fact, I is a two-sided ideal of L. Indeed, it follows from the equalities     [x, x], y = [y, y] + x, [y, y] + x − [x, x], [y, [x, x]] = [[y, x], x] − [[y, x], x] = 0. For a non-Lie Leibniz algebra L the ideal I is non trivial. The quotient algebra L/I is a Lie algebra and I is the smallest ideal in L, satisfying this condition. Therefore, I can be regarded as the “non Lie 41

42  Leibniz Algebras

core” of the Leibniz algebra L. The ideal I can also be described as the linear span of all elements of the form [x, y] + [y, x]. The quotient algebra L/I is called liezation of L and denoted by LLie . For a Leibniz algebra L we define right and left annihilators as follows Annr (L) = {x ∈ L | [L, x] = 0}

and

Annl (L) = {x ∈ L | [x, L] = 0},

respectively. These annihilators can be considered for both left and for right Leibniz algebras. For a right Leibniz algebra L the right annihilator Annr (L) is a two-sided ideal (since [L, [x, y]] = −[L, [y, x]]), while Annl (L) might not be even a subalgebra. For a left Leibniz algebra it is exactly the opposite. In general, the left and right annihilators are different, even they might have different dimensions. Obviously, I ⊆ Annr (L) for a right Leibniz algebra L, whereas I ⊆ Annl (L) for a left Leibniz algebra L. Therefore L/Annr (L) (respectively, L/Annr (L)) is a Lie algebra in the case of right (respectively, left) Leibniz algebra, which is anti-isomorphic to the Lie subalgebra R(L) (respectively, L(L) = {La | a ∈ L}) of Der(L), consisting of all right (left) multiplication operators. We denote by Cent(L) = Annr (L) ∩ Annl (L) the center of a Leibniz algebra L. 2.2 NILPOTENT AND SOLVABLE LEIBNIZ ALGEBRAS

The concepts of nilpotency and solvability for Leibniz algebras are defined similarly to those of Lie algebras. For a given Leibniz algebra (L, [·, ·]) the lower central series and derived series are defined, respectively, as follows: L1 = L, Lk+1 = [Lk , L], k ≥ 1, L[1] = L, L[s+1] = [L[s] , L[s] ], s ≥ 1. Obviously, L1 ⊇ L2 ⊇ · · · ⊇ Li ⊇ · · ·, and L[1] ⊇ L[2] ⊇ · · · ⊇ L[i] ⊇ · · ·. Moreover, L[i] ⊆ Li .

Structure of Leibniz Algebras  43

Definition 2.1. A Leibniz algebra L is said to be nilpotent (respectively, solvable), if there exists n ∈ N (m ∈ N) such that Ln = 0 (respectively, L[m] = 0). The minimal number n (respectively, m) with such property is said to be the index of nilpotency (respectively, of solvability) of the algebra L. Example 2.1. • A nilpotent Lie algebra is nilpotent as a Leibniz algebra; • The sum of nilpotent Leibniz algebras is nilpotent; • On the space of quadratic n×n matrices with complex coefficients Mn (C) we consider a linear map D : Mn (C) → Mn (C) satisfying the condition (see Example in Section 1.4.1) D(AD(B)) = D(A)D(B) = D(D(A)B), for all A, B ∈ Mn (C). Then the space Mn (C) equipped with the multiplication [A, B]D = AD(B)− D(B)A becomes a Leibniz algebra (denoted by Mn (C)D ). Note that Mn (C)D is nilpotent if and only if Card(SpecX) = 1 for every X ∈ ImD; • The algebras with the table of multiplications below on a basis {e1 , e2 , . . . , en } are nilpotent Leibniz algebras: i)

  [ei , e1 ] = ei+1 , 1 ≤ i ≤ n − 3,     [e1 , en−1 ] = e2 + en ,      [e , e ] = e , 2 ≤ i ≤ n − 3, i n−1 i+1

ii) (

[ei , e1 ] = ei+1 , 1 ≤ i ≤ n − 3, [e1 , en−1 ] = en .

Here are examples of solvable Leibniz algebras. Example 2.2. • A solvable Lie algebra is solvable as a Leibniz algebra; • A nilpotent Leibniz algebra is solvable;

44  Leibniz Algebras

• The sum of solvable Leibniz algebras is solvable; • A right Leibniz algebra is solvable if and only if the Lie algebra R(L) is solvable; • A right Leibniz algebra is solvable if and only if the quotient Lie algebra L/Annr (L) is solvable; • An algebra with the following table of multiplications on a basis {e1 , e2 , . . . , en , x}   [e , e ] = ei+1 , 1 ≤ i ≤ n − 1,    i 1  [x, e1 ] = e1 ,      [e , x] = −ie , 1 ≤ i ≤ n. i i is a solvable Leibniz algebra. One of the classical results on finite-dimensional nilpotent Lie algebras is Engel’s theorem on criterion of nilpotency of the Lie algebra in terms of nilpotency of the right multiplications operators. An analogue of Engel’s theorem for Leibniz algebras also holds. The various versions of the theorem have been proved in [23], [93] and [143]. Here we give the following version from [93]. Theorem 2.1. If all operators R x of right multiplication for the right Leibniz algebra L are nilpotent, then the algebra L is nilpotent. In particular, for the right multiplications there is a common eigenvector with zero eigenvalue. Proof. Since L/I is a Lie algebra the assertion of the proposition for L/I holds true. Note that I is the central ideal, its action on L by right multiplications is trivial. Since the Leibniz algebra L is a central extension of the nilpotent Lie algebra L/I, it is also nilpotent.  Corollary 2.1. If for a right Leibniz algebra all right multiplication operators are nilpotent, then all left multiplication operators also are nilpotent. The proof of this corollary is based on the identity (L x )n = (−1)n−1 L x (R x )n−1 to prove that we use induction on n, the case n = 2 being obvious from the identity [x, [x, z]] = −[x, [z, x]]. Assume that (L x )n−1 = (−1)n−2 L x (R x )n−2 .

Structure of Leibniz Algebras  45

Then (L x )n (z) = [x, (L x )n−1 (z)] = [x, (−1)n−2 L x (R x )n−2 (z)] = (−1)n−2 [x, [x, (R x )n−2 (z)]] = (−1)n−1 [x, [(R x )n−2 (z), x]] = (−1)n−1 L x (R x )n−1 (z). For the other nilpotency properties of Leibniz algebras we refer the reader to [143]. Proposition 2.1. Any Leibniz algebra L has a maximal nilpotent ideal containing all nilpotent ideals of L. Proof. Let J be a nilpotent ideal in L. Then the sum J + I is nilpotent, it follows immediately from the right centrality of I. Therefore, in discussing the question of the existence of maximal nilpotent ideal it is sufficient to consider only nilpotent ideals containing I. Consider the epimorphism of liezation π : L → L/I = LLie . For the Lie algebra LLie the existence of a maximal nilpotent ideal is known. Its inverse image under the mapping π is also nilpotent. This completes the proof.  The maximal nilpotent ideal of a Leibniz algebra is said to be the nilradical of the algebra (further denoted by N). The existence of solvable radical or just radical (the maximal solvable ideal), denoted by R is carried out in a similar way as in the case of Lie algebras. The following analogue of the well-known Lie Theorem for solvable Leibniz algebras also holds true (see [93], [144]). Theorem 2.2. A right complex Leibniz algebra L has a complete flag of subspaces which is invariant under the right multiplication operator. In other words, there is a basis with respect to that all right multiplication operators R x , x ∈ L can be simultaneously given in the triangular form. Proof. For Lie algebras this theorem is well known (in one of the variants of its formulation). Now let L be an arbitrary solvable Leibniz algebra. Since the liezator I is central, then the solvability of the Leibniz algebra L is equivalent to the solvability of the Lie algebra L/I. By the classical Lie theorem L/I has a complete flag, which is invariant under the multiplication (both left and right). Let x be an element of L. Its right action on I is trivial and therefore in I there exists a complete flag, which is invariant under right multiplication (for the left multiplication on the right Leibniz algebra such reasoning does not pass).

46  Leibniz Algebras

Extending it to the full invariant flag in the quotient space until the full flag in L, we obtain a complete flag in L, which is invariant under the right multiplication operator.  Proposition 2.2. Let R be the radical of a Leibniz algebra L and let N be the nilradical of L. Then [L, R] ⊆ N. Proof. For Lie algebras the statement above is true. Now let L be an arbitrary Leibniz algebra. Then I ⊆ N ⊆ R and I ⊆ Annr (L). Consider L∗ := LLie the liezation of L. From the definitions of the radical and nilradical it follows that R∗ = R/I and N ∗ = N/I are radical and nilradical of the Lie algebra L∗ , respectively. Since L∗ is a Lie algebra we get [L∗ , R∗ ] ⊂ N ∗ . But then, by the fact that I ⊆ N ⊆ R, we obtain [L, R] ⊆ N.  Corollary 2.2. [R, R] ⊂ N. In particular, [R, R] is nilpotent. We can give the characterization of the solvability of a Leibniz algebra in terms of the nilpotence of its square. Corollary 2.3. A Leibniz algebra L is solvable if and only if [L, L] is nilpotent. Proof. In one direction the assertion of Corollary 2.3 is proved in Corollary 2.2. The converse is proved as follows. Let [L, L] be nilpotent and R be the radical of L. Consider the quotient Lie algebra L/R. Since L/R does not contain a solvable ideal it is a semisimple Lie algebra. Therefore, the quotient algebra coincides with its commutator. Since [L, L] is nilpotent the algebra L/R must be trivial, i.e., L is solvable.  2.3

ON LEVI’S THEOREM FOR LEIBNIZ ALGEBRAS

In the theory of Lie algebras there is well-known Levi’s Theorem, which asserts that any finite-dimensional Lie algebra over a field of characteristic zero is decomposed into a semidirect sum of the solvable radical and a semisimple subalgebra. Moreover, all complements to solvable radical are conjugated. In this section we show that Levi’s Theorem can be extended to Leibniz algebras, but the conjugacy property does not hold (see [32]).

Structure of Leibniz Algebras  47

Theorem 2.3. (Levi’s theorem for Leibniz algebras) For arbitrary finite-dimensional Leibniz algebra L over a field of characteristic zero there exists a subalgebra S (which is a semisimple Lie algebra), such that L = S u R, where R is the radical of L. Proof. Let us consider the Lie algebra L∗ := LLie . By the classical Levi’s theorem there is a semisimple algebra S∗ ⊆ L∗ , which gives the semidirect sum decomposition L∗ = S∗ + R∗ . Here R∗ is the radical of L∗ . Let R be the radical of the Leibniz algebra L and let π : L −→ L∗ be the natural epimorphism. Then it is clear that π(R) = R∗ . Let F = π−1 (S∗ ). Then F is a subalgebra of L containing I, and the quotient algebra F/I is isomorphic to the semisimple Lie algebra S∗ . For the Lie algebra S∗ the abelian Leibniz algebra I is an S∗ - module. For the semisimple Lie algebra S∗ , it follows that the subalgebra F, considered as an extension of S∗ by means of I, is split (by Whitehead’s Lemma for semisimple Lie algebras). But this means that in F there exists a subalgebra (semisimple Lie algebra) which is complementary to the abelian ideal I. So one has S u R = L.  The subalgebra S in the theorem above, similarly to the case of Lie algebras is called a Levi subalgebra of the Leibniz algebra L. In the splitting theorem for Lie algebras over a field of characteristic zero all the semisimple parts are mutually conjugate. But such a conjugacy for Leibniz algebras fails to be true which is illustrated by the following example of D. Barnes [32]. Example 2.3. Let S be a simple Lie algebra and K be isomorphic to S as a right S -module. Denote by x0 the element of K corresponding to x ∈ S under this isomorphism. One can make K into a Leibniz module by defining the left action to be 0. Let L be the split extension of K by S . Then L is a Leibniz algebra and I = K. The space S 1 = {(s, s0 )| s ∈ S } is another subalgebra complementing K since, using the module isomorphism, one has (t, t0 )(s, s0 ) = (ts, t0 s) = (ts, (ts)0 ). For any x = (s, k) ∈ L, the inner derivation d x = d s , d x (S ) ⊆ S and so exp(d x )(S ) ⊆ S . Thus, S and S 1 are not conjugate. Corollary 2.4. Let L be a Leibniz algebra such that its left annihilator lies in the right annihilator (i.e., [L, Annl (L)] = {0}). If the Lie algebra

48  Leibniz Algebras

L/Annl (L) is semisimple, then L is a Lie algebra and its radical is central. Proof. Let us consider the Levi decomposition L = S u R. Since, by the hypothesis, the algebra L/Annr (L) is semisimple, we have R = Annl (L). The right action of the subalgebra S on Annl (L) is always trivial. But due to [Annl (L), L] = {0}, the right action of S on Annl (L) also is trivial. Therefore, S commutes with the radical and then Levi’s decomposition of L is a direct sum. The radical of R = Annl (L) is abelian. Such Lie algebras have been called reductive Lie algebras.  The statement of Corollary 2.4 was proved in [23] for the case where S is a classical simple Lie algebra. 2.3.1

On conjugacy of Levi subalgebras of Leibniz algebras

Here we provide conditions for the Levi-Malcev theorem to hold or not (i.e., for two Levi subalgebras to be conjugate or not by an inner automorphism) in the context of finite-dimensional Leibniz algebras over a field of characteristic zero. Particularly, in the case of the field of complex numbers, we consider all possible cases for Levi subalgebras to be conjugate or not. We make use an adapted version of Schur’s Lemma (see Lemma 1.3, Section 1.4.2) given earlier. Let N be the nilradical of a Leibniz algebra L, a ∈ N and Ra be the right multiplication operator by a. Since Ra is nilpotent there exists n depending on a such that Rn+1 = 0. a We set exp(Ra )(x) = x + Ra (x) +

R2a (x) Rn (x) + ··· + a . 2! n!

Then exp(Ra ) is an automorphism of L. Similarly to the case of Lie algebras exp(Ra ) is called an inner automorphism. It is easy to see that the composition of inner automorphisms again is an inner automorphism and they form a subgroup of the group of all automorphisms of L. Let us start with the following elementary result on invariantness of the ideals I, N and R under automorphisms of a Leibniz algebra L. Proposition 2.3. For any ϕ ∈ Aut(L) we have ϕ(I) = I,

ϕ(N) = N,

ϕ(R) = R.

Structure of Leibniz Algebras  49

Proof. Let ϕ ∈ Aut(L). The inclusion ϕ(I) ⊆ I is obvious. Let us P prove that I ⊆ ϕ(I). Let x = αi [xi , xi ] ∈ I. For P each xi ∈ L there exists y ∈ L such that ϕ(y ) = x . Due to x = αi [xi , xi ] = i i i P P αiP [ϕ(yi ), ϕ(yi )] = ϕ( αi [yi , yi ]), we conclude x = ϕ(y), where y = αi [yi , yi ] ∈ I i.e., I ⊆ ϕ(I). Hence, ϕ(I) = I. Due to isomorphism properties we easily get that for any k ∈ N and an ideal J of L the following equalities are true: ϕ(J)k = ϕ(J k ),

ϕ(J)[k] = ϕ(J [k] ).

Consequently, ϕ(N) and ϕ(R) are nilpotent and solvable radicals, respectively.  ˙ its Levi decomposition and Let L be a Leibniz algebra, L = S+R let ϕ be an automorphism of L. Then due to Proposition 2.3 we get ϕ = ϕS,S + ϕS,R + ϕR,R , where ϕS,S = ProjS (ϕ|S ),

ϕS,R = ProjR (ϕ|S ),

ϕR,R = ϕ|R .

Denote by HomS (S, I) the set of all S-module homomorphisms from S into I. We set Sθ = {x + θ(x) : x ∈ S, θ ∈ HomS (S, I)} .

(2.1)

Since [x + θ(x), y + θ(y)] = [x, y] + [θ(x), y] = [x, y] + θ([x, y]), it follows that the mapping x 7→ x+θ(x) ∈ Sθ , x ∈ S, is an isomorphism from S onto Sθ . Remark. Further we shall consider only automorphisms of the form exp(Ra ) since results can be easily extended for their compositions (inner automorphisms). Theorem 2.4. Let S and S1 be two Levi subalgebras of a Leibniz algebra L. Then there exist τ ∈ HomS (S, I) and an element a ∈ N such that exp(Ra )(Sτ ) = S1 . e where R e = R/I. Due to Levi–Malcev theorem Proof. Let e L = S+˙ R, e = N/I such for Lie algebras (see [123]), there exists an element e a∈N

50  Leibniz Algebras

that exp(Rea )(S) = S1 . For an element a from the class e a, we consider e the automorphism exp(Ra ) of L. Let π : L → L be the quotient map. ]a ) Since any automorphism maps I into itself, the quotient map exp(R is well defined, i.e., ]a )(π(x)) = π exp(Ra )(x)
, x ∈ L. exp(R ]a ) is an automorphism of e Then exp(R L. For x ∈ S we have  n  X Rka (x) 
]a )(x) = exp(R ]a )(π(x)) = π exp(Ra )(x) = π   exp(R k! k=0 =

n X Rkπ(a) (π(x)) k=0

k!

=

n X Rk (x) e a

k=0

k!

= exp(Rea )(x).

]a ) = exp(Rea ) . Therefore, exp(R S S Thus, for any x ∈ S there exists an element τ(x) ∈ I such that
exp(Ra )−1 exp(Rea )(x) = x + τ(x). For x, y ∈ S we consider
[x, y] + τ([x, y]) = exp(Ra )−1 exp(Rea )([x, y]) h

i = exp(Ra )−1 exp(Rea )(x) , exp(Ra )−1 exp(Rea )(y)   = x + τ(x), y + τ(y) = [x, y] + [τ(x), y], i.e., τ ∈ HomS (S, I). Then, exp(Rea )(x) = exp(Ra )(x + τ(x)) for all x ∈ S implies S1 = exp(Ra )(Sτ ).



The following result follows from Theorem 2.4 and it asserts that the question on the conjugacy of Levi subalgebras is reduced to verification of the conjugacy of Levi subalgebras S and Sθ , where Sθ is the algebra form (2.1). Corollary 2.5. A Levi subalgebra S is unique up to conjugation by an inner automorphism of L if and only if for every τ ∈ HomS (S, I) there exists an element b ∈ N such that exp(Rb ) S = idS + τ.

Structure of Leibniz Algebras  51

Proof. Sufficiency. Suppose that S is unique up to conjugation by an inner automorphism of L. Take an arbitrary τ ∈ HomS (S, I) and consider the Levi subalgebra Sτ . Since S is unique up to conjugation by an inner automorphism, there exists an element an element a ∈ N such that exp(Rb )(S) = Sτ and therefore exp(Rb ) S = idS +τ. Necessity. Let S and S1 be two Levi subalgebras. By Theorem 2.4 there exist τ ∈ HomS (S, I) and an element a ∈ N such that exp(Ra )(Sτ ) = S1 . By the assumption there exists an element b ∈ N such that exp(Rb ) S = idS +τ, i.e., exp(Rb )(S) = Sτ . This implies that exp(Ra ) ◦ exp(Rb )(S) = S1 .  In the next two propositions we give sufficient conditions of the conjugacy of Levi subalgebras. Proposition 2.4. Let S be a Levi subalgebra of a Leibniz algebra L and HomS (S, I) = {0}. Then S is unique up to conjugation by an inner automorphism of L. Proof. Taking into account the condition HomS (S, I) = {0} in Theorem 2.4, we obtain Sτ = S and therefore, exp(Ra )(S) = S1 .  Remark. Note that the equality [I, S] = {0} implies HomS (S, I) = {0}. Therefore, due to Proposition 2.4 we conclude that Levi sub˙ satisfying the condition algebras of a Leibniz algebra L = S+R [I, S] = {0} are conjugate via an inner automorphism. Note that an analogue of this result also has been obtained by G. Mason and G. Yamskulma in [126]. Thanks to Proposition 2.4, henceforth we shall consider only the case when HomS (S, I) , {0}. ˙ be a Levi decomposition of a Leibniz algebra L. A Let L = S +R Levi subalgebra S and the S-module I can be uniquely represented as follows S = G ⊕ Q and I = J ⊕ K, where J = {i ∈ I | ∃ ϕ ∈ HomS (S, I) such that i ∈ Imϕ}, K = {i ∈ I | ∀ϕ ∈ HomS (S, I) such that i < Imϕ}, G = {x ∈ S | ∃ ϕ ∈ HomS (S, I) such that ϕ(x) ∈ J},

52  Leibniz Algebras

Q = {x ∈ S | ∀ ϕ ∈ HomS (S, I) such that ϕ(x) = 0}. Clearly, for any ϕ ∈ HomS (S, I) we have ϕ = proj J (ϕ|G ). Therefore, we can identify HomS (S, I) with HomS (G, J) (we denote it further by HomS (S, I) ≡ HomS (G, J)). Let now assume that [J, R] = {0}. For θ ∈ HomS (S, I) we set δθ (xS + xR ) = θ(xS ),

˙ x = xS + xR ∈ S+R.

Then δθ is a derivation of L. Indeed,     δθ (x), y + x, δθ (y) = [θ(xS ), yS + yR ] + [x, θ(yS )] = [θ(xS ), yS ] + [θ(xS ), yR ] = [θ(xS ), yS ] = θ([xS , yS ]) = δθ ([x, y]). It is clear that δ2θ = 0. Hence, exp(δθ )(x) = x + δθ (x), x ∈ L and exp(δθ )(x) = x + θ(x), x ∈ S. Thus, we obtain exp(δθ )(S) = Sθ . Let us define a nilpotent derivation D as follows D = Ra + δθ ,

a ∈ N,

θ ∈ HomS (S, I).

(2.2)

Then one has ˙ be a Levi decomposition of a Leibniz Proposition 2.5. Let L = S+R algebra L such that [J, R] = {0}. Then any two Levi subalgebras S and S1 are conjugate by exp(D), where D has the form (2.2). Proof. By Theorem 2.4 there exist τ ∈ HomS (S, I) and an element a ∈ N such that exp(Ra )(Sτ ) = S1 . Since [J, R] = {0}, it follows that δτ is a derivation with δ2τ = 0. For the derivation D = Ra + δτ , we have exp(D)(S) = exp(Ra + δτ )(S) = (exp(Ra ) ◦ exp(δτ ))(S) = exp(Ra )(exp(δτ )(S)) = exp(Ra )(Sτ ) = S1 . This completes the proof.



A version (when J = I) of Proposition 2.5 above has been proved in [126]. Note that Example 2.3 satisfies the conditions of Proposition 2.5. Now we consider the cases of Leibniz algebras with Levi subalgebras which are not conjugate via automorphisms.

Structure of Leibniz Algebras  53

˙ be a Levi decomposition with condiProposition 2.6. Let L = S+R tions [S, R] = {0} and [J, R] , {0}. Then there exists a Levi subalgebra S1 such that the algebras S and S1 are not conjugate via any automorphism. Proof. Due to [J, R] , {0} we have the existence of elements i ∈ J and y ∈ R such that [i, y] , 0. Consider an S-module homomorphism θ from S to J such that there exists x ∈ S with the condition θ(x) = i. We set S1 = {x + θ(x) : x ∈ S } . Let assume the existence of an automorphism ϕ = ϕS,S + ϕS,R + ϕR,R ∈ Aut(L) such that ϕ(S) = S1 . Then for x ∈ S we have ϕS,S (x)+ϕS,R (x) = ϕ(x) ∈ S1 ⊂ S + J. Hence ϕS,R (x) ∈ J and ϕS,R = ϕS,J . Since ϕ(R) = R there exists z ∈ R such that ϕR,R (z) = y. Due to [S, R] = {0}, we have 0 = ϕ([x, z]) = [ϕ(x), ϕ(z)] = [ϕS,S (x) + ϕS,J (x), ϕR,R (z)] = [ϕS,J (x), ϕR,R (z)] = [i, y] , 0. This contradiction shows that there is no automorphism of L which maps S onto S1 .  ˙ be a Levi decomposition of a Lie algebra K satisfyLet K = S+R ing the following conditions: R = N + Span{p}, for any 0 , x ∈ N. We set

N = [R, R],

I = S,

[S, p] = {0},

[S, x] , {0},

θ = id|S : S → I.

˙ +I, ˙ we define products as follows On the space L = S+R [L, I] = [I, N] = 0,

[θ(x), y] = θ([x, y])

for x, y ∈ S, [i, p] = i, for i ∈ I. A straightforward verification shows that L is a Leibniz algebra. Note that S and I are isomorphic S-modules via the isomorphism θ. This phenomenon is not accidental.

54  Leibniz Algebras

Proposition 2.7. Leibniz algebra L constructed above admits two Levi subalgebras which are non conjugate by any automorphism of L. Proof. Consider the Levi subalgebra Sθ = {x + θ(x) : x ∈ S}. Suppose that there is ϕ = ϕS,S +ϕS,R+I +ϕR+I,R+I ∈ Aut(L) such that ϕ(S) = Sθ . Similarly to that of the proof of Proposition 2.6 we obtain ϕS,R+I = ϕS,I , 0. Taking into account ϕ(N) = N we get ϕ(p) = b + c, where b = αp , 0 and c ∈ N. For an element 0 , x ∈ S we have 0 = ϕ(0) = ϕ([x, p]) = [ϕ(x), ϕ(p)] = [ϕS,S (x), ϕR+I,R+I (p)] + [ϕS,I (x), ϕR+I,R+I (p)] = [ϕS,S (x), b + c] + [ϕS,I (x), b + c] = [ϕS,S (x), c] + [ϕS,I (x), b] = [ϕS,S (x), c] + αϕS,I (x). But [ϕS,S (x), c] + αϕS,I (x) , 0, because [ϕS,S (x), c] ∈ N and 0 , ϕS,I (x) ∈ I. Thus, we get a contradiction with the existence of an automorphism of L mapping S onto Sθ .  Example 2.4. Let us consider the following Leibniz algebra L = Span{e1 , e2 , e3 , y4 , y5 , y6 , x7 , x8 , x9 } with the multiplication table: [e1 , e2 ] [e2 , e3 ] [e1 , y4 ] [e2 , y5 ] [y4 , y6 ]

= = = = =

−[e2 , e1 ] = 2e2 , [e1 , e3 ] = −[e3 , e1 ] = −2e3 , −[e3 , e2 ] = e1 , −[y4 , e1 ] = y4 , [e1 , y5 ] = −[y5 , e1 ] = −y5 , −[y5 , e2 ] = y4 , [e3 , y4 ] = −[y4 , e3 ] = y5 , −[y6 , y4 ] = y4 , [y5 , y6 ] = −[y6 , y5 ] = y5 .

[x7 , e1 ] = −2x7 , [x8 , e3 ] = −2x9 , [x9 , e2 ] = −x8 , [x8 , y6 ] = x8 ,

[x7 , e3 ] = x8 , [x8 , e2 ] = 2x7 , [x9 , e1 ] = 2x9 , [x7 , y6 ] = x7 , [x9 , y6 ] = x9 .

˙ +I), ˙ Then L = S+(R where S = Span{e1 , e2 , e3 }  sl2 , R = Span{y4 , y5 , y6 } and I = Span{x7 , x8 , x9 }. Note that N = Span{x7 , x8 , x9 , y4 , y5 } and [S, y6 ] = {0}. The Leibniz algebra L satisfies the conditions of Proposition 2.7.

Structure of Leibniz Algebras  55

2.3.2

On conjugacy of Levi subalgebras of complex Leibniz algebras

In this subsection we shall discuss conditions for conjugacy of Levi subalgebras of Leibniz algebras over the field C. First we prove the following auxiliary lemma. Lemma 2.1. Let b be an element of R such that [S, b] ⊆ I. Then n Rb S ∈ HomS (S, I) for all n ∈ N. Proof. Since [y, b] ∈ I and [L, I] = {0}, for any x, y ∈ S we have Rb ([x, y]) = [[x, y], b] = [[x, b], y] + [x, [y, b]] = [[x, b], y] = [Rb (x), y]. Now assume that Rn−1 is an S-module homomorphism, n ≥ 2. Then b n−1 n Rnb ([x, y]) = Rb (Rn−1 b ([x, y])) = Rb ([Rb (x), y]) = [Rb (x), y].

This completes the proof.



˙ be a Levi decomposition of a Proposition 2.8. Let L = S+R complex Leibniz algebra L such that [S, E] = I, where E = {b ∈ N : [S, b] ⊆ I} . Then [I, E] = 0. Proof. Let us first consider the case when S and I are simple Smodules. Fix a non-zero element b ∈ E such that [S, b] , 0. Taking into account that S and I are simple S-modules, we get a nonzero module homomorphism Rb |S . Applying Schur’s Lemma (see Lemma 1.3) we derive that the modules S and I are isomorphic and the isomorphism Rb |S is unique up to a constant. Take a non-trivial τ ∈ HomS (S, I). Then by Schur’s Lemma we have τ = α Rb |S , α ∈ C. 2 By Lemma 2.1, Rb S is also an S-module isomorphism and again ap 2 plying Schur’s lemma we get Rb S = λ Rb |S for some λ ∈ C. Since b ∈ N, it follows that Rb is nilpotent, and therefore there exists n ∈ N n−1 n n such that Rb = 0. Then 0 = Rb S = λ Rb |S , which implies λ = 0, due to Rb |S , 0. Thus R2b = 0. Taking an arbitrary y ∈ I, we conclude that there exists an element x ∈ S such that Rb (x) = [x, b] = y. Moreover, [y, b] = [[x, b], b] = R2b (x) = 0. This means that [I, E] = 0.

56  Leibniz Algebras

Let us now consider the case when S is semisimple and I is a simple S-module. Suppose that we have the decomposition S into the n L sum of simple Lie ideals: S = S i . Since I is irreducible, it follows i=1

that there exists an index i such that [Si , E] = I. By the above case we obtain [I, E] = 0. m L Finally, we consider the general case. Let I = I j be a decomj=1

˙ i , where position into sumL of irreducible S-modules. Set Li = S+R Ri = R/Ji , Ji = I j and i = 1, . . . , m. Then [S, Ei ] = Ii , for all j,i

i = 1, . . . , m, where Ei = E/Ji . By the above case we have [Ii , Ei ] = 0, for i = 1, . . . , m. Thus [Ii , E] = [Ii , Ei + Ji ] = [Ii , Ei ] = 0 and

 m  m h M  M i [I, E] =  I j , E = I j , E = 0. j=1

j=1

 Now we provide an example which satisfies the conditions of Proposition 2.8. Example 2.5. Let L be a Leibniz algebra given by the following table of multiplication on a basis {e, f, h, x0 , x1 , x2 , y1 , y2 } (see [44]): [e, h] = 2e, [h, e] = −2e, [x1 , e] = −2x0 , [x1 , f ] = x2 , [e, y1 ] = 2x0 , [y1 , y2 ] = y1 , [x0 , y2 ] = x0 ,

[h, f ] = 2 f, [ f, h] = −2 f, [x2 , e] = −2x1 , [x0 , h] = 2x0 , [ f, y1 ] = x2 , [y2 , y1 ] = −y1 , [x1 , y2 ] = x1 ,

[e, f ] = h, [ f, e] = −h, [x0 , f ] = x1 , [x2 , h] = −2x2 , [h, y1 ] = 2x1 , [x2 , y2 ] = x2 .

˙ It is easy to see that L = S+R, where S = Span{e, f, h}, R = Span{x0 , x1 , x2 , y1 , y2 }, I = Span{x0 , x1 , x2 } and [S , E] = I, with E = Span{x0 , x1 , x2 , y1 }.

Structure of Leibniz Algebras  57

˙ be a Leibniz algebra and Let L = S+R S=G⊕Q

I= J⊕K

and

(2.3)

be the decompositions from Section 2.3.1. Let θ be an S-module homomorphism such that θ(G) = J. Suppose that n M Gi G= i=1

is a decomposition into the sum of simple Lie ideals. Set Ji = θ(Gi ) for all i ∈ {1, . . . , n}. Since Gi is an ideal in G, it follows that Ji is a submodule of J. Take z = θ(xi ) = θ(x j ) ∈ Ji ∩ J j , where i , j. Let yk ∈ Gk . If k , i, then [z, yk ] = [θ(xi ), yk ] = θ([xi , yk ]) = 0. If k , j, then [z, yk ] = [θ(x j ), yk ] = θ([x j , yk ]) = 0. h i Thus Ji ∩ J j , G = 0, and therefore the intersections Ji ∩ J j are trivial submodules. Since θ maps G onto J, it follows that J=

n M

Ji ,

i=1

and each submodule is decomposed as Ji =

ni M

J (i) j ,

j=1

where J (i) j is a simple G i -module. Since Ji = θ(G i ), it follows that G i  (i) J j for all j ∈ {1, . . . , ni } and i ∈ {1, . . . , n}. ˙ be the Levi decomposition of a complex Proposition 2.9. Let L = S+R Leibniz algebra L such that [S, E] = J. Then any two Levi subalgebras of L are conjugate via an inner automorphism.

58  Leibniz Algebras

Proof. Let τ ∈ HomS (S, J). By Lemma 1.3 we have τ =

ni n P P

τi j ,

i=1 j=1

where τi j : Gi → J (i) j are module homomorphisms. For i, j take an element bi, j ∈ N such that 0 , [Gi , bi, j ] ⊆ J (i) exists λi j ∈ C j . Then there 2 such that τi, j = λi, j Rbi, j . By Proposition 2.8 we get Rbi, j S = 0. ni n P P Set b = λi, j bi j . Then i=1 j=1

ni ni n X n Y X Y Rλi, j bi j S = id|S + τ. exp(Rλi, j bi j ) S = id|S + exp(Rb ) S = i=1 j=1

i=1 j=1

Finally, to get the proof we need just to apply Corollary 2.5.



Now we give the criterion on conjugacy of the Levi subalgebras. ˙ be its Levi Theorem 2.5. Let L be a complex Leibniz and L = S+R decomposition. Then the following are equivalent: i) A Levi subalgebra S is unique up to conjugacy via an inner automorphism; ii) [S, E] = J. Proof. The implication (ii) ⇒ (i) follows from Proposition 2.9. Now we prove the implication (i) ⇒ (ii). Let S = G ⊕ Q be a decomposition of the form (2.3). We first consider the case where G is n L a simple Lie algebra. Let J = Ji be a decomposition into simple i=1

modules and G  Ji for all i = 1, . . . , n. Take an arbitrary module isomorphism τ j : G → J j . Since any two subalgebras of L are conjugate, there exists an element b j ∈ N such that exp(Rb j ) G = id|G + τ j . Similarly as in the proof of Proposition 2.8, we obtain R2b j G = 0, i.e., τ j = Rb j G , and therefore [G, b j ] = J j . Thus J ⊇ [G, E] ⊇ [G, b1 ] + · · · + [G, bn ] = J. Hence [G, E] = J.

Structure of Leibniz Algebras  59

Now let us consider the general case. Let G be a semisimple Lie m L algebra and J = J j be the decomposition into the sum of irreducible j=1 L ˙ i , where Ri = R/Ii and Ii = G-modules. Set Li = S +R J j for j,i

i = 1, . . . , m. Then HomS (S, Ji ) ≡ HomS (Gi , Ji ).

(2.4)

Since a Levi subalgebra S of L is unique up to conjugacy via an inner automorphism and the quotient of inner automorphism is also inner, it follows that S is unique up to conjugacy via an inner automorphism as Levi subalgebra of Li , for all i = 1, . . . , m. Taking into account (2.4) and the fact that Gi is a simple algebra, by the above case we obtain [G, Ei ] = Ji , for all i = 1, . . . , m, where Ei = E/Ji . Thus [Si , E] = [Si , Ei + Ii ] = Ji . Further  m  m m h M  M i M   [S, E] =  S j , E = S j, E = J j = J. j=1

j=1

j=1

 Remark. Note that in Proposition 2.7 and Example 2.4, we have E = 0. Therefore, 0 = [S, E] , J. ˙ be a Levi decomposition of a complex Corollary 2.6. Let L = S+R Leibniz algebra L such that [S, N] = J. Then any two Levi subalgebras of L are conjugate via an inner automorphism. Recall that J is the maximal submodule of I such that HomS (S, I) ≡ HomS (S, J). In Figure 2.1 we describe the cases that may occur:

60  Leibniz Algebras

J J=0

J,0

[S, E] = J

[S, R] , 0, [S, E] , J

[S, R] = 0, [J, R] , 0

no conjugation

no conjugation

Prop. 2.7 Exam. 2.4

Prop. 2.6

[J, R] = 0

F=C conjugation by inner automorphisms

conjugation by inner automorphisms

conjugation by noninner automorphisms Prop. 2.5

Prop. 2.4 Figure 2.1

2.4

Thm. 2.5

Conjugacy of Levi’s subalgebras

SEMISIMPLE LEIBNIZ ALGEBRAS

Although Theorem 2.3 reduces the study of finite-dimensional Leibniz algebras to semisimple Lie and solvable Leibniz algebras the study of “simple” and “semisimple” Leibniz algebras are of interest in their own right. In fact, in a non-Lie Leibniz algebra the ideal I is nontrivial. That is why usual notion of simplicity is not applicable to the case of Leibniz algebras. In [63] the authors have proposed the notion of simple Leibniz algebra as follows. Definition 2.2. A Leibniz algebra L is said to be simple if it contains only the following ideals: {0}, I, L and L2 , I. Note that the definition of simplicity of Leibniz algebra above agrees with that of simplicity of Lie algebra, since in this case I = 0. A simple application of Levi’s decomposition for finite-dimensional Leibniz algebras implies that the above definition can be rephrased in terms of quotient Lie algebra as follows: a Leibniz algebra L is simple if and only if its Lie algebra LLie is simple and the ideal I is an irreducible right LLie -module.

Structure of Leibniz Algebras  61

Example 2.6. 1. Let L := sl2 (C) × L(1), where sl2 (C) is the Lie algebra of traceless complex 2 × 2 matrices and L(1) is the two-dimensional left sl2 (C)-module. Then L with the multiplication defined by (X, a)(Y, b) := ([X, Y], X · b) for any X, Y ∈ sl2 (C) and any a, b ∈ L(1) is a simple left Leibniz algebra (see [68]). 2. The example above is generalized as follows. Let G be a (simple) Lie algebra and V a right (irreducible) G-module. Endow the vector space L = G ⊕ V with the bracket product as follows: [(g1 , v1 ), (g2 , v2 )] := ([g1 , g2 ], v1 · g2 ), where v·g (sometimes denoted as [v, g]) is an action of an element g of G on v ∈ V. Then straightforward verification of the Leibniz ˙ is a (simple) Leibniz algebra. identity shows that L = G+V Note that from Levi’s theorem it is immediate that any simple Leibniz algebra can be obtained in the manner of part 2 of Example 2.6. The following definition and results on Lie algebras from [105], given there as Proposition 19.1 and Corollary 21.2, will be applied in the description of semisimple Leibniz algebras. Definition 2.3. A Lie algebra G is said to be reductive, if R = Cent(G), where R is the solvable radical of G. Lemma 2.2. i) Let G be a reductive Lie algebra. Then G = [G, G] ⊕ Cent(G), and [G, G] is either semisimple or 0. ii) Let V be a finite-dimensional space and G ⊂ gl(V) be a non zero Lie algebra acting irreducibly on V. Then G is reductive with dim Cent(G) ≤ 1. If, in addition, G ⊂ sl(V), then G is semisimple. We shall also need the following result from [180, Page 143, Theorem 6]. Theorem 2.6. Let G = G1 ⊕ G2 be a semisimple Lie algebra. Let a Cartan subalgebra H of G be respectively decomposed as H = H1 ⊕ H2 , where Hi is a Cartan subalgebra of Gi , (i = 1, 2). Suppose that the simple root system Π of G with respect to H is decomposed into Π = Π1 ∪ Π2 , where Π1 = {α1 , . . . , αn } and Π2 = {β1 , . . . , βm } are the fundamental root system of Gi with respect to Hi , i = 1, 2.

62  Leibniz Algebras

a) Suppose that %i is an irreducible representation of Gi with the highest weight ωi and the representation space Vi , i = 1, 2. If %(x)(v1 ⊗ v2 ) = %1 (x1 )v1 ⊗ v2 + v1 ⊗ %2 (x2 )v2 , x = x1 + x2 , xi ∈ Gi , i = 1, 2, then % is an irreducible representation of G with the highest weight ω1 + ω2 , where (ω1 + ω2 )(h) = ω1 (h1 ) + ω2 (h2 ), h = h1 + h2 , hi ∈ Hi , i = 1, 2. b) Conversely, every irreducible representation of G is obtained as above. The concept of semisimple Leibniz algebra is defined as follows. Definition 2.4. A Leibniz algebra L is called semisimple if its radical coincides with I. In other words, a Leibniz algebra L is semisimple if its liezation LLie is a semisimple Lie algebra. Note that the definition of semisimple Leibniz algebra agrees with that of Lie algebra (since in that case I = 0). Clearly, a simple Leibniz algebra is semisimple. Application of Theorem 2.3 gives the following Proposition 2.10. Let L be a finite-dimensional semisimple Leibniz ˙ where S is a semisimple Lie subalgebra of L. algebra. Then L = S+I, Moreover [I, S] = I. Now the structure of a simple Leibniz algebra is clearly seen. By Example 2.6 a simple Leibniz algebra is a semidirect sum of a simple Lie algebra S and an irreducible right module over S. Note that a semisimple Leibniz algebra need not necessarily be presented as a direct sum of simple Leibniz algebras as justified by the following example. Example 2.7. Let L be a 10-dimensional Leibniz algebra given on a basis {e1 , h1 , f1 , e2 , h2 , f2 , x1 , x2 , x3 , x4 } by the multiplication table as follows [sli2 , sli2 ] :

[ei , hi ] = 2ei , [ fi , hi ] = −2 fi , [ei , fi ] = hi , [hi , ei ] = −2ei [hi , fi ] = 2 fi , [ fi , ei ] = −hi , i = 1, 2,

Structure of Leibniz Algebras  63

[I, sl12 ] :

[x1 , f1 ] = x2 , [x1 , h1 ] = x1 , [x2 , e1 ] = −x1 , [x2 , h1 ] = −x2 , [x3 , f1 ] = x4 , [x3 , h1 ] = x3 , [x4 , e1 ] = −x3 , [x4 , h1 ] = −x4 ,

[I, sl22 ] :

[x1 , f2 ] = x3 , [x1 , h2 ] = x1 , [x3 , e2 ] = −x1 , [x3 , h2 ] = −x3 , [x2 , f2 ] = x4 , [x2 , h2 ] = x2 , [x4 , e2 ] = −x2 , [x4 , h2 ] = −x4 ,

(omitted products of the basis vectors are equal to zero). It is easy to verify that L is a semisimple Leibniz algebra and I = Span{x1 , x2 , x3 , x4 }. From the table of multiplications we have [I, sl12 ] = [I, sl22 ] = I. Moreover, I splits over sl12 as I = Span{x1 , x2 }⊕ Span{x3 , x4 } and over sl22 as I = Span{x1 , x3 } ⊕ Span{x2 , x4 }. In fact, I is a simple ideal (an irreducible (sl12 ⊕ sl22 )-module). Therefore, ˙ cannot be a direct sum of two non zero ideals. L = (sl12 ⊕ sl22 )+I We give a generalization of Theorem 2.6, which is of independent interest. Theorem 2.7. Let G = G1 ⊕G2 be a direct sum of complex Lie algebras and suppose that V is a finite-dimensional irreducible G-module with dimC V > 1. Then there are finite-dimensional irreducible Gi -modules Vi , i = 1, 2 such that V  V1 ⊗ V2 . Proof. Consider Ann(V) = {x ∈ G : V · x = 0} which is an ideal of G. Then V is a faithful irreducible module over G = G/Ann(V), in particular, for an element x ∈ G from v · x = 0 for all v ∈ V we get x = 0. Thus we have an injective linear map ρ : G → gl(V) given by ρ(¯g)(v) = v · g¯ for any v ∈ V and g¯ ∈ G (a representation of G on V. Therefore, G is also finite-dimensional. Applying part b) of Lemma 2.2 to G we conclude that G is reductive, dim Cent(G) ≤ 1, and G = [G, G] ⊕ Cent(G). Denote the natural projections by ρ1 : G → [G, G] and ρ2 : G → Cent(G), and define π = ρ1 ◦ ρ. Assume that Cent(G) = Cz , 0. Since [[v, x], z] = [[v, z], x] for all x ∈ G and v ∈ V, it follows that the mapping v ∈ V ,→ [v, z] ∈ V is a G-module homomorphism of V. Taking into account that V is an irreducible G-module, by Schur’s Lemma, there exists λ ∈ C such that [v, z] = λv for all v ∈ V. Note that V is a simple module over

64  Leibniz Algebras

π(G) = [G, G]. If π(G1 ) ∩ π(G2 ) , {0}, it has to be semisimple. We note that π(G1 ) ∩ π(G2 ) = [π(G1 ) ∩ π(G2 ), π(G1 ) ∩ π(G2 )}] ⊂ [π(G1 ), π(G2 )}] = 0, which is a contradiction. Thus π(G1 ) ∩ π(G2 ) = 0, and π(G) = π(G1 )⊕π(G2 ). Since V is also an irreducible module over π(G1 )⊕π(G2 ), both π(G1 ) and π(G2 ) are semisimple. By Theorem 2.6, we deduce that there are finite-dimensional irreducible π(Gi )-modules Vi so that V  V1 ⊗ V2 . We can easily extend π(Gi )-module Vi into a Gi -module. This completes the proof.  ˙ is decomWe say that a semisimple Leibniz algebra L = S+I ˙ 2 are ˙ 1 and S2 +I ˙ 2 ) , where S1 +I ˙ 1 ) ⊕ (S2 +I posable, if L = (S1 +I non-trivial semisimple Leibniz algebras. Otherwise, we say that L is indecomposable. Lemma 2.3. Any semisimple Leibniz algebra has the form: L=

n M

˙ i) , (Si +I

i=1

˙ i is an indecomposable semisimple Leibniz algebra for all where Si +I i ∈ {1, . . . , n}. ˙ be a semisimple Leibniz algebra and S = Proof. Let L = S+I

n L

Si

i=1

be a decomposition of simple Lie ideals Si . Note that [I, S] = I. We proceed the proof by induction on n. ˙ is an Let n = 1, i.e., S is a simple Lie algebra. Then L = S+I indecomposable semisimple Leibniz algebra. Suppose that the assertion of the theorem is true for all numbers n L less than n and S = Si . Consider a partition of the set {1, 2, . . . , n} = i=1

A ∪ B into union of disjoint subsets A and B. Set   " # L  L  IA = I, Si and IB = I, S j  . i∈A

j∈B

Case 1. Let IA ∩ IB , {0} for any non trivial partition A ∪ B of the set {1, 2, . . . , n}. In this case L is indecomposable.

Structure of Leibniz Algebras  65

Case 2. Let IA ∩ IB = {0} for some partition A ∪ B of the set {1, 2, . . . , n}. Then L is decomposable and      M  M ˙ A  ⊕  ˙ B  . L =  Si +I S j +I i∈A

j∈B

L ˙ A and By the hypothesis of the induction, the algebras Si +I i∈A L ˙ B can be represented as a direct sum of indecomposable alS j +I j∈B

gebras.



Further we need the following auxiliary result. Lemma 2.4. Let S be a semisimple Lie algebra and I be an irreducible S-module. Then [I, S] = 0 if and only if dim I = 1. Proof. Let ρ be the representation of S on I given by ρ(g)(v) = [v, g] for any v ∈ I and g ∈ S. Then we have the Lie algebra homomorphism ρ : S → gl(I). So ρ(S) is a subalgebra of gl(I). Then the result follows from Lemma 2.2.  Remark. From Lemma 2.4 we conclude that for an indecomposable ˙ the inequality dim I ≥ 2 holds semisimple Leibniz algebra L = S+I true if I , 0. From Lemma 2.3, we need only to determine the structure of indecomposable semisimple Leibniz algebras. Suppose that L is an indecomposable semisimple Leibniz algebra with a given Levi decompo˙ (see Theorem 2.3). We may assume that S = ⊕mi=1 S i sition L = S+I and I = ⊕ni=1 Ii , where each Si is a simple Lie algebra and each Ii is an irreducible S-module. We say that Si and S j are adjacent if there exists Ik such that [Ik , Si ] = [Ik , S j ] = Ik . We say that Si and S j are connected if there exist Sk1 = Si , Sk2 , · · · , Skr = S j such that Skl and Skl+1 are adjacent. Now we can obtain our main result on indecomposable semisimple Leibniz algebras. ˙ be an indecomposable semisimple Leibniz Theorem 2.8. Let L = S+I algebra with I , {0}. Then (a) S = ⊕mi=1 Si where each Si is a simple Lie algebra;

66  Leibniz Algebras

(b) I = ⊕ni=1 Ii where each Ii is an irreducible S-module with [Ii , S] = Ii ; (c) For any 1 ≤ i ≤ m and 1 ≤ j ≤ n there is an irreducible Si module Ii j such that Ii = ⊗mj=1 Ii j ; (d) Any two Si and S j are connected. Proof. Parts (a) and (b) are clear, (c) follows from Theorem 2.6. (d) The proof is carried out by induction on n as follows. Let n = 1. Then I is an irreducible S-module. By Theorem 2.6 for each j ∈ {1, . . . , m} there is an irreducible S j -module J j such that I = ⊗mj=1 J j . ˙ is indecomposable, Since S+I Lemma 2.4 implies that [J j , S j ] = J j h i for all j. Thus I, S j = I for all j. This means that Si and S j are adjacent. Suppose that the assertion (e) is true for all numbers less than n ˙ where J = ⊕n−1 (n > 1). Consider the Leibniz algebra S+J, i=1 Ii . Assume ˙ that S+J is indecomposable. By the hypothesis of the induction, Si and S j are already connected. ˙ is decomposable. Let us consider the deNow assume that S+J ˙ into a direct sum of indecomposable semisimcomposition of S+J p L
ple Leibniz algebras: ⊕i∈At Si +˙ ⊕ s∈Bt I s , where A1 , . . . , A p and t=1

B1 , . . . , B p are partitions of {1, . . . , m} and {1, . . . , n − 1}, respectively. ˙ is indecomposable, for every s ∈ {1, . . . , p} there exists Since L = S+I an index i s ∈ A s such that [In , Sis ] = In . Thus Si1 , . . . , Si p are mutually adjacent. On the other hand, since ⊕i∈At Si +˙ ⊕ s∈Bt I s is indecomposable, by the hypothesis of the induction, for each i, j ∈ A s (1 ≤ s ≤ p) the algebras Si and S j are connected. Thus Si and S j are connected for every i, j ∈ {1, . . . , m}.  Here we present an example that generalizes Example 2.7 (see [45, Theorem 4.2]) of a semisimple Leibniz algebra which can not be decomposed into the direct sum of simple ideals.  L  ˙ be a semisimExample 2.8. Let m, n ∈ N. Let L = sl12 sl22 +I ple Leibniz algebra with a basis {e1 , h1 , f1 , e2 , h2 , f2 , xi ⊗ y j : i = 0, 1, · · · , m; j = 0, 1, · · · , n} and the table of multiplication given as

Structure of Leibniz Algebras  67

follows

[ei , hi ] = −[hi , ei ] = 2ei , [hi , fi ] = −[ fi , hi ] = 2 fi , [ei , fi ] = −[ fi , ei ] = hi , [xi ⊗ y j , h1 ] = (m − 2i)xi ⊗ y j , [xi ⊗ y j , h2 ] = (n − 2 j)xi ⊗ y j , [xi ⊗ y j , e1 ] = ixi+1 ⊗ y j , [xi ⊗ y j , e2 ] = jxi ⊗ y j+1 , [xi ⊗ y j , f1 ] = (m − i)xi−1 ⊗ y j , [xi ⊗ y j , f2 ] = (n − j)xi ⊗ y j−1 .

Consider V1 = span{x0 , x1 , . . . , xm } and V2 = span{y0 , y1 , . . . , yn }. Then V1 and V2 are irreducible (m + 1)-dimensional and (n + 1)-dimensional modules over sl12 and sl22 , respectively. It is obvious that  M  ˙ 1 ⊗ V2 ). L = sl12 sl22 +(V 2.5 ON CARTAN SUBALGEBRAS OF FINITE-DIMENSIONAL LEIBNIZ ALGEBRAS

The subsets and

Norl (H) = {x ∈ L | [x, H] ⊆ H} Norr (H) = {x ∈ L | [H, x] ⊆ H}

of L are said to be left and right normalizers, respectively, of the subalgebra H in L. Definition 2.5. A subalgebra H of a Leibniz algebra L is called a Cartan subalgebra if the following two conditions are satisfied: a) H is nilpotent; b) H coincides with the left normalizer of H in L. It is obvious that if the antisymmetricity property is supposed the sets Norl (H) and Norr (H) coincide. For a Cartan subalgebra H of a Leibniz algebra L we have the inclusion Norl (H) ⊆ Norr (H). It is easy to see that if H contains the ideal generated by squares of elements of

68  Leibniz Algebras

L, then we have Norl (H) = Norr (H). In general, for non Lie Leibniz algebras the equality does not hold, which can be confirmed by the following example. Example 2.9. Let L be three-dimensional Leibniz algebra defined by the following table of multiplications: [x, z] = x, [z, y] = y, [y, z] = −y, [z, z] = x, where {x, y, z} is a basis of L and the omitted products are supposed to be zero. Then H = Span{x − z} = Norl (H) is the Cartan subalgebra of L and Norr (H) = Span{x, z}, i.e., Norl (H) , Norr (H). Let H be a nilpotent subalgebra of a Leibniz algebra L and L = L0 ⊕ L1 be the Fitting decomposition of the algebra L with respect to the nilpotent Lie algebra R(H) of transformations of the vector space L according to Theorem 1.8. Similar to the case of Lie algebras each Cartan subalgebra H of a Leibniz algebra has a close relation with the Fitting null component with respect to the Lie algebra R(H). Proposition 2.11. A nilpotent subalgebra H of a Leibniz algebra L is a Cartan subalgebra if and only if H coincides with L0 in the Fitting decomposition of L with respect to R(H). Proof. Firstly, we note that Norl (H) ⊆ L0 . Indeed, if x ∈ Norl (H) then [x, h] ∈ H for any h ∈ H. Since the subalgebra H is nilpotent, there exists k ∈ N such that [. . . [[x, h], h] = Rkh (x) = 0, this implies | h], {z. . . , } k times

that x ∈ L0 , i.e., Norl (H) ⊆ L0 . Since H ⊆ Norl (H) we have H ⊆ L0 . Suppose that the inclusion H ⊂ L0 is strict, i.e., H , L0 . By Theorem 1.8, the space L0 is invariant with respect to R(H) and for h ∈ H the restriction of the operator Rh on L0 is nilpotent. Moreover, H is an invariant subspace of L0 with respect to R(H). Thus, we obtain the induced Lie algebra H of linear transformations which acts on the non null space L0 /H. Since these transformations are nilpotent by a version of Engel’s theorem we have H(x + H) = 0, where x + H is a non zero vector, i.e., [x, h] ∈ H for any h ∈ H. Hence, x ∈ Norl (H), x < H and H , Norl (H). Thus, H ⊂ Norl (H) if and only if H ⊂ L0 , and the assertion is proved.  Proposition 2.12. Let H be a nilpotent subalgebra of a Leibniz algebra L and L = L0 ⊕ L1 be the Fitting decomposition of L with respect to R(H). Then L0 is a subalgebra and [L1 , L0 ] ⊆ L1 .

Structure of Leibniz Algebras  69

Proof. Let h ∈ H and a ∈ L0 . Then there exists k ∈ N such that [. . . [a, h], . . . , h] | h],{z } = 0. k times

This means that [. . . [[Ra , Rh ], Rh ], . . . , Rh ] = (−1)k R[...[a,h],h],...,h] = 0. Applying the relation above and Lemma 1.7 we derive that the Fitting subspaces L0Rh and L1Rh of the algebra L corresponding to the endomorphism Rh are invariant subspaces with respect to Ra . Since T T L0 = L0Rh and L1 = L1Rh we have Ra (L0 ) ⊆ L0 and Ra (L1 ) ⊆ L1 . h∈H

h∈H

Since a is an arbitrary element in L0 , we obtain [L0 , L0 ] ⊆ L0 and [L1 , L0 ] ⊆ L1 .  Similar to that of Lie algebras case an element h of a Leibniz algebra L is said to be regular if the dimension of the Fitting null component of L with respect to Rh is minimal. This dimension is called the rank of L. Recall that the dimension of the Fitting null component of a linear transformation A is equal to the multiplicity of zero root of the characteristic polynomial of A. Therefore an element h is regular if and only if the multiplicity of zero root of Rh is minimal. Note that in the Lie algebras case the linear transformation Rh is degenerated since [h, h] = 0 for any h, and therefore the rank of the Lie algebra is positive. The following proposition shows that in the case of Leibniz algebras the rank also is positive. Proposition 2.13. Multiplications (right and left) in the Leibniz algebra are degenerate linear operators. Proof. For Lie algebras this is obvious, since [x, x] = 0 for any x ∈ L, i.e., the vector x is the root vector for the zero eigenvalue. Thus for any non Lie but Leibniz algebra L the quotient Lie algebra L/I has positive dimension. It is clear that the subspace I is invariant under operators L x and R x , where 0 , x ∈ L. On the quotient space which is a Lie algebra, the linear operators induced by these multiplications, are degenerated (their determinants equal to zero). Therefore these multiplications operators are degenerate in the whole space L. 

70  Leibniz Algebras

The following theorem establishes relations between regular elements of a Leibniz algebra and its Cartan subalgebras. Theorem 2.9. Let L be a Leibniz algebra over an infinite field F and a be a regular element of L. Then the Fitting null component H of L with respect to Ra is a Cartan subalgebra. Proof. Let L = H ⊕ V1 be the Fitting decomposition of L with respect to Ra . By Proposition 2.12 the subspace H is a subalgebra and [V1 , H] ⊆ H. Now we prove that for any b ∈ H the transformation Rb|H is nilpotent. Indeed, let b ∈ H be an element such that Rb|H is not nilpotent. Let us consider a basis of L consisting of bases of H and V!1 . (ρ1 ) 0 The matrix of Rh with h ∈ H, in this basis has the form , 0 (ρ2 ) where (ρ1 ) and (ρ2 ) are the matrices Rh|H and Rh|V1 , respectively. ! ! (α1 ) 0 (β1 ) 0 Let A = and B = be the matri0 (α2 ) 0 (β2 ) ces of Ra and Rb , respectively. Since (α2 ) is not singular, that is, det(α2 ) , 0, by hypothesis, the matrix (β1 ) is not nilpotent. Therefore if l = rankL, then dim H = l and the characteristic polynomial of the matrix (β1 ) is not divisible by λl . Let λ, µ, ν be algebraic independent variables and let P(λ, µ, ν) be the characteristic polynomial, i.e. P(λ, µ, ν) = det(λI − µA − νB) = det(λI − (µA + νB)). Then the equality P(λ, µ, ν) = P1 (λ, µ, ν)P2 (λ, µ, ν), where Pi (λ, µ, ν) = det(λ1 − µ(αi ) − ν(βi )) = det(λI − (µ(αi ) + ν(βi ))) holds. As it was noted above the polynomial P2 (λ, I, 0) = det(λI − (α2 )) is not divisible by λ and the polynomial P1 (λ, 0, I) = det(λI − (β1 )) is not divisible by λl . Therefore the greatest degree of λ on which the poly0 nomial P(λ, µ, ν) can be divided is λl , where l0 < l. Since the field F is infinite, we can choose µ0 and ν0 such that P(λ, µ0 , ν0 ) is not di0 visible by λl +1 . Put c := µ0 a + ν0 b, then the characteristic polynomial 0 det(λI −Rc ) = det(λI −µ0 A−ν0 B) = P(λ, µ0 , ν0 ) is not divisible by λl +1 . Therefore the multiplicity of zero root for Rc is equal to l0 , (l0 < l). But this contradicts the condition that a is a regular element. Therefore, for any b ∈ H the operator Rb|H is nilpotent. By Engel’s theorem (see Theorem 2.1) we obtain that H is a nilpotent Leibniz algebra. Let now L0 be a Fitting’s null component of L with respect to R(H). Since H is the Fitting null component of the transformation Ra , then L0 ⊆ H. Indeed, T when a ∈ H we have that L0 = L0Rb ⊆ L0Ra . b∈H

Structure of Leibniz Algebras  71

Let a < H. Then one has ak , 0 for any k ∈ N. Consider the following set of vectors {a1 := a, a2 := [a, a], . . . , an+1 := [[[a, a], . . . , a] | a],{z }}, (n+1) times

where n = dim L. This set of vectors is linearly dependent, i.e., there exists a non trivial linear combination: α1 a1 + α2 a2 + · · · + αn+1 an+1 = 0. Let α1 , 0, then using the fact that ai ∈ I for any 2 ≤ i ≤ n + 1, we obtain a1 ∈ I ⊆ Annr (L) and consequently a2 = 0, i.e., we have a contradiction with the condition a2 , 0. Thus, α1 = 0. Let k be the minimal number, such that αk , 0. Then αk ak + · · · + αn+1 an+1 = 0, therefore for the element t = αk a1 + · · · + αn+1 an+1−k we have [[[t, t], . . . , t] | t],{z } = 0 and Rt = αk Ra , i.e., t is a regular element and k times

L0Ra = L0Rt = H. On the T other hand L0 ⊇ H for any nilpotent subalgebra H. In fact, L0 = L0Rb and if h ∈ H then | [[[h, b],{z b], . . . , b] }=0 b∈H

s times

for any b ∈ H (here s is the index of nilpotence of the algebra H) thus h ∈ L0 and L0 = H. Due to Proposition 2.11 we conclude that H is a Cartan subalgebra, which completes the proof of the theorem.  Another useful remark about regular elements and Cartan subalgebras is the following Corollary 2.7. If a Cartan subalgebra H of a Leibniz algebra L contains a regular element a, then H is uniquely defined by the element a as a Fitting null component of the L with respect to Ra , i.e., H = L0Ra . Proof. Let us denote L0Ra by V. It is clear that H ⊆ V, since H is nilpotent (if h ∈ H then using the nilpotency of H and a ∈ H we have [[[h, a], a], . . . , a] = 0 which implies h ∈ L0Ra = V). On the other hand, due to Theorem 2.9 V is nilpotent. And if H , V then there exists z ∈ V \ H. If [z, L] ⊆ H, then since H is a Cartan subalgebra, we have z ∈ H which is a contradiction. Therefore, for any z ∈ V \ H we have [z, L] 1 H. Then there exist l1 , l2 , . . . , lk ∈ H such that [[[z, l1 ], l2 ], . . . , lk ] ∈ V \ H and [[[z, l1 ], l2 ], . . . , lk ] , 0, i.e., V is not nilpotent which is also a contradiction. Hence, H = V. We conclude that two Cartan subalgebras having the same regular element coincide. 

72  Leibniz Algebras

Proposition 2.14. Let L be a complex finite-dimensional Leibniz algebra. Then the image of a regular element of L under π : L → LLie is a regular element of the Lie algebra LLie . Proof. Let a be a regular element of L. We prove that the element a = a + I is a regular element of the Lie algebra LLie . Suppose the contrary, i.e., a = a + I is not a regular element of LLie . Let b = b + I be any regular element of LLie . For any x ∈ L one has R x (I) ⊆ I. Let {i1 , i2 , . . . , in } be a basis of I. We extend the basis {i1 , i2 , . . . , in } to the basis {e1 , e2 , . . . , em , i1 , i2 , . . . , in } of L. Then the matrix of the transformation R x with respect to this basis has the following block form ! X, 0 Rx = , ZX , IX where X is the matrix of the linear transformation R x |Span{e1 ,e2 ,...,em } and IX is the matrix of the transformation R x |I . Let ! ! A, 0 B, 0 Ra = , Rb = Za , Ia Zb , Ib be the matrices of the transformations Ra and Rb , respectively. Let k (respectively k0 ) and s (respectively s0 ) be the multiplicities of the root 0 of the characteristic polynomials of A (respectively B) and Ia (respectively Ib ). Obviously, we have k0 < k, s < s0 . Put !  Y, 0 U = y ∈ L \ I Ry = and Y has the multiplicity of the Zy , Iy  root 0 of the characteristic polynomial less than k , !  Y, 0 V = y ∈ L \ I Ry = and Iy has the multiplicity of the Zy , Iy  root 0 of the characteristic polynomial less than s + 1 . Since b ∈ U and a ∈ V these sets are not empty. Let us show that the set U is an open subset of the set L \ I with respect to the Zariski topology.

Structure of Leibniz Algebras  73

The set Y has the multiplicity of the root 0 of the characteristic polynomial less than k, hence the matrix Y k has the rank greater than n − k. This means that there exists a non-zero minor of the order n − k + 1. In other words, there exists a non zero polynomial of structure constants of the algebra L, hence the set U is an open subset of L \ I. Similarly, one can prove that the set V is open in L \ I. It is not difficult to check that the sets U and V are dense in L\I. Therefore, there exists an element y ∈ U ∩ V such that Y has the multiplicity of the root 0 of the characteristic polynomial less than k and Iy has the multiplicity of the root 0 of the characteristic polynomial less than s + 1. Thus, for this element y the multiplicity of the root 0 of the characteristic polynomial is not greater than k + s − 1, i.e., the rank of the algebra L is less than k + s and we obtain a contradiction to the assumption that a is not a regular element of the Lie algebra LLie .  Let L be a Leibniz algebra with a basis {e1 , e2 , . . . , en } over a field F. Let ξ1 , ξ2 , . . . , ξn be independent variables and P = F(ξ1 , ξ2 , . . . , ξn ) be the field of rational functions at ξi . Consider the extension LP = Pe1 + Pe2 + · · · + Pen of P. The following definition and its comments are step by step modifications of the Lie algebras case and they are included for the sake of completeness. Definition 2.6. An element x =

n P i=1

ξi ei of the algebra LP is called a

generic element of the algebra L and the characteristic polynomial f x (λ) of the transformation R x in LP is called characteristic polynomial of the Leibniz algebra L. If we take the basis {e1 , e2 , . . . , en } of LP then [ei , x] =

n P

ρi j e j ,

j=1

where i = 1, . . . , n and ρi j are homogeneous functions of degree 1 with respect to ξk . Then f x (λ) = det(λI − R x ) = λn − τ1 (ξ)λn−1 + τ2 (ξ)λn−2 − · · · + (−1)l τn−l (ξ)λl ,

(2.5)

where τi (ξ) are homogeneous polynomials of degree i in the variables ξi , ξ = {ξ1 , · · · , ξn } and τn−l (ξ) , 0, where τn−l+k (ξ) = 0 for k > 0. Since x , 0 and R x is a degenerate operator, it follows that l > 0 and detR x = 0.

74  Leibniz Algebras

The value of the characteristic polynomial on an arbitrary element n P a = αi ei of the algebra L is obtained by specialization of ξi = αi , i=1

i = 1, . . . , n in (2.5). Therefore it is evident that the order of zero root of the characteristic polynomial of Ra is not less than l. On the other hand, if F is an infinite field, the polynomial τn−l (x) is a non zero element of F[ξ1 , ξ2 , . . . , ξn ], we can choose ξi = αi such that τn−l (α) , 0. n P Then the transformation Ra for the element a = αi ei has exactly l i=1

characteristic roots which are equal to zero and therefore a is regular. Thus in the case of an infinite field the element a is regular if and only if τn−l (α) , 0. In this sense “almost all” elements of the algebra L are regular (i.e., they form an open set in the Zariski topology). The above statement depends on the choice of the basis {e1 , e2 , . . . , en }. However, it is easy to observe what happens when n P we pass to another basis { f1 , f2 , . . . , fn }, where fi = µi j e j . If η1 , η2 , . . . , ηn are independent variables then y =

n P i=1

j=1 n P

ηi fi =

η i µi j e j .

i=1

Therefore the characteristic polynomial fy (λ) is obtained from polynon P mial f x (λ) by substitution ξ j → ηi µi j in its coefficients. i=1

2.6 SOME PROPERTIES OF WEIGHT SPACES OF LEIBNIZ ALGEBRAS AND CARTAN’S CRITERION OF SOLVABILITY

In order to define a weight module over a Leibniz algebra we need the definition of the right representation for Leibniz algebras. Definition 2.7. A vector space M is said to be a right representation of a Leibniz algebra L if there is an action: [·, ·] : M × L → M such that [m, [x, y]] = [[m, x], y] − [[m, y], x] for all x, y ∈ L, m ∈ M. The definition agrees with that of symmetric representation in [117]. Observe that M has natural right L−module structure (in the Lie sense) and below we shall consider M in this sense. Let M be a

Structure of Leibniz Algebras  75

L-module and α : L → F be linear function. We define: Mα = {a ∈ L | ∃x ∈ M such that (a − α(a)I)k (x) = 0 for some k ∈ N}. The map α : L → F defined above is said to be a weight and Mα is called weight space corresponding to the weight α. Let L be a Leibniz algebra and M be a weight subspace over the algebra L with respect to the weight α. Then for each element x ∈ M one has [x, (a − α(a)I)k ] = 0 if k is sufficiently large. Moreover, if dim M = n, then the polynomial (λ − α(a))n is the characteristic polynomial of the α. Therefore [x, (a − α(a)I)n ] = 0 for any x ∈ M. Consider the contradredient (conjugated) right module M ∗ over a Leibniz algebra satisfying the condition: h[x, a], y∗ i + hx, [y∗ , a∗ ]i = 0, where x ∈ M, y∗ ∈ M ∗ , a∗ from representation which corresponds to the right module M ∗ and hx, y∗ i as usual denotes the value of the linear function y∗ at x. It is clear that hα(a)x, y∗ i + hx, α(a)y∗ i = 0. Adding these equalities we obtain: h[x, (a − α(a)I)], y∗ i + hx, [y∗ , (a∗ + α(a)I)]i = 0. By repeating this procedure, we obtain the equality: h[x, (a − α(a)I)k ], y∗ i + hx, [y∗ , (a∗ + α(a)I)k ]i = 0. If k = n, then [x, (a − α(a)I)n ] = 0 for any x, consequently hx, [y∗ , (a∗ + α(a)I)n ]i = 0. Therefore [y∗ , (a∗ + α(a)I)n ] = 0 for any y∗ ∈ M ∗ . This shows that M ∗ is a weight module with the weight −α. Thus, we have proved the following proposition. Proposition 2.15. If M is a weight module over a Leibniz algebra with the weight α, then the contradredient module M ∗ is a weight module with the weight −α. The next proposition is proved similarly to that in the Lie algebras case (see [106]). Proposition 2.16. If M and R are weight modules over a Leibniz algebra with the weight α and β respectively, then B = M ⊗ R is a weight module with the weight α + β.

76  Leibniz Algebras

A nilpotent Leibniz algebra L of linear transformations is called a split algebra if the characteristic roots of each element A of L are contained in the base field. Let L be a Leibniz algebra, H be a nilpotent subalgebra and let M be a left module over L (and also over H). Suppose that M = L and R(H) is a nilpotent split Lie algebra. From Theorem 1.9 we have that L = Lα ⊕ Lβ ⊕ . . . ⊕ Lδ , where α, β, . . . , δ are maps from the subalgebra R(H) into F such that if xν ∈ Lν , then (Rh − ν(Rh )I)m (xν ) = 0 for some m = m(ν), where ν ∈ {α, β, . . . , δ}. The weights α, β, . . . , δ are called roots of the algebra L with respect to the subalgebra H. Proposition 2.17. [Lα , Lβ ] ⊆ Lα+β if α + β is a root of the Leibniz algebra L with respect to R(H); otherwise [Lα , Lβ ] = 0. P (i) Proof. Elements of [Lα , Lβ ] have the form [xα(i) , y(i) β ] where xα ∈ Lα , i

∈ Lβ . From the characteristic property of the tensor product of two spaces it follows that there a linear map: ϕ : Lα ⊗ Lβ → [Lα , Lβ ] ! exists h P (i) P (i) (i) i xα ⊗ y(i) such that ϕ = xα , yβ . We show that ϕ is actually a β y(i) β

i

i

homomorphism of R(H)-modules. Let Rh ∈ R(H). Then by using the Leibniz identity we obtain the following chain of equalities Rh (xα ⊗ yβ ) = Rh (xα ) ⊗ yβ + xα ⊗ Rh (yβ ) = [xα , h] ⊗ yβ + xα ⊗ [yβ , h], and under ϕ one has Rh ([xα , yβ ]) = [[xα , h], yβ ] + [xα , [yβ , h]]. On the other hand, the image of the element xα ⊗ yβ under the homomorphism ϕ is [xα , yβ ]. So, we prove that [Lα , Lβ ] is a homomorphic image of the module Lα ⊗ Lβ . Moreover, Lα ⊗ Lβ is a weight module with the weight α + β. But from the definition it is clear that the homomorphic image of the weight module with the weight β is either 0 or a weight module with the weight β.  Let L be a finite dimensional Leibniz algebra over an algebraically closed field F and H be a nilpotent subalgebra of L. Let L = Lα ⊕ Lβ ⊕ . . . ⊕ Lδ be a decomposition of the module L into the direct sum of weight submodules with respect to H.

Structure of Leibniz Algebras  77

Suppose that H is a Cartan subalgebra. Then it is not difficult to see that H = L0 , where L0 is the root module P corresponding to the root 0. We have also the equality [L, L] = [Lα , Lβ ], where the sum is taken over all roots α, β. From this we obtain L0 ∩ L2 = H ∩ L2 = P [Lα , L−α ], where summation is made over all α, such that −α is also a root (in particular α = 0). Definition 2.8. The form K(a, b) = tr(Ra Rb ) for a, b ∈ L is called the Killing form of the Leibniz algebra L. A bilinear form K(a, b) on L satisfying the condition: K([a, c], b) + K(a, [b, c]) = 0 is called an invariant form on L. The following equalities show that the Killing form is an invariant form on Leibniz algebra. In fact, K([a, c], b) + K(a, [b, c]) = tr(R[a,c] Rb ) + tr(Ra R[b,c] ) = tr([Rc , Ra ]Rb + Ra [Rc , Rb ]) = tr((Rc Ra − Ra Rc )Rb + Ra (Rc Rb − Rb Rc )) = tr(Rc Ra Rb − Ra Rb Rc ) = tr[Rc , Ra Rb ] = 0. Note that if K(a, b) is the Killing form then the set L⊥ = {z ∈ L | K(a, z) = 0 for any a ∈ L} is an ideal of the algebra L. Theorem 2.10. Let L be a Leibniz algebra over an algebraically closed field of zero characteristic. Then L is solvable if and only if tr(Ra Ra ) = 0 for any a ∈ L2 . Proof. Necessity. Taking into account the solvability of R(L) and applying Proposition 1.4 we obtain [R(L), R(L)] ⊆ N. Since [R(L), R(L)] = R(L2 ) the operator Ra is nilpotent for any a ∈ L2 , therefore tr(Ra Ra ) = 0. Sufficiency. Let us apply Cartan’s criterion for Lie algebras from [106] for the algebra R(L) and consider L as R(L)-module. Then we obtain the solvability of the Lie algebra R(L), but this is equivalent to the solvability of the Leibniz algebra L. 

CHAPTER

3

CLASSIFICATION PROBLEM IN LOW DIMENSIONS

3.1

ALGEBRAIC CLASSIFICATION OF LOW-DIMENSIONAL LEIBNIZ ALGEBRAS

The first part of this chapter concerns with the algebraic classification of complex Leibniz algebras of small dimensions. The second part is devoted to the study of varieties of Leibniz algebras. For the Lie algebras case by using the structure of a graded Lie algebra associated to a Leibniz algebra, and based on observations on the cohomologies of Lie and Leibniz algebras, we give a necessary and sufficient condition for a Lie algebra to be Leibniz rigid. For Leibniz algebras case we give necessary conditions (invariance arguments) for existence of degenerations which is helpful in finding a criterion for a Leibniz algebra to be rigid. 3.1.1

Classification of Lie algebras

The classification of finite-dimensional Lie algebras is a fundamental and very difficult problem. It can be split into three parts: (1) classification of nilpotent Lie algebras; (2) description of solvable Lie algebras with given nilradical; (3) description of Lie algebras with given radical. 79

80  Leibniz Algebras

The third problem reduces to the description of semisimple subalgebras in the algebra of derivations of a given solvable algebra [124]. The classification of semisimple Lie algebras has been known since the works of Cartan and Killing. According to the Cartan-Killing theory the semisimple Lie algebras can be represented as a direct sum of the classical simple Lie algebras from series An (n ≥ 1), Bn (n ≥ 2), Cn (n ≥ 3), Dn (n ≥ 4) and five exceptional simple Lie algebras G2 , F2 , E4 , E6 , E7 , E8 . The second problem reduces to the description of orbits of certain unipotent linear groups [125]. The first problem is most complicated. Just recall that nilpotent complex Lie algebras are classified only in dimension up to 7. In higher dimensions, there are only partial classifications as subclasses of nilpotent Lie algebras. There is no non trivial one-dimensional Lie algebra. On a two-dimensional complex vector space there is only one non trivial Lie algebra structure given by the table of multiplication [e1 , e2 ] = e1 in a fixed basis {e1 , e2 }. The list of representatives of the isomorphism classes of three-dimensional complex Lie algebras can be found in the literature on Lie algebras (for instance see [106]). 3.1.2

Low-dimensional complex Leibniz algebras

There is no non trivial Leibniz algebra in dimension 1. Indeed, let dimF (L) = 1 and e be a non-zero element in L. If [e, e] = 0, then L is an abelian Lie algebra. If [e, e] , 0, then [e, e] = αe for some nonzero α ∈ F. But then the Leibniz identities (both right and left) give us a contradiction. Therefore, on a one-dimensional space there is only one (trivial) Leibniz algebra structure. 3.1.2.1

Two-dimensional Leibniz algebras

Two dimensional Leibniz algebras have been classified by Cuvier [59]. Here is the list of all Leibniz algebras structures on two-dimensional vector space over F. Let dimF (L) = 2, i.e., L = Fe1 + Fe2 . Then it is easy to verify that there are, up to isomorphism, four Leibniz algebra structures on L : 1. L1 : abelian algebra 2. L2 : solvable Lie algebra with the table of multiplication [e1 , e2 ] = −[e2 , e1 ] = e2 .

Classification Problem in Low Dimensions  81

3. L3 : nilpotent Leibniz algebra with the table of multiplication [e2 , e2 ] = e1 . 4. L4 : solvable Leibniz algebra with the table of multiplication [e1 , e2 ] = e1 , [e2 , e2 ] = e1 . Note that the algebra L3 is the left and right Leibniz algebra (obviously), and L4 is the only right Leibniz algebra. One can check it by a direct verification of identities. In fact, let us test left Leibniz algebra’s identity for the triple {e2 , e2 , e2 }. One has [e2 , [e2 , e2 ]] = [e2 , e1 ] = 0 and [[e2 , e2 ], e2 ] + [e2 , [e2 , e2 ]] = [e1 , e2 ] + [e2 , e1 ] = e1 , 0. This means that the intersection of the classes of left and right Liebniz algebras strictly contains the class of Lie algebras. 3.1.2.2

Three-dimensional Leibniz algebras

The list of isomorphism classes of three-dimensional complex Leibniz algebras has been given in [27, 57]. Here we give a list of isomorphism classes over any field F of characteristic not two (see [154]). Let L be a three-dimensional algebra over a field F. To give the list we make use of a case by case consideration with respect to the isomorphism invariants dim Annr (L) and dim Lk . Case 1 : Let dim Annr (L) = 1 and {e1 } be its basis. Since Annr (L) is an ideal of L, the following multiplications may occur: [e1 , e2 ] = α1 e1 , [e2 , e2 ] = α2 e1 , [e3 , e2 ] = β1 e1 + β2 e2 + β3 e3 , [e3 , e3 ] = α4 e1 , [e1 , e3 ] = α3 e1 , [e2 , e3 ] = γ1 e1 − β2 e2 − β3 e3 . Case 1.1 : Let dim L2 = 2. Then (β2 , β3 ) , (0, 0), otherwise dim L2 = 1. Since e2 and e3 are symmetric, one can set β2 , 0. Thus, by taking the basis transformation e01 = e1 , e02 = β2 e2 + β3 e3 , and e03 = β12 e3 we obtain β2 = 1, β3 = 0 and hence the following multiplications occur:

82  Leibniz Algebras

[e1 , e2 ] = α1 e1 , [e1 , e3 ] = α3 e1 , [e3 , e2 ] = β1 e1 + e2 ,

[e2 , e2 ] = α2 e1 , [e3 , e3 ] = α4 e1 , [e2 , e3 ] = γ1 e1 − e2 .

(3.1)

Applying the Leibniz identity as follows [[e1 , e3 ], e2 ] = [[e1 , e2 ], e3 ] + [e1 , [e3 , e2 ]] [[e2 , e3 ], e2 ] = [[e2 , e2 ], e3 ] + [e2 , [e3 , e2 ]] [[e3 , e2 ], e3 ] = [[e3 , e3 ], e2 ] + [e3 , [e2 , e3 ]] we obtain the following constraints for the structure constants:   α1 = 0,     (3.2) α2 (2 + α3 ) = 0,     β + α β + γ = 0. 1 3 1 1 Case 1.1.1 : Suppose that α2 , 0 in (3.2). The application of the α α −β2 basis transformation e01 = α2 e1 , e02 = e2 and e03 = 2 2α4 2 1 e1 − αβ12 e2 + e3 , yields the following algebra: L1 (F) : [e1 , e3 ] = −2e1 , [e2 , e2 ] = e1 , [e2 , e3 ] = −e2 , [e3 , e2 ] = e2 . Case 1.1.2 : Assume that α2 = 0, then we have [e1 , e3 ] = α3 e1 , [e3 , e3 ] = α4 e1 , [e3 , e2 ] = β1 e1 + e2 , [e2 , e3 ] = −β1 (1 + α3 )e1 − e2 . It is observed from (3.2) above that, if (α3 , α4 ) = (0, 0) the algebra obtained is a Lie algebra. Therefore, we may assume that (α3 , α4 ) , (0, 0). By applying the basis transformation e01 = e1 , e02 = β1 + e2 and e03 = e3 we get the following table of multiplications: [e1 , e3 ] = α3 e1 , [e3 , e3 ] = α4 e1 [e3 , e2 ] = e2 , [e2 , e3 ] = −e2 . If α3 , 0, then the basis transformation e01 = e1 , e02 = e2 and e03 = e3 − αα43 e1 yields the algebra L2 (F) : [e1 , e3 ] = αe1 , [e2 , e3 ] = −e2 , [e3 , e2 ] = e2 ,

α , 0.

If α3 = 0, then α4 , 0 and the basis transformation e01 = α4 e1 , e02 = e2 and e03 = e3 yields the algebra L3 (F) : [e3 , e3 ] = e1 , [e3 , e2 ] = e2 , [e2 , e3 ] = −e2 .

Classification Problem in Low Dimensions  83

Case 1.2: Let dim L2 = 1. The table of multiplication of L is written as follows: [e1 , e2 ] = α1 e1 , [e3 , e3 ] = α4 e1 ,

[e2 , e2 ] = α2 e1 , [e3 , e2 ] = β1 e1 ,

[e1 , e3 ] = α3 e1 , [e2 , e3 ] = γ1 e1 .

(3.3)

Applying Leibniz identity as follows [[e2 , e2 ], e3 ] = [[e2 , e3 ], e2 ] + [e2 , [e2 , e3 ]] [[e3 , e2 ], e3 ] = [[e3 , e3 ], e2 ] + [e3 , [e2 , e3 ]] we obtain the following constraints for the structure constants: ( α1 γ1 = α2 α3 α1 α4 = α3 β1 .

(3.4)

Case 1.2.1: Assume that (α1 , α3 ) = (0, 0). Then due to (3.4) we get the following table of multiplications: [e2 , e2 ] = α2 e1 , [e3 , e2 ] = β1 e1 ,

[e3 , e3 ] = α4 e1 [e2 , e3 ] = γ1 e1 .

(3.5)

If (α2 , α4 , β1 + γ1 ) = (0, 0, 0), then L is a Lie algebra. Therefore, we assume that (α2 , α4 , β1 + γ1 ) , (0, 0, 0). The basis transformation e01 = e1 , e02 = Ae2 + Be3 and e03 = e3 with A , 0 reduces (3.5) to the following table of multiplications: [e2 , e2 ] = [A2 α2 + B2 α4 + AB(β1 + γ1 )]e1 , [e3 , e3 ] = (α4 B + γ1 )e1 , [e3 , e2 ] = (α4 B + β1 )e1 , [e2 , e3 ] = γ1 e1 . The condition (α2 , α4 , β1 + γ1 ) , (0, 0, 0) implies that, there exist two numbers A, B ∈ F such that: A2 α2 + B2 α4 + AB(β1 + γ1 ) , 0. Therefore, we presuppose that α2 , 0 in (3.5). The basis transformation e01 = α2 e1 , e02 = e2 and e03 = e3 − αβ12 e2 yields α2 = 1, β1 = 0 and hence the following multiplications table is obtained: [e2 , e2 ] = e1 , [e3 , e3 ] = α4 e1 , [e2 , e3 ] = γ1 e1 ,

(3.6)

84  Leibniz Algebras

where (α4 , γ1 ) , (0, 0), otherwise L is a split algebra. If α4 = 0 in (3.6), then e2 − γ11 e3 ∈ Annr (L) contradicts dim Annr (L) = 1. Therefore we may assume that α4 , 0 which leads to the following cases: Case 1.2.1.1 : Assume that γ1 = 0. This yields the following table of multiplications: [e2 , e2 ] = e1 , [e3 , e3 ] = α4 e1 , with α4 , 0.

(3.7)

Case 1.2.1.1.A : Suppose that F2 = F. Then the equation x2 = α4 has a solution in F. The basis transformation e01 = α4 e1 , e02 = e1 + e3 and √ e03 = e1 + α4 e2 , applied to (3.7) gives the following algebra: L4 (F) : [e2 , e2 ] = e1 , [e3 , e3 ] = e1 . Case 1.2.1.1.B : Suppose that F2 , F. Let the equation x2 = α4 have no solution in F. We have the following sub cases: Case 1.2.1.1.B.1 : The equation x2 = −α4 also has a solution in F. By applying the basis transformation e01 = α4 e1 , e02 = e1 + e3 and √ e03 = e1 + −α4 e2 to (3.7) we get the following algebra: L5 (F) : [e2 , e2 ] = e1 , [e3 , e3 ] = −e1 . Case 1.2.1.1.B.2 : Assume the equation x2 = −α4 has no solution in F. We get the algebra: L6 (F) : [e2 , e2 ] = e1 , [e3 , e3 ] = α4 e1 , α4 , 0. √ √ The basis transformation e01 = α4 e1 , e02 = −α4 e2 and e03 = −α4 e3 reduce L6 (F) to L4 (F). Thus, the two algebras are isomorphic. Case 1.2.1.2 : Let us now consider γ1 , 0 in (3.6). Taking the basis transformation e01 = e1 , e02 = e2 and e03 = γ11 e3 , we obtain γ1 = 1 and replacing the parameter α4 by α, we get the following algebras: L7 (F) : [e2 , e2 ] = e1 , [e2 , e3 ] = e1 , [e3 , e3 ] = αe1 , α , 0. Case 1.2.2 : Let (α1 , α3 ) , (0, 0) in (3.3). We may assume that α1 , 0, otherwise we have the following table of multiplications [e1 , e3 ] = α3 e1 , [e3 , e3 ] = α4 e1 , [e2 , e3 ] = γ1 e1 .

(3.8)

But e2 ∈ Annr (L) in (3.8). This contradicts the claim that dim Annr (L) = 1.

Classification Problem in Low Dimensions  85

Since α1 , 0, then from (3.4) we get γ1 = we write (3.3) as follows:

α2 α3 α1

and α4 =

α3 β1 . α1

Thus

α3 β1 e1 , α1 α2 α3 [e3 , e2 ] = β1 e1 , [e2 , e3 ] = e1 . α1

[e1 , e2 ] = α1 e1 , [e2 , e2 ] = α2 e1 , [e1 , e3 ] = α3 e1 , [e3 , e3 ] =

Similarly, we have e3 − αα31 e2 ∈ Annr (L). This contradicts the claim that dim Annr (L) = 1. Case 2 : Suppose that dim Annr (L) = 2 and {e1 , e2 } is its basis. Let {e1 , e2 , e3 } be the basis of L, then the following table of multiplications occur: [e3 , e3 ] = α1 e1 + α2 e2 , [e2 , e3 ] = γ1 e1 + γ2 e2 .

[e1 , e3 ] = β1 e1 + β2 e2 ,

(3.9)

Case 2.1 : Let dim L2 = 1. This reduces (3.9) to the following table of multiplications: [e3 , e3 ] = α1 e1 , [e1 , e3 ] = β1 e1 , [e2 , e3 ] = γ1 e1 .

(3.10)

Evidently, if γ1 = 0, then (3.10) will be a split algebra. Applying the basis transformation e01 = γ1 e1 , e02 = e2 and e03 = e3 − αγ11 e2 we get α1 = 0, γ1 = 1 and hence the following table of multiplications: [e1 , e3 ] = β1 e1 , [e2 , e3 ] = e1 .

(3.11)

Case 2.1.1 : Assume that β1 = 0. Then from (3.11) we get the following algebra: L8 (F) : [e2 , e3 ] = e1 . Case 2.1.2 : Assume that β1 , 0. The basis transformation e01 = e1 , e02 = e2 − e1 and e03 = e3 yields the following split algebra: [e1 , e3 ] = e1 . Case 2.2 : Let dim L2 = 2. This gives the following table of multiplications: [e3 , e3 ] = α1 e1 + α2 e2 , [e1 , e3 ] = β1 e1 + β2 e2 , (3.12) [e2 , e3 ] = γ1 e1 + γ2 e2 ,

86  Leibniz Algebras

   α1 α2  where the rank of the matrix  β1 β2  is 2. γ1 γ2 Case 2.2.1 : Suppose that det(Re3 |Annr (L) ) , 0, it is equivalently defined as β1 γ2 − γ1 β2 , 0. (3.13) Case 2.2.1.1 : Suppose that β2 , 0. Since Re3 |Annr (L) is nondegenerate, it yields γ1 , 0. The basis transformation ! α1 α1 γ2 0 0 0 − α2 e1 − e2 + e3 e1 = e1 , e2 = e2 and e3 = γ1 γ1 reduces the table (3.12) to [e1 , e3 ] = e2 , [e2 , e3 ] = γ1 e1 + γ2 e2 , γ1 , 0.

(3.14)

If γ2 = 0 one has [e1 , e3 ] = e2 , [e2 , e3 ] = γ1 e1 , γ1 , 0.

(3.15)

Case 2.2.1.1.A : Suppose that F2 = F. Assume that the equation √ x = γ1 has a solution in F. The basis transformation e01 = γ1 e1 , e02 = e2 and e03 = √1γ1 e3 in (3.15) yields the following algebra: 2

L9 (F) : [e1 , e3 ] = e2 , [e2 , e3 ] = e1 . Case 2.2.1.1.B : Suppose that F2 , F. Let the equation x2 = γ1 have no solution in F. Then we consider the following sub cases: Case 2.2.1.1.B.1 : Let x2 = −γ1 have a solution in F. Applying the √ 1 basis transformation e01 = −γ1 e1 , e02 = e2 and e03 = √−γ e3 to (3.15) 1 we get L10 (F) : [e1 , e3 ] = e2 , [e2 , e3 ] = −e1 . Case 2.2.1.1.B.2 : Let x2 = −γ1 have no solution in F. We get the following algebra L11 (F) : [e1 , e3 ] = e2 , [e2 , e3 ] = γ1 e1 , γ1 , 0. Algebras in L11 (F) can be reduced to L9 (F) via basis transformations e01 = γ1 e1 , e02 = γ1 e2 and e03 = e3 . Thus, the algebras L9 (F) and L11 (F) are isomorphic.

Classification Problem in Low Dimensions  87

Suppose that γ1 , 0, then (3.14) yields the following algebra L12 (F) : [e1 , e3 ] = e2 , [e2 , e3 ] = γ1 e1 + e2 , γ1 , 0. Case 2.2.1.2 : Let β2 = 0. Setting the basis transformation e01 = e1 and e02 = e2 − e1 , e03 = β11 e3 , the table of multiplication (3.12) is reduced to [e3 , e3 ] = α1 e1 + α2 e2 , [e1 , e3 ] = e1 , [e2 , e3 ] = γ1 e1 + γ2 e2 . Due to (3.13) we have γ2 , 0. Applying the basis transformation e01 =  e1 , e02 = e2 and e03 = αγ1 γ2 1 − α1 e1 − αγ22 e2 + e3 , we get α1 = α2 = 0 and hence the table of multiplications above is reduced to: [e1 , e3 ] = e1 , [e2 , e3 ] = γ1 e1 + γ2 e2 , γ2 , 0.

(3.16)

If (γ1 , γ2 ) , (0, 1), then there exist numbers A, B ∈ F such that AB(γ2 − 1) − B2 γ1 , 0. The composition of the basis transformation e01 = Ae1 + Be2 , e02 = e2 , e03 = e3 , (A , 0) and 00

e1 =

e01 ,

" !# B2 B  0 00 e2 = 1 + γ1 e1 + Bγ2 − B + γ1 e02 , e3 = e03 , A A 00



brings (3.16) to Case 2.2.1.1 already considered. Furthermore, (γ1 , γ2 ) = (0, 1) yields the following algebra L13 (F) : [e1 , e3 ] = e1 , [e2 , e3 ] = e2 . Case 2.2.2 : Let det(Re3 |Annr (L) ) = 0. In this case, we have the following general table of multiplications: [e3 , e3 ] = α1 e1 + α2 e2 , [e1 , e3 ] = t1 (β1 e1 + β2 e2 ), [e2 , e3 ] = t2 (γ1 e1 + γ2 e2 )

(3.17)

88  Leibniz Algebras

Since in (3.17), the vectors e1 and e2 are “symmetric”, we may assume that t1 , 0. The basis transformation e01 = e1 , e02 = tt21 e1 −e2 and e03 = e3 reduces (3.17) to the following table of multiplications: [e3 , e3 ] = α1 e1 + α2 e2 , [e1 , e3 ] = β1 e1 + β2 e2 .

(3.18)

Applying the basis transformation e01 = α1 e1 + α2 e2 , e02 = e2 and e03 = e3 to the table above we get the following table of multiplications: [e3 , e3 ] = e1 , [e1 , e3 ] = β1 e1 + β2 e2 .

(3.19)

The basis transformation e01 = e1 , e02 = β2 e2 and e03 = e3 reduces (3.19) to the following table of multiplications: [e3 , e3 ] = e1 , [e1 , e3 ] = β1 e1 + e2 .

(3.20)

We consider the following cases in (3.20). Case 2.2.2.1: If β1 = 0, we get the following algebra: L14 (F) : [e1 , e3 ] = e2 , [e3 , e3 ] = e1 .

(3.21)

Case 2.2.2.2 : Let β1 , 0. Taking the basis transformation e01 = 1 e , e02 = β13 e2 , and e03 = β11 e3 , we obtain the following algebra: β2 1 1

1

L15 (F) : [e1 , e3 ] = e1 + e2 , [e3 , e3 ] = e1 . 3.2

APPLICATION

We now apply the algorithm illustrated above to R. In order to make our classification clear herein and after we use the following notation Lij (F) when we list the algebras, where i represents number of algebra from above obtained list L1 (F) − L15 (F) and j stands for the number of the algebra in respective theorem. Theorem 3.1. Up to isomorphism, there exist three one parametric families and eleven explicit representatives of three dimensional nonLie Leibniz algebras over the real field: L11 (R) : [e1 , e3 ] = −2e1 , [e2 , e2 ] = e1 , [e2 , e3 ] = −e2 , [e3 , e2 ] = e2 ;

Classification Problem in Low Dimensions  89

L22 (R) : [e1 , e3 ] = αe1 , [e2 , e3 ] = −e2 , [e3 , e2 ] = e2 , α ∈ R \ {0}; L33 (R) : [e3 , e3 ] = e1 , [e3 , e2 ] = e2 , [e2 , e3 ] = −e2 ; L44 (R) : [e2 , e2 ] = e1 , [e3 , e3 ] = e1 ; L55 (R) : [e2 , e2 ] = e1 , [e3 , e3 ] = −e1 ; L76 (R) : [e2 , e2 ] = e1 , [e2 , e3 ] = e1 , [e3 , e3 ] = αe1 , α ∈ R \ {0}; L87 (R) : [e2 , e3 ] = e1 ; L98 (R) : [e1 , e3 ] = e2 , [e2 , e3 ] = e1 ; 9 L10 (R) : [e1 , e3 ] = e2 , [e2 , e3 ] = −e1 ; 10 L12 (R) : [e1 , e3 ] = e2 , [e2 , e3 ] = αe1 + e2 , α ∈ R \ {0}; 11 L13 (R) : [e1 , e3 ] = e1 , [e2 , e3 ] = e2 ; 12 L14 (R) : [e1 , e3 ] = e2 , [e3 , e3 ] = e1 ; 13 L15 (R) : [e1 , e3 ] = e1 + e2 , [e3 , e3 ] = e1 .

The classification of three-dimensional complex Lie algebras can be found in [106]. Combining Leibniz and Lie algebras cases, we give in Table 3.1 the list of isomorphism classes of all three-dimensional complex Leibniz algebras.

90  Leibniz Algebras

Algebra L1 (α) α , 0, α∈C L2 L3

Multiplication Table [e1 , e3 ] = αe1 , [e2 , e3 ] = e1 + e2 , [e3 , e3 ] = e1 [e3 , e3 ] = e1 , [e2 , e3 ] = e1 + e2 [e1 , e2 ] = e3 , [e1 , e3 ] = −2e3 , [e2 , e1 ] = −e3 , [e2 , e3 ] = 2e3 , [e3 , e1 ] = 2e3 , [e3 , e2 ] = −2e3 [e1 , e3 ] = αe1 , [e2 , e3 ] = −e2 , [e3 , e2 ] = e2 , [e3 , e3 ] = e1 [e1 , e3 ] = e1 , [e2 , e3 ] = e1 , [e3 , e3 ] = e1

Classification Solvable Leibniz Solvable Leibniz Simple Lie

L6

[e1 , e3 ] = e2 , [e3 , e3 ] = e1

L7

[e1 , e2 ] = e1 , [e1 , e3 ] = e1 , [e3 , e2 ] = e1 , [e3 , e3 ] = e1

L8

[e1 , e1 ] = e2 , [e2 , e1 ] = e2

Solvable Leibniz Solvable Leibniz Nilpotent Leibniz Solvable Leibniz Solvable Leibniz

[e1 , e2 ] = e2 , [e1 , e3 ] = αe3 , [e2 , e1 ] = −e2 , [e3 , e1 ] = −αe3

Solvable Lie

[e1 , e2 ] = e2 , [e2 , e1 ] = −e2

Solvable Lie

L4 (α) L5

L9 (α), α , 0, 1, α ↔ α−1 L10 L11 L12 (α) α∈C L13 L14 L15 L16 L17 L18

[e1 , e2 ] = e2 , [e1 , e3 ] = e2 + e3 , [e2 , e1 ] = −e2 , [e3 , e1 ] = −e2 − e3 [e2 , e2 ] = e1 , [e2 , e3 ] = e1 , [e3 , e3 ] = αe1 [e2 , e2 ] = e1 , [e2 , e3 ] = e1 , [e3 , e2 ] = e1 [e1 , e3 ] = e1 , [e2 , e3 ] = e2 , [e3 , e3 ] = e1 [e1 , e1 ] = e2 [e1 , e2 ] = e2 , [e1 , e3 ] = e3 , [e2 , e1 ] = −e2 , [e3 , e1 ] = −e3 [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 -

Solvable Lie Nilpotent Lie Composable, nilpotent Leibniz Solvable Leibniz Composable, nilpotent Leibniz Solvable Lie Nilpotent Lie Abelian

Isomorphism classes of three-dimensional complex Leibniz algebras LB3 (C) Table 3.1

Classification Problem in Low Dimensions  91

3.3

LOW-DIMENSIONAL NILPOTENT LEIBNIZ ALGEBRAS

Further results deal with the nilpotent Leibniz algebras and their subclasses. There are results on classification of p-filiform, quasifiliform Leibniz algebras. They can be found, for instance, in [46] and [48]. The definition of the nilpotency has been given in Section 2.2. Let us specify some subclasses of LNn . Definition 3.1. A Leibniz algebra L is said to be a k-filiform, if dim L2 = n − k − 1, and dim Li = n − k − (i − 1), where 3 ≤ i ≤ n. It has been proved (see [24]) that 0-filiform (null-filiform) Leibniz algebras in each fixed dimension are mutually isomorphic, i.e., there is a unique null-filiform Leibniz algebra in any fixed dimension. This fact is only inherent in Leibniz algebras, i.e., there is no a null-filiform Lie algebra. The class of one-filiform (cited in this book as “filiform” as in [179]) algebras, denoted by Lbn , is most developed part of LNn , which is focused in this book. There are classification results on k-filiform Leibniz algebras as well, but the technique applied [179] is slightly different than one considered here. We begin with assembling a few simple observations on associative and Leibniz algebras. Let L be a non-unital associative algebra and A be an associative algebra, obtained from L by the external adjoining a unit A = L ⊕ C1. Then one has Proposition 3.1. Let L be a finite dimensional nilpotent associative algebra. Then the algebra A = L ⊕ C1 does not contain non trivial idempotent. Proof. Let 1 + a be an idempotent element of A = L ⊕ C1, where a is an element of L with the index of nilpotency m. Obviously, 1 + a = (1 + a)n forany natural n ≥ 2. On the other hand, one has (1 + a)n =   n n P P n k n k 1+ a , this gives a = na + k k a , and multiplying both k=1 k=2 sides of this equality by am−2 we get (n − 1)a(m−1) = 0, i.e., a(m−1) = 0, while the index of nilpotency of a is m. This contradiction shows that A has no non trivial idempotent. 

92  Leibniz Algebras

Note that if L is a Leibniz algebra with L3 = 0, then it is associative. We make use of the following corollary of Proposition 3.1. Corollary 3.1. Let L be a finite dimensional Leibniz algebra with L3 = 0. Then the algebra A = L ⊕ C1 has no non-trivial idempotent. For the further references we state the following simple fact. Proposition 3.2. Let L1 and L2 be finite-dimensional associative algebras without unit. Then L1 ⊕ C1  L2 ⊕ C1 if only if L1  L2 . Another result which we make use in this chapter is Mazzola’s classification result of five-dimensional unital associative algebras [127]. There Mazzola listed, up to isomorphism, all unital associative algebra structures on five-dimensional vector space. However, the list is too long to be given here (there are 59 isomorphism classes there). We suppose that the reader is familiar with the list. For a given n-dimensional nilpotent Leibniz algebra L we define the following isomorphism invariant χ(L) = (dim L1 , dim L2 , . . . , dim Ln−1 , dim Ln ). Evidently, dim L1 > dim L2 > · · · > dim Ln . Proposition 3.3. If for an n-dimensional nilpotent Leibniz algebra L the first two components of the invariant χ(L) are equal to n and n − 1, respectively, then L is a null-filiform Leibniz algebra. Proof. From the hypotheses of the proposition we can easily conclude that L is one-generated algebra. Let x ∈ L\L2 be a generator of L. Then the following set of vectors {x, [x, x], . . . , [[[x, x], . . . , x] | x],{z }} forms a n times

basis of L, with dim Li = n − i + 1, where 2 ≤ i ≤ n + 1.



First we begin with the following description of null-filiform Leibniz algebras.

Classification Problem in Low Dimensions  93

Theorem 3.2. Up to isomorphism, there is only one n-dimensional null-filiform Leibniz algebra. Namely, in every such Leibniz algebra L there exists a basis {x1 , x2 , . . . , xn } with respect to which L has the following table of multiplication. [xi , x1 ] = xi+1 ,

[xi , x j ] = 0, 1 ≤ i ≤ n − 1, j ≥ 2.

(3.22)

Proof. Let L be a null-filiform Leibniz algebra of dimension n and let {e1 , e2 , . . . , en } be a basis of L such that ei ∈ Li \ Li+1 with 1 ≤ i ≤ n (such a basis can be chosen always). Since e2 ∈ L2 , for some elements a2p , b2p of L we have X X e2 = α p [a2p , b2p ] = α2i j [ei , e j ] = α211 [e1 , e1 ] + (∗), where (∗) ∈ L3 ; i.e., e2 = α211 [e1 , e1 ] + (∗). Notice that α211 [e1 , e1 ] , 0 (otherwise e2 ∈ L3 ). Similarly, X e3 = α3i jk [[ei , e j ], ek ] = α3111 [[e1 , e1 ], e1 ] + (∗∗), where (∗∗) ∈ L4 ; i.e., e3 = α3111 [[e1 , e1 ], e1 ] + (∗∗). Notice that α3111 [[e1 , e1 ], e1 ] , 0 (otherwise e3 ∈ L4 ). Continuing likewise, we find the set of non zero elements of L x1 := e1 , x2 := [e1 , e1 ], x3 := [[e1 , e1 ], e1 ], ..., xn := [[[e1 , e1 ], e1 ], ..., e1 ]. It is easy to check that these elements are linearly independent. Hence, they form a basis of L. Thus, [xi , x1 ] = xi+1 for 1 ≤ i ≤ n − 1; moreover, [xi , x j ] = 0 for j ≥ 2. Indeed, if j = 2 then [xi , x2 ] = [xi , [x1 , x1 ]] = [[xi , x1 ], x1 ] − [[xi , x1 ], x1 ] = 0. Assume that it is true for j > 2. The validity for j + 1 follows from the inductive hypothesis and the equality [xi , x j+1 ] = [xi , [x j , x1 ]] = [[xi , x j ], x1 ] − [[xi , x1 ], x j ] = 0.  Henceforth we denote the algebra with multiplication (3.22) by NFn .

94  Leibniz Algebras

3.3.1

Four-dimensional nilpotent complex Leibniz algebras

In this section we shall expose, up to isomorphism, the list of all fourdimensional nilpotent complex Leibniz algebras. Here first we use four-dimensional case of Theorem 4.3 (will be given in Chapter 4), where the description of filiform non Lie Leibniz algebras is reduced to three families of algebras and a verification of the isomorphisms inside each of the families. To find other isomorphic algebras one uses the classification of unital associative algebras of dimension five [127]. From the list in [127], according to some constraints we pick up Leibniz algebras. Recall that the rings of commuting and non-commuting variables x1 , x2 , . . . , xn over a field F are denoted by F[x1 , x2 , . . . , xn ] and F hx1 , x2 , . . . , xn i, respectively. First we give the following auxiliary result. Proposition 3.4. Any four-dimensional nilpotent complex Leibniz algebra L belongs to one of the following types of algebras (i) null-filiform Leibniz algebras, that is χ(L) = (4, 3, 2, 1); (ii) filiform Leibniz algebras, that is χ(L) = (4, 2, 1, 0); (iii) associative algebras, with χ(L) := (4, 2, 0, 0) or (4, 1, 0, 0); (iv) abelian, that is χ(L) = (4, 0, 0, 0). Proof. Let us consider all possible cases for χ(L). The following can occur. a) Let χ(L) = (4, 3, 2, 1). Then in view of Proposition 3.3 the algebra L is null-filiform. b) Let χ(L) = (4, 2, 1, 0). Then L is a filiform. It is clear that in the following three cases the algebra L is associative. c) χ(L) = (4, 2, 0, 0). d) χ(L) = (4, 1, 0, 0). e) Let χ(L) = (4, 0, 0, 0). Then we get an abelian algebra.  Note that Proposition 3.4 implies that the cases χ(L) = (4, 3, 2, 0), (4, 3, 1, 0) and (4, 3, 0, 0) are impossible. From now on as a matter of convenience we assume that the omitted products of basis vectors are zero, we also do not consider the abelian case.

Classification Problem in Low Dimensions  95

Let us consider a Leibniz algebra L with a basis {e1 , e2 , e3 , e4 }. Theorem 3.3. Up to isomorphism all nilpotent Leibniz algebra structures on four-dimensional vector space are given by the following representatives. R1 : [e1 , e1 ] = e2 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 . R2 : [e1 , e1 ] = e3 , [e1 , e2 ] = e4 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 . R3 : [e1 , e1 ] = e3 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 ; R4 (α) : [e1 , e1 ] = e3 , [e1 , e2 ] = αe4 , [e2 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 , α ∈ {0, 1}. R5 : [e1 , e1 ] = e3 , [e1 , e2 ] = e4 , [e3 , e1 ] = e4 . R6 : [e1 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 . R7 : [e1 , e1 ] = e4 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 , [e1 , e2 ] = −e3 , [e1 , e3 ] = −e4 . R8 : [e1 , e1 ] = e4 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 , [e1 , e2 ] = −e3 + e4 , [e1 , e3 ] = −e4 . R9 : [e1 , e1 ] = e4 , [e2 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 , [e1 , e2 ] = −e3 + 2e4 , [e1 , e3 ] = −e4 . R10 : [e1 , e1 ] = e4 , [e2 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 , [e1 , e2 ] = −e3 , [e1 , e3 ] = −e4 . R11 : [e1 , e1 ] = e4 , [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e2 , e2 ] = −2e3 + e4 , R12 : [e1 , e2 ] = e3 , [e2 , e1 ] = e4 , [e2 , e2 ] = −e3 , R13 (α) : [e1 , e1 ] = e3 , [e1 , e2 ] = e4 , [e2 , e1 ] = −αe3 , [e2 , e2 ] = −e4 , α ∈ C, R14 (α) : [e1 , e1 ] = e4 , [e1 , e2 ] = αe4 , [e2 , e1 ] = −αe4 , [e2 , e2 ] = e4 , [e3 , e3 ] = e4 , α ∈ C, R15 : [e1 , e2 ] = e4 , [e1 , e3 ] = e4 , [e2 , e1 ] = −e4 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 , R16 : [e1 , e1 ] = e4 , [e1 , e2 ] = e4 , [e2 , e1 ] = −e4 , [e3 , e3 ] = e4 , R17 : [e1 , e2 ] = e3 , [e2 , e1 ] = e4 , R18 : [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e2 , e2 ] = e4 , R19 : [e2 , e1 ] = e4 , [e2 , e2 ] = e3 , R20 (α) : [e1 , e2 ] = e4 , [e2 , e1 ] = (1 + α)/(1 − α)e4 , [e2 , e2 ] = e3 , α ∈ C \ {1}, R21 : [e1 , e2 ] = e4 , [e2 , e1 ] = −e4 , [e3 , e3 ] = e4 , R22 : [e1 , e1 ] = e2 , R23 : [e1 , e1 ] = e2 , [e3 , e3 ] = e4 , R24 : [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , R25 : [e1 , e1 ] = e3 , [e1 , e2 ] = e3 , [e2 , e2 ] = βe3 ,

96  Leibniz Algebras

R26 : [e1 , e1 ] = e3 , [e1 , e2 ] = e3 , [e2 , e1 ] = e3 , R27 : [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e1 , e3 ] = e4 , [e3 , e1 ] = −e4 , R28 : [e1 , e1 ] = e2 , [e2 , e1 ] = e3 . Proof. According to Proposition 3.4 any four-dimensional nilpotent Leibniz algebra is either null-filiform or filiform or associative. Let us consider each of these classes separately. Let L be null-filiform. Then in view of Theorem 3.2 there is only one null-filiform Leibniz algebra and it can be given by the table of multiplication [e1 , e1 ] = e2 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 . This is R1 in the list. Let now L be a filiform Leibniz algebra and {e1 , e2 , e3 , e4 } be a basis such that e1 , e2 ∈ L \ L2 , e3 ∈ L2 \ L3 and e4 ∈ L3 . We choose the basis {e1 , e2 , e3 , e4 } so that [e2 , e1 ] = e3 , [e3 , e1 ] = e4 and [e4 , e1 ] = 0. Let  [e1 , e1 ] = α1 e3 + α2 e4 , [e2 , e2 ] = β1 e3 + β2 e4 ,     [e , e ] = γ e + γ e , [e , e ] = δ e , 1 2 1 3 2 4 3 2 1 4     [e1 , e3 ] = δ2 e4 , [e2 , e3 ] = δ3 e4 . Applying the following equalities 0 = [e1 , [e1 , e1 ]] = [e1 , α1 e3 + α2 e4 ] = α1 δ2 e4 , 0 = [e2 , [e1 , e1 ]] = [e2 , α1 e3 + α2 e4 ] = α1 δ3 e4 , we derive α1 δ2 = α1 δ3 = 0. Case i). Let α1 , 0. Then δ2 = δ3 = 0 and taking the basis transformation as follows: e0i = αi1 ei ,

1 ≤ i ≤ 4,

one can assume that α1 = 1. Putting e01 = e1 − α2 e3 , we derive α2 = 0. From the equalities 0 = [e1 , [e2 , e1 ]] = [[e1 , e2 ], e1 ] − [[e1 , e1 ], e2 ] = γ1 e4 − δ1 e4 , 0 = [e2 , [e2 , e1 ]] = [[e2 , e2 ], e1 ] − [[e2 , e1 ], e2 ] = β1 e4 − δ1 e4 , we get β1 = γ1 = δ1 .

Classification Problem in Low Dimensions  97

So, the table of multiplication of L has the following form:  [e1 , e1 ] = e3 , [e2 , e2 ] = β1 e3 + β2 e4 ,     [e , e ] = e , [e3 , e1 ] = e4 , 2 1 3     [e1 , e2 ] = β1 e3 + γ2 e4 , [e3 , e2 ] = β1 e4 . Taking e02 = e2 − β1 e1 we transform the table of multiplication to the following form:  [e2 , e1 ] = (1 − β1 )e3 , [e3 , e1 ] = e4 ,     [e , e ] = e , [e2 , e2 ] = β2 e4 , 1 1 3     [e1 , e2 ] = γ2 e4 . If β1 , 1, then taking the basis transformation e01 = (1 − β1 )e1 , e02 = e2 , e03 = (1 − β1 )2 e3 , e04 = (1 − β1 )3 e4 we obtain the family of algebras F1 (α, β) : [e1 , e1 ] = e3 , [e1 , e2 ] = αe4 , [e2 , e1 ] = e3 , [e2 , e2 ] = βe4 , [e3 , e1 ] = e4 . If β1 = 1, then we obtain the family of algebras F2 (α, β) : [e1 , e1 ] = e3 , [e1 , e2 ] = αe4 , [e2 , e2 ] = βe4 , [e3 , e1 ] = e4 . Case ii). Let α1 = 0. Then taking the base change e02 = e2 − δ1 e1 ,

e03 = e3 − α2 δ1 e4 ,

we derive δ1 = 0. From the equalities γ1 δ2 e3 = [e1 , γ1 e3 + γ2 e4 ] = [e1 , [e1 , e2 ]] = [[e1 , e1 ], e2 ] − [[e1 , e2 ], e1 ] = −γ1 e4 , γ1 δ3 e3 = [e2 , γ1 e3 + γ2 e4 ] = [e2 , [e1 , e2 ]] = [[e2 , e1 ], e2 ] − [[e2 , e2 ], e1 ] = −β1 e4 , we get

γ1 (δ2 + 1) = 0,

β1 = −γ1 δ3 .

Further, the equalities δ2 e4 = [e1 , e3 ] = [e1 , [e2 , e1 ]] = [[e1 , e2 ], e1 ] − [[e1 , e1 ], e2 ] = γ1 e4 ,

98  Leibniz Algebras

δ3 e4 = [e2 , e3 ] = [e2 , [e2 , e1 ]] = [[e2 , e2 ], e1 ] − [[e2 , e1 ], e2 ] = β1 e4 , 0 = [e1 , [e2 , e2 ]] = [e1 , β1 e3 + β2 e4 ] = δ2 β1 e4 , imply

δ2 β1 = 0,

γ1 = δ2 ,

β1 = δ3 .

Therefore, we get β1 = 0 and the table of multiplication of L has the following form  [e2 , e1 ] = e3 , [e3 , e1 ] = e4 ,     [e , e ] = α e , [e2 , e2 ] = β2 e4 , 1 1 2 4     [e1 , e2 ] = γ1 e3 + γ2 e4 , [e1 , e3 ] = γ1 e4 with the restriction on γ1 as follows γ1 (γ1 + 1) = 0. If γ1 = 0, then the base change e01 = e1 + e2 − (γ2 + α2 )e3 , e03 = e3 + β2 e4 gives the family of algebras F1 (α, β). If γ1 = −1, then we obtain the family of algebras F3 (α, β, γ) : [e1 , e1 ] = αe4 , [e2 , e2 ] = βe4 , [e2 , e1 ] = e3 , [e3 , e1 ] = e4 , [e1 , e2 ] = −e3 + γe4 , [e1 , e3 ] = −e4 , where in the latter case at least one of α, β, γ is not zero. Let us consider isomorphisms inside each of the families F1 (α, β) F1 (α, β) and F3 (α, β, γ). We start with F1 (α, β). The following cases may occur: Case 1.1: β = 0 and α , 0. Then the following base change reduces F1 (α, β) to F1 (1, 0): e01 = αe1 , e02 = αe2 , e03 = α2 e3 , e04 = α3 e4 . In this case we get R2 . Case 1.2: β = 0 and α = 0. Then F1 (0, 0) is R3 . Case 2: β , 0. By changing the basis {e1 , e2 , e3 , e4 } as follows 0

0

0

0

e1 = βe1 , e2 = βe2 , e3 = β2 e3 , e4 = β3 e4 the algebra F1 (α, β) can be reduced to F1 (α, 1).

Classification Problem in Low Dimensions  99

Thus we get the table of multiplication for the family of algebras F1 (α, 1) with parameter α as follows [e1 , e1 ] = e3 , [e1 , e2 ] = αe4 , [e2 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 . Now we specify isomorphisms within F1 (α, 1). It is not difficult to see that the following change of the basis e01 = A1 e1 + A2 e2 , e02 = (A1 + A2 )e2 + A2 (α − 1)e3 , e03 = A1 (A1 + A2 )e3 + A2 (A1 α + A2 )e4 , e04 = A21 (A1 + A2 )e4 where A1 , 0, ±A2 transforms algebras from F1 (α, 1) either to R4 (1) : [e1 , e1 ] = e3 , [e2 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 , [e1 , e2 ] = e4 or to R4 (0) : [e1 , e1 ] = e3 , [e2 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 . Indeed, for α, α , A1 , A2 we obtain the following relations α = 0

0

A1 α+A2 A21

and A2 = A21 − A1 . Clearly at α = 1 we get α = 1. But if α , 1 then 0 setting A1 = 1 − α we obtain α = 0. We show that the exposed algebras are not isomorphic. Note that the dimensions of maximal abelian subalgebras of the algebras R2 and R4 (α) are different and therefore, R2 can not be isomorphic to the algebra R4 (α) for any α ∈ {0, 1}. The algebra R3 is not isomorphic to the algebras R2 and R4 (α) for any α ∈ {0, 1}, by dimension reasons of the right annihilators. Let us now consider the family of algebras F2 (α, β). We treat again a few cases. Let β = 0 and α , 0. The transformation 0

0

0

0

0

e1 = e1 , e2 = α−1 e2 , e3 = e3 , e4 = e4 gives F2 (1, 0) that is R5 . Suppose that β , 0. Then it is easy to see that the transformation e01 = βe1 , e02 = βe2 , e03 = β2 e3 , e04 = β3 e4 leads to the type F2 (α, 1) : [e1 , e1 ] = e3 , [e1 , e2 ] = αe4 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 .

100  Leibniz Algebras

Note that the possible base change inside of F2 (α, 1) is as follows e01 = A1 e1 + A2 e2 ,

e02 = B2 e2 − A1 A2 B2 −1 e3 ,

e03 = A21 e3 + A2 (A1 α + A2 )e4 ,

e04 = A31 e4 ,

2) and B22 = A31 . Now we put where A1 B2 , 0. This yields α0 = B2 (A1Aα+A 3 1 A2 = −A1 α to obtain α0 = 0. Therefore the algebras F2 (α, 1) are isomorphic to the algebra R6 with the following table of multiplications:

[e1 , e1 ] = e3 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 . We note that the algebra F2 (0, 0) is split. It just remains to show that the obtained algebras are not isomorphic to each other. Indeed, the algebra R5 is not isomorphic to the algebra R6 , by dimension reasons of the left annihilators. Let us consider the last family of filiform Leibniz algebras: F3 (α, β, γ). An algebra of the family F3 (α, β, γ) is non Lie if and only if at least one of α, β, γ is not zero. Since we know already the classification of four-dimensional filiform Lie algebras (the non split filiform Lie algebra of dimension four has the table of multiplication [e2 , e1 ] = e3 , [e3 , e1 ] = e4 , [e1 , e2 ] = −e3 , [e1 , e3 ] = −e4 ), we shall consider only non Lie algebras of the family F3 (α, β, γ), that is, at least one of α, β, γ is not zero. Without loss of generality, one can assume that α , 0. Taking the transformation e01 = e1 , e02 = αe2 , e03 = αe3 , e04 = αe4 we obtain α = 1. To treat the family F3 (1, β, γ) we consider the general base change e01 = A1 e1 + A2 e2 + A3 e3 , e02 = B1 e1 + B2 e2 + B3 e3 . Then express the new basis {e01 , e02 , e03 , e04 } of the algebras from 0 0 F3 (1, β , γ ) with respect to the old basis {e1 , e2 , e3 , e4 } and comparing the coefficients we obtain the following relations 0

A21 + A1 A2 γ + A22 β = A21 B2 , B1 = 0, A21 B2 , 0, 0

β =

B2 β A1 γ + 2A2 β 0 , γ = . 2 A1 A21

Classification Problem in Low Dimensions  101

Note that 02

0

γ − 4β =

1 2 (γ − 4β). A21

Consider the case β = 0. Then β = 0. 0 So, in this case if γ = 0, then γ = 0 and we get R7 . But if γ , 0, 0 then setting A1 = γ, B2 = 1 and A2 = 0 we obtain γ = 1, and we get R8 . 0 A2 Consider the case β , 0. Then putting B2 = β1 we get β = 1. 0

−2A γ+4A2

1 1 and as A1 any non zero If γ2 − 4β = 0, then taking A2 = γ2 0 complex number we obtain γ = 2. Thus, q in this case we getqR9 .

γ 4β−γ If γ2 − 4β , 0, then putting A1 = 4β−γ and A2 = − 2β we 4 4 0 obtain γ = 0 and it is R10 . Now we suppose that the algebra L has the type iii) in Proposition 3.4, that L is a Leibniz algebra with χ(L) := (4,2,0,0) or (4,1,0,0), in particular, L is associative. Then all the results on associative algebras are applicable to this case and we deal with associative algebras. We consider the associative algebra A = L ⊕ C1. It is fivedimensional and unital. Now by using properties of L we remove from Mazzola’s list [127] inappropriate algebras. The first condition concerns the number of central idempotents. In view of this condition A is not isomorphic to the algebras with numbers 1–15, 25, 26, 38, 39, 55. Then in view of Corollary 3.1 and due to the fact that the image of an idempotent element under isomorphism is an idempotent element we obtain that A is not isomorphic to each of the algebras 16–22, 27, 28, 29, 40, 41, 46, 47, 52, 58 in the list of Mazzola. Moreover, the condition L3 = 0 implies that A is not isomorphic to the algebras 23, 24, 33, 34, 44. Now it remains to pick up the remaining algebras. 2

2

N 30. Let A = C hx, yi /(xy + yx, xy − yx + y2 − x2 ) + (x, y)3 . Choose the following basis of A: e0 = 1, e1 = x, e2 = y, e3 = xy, e4 = x2 . Then the subalgebra L = Span{e1 , e2 , e3 , e4 } has the composition law: [e1 , e1 ] = e4 , [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e2 , e2 ] = −2e3 + e4 . This is the algebra R11 . N 31. Let A = C hx, yi /(x2 , xy + y2 ) + (x, y)3 . Choose the basis of A

102  Leibniz Algebras

as e0 = 1, e1 = x, e2 = y, e3 = xy, e4 = yx. Then the subalgebra L = Span{e1 , e2 , e3 , e4 } has the following composition law: [e1 , e2 ] = e3 , [e2 , e1 ] = e4 , [e2 , e2 ] = −e3 . This coincides with R12 . N 32. Let A = C hx, yi /(xy + y2 , αx2 + yx) + (x, y)3 . As a basis of A we take the vectors e0 = 1, e1 = x, e2 = y, e3 = x2 , e4 = xy of A. Consider the subalgebra L = Span{e1 , e2 , e3 , e4 }. It has the following table of multiplication: [e1 , e1 ] = e3 , [e1 , e2 ] = e4 , [e2 , e1 ] = −αe3 , [e2 , e2 ] = −e4 . For distinct values of the parameter α these algebras are not isomorphic. In this case L is R13 (α). N 35. Let A = C hx, y, zi /(xz, yz, zx, zy, x2 − y2 , x2 − z2 , xy + yx, αx2 + yx), where α , 0. Choose the following basis for A : e0 = 1, e1 = x, e2 = y, e3 = z, e4 = x2 . Then the subalgebra L = Span{e1 , e2 , e3 , e4 } has the following composition law: [e1 , e1 ] = e4 , [e1 , e2 ] = αe4 , [e2 , e1 ] = −αe4 , [e2 , e2 ] = e4 , [e3 , e3 ] = e4 and it is R14 (α). We note that the algebras R14 (α1 ) and R14 (α2 ) (α1 , α2 ) are not isomorphic except of the case, where α2 = −α1 , in the latter case they are isomorphic [127]. N 36. Let A = C hx, y, zi /(x2 , yz, zy, z2 , xy− xz, xy+yx, yx+y2 , yx+zx). As the required basis here we choose e0 = 1, e1 = x, e2 = y, e3 = z, e4 = xy. Then the subalgebra L = Span{e1 , e2 , e3 , e4 } has the following law of multiplication [e1 , e2 ] = e4 , [e1 , e3 ] = e4 , [e2 , e1 ] = −e4 , [e2 , e2 ] = e4 , [e3 , e1 ] = e4 and it is isomorphic to R15 from the theorem. N 37. Let A = C hx, y, zi /(xz, y2 , yz, zx, zy, x2 − z2 , x2 − xy, x2 + yx). As a basis we take e0 = 1, e1 = x, e2 = y, e3 = z, e4 = xy. Then the subalgebra L = Span{e1 , e2 , e3 , e4 } is isomorphic to the algebra: R16 : [e1 , e1 ] = e4 , [e1 , e2 ] = e4 , [e2 , e1 ] = −e4 , [e3 , e3 ] = e4 .

Classification Problem in Low Dimensions  103

N 42. Let A = C hx, yi /(x2 , y2 )+(x, y)3 . As a basis we take e0 = 1, e1 = x, e2 = y, e3 = xy, e4 = yx. Then the table of multiplication of the subalgebra L = Span{e1 , e2 , e3 , e4 } is the same as: R17 : [e1 , e2 ] = e3 , [e2 , e1 ] = e4 . N 43. Let A = C[x, y]/(y3 , xy, x3 ) and choose the basis e0 = 1, e1 = x, e2 = y, e3 = x2 , e4 = y2 of A. Then the subalgebra L = Span{e1 , e2 , e3 , e4 } with the table of multiplication: [e1 , e1 ] = e3 , [e2 , e2 ] = e4 is split and it is isomorphic to the algebra R23 . N 48. Let A = C hx, yi /(x2 , xy + yx) + (x, y)3 . As a basis we choose e0 = 1, e1 = x, e2 = y, e3 = xy, e4 = y2 . Then the subalgebra L = Span{e1 , e2 , e3 , e4 } coincides with the algebra: R18 : [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e2 , e2 ] = e4 . N 49 (α = 1). Let A = C hx, yi /(x2 , xy) + (x, y)3 . As a basis we take e0 = 1, e1 = x, e2 = y, e3 = y2 , e4 = yx. Then the subalgebra L = Span{e1 , e2 , e3 , e4 } is isomorphic to: R19 : [e2 , e1 ] = e4 , [e2 , e2 ] = e3 . N 49 (α , 1). Let us now consider N49(α , 1) i.e.,   A = C hx, yi / x2 , (1 + α)xy + (1 − α)yx + (x, y)3 . As a basis of A we can choose the vectors e0 = 1, e1 = x, e2 = y, e3 = y2 , e4 = xy and the subalgebra L = Span{e1 , e2 , e3 , e4 } is isomorphic to R20 (α) : [e1 , e2 ] = e4 , [e2 , e1 ] =

(1 + α) e4 , [e2 , e2 ] = e3 . (1 − α)

It is easy to see that for different values of α we obtain non isomorphic algebras.

104  Leibniz Algebras

N 50. Let A = C hx, y, zi /(x2 , xz, y2 , yz, zx, zy, xy+yx, yx+z2 ). As a basis of A we choose the vectors e0 = 1, e1 = x, e2 = y, e3 = z, e4 = xy. Then the subalgebra L = Span{e1 , e2 , e3 , e4 } is isomorphic to the algebra: R21 : [e1 , e2 ] = e4 , [e2 , e1 ] = −e4 , [e3 , e3 ] = e4 . N 51. Let A = C hx, y, zi /(xz, yz, zx, zy, x2 − y2 , x2 − z2 , xy, yx). As a basis we choose the set of vectors e0 = 1, e1 = x, e2 = y, e3 = z, e4 = x2 . Then the subalgebra L = Span{e1 , e2 , e3 , e4 } has the following table of multiplication: [e1 , e1 ] = e4 , [e2 , e2 ] = e4 , [e3 , e3 ] = e4 . But this algebra can be included into the family of algebras R14 (α) with α = 0. Due to Proposition 3.2 the obtained algebras are not pairwise isomorphic. By adding Lie algebras and split Leibniz algebras, i.e., Leibniz algebras which are direct sums of proper ideals we get the algebras R22 − R28 .  Note 3.1. All the other algebras of Mazzola’s list with L3 = {0} are either Lie algebras or split Leibniz algebras. Summarizing the classification result of the above theorem, the result on classifications of complex nilpotent Lie algebras with dimensions at most four and complex nilpotent Leibniz algebras of dimensions at most three, we obtain the complete classification of complex nilpotent Leibniz algebras of the dimension at most four. 3.4 FOUR-DIMENSIONAL SOLVABLE LEIBNIZ ALGEBRAS

Here we shall consider only non nilpotent and non split complex solvable Leibniz algebras. This section along with [50], [54] and [55] completes the classification of four-dimensional solvable Leibniz algebras. We divide this section into two subsections. In the first subsection we study four-dimensional solvable Leibniz algebras with a twodimensional nilradical, while the second part is devoted to the study of solvable Leibniz algebras with a three-dimensional nilradical.

Classification Problem in Low Dimensions  105

Let L be a solvable Leibniz algebra. Then it can be decomposed as a direct sum of vector spaces L = N + Q, where N is the nilradical and Q is the complementary vector space. Since the square of a solvable algebra is a nilpotent ideal (due to Proposition 2.2) we obtain that L2 is nilpotent, i.e., L2 ⊆ N and consequently, Q2 ⊆ N. Definition 3.2. Let d1 , d2 , . . . , dn be derivations of a Leibniz algebra L. The derivations d1 , d2 , . . . , dn are said to be nil-independent if any non trivial linear combination a1 d1 + a2 d2 + · · · + an dn is not nilpotent, where a1 , a2 , . . . , an ∈ C. In other words, if for a1 , a2 , . . . , an ∈ C there exists a natural number k such that (a1 d1 + a2 d2 + · · · + an dn )k = 0 then a1 = a2 = . . . = an = 0. Lemma 3.1. Suppose that for x ∈ Q the operator R x |N is nilpotent. Then the subspace V = Span{x + N} is a nilpotent ideal of the algebra L. Proof. Since L2 ⊆ N, the subspace V is an ideal of L. We claim that it is nilpotent as well. If a ∈ N, then Ra |N is a nilpotent operator. Let us suppose that k ∈ N is such that (Ra |N )k = 0, then (Ra |V )k+1 = 0. Hence Ra |V is nilpotent. Since V is an ideal of the solvable Leibniz algebra L, then the space of inner derivations Inner(V) of V is a solvable Lie subalgebra of End(V), and thus by Lie theorem for Lie algebras there exists a basis such that Ra |V and R x |V are upper triangular; moreover, Ra |V is nilpotent, which means that Ra |V has zero diagonal elements. On the other hand, by the assumption, R x |N is nilpotent, then with the similar argument as the previous one, there exists s ∈ N such that (R x |N ) s = 0, then (R x |V ) s+1 = 0. Summarizing, we obtain that Ra |V and R x |V are nilpotent and upper triangular, hence Ra |V + R x |V also is nilpotent. Thus, by Engel’s theorem for Leibniz algebras, V is a nilpotent ideal.  In [54], Casas et al. proved the following theorem, which is applied in this section. Theorem 3.4. Let L be a solvable Leibniz algebra and N be its nilradical. Then the dimension of the complementary vector space to N is not greater than the maximal number of nil-independent derivations of N.

106  Leibniz Algebras

Proof. We claim that for x ∈ Q, the operator R x |N is a non nilpotent outer derivation of N. Indeed, if there exists x ∈ Q such that the operator R x |N is nilpotent, then the subspace V = Span{x + N} is a nilpotent ideal of the algebra L by Lemma 3.1 contradicting the maximality of N. Let {x1 , . . . , xm } be a basis of Q. Then the operators R x1 |N , R x2 |N , . . . , R xm |N are nil-independent, since if for some scalars !k m P αi R xi |N = 0, then Rky|N = 0, where α1 , . . . , αm ∈ C we have i=1

y=

m P i=1

αi xi . Hence y = 0, and so αi = 0 for all i = 1, . . . , m. Therefore,

we see that the dimension of Q is bounded by the maximal number of nil-independent derivations of the nilradical N. Moreover, similar to the case of Lie algebras, for a solvable Leibniz algebra L we also have the inequality dim N ≥ dim2 L .  According to the theorem above the dimension of the nilradical of four-dimensional solvable Leibniz algebra can be equal to two or three. The classifications of two and three-dimensional nilpotent complex Leibniz algebras have been given in Section 3.1.2. We give the results in the following two theorems. Theorem 3.5. Let L be a two-dimensional nilpotent Leibniz algebra. Then L either is C2 (abelian) or isomorphic to µ1 : [e1 , e1 ] = e2 . Theorem 3.6. Let L be a three-dimensional nilpotent Leibniz algebra. Then L is isomorphic to one of the following pairwise non-isomorphic algebras: λ1 := λ2 := λ3 := λ4 (α) := λ5 := λ6 :=

C3 − abelian, µ1 ⊕ C, [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e1 , e1 ] = e3 , [e2 , e2 ] = αe3 , [e1 , e2 ] = e3 , [e2 , e1 ] = e3 , [e1 , e2 ] = e3 , [e1 , e1 ] = e2 , [e2 , e1 ] = e3 .

Further for a convenience we shall consider another form of the family of λ4 (α). In fact, we shall transform the family of algebras λ4 (α)

Classification Problem in Low Dimensions  107

to the form in which nil-independent derivations of new algebras λ04 , λ04 (β) have diagonal forms. Namely, by choosing an appropriate basis we represent the parametric familyλ4 (α) as two non isomorphic algebras:   [e2 , e1 ] = e3 , √ λ04 (β) :=   [e1 , e2 ] = βe3 , with β = √1−4α−1 , α , 14 , 1−4α+1   [e1 , e1 ] = e3 ,     0 λ4 :=  [e2 , e1 ] = e3 ,    [e , e ] = −e . 1 2 3 Later we shall use a description of the derivation algebras of threedimensional nilpotent complex Leibniz algebras given above. The required description is given in a matrix form as follows. Proposition 3.5. There exists a basis such that the derivations of the algebras λ1 , λ2 , λ3 , λ04 , λ04 (β), λ5 and λ6 are given as follows Der(λ1 ) = ( M3 (C); a1 a2 a3 !) Der(λ2 ) = A ∈ M3 (C)| A = 0 b2 b3 ; 0 0 2a 1 ( a1 a2 a3 !) b3 Der(λ3 ) = A ∈ M3 (C)| A = b1 b2 ; 0 0 a + b 1 2 ( a1 a2 a3 !) 0 Der(λ4 ) = A ∈ M3 (C)| A = 0 2a1 b3 ; 0 0 3a 1     a3     a1 0  0    0 b b Der(λ4 (β)) =  A ∈ M (C)| A = 2 3   3  ;  0 0 a1 + b2      a3     a1 0   b3  Der(λ5 ) =  A ∈ M3 (C)| A =  0 b2 ;     0 0 a + b 1 2     a3     a1 0   Der(λ6 ) =  A ∈ M3 (C)| A =  0 2a1 a2   ,  0 0 3a1  where ai , b j , ck ∈ C with i, j, k = 1, 2, 3. Proof. The proof is carried out by straightforward checking the derivation property and using the table of multiplications of the algebras. 

108  Leibniz Algebras

3.4.1

Four-dimensional solvable Leibniz algebras with two-dimensional nilradical

Let us now consider Leibniz algebras whose nilradical is twodimensional. Here is the classification result of these algebras. Theorem 3.7. Let L be a four-dimensional solvable Leibniz algebra with a two-dimensional nilradical. Then, it is isomorphic to one of the following pairwise non-isomorphic algebras with the basis {e1 , e2 , x, y} :    [e1 , x] = e1 ,      [e1 , x] = e1 ,         [e1 , x] = e1 , [e2 , y] = e2 ,  R2 :=  R1 :=  R3 :=  [e2 , y] = e2 ,    [e2 , y] = e2 ,   [x, e1 ] = −e1 ,      [y, e2 ] = −e2 .  [y, e ] = −e , 2 2 Note that the basis {e1 , e2 , x, y} is the extension of the basis {e1 , e2 } of the corresponding nilradicals. Proof. Recall that the classification of two-dimensional nilpotent Leibniz algebras is given by Theorem 3.5. Let L be a four-dimensional solvable Leibniz algebra with two-dimensional nilradical µ. Since the number of nil-independent derivations of µ1 is equal to 1, we conclude that µ must be an abelian algebra. Therefore, considering the basis {x, y, e1 , e2 }, we write the table of L as follows    [e1 , x] = a1 e1 + a2 e2 , [e2 , x] = a3 e1 + a4 e2 ,   [e , y] = b e + b e , [e , y] = b e + b e . 1

1 1

2 2

2

3 1

4 2

Since R x and Ry are nil-independent derivations of µ, we conclude that a1 , 0. Without loss of generality, we can assume that a1 = 1, and taking the basis transformation y0 = y − b1 x we conclude that b1 = 0. It is easy to prove that the matrix of R x may have one of the following forms:     1 0 1 1 or 0 a4 0 1 . Therefore, we consider the following cases:     1 0 1 a 2 Case 1. If a a ' 0 a , then we can write: 3 4 4    [e2 , x] = a4 e2 , [e1 , x] = e1 ,   [e1 , y] = b2 e2 , [e2 , y] = b3 e1 + b4 e2 .

Classification Problem in Low Dimensions  109

Applying the Leibniz identity to [e1 , [x, y]] and [e2 , [x, y]], we get the restrictions b2 (1 − a4 ) = 0 and b3 (1 − a4 ) = 0. Thus, we have to consider the following subcases:   0 b Case 1.1. Let a4 = 1. It is clear that Ry :' b b2 is congruent 3 4 to one of the following matrices:     1 0 1 1 or 0 1 . 0 b 4

  1 0 It is easy to see that Ry :' 0 b , with b4 , 1, other4 wise R x − Ry would be nilpotent, which is impossible. 4 Taking the basis transformation x0 = b4b−1 x − b41−1 y and y0 = b41−1 y − b41−1 x, we obtain the following products [e1 , x] = e1 ,

[e2 , x] = 0,

[e1 , y] = 0,

[e2 , y] = e2 .

Case 1.2. If a4 , 1, then we get b2 = b3 = 0. Since R x and Ry are nil-independent derivations of µ, we obtain b4 , 0. Considering the base change x0 = x − ab44 y, y0 = b14 y, we again get [e1 , x] = e1 , [e2 , x] = 0, [e1 , y] = 0, [e2 , y] = e2 .   1 1 Case 2. If R x :' 0 1 , applying the Leibniz identity to [e1 , [x, y]], we get b3 = b4 = 0, giving rise to a contradiction with the assumption that Ry is not nilpotent. Therefore, [e1 , x] = e1 ,

[e2 , x] = 0,

[e1 , y] = 0,

[e2 , y] = e2 .

Let us consider the remaining products in L. Suppose that   [x, e1 ] = α1 e1 + α2 e2 , [x, e2 ] = α3 e1 + α4 e2       [y, e1 ] = β1 e1 + β2 e2 , [y, e2 ] = β3 e1 + β4 e2 ,    [x, x] = c1 e1 + c2 e2 , [x, y] = c3 e1 + c4 e2 ,     [y, x] = d e + d e , [y, y] = d3 e1 + d4 e2 . 1 1 2 2

110  Leibniz Algebras

By using the basis change x0 = x − c1 e1 − c4 e2 , y0 = y − d1 e1 − d4 e2 , we get c1 = c4 = d1 = d4 = 0. Applying the Leibniz identity we obtain α2 = α3 = α4 = β1 = β2 = β3 = c2 = c3 = d2 = d3 = 0, and

α21 + α1 = 0,

β23 + β3 = 0.

As a result we have the following cases: Case 2.1. If α1 = β3 = 0, we have R1 . Case 2.2. If α1 = β3 = −1, we get R2 . Case 2.3. If (α1 , β3 ) = (0, −1) or (α1 , β3 ) = (−1, 0), we obtain R3 .  3.4.2

Four-dimensional solvable Leibniz algebras with three-dimensional nilradical

In this section we give a complete list of four-dimensional solvable complex Leibniz algebras with three-dimensional nilradical. We make use of the list of algebras given in Theorem 3.6. Note that for a four-dimensional solvable Leibniz algebra L with three-dimensional nilradical N, there exists a basis {e1 , e2 , e3 , x} such that the right multiplication operator R x is non-nilpotent derivation on N, where {e1 , e2 , e3 } is a basis of N. Proposition 3.6. Let L be a four-dimensional solvable Leibniz algebra, whose nilradical is three-dimensional abelian algebra. Then L is isomorphic to one algebras   of the following pairwise non isomorphic     [e1 , x] = e1 , [e1 , x] = e1 ,             [e2 , x] = µ2 e2 , [e2 , x] = µ2 e2 ,             [e3 , x] = µ3 e3 , [e3 , x] = µ3 e3 , L1 (µ2 , µ3 ) :=  L2 (µ2 , µ3 ) :=      [x, e1 ] = −e1 , [x, e1 ] = −e1 ,             [x, e2 ] = −µ2 e2 , [x, e2 ] = −µ2 e2 ,           where µ3 , 0, [x, e3 ] = −µ3 e3 ,

Classification Problem in Low Dimensions  111

   [e1 , x] = e1 ,       [e2 , x] = µ2 e2 ,     L3 (µ2 , µ3 ) :=  [e3 , x] = µ3 e3 ,      [x, e1 ] = −e1 ,      where µ2 , µ3 , 0,    [e1 , x] = e1 ,       [e2 , x] = µ2 e2 ,     L5 (µ2 ) :=  [x, e1 ] = −e1 ,      [x, e2 ] = −µ2 e2 ,      [x, x] = e3 .   [e1 , x] = e1 ,     L7 (µ2 ) :=  [e2 , x] = µ2 e2 ,    [x, x] = e . 3

  [e , x] = e1 ,    1  L4 (µ2 , µ3 ) :=  [e2 , x] = µ2 e2 ,    [e , x] = µ e , 3 3 3

   [e1 , x] = e1 ,       [e2 , x] = µ2 e2 ,     L6 (µ2 ) :=  [x, e1 ] = −e1 ,      [x, x] = e3 ,      where µ2 , 0,   [e1 , x] = e1 ,     L8 :=  [x, e1 ] = −e1 ,    [x, e ] = e , 2 3    [e1 , x] = e1 + e2 ,        [e2 , x] = e2 ,      [e1 , x] = e1 ,  L9 :=  L10 (µ3 ) :=  [e3 , x] = µ3 e3 ,   [x, e2 ] = e3 .     [x, e1 ] = −e1 − e2 ,      [x, e2 ] = −e2 ,     [e1 , x] = e1 + e2 ,      [e , x] = e + e ,   1 1 2     [e2 , x] = e2 ,       [e2 , x] = e2 ,        [e3 , x] = µ3 e3 , L11 :=  L12 (µ3 , 0) :=  [x, e1 ] = −e1 − e2 ,     [x, e1 ] = −e1 − e2 ,       [x, e ] = −e ,   2 2     [x, e2 ] = −e2 ,     [x, x] = e3 ,    [x, e3 ] = −µ3 e3 ,     [e1 , x] = e1 + e2 , [e1 , x] = e1 + e2 ,         L13 (µ3 ) :=  L14 :=  [e2 , x] = e2 , [e2 , x] = e2 ,       [e , x] = µ e , [x, x] = e , 3 3 3 3     [e1 , x] = e2 , [e1 , x] = e1 + e2 ,             [e2 , x] = e2 , [e3 , x] = e3 , L16 (α) :=  L15 (µ3 , 0) :=      [e3 , x] = µ3 e3 , [x, e1 ] = αe2 ,         [x, e ] = −e , [x, e ] = −µ e , 3 3 3 3 3

112  Leibniz Algebras

   [e1 , x] = e2 ,       [e3 , x] = e3 ,     L17 :=  [x, e1 ] = −e2 ,      [x, e3 ] = −e3 ,      [x, x] = e2 .   [e , x] = e2 ,    1  L19 (α) :=  [e3 , x] = e3 ,    [x, e ] = αe , 1 2   [e , x] = e2 ,    1  L21 :=  [e3 , x] = e3 ,    [x, x] = e . 1    [e1 , x] = e1 + e2 ,       [e2 , x] = e2 + e3 ,       [e3 , x] = e3 , L23 :=    [x, e1 ] = e1 − e2 ,       [x, e2 ] = e2 − e3 ,      [x, e3 ] = e3 , where α, µi ∈ C.

  [e1 , x] = e2 ,       [e3 , x] = e3 , L18 :=    [x, e3 ] = −e3 ,     [x, x] = e . 1   [e1 , x] = e2 ,       [e3 , x] = e3 , L20 :=    [x, e1 ] = −e2 ,     [x, x] = e . 2   [e1 , x] = e1 + e2 ,     L22 :=  [e2 , x] = e2 + e3 ,    [e , x] = e , 3 3

Proof. Let {x, e1 , e2 , e3 } be the basis of L extended from a basis {e1 , e2 , e3 } of N. It is easy to see that the matrix of R x may have one of the following forms:        µ1 0 0   µ1 1 0   µ1 1 0   0 µ2 0  ,  0 µ1 0  ,  0 µ1 1  . 0 0 µ3 0 0 µ3 0 0 µ1    µ1 0 0  Let R x :'  0 µ2 0  . Then the multiplication in L can be writ0 0 µ3 ten as follows:   [e1 , x] = µ1 e1 ,       [e2 , x] = µ2 e2 ,    [e3 , x] = µ3 e3 ,     [x, x] = δ e + δ e + δ e . 1 1 2 2 3 3

[x, e1 ] = α1 e1 + α2 e2 + α3 e3 , [x, e2 ] = β1 e1 + β2 e2 + β3 e3 , [x, e3 ] = γ1 e1 + γ2 e2 + γ3 e3 ,

Classification Problem in Low Dimensions  113

Applying the Leibniz identity to [x, [e1 , x]], [x, [e2 , x]] and [x, [e3 , x]], we obtain the following restrictions for the structure constants α2 (µ1 − µ2 ) = 0, α3 (µ1 − µ3 ) = 0, β1 (µ1 − µ2 ) = 0, β3 (µ2 − µ3 ) = 0, (3.23) γ1 (µ1 − µ3 ) = 0, γ2 (µ2 − µ3 ) = 0. Applying the Leibniz identity to [x, [x, e1 ]], [x, [x, e2 ]] and [x, [x, e3 ]], we get (µ1 + α1 )α1 + α2 β1 + α3 γ1 = 0, (µ1 + α1 )α2 + α2 β2 + α3 γ2 = 0, (µ1 + α1 )α3 + α2 β3 + α3 γ3 = 0, β1 α1 + (β2 + µ2 )β1 + β3 γ1 = 0, β1 α2 + (β2 + µ2 )β2 + β3 γ2 = 0, β1 α3 + (β2 + µ2 )β3 + β3 γ3 = 0, γ1 α1 + γ2 β1 + (γ3 + µ3 )γ1 = 0, γ1 α2 + γ2 β2 + (γ3 + µ3 )γ2 = 0, γ1 α3 + γ2 β3 + (γ3 + µ3 )γ3 = 0,

(3.24)

and the Leibniz identity applied to [x, [x, x]] gives α1 δ1 + β1 δ2 + γ1 δ3 = 0, α2 δ1 + β2 δ2 + γ2 δ3 = 0, α3 δ1 + β3 δ2 + γ3 δ3 = 0.

(3.25)

To specify the structure constants, we distinguish the following cases: Case 1. Let µ1 , µ2 , µ1 , µ3 , µ2 , µ3 . Thanks to the restrictions (3.23) and (3.24) we get α2 = α3 = β1 = β3 = γ1 = γ2 = 0, (µ1 + α1 )α1 = 0, (β2 + µ2 )β2 = 0, (γ3 + µ3 )γ3 = 0, α1 δ1 = 0, β2 δ2 = 0, γ3 δ3 = 0.

(3.26)

Case 1.1. Let µ1 µ2 µ3 , 0. Then, without loss of generality, we can assume that µ1 = 1. Therefore, we have to consider the following cases: • If α1 , 0, β2 , 0, γ3 , 0, then according to the restriction (3.26), we have α1 = −1,

β2 = −µ2 ,

γ3 = −µ3 ,

δ1 = δ2 = δ3 = 0.

Hence we obtain L1 (µ2 , µ3 ) for µ2 , 1, µ3 , 1, µ2 , µ3 and µ2 µ3 , 0.

114  Leibniz Algebras

• If one of the parameters α1 , β2 , γ3 equals to zero and the other two do not, then without loss of generality we may assume that α1 , 0, β2 , 0 and γ3 = 0. Due to the restriction (3.26), we come to α1 = −1,

β2 = −µ2 ,

δ1 = δ2 = 0.

It suffices to make the transformation x0 = x − µδ33 e3 , to get δ3 = 0. As a result we obtain L2 (µ2 , µ3 ) for µ2 , 1, µ3 , 1, µ2 , µ3 and µ2 µ3 , 0. • If two of the parameters α1 , β2 , γ3 are equal to zero and another one is not zero, then, without loss of generality, we can suppose that α1 , 0, β2 = 0 and γ3 = 0, . Due to the restriction (3.26), we get α1 = −1,

δ1 = 0.

Considering the base change x0 = x − µδ22 e2 − µδ33 e3 , we get δ2 = δ3 = 0. This gives L3 (µ2 , µ3 ) for µ2 , 1, µ3 , 1, µ2 , µ3 and µ2 µ3 , 0. • If α1 = 0, β2 = 0 and γ3 = 0, then applying the base change x0 = x − µδ12 e1 − µδ22 e2 − µδ33 e3 we get δ1 = δ2 = δ3 = 0. Hence we obtain L4 (µ2 , µ3 ) with µ2 , 1, µ3 , 1, µ2 , µ3 and µ2 µ3 , 0. Case 1.2. Let one of the parameters µ1 , µ2 , µ3 be equal to zero. Then, without loss of generality, we can assume that µ1 = 1, µ3 = 0. Moreover, according to the restriction (3.26), we have γ3 = 0. • Let α1 , 0, β2 , 0. Then, by the restriction (3.26), we get α1 = −1, β2 = −µ2 , δ1 = δ2 = 0. Therefore the following cases occur: · If δ3 = 0, we obtain L1 (µ2 , 0) with µ2 < {0, 1}. · If δ3 , 0, by taking the basis transformation e03 = δ3 e3 , we get L5 (µ2 ) with µ2 < {0, 1}. • Let only one of the parameters α1 , β2 equal zero. Then, without loss of generality, we can assume that α1 , 0

Classification Problem in Low Dimensions  115

and β2 = 0. Moreover, due to the restriction (3.26) we conclude that α1 = −1,

δ1 = 0.

δ2 e µ2 2

The base change x = x − gives δ2 = 0. Again the following cases may occur: · If δ3 = 0, we derive the algebra which is isomorphic to L2 (0, µ3 ) with µ3 < {0, 1}. · If δ3 , 0, by taking the basis transformation e03 = δ3 e3 , we obtain L6 (µ2 ) with µ2 < {0, 1}. • Let α1 = 0, β2 = 0. We apply the basis transformation x0 = x − µδ11 e2 − µδ22 e2 , to obtain δ1 = δ2 = 0. It is sufficient here to consider the following cases: · If δ3 = 0, then we get L4 (µ2 , 0) with µ2 < {0, 1}. · If δ3 , 0, then taking the basis transformation e03 = δ3 e3 , we get L7 (µ2 ) with µ2 < {0, 1}. 0

Case 2. Let two of the structure constants µ1 , µ2 , µ3 be equal. Then, without loss of generality, we can assume that µ1 = µ2 . We have to distinguish the following cases: Case 2.1. Let µ1 = µ2 , 0. Then it is clear that we can assume that µ1 = µ2 = 1 and µ3 , 1. Moreover, due to (3.23), (3.24) and (3.25), we come to the following restrictions: α3 = β3 = γ1 = γ2 = 0, (1 + α1 )α1 + α2 β1 = 0, (1 + α1 )α2 + α2 β2 = 0, β1 α1 + (1 + β2 )β1 = 0, β1 α2 + (1 + β2 )β2 = 0, (3.27) (γ3 + µ3 )γ3 = 0, α1 δ1 + β1 δ2 = 0, α2 δ1 + β2 δ2 = 0, γ3 δ3 = 0. Note that for y ∈ {e1 , e2 }, we have [y, x] = y. Thus, by any change of the basis {e1 , e2 } the products [e1 , x] = e1 and [e2 , x] = e2 are  unchanged.  α α The matrix β 1 β 2 may have the following two Jor 1 2   α 0 α 1 dan forms 01 β and 01 α . 2

1

116  Leibniz Algebras

    α α α 0 Case 2.1.1. If β 1 β 2 :' 01 β , i.e., α2 = β1 = 0, then 1 2 2 thanks to (3.27), we have the following restrictions: (1 + α1 )α1 ,

(1 + β2 )β2 = 0,

α1 δ1 = 0,

β2 δ2 = 0,

(γ3 + µ3 )γ3 = 0, γ3 δ3 = 0.

It is clear that this case is similar to that of Case 1, with µ1 = µ2 = 1. Hence, similarly to Case 1 we get the following algebras: • If µ3 , 0, · We get L1 (1, µ3 ) with µ3 , 1, by considering α1 , 0, β2 , 0 and γ3 , 0. · We obtain L2 (1, µ3 ) with µ3 , 1, by considering α1 , 0, β2 , 0 and γ3 = 0. · We derive L2 (µ2 , 1) with µ2 , 1, by considering α1 , 0, β2 = 0 and γ3 , 0. · We get L3 (1, µ3 ) with µ3 , 1, by considering α1 , 0, β2 = 0 and γ3 = 0. · And finally we obtain L4 (1, µ3 ) with µ3 , 1, by considering α1 = 0, β2 = 0 and γ3 = 0. • If µ3 = 0, we obtain · L5 (1) by considering α1 , 0 and β2 , 0. · L6 (1) by considering α1 , 0 and β2 = 0. · L7 (1) by considering α1 = 0 and β2 = 0.     α α α1 1 Case 2.1.2. If β 1 β 2 :' 0 α1 , i.e., α2 = 1, β1 = 1 2 0, β2 = α2 . Due to the restriction (3.27) we obtain the system of equations (1 + α1 )α1 = 0, 1 + 2α1 = 0, which has no solution. Therefore in this case such algebra does not exist. Case 2.2. Let µ1 = µ2 = 0. Then, without loss of generality, we can take µ3 = 1. A simple basis transformations shows that µ1 = 1, µ2 = µ3 = 0.

Classification Problem in Low Dimensions  117

Due to the restrictions (3.23,) (3.24) and (3.25), we get α2 = α3 = β1 = γ1 = 0 and β2 β3 + β3 γ3 = 0, (1 + α1 )α1 = 0, β22 + β3 γ2 = 0, γ2 β2 + γ3 γ2 = 0, γ2 β3 + γ32 = 0, α1 δ1 = 0, β2 δ2 + γ2 δ3 = 0, β3 δ2 + γ3 δ3 = 0. (3.28) Similar to Case 2.1, we have to consider the following two subcases:     β2 β3 β2 0 :' 0 γ , i.e., β3 = γ2 = 0. By Case 2.2.1. Let γ γ 3 2 3 (3.28), we obtain the following restrictions β2 = γ3 = 0, (1 + α1 )α1 = 0 and α1 δ1 = 0. · If α1 = −1, then δ1 = 0. If (δ2 , δ3 ) = (0, 0), we obtain L1 (0, 0). If (δ2 , δ3 ) , (0, 0), we make the base change e03 = δ2 e2 + δ3 e3 , to obtain the algebra L5 (0). · If α1 = 0, then the base change x0 = x − δ1 e1 , gives [x, x] = δ2 e2 + δ3 e3 . If (δ2 , δ3 ) = (0, 0), we get L4 (0, 0). If (δ2 , δ3 ) , (0, 0), it suffices to make the basis transformation e03 = δ2 e2 + δ3 e3 , to obtain L7 (0).     β β β2 1 Case 2.2.2. If γ2 γ3 :' 0 β2 , i.e., β3 = 1, γ2 = 0 2 3 and γ3 = β2 . Due to (3.28), we have the following restrictions β2 = 0, (1 + α1 )α1 = 0, α1 δ1 = 0 and δ2 = 0. · If α1 = −1, hence δ1 = 0 and the structural constant δ3 is determined by taking the change x0 = x − δ3 e2 . Thus, we obtain the algebra L8 . · If α1 = 0, then making the change of basis x0 = x − δ1 e1 − δ3 e2 , the algebra L9 is obtained. Case 3. Let µ1 = µ2 = µ3 = 1, then we consider the following subcases.   α1 α2 α3 !  α1 0 0  Case 3.1. Let β1 β2 β3 :'  0 β2 0  , i.e., α2 = α3 = γ1 γ2 γ3 0 0 γ3 β1 = β3 = γ1 = γ2 = 0.

118  Leibniz Algebras

In view of the restrictions (3.24) and (3.25), we have (1 + α1 )α1 = 0, (1 + β2 )β2 = 0, (1 + γ3 )γ3 = 0, α1 δ1 = 0, β2 δ2 = 0 and γ3 δ3 = 0. It is clear that this case is similar to Case 1, with µ1 = µ2 = µ3 = 1. Hence, by treating similarly, we get the following algebras: L1 (1, 1) by considering α1 , 0, β2 , 0 and γ3 , 0. L2 (1, 1) by considering α1 , 0, β2 , 0 and γ3 = 0. L3 (1, 1) by considering α1 , 0, β2 = 0 and γ3 = 0. L4 (1, 1) by considering α1 = 0, β2 = 0 and γ3 = 0.   α1 α2 α3 !  α1 1 0  Case 3.2. Let β1 β2 β3 :'  0 α1 0  , i.e., α2 = 1, γ1 γ2 γ3 0 0 γ3 α3 = β1 = β3 = γ1 = γ2 = 0, and β2 = α1 . Due to the restrictions (3.24) and (3.25), we obtain the following system of equations: • • • •

(1 + α1 )α1 = 0,

1 + 2α1 = 0,

which has no solution. Therefore this case is impossible.   α1 α2 α3 !  α1 1 0   Case 3.3. Let β1 β2 β3 :'  0 α1 1  , i.e., α2 = β3 = γ1 γ2 γ3 0 0 α1 1, α3 = β1 = γ1 = γ2 = 0, β2 = γ3 = α1 . Due to the restrictions (3.24) and (3.25), we have the following system of equations: (1 + α1 )α1 = 0,

1 + 2α1 = 0,    µ1 1 0  which has no solution. Similarly, if R x :'  0 µ1 0  , 0 0 µ3 we obtain the algebras L (µ ) − L . On the other hand, if 10 3 21    µ1 1 0  R x :'  0 µ1 1  , the algebras L22 , L23 are obtained. 0 0 µ1  Proposition 3.7. Let L be a four-dimensional solvable Leibniz algebra, whose nilradical is isomorphic to λ2 . Then, L is isomorphic to one

Classification Problem in Low Dimensions  119

of the following pairwise non isomorphic algebras:     [e1 , e2 ] = e3 ,      [e1 , e1 ] = e3 ,       [e , x] = e ,   1 1     [e1 , x] = e1 ,       [e2 , x] = γe2 ,  L24 (γ) :=  L25 (δ) :=  [e2 , x] = δe2 ,     [e3 , x] = 2e3 ,       [e3 , x] = 2e3       [x, e ] = −e ,   1 1     [x, e1 ] = −e1 .   [x, e2 ] = −γe2 .       [e , e ] = e , [e1 , e1 ] = e3 ,   1 1 3           [e1 , x] = e1 , [e1 , x] = e1 ,         L26 :=  L27 :=  [e3 , x] = 2e3 , [e2 , x] = 2e2 + e3 ,           [x, e1 ] = −e1 , [e3 , x] = 2e3 ,           [x, x] = e2 . [x, e1 ] = −e1 .    [e1 , e1 ] = e3 ,        [e1 , x] = e1 + e2 ,     [e1 , e1 ] = e3 ,       [e2 , x] = e2 ,  L28 :=  L29 :=  [e2 , x] = e2 ,     [e3 , x] = 2e3 ,      [x, e2 ] = −e2 .    [x, e1 ] = −e1 − e2 ,      [x, e2 ] = −e2 .     [e1 , e1 ] = e3 ,     [e1 , e1 ] = e3 ,       [e2 , x] = e2 ,        [e2 , x] = e2 , L30 (λ) :=  L31 :=  [x, e1 ] = e3 ,     [x, e2 ] = −e2 ,       [x, e2 ] = −e2 ,      [x, x] = −2e3 .   [x, x] = λe3 .   [e1 , e1 ] = e3 ,          [e1 , e1 ] = e3 , [e2 , x] = e2 , L32 :=  L33 (µ) :=    [e2 , x] = e2 .  [x, e1 ] = e3 ,     [x, x] = µe . 3

120  Leibniz Algebras

  [e1 , e1 ] = e3 ,       [e2 , x] = e2 , L34 :=    [x, e1 ] = e3 ,     [e , x] = e . 1 3 where γ, λ, µ ∈ C and δ ∈ C \ {0}. Proof. The proof is similar to that of the previous propositions.



Below we present the classification of four-dimensional solvable Leibniz algebras whose nilradical is λ3 . Proposition 3.8. Let L be a four-dimensional solvable Leibniz algebra, whose nilradical is isomorphic to λ3 . Then, L is isomorphic to one of the following pairwise non-isomorphic algebras:     [e1 , e2 ] = e3 ,      [e1 , e2 ] = e3 ,       [e , e ] = −e ,   2 1 3     [e2 , e1 ] = −e3 ,       [e1 , x] = e1 ,       [e1 , x] = e1 ,       [e2 , x] = γe2 ,  L35 (γ) :=  L36 :=  [e2 , x] = −e2 ,     [e3 , x] = (1 + γ)e3 ,       [x, e1 ] = −e1 ,       [x, e1 ] = −e1 ,       [x, e2 ] = e2 ,       [x, e ] = −γe ,   2 2     [x, x] = e3 .   [x, e3 ] = −(1 + γ)e3 .    [e1 , e2 ] = e3 ,       [e2 , e1 ] = −e3 ,       [e1 , x] = e1 + e2 ,       [e2 , x] = e2 , L37 :=    [e3 , x] = 2e3 ,       [x, e1 ] = −e1 − e2 ,       [x, e2 ] = −e2 ,      [x, e3 ] = −2e3 .

Classification Problem in Low Dimensions  121

Proof. According to Proposition 3.5 and the multiplication law of λ3 , the multiplication of L can be written as follows: [e1 , e2 ] = e3 , [e2 , e1 ] = −e3 , [e1 , x] = a1 e1 + a2 e2 + a3 e3 , [e2 , x] = b1 e1 + b2 e2 + b3 e3 , [e3 , x] = (a1 + b2 )e3 . Since e1 , e2 < Annr (L), applying properties of the right annihilator we write: [x, e1 ] = −a1 e1 − a2 e2 + α3 e3 , [x, e2 ] = −b1 e1 − b2 e2 + β3 e3 , [x, e3 ] = −(a1 + b2 )e3 , [x, x] = γ3 e3 . Let us distinguish the following cases: Case 1. If (a1 , b2 ) , (0, 0), without loss of generality, we may assume that a1 , 0. Also note that b1 = 0. Otherwise, we could consider two cases: a2 = 0 or a2 , 0. In the first case, considering the base change e01 = e2 , e02 = e1 , e03 = −e3 , we get b1 = 0. On the other hand, √ if a2 , 0, applying the basis transformation −(b −a )+

(b −a )2 +4a b

2 1 e1 , we come to b1 = 0. It suffices e02 = e2 + 2 1 2a22 1 to make the basis transformation x0 = a11 x to obtain a1 = 1.

Let us consider the following cases: Case 1.1. If b2 < {0, 1}, then applying the basis transformation: e01 = e1 −

a3 (b2 − 1) − a2 b3 a2 e2 − e3 , b2 − 1 b2 (b2 − 1)

e02 = e2 − b3 e3 ,

we obtain a2 = a3 = b3 = 0. The Leibniz identity applied to [x, [e1 , x]], [x, [e2 , x]] and [x, [x, x]] gives α3 = 0,

β3 = 0,

γ3 (1 + b2 ) = 0.

Therefore there are the following options:

122  Leibniz Algebras

• If γ3 = 0, we get L35 (b2 ), where b2 < {0, 1}. • If γ3 , 0, then b2 = −1. We make the base change e01 = γ3 e1 ,

e03 = γ3 e3 ,

to obtain the algebra L36 . Case 1.2. If b2 ∈ {0, 1}, then e3 < Annr (L) and in view of [x, x], [ei , x] + [x, ei ] ∈ Annr (L) for 1 ≤ i ≤ 2, we get γ3 = 0, α3 = −a3 and β3 = −b3 . • If b2 = 0, the base change e01 = e1 + a2 e2 ,

e02 = e2 − b3 e3 ,

x0 = x − (a3 + a2 b3 )e2

gives a2 = a3 = b3 = 0. The result is L35 (0). • If b2 = 1, and we make the following change of basis: e01 = e1 − (a3 + α2 b3 )e3 ,

e02 = e2 − b3 e3 ,

we come to a3 = b3 = 0. Finally if a2 = 0, we obtain the algebra L35 (1). Otherwise, if a2 , 0, we assert a2 = 1 by considering the basis transformation e02 = a2 e2 , e03 = a2 e3 . Therefore, we get L37 . Case 2. If (a1 , b2 ) = (0, 0), due to the non nilpotency of R x , we have a2 b1 , 0. Taking the base change e01 = e1 + e2 , we obtain [e01 , x] = [e1 + e2 , x] = b1 (e1 + e2 ) + (a2 − b1 )e2 + (a3 + b3 )e3 = b1 e01 + a02 e2 + a03 e3 . Since b1 , 0, we come to Case 1.  Proposition 3.9. Let L be a four-dimensional solvable Leibniz algebra, whose nilradical is isomorphic to λ4 (α). Then, L is isomorphic to one of the following pairwise non isomorphic algebras:       [e2 , e1 ] = e3 , [e , e ] = e ,   2 1 3           [e1 , x] = e1 , [e1 , x] = e1 ,         L39 :=  L38 (γ) :=  [e2 , x] = −e2 , [e2 , x] = γe2 , γ ∈ C,           [x, e1 ] = −e1 , [e3 , x] = (1 + γ)e3 ,           [x, x] = e3 . [x, e1 ] = −e1 .

Classification Problem in Low Dimensions  123

   [e2 , e1 ] = e3 ,         [e1 , x] = e1 + e3 , [e2 , e1 ] = e3 ,         L40 :=  L41 :=  [e3 , x] = e3 , [e2 , x] = e2 ,          [x, e1 ] = −e1 , [e3 , x] = e3 .      [x, x] = −e2 .    [e2 , e1 ] = e3 ,       [e1 , e2 ] = βe3 ,       [e1 , x] = e1 ,     L42 (β) :=  [e2 , x] = βe2 ,      [e3 , x] = (β + 1)e3 ,       [x, e1 ] = −e1 ,      [x, e2 ] = −βe2 , √ n o , with α < 0, 14 where β = √1−4α−1 1−4α+1 Proof. We have the following options: 1. If N is isomorphic to λ04 (0) then applying the same approach as in Proposition 3.8, we obtain the algebras L38 (γ) – L41 . 2. If N is isomorphic to one of λ04 (α) (α , 0) then we get the family L42 (β).  The proofs of the following two propositions are similar to those of the above propositions. Proposition 3.10. Let L be a four-dimensional solvable Leibniz algebra, whose nilradical is isomorphic to λ5 . Then L is isomorphic to the following algebra   [e1 , e2 ] = e3 , [e1 , x] = e1 ,       [e2 , e1 ] = e3 , [e2 , x] = e2 , L43 :=    [x, e1 ] = −e1 , [e3 , x] = 2e3 ,     [x, e ] = −e . 2 2

124  Leibniz Algebras

Proposition 3.11. Let L be a four-dimensional solvable Leibniz algebra, whose nilradical is isomorphic to λ6 . Then L is isomorphic to the following algebra   [e , e ] = e2 , [e1 , x] = e1 ,    1 1  L44 :=  [e2 , e1 ] = e3 , [e2 , x] = 2e1 ,    [x, e ] = −e , [e , x] = 3e . 1 1 3 1 Remark. To verify that the above algebras are non isomorphic, we use a computer program, implemented in Mathematica. The computer code is available from the authors. 3.5 RIGIDITY OF LIE AND LEIBNIZ ALGEBRAS

In 1951 I.E.Segal [168] introduced the notion of contractions of Lie algebras on physical grounds: if two physical theories (like relativistic and classical mechanics) are related by a limiting process, then the associated invariance groups (like the Poincar´e and Galilean groups) should also be related by some limiting process. If the velocity of light is assumed to go to infinity, relativistic mechanics “transforms” into classical mechanics. This also induces a singular transition from the Poincar´e algebra to the Galilean one. Another example is a limiting process from quantum mechanics to classical mechanics under ~ −→ 0, that corresponds to the contraction of the Heisenberg algebras to the abelian ones of the same dimensions [62]. There are two approaches to the contraction problems of algebras. The first of them is based on physical considerations that are mainly oriented to applications of contractions. Contractions were used to establish connections between various kinematical groups and to shed a light on their physical meaning. In this way relationships between the conformal and Schr¨odinger groups were elucidated and various Lie algebras including a relativistic position operator were interrelated. Under dynamical group description of interacting systems, contractions corresponding to the coupling constant going to zero give non interacting systems. Application of contractions allows us to derive interesting results in the special function theory as well. The second consideration is purely algebraical, dealing with abstract algebraic structures. We will deal with this case and focus mainly on algebraic aspects of the contractions.

Classification Problem in Low Dimensions  125

Let V be a vector space of dimension n over an algebraically closed field F (charF=0). The set of bilinear maps V × V → V forms a vector space Hom(V ⊗ V, V) of dimension n3 , which can be considered together with its natural structure of an affine algebraic variety over F. It is denoted by Algn (F). An n-dimensional algebra A over F may be regarded as an element λ of Hom(V ⊗ V, V) via the bilinear mapping λ : A ⊗ A → A defining a binary algebraic operation on A. In this section, we have occasion to write its composition law λ instead of the algebra A for the sake of simplicity. In fact, in terms of λ the Leibniz identity is as follows λ((x, λ(y, z)) = λ(λ(x, y), z) − λ(λ(x, z), y). Recall, that the set of all n-dimensional Leibniz algebras over a field F is denoted by LBn (F). The set LBn (F) can be included in the above mentioned n3 -dimensional affine space as follows: let {e1 , e2 , . . . , en } be a basis of the Leibniz algebra L. Then the table of multiplication of L is represented by a point (γikj ) of this affine space as follows: n X λ(ei , e j ) = γikj ek . k=1

Thus, the algebra L corresponds to the point (γikj ) ∈ Fn . The Leibniz identity gives polynomial relations among γikj . Hence we regard 3 LBn (F) as a subvariety of Fn . The linear reductive group GLn (F) acts on Algn (F) by 3

(g ∗ λ)(x, y) = g(λ(g−1 (x), g−1 (y)))

[transport of structure].

Two algebras λ1 and λ2 are isomorphic if and only if they belong to the same orbit under this action. For given two algebras λ and µ we say that λ degenerates to µ, if µ lies in the Zariski closure of the orbit of λ. We denote this by λ → µ. In this case entire orbit Orb(µ) lies in the closure of Orb(λ). We denote this, as has been mentioned above, by λ → µ, i.e., µ ∈ Orb(λ). Degeneration is transitive, that is if λ → µ and µ → ν then λ → ν. 3 From now on we shall consider LBn (F) as a subvariety of Fn .

126  Leibniz Algebras

It is easy to see that the scalar matrices of GLn (F) act on LBn (F) scalarly. Therefore, the orbits Orb (L) are cones with the deleted vertex {0} that corresponds to the abelian complex algebra (denoted by Fn ). Thus, Fn belongs to Orb (L) for all L ∈ LBn (F). In particular, among the orbits Orb (L) only one is closed and that is the orbit of Fn . A Leibniz algebra λ is said to be degenerate to a Leibniz algebra µ, if µ is represented by a structure which lies in the Zariski closure of the GLn (F)-orbit of the structure which represents λ. There are algebras the orbits of which are open in LBn (F). These algebras are called rigid. In that case the corresponding algebra does not admit any non trivial deformation. The orbits of rigid algebras give irreducible components of the variety LBn (F). Hence to find the rigid orbits of variety of algebras is of great interest. By the Noetherian consideration they are of finite number. However, in general, not every irreducible component is generated by rigid algebras. Flanigan [78] has shown that there exists a component in Alg3 which consists of union of infinitely many orbits of non isomorphic algebras having the same dimension (so-called rigid families). If L is a rigid algebra in LBn (F) then there exists an irreducible component C of LBn (F) such that Orb(L) ∩ C is non empty open subset of C. The closure of Orb(L) is contained in C. Then the dimension of the irreducible component C of LBn (F) is given by dim C = n2 − dim AutL. Here is a proposition from [53] in Lie algebras case counting the number of irreducible components and the number of open orbits. Proposition 3.12. Let r(n) and s(n) be the number of irreducible components and the number of open orbits, respectively, in the variety Ln (C) of n-dimensional complex Lie algebras. Then one has (r(1), r(2), . . . , r(7)) = (1, 1, 2, 4, 7, 17, 49) and (s(1), s(2), . . . , s(7)) = (1, 1, 1, 2, 3, 6, 14). Let us recall a few useful facts from the algebraic groups theory, concerning the degenerations. The first of them concerns constructive subsets of algebraic varieties over C, the closures of which relative to Euclidean and Zariski topologies coincide. Since GLn (C)-orbits are 3 constructive sets, the usual Euclidean topology on Cn leads to the same

Classification Problem in Low Dimensions  127

degenerations as does the Zariski topology. Now we may express the concept of degeneration in a slightly different way, that is the following condition will imply that λ → µ : ∃gt ∈ GLn (C(t)) such that lim gt ∗ λ = µ, t→0

where C(t) is the field of fractions of the polynomial ring C[t]. The second fact concerns the closure of GLn (C)-orbits stating that the boundary of each orbit is a union of finitely many orbits with dimensions strictly less than the dimension of the given orbit. It follows that each irreducible component of the variety, on which algebraic group acts, contains only one open orbit that has a maximal dimension. It is obvious that in the content of variety of algebras the representatives of this kind orbits are rigid. It is an interesting and at the same time difficult problem to determine the number of irreducible components of an algebraic variety. The dimension of the algebraic variety can be found by using degeneration approach. In this case no need to find all the degenerations, just to find the rigid algebras and rigid families. The closure of the orbit of a rigid algebra gives a component of the variety. In order to find the dimension of the variety it is sufficient to find rigid algebras and rigid families having a maximal orbit dimension. Here are some cases where the degenerations of classes of algebras have been studied. For associative algebras Algn (C) : Alg4 (C) see [81], Alg5 (C) see [128] and [121], [127]; for nilpotent associative algebras case (see [122]); for the class of Jordan algebras see [107], [108], for nilpotent Lie algebras NLn (C) : at n ≤ 5 it can be found in [41], [99], [100] and NL6 was described by Seeley [166], NL7 and NL8 were investigated by Goze, Ancochea-Berm´udez [95]; the variety of filiform Lie algebras were investigated by Goze, Khakimdjanov Yu. [97]; for nilpotent Leibniz algebras in dimensions less than five, the geometric classification can be found in [13]. A slightly different approach to the geometric classification problem of algebras can be found in [70] [72], [73], [74], [75] and [76]. Let A be a complex n-dimensional algebra (underlying vector space is denoted by V) with the binary operation λ : V ⊗ V → V. Consider a continuous function gt : (0, 1] → GL(V). In other words, gt is a non singular linear operator on V for all t ∈ (0, 1]. Define a parameterized family λt of new algebra structures on V via the old binary operation λ

128  Leibniz Algebras

as follows: λt (x, y) = (gt ∗ λ)(x, y) = g−1 t λ(gt (x), gt (y)), x, y ∈ V. Definition 3.3. If the limit lim λt (x, y) = λ0 (x, y) exists for all x, y ∈ V, t→+0 then the algebraic structure λ0 defined by this way on V is said to be a contraction of the algebra A. Note. Obviously, the contractions can be considered in the basis level. Namely, let {e1 , e2 , . . . , en } be a basis of an n-dimensional algebra A. If the limit lim λt (ei , e j ) = λ0 (ei , e j ) exists then the algebra (V, λ0 ) is a t→+0 contraction of A. Definition 3.4. A contraction from an algebra A to algebra A0 is said to be trivial if A0 is abelian, and to be improper if A0 is isomorphic to A. Note that both trivial and improper contractions always exist. Here is an example of the trivial and the improper contractions. Example 3.1. Let A = (V, λ) be a complex n-dimensional algebra. If we take gt = diag(t, t, . . . , t) then gt ∗ λ is abelian and for gt = diag(1, 1, . . . , 1) we get gt ∗ λ = A. We equally use the notions of rigidity with respect to degeneration and deformation as the same, since the rigidity with respect to degeneration follows from the rigidity with respect to deformation, but the converse does not always hold. In the case of Lie algebras over a field characteristic zero, these two definitions of rigidity are equivalent [70],[122], [182]. By using the results of [70] this can be extended to Leibniz algebras case. Algebraic deformation theory was introduced, for associative algebras, in 1963 by Gerstenhaber [84], and it was extended to Lie algebras by Nijenhuis and Richardson [133], [134]. Following Gerstenhaber, Nijenhuis and Richardson, many authors have published papers on some aspect of the deformations of given type of algebraic structures (associative, commutative, Lie, Jordan, etc.). One of the fundamental concepts in deformation theory is the notion of rigidity. Loosely speaking, an algebra A is rigid if an arbitrary “infinitesimal” deformation of A produces an algebra A0 isomorphic

Classification Problem in Low Dimensions  129

to A. Focusing on Lie algebras, a geometric formulation of rigidity is given as follows. Let V be a complex vector space of dimension n and Ln denote the set of all Lie algebra structures on V. The set Ln has a natural structure of an affine variety. There is an action of the group GLn , of all automorphisms of V, on Ln . Under this action each orbit represents an isomorphism class of Lie algebra structures. A Lie algebra L is rigid if its orbit under the action of GLn is Zariski open in Ln . A theorem of Nijenhuis and Richardson [133] shows that the vanishing of the second cohomology of a Lie algebra, with coefficients in the adjoint representation, is a sufficient condition for a Lie algebra to be rigid. Lie algebras are part of a bigger class of Leibniz algebras. Balavoine [28] published a paper on the deformations of Leibniz algebras. In particular, using the theory of Fox [79], he obtained a result similar to that of Nijenhuis and Richardson, i.e., the vanishing of the second Leibniz cohomology is a sufficient condition for the rigidity of a Leibniz algebra. Since a Lie algebra is also a Leibniz algebra, we need to consider the rigidity of Lie algebras both as Lie algebras and as Leibniz algebras. 3.6

LEIBNIZ COHOMOLOGY COMPUTATIONS

This section is dedicated to the cohomology of Leibniz algebras, built on the computational methods of [69]; a large part of the material is due to [141]. 3.6.1

Leibniz cohomology

Let g be a Lie algebra over a field F, and M a left g-module. Cohomology theory associates to g two complexes, namely the ChevalleyEilenberg complex C ∗ (g, M) := (Hom(Λ∗ g, M), d), and the Leibniz or Loday complex CL∗ (g, M) := (Hom(⊗∗ g, M), d), for the Leibniz cohomology of g with values in the symmetric (or antisymmetric) Leibniz g-bimodule M. The coboundary operator on the complex C ∗ (g, M) is the standard Chevalley-Eilenberg coboundary operator, see e.g. [104]. The coboundary operator on CL∗ (g, M) is the

130  Leibniz Algebras

Leibniz or Loday coboundary operator d : CLn (g, M) → CLn+1 (g, M) defined by (d f )(x , . . . , xn+1 ) = x1 · f (x2 , . . . , xn+1 ) P1n+1 + Pi=1 (−1)i f (x1 , . . . , xˆi , . . . , xn+1 ) · xi + 1≤i< j≤n+1 (−1) j+1 f (x1 , . . . , xi−1 , [xi , x j ], xi+1 , . . . , xˆ j , . . . , xn+1 ) for any f ∈ CLn (g, M) and all elements x1 , . . . , xn+1 ∈ g. Leibniz cohomology is more generally defined for any Leibniz algebra h and any Leibniz h-bimodule M with the same coboundary operator d, see e.g. [69] for further details. With in the symmetric g-bimodule M, the natural epimorNvalues ∗ phism g → Λ∗ g induces a monomorphism of complexes ϕ : C ∗ (g, M) ,→ CL∗ (g, M), which is an isomorphism in degree 0 and 1. There is then a short exact sequence of complexes inducing a long exact sequence in cohomology which mediates between Chevalley-Eilenberg and Leibniz cohomology. It is shown in Lemma 1.5 in [69] that Leibniz cohomology of a Leibniz algebra h with values in an antisymmetric h-bimodule M a reduces to lower degree cohomology with values in a symmetric hbimodule: HL p (h, M a )  HL p−1 (h, Hom(h, M) s ) (3.29) for all p ≥ 1. As usual in cohomology, for a short exact sequence of Leibniz hbimodules 0 → M 0 → M → M 00 → 0, there is a long exact sequence in cohomology . . . → HLn (h, M 0 ) → HLn (h, M) → HLn (h, M 00 ) → HLn+1 (h, M 0 ) → . . . ,

for all n ≥ 0 and starting with a monomorphism HL0 (h, M 0 ) → HL0 (h, M). 3.6.2

Leibniz cohomology of rigid Lie algebras

In this subsection, fix the base field F to be the field C of complex numbers.

Classification Problem in Low Dimensions  131

Observe that in general H n (g, M) = 0 for a certain n does not necessarily imply that HLn (g, M) = 0. For example, the trivial Lie algebra g = F has H 2 (F, F) = 0, but HL2 (F, F) = F , 0, (see [71]). For another counter-example, a short computation with the Hochschild-Serre spectral sequence [104] shows that we have also H 2 (g, g) = 0 for the direct sum g = sl2 (C) ⊕ C, while HL2 (g, g)  C (see [71]). We will consider in this subsection the question for which finitedimensional Lie algebras g , F the hypothesis H 2 (g, g) = 0 implies that HL2 (g, g) = 0. This assertion is true for nilpotent Lie algebras of dimension ≥ 2, since Theorem 2 in [60] shows that for a non-trivial nilpotent Lie algebra dim H 2 (g, g) ≥ 2 as its center is non-trivial. The assertion is also true for (non-nilpotent) solvable Lie algebras. In order to show this, let us elaborate a little on an article of Carles [51]. Carles [51] investigates Lie algebras g possessing a onecodimensional ideal. For these, Carles shows that the dimension dim Der(g) of the Lie algebra of derivations Der(g) of g is greater or equal to the dimension dim g of g. As a corollary, this is true for Lie algebras g with [g, g] , g, because in this case g admits an ideal of codimension one. In a later section, Carles investigates Lie algebras g which satisfy dim Der(g) = dim g. He shows that any such a Lie algebra is algebraic and therefore admits a decomposition S ⊕ U ⊕ N where S is a Levi’s subalgebra, N is the nilradical and U is a subalgebra consisting of elements (called ad-semisimple) with [S + U, U] = 0 and such that the exterior torus (denoted by ad(U)) is algebraic. In addition, Carles shows that if codim [g, g] > 1, then g is complete, i.e. H 0 (g, g) = H 1 (g, g) = {0}. In [51] (Lemma 5.1), Carles shows that if [g, g] , g, dim Der(g) ≤ dim g + dim H 2 (g, g). Let us draw conclusions from these results with respect to the above question. Suppose that g is a finite-dimensional solvable Lie algebra with H 2 (g, g) = 0 and codim [g, g] > 1. By [51], we have on the one hand dim Der(g) ≥ dim g (by Corollary 2.20, because g is solvable it implies [g, g] , g and thus g admits an ideal of codimension one). On the other hand, we have dim Der(g) ≤ dim g (by Lemma 5.1, because H 2 (g, g) = 0). Thus we conclude that dim Der(g) = dim g, which implies by Proposition 3.1 in [51] that g is algebraic and, thanks to codim [g, g] > 1, g is complete, i.e., H 0 (g, g) = H 1 (g, g) = {0}. This implies that Center(g) = {0} and therefore by Theorem 3.8 below that HL2 (g, g) = {0}.

132  Leibniz Algebras

Let us cite a part of Theorem 2 from [71]: Theorem 3.8. Let g be a finite-dimensional complex Lie algebra. Then H 2 (g, g) is a direct factor of HL2 (g, g). Furthermore, the supplementary subspace vanishes in the case when the center of g is zero. Proof. A proof of this result (as well as further discussion and extensions) is available in [69].  Observe that this implies in particular that Richardson’s example ˙ ˙ M +g). ˙ M +g (see [159]) has non-trivial HL2 (M +g, Corollary 3.2. A finite dimensional solvable non-nilpotent Lie algebra g with H 2 (g, g) = 0 is the semidirect product of its nilradical N and ˙ an exterior torus of derivations Q, i.e., g = N +Q. Furthermore, if 2 dim Q > 1, then HL (g, g) = {0}. Proof. By the above discussion of the results of Carles in [51], it follows that any solvable Lie algebra g with H 2 (g, g) = 0 is algebraic, i.e., it is isomorphic to the Lie algebra of an algebraic group. As the algebraicity implies (by Proposition 1.5 in [51]) the decomposability of the algebra, it follows that for solvable Lie algebras g with H 2 (g, g) = 0, ˙ where N is the nilradical of g and we have a decomposition g = N +Q, Q is an exterior torus of derivations in the sense of Malcev; that is, Q is an abelian subalgebra of g such that ad(x) is semisimple for all x ∈ Q. If dim Q > 1, it follows from the above discussion before Theorem 3.8 that HL2 (g, g) = {0}.  3.6.3

Non-triviality of Leibniz cohomology

Let f : h → q be the quotient morphism which sends a Leibniz algebra h onto its quotient by some two-sided ideal I, and let M be a Leibniz q-bimodule. Then M is also a Leibniz h-bimodule via f . There is a monomorphism of cochain complexes f ∗ : CL∗ (q, M) → CL∗ (h, M), and a quotient complex, called the relative complex. The cohomology spaces are denoted accordingly HL∗ (h; q, M). The corresponding short exact sequence of complexes induces a long exact sequence in cohomology:

Classification Problem in Low Dimensions  133

Proposition 3.13. The map f induces a long exact sequence 0 → HL1 (q, M) → HL1 (h, M) → HL1 (h; q, M) → HL2 (q, M) → . . . Proof. This is Proposition 3.1 in [69].



˙ Let us consider the semidirect product Lie algebra b g = M +g where g is a semisimple Lie algebra (over C) and M is a non-trivial finite-dimensional irreducible g-module. From b g, we construct as our main object the Leibniz algebra h which is the hemisemidirect prod˙ g of b uct h := I +b g with ideal of squares I which is another non-trivial finite-dimensional irreducible g-module. We will need now the above observation that the hemisemidirect product is a split extension. Proposition 3.14. Let g be a semisimple Lie algebra with non-trivial finite-dimensional irreducible g-modules M and I. Then the Leibniz ˙ +g) ˙ satisfies algebra h = I +(M H 2 (b g,b g) ,→ HL2 (h, h), ˙ where b g = M +g. Proof. By Theorem 3.8, we have that H 2 (b g,b g) is a direct factor of 2 HL (b g,b g). On the other hand, the long exact sequence for f : h → b g of Proposition 3.13 splits, because h is the hemisemidirect product of b g and I. Therefore all connecting homomorphisms are zero and we have a monomorphism for all n ≥ 1 HLn (b g,b g) ,→ HLn (h,b g). Now consider the long exact sequence for the Leibniz cohomology of h associated with values in the short exact sequence 0 → I → h →b g → 0. Recall the construction of the connecting homomorphism ∂ : HL2 (h,b g) → HL3 (h, I a ) in the long exact sequence. We claim that the subspace HL2 (b g,b g) ⊂ HL2 (h,b g) is in the kernel of ∂. This is clear, because lifting a 2-cocycle c ∈ CL2 (b g,b g) to a cochain in CL2 (h, h), it will remain a cocycle and thus the preimage of its coboundary in CL3 (h, I a ) is zero. Therefore we have an epimorphism HL2 (h, h)  HL2 (b g,b g). This ends the proof of the proposition, because all cohomology spaces are C-vector spaces and thus the above epimorphism splits. 

134  Leibniz Algebras

3.7

RIGID LEIBNIZ ALGEBRA WITH NON-TRIVIAL HL2

In this section, we finally obtain an analogue of Richardson’s theorem [159] for Leibniz algebras. Namely, we give an example of a finitedimensional Leibniz algebra h over a field F which is geometrically rigid, but h has HL2 (h, h) , 0. It is showed that h is geometrically rigid, but satisfies HL2 (h, h) , 0. Let V be a finite dimensional complex vector space. Let LB be the algebraic variety of all Leibniz algebra structures on V, defined in Hom(V ⊗2 , V) by the quadratic equations corresponding to the left Leibniz identity, i.e., for µ ∈ Hom(V ⊗2 , V), we require µ ◦ µ(x, y, z) := µ(x, µ(y, z)) − µ(µ(x, y), z) + µ(µ(x, z), y) = 0 for all x, y, z ∈ V. Recall that two Leibniz algebra structures µ and µ0 give rise to isomorphic Leibniz algebras (V, µ) and (V, µ0 ) in case there exists g ∈ GL(V) such that g · µ = µ0 , where the action of GL(V) on LB is defined by g · µ(x, y) := g(µ(g−1 (x), g−1 (y))) for all x, y ∈ V. Let h = (V, µ) be a Leibniz algebra on V, and consider a Leibniz subalgebra s of h. We will denote the complex subspace of V corresponding to s by S . Let W be a supplementary subspace of S in V. For an element φ ∈ CL2 (h, V), we denote by r(φ) the restriction to the union of S ⊗ S , S ⊗ W and W ⊗ S . Let N1 := {(g, m) ∈ GL(V) × LB | r(g · m) = r(m)}, denote by projLB : GL(V) × LB → LB the projection map and let p1 : N1 → LB be the restriction of projLB to N1 . Definition 3.5. The subalgebra s is called a stable subalgebra of h if p1 maps every neighborhood of (1, µ) ∈ N1 onto a neighborhood of µ in LB. Remark. (a) This is not the original definition of stable subalgebra in [159], but the strong version of stability which permits Page and Richardson to show the strengthened stability theorem at the end of their paper.

Classification Problem in Low Dimensions  135

(b) The definition implies that if h1 = (V, µ1 ) is a Leibniz algebra sufficiently near to h, then h1 is isomorphic to a Leibniz algebra h2 = (V, µ2 ) with the following property: for all s ∈ s and all x ∈ V, we have µ2 (x, s) = µ(x, s) and µ2 (s, x) = µ(s, x). The following theorem was proved in [141]. Theorem 3.9 (Stability Theorem). Let h = (V, µ) be a finitedimensional complex Leibniz algebra, and let s be a subalgebra of h such that E 2 (h, s, V) = 0. Then s is a stable subalgebra of h. Let g be a semisimple Lie algebra and M be an irreducible (left) ˙ semidirect product of g-module (of dimension ≥ 2). We put b g = g+M g and M (with [M, M] = 0). ˙ as above. Then Theorem 3.10 (Richardson’s Theorem). Let b g = g+M b g is not rigid if and only if there exists a semisimple Lie algebra b g0 which satisfies the following conditions: (a) there exists a semisimple subalgebra g0 of b g0 which is isomorphic to g, (b) if we identify g0 with g, then b g0 / g0 is isomorphic to M as a gmodule. Proof. This is Theorem 2.1 in [159].



Richardson shows in [159] that for g = sl2 (C) and for M = Mk the standard irreducible sl2 (C)-module of dimension k + 1 and highest weight k, the Lie algebra b g is not rigid if and only if k = 2, 4, 6, 10. The semisimple Lie algebrasb g0 are in this case the standard semisimple Lie algebras of dimension 6, 8, 10 and 14. It turns out thatb g has necessarily rank 2 (see [159]), and these are all rank 2 semi-simple Lie algebras (A1 × A1 , A2 , B2 and G2 ). We will extend Richardson’s theorem to Leibniz algebras in the following sense. First of all, we will restrict to simple Lie algebras g. Let I be another irreducible (right) g-module (of dimension ≥ 2). We ˙ g the hemisemidirect product with the g-module I (in also set h = I +b particular [g, I] = 0, [I, g] = I and [M, I] = 0). Observe that h is a non-Lie Leibniz algebra with ideal of squares I and with quotient Lie algebra b g.

136  Leibniz Algebras

So, we have h = g ⊕ M ⊕ I as vector spaces. In all the following, we fix the complex vector space g ⊕ M ⊕ I and we will be interested in the different Leibniz algebra structures on g ⊕ M ⊕ I. Theorem 3.11. Let h = (V, µ) be a finite-dimensional complex Leibniz algebra, and let g be a subalgebra of h such that E 2 (h, g, V) = 0. Then the Leibniz algebra h is not rigid if and only if there exists a Leibniz algebra h0 which satisfies the following conditions: a) There exists a semisimple Lie subalgebra s0 of h0 with a semisimple Lie subalgebra g0 ⊂ s0 which is isomorphic to g, i.e. g  g0 ; b) if we identify the subalgebra g0 with g by this isomorphism, then s0 /g is isomorphic to M as a g-module and h0 /s0 is isomorphic to I as a Leibniz antisymmetric g-module. Proof. Let V := g ⊕ M ⊕ I. We will consider Leibniz algebras h0 on this fixed vector space V, i.e. h = (V, µ) and h0 = (V, µ0 ). “⇐” Suppose that there exists a Leibniz algebra h0 which satisfies the conditions a) and b). We may assume that µ(x, x0 ) = µ0 (x, x0 ), µ(x, m) = µ0 (x, m), µ(m, x) = µ0 (m, x), µ(i, x) = µ0 (i, x) for all x, x0 ∈ g, all m ∈ M and all i ∈ I. Putting gt (x) = x, gt (m) = tm, gt (i) = ti, we have that gt ∈ GL(V) for all t , 0 and lim gt · µ0 = µ. t→0

Therefore, L is not rigid. “⇒” Let LB be the set of all Leibniz algebras defined on the vector space V. We are considering the Leibniz subalgebra s := g of h. It satisfies the cohomological condition in order to apply the stability theorem. From Theorem 3.9, we therefore get the existence of a neighborhood U of µ in LB such that if µ1 ∈ U, the Leibniz algebra L1 = (V, µ1 ) is isomorphic to a Leibniz algebra L0 = (V, µ0 ) which satisfies the following conditions: 1) µ(x, x0 ) = µ0 (x, x0 ), 2) µ(x, m) = µ0 (x, m), µ(m, x) = µ0 (m, x),

Classification Problem in Low Dimensions  137

3) µ(x, i) = µ0 (x, i) = 0, µ(i, x) = µ0 (i, x), for all x, x0 ∈ g, all m ∈ M and all i ∈ I. Since h is a non-Lie algebra, the Leibniz algebra h0 is also a nonLie algebra. Therefore, the ideal of squares of the algebra h0 is also non zero. We denote it by I 0 . From conditions 1) and 2) we conclude that I 0 ∩ (g + M) = {0}, and thus J := I 0 ∩ I , {0}. The condition 3) implies that J is left module over g. Since I is an irreducible left g-module, we have J = I and thus I 0 = I as a vector spaces. We conclude that the restriction to g ⊕ M places us in the Lie situation, i.e. in exactly the same situation as in Richardson’s theorem. By Theorem 3.10, we thus obtain that there exists a semisimple Lie algebra s0 with a semisimple subalgebra g0 , isomorphic to g and (when g is identified with g0 ) an isomorphism of g-modules s0 /g  M. Furthermore, the conditions 1)-3) imply that a) and b) are satisfied.  ˙ k +sl ˙ 2 (C)) for two Corollary 3.3. The Leibniz algebra h := Il +(M standard irreducible left sl2 (C)-modules Mk and Il of highest weights k = 2n and l respectively with odd integer n > 5 and odd l > 2 is rigid and satisfies HL2 (h, h) , 0. Proof. As discussed earlier, Richardson shows in [159] that for g = sl2 (C) and for Mk the standard irreducible sl2 (C)-module of dimension ˙ is not rigid if k + 1 and highest weight k, the Lie algebra b g = Mk +g and only if k = 2, 4, 6, 10. He also shows that in case the half-highestweight 2k =: n is an odd integer n > 5, the Lie algebra cohomology of the Lie algebra b g is not zero. As a candidate for our Leibniz algebra ˙ g for h satisfying the claim of the corollary, we take as before h = Il +b some irreducible sl2 (C)-modules Il , sl2 (C) and Mk such that the halfhighest-weight n := 2k > 5 is an odd integer and l > 3 is odd. By Theorem 3.11, h is then rigid. On the other hand, by Proposition 3.14, H 2 (b g,b g) injects into 2 HL (h, h), thus we obtain that this Leibniz cohomology space is not zero. 

138  Leibniz Algebras

3.8

LIE-RIGIDITY VERSUS LEIBNIZ-RIGIDITY

We record in this section further results on the question whether H 2 (g, g) = 0 implies HL2 (g, g) = 0. The base field is fixed to be the field C of complex numbers. We have seen in Corollary 3.2 that all (non nilpotent) solvable Lie algebras g with H 2 (g, g) = 0 and dim Q > 1 satisfy HL2 (g, g) = 0. In the case where the Lie algebra g is only (Lie-)rigid, but does not necessarily satisfy H 2 (g, g) = 0, we cannot conclude that HL2 (g, g) = 0. But we will see in Theorems 3.12 and 3.13 below that for Lie algebras of a special form, one can still conclude in this situation that Cent(g) = 0. Recall that by Carles [51], any rigid Lie algebra g is algebraic, i.e., it is isomorphic to the Lie algebra of an algebraic group. As the algebraicity implies the decomposability of the algebra [51], it follows that ˙ where S for a rigid Lie algebra g, we have a decomposition g = S+R, ˙ is the solvable radical of g, N is the nilis a Levi subalgebra, R = N +Q radical and Q is an exterior torus of derivations in the sense of Malcev; that is, Q is an abelian subalgebra of g such that ad(x) is semisimple for all x ∈ Q. Note that over the complex numbers, the semisimplicity of ad(x) means that there exists a basis of N such that for any x ∈ Q, the operator ad(x)|N has diagonal form and ad(x)|Q = 0. ˙ we have [S, R] ⊆ R, Remark. Since in the decomposition g = S+R we can view R as an S-module. The fact that S is semisimple implies that R can be decomposed into a direct sum of irreducible submodules. Let r0 be the sum of all one-dimensional submodules (so these are trivial submodules) and R1 be the sum of the non-trivial irreducible submodules. Then [S, R0 ] = 0,

[S, R1 ] = R1 ,

[R0 , R0 ] ⊆ R0 ,

R1 ⊆ N,

Q ⊆ R0 .

Remark. Above, we have already considered the direct sum Lie algebra g = sl2 (C) ⊕ C. A short computation with the Hochschild-Serre spectral sequence shows that H 2 (g, g) = 0, but (cf. proof of Corollary 3 in [71]) dim HL2 (g, g) = 1. Therefore, the Lie algebra g is an ex˙ where ample of a complex rigid Lie algebra of the form g = S+N dim(g/[g, g]) = 1 and where Cent(g) is non-trivial. According to Theorem 3.8, the center is always non-trivial in the case where H 2 (g, g) and HL2 (g, g) differ.

Classification Problem in Low Dimensions  139

Proposition 3.15. Let R = R1 ⊕ Ck be a split solvable Lie algebra. Then R is not rigid. Proof. Since R is solvable, R1 is also solvable and for this algebra we have R1 = N1 ⊕ Q1 , where N1 is the nilradical of R1 and Q1 is a supplementary subspace. Let us fix elements x ∈ Q1 and c = (c1 , c2 , . . . , ck ) ∈ Ck and set ϕ(x, ci ) = −ϕ(ci , x) = ci , 1 ≤ i ≤ k. Clearly, ϕ ∈ Z 2 (R, R). Consider the infinitesimal deformation Rt = R + tϕ of the algebra R and let us show that ϕ < B2 (R, R). Consider (d f )(x, ci ) = [ f (x), ci ] + [x, f (ci )] − f ([x, ci ]) = [x, f (ci )], 1 ≤ i ≤ k. This equation implies that ∀ f ∈ C 1 (R, R), we have (d f )(x, ci ) , ci , because [R, R] ⊂ N1 . Therefore, ϕ < B2 (R, R) and Rt = R + tϕ is a non-trivial deformation of the Lie algebra R. Therefore the Lie algebra R is not rigid.  ˙ Corollary 3.4. Let g = S+(N ⊕ Q) ⊕ Ck be a Lie algebra. Then g is not rigid. Proof. Due to Remark 3.8, we have that Q ∩ [g, g] = 0. So, taking gt = g + tϕ with ϕ(x, ci ) = −ϕ(ci , x) = ci , 1 ≤ i ≤ k similarly as in Proposition 3.15, we obtain a non-trivial deformation of g. Therefore, g is not rigid.  Let R = N ⊕ Q be a rigid solvable Lie algebra with basis {x1 , x2 , . . . , xk } of Q and basis N := {e1 , e2 , . . . , en } of N. Due to the arguments above, we can assume that its the table of multiplication has the following form:  n P     [ei , e j ] = γi,t j et , 1 ≤ i, j ≤ n,  t=1    [ei , x j ] = αi, j ei , 1 ≤ i ≤ n, 1 ≤ j ≤ k, 1 ≤ k. Note that for any j ∈ {1, . . . , k}, there exists i such that αi, j , 0. In the first part, we consider the center of rigid solvable Lie algebras of the form R = N ⊕ Q. This study will cumulate in Theorem 3.12 below. Proposition 3.16. Let R = N ⊕ Q be a rigid solvable Lie algebra such that dim Q > 1. Then Cent(R) = {0}.

140  Leibniz Algebras

Proof. First of all, we divide the basis N of N into two subsets. Namely, N1 = {ei ∈ N | ∃ j ∈ {1, . . . , k} such that αi, j , 0}, N2 = {ei ∈ N | αi, j = 0 ∀ j ∈ {1, . . . , k}}. For convenience, we shall denote the bases N1 and N2 as follows: N1 = {e1 , e2 , . . . , en1 },

N2 = { f1 , f2 , . . . , fn2 }.

Evidently, n = n1 + n2 . Now let us suppose that Cent(R) , {0}. Then there exists an element 0 , c ∈ Cent(R). We put c=

n1 X

βt et +

t=1

n2 X t=1

ct ft +

k X

bt xt .

t=1

Then adc ≡ 0 and x1 , . . . , xk act diagonally on N, we derive bt = 0, 1 ≤ t ≤ k. Case 1. Let N2 = ∅. Then n = n1 . Consider   n1 n  X X βt αt,i et . 0 = [c, xi ] =  βt et , xi  = t=1

t=1

From this we deduce that βt = 0 for any 1 ≤ t ≤ n1 , as all αt,i are nonzero. Therefore, c = 0. So, we have a contradiction with Cent(R) , {0}. Case 2. Let N2 , ∅. We set ϕ(x1 , x2 ) = c, which is possible as by our hypothesis dim Q > 1. It is easy to see that ϕ is a 2-cocycle. Now we consider the associated infinitesimal deformation of the solvable Lie algebra Rt = R + tϕ. Let us check that this deformation Rt is not equivalent to R. In fact, it is enough to check that ϕ is not 2-coboundary of R. Consider (d f )(x1 , x2 ) = [ f (x1 ), x2 ] + [x1 , f (x2 )] − f ([x1 , x2 ]) = [ f (x1 ), x2 ] + [x1 , f (x2 )]. Since ad(x1 ), ad(x2 ) are diagonal and c ∈ Cent(R), we conclude that (d f )(x1 , x2 ) , c for any f ∈ C 1 (R, R). This means, ϕ < B2 (R, R). Thus, the deformation Rt = R + ϕ is a non-trivial deformation of the rigid algebra R. This is a contradiction to the assumption that Cent(R) , {0}. 

Classification Problem in Low Dimensions  141

Remark. Note that the rigidity of the solvable Lie algebra g = N ⊕ Q does not always imply that dim Q > 1. Indeed, there are examples of rigid solvable Lie algebras with dim Q = 1 (see [94]). Recall the commutation relations and notations which we introduced earlier: [ei , x j ] = αi, j ei for all 1 ≤ i ≤ n, all 1 ≤ j ≤ k, and all 1 ≤ k, and {e1 , . . . , e s } are chosen Lie algebra generators of N. In the following proposition, we neglect the index j in αi, j as Q is one dimensional and we have thus always j = 1. Proposition 3.17. Let R = N ⊕ Q be a rigid solvable Lie algebra such that dim Q = 1. Then αi , 0 for any s + 1 ≤ i ≤ n. Proof. Let hxi = Q. Without loss of generality, one can assume that the basis elements {e1 , e2 , . . . , en } of N are homogeneous products of the generator basis elements, that is, for any e j ∈ [N, N], we have e j = [. . . [e j1 , e j2 ], . . . e jt j ]

s + 1 ≤ j ≤ n,

1 ≤ ji ≤ s.

Then because of [e j , x] = (α j1 + α j2 + · · · + α jt j )e j , we obtain α j = α j1 + α j2 + · · · + α jt j

s + 1 ≤ j ≤ n,

1 ≤ ji ≤ s.

To the algebra R as above and its regular element x, we associate a linear system S (x) of equations of n−1 variables z1 , . . . , zn−1 consisting of equalities zi + z j = zk if and only if the vector [ei , e j ] contains ek with a non-zero coefficient. It is clear that {α1 , α2 , . . . , αn } is one of the solutions of this system. If the system S (x) has a unique fundamental solution, then we can obtain that z p = k p zi0 with k p > 0. Therefore, in this case all elements of the solution are strictly positive. Since {α1 , α2 , . . . , αn } is a solution, αi , 0 for any s + 1 ≤ i ≤ n. So it remains to consider the case of a system of equations which has a space of fundamental solutions which is of dimension at least two. Without loss of generality, we can assume α1 , 0 and let us assume that there exists some p ≥ s + 1 such that α p = 0. It follows then that α p = α p1 + α p2 + · · · + α pt p = 0.

142  Leibniz Algebras

Let {z1 , z2 , . . . , zh } be another solution of S (x), linearly independent from {α1 , α2 , . . . , αn }. As α1 , 0, we can assume z1 = 0. Consider the following solution {0, 1, 1, . . . , 1, zh+1 , . . . , z s , z s+1 , zn }, that is, here we get z1 = 0, z2 = 1, . . . , zh = 1. Then we consider the following cochain: ϕ(e2 , x) = e2 ,

ϕ(e3 , x) = e3 ,

...,

ϕ(eh , x) = eh ,

ϕ(eh+1 , x) = zh+1 eh+1 ,

ϕ(eh+2 , x) = zh+2 eh+2 ,

...,

ϕ(e s , x) = z s e s ,

ϕ(e s+1 , x) = z s+1 e s+1 ,

ϕ(e s+2 , x) = z s+2 e s+2 ,

...,

ϕ(en , x) = zn en .

It is easy to check that ϕ ∈ Z 2 (R, R). However, ϕ < B2 (R, R). Indeed, if we had ϕ ∈ B2 (R, R), then ϕ = d f for some f ∈ C 1 (R, R). Consider e2 = ϕ(e2 , x) = f ([e2 , x]) − [ f (e2 ), x] − [e2 , f (x)] = f (α2 e2 ) − [c2 x +

n X

a2,i ei , x] − [e2 , d0 x +

i=1

= α2 (c2 x +

n X

a2,i ei ) −

n X

i=1

n X

di ei ] =

i=1

αi a2,i ei + d0 α2 e2 + (∗) = d0 α2 e2 + (∗∗).

i=1

Hence, we obtain d0 α2 = 1 On the other hand, we have 0 = ϕ(e1 , x) = f ([e1 , x]) − [ f (e1 ), x] − [e1 , f (x)] = f (α1 e1 ) − [c1 x +

n X i=1

= α1 (c1 x +

n X i=1

a1,i ei ) −

n X

a1,i ei , x] − [e1 , d0 x +

n X

di e i ] =

i=1

αi a1,i ei + d0 α1 e1 + (∗) = d0 α1 e1 + (∗∗).

i=1

Therefore, d0 α1 = 0. Since α1 , 0, we have d0 = 0. This is a contradiction to the assumption that ϕ ∈ B2 (R, R). Finally, the deformation Rt = R + tϕ is a non-trivial deformation of R, which contradicts the rigidity of R. Therefore, we obtain that αi , 0 for any 1 ≤ i ≤ n.  As a synthesis of Propositions 3.16 and 3.17, we have the following theorem.

Classification Problem in Low Dimensions  143

Theorem 3.12. Let R be a rigid solvable Lie algebra of the form R = N ⊕ Q. Then Cent(R) = 0. Proof. The assertion of the theorem for the case when dim Q > 1 follows from Proposition 3.16. Consider therefore the case dim Q = 1. Let us assume that Cent(R) , 0. Then there exists 0 , c ∈ Cent(R). If c ∈ N \ [N, N], then we obtain that R = R1 ⊕ C and due to Proposition 3.15, we conclude that R is not rigid. Let now c ∈ [N, N]. We set c=

n X

βi ei .

i=s+1

Taking into account Proposition 3.17 and the following equality 0 = [c, x] =

n X i=s+1

βi [ei , x] =

n X

βi αi ei ,

i=s+1

we conclude that c = 0. The proof is complete.



Conclusion 1. Let R be a solvable Lie algebra of the form R = N ⊕ Q such that H 2 (R, R) = 0, then R is rigid. By Theorem 3.12, we have Cent(R) = {0}. Now applying Theorem 3.8, we conclude that HL2 (R, R) = 0, which implies by [29] that R is rigid as a Leibniz algebra. In a second part, we will now study the center of general rigid Lie ˙ ⊕ Q) using the same methods as before. algebras of the form g = S+(N ˙ Proposition 3.18. Let g = S+(N ⊕ Q) be a rigid Lie algebra such that dim Q > 1. Then Cent(g) = {0}. Proof. Let us suppose that Cent(g) , 0 and 0 , c ∈ Cent(g). Clearly, c ∈ N. From Remark 3.8 we conclude that [g, g] ∩ Q = 0. Then similarly as in the Proposition 3.16 we conclude that gt = g + tϕ with ϕ(x1 , x2 ) = c is a non-trivial deformation of g. This implies that Cent(g) = {0}.  ˙ As recalled in the beginning, if g = S+(N ⊕ Q) is rigid, then S is a Levi subalgebra of g, N is the nilradical and Q is an abelian subalgebra with diagonal operators ad(x)|N for any x ∈ Q. Moreover,

144  Leibniz Algebras

[h, h] = [Q, Q] = 0 for a Cartan subalgebra h of S. Now due to Remark 3.8 (indeed, since Q ⊆ R0 , we obtain [S, Q] ⊆ [S, R0 ] = 0), we conclude that h ⊕ Q is toroidal subalgebra in g which acts on N ⊕ (⊕i lβi ) ⊕ (⊕i l−βi ), where

S = h ⊕ (⊕i lβi ) ⊕ (⊕i l−βi ).

Therefore, we have the following toroidal decomposition: g = (h ⊕ Q) ⊕ (gγ1 ⊕ gγ2 ⊕ · · · ⊕ gγt ) such that [g, h + x] = γ(h + x)g for any h ∈ h, x ∈ Q and g ∈ gγ . Let dim Q = 1 with Q = hxi, dim h = q and lβi = hsβi i and l−βi = hs−βi i. Then we have the following products in g:  n P    [e , e ] = γi,t j et , 1 ≤ i, j ≤ n,  i j    t=1   [ei , x] = αi ei , 1 ≤ i ≤ n,     [e , h ] = θ e , 1 ≤ i ≤ n, 1 ≤ p ≤ q,  i p i,p i     [x, S] = 0. ˙ Proposition 3.19. Let g = S+(N ⊕ Q) be a rigid Lie algebra such that dim Q = 1. Then Cent(g) = {0}. Proof. Similarly as in the proof of Proposition 3.17 we consider the system of linear equations with respect to α1 , α2 , . . . , α s : α j = α j1 + α j2 + · · · + α jt j

s + 1 ≤ j ≤ n,

1 ≤ ji ≤ s.

Let us assume that there exists some p ≥ s + 1 such that α p = 0. It follows that α p = α p1 + α p2 + · · · + α pt p = 0. To the above relations, we associate the system S (z) of linear equations with respect to z1 , z2 , . . . , z s . That is, the system contains the relations z j = z j1 + z j2 + · · · + z jt j

s + 1 ≤ j ≤ n,

1 ≤ ji ≤ s.

Note that the fundamental solution of the system S (z) does not consist of a unique solution. Indeed, if it had a unique fundamental solution, then we would obtain that 0 = z p = k p z1 with k p , 0, which

Classification Problem in Low Dimensions  145

implies z1 = 0 and hence, α1 = α2 = · · · = αn = 0 which is impossible because in this case ad(x) acts trivially to N. So, we conclude that the system S (z) of equations has a fundamental solution which is a vector space of dimension at least two. Let us suppose that {z1 , z2 , . . . , zh } is another solution of S (z), linearly independent of {α1 , . . . , αh }. Moreover, we may assume that α1 appears in the set {α p1 , α p2 , . . . , α pt p }, i.e., e1 appears in the following product e p = [. . . [e p1 , e p2 ], . . . e pt p ] Consider following solution {0, 1, 1, . . . , 1, zh+1 , . . . , z s , z s+1 , . . . , zn }, that is, here we get z1 = 0, z2 = 1, . . . , zh = 1. We have [[ei , sβ ], x] = [[ei , x], sβ ] + [ei , [sβ , x]] = [[ei , x], sβ ] = αi [ei , sβ ]. Then we consider the following cochain: ϕ(e2 , x) = e2 ,

ϕ(e3 , x) = e3 ,

...,

ϕ(eh , x) = eh ,

ϕ(eh+1 , x) = zh+1 eh+1 ,

ϕ(eh+2 , x) = zh+2 eh+2 ,

...,

ϕ(e s , x) = z s e s ,

ϕ(e s+1 , x) = z s+1 e s+1 ,

ϕ(e s+2 , x) = z s+2 e s+2 ,

...,

ϕ(en , x) = zn en .

ϕ(ei , h p ) = θi,p ei ,

ϕ(h p , x) = 0,

ϕ(ei , sβ ) = ϕ(x, sβ ) = ϕ(ei , e j ) = 0,

1 ≤ i ≤ n, 1 ≤ p ≤ q, ϕ([ei , sβ ], x) = zi [ei , sβ ],

1 ≤ i, j ≤ n, 1 ≤ p ≤ q. By straightforward computations we check that ϕ is a 2-cocycle of g. However, ϕ < B2 (g, g). Indeed, if ϕ ∈ B2 (g, g), then ϕ = d f for some f = fR + fS ∈ C 1 (g, g) with fR : g → R and fS : g → S. Consider 0 = ϕ(S, x) = f ([S, x]) − [ f (S), x] − [S, f (x)] = −[ f (S), x] − [S, fR (x)] − [S, fS (x)]. This implies [ f (S), x] + [S, fR (x)] = 0 and [S, fS (x)] = 0. Taking into account that for any s ∈ S, there exists s0 ∈ S such that [s, s0 ] , 0, we conclude that fS (x) = 0.

146  Leibniz Algebras

Consider e2 = ϕ(e2 , x) = f ([e2 , x]) − [ f (e2 ), x] − [e2 , f (x)] = n n X X di ei ] = a2,i ei + fS (e2 ), x] − [e2 , d0 x + f (α2 e2 ) − [c2 x + i=1

i=1

= α2 (c2 x+

n X

a2,i ei + fS (e2 ))−

n X

αi a2,i ei +d0 α2 e2 +(∗) = d0 α2 e2 +(∗∗).

i=1

i=1

Hence, we obtain d0 α2 = 1 On the other hand, we have 0 = ϕ(e1 , x) = f ([e1 , x]) − [ f (e1 ), x] − [e1 , f (x)] = n n X X f (α1 e1 ) − [c1 x + a1,i ei + fS (e1 ), x] − [e1 , d0 x + di ei ] = i=1

= α1 (c1 x+

n X i=1

a1,i ei + fS (e1 ))−

i=1 n X

αi a1,i ei +d0 α1 e1 +(∗) = d0 α1 e1 +(∗∗).

i=1

Therefore, d0 α1 = 0. Since α1 , 0, we have d0 = 0. This is a contradiction to assumption that ϕ ∈ B2 (g, g). Finally, the deformation gt = g + tϕ is a non-trivial deformation of g, which contradicts the rigidity of g. Thus, we have proved that if [e p , x] = 0 for some e p ∈ N, then g is non-rigid. Similar as in the proof of Theorem 3.12, we obtain that Cent(g) = {0}.  Again, we perform a synthesis of the preceding two propositions in the following theorem. ˙ Theorem 3.13. Let g = S+(N ⊕ Q) be a rigid Lie algebra such that Q , 0. Then Cent(g) = 0. Proof. The assertion of the theorem for the case when dim Q > 1 follows from Proposition 3.18. In case dim Q = 1, we have that Cent(g) = 0 due to Proposition 3.19.  ˙ Conclusion 2. Let g = S+(N ⊕ Q) be a Lie algebra such that H 2 (g, g) = 0, then g is rigid. From Theorem 3.13 we obtain that Cent(g) = {0}. Now applying Theorem 3.8, we conclude that HL2 (g, g) = 0, which implies by [29] that g is rigid as a Leibniz algebra.

Classification Problem in Low Dimensions  147

3.8.1

Necessary criteria for rigidity of Leibniz algebras

An optimal way of exhaustive study of degenerations in a set of algebras includes intensive usage of necessary criteria based on degeneration invariants. The invariants are preserved under degenerations. In the next section for the further references we collect some degeneration invariants. The proofs can be found more or less in the literature, see [166]. For a given Leibniz algebra L we define: 1. SA(L)− the maximal abelian subalgebra of L; 2. Com(L)− the maximal commutative subalgebra of L; 3. SLie(L)− the maximal Lie subalgebra of L; 4. HLi (L, L)− the i-th Leibniz cohomology group with coefficients itself. 3.8.1.1

Invariance Argument 1

Theorem 3.14. For any m, r ∈ N the following subsets of LBn are closed with respect to the Zariski topology: 1. {L ∈ LBn | dim Lm ≤ r}; 2. {L ∈ LBn | dim Annr (L) ≥ m}; 3. {L ∈ LBn | dim Annl (L) ≥ m}; 4. {L ∈ LBn | dim Cent(L) ≥ m}; 5. {L ∈ LBn | dim Aut(L) > m}; 6. {L ∈ LBn | dim SA(L) ≥ m}; 7. {L ∈ LBn | dim Com(L) ≥ m}; 8. {L ∈ LBn | dim SLie(L) ≥ m}; 9. {L ∈ LBn | dim HLi (L, L) ≥ m}. The proof is an easy consequence of the following fact from algebraic group theory. Let G be a complex reductive algebraic group acting rationally on an algebraic set X and let B be a Borel subgroup of G. Then G = G ∗ B (see [13], also [99]).

148  Leibniz Algebras

Corollary 3.5. An algebra L does not degenerate to an algebra L0 if one of the following conditions is valid: 1. dim Lm < dim L0m for some m, 2. dim Annr (L) > dim Annr (L0 ), 3. dim Annl (L) > dim Annl (L0 ), 4. dim Cent(L) > dim Cent(L0 ), 5. dim Aut(L) ≥ dim Aut(L0 ), 6. dim SA(L) > dim SA(L0 ), 7. dim Com(L) > dim Com(L0 ), 8. dim SLie(L) > dim SLie(L0 ). 9. dim HLi (L, L) > dim HLi (L0 , L0 ). The invariance arguments below are stated in a general setting and the Leibniz algebras case is deduced from these as a special case. 3.8.1.2

Invariance Argument 2

Let A be an n-dimensional algebra over a field F and {e1 , e2 , . . . , en } be a basis on it. Then the element x = x1 ⊗ e1 + x2 ⊗ e2 + · · · + xn ⊗ en ∈ F[x1 , x2 , . . . , xn ] ⊗F A, where x1 , x2 , . . . , xn are independent variables, is called the generic element of A. Denote by fA (R x ) a Cayley-Hamilton polynomial of the right multiplication operator to the generic element b = F[x1 , x2 , . . . , xn ] ⊗F A. It is known that fA (R x ) does x in the algebra A not depend on choosing of a basis in A. Proposition 3.20. If an algebra A degenerates to algebra B, then fA (R x ) = 0 in b B. 3.8.1.3

Invariance Argument 3

Let {e1 , e2 , . . . , en } be a basis of A and tr(Rei ) = 0 for all 1 ≤ i ≤ n. If there exists a basis { f1 , f2 , . . . , fn } of B such that tr(R fi ) , 0 for some i ∈ {1, . . . , n}, then A does not degenerate to B.

Classification Problem in Low Dimensions  149

3.8.1.4

Invariance Argument 4

Let an algebra A be given by the structure constants γ1 , γ2 , . . . , γr and for a pair (i, j) of positive integers define ci, j =

tr(R x )i tr(Ry ) j . tr((R x )i ◦ (Ry ) j )

Then ci, j is a polynomial of γ1 , γ2 , . . . , γr and it does not depend on the elements x, y of A. If neither of these polynomials is zero, we call ci, j as (i, j)-invariant of A. Suppose that A has an (i, j)-invariant ci, j . Then all B ∈ Orb(A) have the same (i, j)-invariant ci, j . 3.8.1.5

Invariance Argument 5

Assume that in the previous invariance argument either tr(R x )i tr(Ry ) j = 0 or tr((R x )i ◦ (Ry ) j ) = 0 for all x, y ∈ A and some pair (i, j). Then these equations hold for all B ∈ Orb(A). The proofs, except for the argument on dimension of the group of cohomology, can be found in the literature, see for example [39], [40], [41], [99]. The main argument relies on the following more general fact from the theory of algebraic groups. Let G be a complex reductive algebraic group acting rationally on an algebraic set X. Let B be a Borel subgroup of G. Then G · x = G · B · x for all x ∈ X. We shall use it in the following context. Let B be a Borel subgroup of GLn and L and L0 be two elements of a class of algebras A. If L → L0 and L lies in a B-stable closed subset R of A, then L0 must also be represented by a structure in R. Let us now consider the argument on the group of cohomologies. It is clear that dim ZLi (L, L) ≤ dim ZLi (L0 , L0 ) for a natural i. One has dim HLi (L, L) = dim ZLi (L, L) − dim CLi+1 (L, L) + dim ZLi+1 (L, L). This implies that dim HLi (L, L) ≤ dim HLi (L0 , L0 ), since dim CLi+1 (L, L) = dim CLi+1 (L0 , L0 ).

150  Leibniz Algebras

3.8.2

Applications of invariance arguments to varieties of low-dimensional Leibniz algebras

3.8.2.1

Two-dimensional Leibniz algebras

In Table 3.2 we give the orbit closures of two-dimensional Leibniz algebras under the action of GL2 as above. Notation L1 L2 L3 L4 Table 3.2

Leibniz brackets [e1 , e2 ] = e1 , [e2 , e2 ] = e1 [e2 , e2 ] = e1 [e1 , e2 ] = −[e2 , e1 ] = e2 -

Orbit closure L1 , L2 , L4 L2 , L4 L3 , L4 L4

Orbit closures of two-dimensional Leibniz algebras

It is easy to see here that the algebra L1 is rigid. Non degeneracy  L21 9 L3 and L2 9 L3 occur due to Corollary 3.5 item 2). The matrix 0t 0t is the base change giving the degeneration L1 → L2 and the algebra L3 being a Lie algebra does not degenerate to the Leibniz algebra L1 . The Lie algebra L3 itself also is Leibniz rigid, the fact which follows from a cohomological version of the result given in [146]. Therefore, the variety of two-dimensional Leibniz algebras has two irreducible components, they are generated by the algebras L1 and L3 . 3.8.3

Variety of three-dimensional nilpotent Leibniz algebras

Let LNn be the variety of n-dimensional nilpotent Leibniz algebras. In this section we describe LNn for n ≤ 3. The varieties of one and twodimensional Leibniz algebras easily can be described as follows. The variety LN1 consists of one point (it is the trivial Leibniz algebra). The variety LN2 consists of two points: one of them is the trivial algebra and another one is the Leibniz algebra L given by the table of multiplication [e1 , e1 ] = e2 , on a basis {e1 , e2 }. The algebra L degenerates to abelian. Hence, LN2 is irreducible and its irreducible component is generated by the rigid algebra L, i.e., LN2 = Orb(L) and dim LN2 = 2.

Classification Problem in Low Dimensions  151

In the dimension three there are five isolated and one parametric family of pairwise non isomorphic algebras. The values of invariants of these algebras are given in Table 3.1. By using degeneration arguments stated in the table we write down all possible degenerations for nilpotent three-dimensional Leibniz algebras: λ4 (α = 0) → λ1 , λ2 , λ4 (α , 0) → λ1 , λ2 , λ3 , λ5 → λ1 , λ2 , λ6 → λ1 , λ2 , λ4 (α = 0). Later we shall prove (see Proposition 3.21) that all non Lie Leibniz algebras degenerate to λ2 . Moreover, the algebra λ6 degenerates to the algebra λ4 (α = 0) via the following family of matrices: gt (e1 ) = e2 −e3 , gt (e2 ) = t−1 e1 , gt (e3 ) = t−1 e3 . Algebra Multiplication table of λ Orbit closure λ1 abelian λ1 λ2 [e1 , e1 ] = e2 λ1 , λ2 [e2 , e3 ] = e1 λ3 [e3 , e2 ] = −e1 λ1 , λ3 [e2 , e2 ] = e1 [e3 , e3 ] = αe1 λ4 (α , 0) [e2 , e3 ] = e1 λ1 , λ2 , λ3 , λ4 (α , 0) [e2 , e2 ] = e1 λ4 (α = 0) [e2 , e3 ] = e1 λ1 , λ2 , λ4 (α = 0) [e2 , e2 ] = e1 λ5 [e3 , e2 ] = e1 λ1 , λ2 , λ5 [e2 , e3 ] = e1 λ6 [e3 , e3 ] = e1 λ1 , λ2 , λ4 (α = 0), λ6 [e1 , e3 ] = e2 Table 3.3

Geometric descriptions of the algebras

LN3 (C) :

λ4 (α , 0) . & λ3 &

. λ1

λ5 λ6 . . λ2 ←− λ4 (α = 0)

152  Leibniz Algebras

The following Hasse diagram gives us the complete description of the set of the algebras given in Table 3.3. Summing up all the above, we conclude that the variety LN3 (C) is the union of three irreducible components Ci , i = 1, 2.3 as follows [ C1 = Orb(λ4 ), C2 = Orb(λ5 ), C3 = Orb(λ6 ) α,0

and dim LN3 (C) = 6. 3.8.4

Three-dimensional rigid Leibniz algebras

This section deals with Leibniz algebras in dimension three. We explore rigidity problems here. One needs the algebraic classification of all three-dimensional Leibniz algebras from Table 3.1. By using the invariance arguments, we find all possible degenerations of threedimensional complex Leibniz algebras that may occur: L1 → L2 , L5 , L6 , L7 , L8 , L12 (α = 0), L14 , L15 , L18 ; L2 → L7 , L8 , L12 (α = 0), L14 , L15 , L18 ; L3 → L10 , L17 , L18 ; L4 (α = 0) → L7 , L8 , L10 , L12 (α = 0), L13 , L14 , L15 , L17 , L18 ; L4 (α , 0) → L4 (α = 0), L5 , L6 , L7 , L8 , L10 , L12 (α = 0), L13 , L14 , L15 , L17 , L18 ; L5 → L7 , L8 , L10 , L12 (α = 0), L15 , L18 ; L6 → L7 , L8 , L12 (α = 0), L14 , L15 , L18 L7 → L8 , L12 (α = 0), L15 , L18 ; L8 → L8 , L12 (α = 0), L15 , L18 ; L9 → L10 , L11 , L16 , L17 , L18 ; L10 → L17 , L18 ; L11 → L16 , L17 , L18 ; L12 (α = 0) → L15 , L18 L12 (α , 0) → L12 (α = 0), L15 , L18 ; L13 → L15 , L17 , L18 ; L14 → L17 , L18 ; L15 → L18 L16 → L17 , L18 L17 → L18 L18 → L18 Some algebras do not appear on the right hand side of this list after arrows, this means that the algebra L3 is rigid and the parametric family

Classification Problem in Low Dimensions  153

of algebras L1 (α), L4 (α), L9 (α), L12 (α) are rigid families, i.e., they are not degeneration of other Leibniz algebras in dimension three. The final result can be written as follows: Theorem 3.15. The algebra L3 and the continuous parametric families of algebras L1 (α), (α , 0), L4 (α), L9 (α), (|α| < 1, α , 0), L12 (α), (α , 0) generate rigid irreducible components of LB3 (C) with the dimensions: C1 = Orb(L3 ), dim C1 = 6, C2 = ∪α Orb(L1 (α)), dim C2 = 7, C3 = ∪|α| n/2 it is isomorphic to one of the two non-isomorphic algebras: [ei , e1 ] = ei+1 , 1 ≤ i ≤ m − 1, [ei , e1 ] = ei+1 , 1 ≤ i ≤ n−m−1,

[em+i , e1 ] = em+i+1 , 1 ≤ i ≤ n − m − 1, [en−m+i , e1 ] = en−m+i+1 , 1 ≤ i ≤ m−1.

Proof. Let L ∈ Nn Z and C(L) = (m, n − m). Then there is a basis {e1 , . . . , en } of L such that for x ∈ L \ L2 the matrix form of the operator R x can be represented as follows:     J 0 Jn−m 0 R x :' 0m J , R x :' 0 J , n−m

m

On Some Classes of Leibniz Algebras  161

i.e., in the first case we have products: [ei , x] = ei+1 , 1 ≤ i ≤ m − 1, [em+i , x] = em+i+1 , 1 ≤ i ≤ n − m − 1, [em , x] = [en , x] = 0, whereas in the second case we have [ei , x] = ei+1 , 1 ≤ i ≤ n − m − 1, [en−m+i , x] = en−m+i+1 , 1 ≤ i ≤ m − 1, [en−m , x] = [en , x] = 0. Consider the first case. Since L2 ⊆ Annr (L), the basis vectors e2 , . . . , em , em+2 , . . . , en are in Annr (L). Consequently, the element x is not in the linear span of the basis vectors {e2 , . . . , em , em+2 , . . . , en }. Therefore, instead of x we can chose either e1 or em+1 (due to dim Annr (L) = n − 1). If x = e1 , we get the algebra L1 : [ei , e1 ] = ei+1 , 1 ≤ i ≤ m − 1, [em+i , e1 ] = em+i+1 , 1 ≤ i ≤ n − m − 1. If x = em+1 , we obtain L2 : [ei , em+1 ] = ei+1 , 1 ≤ i ≤ m − 1, [em+i , em+1 ] = em+i+1 , 1 ≤ i ≤ n − m − 1. Consider now the second case: [ei , x] = ei+1 , 1 ≤ i ≤ n − m − 1, [en−m+i , x] = en−m+i+1 , 1 ≤ i ≤ m − 1, [en−m , x] = [en , x] = 0. Applying the similar arguments as in the first case we can suppose that either x = e1 or x = en−m+1 . So, we derive the following two algebras L10 and L20 : L10 : [ei , e1 ] = ei+1 , 1 ≤ i ≤ n − m − 1, [en−m+i , e1 ] = en−m+i+1 , 1 ≤ i ≤ m − 1, 0 L2 : [ei , en−m+1 ] = ei+1 , 1 ≤ i ≤ n − m − 1, [en−m+i , en−m+1 ] = en−m+i+1 , 1 ≤ i ≤ m − 1. It is not difficult to check that the algebras L1 and L10 are isomorphic to L20 and L2 , respectively. It is easy to see that in the case m , n2 the algebras L1 and L10 are non isomorphic, otherwise they coincide.  For a convenience, in the case of dim Annr (L) = n − 1 we henceforth express the table of multiplication of L via the right multiplication operator of an basis element e1 ∈ L \ Annr (L).

162  Leibniz Algebras

Lemma 4.2. Assume that L ∈ Nn Z and C(L) = (n1 , . . . , n s ). Then L is isomorphic to one of the algebras  J 0 . . . 0 0   nσ(1)  0 Jnσ(2) . . . 0 0   Re1 ,σ :=  .. .. ..  , . . . ..  . . . .  0 0 . . . 0 Jnσ(s) where σ is an element of the symmetric group S s and Jn1 , . . . , Jns are Jordan blocks with dimensions n1 , . . . , n s , respectively. In particular, Re1 ,σ1  Re2 ,σ2 if and only if nσ1 = nσ2 . Proof. Suppose that L satisfies the conditions of the lemma. By permutation of corresponding basis elements it is easy to see that the interchange of the second and the third Jordan blocks do not change the multiplication of the algebra L. Continuing this procedure, we conclude that permutations of Jordan blocks Jnσ(2) , . . . , Jnσ(s) do not change the multiplication of the algebra Re1 ,σ1 . The arguments similar to those in Lemma 4.1 complete the proof of the corollary.  Under the assumptions of Lemma 4.2, we also have Corollary 4.1. The number of non isomorphic algebras in Nn Z with characteristic sequence (n1 , . . . , n s ) equals the cardinality of different numbers in the set {n1 , . . . , n s }. Lemma 4.3. The set Orbn (NFn ) consists of the algebras of the form:  J   n1 0 . . . 0 0   0 Jn2 . . . 0 0  Re1 :=  .. .. . . .. ..  ,  . . . .  . 0 0 . . . 0 Jn s where n1 + n2 + · · · + n s = n. Proof. Since NFn ∈ LNn ∩ {L ∈ LBn (K) : dim Annr (L) ≥ n − 1} = Nn Z ∪ Cn and Nn Z ∪ Cn is a closed set in Zariski topology we have Orbn (NFn ) ⊆ Nn Z ∪ Cn . Therefore, Orbn (NFn ) is in the set of the algebras of Lemma 4.2. We show that the set Orbn (NFn ) coincides with Nn Z ∪ Cn .

On Some Classes of Leibniz Algebras  163

Let L be an element of Nn Z ∪ Cn and  J  n1 0 . . .  0 Jn2 . . . Re1 :=  .. .. . .  . . . 0 0 ...

0 0 0 0 .. .. . . 0 Jns

    . 

Consider the family of the matrices (gt )t∈R\{0} defined as follows: gt (ei ) = t−i ei gt (ei ) = t−(i−1) ei

for 1 ≤ i ≤ n1 , for n1 + 1 ≤ i ≤ n.

Passing to the limit of this family as t → 0, i.e., lim g−1 t [gt (ei ), gt (e j )], t→0 we obtain   Jn1 0 . gt ∗ NFn −→ Re1 : = 0 J t→0

n−n1

Now, take the family of matrices ( ft )t∈R\{0} defined by ft (ei ) = t−i ei , for 1 ≤ i ≤ n1 + n2 , ft (ei ) = t−(i−1) ei , for n1 + n2 + 1 ≤ i ≤ n. Taking the limit of this family as t → 0, i.e., lim ft−1 [ ft (ei ), ft (e j )], t→0 we obtain  J  0  n1 0   . 0 ft ∗ NFn −→ Re1 :=  0 Jn2 t→0 0 0 Jn−n1 −n2 Repeating this procedure s times, we conclude that the algebra defined by the operator  J   n1 0 . . . 0 0   0 Jn2 . . . 0 0  Ren :=  .. .. . . .. ..  .  . . . .  . 0 0 . . . 0 Jns is in Orbn (NFn ).



Since the orbit of a null-filiform algebra is an open set in LNn , we conclude that its closure is an irreducible component of LNn (see [170]) and the following theorem holds.

164  Leibniz Algebras

Theorem 4.1. An irreducible component of the variety LNn , containing the null-filiform algebra, consists of the following algebras:  J   n1 0 . . . 0 0   0 Jn2 . . . 0 0  Re1 :'  .. .. . . .. ..   . . . .  . 0 0 . . . 0 Jns where n1 + · · · + n s = n. Moreover, two algebras  J  n1 0 . . . 0 0  0 Jn2 . . . 0 0 Re1 :'  .. .. . . .. ..  . . . . . 0 0 . . . 0 Jns

    , 

   0 Re1 :'  

Jm1 0 . . . 0 0 0 Jm2 . . . 0 0 .. .. . . .. .. . . . . . 0 0 . . . 0 Jms

    

are isomorphic if and only if n1 = m1 and the set {m2 , m3 , . . . , m s } is obtained by permutation of {n2 , n3 , . . . , n s }. Proof. The proof follows from Lemmas 4.3 and 4.2.



Remark 1. Theorem 4.1 implies that the number of non isomorphic algebras in the irreducible component of LNn containing the algebra NFn equals p(n), where p(n) is the number of integer solutions of the equation x1 + x2 +· · ·+ xn = n, with x1 ≥ x2 ≥ · · · ≥ xn ≥ 0. The asymp√ totic value of p(n), given in [102] by the expression p(n) ≈ 4n1√3 eA n q with A = π 23 , (where a(n) ≈ b(n) means that lim a(n) = 1) shows how n→∞ b(n) small is the set of non isomorphic Leibniz algebras in the irreducible component of LNn containing the algebra NFn , i.e., the number of orbits in this component is finite for every value of n. 4.2

CLASSIFICATION OF NATURALLY GRADED COMPLEX FILIFORM LEIBNIZ ALGEBRAS

Let L be a nilpotent Leibniz algebra with nilindex s. Consider Li = Li /Li+1 , 1 ≤ i ≤ s − 1, and grL = L1 ⊕ L2 ⊕ · · · ⊕ L s−1 . Then [Li , L j ] ⊆ Li+ j and we obtain the graded algebra grL. Definition 4.2. If a Leibniz algebra L0 is isomorphic to a naturally graded filiform algebra grL, then L0 is said to be naturally graded filiform Leibniz algebra.

On Some Classes of Leibniz Algebras  165

Lemma 4.4. Let L be an n-dimensional Leibniz algebra. Then the following statements are equivalent: (a) C(L) = (n − 1, 1); (b) L is a filiform Leibniz algebra; (c) Ln−1 , 0 and Ln = 0. Proof. The implications (a) ⇒ (b) ⇒ (c) are obvious. (b) ⇒ (a): Let {e1 , . . . , en } be a basis for a filiform algebra L such that {e3 , . . . , en } ⊆ L2 , {e4 , . . . , en } ⊆ L3 , . . . , {en } ⊆ Ln−1 . Consider the products [x, e1 + αe2 ] = γ1 e3 + αβ1 e3 , [e3 , e1 + αe2 ] = γ2 e4 + αβ2 e4 , [e4 , e1 + αe2 ] = γ3 e5 + αβ3 e5 , . . . , [en , e1 + αe2 ] = 0, where x is an arbitrary element of L and (γi , βi ) , (0, 0) for any i. Choose α such that γi + αβi , 0 for any i. Then z = e1 + αe2 ∈ L \ [L, L] and C(z) = (n − 1, 1). (c) ⇒ (b): Assume that Ln = 0. Then we obtain a decreasing chain of subalgebras L ⊃ L2 ⊃ L3 ⊃ · · · ⊃ Ln−1 ⊃ Ln = 0 of the length n. Obviously, dim L2 = n − 1 or dim L2 = n − 2 (otherwise Ln−1 = 0). The case dim L2 = n − 1 is not possible because of Proposition 3.3. Therefore, dim L2 = n − 2. Obviously, for i = 2, 3, . . . , n we have dim Li = n − i, i.e., L is a filiform Leibniz algebra.  Let L be an (n + 1)-dimensional complex filiform Leibniz algebra. By the arguments similar to those are in [178], over an infinite field we can find a basis e0 , e1 ∈ L1 , ei ∈ Li (i ≥ 2) for L such that [ei , e0 ] = ei+1 and [en , e0 ] = 0, 1 ≤ i ≤ n. Case 1. Assume that [e0 , e0 ] = αe2 , α , 0. Then e2 ∈ Annr (L). Hence, {e3 , . . . , en } ⊆ Annr (L). Changing the basis as below e1 = αe1 , e2 = αe2 , e3 = αe3 , . . . , en = αen , we may assume that α is equal to 1. Thus, [e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , and [en , e0 ] = 0. Suppose that [e0 , e1 ] = βe2 and [e1 , e1 ] = γe2 . Then [e0 , [e1 , e0 ]] = [[e0 , e1 ], e0 ] − [[e0 , e0 ], e1 ] ⇒ βe3 = [e2 , e1 ]

166  Leibniz Algebras

and [e1 , [e0 , e1 ]] = [[e1 , e0 ], e1 ] − [[e1 , e1 ], e0 ] ⇒ γe3 = [e2 , e1 ]. It follows that β = γ. Applying the induction with respect to the dimension and using the equality [ei , [e0 , e1 ]] = [[ei , e0 ], e1 ] − [[ei , e1 ], e0 ], one can easily prove that [ei , e1 ] = βei+1 , i.e., in Case 1 we obtain the algebra [e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , [e1 , e1 ] = βe2 , [ei , e1 ] = βei+1 , [e0 , e1 ] = βe2 . Case 2. Let [e0 , e0 ] = 0 and [e1 , e1 ] = αe2 , α , 0. In this case e2 ∈ Annr (L). Hence, {e3 , . . . , en } ⊆ Annr (L). Putting e0 = αe0 , e2 = αe2 , e3 = α2 e3 , . . . , en = αn−1 en , we may assume that α = 1, i.e., [e1 , e1 ] = e2 , [ei , e0 ] = ei+1 . Put [e0 , e1 ] = βe2 . Then [e0 , [e1 , e0 ]] = [[e0 , e1 ], e0 ] − [[e0 , e0 ], e1 ] ⇒ [[e0 , e1 ], e0 ] = 0, i.e., β[e2 , e0 ] = βe3 = 0 ⇒ β = 0. Again applying the induction with respect to the dimension and using the equality [ei , [e0 , e1 ]] = [[ei+1 , e0 ], e1 ] − [[ei , e1 ], e0 ], we can easily show that [ei , e1 ] = ei+1 , i.e., in Case 2 we obtain the algebra [ei , e0 ] = ei+1 , [ei , e1 ] = ei+1 (i > 1). Changing the basis as e0 := e0 − e1 , e1 := e1 and others are left unchanged, we get the algebra [ei , e1 ] = ei+1 . It is easy to see that this algebra is isomorphic to the algebra of Case 1 for β = 1 (e00 := e0 − e1 , e01 := e1 ). Case 3. Let [e0 , e0 ] = 0 and [e1 , e1 ] = 0. We set [e0 , e1 ] = αe2 . Subcase 1. Assume that [e0 , e1 ] = αe2 with α , 1. Then e2 ∈ Annr (L). Again {e3 , . . . , en } ⊆ Annr (L). Since α , 1, on putting e1 = e1 + e0 we obtain e21 = (α + 1)e2 and [e1 , e0 ] = e2 , i.e., we come to Case 2.

On Some Classes of Leibniz Algebras  167

Subcase 2. [e0 , e1 ] = −e2 . Before treating this subcase, we prove the following Lemma 4.5. Let L be an (n + 1)-dimensional naturally graded filiform Leibniz algebra with a basis {e0 , e1 , . . . , en } satisfying the conditions: [e1 , e1 ] = [e0 , e0 ] = 0, [e0 , e1 ] = −e2 , [ei , e0 ] = ei+1 . Then L is a Lie algebra. Proof. Again applying the induction with respect to the dimension and using the relation [e0 , [ei , e0 ]] = [[e0 , ei ], e0 ]−[[e0 , e0 ], ei ], one can show that [e0 , ei ] = −[ei , e0 ] for 1 ≤ i ≤ n. The equality [e1 , [e1 , e0 ]] = [[e1 , e1 ], e0 ] − [[e1 , e0 ], e1 ] implies [e1 , e2 ] = −[e2 , e1 ]. Applying the chain of the equalities [e1 , ei+1 ] = [e1 , [ei , e0 ]] = [[e1 , ei ], e0 ] − [[e1 , e0 ], ei ] = −[[ei , e1 ], e0 ] − [e2 , ei ] = [e0 , [ei , e1 ]] − [e2 , ei ] = [[e0 , ei ], e1 ] − [[e0 , e1 ], ei ] − [e2 , ei ] = [[e0 , ei ], e1 ] + [e2 , ei ] − [e2 , ei ] = −[[ei , e0 ], e1 ] = −[ei+1 , e1 ] and the induction hypothesis we obtain [e1 , ei ] = −[ei , e1 ] for 1 ≤ i ≤ n. Thus, [e1 , ei ] = −[ei , e1 ] and [e0 , ei ] = −[ei , e0 ] for 0 ≤ i ≤ n. Let us prove the equality [ei , e j ] = −[e j , ei ] for all i, j. We proceed by the induction on i for a fixed j. Observe that j may be assumed to be greater than i. Using the chain of the equalities [ei+1 , e j ] = [[ei , e0 ], [e j−1 , e0 ]] = [[[ei , e0 ], e j−1 ], e0 ] − [[[ei , e0 ], e0 ], e j−1 ] = −[e0 , [[ei , e0 ], e j−1 ]] + [[e0 , [ei , e0 ]], e j−1 ] = [e0 , [[e0 , ei ], e j−1 ] − [[e0 , [e0 , ei ]], e j−1 ] = [[e0 , [e0 , ei ]], e j−1 ] − [[e0 , e j−1 , [e0 , ei ]] −[[[e0 , e0 ], ei ], e j−1 ] + [[e0 , ei ], e0 ], e j−1 ] = [[[e0 , e0 ], ei ], e j−1 ] −[[[e0 , ei ], e0 ], e j−1 ] − [[[e0 , e0 ], ei ], e j−1 ] −[[e j−1 , e0 ], [ei , e0 ]] + [[[e0 , ei ], e0 ], e j−1 ] = −[e j , ei+1 ], we obtain anti-commutativity of the basis elements of the algebra L. 

168  Leibniz Algebras

Thus, the naturally graded filiform Leibniz algebras which are not Lie algebras are given as follows: [e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , [ei , e1 ] = βei+1 , [e0 , e1 ] = βe2 . Let us suppose that β , 1. Changing the basis as follows e0 = (1 − β)e0 , e1 = −βe0 + e1 , e2 = (1 − β)2 e2 , . . . , en = (1 − β)n en , we may assume that β = 0. Now, consider the case β = 1, i.e., [e0 , e0 ] = e2 , [ei , e1 ] = ei+1 , [e0 , e1 ] = e2 (1 ≤ i ≤ n). Making the basis change e1 = e1 − e0 , we obtain [e0 , e0 ] = e2 , [ei , e0 ] = ei+1 (1 ≤ i ≤ n). We show that the obtained algebras L1 :=

[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 (1 ≤ i ≤ n − 1),

L2 :=

[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 (2 ≤ i ≤ n − 1)

and

are not isomorphic. Assume the contrary and let ϕ be such an isomorphism, i.e., ϕ : L1 → L2 and n X ϕ(ei ) = αi j e j . j=0

We  have  P   n ϕ(e0 ), ϕ(e0 ) = α0 j e j , α00 e0 = α00 α00 e2 + α02 e3 + · · · + j=0  α0,n−1 en . On the other hand, ϕ([e0 , e0 ]) = ϕ(e2 ) =

n P

α2 j e j .

j=0

Comparing the two equalities above, we conclude that α20 = α21 = 0, α22 = α200 , α2,k = α00 α0,k−1 for 3 ≤ k ≤ n.

(4.1)

On Some Classes of Leibniz Algebras  169

Consider the product P  n n P αi j e j , α00 e0 = α00 αi j [e j , e0 ] [ϕ(ei ), ϕ(e0 )] = j=0

j=0

= α00 (αi,0 e2 + αi,2 e3 + · · · + αi,n−1 en )

and the element n P ϕ([ei , e0 ]) = ϕ(ei+1 ) = αi+1, j x j for 1 ≤ i ≤ n − 1. j=0

Comparing the two equalities, we deduce that αi+1,0 = αi+1,1 = 0, αi+1,2 = α00 αi,0 , αi+1,k = α00 αi,k−1

(4.2)

for 3 ≤ k ≤ n, 1 ≤ i ≤ n − 1. It follows from (4.2) that α22 = α00 α10 . Since α00 , 0 (otherwise ϕ is degenerate) the relation (4.1) implies that α00 = α10 . We have ϕ([e0 , e1 ]) = ϕ(0) = 0. On the other hand, P  n n P [ϕ(e0 ), ϕ(e1 )] = α0 j e j , α10 e0 = α10 α0 j [e j , e0 ] j=0

j=0

= α10 (α00 e0 + α02 e3 + · · · + α0,n−1 en ) = 0. Hence, α10 α00 = 0 and so α10 = 0, i.e., the first column of the matrix [ϕ] of ϕ is zero. Therefore, ϕ is degenerate. Thus we have proved the following Proposition 4.1. There are exactly two non isomorphic naturally graded complex non-Lie filiform Leibniz algebras NGF1 and NGF2 of dimension n + 1, where NGF1 :=

[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 for 1 ≤ i ≤ n − 1,

NGF2 :=

[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 for 2 ≤ i ≤ n − 1,

the other products vanish. Remark. The naturally graded complex filiform Lie algebras have been described by M. Vergne in [178]. The following theorem summarizes the results of Proposition 4.1 and Vergne [178].

170  Leibniz Algebras

Theorem 4.2. Any complex (n + 1)-dimensional naturally graded filiform Leibniz algebra is isomorphic to one of the following pairwise non isomorphic  algebras: [e0 , e0 ] = e2 , NGF1 := ( [ei , e0 ] = ei+1 , 1 ≤ i ≤ n − 1 [e0 , e0 ] = e2 , NGF2 := [e , e ] = e , 2 ≤ i ≤ n − 1 i 0 i+1  [e , e ] = −[e 1≤i≤n−1  0 , ei ] = ei+1 ,   i 0 i+1 NGF3 :=   [ei , en−i ] = −[en−i , ei ] = α(−1) en , 1 ≤ i ≤ n − 1,  α ∈ {0, 1} for odd n and α = 0 for even n. It is clear that NGF3 is a Lie algebra, but NGF1 and NGF2 are non Lie algebras. Theorem 4.3. An arbitrary (n+1)−dimensional complex filiform Leibniz algebra admits a basis {e0 , e1 , . . . , en } called adapted, such that the table of multiplication of the algebra has one of the following forms:  [e0 , e0 ] = e2 ,      [e , e ] = ei+1 , 1 ≤ i ≤ n − 1,    i 0 FLbn+1 :=  [e 0 , e1 ] = α3 e3 + α4 e4 + · · · + αn−1 en−1 + θen ,     [e j , e1 ] = α3 e j+2 + α4 e j+3 + · · · + αn+1− j en ,    1≤ j ≤ n − 2, and α3 , α4 , . . . , αn , θ ∈ C.

SLbn+1

           :=          

[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , 2 ≤ i ≤ n − 1, [e0 , e1 ] = β3 e3 + β4 e4 + · · · + βn en , [e1 , e1 ] = γen , [e j , e1 ] = β3 e j+2 + β4 e j+3 + · · · + βn+1− j en , 2 ≤ j ≤ n − 2 and β3 , β4 , . . . , βn , γ ∈ C.

[e0 , e0 ] = γen , [e1 , e1 ] = αen , [ei , e0 ] = ei+1 , 1 ≤ i ≤ n − 1 [e0 , e1 ] = −e2 + βen , TLbn+1 [e0 , ei ] = −ei+1 , 2 ≤ i ≤ n − 1 [ei , e j ] = −[e j , ei ] ∈ Span{ei+ j+1 , ei+ j+2 , . . . , en }, 1 ≤ i ≤ n − 3, and 2 ≤ j ≤ n − 1 − i [en−i , ei ] = −[ei , en−i ] = (−1)i δen , 1 ≤ i ≤ n − 1 where δ ∈ {0, 1} for odd n, δ = 0 for even n and the bracket of TLbn+1 must satisfy the Leibniz identity.                   :=                 

On Some Classes of Leibniz Algebras  171

Proof. From Theorem 4.2 we have that the natural gradation of a filiform Leibniz algebra may be an algebra from one of NGFi for i = 1, 2, 3. By straightforward verification we can make sure that all the above algebras are Leibniz algebras. By Proposition 4.1, every (n + 1)dimensional complex non-Lie filiform Leibniz algebra (L, µ) is isomorphic to the algebra µn0 + β, where NGF1 = (V, µn0 ) and β(e0 , e0 ) β(ei , e0 ) β(ei , e j ) β(e0 , e j )

= 0, = 0 for 1 ≤ i ≤ n − 1, ∈ Span{ei+ j+1 , . . . , en } for i , 0, ∈ Span{e j+2 , . . . , en } for 1 ≤ j ≤ n − 2,

or to the algebra µn1 + β, where NGF2 = (V, µn1 ) and β(e0 , e0 ) β(ei , e0 ) β(ei , e j ) β(e0 , e j )

= 0, = 0, for 2 ≤ i ≤ n − 1, ∈ Span{ei+ j+1 , . . . , en } for i, j , 0, ∈ Span{e j+2 , . . . , en } for 1 ≤ j ≤ n − 2.

Case 1. Assume that µ : µn0 + β. Then [e0 , e0 ] = e2 and [ei , e0 ] = ei+1 for 1 ≤ i ≤ n − 1; hence Span{e2 , e3 , . . . , en } = Annr (L), so that [ei , e j ] = 0 for 2 ≤ j ≤ n, 0 ≤ i ≤ n. Put [e1 , e1 ] = α3 e3 + α4 e4 + · · · + αn en . Consider [ei , [e0 , e1 ]] = [[ei , e0 ], e1 ] − [[ei , e1 ], e0 ]. Since [e0 , e1 ] ∈ Annr (L), we have [ei , [e0 , e1 ]] = 0 and so [[ei , e0 ], e1 ] = [[ei , e1 ], e0 for all i ≥ 1. Thus, [ei , e1 ] = α3 ei+2 + α4 ei+3 + · · · + αn+1−i en for 1 ≤ i ≤ n. Let [e0 , e1 ] = θ3 e3 + θ4 e4 + · · · + θn en . Consider [e0 , [e1 , e0 ]] = [[e0 , e1 ], e0 ] − [[e0 , e0 ], e1 ]. We have

[[e0 , e1 ], e0 ] = [[e0 , e0 ], e1 ].

However, [e0 , e0 ] = e2 and [ei , e0 ] = ei+1 . Therefore, θ3 e4 + θ4 e5 + · · · + θn−1 en = α3 e4 + α4 e5 + · · · + αn−1 en ; whence [e0 , e1 ] = α3 e3 + α4 e4 + · · · + αn−1 en−1 + αn en .

172  Leibniz Algebras

Thus, in Case 1 we obtain the following family: [e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , [e0 , e1 ] = α3 e3 +α4 e4 +· · ·+αn−1 en−1 +θn en , [ei , e1 ] = α3 ei+2 + α4 ei+3 + · · · + αn+1−i en for 1 ≤ i ≤ n. Case 2. µ : µn1 + β. In this case [e0 , e0 ] = e2 and [ei , e0 ] = ei+1 for 2 ≤ i ≤ n − 1, whence {e2 , e3 , . . . , en } ⊆ Annr (L) and so [ei , e j ] = 0 for 2 ≤ j ≤ n, 0 ≤ i ≤ n. Let β(e1 , e0 ) = α3 e3 + α4 e4 + · · · + αn en . Making the change e1 := e1 − α3 e2 − α4 e3 − · · · − αn en−1 , we may assume that [e1 , e0 ] = 0. Let [e0 , e1 ] = β3 e3 + β4 e4 + · · · + βn en . Consider the product [e0 , [e1 , e0 ]] = [[e0 , e1 ], e0 ] − [[e0 , e0 ], e1 ]. Since [e1 , e0 ] ∈ Annr (L), we have [[e0 , e1 ], e0 ] = [[e0 , e0 ], e1 ]. Therefore, [[e0 , e1 ], e0 ] = [e2 , e1 ], i.e., [e2 , e1 ] = β3 e4 + β4 e5 + · · · + βn−1 en . Consider the equality [e1 , [e0 , e1 ]] = [[e1 , e0 ], e1 ] − [[e1 , e1 ], e0 ]. In view of [e0 , e1 ] ∈ Annr (L) and [e1 , e0 ] = 0, we have [[e1 , e1 ], e0 ] = 0. However, e0 annihilates from the left only en . Therefore, [e1 , e1 ] = en . Considering the equality [ei , [e0 , e1 ]] = [[ei , e0 ], e1 ] − [[ei , e1 ], e0 ] for 2 ≤ i ≤ n − 1 and taking into account [e0 , e1 ] ∈ Annr (L) we derive [[ei , e0 ], e1 ] = [[ei , e1 ], e0 ]. Therefore [ei+1 , e1 ] = [[ei , e1 ], e0 ], i.e., [ei , e1 ] = β3 ei+2 +β4 ei+3 +· · ·+βn+1−i en for 2 ≤ i ≤ n − 1. Thus, in Case 2 we obtain the following family

On Some Classes of Leibniz Algebras  173

[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , [e0 , e1 ] = β3 e3 + β4 e4 + · · · + βn en = γen . [ei , e1 ] = β3 ei+2 + β4 ei+3 + · · · + βn+1−i en , 2 ≤ i ≤ n. Case 3. Let L ∈ TLbn+1 and let {e0 , e1 , . . . , en } be a basis of L. Then due to Theorem 4.3 [ei , e j ] ∈ Span{ei+ j+1 , . . . , en } for any i, j , 0. Then [ei , e0 ] = ei+1 + (∗)ei+2 + · · · + (∗)en , 1 ≤ i ≤ n − 1. Putting e01 = e1 , e00 = e0 , e0i+1 := [e0i , e00 ] we may assume that [ei , e0 ] = ei+1 , 1 ≤ i ≤ n − 1. Now consider i+3 n [e0 , ei ] = −ei+1 + αi+2 0,i ei+2 + α0,i ei+3 + · · · + α0,i en , 1 ≤ i ≤ n − 1.

Then we get i+3 n [ei , e0 ] + [e0 , ei ] = αi+2 0,i ei+2 + α0,i ei+3 + · · · + α0,i en , 1 ≤ i ≤ n − 1. (4.3) From Leibniz identity we have [x, y]+[y, x] ∈ Annr (L), for any x, y ∈ L. Therefore, if we multiply by e0 the both sides of (4.3) from the right-hand side (n − i − 2) times, we obtain αi+2 0,i = 0. Substituting it in (4.3) and repeating this action sufficient times, we get

αi+k 0,i = 0, 2 ≤ k ≤ n − 1 − i. Applying the above to [ei , ei ] for i : 0 ≤ i ≤ that [ei , ei ] = αni,i en . The following chain of equalities

h i n 2

we conclude

[e0 , ei ] = [e0 , [ei−1 , e0 ]] = [[e0 , ei−1 ], e0 ] − [[e0 , e0 ], ei−1 ] = = [−ei + αn0,i−1 en , e0 ] = −[ei , e0 ] = −ei+1

174  Leibniz Algebras

leads to [ei , e0 ] = −[e0 , ei ] = ei+1 for 2 ≤ i ≤ n − 1, i.e., [e0 , x] = −[x, e0 ] for any x ∈ L2 . We claim that [ei , e j ] = −[e j , ei ], 1 ≤ i < j ≤ n − 1.

(4.4)

Indeed, the induction by i for any j and the following chain of equalities: [ei , e j+1 ] = [ei , [e j , e0 ]] = [[ei , e j ], e0 ] − [[ei , e0 ], e j ] (since [ei , e j ] ∈ L2 ) = −[e0 , [ei , e j ]] + [[e0 , ei ] − αn0,i en , e j ] = −[e0 , [ei , e j ]] + [[e0 , ei ], e j ] = −[[e0 , ei ], e j ] + [[e0 , e j ], ei ] + [[e0 , ei ], e j ] = −[e j+1 , ei ], 1 ≤ j ≤ n − 1 gives (4.4). The above observations lead to the required table of multiplication of L ∈ TLbn+1 .  Remark. It should be noted that a Leibniz algebra from the family TLbn+1 is non Lie algebra if and only if (α, β, γ) , (0, 0, 0). In fact, this inequality can be replaced by the condition γ = 1. For L ∈ TLbn+1 the subspace Span{en } spanned by {en } is an ideal of L and the quotient algebra L/Span{en } is the n-dimensional filiform Lie algebra with the composition law [ei , e0 ] = ei+1 , i = 1, 2, . . . , n − 1, j+1) [ei , e j ] = a1i, j ei+ j+1 + · · · + an−(i+ en−1 , 1 ≤ i < j ≤ n − 1. i, j Lemma 4.6. Let L ∈ TLbn+1 . Then n−(i+ j+k+1) X s=1

a sj,k bi, j+k+s

=

n−(i+ j+k+1) X s=1

s (ai,s j bi+ j+s,k − ai,k bi+k+s, j ).

(4.5)

On Some Classes of Leibniz Algebras  175

Proof. The Leibniz identity for ei , e j and ek gives the required relations between htheP structure constants. Indeed, i j+k+1) s [ei , [e j , ek ]] = ei , n−( a j,k e j+k+s + b j,k en s=1   P j+k+1) s Pn−(i+ j+k+s+1) t e + b e , = n−(i+ a a i+ j+k+s+t i, j+k+s n i,ij+k+s t=1 j,k hs=1 Pn−(i+ j+1) s [[ei , e j ], ek ] = s=1 ai, j ei+ j+s + bi, j en , ek  Pn−(i+ j+k+1) s Pn−(i+ j+k+s+1) t = s=1 ai, j t=1 ai+ j+s,k ei+ j+k+s+t + bi+ j+s,k en , i hP s e + b e , e a [[ei , ek ], e j ] = n−(i+k+1) i+k+s i,k n j s=1 i,k  P j+k+1) s Pn−(i+ j+k+s+1) t = n−(i+ a ai+k+s, j ei+ j+k+s+t + bi+k+s, j en , s=1 t=1 i,k and this implies that n−(i+ j+k+1) X

a sj,k bi, j+k+s

=

n−(i+ j+k+1) X

s=1

s (ai,s j bi+ j+s,k − ai,k bi+k+s, j ).

s=1

 Here are several useful remarks regarding (4.5) which can significantly simplify the multiplication table of TLbn+1 : 1. It is symmetric with respect to i, j, k (since aks,t = −akt,s and b s,t = −bt,s for any s and t, except for (s, t) = (0, 0), (1, 1), (0, 1), (1, 0)). 2. In the case when (i, j, k) = (0, j, k) we get n−(X j+k+1)

a sj,k b0, j+k+s

s=1

=

n−(X j+k+1)

s (a0,s j b j+s,k − a0,k bk+s, j ),

s=1

where j , 0, k , 0. s 3. Since a0,t = 0 as s , 1 and a10,t = −1, we get j+k+1) a1j,k b0, j+k+1 + a2j,k b0, j+k+2 + · · · + an−( b0,n−1 = −b j+1,k + bk+1, j . j,k

4. Since b0,t = 0 as t = 2, . . . , n − 2 and b0,n−1 = −1, one has that j+k+1) an−( = b j+1,k − bk+1, j , j,k

for k = j + 1, j + 2, . . . , n − j − 2 and j = 1, 2, . . . ,

h

n−3 2

i

.

Lemma 4.7. Let L ∈ TLbn+1 . Then k   X k−s k s [ei , e j+k ] = (−1) s [ei+k−s , e j ]Re0 , s=0

where 1 ≤ i, j, k ≤ n.

(4.6)

176  Leibniz Algebras

Proof. We proceed by the induction on k. Let k = 1. Then [ei , e j+1 ] = [ei , [e j , e0 ]] = −[ei+1 , e j ] + [[ei , e j ], e0 ], i.e., the equality (4.6) holds at k = 1. This is the base of the induction. Then the following chain of equalities lead to the claim: [ei , e j+k+1 ] = [ei , [e j+k , e0 ]] = [[ei , e j+k ], e0 ] − [[e  i , e0 ], e j+k ] Pk = s=0 (−1)k−s ks [ei+k−s , e j ]Res+1 0   Pk − s=0 (−1)k−s ks [ei+k+1−s , e j ]Res0   P k k−s = − k+1 (−1) [ei+k+1−s , e j ]Res0 s=1 s − 1   P − ks=0 (−1)k−s ks [ei+k+1−s , e j ]Res0     P k k s = ks=1 (−1)k+1−s s − 1 + s [ei+k+1−s , e j ]Re0 k +[ei+k+1 , e j ]Rk+1 [ei+1+k , ek ] e0 − (−1)  Pk+1 = s=0 (−1)k+1−s k +s 1 [ei+k+1−s , e j ]Res0 .



4.3 CLASSIFICATION OF SOME SOLVABLE LEIBNIZ ALGEBRAS

In this section we give classification of solvable Leibniz algebras with given nilradicals. Let us first consider a solvable Leibniz algebra whose nilradical is the null-filiform Leibniz algebra NFn . Due to Theorem 3.4 we need to describe the derivations of NFn . Their description is given below. Proposition 4.2. Any derivation of the algebra NFn has the following matrix form: a a a3 . . . an  2  1  0 2a1 a2 . . . an−1   0 0 3a1 . . . an−2   . .. .. .. ..  .  .. . . . .  0 0 0 . . . na1 Proof. The proof is straightforward verification using the table of multiplication of NFn . 

On Some Classes of Leibniz Algebras  177

Corollary 4.2. The maximal number of nil-independent derivations of the n-dimensional null-filiform Leibniz algebra NFn is 1. Proof. Let  . . . ain   . . . ain−1   . . . ain−2  , i = 1, 2, . . . , p, .. ..  . .  . . . nai1 n  ai be derivations of NFn . If p > 1, then Di − a11 D1 = 0 with non-trivial 1 scalars. Hence the system {D1 , D2 , . . . , D p } is not nil-independent.   i a1 ai2 ai3  0 2ai ai 1 2  i Di :=  0 0 3a1  .. .. ..  . . . 0 0 0

Corollary 4.3. The dimension of a solvable Leibniz algebra with nilradical NFn is equal to n + 1. Proof. Let us assume that the solvable Leibniz algebra is decomposed as R = NFn ⊕ Q. Then by Corollary 4.2 and Theorem 3.4 we have 1 ≤ dim Q ≤ 1. Hence, dimK Q = 1.  Theorem 4.4. Let R be a solvable Leibniz algebra whose nilradical is NFn . Then there exists a basis {e1 , e2 , . . . , en , x} of the algebra R such that the multiplication table of R with respect to this basis has the following form:   [e , e ] = ei+1 , 1 ≤ i ≤ n − 1,    i 1  [x, e1 ] = e1 ,      [e , x] = −ie , 1 ≤ i ≤ n. i i Proof. According to Theorem 3.2 and Corollary 4.3 there exists a basis {e1 , e2 , . . . , en , x} such that all products of elements of the basis are known, except of the products [ei , x], [x, e1 ] and [x, x]. The products [ei , x] can be derived from the equalities [ei+1 , x] = [[ei , e1 ], x] = [ei , [e1 , x]] + [[ei , x], e1 ] , 1 ≤ i ≤ n − 1 and [e1 , x]. Let us introduce the notations n n X X [x, e1 ] = αi ei , [e1 , x] = βi ei , i=1

i=1

[x, x] =

n X i=1

γi ei ,

178  Leibniz Algebras

where {e1 , e2 , . . . , en } is a basis of NFn and {x} is a basis of Q. Now we consider the following two possible cases. Case 1. Let α1 , 0. Then taking the change of basis: e0i

n 1 X α j−i+1 e j , = α1 j=i

x0 =

1 ≤ i ≤ n,

1 x, α1

we can assume that [x, e1 ] = e1 and other products by redesignation of parameters can be assumed not changed. From the products  n  n  X  X 0 = [x, [x, x]] =  x, γi ei  = γi [x, ei ] = γ1 e1 , i=1

i=1

we can deduce that γ1 = 0. On the other hand, from the Leibniz identity [x, [e1 , x]] = [[x, e1 ], x] − [[x, x], e1 ] we get β1 [x, e1 ] = [e1 , x] −

n X

γi−1 ei , i.e.,

i=3

β1 e1 =

n X

βi ei −

i=1

n X

γi−1 ei .

i=3

Comparing the coefficients at the elements of the basis, we obtain β2 = 0 and γi = βi+1 for 2 ≤ i ≤ n − 1. From the equality [e1 , [e1 , x]] = − [e1 , [x, e1 ]], we derive that β1 = −1. Thus, we have [e1 , x] = −e1 +

n X

βi ei ,

[x, x] =

i=3

n−1 X

βi+1 ei + γn en .

i=2

Now we prove the following identity [ei , x] = −iei +

n X j=i+2

β j−i+1 e j ,

(4.7)

On Some Classes of Leibniz Algebras  179

for 1 ≤ i ≤ n. It is obvious that (4.7) is true for i = 1. Assume that (4.7) holds for each i, 1 ≤ i < k ≤ n. Then [ek , x] = [[ek−1 , e1 ], x] = [ek−1 , [e1 , x]] + [[ek−1 , x], e1 ]   n X   = [ek−1 , −e1 ] + −(k − 1)ek−1 + β j−k+2 e j , e1  j=k+1

n X

= −ek − (k − 1)ek +

n X

β j−k+2 [e j , e1 ] = −kek +

β j−k+1 e j .

j=k+2

j=k+1

By induction, we see that indeed (4.7) holds for all i, 1 ≤ i ≤ n. Thus, the multiplication table of the algebra R is written as follows  [ei , e1 ] = ei+1 , 1 ≤ i ≤ n − 1,        [x, e1 ] = e1 ,     n  X     [ei , x] = −iei + β j−i+1 e j , 1 ≤ i ≤ n, (4.8)    j=i+2      n−1  X      βi+1 ei + γn en .   [x, x] = i=2

Let us make the following change of basis: e0i

= ei +

n X

A j−i+1 e j , 1 ≤ i ≤ n,

x = 0

j=i+2

n−1 X

Ai+1 ei + Bn en + x,

i=2

where Ai , Bn are given as follows A3 = 12 β3 , A4 = 31 β4 Ai

=

Bn =

1 i−1 1 n

i−2 P

!

Ai− j+1 β j + βi , ! An− j+2 β j + γn .

5 ≤ i ≤ n,

j=3 n−1 P j=3

Then taking into account the multiplication table (4.8) we compute the products in the new basis " # n n P P 0 0 [ei , e1 ] = ei + A j−i+1 e j , e1 = ei+1 + A j−i e j = e0i+1 , j=i+2

j=i+3

where 1 ≤ i ≤ n − 1,

180  Leibniz Algebras

#

" [x , e1 ] = 0

n−1 P

0

i=2 n P

Ai+1 ei + Bn en + x, e1

n P = Ai ei + [x, e1 ] = e1 + Ai ei = e01 , #i=3 "i=3 n−1 P Ai+1 ei + Bn en + x, x [x0 , x0 ] =

= =

i=2 n−1 P

i=2 n−1 P i=2

Ai+1 [ei , x] + Bn [en , x] + [x, x] ! n P Ai+1 −iei + β j−i+1 e j − nBn en j=i+2

+

=−

n−1 P i=2

iAi+1 ei +

n−3 P i=2

j=i+2

+ =

n−1 P i=2

n−1 P

Ai+1 n−2 P i=2

β j−i+1 e j +

n−1 P

i=2 n−1 P

βi+1 ei + γn en

βi+1 ei

i=2

Ai+1 βn−i+1 en − Bn en + γn en

(−iAi+1 + βi+1 )ei +

n−1 P i−1 P

Ai− j+2 β j ei

i=4 j=3

+ −nBn + γn +

n−1 P

! Ai+1 βn−i+1 en

i=2

= (−2A3 + β3 )e2 + (−3A4 + β4 )e3 n−1 P i−1 P + (−iAi+1 + βi+1 + Ai− j+2 β j )ei = 0, i=4 j=3 " # n n P P 0 0 [e1 , x ] = e1 + Ai ei , x = [e1 , x] + Ai [ei , x] i=3 i=3 ! n n n P P P = −e1 + βi ei + Ai −iei + β j−i+1 e j = −e1 + = −e1 − = −e1 −

i=3 n P i=3 n P

i=3 n P i=3 n P

βi ei − Ai ei − Ai ei +

i=3 n P

i=3 n P

iAi ei +

n P

j=i+2 n P

Ai

i=3

(i − 1)Ai ei +

i=3 n P i=3

β j−i+1 e j

j=i+2 n P i=3

βi ei +

(−(i − 1)Ai + βi ) ei +

n i−2 P P

! Ai− j+1 b j ei

i=3 j=3 n i−2 P P

Ai− j+1 β j ei

i=5 j=3

= −e1 − Ai ei + (−2A3 + β3 )e3 + (−3A4 + β4 )e4 i=3  n i−2 n P P P + −(i − 1)Ai + βi + Ai− j+1 β j ei = −e1 − Ai ei = −e01 . i=5 j=3

i=3

On Some Classes of Leibniz Algebras  181

Similarly to that in (4.7) we obtain [e0i , x0 ] = −ie0i , 1 ≤ i ≤ n. Thus, we get the required multiplication table for R. Case 2. Let α1 = 0. Then from the equalities [e1 , [e1 , x]] = − [e1 , [x, e1 ]] and 0 = [x, [x, x]] we get β1 = 0 and γ1 = 0, respectively. Thus, we have the following products:   [ei , e1 ] = ei+1 , 1 ≤ i ≤ n − 1,      n  X     αi ei , [x, e ] =  1     i=2     n X    [e , x] = βi ei ,  1     i=2     n X      [x, x] = γi ei .    i=2

Similarly to that in (4.7) we can prove the equality: [ei , x] = n P β j−i+1 e j . Consequently, we have [ei , x] ∈ Span{ei+1 , ei+2 , . . . , en }, j=i+1

i.e., Ri ⊆ Span{ei , ei+1 , . . . , en }. Thus, Rn+1 = 0 which contradicts the assumption of non nilpotency of the algebra R. This implies that, in the case of α1 = 0, there is no non nilpotent solvable Leibniz algebra with nilradical NFn .  Now we clarify the situation when the nilradical is represented as a direct sum of its two null-filiform ideals and the complementary space to the nilradical is one-dimensional. Theorem 4.5. Let R be a solvable Leibniz algebra such that R = NFk ⊕ NF s + Q, where NFk ⊕ NF s is the nilradical of R, NFk and NF s are ideals of the nilradical and dim Q = 1. Then NFk and NF s are also ideals of the algebra R. Proof. Let {e1 , e2 , . . . , ek } be a basis of NFk and { f1 , f2 , . . . , f s } a basis of NF s and let {x} be a basis of Q. We can assume, without loss of generality, that k ≥ s. Theorem 3.2 implies the inclusion {e2 , e3 , . . . , ek , f2 , f3 , . . . , f s } ⊆ Annr (R)

182  Leibniz Algebras

and the following equalities: [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1, Let us introduce the notations:  s k X X      βi fi , α e + [x, e ] = i i 1     i=1 i=1      s k  X X   σi fi , λ e + [e , x] =  i i 1     i=1 i=1     k s  X X      ρi ei + ξi fi .   [x, x] = i=1

[ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1 .

[x, f1 ] = [ f1 , x] =

k X

δi ei +

s X

i=1

i=1

k X

s X

i=1

τi ei +

γi f i , µi f i ,

i=1

i=1

From the products s

k

X      X 0 = x, [e1 , f1 ] = [x, e1 ], f1 − [x, f1 ], e1 = βi−1 fi − δi−1 ei , 

i=2

i=2

we obtain βi = 0, 1 ≤ i ≤ s − 1 and δi = 0, 1 ≤ i ≤ k − 1.   [e1 , [e1 , x]] = − [e1 , [x, e1 ]] and f1 , [ f1 , x] =  The equalities  − f1 , [x, f1 ] imply that λ1 = −α1 , µ1 = −γ1 .   From the equalities 0 = [e1 , [x, x]] = ρ1 e2 and 0 = f1 , [x, x] = ξ1 f2 , we get ρ1 = ξ1 = 0. In a similar way as in the proof of Theorem 4.4, the following equalities can be proved: [ei , x] = −iα1 ei +

k X

λ j−i+1 e j ,

2 ≤ i ≤ k,

µ j−i+1 f j ,

2 ≤ i ≤ s.

j=i+1

[ fi , x] = −iγ1 fi +

s X j=i+1

Summarizing, we obtain the following multiplication table for the

On Some Classes of Leibniz Algebras  183

algebra R:   [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,       [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,       k X      αi ei + β s f s , [x, e1 ] =      i=1    s  X      [x, f1 ] = δk ek + γi f i ,      i=1     k s  X X     [e1 , x] = −α1 e1 + λi ei + σi fi ,      i=2 i=1    k s X X      [ f1 , x] = τi ei − γ1 f1 + µi fi ,      i=1 i=2     k  X     [e , x] = −iα e + λ j−i+1 e j , 2 ≤ i ≤ k,  i 1 i     j=i+1      s X      [ f , x] = −iγ e + µ j−i+1 f j , 2 ≤ i ≤ s,  i 1 i     j=i+1      k s  X X     [x, x] = ρi ei + ξi fi .    i=2

(4.9)

i=2

Below, we analyze the different cases that can appear in terms of the possible values of α1 and γ1 . Case 1. Let α1 = γ1 = 0. Then the multiplication table (4.9) implies [ei , x] [ fi , x] [e1 , x] [ f1 , x]

∈ Span{ei+1 , ei+2 , . . . , ek }, ∈ Span{ fi+1 , fi+2 , . . . , f s }, ∈ Span{e2 , e3 , . . . , ek , f1 , f2 , . . . , f s }, ∈ Span{e1 , e2 , . . . , ek , f2 , f3 , . . . , f s }.

The above relations mean that the algebra R is nilpotent, so we get a contradiction with the assumption of non nilpotency of R. Therefore, this case is impossible. Case 2. Let α1 , 0 and γ1 = 0. Using the following change of basis:

184  Leibniz Algebras

=

e0i =

1 α1

x0 =

1 x, α1

we may assume that

!

k P

1 α1

e01

i=1 k P

αi ei + β s f s ,

α j−i+1 e j , 2 ≤ i ≤ k,

j=i

[x, e1 ] = e1 .

From the identity [x, [x, e1 ]] = [[x, x], e1 ] − [[x, e1 ], x] we have that e1 =

k X

ρi [ei , e1 ] − [e1 , x] =

i=2

k X

ρi−1 ei + e1 −

i=3

k X

λi ei −

i=2

s X

σi fi .

i=1

Consequently, λ2 = σi = 0 for 1 ≤ i ≤ s and ρi = λi+1 for 2 ≤ i ≤ k − 1. From the identity       f1 , [x, e1 ] = [ f1 , x], e1 − [ f1 , e1 ], x k   P we conclude that 0 = [ f1 , x], e1 = τi−1 ei ⇒ τi = 0, 1 ≤ i ≤ k − 1. i=2

From the identity       x, [x, f1 ] = [x, x], f1 − [x, f1 ], x , we obtain 0=

s X i=3

ξi−1 fi −

s X

γi [ fi , x] + δk [ek , x]

i=2

 s    X = ξi−1 fi − γi  µ j−i+1 f j  − kδk ek i=3 i=2 j=i+1   s s X X X   i  γ j−1 µi− j+2  fi − kδk ek = ξi−1 fi − i=3 i=3 j=3   s i X  X  ξi−1 − γ j−1 µi− j+2  fi − kδk ek . = s X

s X

i=3

j=3

On Some Classes of Leibniz Algebras  185

By comparison of coefficients at the elements of the basis we deduce that: i+1 X γ j−1 µi− j+3 , 2 ≤ i ≤ s − 1 and δk = 0. ξi = j=3

Now we consider the following change of basis: f10 = f1 + τkk ek , fi0 = fi , 2 ≤ i ≤ s. Then we obtain s

[ f10 ,

s

s

X X X τk x] = [ f1 + ek , x] = µi fi + τk ek − τk ek = µi fi = µi fi0 k i=2 i=2 i=2

and

s

[x, f10 ] = [x, f1 +

s

X X τk ek ] = [x, f1 ] = γi fi = γi fi0 . k i=2 i=2

Thus, we have the following multiplication table of the algebra R:   [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,       [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,       [x, e1 ] = e1 ,      s X      [x, f1 ] = γi fi ,      i=2     k  X      [e1 , x] = −e1 + λi ei ,     i=2     s  X   [ f , x] = µi fi ,  1     i=2      k X      [e , x] = −ie + λ j−i+1 e j , 2 ≤ i ≤ k,  i i     j=i+2     s  X     [ fi , x] = µ j−i+1 f j , 2 ≤ i ≤ s,      j=i+1      k s  X X      ρi ei + ξi fi .   [x, x] = i=2

i=2

186  Leibniz Algebras

From the above multiplication table the following inclusions can be immediately derived: [x, NFk ] ⊆ NFk ,

[NFk , x] ⊆ NFk ,

[x, NF s ] ⊆ NF s ,

[NF s , x] ⊆ NF s .

This completes the proof of the assertion of the theorem for this case. Case 3. Let α1 = 0 and γ1 , 0. Due to symmetry of Cases 2 and 3, the proof of the theorem follows similar arguments as in Case 2. Case 4. Let α1 , 0 and γ1 , 0. Consider the following change of basis:  k  k  1 X 1 X 0 0  e1 = α j−i+1 e j , 2 ≤ i ≤ k,  αi ei + β s f s  , ei = α1  i=1 α1 j=i  s  s  1 X 1 1 X 0 0  f1 = γ j−i+1 f j , 2 ≤ i ≤ s, x0 = x.  γi fi + δk ek  , fi = γ1 i=1 γ1 j=i α1 Then we derive   k   1 X  1 0 0  αi ei + β s f s  = 1 α1 [x, e1 ] = 1 [x, e1 ] = e01 , [x , e1 ] =  x, α1 α1 i=1 α1 α21  s    1  1 X γ1 0 1 γ1 [x, f1 ] = f . [x0 , f10 ] =  x,  γi fi + δk ek  = α1 γ1 i=1 α1 γ1 α1 1 From the identity [x, [x, e1 ]] = [[x, x], e1 ] − [[x, e1 ], x] we deduce: e1 =

k X i=2

ρi [ei , e1 ] − [e1 , x] =

k X i=3

ρi−1 ei + α1 e1 −

k X i=2

λi ei −

s X

σi fi .

i=1

Therefore, α1 = 1, λ1 = −1, λ2 = σi = 0, 1 ≤ i ≤ s and ρi = λi+1 , 2 ≤ i ≤ k − 1.       Expanding the identity x, [x, f1 ] = [x, x], f1 − [x, f1 ], x , we derive the equalities: !2 s s s k X X γ1 γ1 γ1 X γ1 X f1 = ξi [ fi , f1 ]− [ f1 , x] = ξi−1 fi − µi f i − τi ei α1 α1 α1 i=1 α1 i=1 i=2 i=3 from which we have µ1 = − αγ11 , µ2 = τi = 0, 1 ≤ i ≤ k and ξi = γ1 µ , 2 ≤ i ≤ s − 1. α1 i+1

On Some Classes of Leibniz Algebras  187

Finally, we obtain the following products of basis elements in the algebra R:  [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,       [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,        [x, e1 ] = e1 ,      γ1    f1 , [x, f1 ] =    α 1     k  X   λi ei , [e1 , x] = −e1 +      i=3     s  X  γ1    [ f , x] = − f + µi f i ,  1 1   α1   i=3     k s  X X      [x, x] = ρ e + ξi fi . i i   i=2

i=2

These products are sufficient in order to check the inclusions: [x, NFk ] ⊆ NFk ,

[NFk , x] ⊆ NFk ,

[x, NF s ] ⊆ NF s ,

[NF s , x] ⊆ NF s .

Thus, the ideals NFk and NF s of the nilradical are also ideals of the algebra.  Now we study the solvable Leibniz algebras with nilradical NFk ⊕ NF s and with one-dimensional complementary vector space. Due to Theorem 4.5 we can assume that NFk and NF s are ideals of the algebra. Theorem 4.6. Let R be a solvable Leibniz algebra such that R = NFk ⊕ NF s + Q, where NFk ⊕ NF s is the nilradical of R and dim Q = 1. Let us assume that {e1 , e2 , . . . , ek } is a basis of NFk , { f1 , f2 , . . . , f s } is a basis of NF s and {x} is a basis of Q. Then the algebra R is isomorphic to one

188  Leibniz Algebras

of the following pairwise non-isomorphic algebras:  [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,        [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,        [x, e1 ] = e1 , R(α) :=    [x, f1 ] = α f1 , α , 0,       [ei , x] = −iei , 1 ≤ i ≤ k,      [ f , x] = −iα f , 1 ≤ i ≤ s, i i   [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,       [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,        [x, e1 ] = e1 ,     s  X R(β2 , β3 , . . . , β s , γ) :=    [ fi , x] = β j−i+1 f j , 1 ≤ i ≤ s,     j=i+1        [ei , x] = −iei , 1 ≤ i ≤ k,      [x, x] = γ f s . In the second family of algebras the first non-zero element of the vector (β2 , β3 , . . . , β s , γ) can be assumed to be equal to 1. Proof. Firstly, we note that the algebras NFk + Q and NF s + Q are not simultaneously nilpotent. Indeed, if they are both nilpotent, then we have: [ei , e1 ] ∈ Span{ei+1 , . . . , ek }, [ fi , f1 ] ∈ Span{ fi+1 , . . . , f s }, [x, e1 ] ∈ Span{e2 , e3 , . . . , ek }, [ei , x] ∈ Span{ei+1 , . . . , ek }, [ f j , x] ∈ Span{ f j+1 , . . . , f s },

1 ≤ i ≤ k − 1, 1 ≤ i ≤ s − 1, [x, f1 ] ∈ Span{ f2 , f3 , . . . , f s }, 1 ≤ i ≤ k − 1, 2 ≤ i ≤ s − 1.   From the equalities 0 = [e1 , [x, x]] , 0 = f1 , [x, x] we conclude that: [x, x] ∈ Span{e2 , e3 , . . . , ek , f2 , f3 , . . . , f s }. Therefore, R2 ⊆ Span{e2 , e3 , . . . , ek , f2 , f3 , . . . , f s }. Moreover, we have Ri ⊆ Span{ei , ei+1 , . . . , ek , fi , fi+1 , . . . , f s },

On Some Classes of Leibniz Algebras  189

which implies that Rmax{k,s}+1 = {0}. Thus, we have a contradiction to the assumption that R is not nilpotent. Hence, the algebras NFk + Q and NF s + Q cannot be nilpotent at the same time. Without loss of generality, we can assume that the algebra NFk + Q is non-nilpotent. We take the quotient algebra by the ideal NF s , then R/NF s  NF k + Q. By Theorem 4.4 the structure of the algebra NF k + Q is known. Namely,   [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,     [x, e1 ] = e1 , (4.10)      [e , x] = −i e , 1 ≤ i ≤ k. i i Using the fact that NFk and NF s are ideals of R and considering the multiplication table (4.10), we have that:   [ei , e1 ] = ei+1 , 1 ≤ i ≤ k − 1,       [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,       [x, e1 ] = e1 ,      s  X     [x, f1 ] = αi fi ,      i=1  (4.11)   [ei , x] = −iei , 1 ≤ i ≤ k,      s  X     [ f , x] = βi fi ,  1     i=1     s  X     γi fi .    [x, x] = i=1

If α1 , 0, then in a similar way as in the Case 1 of Theorem 4.4 we obtain the family of algebras R(α), where α , 0. The fact that two algebras in the family R(α) with different values of parameter α are not isomorphic can be easily established by a general change of basis and considering the expansion of the product [x0 , f10 ] in both bases. Now consider α1 = 0. Then by the change of basis x0 = x − (α2 f1 + α3 f2 + · · · + α s f s−1 ), we can suppose that [x, f1 ] = 0.

190  Leibniz Algebras

      From the identity f1 , [ f1 , x] = [ f1 , f1 ], x − [ f1 , x], f1 we get β1 = 0. Similarly to the proof of (4.7), we can show that [ fi , x] =

s X

βm−i+1 f j , 1 ≤ i ≤ s.

m=i+1

      The identity x, [ f1 , x] = [x, f1 ], x − [x, x], f1 implies the following equalities: s

X   0 = − [x, x], f1 = − γm−1 fm . m=3

Consequently, γi = 0, 2 ≤ i ≤ s − 1. Thus, we obtain the products of the family R(β2 , β3 , . . . , β s , γ)   [ fi , f1 ] = fi+1 , 1 ≤ i ≤ s − 1,     s  X    [ f , x] = βm−i+1 fm , 1 ≤ i ≤ s, i      m=i+1     [x, x] = γ f . s s

Now we are going to study the isomorphism inside the family R(β2 , β3 , . . . , β s , γ). Taking into account that, under general basis transformation, the products (4.11) should be the same, we conclude that it is sufficient to take the following change of basis: fi0

=

Ai−1 1

s X

A j−i+1 f j , (A1 , 0),

x0 = x.

1 ≤ i ≤ s,

j=i

Then we have [ f10 , x0 ] =

s X

Ai [ fi , x] =

i=1

s−1 X i=1

 s   i−1  s X  X  X      fi . Ai  β j−i+1 f j  = A B j i− j+1    j=i+1

i=2

j=1

On the other hand [ f10 , x0 ] =

s X i=2

β0i fi0 =

s−1 X i=1

  i−1   s−i s X X  X    j 0    fi . Ai1 β0i+1  A j fi+ j  = A A β 1 i− j j+1    j=1

i=2

j=1

On Some Classes of Leibniz Algebras  191

Comparing coefficients at the elements of the basis we deduce that: k−1 X

Ai βk−i+1 =

k−1 X

Ai1 Ak−i β0i+1 ,

k = 2, 3, . . . , s .

i=1

i=1

From these systems of equations it follows that β0i =

βi , Ai−1 1 , Ai−1 1

2 ≤ i ≤ s.

If we consider γ0s A1s f s = γ0s f s0 = [x0 , x0 ] = [x, x] = γ s f s , then we obtain

γ0s =

γs . A1s

It is easy to see that by choosing an appropriate value for the parameter A1 , the first non-zero element of the vector (β2 , β3 , . . . , β s , γ) can be assumed to be equal to 1. Therefore, two algebras R(β2 , β3 , . . . , β s , γ) and R(β02 , β03 , . . . , β0s , γ0 ) with different set of parameters are not isomorphic. For given parameters α and β2 , β3 , . . . , β s , γ, the algebras R(α) and R(β2 , β3 , . . . , β s , γ) are not isomorphic because k + s = dim R(α)2 , dim R(β2 , β3 , . . . , β s , γ)2 = k + s − 1.  Remark. In the case when all coefficients (β2 , β3 , . . . , β s , γ) are equal to zero we have the split algebra (NFk + Q) ⊕ NF s . Therefore, in the non split case, we can always assume that (β2 , β3 , . . . , β s , γ) , (0, 0, 0, . . . , 0). Now, by an induction process, we are going to generalize Theorem 4.6 to the case when the nilradical is a direct sum of several (greater than 2) copies of null-filiform ideals. Theorem 4.7. Let R be a solvable Leibniz algebra such that R = NFn1 ⊕ NFn2 ⊕ · · · ⊕ NFns + Q, where NFn1 ⊕ NFn2 ⊕ · · · ⊕ NFns is the nilradical of R and dim Q = 1. There exist p, q ∈ N with p , 0 and p + q = s, a basis {ei1 , ei2 , . . . , eini } of NFni , for 1 ≤ i ≤ p, a basis

192  Leibniz Algebras

{ f1k , f2k , . . . , fnkk } of NFn p+k , for 1 ≤ k ≤ q, and a basis {x} of Q such that the multiplication table of the algebra R is given by:  j j j   , 1 ≤ i ≤ n j − 1, [ei , e1 ] = ei+1     k k k   [ fi , f1 ] = fi+1 , 1 ≤ i ≤ nk − 1,        [x, e1j ] = δ j e1j , δ j , 0      nk  X    [ f k , x] = βkm−i+1 fmk , 1 ≤ i ≤ nk , R p,q :=  (4.12) i    m=i+1       [eij , x] = −iδ j eij , 1 ≤ i ≤ n j ,       k  X     [x, x] = γm fnm ,    m=1

where 1 ≤ j ≤ p, 1 ≤ k ≤ q and δ1 = 1. Moreover, the first nonzero component of the vectors (βk2 , βk3 , . . . , βknk , γk ) can be assumed to be equal to 1. Moreover, the above algebras are pairwise non-isomorphic. Proof. We shall prove the theorem by the induction on s: If s = 1, then p = 1, q = 0, so R1,0 is the algebra given in Theorem 4.4. If s = 2, then we have two cases: either p = 2, q = 0 or p = 1, q = 1, which were considered in Theorem 4.5. Namely, we have two families of algebras: R(α), which corresponds to R2,0 , and R(β2 , β3 , . . . , β s , γ), which corresponds to R1,1 . Let us assume that the theorem is true for s and we shall prove it for s + 1. Let R = NFn1 ⊕ NFn2 ⊕ · · · ⊕ NFns ⊕ NFns+1 + Q. We consider the quotient algebra by NFns+1 , i.e. R/NFns+1  NF n1 ⊕NF n2 ⊕· · ·⊕NF ns +Q. Then we get the multiplication table given in (4.12). Note that the multiplication table for the algebra R can be obtained from (4.12) by adding the products s+1 [eis+1 , e1s+1 ] = ei+1 , n s+1 X [x, e1s+1 ] = αms+1 ems+1 , m=1

[e1s+1 , x] =

n s+1 X m=1

βms+1 ems+1 ,

1 ≤ i ≤ n s+1 − 1,

On Some Classes of Leibniz Algebras  193

[x, x] =

n s+1 X

γms+1 ems+1 .

m=1

If α1s+1 , 0, then in a similar way as in proof of Theorem 4.4, we deduce that s+1 [eis+1 , e1s+1 ] = ei+1 ,

1 ≤ i ≤ n s+1 − 1,

s+1 [x, e1s+1 ] = α s+1 s+1 e1 ,

[eis+1 , x] = −iα s+1 eis+1 ,

1 ≤ i ≤ n s+1 .

Therefore we get the algebra R p+1,q . If α1s+1 = 0, then by similar arguments as in Theorem 4.6, we obtain s+1 , [eis+1 , e1s+1 ] = ei+1 n s+1 X s+1 βm−i+1 fms+1 , [eis+1 , x] =

[x, x] =

1 ≤ i ≤ n s+1 − 1, 1 ≤ i ≤ n s+1 ,

m=i+1 k X

γm fnm + γ s+1 fns+1 . s+1

m=1 q+1 s+1 Setting fi−1 = ei−1 we get the family of algebras R p,q+1 . The proof that two algebras of the family R p,q with different values of parameters are not isomorphic is carried on in the same way as in the proof of Theorem 4.6. 

In fact, due to Theorem 3.4, when the nilradical of a solvable Leibniz algebra is a direct sum of s copies of null-filiform ideals, the complementary vector space has dimension not greater than s. By taking the direct sum of ideals NFi + Qi and NFk ⊕ · · · ⊕ NF s , where 1 ≤ i ≤ k − 1, k ≤ s, we can construct a solvable Leibniz algebra whose nilradical is NF1 ⊕ · · · ⊕ NF s and whose complementary vector space is k-dimensional.

CHAPTER

5

ISOMORPHISM CRITERIA FOR FILIFORM LEIBNIZ ALGEBRAS

5.1

ON BASE CHANGES IN COMPLEX FILIFORM LEIBNIZ ALGEBRAS

Since an arbitrary filiform Leibniz algebra, up to isomorphism, belongs to one of the classes from Theorem 4.3, we conclude that in order to investigate the problem of isomorphisms inside the classes, we need to study the behavior of the parameters (structure constants) under base change inside each family. Let L be an (n + 1)-dimensional complex filiform Leibniz algebra obtained from a naturally graded filiform non Lie Leibniz algebra and let {e0 , e1 , . . . , en } be its basis. Definition 5.1. The basis {e0 , e1 , . . . , en } of L is said to be adapted if the table of multiplications of the algebra with respect to {e0 , e1 , . . . , en } has one of the forms FLbn+1 , SLbn+1 and TLbn+1 . Let L be a Leibniz algebra defined on a vector space V and let {e0 , e1 , . . . , en } be an adapted basis of L.

195

196  Leibniz Algebras

Definition 5.2. A basis transformation f ∈ GL(V) is said to be adapted with respect to the table of multiplications of L if the basis { f (e0 ), f (e1 ), . . . , f (en )} also is adapted. Obviously, the composition of the adapted base changes is an adapted transformation. The closed subgroup of GL(V) consisting of the adapted transformations is denoted by GLad (V). Further we shall need the following identity, the proof of which is straightforward. n P i=k

n P

ai

j=i+p

bi, j e j =

j−p n P P

ai bi, j e j ,

j=k+p i=k

(5.1)

0 ≤ p ≤ n − k, 3 ≤ k ≤ n. The proposition below describes the adapted basis transformations in algebras from the classes FLbn+1 and SLbn+1 . Proposition 5.1. Let f ∈ GLad (V) and {e0 , e1 , . . . , en } be the adapted basis. a) If L ∈ FLbn+1 then f : L −→ L has the following form:  n P    f (e0 ) = ai ei ,    i=0    n−2  P  f (e ) = (a + a )e + ai ei + (an−1 + a1 (θ − αn ))en−1 + bn en , 1 0 1 1    i=2     (ei+1 ) = [ f (ei ), f (e0 )], 1 ≤ i ≤ n − 1   ff (e 2 ) = [ f (e0 ), f (e0 )]. b) If L ∈ SLbn+1 then f : L −→ L has the following form:  n P    f (e ) = ai ei , 0    i=0    f (e1 ) = b1 e1 − a1ab01 γ en−1 + bn en      (ei+1 ) = [ f (ei ), f (e0 )], 2 ≤ i ≤ n − 1   ff (e ) = [ f (e ), f (e )]. 2

0

0

Proof. Note that an adapted basis change for filiform Lie and Leibniz algebras is defined by its action on the basis vectors e0 and e1 . Therefore we set n n X X f (e0 ) = ai ei and f (e1 ) = b je j. i=0

j=0

Isomorphism Criteria for Filiform Leibniz Algebras  197

Case a) Consider the product f (e2 ) = [ f (e0 ), f (e0 )]. Then by (5.1) we have n P [ f (e0 ), f (e0 )] = a0 (a0 + a1 )e2 + a0 ai−1 ei ! i=3 n n−1 P P αi ei + θen + a21 αi ei +a0 a1 i=3

i=3

+a1

n−2 P

ai

i=2

n P

αk+1−i ek

k=i+2 n P

n−1 P ai−1 ei + a1 (a0 + a1 ) αi ei i=3 i=3 n n P P +a1 (a0 θ + a1 αn )en + a1 ai−2 αk+3−i ek i=4 k=i n P = a0 (a0 + a1 )e2 + a0 ai−1 ei

= a0 (a0 + a1 )e2 + a0

i=3

+a1 (a0 + a1 ) +a1

=

n−1 P i=3

αi ei + a1 (a0 θ + a1 αn )en

n P k P

(ai−2 αk+3−i ek ) i=4 i=4 a0 (a0 + a1 )e2 + (a0 a2 + a1 (a0 n−1 P +

t=4

+a1 +a1

+ a1 )α3 )e3

(a0 at−1 + a1 (a0 + a1 )αt

t P

i=4 n P

ai−2 αt+3−i )et + (a0 an−1 + a1 (a0 θ + a1 αn ) ai−2 αn+3−i )en

i=4

= f (e2 ). Due to [ f (e0 ), f (e0 )] ∈ L2 , we get a0 (a0 + a1 ) , 0. Consider the product n X [ f (e0 ), f (e1 )] = b0 (a0 + a1 )e2 + ci ei . i=3

Since [ f (e0 ), f (e1 )] < L2 and a0 + a1 , 0, this implies that b0 = 0. The properties of the adapted transformation give f (e2 ) = [ f (e1 ), f (e0 )]. The product [ f (e1 ), f (e0 )] in the basis {e0 , e1 , . . . , en } is expanded as follows [ f (e1 ), f (e0 )] = a0 b1 e2 + (a0 b2 + a1 b1 α3 )e3

198  Leibniz Algebras

+

n−1 P

!

bi−2 αt+3−i et ! n P + a0 bn−1 + a1 b1 αn + a1 bi−2 αn+3−i en . t=4

a0 bt−1 + a1 b1 αt + a1

t P

i=4

i=4

Comparing the coefficients at the basis vectors we get the conditions to the coefficients of f :  a0 + a1 = b1 , a2 = b2 ,     t t P P    a a + a a α = a b + a bi−2 αt+3−i ,  0 t−1 1 i−2 t+3−i 0 t−1 1    i=4 i=4   n P   a0 an−1 + a1 (a0 θ + a1 αn ) + a1 ai−2 αn+3−i    i=4    n P    = a b + a b α + a bi−2 αn+3−i .  0 n−1 1 1 n 1  i=4

Simplifying we obtain   = a0 + a1 ,   b1 b = a, 2≤i≤n−2 i    bn−1 = ain−1 + a1 (θ − αn ). Case b) is proved similarly.  We introduce the notion of elementary transformations for algebras from families FLbn+1 and SLbn+1 as follows (for Lie algebras case see [87]). Definition 5.3. The following types of the adapted transformations are said to be elementary:  f (e0 ) = e0 + aek , 2 ≤ k ≤ n     f (e1 ) = e1 + bek , 2 ≤ k ≤ n first type − τ(a, b, k) =     f (ei+1 ) = [ f (ei ), f (e0 )], 1 ≤ i ≤ n − 1, = [ f (e0 ), f (e0 )] 2)  f (e f (e ) = ae0 + be1 0     f (e ) = (a + b)e1 + b(θ − αn )en−1 ,  1  a(a + b) , 0 second type − ϑ(a, b) =     f (e ) = [ f (e ), f (e )], 1 ≤ i ≤ n − 1,  i+1 i 0  f (e2 ) = [ f (e0 ), f (e0 )]

Isomorphism Criteria for Filiform Leibniz Algebras  199

 f (e ) = e  0 0    f (e1 ) = e1 + ben , third type − σ(b, n) =     f (ei+1 ) = [ f (ei ), f (e0 )], 2 ≤ i ≤ n − 1, 0 ), f (e0 )]  f f(e(e2 )) = [=f (e e + aek  0 0    f (e1 ) = e1 fourth type − η(a, k) =     f (ei+1 ) = [ f (ei ), f (e0 )], 2 ≤ i ≤ n − 1, f (e2 ) = [ f (e0 ), f (e0 )] 2≤k≤n  f (e0 ) = ae0 + be1      f (e1 ) = de1 − bdγ en−1 , ad , 0 a fifth type − δ(a, b, d) =    (ei+1 ) = [ f (ei ), f (e0 )], 2 ≤ i ≤ n − 1,   ff (e 2 ) = [ f (e0 ), f (e0 )] a, b, d ∈ C. Let f be an arbitrary element of the group GLad (V), then f can be expressed as compositions of the elementary transformations as follows. Proposition 5.2. i) Let f be of the form a) of Proposition 5.1. Then f = τ(an , bn , n) ◦ τ(an−1 , an−1 , n − 1) ◦ . . . . ◦ τ(a2 , a2 , 2) ◦ ϑ(a0 , a1 ) ii) Let f be of the form b) of Proposition 5.1. Then f = σ(bn , n) ◦ η(an , n) ◦ η(an−1 , n − 2) ◦ · · · ◦ η(a2 , 2) ◦ δ(a0 , a1 , b1 ) Proof. Straightforward.



One has the following proposition. Proposition 5.3. 1) The basis transformation φ = τ(an , bn , n) ◦ τ(an−1 , an−1 , n − 1) ◦ · · · ◦ τ(a2 , a2 , 2) does not change the structure constants of algebras from FLbn+1 . 2) The basis transformation ϕ = σ(bn , n) ◦ η(an , n) ◦ η(an−1 , n − 2) ◦ · · · ◦ η(a2 , 2) does not change the structure constants of algebras from SLbn+1 .

200  Leibniz Algebras

Proof. Let us prove the first assertion. Consider the basis transformation τ(a, b, k) :  f (e ) = e + ae ,  0 0 k    f (e1 ) = e1 + bek , τ(a, b, k) =     f (ei+1 ) = [ f (ei ), f (e0 )], 1 ≤ i ≤ n − 1 f (e2 ) = [ f (e0 ), f (e0 )]

2≤k≤n

For 2 ≤ k ≤ n − 1 we put a = b and consider the products which involve the parameters: [ f (e0 ), f (e1 )] = = [ f (e1 ), f (e1 )] = =

n−k+1 P i=3 n−1 P i=3 n P

αi (ei + aek+i−1 ) +

n−1 P i=n−k+2

αi ei + θen

αi f (ei ) + θ f (en ),

αi ei + a

i=3 n−k+1 P i=3

n−k+1 P

αi ek+i−1

i=3

αi (ei + aek+i−1 ) +

n P i=n−k+2

αi ei =

n P

αi f (ei ).

i=3

Therefore, the basis transformation τ(a, a, k), 2 ≤ k ≤ n − 1, for any a ∈ C does not change the structure constants αi , θ. Similarly, one can check that τ(a, b, n) ∈ GLad (V) also does not change the structure constants αi , θ for any value of a. Since a composition of adapted transformations is again an adapted transformation, we conclude that the transformation φ = τ(an , bn , n) ◦ τ(an−1 , an−1 , n − 1) ◦ · · · ◦ τ(a2 , a2 , 2) also does not change the structure constants of the family FLbn+1 . The proof of the second part is carried out similarly.



Thus, the study of all basis transformations is reduced to the second and fifth types of elementary transformations for the families FLbn+1 and SLbn+1 , respectively. 5.2 CRITERIA OF ISOMORPHISMS OF COMPLEX FILIFORM NON-LIE LEIBNIZ ALGEBRAS

We recall that for any element a of a Leibniz algebra L the right multiplication operator is defined as follows Ra (x) = [x, a]. Set 0 Rma (x) := [[. . . [x, a], . . . , a] | a],{z } and Ra (x) := x. m times

Isomorphism Criteria for Filiform Leibniz Algebras  201

It is easy to check that for an algebra from the first two families of Theorem 4.3 the following equality is valid: [[e s , e1 ], e0 ] = [e s+1 , e1 ], 2 ≤ s ≤ n.

(5.2)

Let L belong to the family FLbn+1 (respectively, of the family SLbn+1 ). Then (5.2) implies that for m ∈ N, 0 ≤ p ≤ n (respectively, 0 ≤ p ≤ n, p , 1) the following identity holds true: Rme1 (e p ) = Rep−1 (Rme1 (e0 )). 0

(5.3)

To prove the main result (Theorem 5.1) we need the following lemma. Lemma 5.1. Let L belong to the family the first two families from Theorem 4.3. Then for 2 ≤ m ≤ n−1 the following identities hold true 2 Rme1 (e0 ) =

im X

n X

im =2m+1 im−1 =2m+1

where ηi =



i2 X

···

ηim +3−im−1 . . . ηi2 +3−i1 ηi1 +3−2(m−1) eim ,

i1 =2m+1

αi , if L ∈ FLbn+1 βi , if L ∈ SLbn+1 .

Proof. Let us consider the case ηi = αi , whereas the case ηi = βi is proved similarly. We use the method of mathematical induction with respect to m. For m = 2 we have " # n−1 n P P 2 Re1 (e0 ) := αi ei + θen , e1 = αi−2 [ei−2 , e1 ] i=3

=

n P i=5

αi−2

n P

α j+3−i e1

j=i

=

i=5 n P

Pj

αi−2 α j+3−i e1 .

j=5 i=5

Assume that for m the equality of the lemma is true. Then the following equalities complete the proof m Rm+1 e1 (e0 ) = [Re1 (e0 ), e1 ] " # im i2 n P P P = ··· αim +3−ik−1 . . . αi2 +3−i1 αi1 +3−(2m+1) eim , e1 ik =2m+1 im−1 =2m+1

i1 =2m+1

202  Leibniz Algebras

" = =

im P

n P

···

im =2m+3 im−1 =2m+3 im n P P

αim +3−im−1 . . . αi2 +3−i1 αi1 +3−(2m+3) eim −2 , e1

i1 =2m+3 i2 P

αim +3−im−1 . . . αi2 +3−i1 αi1 +3−(2m+3)

···

im =2m+3 im−1 =2m+3

#

i2 P

i1 =2m+3

n P im+1 =im

= =

n P

n P

im P

im =2m+3 im+1 =im im−1 =2m+3 iP im n m+1 P P

i2 P

···

αim +3−im−1 . . . αi2 +3−i1 αi1 +3−(2m+3) eim +1

i1 =2m+3 i2 P

···

im+1 =2m+3 im =2m+3 im−1 =2m+3

αim+1 +3−im eim +1

i1 =2m+3

αim +3−im−1 . . . αi2 +3−i1 αi1 +3−(2m+3) eim +1 . 

The study of adapted transformations for the family FLbn+1 is reduced to the study of its action on e0 and e1 as follows ( 0 e0 = f (e0 ) = Ae0 + Be1 , A(A + B) , 0, 0 (5.4) e1 = f (e2 ) = (A + B)e1 + B(θ − αn )en−1 . Let us find out the action on whole basis. It is given in the corollary below. Corollary 5.1. The adapted base change for the class FLbn+1 is given as follows e0 = Ae0 + Be1 , A(A + B) , 0, 0 e1 = (A + B)e1 + B(θ − αn )en−1 n−1 P 0 e2 = A(A + B)e2 + B(A + B) αi ei + B(Aθ + Bαn ), i=3 k−2 P  k − 1  k−1−i i i 0 ek = (A + B) B Re1 (ek−i ) k−1−i A i=0  +Bk−1 Rk−1 (e ) , 0 e1 0

 where 3 ≤ k ≤ n and

s t



=

(5.5)

s! . (s−t)!t!

Proof. We prove the corollary by induction on k. For k = 2, 3 we have n−1 P 0 0 0 e2 = [e1 , e0 ] = A(A + B)e2 + B(A + B) αi ei + B(Aθ + Bαn ), i=3

e3 = [e2 , e0 ] = A2 (A + B)e3 + 2AB(A + B)[e2 , e1 ] +B2 (A + B)[[e0 , e1 ], e1 ] 2 2 2 = (A + B)(A e3 + 2ABRe1 (e2 ) + B Re1 (e0 )). 0

0

0

Isomorphism Criteria for Filiform Leibniz Algebras  203

Suppose that equality (5.5) is true for k. Taking into account the equality (5.2) and the following equalities 0 ek+1 0 = [ek , e0 ] ! h i k−2 P  k − 1  k−1−i i i k−1 k−1 = (A + B) A B R (e ) + B R (e ) , Ae + Be k−i 0 0 1 e1 e1 k−1−i i=0   k−2 P k−1 k−i i i k−1 k−1 = (A + B) k − 1 − i A B Re1 (ek+1−i ) + AB Re1 (e2 ) i=0 ! k−2 P  k − 1  k−1−i i+1 i+1 k k + B Re1 (ek−i ) + B Re1 (e0 ) k−1−i A i=0   k−2 P k−1 k−1 k−1 k−i i i = (A + B) k − 1 − i A B Re1 (ek+1−i ) + AB Re1 (e2 ) i=0 ! k−1 P  k − 1  k−i i i k k + k − i A B Re1 (ek+1−i ) + B Re1 (e0 ) i=1 k−2 P  k − 1   k − 1  k−i i i = (A + B) k − 1 − i + k − i A B Re1 (ek+1−i ) i=1 

k k k−1 k−1 (e ) (e ) + B R (e ) + AB R + kk −− 11 Ak ek+1 + k −1 1 ABk−1 Rk−1 0 2 2 e1 e1 e1   k−2 P  k  k−i i i k k = (A + B) k − i A B Re1 (ek+1−i ) + k A ek+1 i=1   1 ABk−1 Rk−1 (e ) + Bk Rk (e ) + k− 2 e1 0 e1 1 ! k−1 P  k  k−i i i k k = (A + B) k − i A B Re1 (ek+1−i ) + B Re1 (e0 ) i=0 we complete the proof of the equality (5.5) for k + 1.  Similarly, for the family SLbn+1 the result is given by the following corollary. Corollary 5.2. The adapted base change for the class SLbn+1 is given as  follows 0  e0 = Ae0 + Be1 , AD , 0,    0 BDγ   e  1 − A en−1 ,  !  1 = Dek−2   P  0 k − 1  k−1−i i i  ek = A B Re1 (ek−i ) + Bk−1 Rk−1  e1 (e0 ) ,  k−1−i A   i=0   3 ≤ k ≤ n. Proof. The proof is the similar to that of Corollary 5.1.



We shall denote an algebra from family FLbn+1 (respectively, SLbn+1 ) by L(α3 , α4 , . . . , αn , θ) (respectively, L(β3 , β4 , . . . , βn , γ)).

204  Leibniz Algebras

Theorem 5.1. a) Two algebras L(α3 , α4 , . . . , αn , θ) and 0 L (α03 , α04 , . . . , α0n , θ0 ) are isomorphic if and only if there exist A, B ∈ C such that A(A + B) , 0 and the following conditions hold: α03 = (A+B) α3 A2 t−1 P  k − 1  k−2 1 0 αt = At−1 (A + B)αt − k − 2 A Bαt+2−k k=3   t P k−1 αt+3−i1 αi1 +1−k + k − 3 Ak−3 B2 i1 =k+2   i2 t P P k−1 + k − 4 Ak−4 B3 αt+3−i2 αi2 +3−i1 · αi1 −k . . . i2 =k+3 i1 =k+3   iP i2 t k−3 P 1 ABk−2 P + k− · · · αt+3−ik−3 αik−3 +3−ik−4 1 ik−3 =2k−2 ik−4 =2k−2

. . . αi2 +3−i1 αi1 +5−2k iP t k−2 P + Bk−1

i2 P

···

ik−2 =2k−1 ik−3 =2k−1

i1 =2k−1

i1 =2k−2

αt+3−ik−2 αik−2 +3−ik−3 . . .


 αi2 +3−i1 αi1 +4−2k α0k , where 4 ≤ t ≤ n. n−1 P  k − 1  k−2 1 Aθ + Bαn − θ0 = An−1 k − 2 A Bαn+2−k k=3     n P k−1 k−1 + k − 3 Ak−3 B2 αn+3−i1 αi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 n P P 1 ABk−2 αn+3−i2 αi2 +3−i1 αi1 −k + . . . + k − 1 i2 =k+3 i1 =k+3 iP n k−3 P

ik−3 =2k−2 ik−4 =2k−2 n P k−1

+ B

i2 P

...

i1 =2k−2

iP k−2

ik−2 =2k−1 ik−3 =2k−1

...

αn+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k i2 P i1 =2k−1

αn+3−ik−2 αik−2 +3−ik−3 . . .
 αi2 +3−i1 αi1 +4−2k α0k ,

b) Two algebras L(β3 , β4 , . . . , βn , γ) and L (β03 , β04 , . . . , β0n , γ0 ) are isomorphic if and only if there exist A, B, D ∈ C such that AD , 0 and the following conditions hold: 0

γ0 = β03 =

D2 γ, An D β, A2 3

β0t =

At−1

1

Dβt −

t−1 P k=3



   k−1 k−1 k−2 k−3 2 A Bβ + t+2−k k−2 k−3 A B

Isomorphism Criteria for Filiform Leibniz Algebras  205 t P i1 =k+2 t P



βt+3−i1 βi1 +1−k + i2 P

k−1 k−4



Ak−4 B3

βt+3−i2 βi2 +3−i1 βi1 −k + . . . +

i2 =k+3 i1 =k+3 iP t k−3 P

ik−3 =2k−2 ik−4 =2k−2 iP t k−2 P ik−2 =2k−1 ik−3 =2k−1

···

i2 P i1 =2k−2 i2 P

....



k−1 1



ABk−2

βt+3−ik−3 βik−3 +3−ik−4 . . . βi2 +3−i1 βi1 +5−2k + Bk−1 ! ! βt+3−ik−2 βik−2 +3−ik−3 . . . .βi2 +3−i1 βi1 +4−2k β0k

i1 =2k−1

where 4 ≤ t < n,

n−1 P 

  k−1 k−1 k−2 + An−1 Dβn − = k − 2 A Bβn+2−k + k − 3 k=3   n k−1 k−3 2 P ·A B βn+3−i1 · βi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 n P P k − 1 · βn+3−i2 βi2 +3−i1 βi1 −k + . . . + ABk−2 1 β0n

·

BDγ An

1

i2 =k+3 i1 =k+3 iP n k−3 P

···

i2 P

ik−3 =2k−2 ik−4 =2k−2 i1 =2k−2 i n k−2 P P k−1

+B

···

ik−2 =2k−1 ik−3 =2k−1
 ·βi2 +3−i1 βi1 +4−2k β0k .



βn+3−ik−3 βik−3 +3−ik−4 . . . βi2 +3−i1 βi1 +5−2k i2 P i1 =2k−1

βn+3−ik−2 βik−2 +3−ik−3 . . .

Proof. Consider the class FLbn+1 . Let {e0 , e1 , . . . , en } be a basis of L(α3 , α4 , . . . , αn , θ), and let {e00 , e01 , . . . , e0n } be a basis of 0 0 0 0 0 L (α3 , α4 , . . . , αn , θ ). It is easy to see that in L(α3 , α4 , . . . , αn , θ) the following holds [[e0 , e1 ], e1 ] = [[e1 , e1 ], e1 ]. Lemma 5.1 and the identity (5.3) imply Rme1 (ek−m ) im n P P =

im =k+m im−1 =k+m

···

i2 P

αim +3−im−1 . . . αi2 +3−i1 αi1 +3−(k+m) eim ,

i1 =2m+1

where m ≤ n − k and m ≤ k ≤ n.

(5.6)

206  Leibniz Algebras

Now we substitute (5.6) in (5.5) and use the equalities (5.1), (5.3) to get ! k−2 P  k − 1  k−1−i i i 0 k−1 k−1 ek = (A + B) B Re1 (ek−1 ) + B Re1 (e0 ) k−1−i A i=0     n P k−1 k−1 k−1 k−2 = (A + B)(A ek + k − 2 A B αi+2−k ei + k − 3 Ak−3 B2 i=k+1   n P Pi k−1 αi+3−i1 · αi1 +1−k ei + k − 4 Ak−4 B3 · i=k+2 i1 =k+2   i2 n P Pi P 1 ABk−2 · αi+3−i2 αi2 +3−i1 αi1 −k ei + · · · + k − 1 i=k+3 i2 =k+3 i1 =k+3 i2 n P Pi P · ... αi+3−ik−3 . . . αi2 +3−i1 · αi1 +5−2k e1 i=2k−2 ik−3 =2k−2 n Pi k−1 P

i1 =2k−2

i2 P ... αi+3−ik−2 . . . αi2 +3−i1 αi1 +4−2k ei ) i=2k−1 ik−2 =2k−1  i1 =2k−1    k−1 k−1 = (A + B) ( Ak−1 ek + k − 2 Ak−2 Bα3 ek+1 + k − 2 Ak−2 Bα4 !   k+2 k−1 k−3 2 P + k−3 A B αk+5−i1 · αi1 +1−k ek+2 + . . . i1 =k+2     t P k−1 k−1 + k − 2 Ak−2 Bαt+2−k + k − 3 Ak−3 B2 αt+3−i1 αi1 +1−k i1 =k+1     i2 t P k−1 k − 1 k−4 3 P + k−4 A B αt+3−i2 αi2 +3−i1 αi1 −k + · · · + 1

+B

·ABk−2 + Bk−1

t P

i2 =k+3 i1 =k+3 iP i2 k−3 P

ik−3 =2k−2 ik−4 =2k−2 iP t k−2 P

...

αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k !

αt+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k et     n P k−1 k−1 + . . . + k − 2 Ak−2 Bαn+2−k + k − 3 Ak−3 B2 αn+3−i1 αi1 +1−k i1 =k+1     i2 n P P k−1 1 + k − 4 Ak−4 B3 αn+3−i2 αi2 +3−i1 αi1 −k + . . . + k − 1 i2 =k+3 i1 =k+3 ·ABk−2 +Bk−1

ik−2 =2k−1 ik−3 =2k−1

n P

iP k−3

ik−3 =2k−2 ik−4 =2k−2 iP n k−2 P ik−2 =2k−1 ik−3 =2k−1

...

i1 =2k−2 i2 P

...

...

i1 =2k−1

i2 P

αn+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k ! αn+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k en )

i1 =2k−2 i2 P

i1 =2k−1

Isomorphism Criteria for Filiform Leibniz Algebras  207

   n  P k−1 k−1 k−2 k−3 2 A Bα + t+2−k k−2 k−3 A B t=k+1   t P k−1 · αt+3−i1 αi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 t P P 1 ABk−2 · αt+3−i2 αi2 +3−i1 αi1 −k + . . . + k − 1 = (A + B) Ak−1 ek +

i2 =k+3 i1 =k+3 iP t k−3 P

·

ik−3 =2k−2 ik−4 =2k−2 iP t k−2 P

i2 P

...

·

ik−2 =2k−1 ik−3 =2k−1

αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k + Bk−1 i1 =2k−2 ! ! i2 P ... αt+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k et . i1 =2k−1

Consider the following products in the algebra L (α3 , α4 , . . . , αn , θ ): 0

0

0

[e0 , e1 ] =

n−1 X

0

0

0

0

αk ek + θ en ,

k=3

0

0

[e1 , e1 ] =

0

n X

0

0

0

0

0

αk ek .

k=3

Substituting the expression for e0k , obtained above, into the product using the equality (5.1) with p = 1, we derive:

[e00 , e01 ] and [e00 , e01 ] n−1 P 0

   n  P k−1 k−1 k−2 = αk (A + B) A ek + k − 2 A Bαt+2−k + k − 3 k=3 t=k+1   t P k−1 ·Ak−3 B2 αt+3−i1 αi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 t P P k − 1 · αt+3−i2 αi2 +3−i1 αi1 −k + · · · + ABk−2 1

·

i2 =k+3 i1 =k+3 iP t k−3 P

ik−3 =2k−2 ik−4 =2k−2 iP t k−2 P

k−1

...

·

ik−2 =2k−1 ik−3 =2k−1

i2 P i1 =2k−2

...

αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k + Bk−1 ! ! i2 P αt+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k et

i1 =2k−1

+θ An−1 (A + B)en 0

 n−1 n−1 n  P k−1 0 P P k−1 Ak−2 Bαt+2−k = (A + B) A αk ek + k − 2 k=3 t=k+1     k=3 t k−1 k−1 k−3 2 P + k−3 A B αt+3−i1 αi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 t P P k − 1 ABk−2 αt+3−i2 αi2 +3−i1 αi1 −k + · · · + 1 i2 =k+3 i1 =k+3

208  Leibniz Algebras t P

iP k−3

i2 P

αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k + Bk−1 ! ... αt+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k · i1 =2k−1 ik−2 =2k−1 ik−3 =2k−1  0 ·αk et +θ0 An−1 en n−1 n−1 P t−1 0 P t−1 P  k − 1  k−2 = (A + B) A2 α03 e3 + A αt et + k − 2 A Bαt+2−k t=3 t=3 k=3     t P k−1 k−1 αt+3−i1 αi1 +1−k + k − 4 Ak−4 B3 + k − 3 Ak−3 B2 i1 =k+2   i2 t P P 1 ABk−2 · αt+3−i2 αi2 +3−i1 αi1 −k + · · · + k − 1 i2 =k+3 i1 =k+3 i i t k−3 2 P P P · ... αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k + Bk−1 ik−3 =2k−2 ik−4 =2k−2 i1 =2k−2 ! ! iP i2 t k−2 P P 0 · ... αt+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k αk et ik−2 =2k−1 ik−3 =2k−1 i1 =2k−1   n−1 P  k − 1  k−2 k−1 +(A + B) θ0 An−1 + A Bα + Ak−3 B2 n+2−k k − 2 k − 3 k=3  n P k−1 · αn+3−i1 αi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 n P P 1 ABk−2 · αn+3−i2 αi2 +3−i1 αi1 −k · · · + k − 1 ik−3 =2k−2 ik−4 =2k−2 iP t k−2 P

...

i2 =k+3 i1 =k+3 iP n k−3 P

i1 =2k−2 i2 P

...

i2 P

αn+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k + Bk−1 ! ! · ... αn+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k α0k en ik−2 =2k−1 ik−3 =2k−1 i1 =2k−1 n−1 t−1 P P  k − 1  k−2 = (A + B) A2 α03 e3 + At−1 α0t et + k − 2 A Bαt+2−k t=3 k=3     t P k−1 k−1 + k − 3 Ak−3 B2 αt+3−i1 αi1 +1−k + k − 4 Ak−4 B3 i1 =k+2   i2 t P P 1 ABk−2 · αt+3−i2 αi2 +3−i1 αi1 −k + · · · + k − 1 ·

ik−3 =2k−2 ik−4 =2k−2 iP n k−2 P

·

i2 =k+3 i1 =k+3 iP t k−3 P

ik−3 =2k−2 ik−4 =2k−2 iP t k−2 P

...

·

ik−2 =2k−1 ik−3 =2k−1

i1 =2k−2 i2 P

i2 P i1 =2k−2 i2 P

...

αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k + Bk−1 ! ! ! 0 αt+3−ik−2 αik−2 +3−ik−3 . . . αi2 +3−i1 αi1 +4−2k αk et

i1 =2k−1

Isomorphism Criteria for Filiform Leibniz Algebras  209 n−1 P  k − 1  k−2 0 + θ An−1 + k − 2 A Bαn+2−k     k=3 P n k−1 k−1 2 αn+3−i1 αi1 +1−k + k − 4 Ak−4 B3 + k−3 B i1 =k+2   i2 n P P 1 ABk−2 · αn+3−i2 αi2 +3−i1 αi1 −k + · · · + k − 1

·

i2 =k+3 i1 =k+3 iP n k−3 P

...

i2 P

ik−3 =2k−2 ik−4 =2k−2 i1 =2k−2 i n k−2 P P k−1

i2 P

αn+3−ik−2 αik−2 +3−ik−3 . . .
  αi2 +3−i1 αi1 +4−2k α0k en . The similar expression for [e01 , e01 ] can be easily obtained by substitution in [e00 , e01 ] the coefficient α0n instead of θ0 : n t−1 P P  k − 1  0 0 [e01 , e01 ] = (A + B) A2 α3 e3 + At−1 αt + k−2 t=4 k=3   t P k−1 ·Ak−2 Bαt+2−k + k − 3 Ak−3 B2 αt+3−i1 αi1 +1−k i1 =k+2     i2 t P k−1 k − 1 k−4 3 P + k−4 A B αt+3−i2 αi2 +3−i1 αi1 −k + ABk−2 1 i2 =k+3 i1 =k+3 iP i2 t k−3 P P αt+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k · ... +B

ik−2 =2k−1 ik−3 =2k−1

...

αn+3−ik−3 αik−3 +3−ik−4 . . . αi2 +3−i1 αi1 +5−2k i1 =2k−1

ik−3 =2k−2 ik−4 =2k−2 i1 =2k−2 iP t k−2 P k−1

+B

ik−2 =2k−1 ik−3 =2k−1

...

i2 P i1 =2k−1

αt+3−ik−2 αik−2 +3−ik−3 . . .
0  αi2 +3−i1 αi1 +4−2k αk et .

On the other hand, we have 0 0 [e0 , e1 ] = [Ae0 + Be1 , (A + B)e1 + B(θ − αn )en−1 ] n−1 P = (A + B)2 αt et + (A + B)(Aθ + Bαn )en , t=3

[e01 , e01 ] = [(A + B)e1 + B(θ − αn )en−1 , (A + B)e1 + B(θ − αn )en−1 ] = (A + B)

n P

αt et .

t=3

Comparing the coefficients of the basis elements et and keeping in mind that the coefficient of A + B should not be zero, we get the restrictions, that were outlined in the first assertion of the theorem. Using Corollary 5.2, the assertion b) of the theorem is proved by applying similar arguments. 

210  Leibniz Algebras

Remark. From Theorem 5.1, we have that α0k is a polynomial of the form Pk (A, B, α3 , α4 , . . . , αk , α03 , α04 , . . . , α0k−1 ), where the parameters α3 , α4 , . . . , αk , α03 , α04 , . . . , α0k−1 are given and coefficients A, B are unknown, but satisfy the condition A(A + B) , 0. And β0k is also a polynomial of the form Qk (A, B, D, β3 , β4 , . . . , βk , β03 , β04 , . . . , β0k−1 ), where the parameters β3 , β4 , . . . , βk , β03 , β04 , . . . , β0k−1 are given and coefficients A, B, D are unknown, but satisfy the condition AD , 0. Therefore, the calculation of parameters α0k and β0k is a recursive procedure. Consequently, we conclude that in any given dimension the problem of the classification (up to an isomorphism) of complex filiform Leibniz algebras, which are obtained from the naturally graded filiform non-Lie algebras, is an algorithmically solvable task. 5.2.1

Isomorphism criteria

In this section we give a few versions of the isomorphism criteria discussed in the previous section. We write them in an invariant form adapted for application to the classification problem of filiform Leibniz algebras in low dimensions. Let us first treat the classes FLbn+1 and SLbn+1 . The action of GLad (V) on LBn induces the actions on FLbn+1 and SLbn+1 . Due to Proposition 5.3 the action can be written as a composition of elementary transformations of the second and the fifth types, respectively. The next two theorems are reformulations of the corresponding results of [88] on isomorphism criteria for filiform Leibniz algebras appearing from the naturally graded non Lie filiform Leibniz algebras. Introduce the following series of functions: ϕt (y; z) = ϕt (y; z3 , z4 , . . . , zn , zn+1 )   t−1 t P  k−1  k−1 2 P = (1 + y) zt − yz + y zt+3−i1 zi1 +1−k + t+2−k k−2 k−3 k=3 i1 =k+2     i2 t P k−1 3 P k − 1 + k−4 y zt+3−i2 zi2 +3−i1 zi1 −k + . . . + yk−2 1 i2 =k+3 i1 =k+3 iP i2 t k−3 P P · ... zt+3−ik−3 zik−3 +3−ik−4 . . . zi2 +3−i1 zi1 +5−2k + yk−1 ik−3 =2k−2 ik−4 =2k−2 i1 =2k−2 ! iP i2 t k−2 P P · ... zt+3−ik−2 zik−2 +3−ik−3 . . . zi2 +3−i1 zi1 +4−2k ϕk (y; z), ik−2 =2k−1 ik−3 =2k−1

i1 =2k−1

for 3 ≤ t ≤ n and   ϕn+1 (y; z) = (zn+1 − zn ) + (1 + y)ϕn (y; z) .

Isomorphism Criteria for Filiform Leibniz Algebras  211

Theorem 5.2. Two algebras L(α) and L(α0 ) from FLbn+1 , where α = (α3 , α4 , . . . , αn , θ), and α0 = (α03 , α04 , . . . , α0n , θ0 ), are isomorphic if and only if there exist complex numbers A and B such that A(A + B) , 0 and the following conditions hold:  1 B 0 αt = At−2 ϕt A ; α , 3 ≤ t ≤ n,   1 θ0 = An−2 ϕn+1 AB ; α , (see Theorem 5.1 a)). Let ψt (y; z) = ψt (y; z3 , z4 , . . . , zn , zn+1 )   t−1 t P  k−1  k−1 2 P = zt − yz + y zt+3−i1 zi1 +1−k t+2−k k−2 k−3 k=3 i1 =k+2     i2 t P P k−1 1 yk−2 + k − 4 y3 zt+3−i2 zi2 +3−i1 zi1 −k + · · · + k − 1 i2 =k+3 i1 =k+3 iP i2 t k−3 P P · ··· zt+3−ik−3 zik−3 +3−ik−4 . . . zi2 +3−i1 zi1 +5−2k + yk−1 ik−3 =2k−2 ik−4 =2k−2 i1 =2k−2 ! iP i2 t k−2 P P · ··· zt+3−ik−2 zik−2 +3−ik−3 . . . zi2 +3−i1 zi1 +4−2k ψk (y; z) , ik−2 =2k−1 ik−3 =2k−1

i1 =2k−1

where 3 ≤ t ≤ n, and ψn+1 (y; z) = zn+1 . Theorem 5.3. Two algebras L(β) and L(β0 ) from SLbn+1 , where β = (β3 , β4 , . . . , βn , γ), and β0 = (β03 , β04 , . . . , β0n , γ0 ), are isomorphic if and only if there exist complex numbers A, B and D such that AD , 0 and the following conditions  hold:  B 1 D 0 βt = At−2 A ψt A ; β , 3 ≤ t ≤ n − 1,   1 DB β0n = An−2 γ + ψn AB ; β , AA and  2   1 D B γ0 = An−2 ψ ; β , n+1 A A (see Theorem 5.1 b)). Now we shall present the classification procedure for Lbn+1 . To simplify notation let us agree that in the above case for transition   from L(α) to L(α0 ) and from L(β) to L(β0 ) we write as α0 = ρ A1 , AB ; α   and β0 = ν A1 , AB , DA ; β , respectively, where          ρ A1 , AB ; α = ρ1 A1 , AB ; α , ρ2 A1 , AB ; α , . . . , ρn−1 A1 , AB ; α ,

212  Leibniz Algebras

with ρt (x, y; z) = xt ϕt+2 (y; z) for 1 ≤ t ≤ n − 2, ρn−1 (x, y; z) = xn−2 ϕn+1 (y; z)  and         ν A1 , AB , DA ; β = ν1 A1 , AB , DA ; β , ν2 A1 , AB , DA ; β , . . . , νn−1 A1 , AB , DA ; β , with νt (x, y, v; z) = xt vψt+2 (y; z) for 1 ≤ t ≤ n − 3, νn−2 (x, y, v; z) = xn−2 v(yzn+1 + ψn (y; z)), . νn−1 (x, y, v; z) = xn−2 ψn+1 (y; z) , Here are main properties of the operators ρ and ν used in what follows. Proposition 5.4. 1. ρ(1, 0; ·) is  the identity  operator.   2. ρ A12 , AB22 ; ρ A11 , AB11 ; α = ρ A11A2 , A1 B2 +AA21 AB21 +B1 B2 ; α .     B ; α0 . 3. If α0 = ρ A1 , AB ; α then α = ρ A, − A+B 4. ν (1, is the identity operator.  0,B1; ·)    1 2 D2 2 D1 D1 D2 5. ν A2 , A2 , A2 ; ν A11 , AB11 , DA11 ; β = ν A11A2 , B1 AA21+B , ; β . A2  A1 A2    6. If β0 = ν A1 , AB , DA ; β then β = ν A, − DB , DA ; β0 . From now on we assume that n ≥ 4 since there are complete classifications of complex nilpotent Leibniz algebras of dimension at most four in Albeverio et al. [13]. 5.2.2

Classification procedure

We remind the reader of the following definition of the action of an algebraic group on an algebraic variety. Definition 5.4. An action of an algebraic group G on a variety X is a morphism σ : G × X −→ X with (i) σ(e, x) = x, where e is an identity element of G and x ∈ X, (ii) σ(g, σ(h, x)) = σ(gh, x), for all g, h ∈ G and x ∈ X. We write gx for σ(g, x), and call X a G-variety. Definition 5.5. A function f : X −→ F is said to be invariant if f (gx) = f (x) for any g ∈ G and x ∈ X. We consider the case when G = Gad and X = Lbn+1 . Then the orbits with respect to the action of G = Gad on X = Lbn+1 consist of mutually isomorphic algebras.

Isomorphism Criteria for Filiform Leibniz Algebras  213

5.2.2.1

Classification algorithm and invariants for FLbn+1

Consider the following representation of FLbn+1 = U ∪ F, where U = {L(α) ∈ FLbn+1 : α3 , 0} and F = {L(α) ∈ FLbn+1 : α3 = 0}. Then U can be represented as a disjoint union of the subsets U1 = {L(α) ∈ U : α4 , −2α23 } and F1 = {L(α) ∈ U : α4 = −2α23 }. Theorem 5.4. i) Two algebras L(α) and L(α0 ) from U1 are isomorphic if and only if ! ! 2α03 α04 α4 2α3 0 , ; α = ρi 0 , 02 ; α , ρi α4 + 2α23 2α23 α4 + 2α02 3 2α3 whenever i = 3, 4, . . . , n − 1. ii) For any (a3 , a4 , . . . , an−1 ) ∈ Cn−3 there is an algebra L(α) from U1 such that ! 2α3 α4 ρi , ; α = ai , i = 3, 4, . . . , n − 1. α4 + 2α23 2α23 Proof. i). “If” part. Let two algebras L(α) and L(α0 ) be isomorphic.   Then there exist A, B ∈ C such that A(A + B) , 0 and α0 = ρ A1 , AB ; α . Hence, making use of Properties 5.4 we have   −B 0 α = ρ A, ;α . A+B Consider the algebra L(α0 ), where ! α4 + 2α23 α4 (α4 + 2α23 ) 1 B0 0 , ; α and A0 = , B0 = . α =ρ A0 A0 2α3 4α33 Then again due to Properties 5.4 we get   2α3 α4 0 α = ρ α4 +2α2 , 2α2 ; α 3  3  −B = ρ A10 , AB00 ; ρ A, A+B ; α0   0 A−A0 B = ρ AA0 , BA(A+B) ; α0 .

214  Leibniz Algebras

It is easy to check that 2α0 α0 A B0 A − A0 B = 0 3 02 and = 402 . A0 α4 + 2α3 A(A + B) 2α3 Therefore, ! ! 2α03 α04 α4 2α3 0 , ;α = ρ 0 , 02 ; α ρ α4 + 2α23 2α23 α4 + 2α02 3 2α3 and, hence, ! ! 2α03 α04 α4 2α3 0 , ; α = ρi 0 , 02 ; α , ρi α4 + 2α23 2α23 α4 + 2α02 3 2α3 for all i = 3, 4, . . . , n − 1. This procedure can be shown schematically as follows: α (A, B) &

(A0 ,B0 )

−→

α0

0 α  0B % A0 A−1 , BA00A−A (A+B)

 “Only if” part.  Letthe 0equalities  2α α0 2α3 α4 ρi α4 +2α2 , 2α2 ; α = ρi α0 +2α3 02 , 2α402 ; α0 , i = 3, 4, . . . , n − 1 3 3 4 3 3 hold. Then it is easy to see that ! ! 2α03 α04 2α3 α4 0 ρi , ; α = ρi 0 , 02 ; α for i = 1, 2 α4 + 2α23 2α23 α4 + 2α02 3 2α3 as well and therefore, ! ! 2α03 α04 α4 2α3 0 , ;α = ρ 0 , 02 ; α ρ α4 + 2α23 2α23 α4 + 2α02 3 2α3 that means the algebras L(α) and L(α0 ) are isomorphic. ii). The system of equations ! 2α3 α4 ρi , ; α = ai , 3 ≤ i ≤ n − 1, α4 + 2α23 2α23

(5.7)

where (a3 , a4 , . . . , an−1 ) is given and α = (α3 , α4 , . . . , αn−1 , θ) is un- 3 known, has a solution as far as for any 3 ≤ i ≤ n−1 in ρi α42α , α4 ; α +2α2 2α2 3

3

Isomorphism Criteria for Filiform Leibniz Algebras  215

only variables α3 , α4 , . . . , αi occur and each of these equations is a linear equation with respect to its last variable. Hence, making each of αi the subject of (5.7) for i = 3, . . . , n − 1, one can find the required algebra L(α).  Here are the corresponding invariants for some low-dimensional cases. Case of n = 4 i.e., dim L=5:    2 2α3 α4 3 (θ − α4 ) . ρ3 α42α , ; α = +2α2 2α2 α4 +2α2 3

3

3

Case of n = 5 i.e., dim L=6:   4α3 α5 −5α24 α4 3 , ; α = ρ3 α42α 2 2 +2α3 2α3 (α +2α2 )2     4 3 3  2α3 2α3 2α3 α4 α4 ρ4 α4 +2α2 , 2α2 ; α = α4 +2α2 (θ − α5 ) + ρ3 α4 +2α2 , 2α2 ; α . 3

3

3

3

3

Let us now consider the isomorphism criterion for F1 . This set in its turn can be written as a disjoint union of the subsets V1 = {L(α) ∈ F1 : α5 , 5α33 } and G1 = {L(α) ∈ F1 : α5 = 5α33 }. Further, V1 can be represented as a disjoint union of the subsets U2 = {L(α) ∈ V1 : α6 + 6α3 α5 − 16α43 , 0} and

G2 = {L(α) ∈ V1 : α6 + 6α3 α5 − 16α43 = 0}.

Then the isomorphism criterion for U2 can be spelled out as follows. Theorem 5.5. i) Two algebras L(α) and L(α0 ) from U2 are isomorphic if and only if   5α3 −α5 α6 +7α3 α5 −21α43 ρi α +6α33α −16α , ; α 4 α3 (5α33 −α5 ) 6 5 3   5α03 −α05 α06 +7α03 α05 −21α04 0 3 ; α = ρi α0 +6α03α0 −16α , 04 α0 (5α03 −α0 ) 6

3 5

3

3

3

5

for i = 4, . . . , n − 1.

216  Leibniz Algebras

ii) For any (a4 , . . . , an−1 ) ∈ Cn−4 there is an algebra L(α) from U2 such that   3 4   5α − α α + 7α α − 21α 5 6 3 5 3 3  = ai   ; α ρi  ,  α6 + 6α3 α5 − 16α43 α3 5α3 − α5 3

for all i = 4, 5, . . . , n − 1. The proof can be carried out with minor changes in the proof of Theorem 5.4. Here are the corresponding invariants for n = 7 case:    5α33 −α5 α6 +7α3 α5 −21α43 (5α33 −α5 )3    , ; α = ρ 4 α +6α α −16α4 2  α3 (5α33 −α5 )  3 5 6 α3 (α6 +6α3 α5 −16α43 ) 3       4      α7 +7α3 α6 −14α53 5α33 −α5 5α33 −α5 α6 +7α3 α5 −21α43    , ρ ; α =  5 4 4 3  α3 α +6α α −16α3 α6 +6α3 α5 −16α3 α 5α −α   3 3 ( 3 5 ) 3 2  6 3 35    4 3 −α  42α (α −5α 28(α5 −5α33 ) 5α )+7(α +14α 5α −α 3 5 6 5 5  3 3 3 3  − +  4 4  α3 α3 α6 +6α3 α5 −16α3 α6 +6α3 α5 −16α3         5α33 −α5 5α33 −α5 α6 +7α3 α5 −21α43  5  , ; α = ( ρ  6 α +6α α −16α4 3 −α ) 4 ) (θ − α7 )  α +6α α (5α 3 5 3 α5 −16α3 6 6 3  5 3 3   4  3    5α33 −α5 α7 +7α3 α6 −14α53 5α33 −α5 42α3 (α5 −5α33 )+7(α6 +14α43    + − 4 4  α α3 α6 +6α3 α5 −16α3 α6 +6α3 α5 −16α3  3      2 3  5α −α5 28(α5 −5α33 )    . + α +6α33α −16α 4 α3 6

5

3

As for the subsets F, G1 and G2 the isomorphism criteria for them can be treated likewise. 5.2.2.2

Classification algorithm and invariants for SLbn+1

In this section we treat SLbn+1 . The classification algorithm in this case works effectively as well. However, in this case instead of the representation ρ we have to use the representation ν (see Proposition 5.4). Introduce the following notations. Let x = (x3 , x4 , x5 , x6 ) be a vector variable. χ1 (x) = 4x32 x6 − 12x3 x4 x6 + x43 ; χ2 (x) = 4x3 x5 − 5x42 ; χ3 (x) = x32 x6 − 3x3 x4 x5 + 2x43 ; χ4 (x) = 4x5 x32 − 5x42 x3 + 2x4 γ; χ5 (x) = 2x32 x6 − 6x3 x4 x5 + x4 γ + 4x43

Isomorphism Criteria for Filiform Leibniz Algebras  217

Consider the following presentation of SLbn+1 : SLbn+1 = U ∪ F, where U = {L(β) : β3 χ1 (β)χ2 (β) , 0}, and F = {L(β) : β3 χ1 (β)χ2 (β) = 0}. Theorem 5.6. i) Two algebras L(β) and L(β0 ) from U are isomorphic if and only if ! ! β03 χ2 (β0 ) β04 4χ3 (β0 ) 0 β3 χ2 (β) β4 4χ3 (β) , , νi , ; β = νi , 02 ;β , 0 4χ3 (β) 2β23 β23 χ2 (β) 4χ3 (β0 ) 2β02 3 β3 χ2 (β ) whenever i = 3, 4, . . . , n − 1. ii) For any (λ3 , λ4 , . . . , λn−1 ) ∈ Cn−3 there is an algebra L(β) from U such that ! β3 χ2 (β) β4 4χ3 (β) νi , , ; β = λi for all i = 3, 4, . . . , n − 1. 4χ3 (β) 2β23 β23 χ2 (β) Proof. i). Let L(β) and L(β0 ) be isomorphic. Then there exist A, B, D ∈ C such that AD , 0 and β0 = ν( A1 , AB , DA ; β). Consider the algebra L(β0 ), where ! 1 B0 D0 0 β =ν , , ;β A0 A0 A0 with A0 =

4χ3 (β) , β3 χ2 (β)

B0 =

2β4 χ3 (β) 4χ3 (β) and D = . 0 β23 χ2 (β) β33 χ2 (β)

  A 0 Since β = ν A, −B , ; β and D D   β0 = ν A10 , AB00 , DA00 ; β    A 0 = ν A10 , AB00 , DA00 ; ν A, −B , ; β D D  0 B D0 A , ; β0 . = ν AA0 , B0 A−A A0 D A0 D

218  Leibniz Algebras

One can easily check that

A A0

=

β03 χ2 (β0 ) B0 A−A0 B , A0 D 4χ3 (β0 )

=

(β0 )

4χ3 0 . β02 3 χ2 (β )

β04 , 2β02 3

and

D0 A A0 D

=

Therefore ! ! β03 χ2 (β0 ) β04 4χ3 (β0 ) 0 β3 χ2 (β) β4 4χ3 (β) , , ν ;β = ν ;β , , , 02 0 4χ3 (β) 2β23 β23 χ2 (β) 4χ3 (β0 ) 2β02 3 β3 χ2 (β )

and, in particular, ! ! β03 χ2 (β0 ) β04 4χ3 (β0 ) 0 β3 χ2 (β) β4 4χ3 (β) νi , , , , 02 ; β = νi ;β , 0 4χ3 (β) 2β23 β23 χ2 (β) 4χ3 (β0 ) 2β02 3 β3 χ2 (β ) for all i = 3, 4, . . . , n − 1. This procedure can be shown schematically as follows B

D

( A1 , A0 , A 0 )

−→ β0 β 1 B D 0 B D0 A (A, A, A ) & % ( AA0 , B0 A−A , A0 D ) A0 D β0 Conversely, let the equalities ! ! β03 χ2 (β0 ) β04 4χ3 (β0 ) 0 β3 χ2 (β) β4 4χ3 (β) , , ; β = νi , , 02 ;β , νi 0 4χ3 (β) 2β23 β23 χ2 (β) 4χ3 (β0 ) 2β02 3 β3 χ2 (β ) 0

0

0

hold for i = 3, 4, . . . , n − 1. Then it is easy to see that ! ! β03 χ2 (β0 ) β04 4χ3 (β0 ) 0 β3 χ2 (β) β4 4χ3 (β) νi , , , ; β = νi , 02 ;β , 0 4χ3 (β) 2β23 β23 χ2 (β) 4χ3 (β0 ) 2β02 3 β3 χ2 (β ) for i = 1, 2 as well and, therefore ! ! β03 χ2 (β0 ) β04 4χ3 (β0 ) 0 β3 χ2 (β) β4 4χ3 (β) ν , , ;β = ν , , 02 ;β , 0 4χ3 (β) 2β23 β23 χ2 (β) 4χ3 (β0 ) 2β02 3 β3 χ2 (β ) that means that the algebras L(β) and L(β0 ) are isomorphic. The proof of Part ii), is similar to that of Theorem 5.4. 

Isomorphism Criteria for Filiform Leibniz Algebras  219

Here are examples of invariants for some low-dimensional cases. Let dim L = 6 : ! β23 β4 1 β3 χ4 (β) ν3 , 2, 3;β = . γ 2β3 β3 4γ2 Let dim L = 7 :     β4 2χ5 (β) β3 χ2 (β) β4 2χ5 (β) 3 χ2 (β) ν3 β2χ , , ; β = ν , , ; β = 4 2χ5 (β) 2β2 β2 χ (β) 2β2 β2 χ2 (β) 5 (β) 2 3

3

ν5

3

3

! β23 β4 1 γχ2 (β) , 2 , 3 ; β = 22 . γ 2β3 β3 4χ5 (β)

Let dim L = 8 :     β3 χ2 (β) β4 2χ5 (β) β4 2χ5 (β) 3 χ2 (β) , ; β = ν , ; β = ν3 β2χ , , 4 2χ5 (β) 2β2 β2 χ (β) 2β2 β2 χ2 (β) 5 (β) 2 3

3

χ32 (β) 16χ25 (β)

3

3

χ32 (β) . 16χ25 (β)

! χ4 (β)χ6 (β) β3 χ2 (β) β4 2χ5 (β) ν5 , 2, 2 ;β = 2 . 2χ5 (β) 2β3 β3 χ2 (β) (χ5 (β) − β4 γ) ! β3 γχ32 (β) β3 χ2 (β) β4 2χ5 (β) ν6 , 2, 2 ;β = . 2χ5 (β) 2β3 β3 χ2 (β) (χ5 (β) − β4 γ))3 In regard to the set F, it can be split into its subsets and the algorithm can be applied with ν, instead of ρ, by using the properties of ν (see Proposition 5.4). Applications of the procedure are demonstrated in Chapter 6. 5.2.2.3

Classification algorithm and invariants for TLbn+1

In this section we treat filiform Leibniz algebras whose natural gradation is an algebra from NGF3 . This class has been denoted by TLbn+1 . Here we investigate the behavior of structure constants under the base change in TLbn+1 . Recall that (n+1)−dimensional filiform Lie algebras are also in TLbn+1 . Further, j+1) L(¯a) = L(b0,0 , b0,1 , b1,1 , a1i, j , . . . , an−(i+ , bi, j ) i, j

or sometimes just j+1) a¯ = (b0,0 , b0,1 , b1,1 , a1i, j , . . . , an−(i+ , bi, j ) i, j

220  Leibniz Algebras

stand for an algebra from TLbn+1 , with the structure constants j+1) , bi, j . b0,0 , b0,1 , b1,1 , a1i, j , . . . , an−(i+ i, j j+1) 0 0 0 0 0 0 , bi, j ) be the image of Let L(¯a ) = L(b0,0 , b0,1 , b1,1 , a1i, j0 , . . . , an−(i+ i, j L(¯a) under an adapted basis change. The proof of the following proposition with specified elements of Gad , is straightforward. Proposition 5.5. Let f ∈ Gad . Then f can be represented as follows: f (e0 ) = e00 = A0 e0 + A1 e1 + · · · + An en , f (e1 ) = e01 =

B1 e1 + · · · + Bn en ,

f (ei ) = e0i = [ f (ei−1 ), f (e0 )],

2 ≤ i ≤ n,

where A0 , Ai , B j , (i, j = 1, . . . , n) are complex numbers and A0 B1 (A0 + A1 b) , 0. The following elements of Gad are said to be elementary with respect to the structure of L ∈ TLbn+1 :  f (e0 ) = e0 ,     f (e ) = e + b e , b ∈ C, 2 ≤ k ≤ n, 1 1 k σ(b, k) =     f (ei+1 ) = [ f (ei ), f (e0 )], 1 ≤ i ≤ n − 1,  f (e0 ) = e0 + a ek , a ∈ C, 1 ≤ k ≤ n,     f (e ) = e , 1 1 τ(a, k) =     f (ei+1 ) = [ f (ei ), f (e0 )], 1 ≤ i ≤ n − 1,   f (e0 ) = a e0 ,    f (e1 ) = b e1 , a, b ∈ C∗ , υ(a, b) =     f (ei+1 ) = [ f (ei ), f (e0 )], 1 ≤ i ≤ n − 1. Proposition 5.6. Let f be an adapted transformation of L. Then it can be represented as the following composition: f = τ(An , n) ◦ τ(An−1 , n − 1) ◦ · · · ◦ τ(A2 , 2) ◦ σ(Bn , n) ◦ σ(Bn−1 , n − 1) ◦ · · · ◦ σ(B2 , 2) ◦ τ(A1 , 1) ◦ υ(A0 , B1 ). Proof. The proof is straightforward.  Proposition 5.7. The transformations g = τ(An , n) ◦ τ(An−1 , n − 1) ◦ · · · ◦ τ(An−4 , n − 4) ◦σ(Bn , n) ◦ σ(Bn−1 , n − 1) ◦ · · · ◦ σ(Bn−3 , n − 3)

Isomorphism Criteria for Filiform Leibniz Algebras  221

if n is even, and g = τ(An , n) ◦ τ(An−1 , n − 1) ◦ · · · ◦ τ(An−3 , n − 3) ◦σ(Bn , n) ◦ σ(Bn−1 , n − 1) ◦ · · · ◦ σ(Bn−2 , n − 2) for odd n, do not change the structure constants of these cases. Lemma 5.2. For b00,0 , b00,1 and b01,1 the following hold: b00,0 = b00,1 = b01,1 =

A20 b0,0 +A0 A1 b0,1 +A21 b1,1 An−2 0 B1 (A0 +A1 b) A0 b0,1 +2A1 b1,1 , An−2 0 (A0 +A1 b) B1 b1,1 . An−2 0 (A0 +A1 b)

Proof. Consider the product [ f (e0 ), f (e0 )] = b00,0 f (en ). Equating the coefficients of en in it we get A20 b0,0 + A0 A1 b0,1 + A21 b1,1 = b00,0 An−2 0 B1 (A0 + A1 b). Then b00,0 =

A20 b0,0 + A0 A1 b0,1 + A21 b1,1 An−2 0 B1 (A0 + A1 b)

.

The product [ f (e1 ), f (e1 )] = b01,1 f (en ) yields b01,1 =

B1 b1,1 . n−2 A0 (A0 + A1 b)

Consider the equality b00,1 f (en ) = [ f (e1 ), f (e0 )] + [ f (e0 ), f (e1 )]. Then b00,1 An−2 0 B1 (A0 + A1 b) = A0 B1 b0,1 + 2A1 B1 b1,1 and this implies that b00,1 =

A0 b0,1 + 2A1 b1,1 . An−2 (A + A b) 0 1 0 

The complete implementation of the procedure for some lowdimensional cases is given in the next chapters.

CHAPTER

6

CLASSIFICATION OF FILIFORM LEIBNIZ ALGEBRAS IN LOW DIMENSIONS

6.1

ISOMORPHISM CRITERIA FOR FIRST CLASS

We are now ready to obtain a complete classification of filiform Leibniz algebras in low dimensions. In this chapter we start with the class FLbn . Considering the natural gradation on algebras from FLbn we obtain the algebra NGF1 (see [148]). In accordance with results of [88] the class FLbn is represented by the following table of multiplication in the so-called adapted basis {e0 , e1 , . . . , en } : [e0 , e0 ] = e2 ,      [e , e ] = ei+1 , 1 ≤ i ≤ n − 1,    i 0 FLbn+1 :=  [e 0 , e1 ] = α3 e3 + α4 e4 + · · · + αn−1 en−1 + θen ,     [e j , e1 ] = α3 e j+2 + α4 e j+3 + · · · + αn+1− j en ,    where 1 ≤ j ≤ n − 2 and α3 , α4 , . . . , αn , θ ∈ C. Recall the isomorphism criteria for FLbn+1 (see Theorem 5.5).

223

224  Leibniz Algebras

Criterion 1. Introduce the following series of functions: ϕt (y; z) = ϕt (y; z3 , z4 , . . . , zn , zn+1 )   t t−1 P  k−1  k−1 2 P zt+3−i1 zi1 +1−k = zt − yz + y t+2−k k−2 k−3 i1 =k+2 k=3     i2 t P P k−1 1 yk−2 zt+3−i2 zi2 +3−i1 zi1 −k + . . . + k − + k − 4 y3 1 i2 =k+3 i1 =k+3 iP i2 t k−3 P P · ... zt+3−ik−3 zik−3 +3−ik−4 . . . zi2 +3−i1 zi1 +5−2k + yk−1 ·

ik−3 =2k−2 t P

ik−4 =2k−2 iP k−2

ik−2 =2k−1

ik−3 =2k−1

...

i1 =2k−2 i2 P

i1 =2k−1

zt+3−ik−2 zik−2 +3−ik−3 . . .
zi2 +3−i1 zi1 +4−2k ϕk (y; z) , for 3 ≤ t ≤ n,

and   ϕn+1 (y; z) = (zn+1 − zn ) + (1 + y)ϕn (y; z) . Theorem 6.1. Two algebras L(α) and L(α0 ) from FLbn+1 , where α = (α3 , α4 , . . . , αn , θ), and α0 = (α03 , α04 , . . . , α0n , θ0 ), are isomorphic if and only if there exist complex numbers A and B such that A(A + B) , 0 and the following conditions hold: B  1  B  B  1 0 0 αt = t−2 1 + ϕt ; α , 3 ≤ t ≤ n, θ = n−2 ϕn+1 ; α . A A A A A We now demonstrate the systems of equalities for the cases of FLb5 and FLb6 . Case of n = 4 i.e., dim L = 5 :       α03 = A1 1 + AB α3 = A1 1 + AB ϕ3 AB ; α ,        α04 = A12 1 + AB α4 − 2 AB α23 = A12 1 + AB ϕ4 AB ; α ,       θ0 = A12 θ − α4 + 1 + AB α4 − 2 AB α23 = A12 ϕ5 AB ; α . Case of n = 5 i.e., dim L = 6 :     α03 = A1 (1 + AB )α3 = A1 1 + AB ϕ3 AB ; α ,     α04 = A12 (1 + AB )(α4 − 2 AB α23 ) = A12 1 + AB ϕ4 AB ; α ,

Classification of Filiform Leibniz Algebras in Low Dimensions  225

α05 = θ0 =

1 (1 A3 1 [θ A3

+ AB )(α5 − 5 AB (α4 − AB α23 )α3 =

1 A3



1+

B A



− α5 + (1 + AB )(α5 − 5 AB (α4 − AB α23 )α3 )] =

ϕ5





B ;α A

1 ϕ A2 6



,

B ;α A



.

Based on this isomorphism criterion we create another two equivalent criteria (Theorem 3.9 and Theorem 6.3) for FLbn+1 . Criterion 2. We introduce the notations: ∆3 = α3 , 0 ∆04 = α04 + 2α32 , ∆6 = α6 + 14α43 , 0 ∆07 = α07 − 42α35 , Θ8 = θ − α8 ,

∆03 = α03 , ∆5 = α5 − 5α33 , 0 ∆06 = α06 + 14α34 , ∆8 = α8 + 132α63 , Θ08 = θ0 − α08 .

∆4 ∆05 ∆7 ∆08

= α4 + 2α23 , 0 = α05 − 5α33 , = α7 − 42α53 , 0 = α08 + 132α36 ,

(6.1)

In general, 0

∆ s = α s + (−1) sC s−2 α3s−2 , and ∆0s = α0s + (−1) s−2C s−2 α3s−2 , Θ s = θ − α s , and Θ0s = θ0 − α0s , (2k)! where s = 4, 5, . . . . and Ck are Catalan numbers, i.e., Ck = (k+1)!k! , k = 0, 1, 2, . . . . Remark. Under these notations the structure constants α3 , α4 , . . . , αn , θ are transformed into ∆3 , ∆4 , ∆5 , . . . , ∆n , Θn and this gives rise to a transition from the adapted basis {e0 , e1 , . . . , en } into another adapted basis. We shall keep the notation L(∆3 , ∆4 , ∆5 , . . . , ∆n , Θn ) for algebras from FLbn+1 . Let us define the following functions

ft (y; z) = ft (y; z3 , z4 , . . . , zn , zn+1 ) t−1     X k−1 2 k−1 t+1−k t−k = zt − Ct−k z3 ) + k − 3 y k − 2 y(zt+2−k + (−1) k=3 t X

·

1 (zt+3−i1 + (−1)t−i1 Ct+1−i1 zt+1−i ) (zi1 +1−k + (−1)i1 −kCi1 −1−k zi31 −1−k ) 3

i1 =k+3

+



t  X k−1 3 k−4 y

i2 =k+3

i2 X i1 =k+3

2 )(zi2 +3−i1 (zt+3−i2 + (−1)t−i2 Ct+1−i2 zt+1−i 3

226  Leibniz Algebras

  +(−1)i2 −i1 Ci2 +1−i1 zi32 +1−i1 )(zi1 −k + (−1)i1 −k−1Ci1 −k−2 zi31 −k−2 ) . . . + k −1 1 ik−3 i2 t X X X k−2 k−3 ... (zt+3−ik−3 + (−1)t−ik−3 Ct+1−ik−3 zt+1−i ) ·y 3 ik−3 =2k−2

ik−4 =2k−2

· (zik−3 +3−ik−4 + (−1)

ik−3 −ik−4

i1 =2k−2

Cik−3 +1−ik−4 zi3k−3 +1−ik−4 ) . . . (zi2 +3−i1

+ (−1)i2 −i1 Ci2 +1−i1 zi32 +1−i1 )(zi1 +5−2k + (−1)i2 −i1 Ci2 +3−i1 zi32 +3−i1 ) ik−2 i2 t X X X ... (zt+3−ik−2 + (−1)t−ik−2 Ct+1−ik−2 z3t+1−ik−2 ) + yk−1 ik−2 =2k−1 ik−3 =2k−1

· (zik−2 +3−ik−3 + (−1) · (zi2 +3−i1 + (−1)

i2 −i1

ik−2 −ik−3

i1 =2k−1

Cik−2 +1−ik−3 zi3k−2 +1−ik−3 ) . . .

Ci2 +1−i1 zi32 +1−i1 )(zi1 +4−2k

+ (−1)

i1 +1−2k

Ci1 +2−2k zi31 +2−2k )



t−2 t−2 · fk (y; z) + (−1)t (1 + y)t−3 zt−2 3 − (−1) C t−2 z3 ,

for 3 ≤ t ≤ n, and

fn+1 (y; z) = zn+1 .

Then one has the following theorem. Theorem 6.2. Two algebras L(∆) and L(∆0 ) from FLbn+1 , where ∆ = (∆3 , ∆4 , . . . , ∆n , Θn ), and ∆0 = (∆03 , ∆04 , . . . , ∆0n , Θ0n ), are isomorphic if and only if there exist complex numbers A and B such that A(A + B) , 0 and the following conditions hold: 1  B  B  1 0 ∆t = t−2 1 + ft ; ∆ , 3 ≤ t ≤ n, Θ0n = n−2 Θn . (6.2) A A A A Criterion 3. In order to state the Criterion 3 let us consider the following functions ft (z) = ft (z3 , z4 , . . . , zn , zn+1 ) = ft (−1; z) = ft (−1; z3 , z4 , . . . , zn , zn+1 ) t−1     X k−1 k−1 t+1−k t−k = zt + (z + (−1) C z ) + t+2−k t−k 3 k−3 k−2 k=3

·

t X i1 =k+2

1 ) (zi1 +1−k + (−1)i1 −kCi1 −1−k zi31 −1−k ) (zt+3−i1 + (−1)t−i1 Ct+1−i1 zt+1−i 3

Classification of Filiform Leibniz Algebras in Low Dimensions  227



k−1 + k−4

i2 X

t  X i2 =k+3

i1 =k+3

· Ci2 +1−i1 zi32 +1−i1 )(zi1 −k ·

2 (zt+3−i2 + (−1)t−i2 Ct+1−i2 zt+1−i )(zi2 +3−i1 + (−1)i2 −i1 3

t X

ik−3 X

ik−3 =2k−2

ik−4 =2k−2

...

+ (−1) i2 X

i1 −k−1

Ci1 −k−2 zi31 −k−2 )

+ . . . + (−1)

k−2



k−1 1



k−3 ) (zt+3−ik−3 + (−1)t−ik−3 Ct+1−ik−3 zt+1−i 3

i1 =2k−2

· (zik−3 +3−ik−4 + (−1)ik−3 −ik−4 Cik−3 +1−ik−4 z3ik−3 +1−ik−4 ) . . . (zi2 +3−i1 + (−1)i2 −i1 Ci2 +1−i1 zi32 +1−i1 )(zi1 +5−2k + (−1)i2 −i1 Ci2 +3−i1 zi32 +3−i1 ) ik−2 i2 t X X X k−1 + (−1) ... (zt+3−ik−2 + (−1)t−ik−2 Ct+1−ik−2 z3t+1−ik−2 ) ik−2 =2k−1

ik−3 =2k−1

· (zik−2 +3−ik−3 + (−1)

ik−2 −ik−3

i1 =2k−1

Cik−2 +1−ik−3 zi3k−2 +1−ik−3 ) . . .

· (zi2 +3−i1 + (−1)i2 −i1 Ci2 +1−i1 zi32 +1−i1 )(zi1 +4−2k + (−1)i1 +1−2kCi1 +2−2k zi31 +2−2k )



· fk (z) + (−1)t−1Ct−2 zt−2 3 , for 3 ≤ t ≤ n, and

fn+1 (z) = zn+1 .

Criterion 3 is now stated as follows. Theorem 6.3. Two algebras L(∆) and L(∆0 ) from FLbn+1 , where ∆ = (∆3 , ∆4 , . . . , ∆n , Θn ), and ∆0 = (∆03 , ∆04 , . . . , ∆0n , Θ0n ), are isomorphic if and only if there exist complex numbers A and B such that A(A + B) , 0 and the following conditions hold: 1  B 1 ft (∆0 ) = t−2 1 + ft (∆), 3 ≤ t ≤ n and Θ0n = 2 Θn . A A A The equivalence of the criteria 1 and 2 can be shown by substitutions. The equivalence of the criteria 2 and 3 is based on the following properties of the Catalan numbers, which are new to the authors’ best knowledge. These properties are of interest in their own right. • A1 (t) =

t−1 P k=3



   t−3 k−1 C C = Ct−2 ; t−k k−2 1 k−2

228  Leibniz Algebras

• A2 (t, j1 ) =

t−1 P



k=3

 k−1 P  l−1  k−1 k − 2 Ct−k l=3 l − 2 Ck−lCl−2 t t−1 P  k−1  P Ct+1−iCi−1−kCk−2 − k−3 i=k+2

k=3

= • A3 (t, j1 , j2 ) =

t−1 P



k−1 k−2



k−1 k−3



k−1 k−4

k=3

−

t−1 P k=3

+

t−1 P k=3

 t−3 Ct−2 ; 2



 k−3 C C t−k k−2 2   P t k−3 Ct+1−i1 Ci1 −1−kCk−2 1 i1 =k+2  P i2 t P Ct+1−i2 Ci2 +1−i1 Ci1 −k−2Ck−2 i2 =k+3 i1 =k+3

= t−1 P





 t−3 Ct−2 ; 3



  k−1 k−3 C C t−k k−2 3 k−2 k=3 t−1 t P  k − 1  k − 3  P Ct+1−i1 Ci1 −1−kCk−2 − 2 k−3 k=3 i1 =k+2 i2 t−1 t P  k − 1  k − 3  P P + Ct+1−i2 Ci2 +1−i1 Ci1 −k−2Ck−2 1 k−4

• A4 (t, j1 , j2 , j3 ) =

i2 =k+3 i1 =k+3

k=3

−

t−1 P k=3

 P i3 i2 t P P k−1 k − 5 i3 =k+4 i2 =k+4 i1 =k+4 Ct+1−i3 Ci3 +1−i2 Ci2 +1−i1 Ci1 −k−2Ck−2   t−3 = Ct−2 . 4

In general, • Ak (t, j1 , j2 , . . . , jk−1 ) =

t−1 P j1 =3



 j1 − 1 j1 − 2 Ct− j1 Ak−1 (t, j1 , j2 , . . . , jk−2 )

t t−1 P  j1 − 1  P − j1 − 3 i1 = j1 +2 Ct+1−i1 Ci1 −1− j1 Ak−2 (t, j1 , j2 , . . . , jk−3 ) j1 =3 i2 t−1 t P  j1 − 1  P P + j1 − 4 i2 = j1 +3 i1 = j1 +3 Ct+1−i2 Ci2 +1−i1 Ci1 − j1 −2 j1 =3

Classification of Filiform Leibniz Algebras in Low Dimensions  229

·Ak−3 (t, j1 , j2 , . . . , jk−4 ) + · · · + (−1)k ...

i2 P i1 = j1 +(k−1)

+(−1)k+1 ...

i2 P i1 = j1 +k

t−1 P j1 =3



j1 − 1 j1 − k

iP k−2

t P



ik−2 = j1 +k−1 ik−3 = j1 +k−1

Ct+1−ik−2 Cik−1 +1−ik−3 . . . Ci2 +1−i1 Ci1 − j1 −2 A1 ( j1 )

t−1 P j1 =3



j1 − 1 j1 − (k + 1)

t P



iP k−2

ik−2 = j1 +k ik−3 = j1 +k

Ct+1−ik−2 Cik−1 +1−ik−3 . . . Ci2 +1−i1 Ci1 − j1 −2C j1 −2 =



 t−3 Ct−2 . k

The details in dimensions ≤ 9 can be found in [174]. Remark. Further we equally use all the three criteria in our research. From now on we assume that n ≥ 4 is a positive integer, since there are complete classifications of complex nilpotent Leibniz algebras of dimension at most four [13], [27], [50], [57], [160] (see previous Chapters as well).   Remark. The same notation ∆0 = ρ A1 , AB ; ∆ will be used for the transition from an (n + 1)-dimensional filiform Leibniz algebra L(∆) to (n + 1)-dimensional filiform Leibniz algebra L(∆0 ). Introduce the notation: Φ(z) = Φ(z3 , z4 , z5 ) = z5 + 5z3 z4 . 6.2 6.2.1

CLASSIFICATION OF FIRST CLASS IN LOW DIMENSIONS Classification in dimension five

According to (5.2) the isomorphism criterion for FLb5 stands as follows. Theorem 6.4. Two algebras L(∆) and L(∆0 ) from FLb5 , where ∆ = (∆3 , ∆4 , Θ4 ) and ∆0 = (∆03 , ∆04 , Θ04 ), are isomorphic if and only if there exist complex numbers A and B such that A(A + B) , 0 and the following conditions hold: 1 B 1  B 1 ∆03 = 1+ ∆3 , ∆04 = 2 1 + ∆4 , Θ04 = 2 Θ4 . A A A A A

230  Leibniz Algebras

In order to describe the orbits of the adapted base change on FLb5 we split FLb5 into the following subsets: U1 U2 U3 U4 U5 U6 U7

= = = = = = =

{L(∆) ∈ FLb5 {L(∆) ∈ FLb5 {L(∆) ∈ FLb5 {L(∆) ∈ FLb5 {L(∆) ∈ FLb5 {L(∆) ∈ FLb5 {L(∆) ∈ FLb5

: ∆3 : ∆3 : ∆3 : ∆3 : ∆3 : ∆3 : ∆3

, 0, ∆4 , 0, ∆4 , 0, ∆4 = 0, ∆4 = 0, ∆4 = 0, ∆4 = 0, ∆4

, 0}, = 0, Θ4 = 0, Θ4 , 0, Θ4 , 0, Θ4 = 0, Θ4 = 0, Θ4

, 0}, = 0}, , 0}, = 0}, , 0}, = 0}.

It is obvious that {Ui }, i = 1, . . . , 7, is a partition of FLb5 . The following proposition shows that U1 is a union of infinitely many orbits and these orbits can be parameterized by C. Proposition 6.1. i) Two algebras L(∆) and L(∆0 ) from U1 are isomorphic if and only if ! !2 ∆03 2 0 ∆3 Θ4 = 0 Θ4 . ∆4 ∆4 ii) Orbits in U1 can be parameterized by C and L(1, 1, λ), λ ∈ C, are representatives of the orbits. The following proposition is a description of the subsets Ui , i = 2, . . . , 7. Proposition 6.2. The subsets U2 , U3 , U4 , U5 , U6 and U7 are single orbits with representatives L(1, 0, 1), L(1, 0, 0), L(0, 1, 1), L(0, 1, 0), L(0, 0, 1) and L(0, 0, 0), respectively. We summarize the above observations in the following classification theorem. Theorem 6.5. Let L be a non-Lie complex filiform Leibniz algebra in FLb5 . Then it is isomorphic to one of the following pairwise nonisomorphic Leibniz algebras:

Classification of Filiform Leibniz Algebras in Low Dimensions  231

 1. L(0, 0, 0) := L5s = [e0 , e0 ] = e2 , [ei , e0 ] = ei+1, 1 ≤ i ≤ 3 . 2. L(0, 0, 1) := L5s , [e0 , e1 ] = e4 . 3. L(0, 1, 0) := L5s , [e0 , e1 ] = e4 , [e1 , e1 ] = e4 . 4. L(0, 1, 1) := L5s , [e0 , e1 ] = 2e4 , [e1 , e1 ] = e4 . 5. L(1, 0, 0) := L5s , [e0 , e1 ] = e3 − 2e4 , [e1 , e1 ] = e3 − 2e4 , [e2 , e1 ] = e4 . 6. L(1, 0, 1) := L5s , [e0 , e1 ] = e3 − e4 , [e1 , e1 ] = e3 − 2e4 , [e2 , e1 ] = e4 . 7. L(1, 1, λ) := L5s , [e0 , e1 ] = e3 + (λ − 1)e4 , [e1 , e1 ] = e3 − e4 , [e2 , e1 ] = e4 , λ ∈ C. Conclusion. The class FLb5 consists of one parametric family and six isolated isomorphism classes of filiform Leibniz algebras. 6.2.2

Classification in dimension six

The isomorphism criterion for FLb6 in terms of ∆-notations is written as follows. Theorem 6.6. Two algebras L(∆) and L(∆0 ) from FLb6 , where ∆ = (∆3 , ∆4 , ∆5 , Θ5 ) and ∆0 = (∆03 , ∆04 , ∆05 , Θ05 ), are isomorphic if and only if there exist complex numbers A and B such that A(A + B) , 0 and the following conditions hold:     ∆03 = A1 1 + AB ∆3 , ∆04 = A12 1 + AB ∆4 ,   Φ(∆0 ) = A13 1 + AB Φ(∆), Θ05 = A13 Θ5 . We represent FLb6 as a disjoint union of the following subsets: U1 U2 U3 U6 U5 U6 U7 U8

= = = = = = = =

{L(∆) ∈ FLb6 {L(∆) ∈ FLb6 {L(∆) ∈ FLb6 {L(∆) ∈ FLb6 {L(∆) ∈ FLb6 {L(∆) ∈ FLb6 {L(∆) ∈ FLb6 {L(∆) ∈ FLb6

: ∆3 : ∆3 : ∆3 : ∆3 : ∆3 : ∆3 : ∆3 : ∆3

, 0, ∆4 , 0, ∆4 = 0, ∆4 , 0, ∆4 , 0, ∆4 , 0, ∆4 = 0, ∆4 = 0, ∆4

, 0}, = 0, ∆5 , 0, ∆5 = 0, ∆5 = 0, ∆5 = 0, ∆5 , 0, ∆5 , 0, ∆5

, 0, Θ5 , 0}, , 0, Θ5 = 0, Θ5 = 0, Θ5 = 0, Θ5 = 0, Θ5

, 0}, = 0}, , 0}, = 0}, , 0}, = 0},

232  Leibniz Algebras

U9 = {L(∆) ∈ FLb6 : ∆3 = 0, ∆4 = 0, ∆5 , 0, Θ5 , 0}, U10 = {L(∆) ∈ FLb6 : ∆3 = 0, ∆4 = 0, ∆5 , 0, Θ5 = 0}, U11 = {L(∆) ∈ FLb6 : ∆3 = 0, ∆4 = 0, ∆5 = 0, Θ5 , 0}, U12 = {L(∆) ∈ FLb6 : ∆3 = 0, ∆4 = 0, ∆5 = 0, Θ5 = 0}. The following propositions show that each of these subsets is either a single orbit or a union of infinitely many orbits of the group Gad . Indeed, the subsets U1 , U2 and U3 turn out to be a union of infinitely many orbits and they can be described as follows: Proposition 6.3. i) Two algebras L(∆) and L(∆0 ) from U1 are isomorphic if and only if 0 ∆3 Φ(∆) ∆03 Φ(∆0 ) ∆33 Θ5 ∆03 3 Θ5 = , = . ∆24 ∆02 ∆34 ∆03 4 4 ii) Orbits in U1 can be parameterized as L(1, 1, λ1 , λ2 ), λ1 , λ2 ∈ C. Proposition 6.4. i) Two algebras L(∆) and L(∆0 ) from U2 are isomorphic if and only if ∆35 ∆03 5 = . 3 2 03 02 ∆3 Θ5 ∆3 Θ5 ii) Orbits in U2 can be parameterized as L(1, 0, λ, λ), λ ∈ C∗ . Proposition 6.5. i) Two algebras L(∆) and L(∆0 ) from U3 are isomorphic if and only if 0 ∆34 Θ5 ∆03 4 Θ5 = . ∆35 ∆03 5 ii) Orbits in U3 can be parameterized as L(0, 1, 1, λ), λ ∈ C. Each of the sets U4 − U12 is a single orbit and here is their description. Proposition 6.6. The subsets U4 , U5 , U6 , U7 , U8 , U9 , U10 , U11 and U12 are single orbits with representatives L(1, 0, 1, 0), L(1, 0, 0, 1), L(1, 0, 0, 0), L(0, 1, 0, 1), L(0, 1, 0, 0), L(0, 0, 1, 1), L(0, 0, 1, 0), L(0, 0, 0, 1) and L(0, 0, 0, 0), respectively.

Classification of Filiform Leibniz Algebras in Low Dimensions  233

The result of all the observation above can be written down as follows: Theorem 6.7. Let L be a non-Lie complex filiform Leibniz algebra in FLb6 . Then it is isomorphic to one of the following pairwise nonisomorphic Leibniz algebras: 1. L(0, 0, 0, 0) := L6s = {[e0 , e0 ] = e2 , [ei , e0 ] = ei+1 , 1 ≤ i ≤ 4} . 2. L(0, 0, 0, 1) := L6s , [e0 , e1 ] = e5 . 3. L(0, 0, 1, 0) := L6s , [e0 , e1 ] = e5 , [e1 , e1 ] = e5 . 4. L(0, 0, 1, 1) := L6s , [e0 , e1 ] = 2e5 , [e1 , e1 ] = e5 . 5. L(0, 1, 0, 0) := L6s , [e0 , e1 ] = e4 , [e1 , e1 ] = e4 , [e2 , e1 ] = e5 . 6. L(0, 1, 0, 1) := L6s , [e0 , e1 ] = e4 + e5 , [e1 , e1 ] = e4 , [e2 , e1 ] = e5 . 7. L(1, 0, 0, 1) := L6s , [e0 , e1 ] = e3 − 2e4 + 6e5 , [e1 , e1 ] = e3 − 2e4 + 5e5 , [e2 , e1 ] = e4 − 2e5 , [e3 , e1 ] = e5 . 8. L(1, 0, 1, 0) := L6s , [e0 , e1 ] = e3 − 2e4 + 6e5 , [e1 , e1 ] = e3 − 2e4 + 6e5 , [e2 , e1 ] = e4 − 2e5 , [e3 , e1 ] = e5 . 9. L(0, 1, 1, λ) := L6s , [e0 , e1 ] = e4 + (λ + 1)e5 , [e1 , e1 ] = e4 + e5 , [e2 , e1 ] = e5 , λ ∈ C. 10. L(1, 0, λ, λ) := L6s , [e0 , e1 ] = e3 − 2e4 + (2λ + 5)e5 , [e1 , e1 ] = e3 − 2e4 + (λ + 5)e5 , [e2 , e1 ] = e4 − 2e5 , [e3 , e1 ] = e5 , λ ∈ C. 11. L(1, 1, λ1 , λ2 ) := L6s , [e0 , e1 ] = e3 − e4 + (λ1 + λ2 + 5)e5 , [e1 , e1 ] = e3 − e4 + (λ1 + 5)e5 , [e2 , e1 ] = e4 − e5 , [e3 , e1 ] = e5 , λ1 , λ2 ∈ C. Note 6.1. The algebra L(1, 0, 0, 0) from U6 can be included into the parametric family L(1, 0, λ, λ) at λ = 0. There are 11 isomorphism classes in FLb6 , three of them are parametric families and the others are isolated orbits.

234  Leibniz Algebras

6.3

ISOMORPHISM CRITERIA FOR SECOND CLASS

According to Theorem 4.3 there exists an adapted basis {e0 , e1 , . . . , en } for algebras from the class SLbn+1 such that the multiplication table can be represented as  [e , e ] = e , 0 0 2      [e , e ] = e i 0 i+1 , 2 ≤ i ≤ n − 1      [e0 , e1 ] = β3 e3 + β4 e4 + · · · + βn en , SLbn+1 :=    [e1 , e1 ] = γen ,        [e j , e1 ] = β3 e j+2 + β4 e j+3 + · · · + βn+1− j en , 2≤ j≤n−2 Further an algebra from SLbn+1 will be denoted as L(β), where β = (β3 , β4 , . . . , βn , γ) are structure constants defined above. Recall that the isomorphism criterion for SLbn+1 is written as follows Theorem 6.8. Two algebras L(β) and L(β0 ) from SLbn+1 , where β = (β3 , β4 , . . . , βn , γ), and β0 = (β03 , β04 , . . . , β0n , γ0 ), are isomorphic if and only if there exist complex numbers A, B and D such that AD , 0 and the following conditions hold:   B 1 D ψ ; β , 3 ≤ t ≤ n − 1, β0t = At−2 A t A   1 D B 0 βn = An−2 A A γ + ψn AB ; β , and γ0 = 6.4

1 An−2

 2 D A

ψn+1





B ;β A

.

CLASSIFICATION OF SECOND CLASS IN LOW DIMENSIONS

This section deals with finding the isomorphism classes of SLbn for n = 5, 6. Every filiform Leibniz algebra admits an adapted basis with respect to which the table of multiplication can be represented as in Theorem 6.8. Recall that a filiform Leibniz algebra is two-generated, therefore it is sufficient to give nonsingular transformation for the elements e0 and e1 of the adapted basis. Note that in each specified dimension we split the class SLbn into its subsets. Meanwhile some of the subsets turn out to be a union of infinitely many orbits. In this case we give a proposition on structure of these subsets. The propositions

Classification of Filiform Leibniz Algebras in Low Dimensions  235

consist of two parts; the first part gives orbit functions (invariants), another part guaranties that the algebras with the different invariants are not isomorphic. For the simplification purpose, we use the following notations: Λ1 = 4β3 β5 − 5β24 , and Λ01 = 4β03 β05 − 5β0 24 . 6.4.1

Classification in dimension five

As we have seen above, the table of multiplication of SLb5 with respect to the adapted basis is as follows  0 1D  β3 = A A β3 ,             β04 = A12 DA AB γ + β4 − 2 AB β23 ,      2     γ0 = A12 DA γ.    The class SLb5 can be represented as a disjoint union of its subsets as follows: [ [ [ [ [ SLb5 = U1 U2 U3 U4 U5 F, where U1 = {L(β) ∈ SLb5 : β3 , 0 and γ − 2β23 , 0}, U2 = {L(β) ∈ SLb5 : β3 , 0, γ − 2β23 = 0 and β4 , 0}, U3 = {L(β) ∈ SLb5 : β3 , 0, γ − 2β23 = 0 and β4 = 0}, U4 = {L(β) ∈ SLb5 : β3 = 0, γ , 0}, U5 = {L(β) ∈ SLb5 : β3 = 0, γ = 0 and β4 , 0}, F = {L(β) ∈ SLb5 : β3 = 0, γ = 0 and β4 = 0}. Proposition 6.7. i) Two algebras L(β) and L(β0 ) from U1 are isomorphic if and only if γ γ0 = . β23 β02 3 ii) For any λ ∈ C there is an algebra L(β) from U1 such that

γ β23

= λ.

Proof. If this system of equalities holds true, then the base change β4 1 β4 e00 = e0 − e , e01 = e1 − e3 2 1 β3 γ − 2β3 β3 (γ − 2β23 ) transforms the algebra L(β) to the algebra L(β0 ).

236  Leibniz Algebras

Conversely, according to Theorem 5.6 there is a base change that brings L(β) = L(β3 , β4 , γ) and L(β0 ) = L(β03 , β04 , γ0 ) into L(1, 0, βγ2 ) and 3

L(1, 0, βγ02 ), respectively. But due to the condition the last two algebras 3 are the same. The second part is obvious if we take into account that the equation γ = λ is solvable with respect to β3 and γ in U1 .  β2 0

3

Proposition 6.8. The subsets U2 , U3 , U4 , U5 and F are single orbits with representatives L(1, 1, 2), L(1, 0, 2), L(0, 0, 1), L(0, 1, 0) and L(0, 0, 0), respectively. Proof. We show base changes transforming algebras from U2 to the algebra L(1, 1, 2). The others can be shown similarly. Indeed, it can be easily verified that e00 = e0 and e01 = β13 e1 is the required base change.  Theorem 6.9. Any five-dimensional complex filiform Leibniz algebra from SLb5 is isomorphic to one of the following pairwise nonisomorphic non-Lie filiform complex Leibniz algebras, whose tables of multiplication with respect to the adapted basis {e0 , e1 , e2 , e3 , e4 } are as follows: 1. L(1, 0, λ) :=

[e0 , e0 ] = e3 , [e2 , e0 ] = e4 , [e3 , e0 ] = e5 , [e0 , e1 ] = e3 , [e1 , e1 ] = λe4 , [e2 , e1 ] = e5 , λ ∈ C.

2. L(1, 1, 2) :=

[e0 , e0 ] = e3 , [e2 , e0 ] = e4 , [e3 , e0 ] = e5 , [e0 , e1 ] = e3 + e4 , [e1 , e1 ] = 2e4 , [e2 , e1 ] = e5 .

3. L(0, 0, 1) := [e0 , e0 ] = e3 , [e2 , e0 ] = e4 , [e3 , e0 ] = e5 , [e1 , e1 ] = e4 . 4. L(0, 1, 0) := [e0 , e0 ] = e3 , [e2 , e0 ] = e4 , [e3 , e0 ] = e5 , [e0 , e1 ] = e4 . 5. L(0, 0, 0) := [e0 , e0 ] = e3 , [e2 , e0 ] = e4 , [e3 , e0 ] = e5 .

Classification of Filiform Leibniz Algebras in Low Dimensions  237

6.4.2

Classification in dimension six

This section concerns the six-dimensional case of SLbn . Let {e0 , e1 , e2 , e3 , e4 , e5 } be an adapted basis. Then the multiplication table of the class SLb6 with respect to this basis is as follows:  [e , e ] = e , 0 0 2      [e , e ] = e , i = 2, 3, 4, i 0 i+1    [e , e ] = β e + β e + β e , SLb6 :=  0 1 3 3 4 4 5 5     [e , e ] = γe5 ,    1 1 [e j , e1 ] = β3 e j+2 + β4 e j+3 + · · · + β6− j e5 , j = 2, 3. The representative of the class SLb6 is written as L(β3 , β4 , β5 , γ). Here is the isomorphism criterion for algebras from SLb6 :  0 1D  β3 = A A β3 ,          1 D B 2 0   β = β β − 2 ,  4 4  A 3 A2 A      2    B 1 D B 0   β5 = A3 A A γ + β5 − 5 A β3 β4 + 5 AB β33 ,              γ0 = 13 D 2 γ. A A The set SLb6 can be represented as a disjoint union of its subsets as follows: 9 [ SLb6 = Ui , i= 1

where U1 U2 U3 U4 U5 U6 U7 U8 U9

= = = = = = = = =

{L(β) ∈ SLb6 : β3 , 0, γ , 0}, {L(β) ∈ SLb6 : β3 , 0, γ = 0, Λ1 , 0}, {L(β) ∈ SLb6 : β3 , 0, γ = 0, Λ1 = 0}, {L(β) ∈ SLb6 : β3 = 0, β4 , 0, γ , 0}, {L(β) ∈ SLb6 : β3 = 0, β4 , 0, γ = 0, β5 , 0}, {L(β) ∈ SLb6 : β3 = 0, β4 , 0, γ = 0, β5 = 0}, {L(β) ∈ SLb6 : β3 = 0, β4 = 0, γ , 0}, {L(α) ∈ SLb6 : β3 = 0, β4 = 0, γ = 0, β5 , 0}, {L(α) ∈ SLb6 : β3 = 0, β4 = 0, γ = 0, β5 = 0}.

“If” part of the following proposition is just a substitution and “Only if” part is a particular case of Theorem 5.6.

238  Leibniz Algebras

Proposition 6.9. i) Two algebras L(β) and L(β0 ) from U1 are isomorphic, if and only if 0 2β3 β4 γ + β23 Λ1 2β03 β04 γ0 + β32 Λ01 = . γ2 γ0 2 ii) Orbits in U1 can be parameterized as L(1, 0, λ, 1), λ ∈ C. The proof of the proposition below can carried out by showing the base change leading to the specified representatives. Proposition 6.10. The subsets U2 , U3 , U4 , U5 , U6 , U7 , U8 , and U9 are single orbits under the action of Gad with the representatives L(1, 0, 1, 0), L(1, 0, 0, 0), L(0, 1, 0, 1), L(0, 1, 1, 0), L(0, 1, 0, 0), L(0, 0, 0, 1), L(0, 0, 1, 0), and L(0, 0, 0, 0), respectively. Theorem 6.10. Let L be algebra from SLb6 . Then, it is isomorphic to one of the following pairwise non-isomorphic Leibniz algebras:  1. L(0, 0, 0, 0) := G6s = [e0 , e0 ] = e2 , [ei , e0 ] = ei+1, 2 ≤ i ≤ 4 . 2. L(0, 0, 1, 0) := G6s , [e0 , e1 ] = e5 . 3. L(0, 0, 0, 1) := G6s , [e1 , e1 ] = e5 . 4. L(0, 1, 0, 0) := G6s , [e0 , e1 ] = e4 , [e2 , e1 ] = e5 . 5. L(0, 1, 1, 0) := G6s , [e0 , e1 ] = e4 + e5 , [e2 , e1 ] = e5 . 6. L(0, 1, 0, 1) := G6s , [e0 , e1 ] = e4 , [e1 , e1 ] = e5 , [e2 , e1 ] = e5 . 7. L(1, 0, 0, 0) := G6s , [e0 , e1 ] = e3 , [e2 , e1 ] = e4 , [e3 , e1 ] = e5 . 8. L(1, 0, 1, 0) := G6s , [e0 , e1 ] = e3 + e5 , [e2 , e1 ] = e4 , [e3 , e1 ] = e5 .

Classification of Filiform Leibniz Algebras in Low Dimensions  239

9. L(1, 0, λ, 1) := G6s , [e0 , e1 ] = e3 + λe5 , [e1 , e1 ] = e5 , [e2 , e1 ] = e4 , [e3 , e1 ] = e5 , λ ∈ C. 6.5 SIMPLIFICATIONS AND NOTATIONS IN THIRD CLASS

In this section we treat filiform Leibniz algebras whose natural gradation is an algebra from NGF3 . This class has been denoted by TLbn+1 in dimension n + 1. We recall that (n + 1)−dimensional filiform Lie algebras are in TLbn+1 . The study of TLbn+1 has been initiated in Omirov et al. [140]. The multiplication table of TLbn+1 on an adapted basis is as follows.  [ei , e0 ] = ei+1 , 1 ≤ i ≤ n − 1,      [e , e ] = −e , 2 ≤ i ≤ n − 1,  0 i i+1     [e0 , e0 ] = b0,0 en ,      [e  0 , e1 ] = −e2 + b0,1 en ,     [e   1 , e1 ] = b1,1 en , TLbn+1 :=  j+1)   [ei , e j ] = a1i, j ei+ j+1 + · · · + an−(i+ en−1 + bi, j en ,  i, j    1 ≤ i < j ≤ n − 1,      [ei , e j ] = −[e j , ei ],      1 ≤ i < j ≤ n − 1,      [e , e ] = −[e , e ] = (−1)i b e , i

n−i

n−i

i

i,n−i n

where aki, j , bi, j ∈ C and bi,n−i = b whenever 1 ≤ i ≤ n − 1, and b = 0 for even n. An element of TLb( n + 1) is denoted by j+1) L(α) = L(b00 , b01 , b11 , a1i j , . . . , an−(i+ , bi j ). ij

Recall that, in general, the structure constants b00 , b01 , b11 , a1i j , . . . , j+1) an−(i+ , bi j are not free and in each fixed dimensional case we reij arrange them taking into account the relations between them from Lemma 4.6. Mainly, we keep lexicographical order with respect to the indices. Moreover, we express the akij via b st , s, t = 0, 1, . . . . The complete implementation of the classification procedure for TLbn+1 in low dimensional cases will be given in the next sections. For the purpose of simplification, we introduce the following notations: Π1 (Z) = 4z00 z11 − z201 ; Π3 (Z) = z01 − z00 z23 ; Π5 (Z) = z01 z212 − z11 z13 ;

Π2 (Z) = 2z11 − z01 z23 ; Π4 (Z) = 4z00 z412 − 2z01 z212 z13 + z11 z213 ; Π6 (Z) = 2z00 z212 − z01 z13 ;

240  Leibniz Algebras

Introduce functions: χ0 (X; Z) = 1 + x1 z23 ; χ1 (X; Z) = z00 + x1 z01 + x12 z11 ; χ2 (X; Z) = z01 + 2x1 z11 ; χ3 (X; Y; Z) = z13 + (y22 − 2y3 )z23 + x12 z212 z23 ; χ4 (X; Z) = z13 + 2x1 z212 , where X = (x1 , x2 , . . . , xn ), Y = (y2 , . . . , yn ) and Z = (z00 , z01 , z02 , . . . , z11 , z12 , . . . ) Let

and

! A1 A2 An A= , ,..., , A0 A0 A0 ! B2 B3 Bn B= , ,..., , B1 B1 B1 α = (b00 , b01 , b11 , b12 , b13 , . . . ).

6.6 CLASSIFICATION IN DIMENSION FIVE

In this section we deal with the class TLb5 . By virtue of Theorem 4.3 we can represent TLb5 as follows:  [ei , e0 ] = ei+1 , 1 ≤ i ≤ 3,      [e , e ] = −e , 2 ≤ i ≤ 3,  0 i i+1     [e , e ] = b00 e4 ,    0 0 TLb5 :=  [e 0 , e1 ] = −e2 + b01 e4 ,     [e1 , e1 ] = b11 e4 ,      [e  2 , e1 ] = b12 e4 ,   b 1 ,, eb2 ] ,=b −[e , b ∈ C. 00

01

11

12

Further, the elements of TLb5 will be denoted by L(α) L(b00 , b01 , b11 , b12 ).

=

Classification of Filiform Leibniz Algebras in Low Dimensions  241

Theorem 6.11. (I somorphism criterion f or TLb5 ) Two algebras L(α) and L(α0 ) from TLb5 are isomorphic, if and only if there exist complex numbers A0 , A1 and B1 such that A0 B1 , 0 and the following conditions hold: b000 = A01B1 χ1 (A; α); b001 = A12 χ2 (A; α); b011 = b012 =

0

B1 b ; A30 11 B1 b . A20 12

Proof. Necessity. Let L1 and L2 from TLb5 be isomorphic: f : L1  L2 . We choose the corresponding adapted bases {e0 , e1 , e2 , e3 , e4 } and {e00 , e01 , e02 , e03 , e04 } in L1 and L2 , respectively. Then, in these bases the algebras will be presented as L(α) and L(α0 ), where α = (b00 , b01 , b11 , b12 ), and α0 = (b000 , b001 , b011 , b012 ). The change of the generators e0 and e2 is as follows: e00 = f (e0 ) = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = f (e1 ) = B1 e1 + B2 e2 + B3 e3 + B4 e4 .

(6.3)

Then we obtain e02 = f (e2 ) = [ f (e1 ), f (e0 )] = A0 B1 e2 + A0 B2 e3 + (A0 B3 + A1 B1 b11 + (A2 B1 − A1 B2 )b12 ) e4 ,

(6.4)

e03 = f (e3 ) = [ f (e2 ), f (e0 )] = A20 B1 e3 + (A20 B2 − A0 A1 B1 b12 )e4 , e04 = f (e4 ) = [ f (e3 ), f (e0 )] = A30 B1 e4 . By using the adapted bases {e0 , e1 , e2 , e3 , e4 } and {e00 , e01 , e02 , e03 , e04 } one finds the relations between the structure constants b00 , b01 , b11 , b12 and b000 , b001 , b011 , b01,2 . First, we consider the equality [ f (e0 ), f (e0 )] = b000 f (e4 ), and get the equation (7.1) and from the equality [ f (e1 ), f (e0 )]+ [ f (e0 ), f (e1 )] = b001 f (e4 ) we obtain (7.2), and [ f (e1 ), f (e1 )] = b011 f (e4 ) gives (7.3). Finally, the equality (7.4) comes out from [ f (e1 ), f (e2 )] = b012 f (e4 ). Sufficiency. Let the equalities of Theorem 6.11 hold. Then, the base change (6.3) above is adapted and it transforms L(α) into L(α0 ).

242  Leibniz Algebras

Indeed,  4  4 X X    Ai ei  [e00 , e00 ] =  Ai ei , i=0

i=0

= A20 [e0 , e0 ] + A0 A1 [e0 , e1 ] + A0 A1 [e1 , e0 ] + A21 [e1 , e1 ]   = A20 b00 + A0 A1 b01 + A21 b11 e4 = b000 A30 B1 e4 = b000 e04 .  4  4 X X    [e00 , e01 ] =  Ai ei , Bi ei  i=0

i=1

= −(A0 B1 e2 + A0 B2 e3 + (A1 B1 b11 + A2 B1 b12 −A1 B2 b1,2 + A0 B3 )e4 ) + B1 (b01 A0 + 2 A1 b11 ) e4 = −e02 + A30 B1 b001 e4 = −e02 + b001 e04 . In the same way one can prove that [e01 , e01 ] = b011 e04 , [e01 , e02 ] = b012 e04 and the other products to be zero.  6.6.1

Isomorphism classes in TLb5

Now, we list the isomorphism classes of algebras from TLb5 . First we represent TLb5 as a disjoint union of its subsets as follows: TLb5 = 9 S U5i , where i=1

U51 U52 U53 U54 U55 U56 U57 U58 U59

= {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5 = {L(α) ∈ TLb5

: b11 : b11 : b11 : b11 : b11 : b11 : b11 : b11 : b11

, 0, b12 , 0}; , 0, b12 = 0, Π1 (α) , 0}; , 0, b12 = Π1 (α) = 0}; = 0, b01 , 0, b12 , 0}; = 0, b01 , 0, b12 = 0}; = b01 = 0, b00 , 0, b12 , 0}; = b01 = 0, b00 , 0, b12 = 0}; = b01 = b00 = 0, b12 , 0}; = b01 = b00 = b12 = 0}.

For each of the subsets we state an isomorphism criterion.

Classification of Filiform Leibniz Algebras in Low Dimensions  243

Proposition 6.11. 1. Two algebras L(α) and L(α0 ) from U51 are isomorphic, if and only if ! !4 b012 4 b12 0 Π1 (α ) = Π1 (α). b011 b11 2. For any λ from C, there exists L(α) ∈ U51 such that λ.

 b 4 12

b11

Π1 (α) =

Necessity. Let L(α) and L(α0 ) be isomorphic. Then  0 4  4 b by a substitution it is easy to see that b012 Π1 (α0 ) = bb12 Π1 (α). 11 11  0 4  4 b Π1 (α) Sufficiency. Let the equality b012 Π1 (α0 ) = bb12 11

Proof.

1.

11

hold. Consider the base change (6.3) above with A0 = b2

b11 , b12

A1 = − 2bb0112 and B1 = b11 3 . This base change leads L(α) 12   4  b12 to L b11 Π1 (α), 0, 1, 1 . A similar base change with A0 = b0

b011 , b012

A1 = − 2b010 and 12 !  0 4 b Π1 (α0 ), 0, 1, 1 . L b12 0

B1

=

b02 11 b03 12

transforms L(α0 ) to

11

 Since L(α0 ). 2.

b012 b011

4

Π1 (α0 ) =

 b 4 12

b11

Π1 (α), then L(α) is isomorphic to

Obvious. 

Proposition 6.12. The subsets U52 , U53 , U54 , U55 , U56 , U57 , U58 and U59 are single orbits with representatives L(1, 0, 1, 0), L(0, 0, 1, 0), L(0, 1, 0, 1), L(0, 1, 0, 0), L(1, 0, 0, 1), L(1, 0, 0, 0), L(0, 0, 0, 1) and L(0, 0, 0, 0), respectively. Proof. To prove this, we give the appropriate values of A0 , A1 and B1 in the base change (the other Ai , B j , i, j = 2, 3, 4 are arbitrary, except where specified otherwise).

244  Leibniz Algebras

For U52 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + (A1 B1 b11 + A0 B3 ) e4 , e03 = A20 B1 e3 + A20 B2 e4 , e04 = A0 3 B1 e4 , where 3 b4 Π (α) 1 (α) A40 = Π14(α) , A41 = 0164b14 and B41 = Π64b 4 . 11

11

For U53 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + (A1 B1 b11 + A0 B3 ) e4 , e03 = A20 B1 e3 + A0 2 B2 e4 , e04 = A0 3 B1 e4 , A3

where A0 ∈ C∗ , A1 = − A20bb1101 and B1 = b110 . For U54 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + (A0 B3 + (A2 B1 − A1 B2 )b12 )e4 , e03 = A20 B1 e3 + (A20 B2 − A1 A0 B1 b12 )e4 , e04 = A0 3 B1 e4 , A2

where A20 = b01 , A1 = − Ab001b00 and B1 = b120 . For U55 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + A0 B3 e4 , e03 = A20 B1 e3 + A20 B2 e4 , e04 = A30 B1 e4 , where b2 and B1 ∈ C∗ . A20 = b01 , A21 = b00 01 For U56 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + (A0 B3 + (A2 B1 − A1 B2 )b12 )e4 , e03 = A20 B1 e3 + (A20 B2 − A1 A0 B1 b12 )e4 , e04 = A0 3 B1 e4 , b2 where A30 = b00 b12 , A1 ∈ C and B31 = b1,20,0 .

Classification of Filiform Leibniz Algebras in Low Dimensions  245 b2

A30 = b00 b12 , A1 ∈ C and B31 = b0,0 . 12 7 For U5 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + A0 B3 e4 , e03 = A20 B1 e3 + A20 B2 e4 , e04 = A30 B1 e4 , where A0 ∈ C∗ , A1 ∈ C and B1 = bA000 . For U58 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + (A0 B3 + (A2 B1 − A1 B2 )b12 )e4 , e03 = A20 B1 e3 + (A20 B2 − A1 A0 B1 b12 )e4 , e04 = A0 3 B1 e4 , A2

where A0 ∈ C∗ , A1 ∈ C and B1 = b120 . For U59 : e00 = A0 e0 + A1 e1 + A2 e2 + A3 e3 + A4 e4 , e01 = B1 e1 + B2 e2 + B3 e3 + B4 e4 , e02 = A0 B1 e2 + A0 B2 e3 + A0 B3 e4 , e03 = A20 B1 e3 + A20 B2 e4 , e04 = A30 B1 e4 , where A0 , B1 ∈ C∗ and A1 ∈ C. 6.7



CLASSIFICATION IN DIMENSION SIX

This section is devoted to the description of isomorphism classes in TLb6 . The class TLb6 is represented by the following table of multiplication:  [ei , e0 ] = ei+1 , 1 ≤ i ≤ 4,      [e , e ] = −e , 2 ≤ i ≤ 4,  0 i i+1     [e0 , e0 ] = b00 e5 ,      [e0 , e1 ] = −e2 + b01 e5 , TLb6 :=  [e1 , e1 ] = b11 e5 ,      [e1 , e2 ] = −[e2 , e1 ] = b12 e4 + b13 e5 ,      [e1 , e3 ] = −[e3 , e1 ] = b12 e5 ,     [e , e ] = −[e , e ] = −[e , e ] = [e , e ] = −b e . 1 4 4 1 2 3 3 2 23 5 Further elements of TLb6 will be denoted by L(α), where α = (b01 , b11 , b12 , b13 , b23 ).

246  Leibniz Algebras

Theorem 6.12. (Isomorphism criterion for TLb6 ) Two filiform Leibniz algebras L(α0 ) and L(α) from TLb6 are isomorphic, if and only if there exist A0 , A1 , B1 , B2 , B3 ∈ C such that A0 B1 (A0 + A1 b23 ) , 0 and the following equalities hold: b000

=

1 χ1 (A;α) ; A20 B1 χ0 (A;α)

b001 =

1 χ2 (A;α) ; α); A30 χ0 (A;α)

b011 =

B1 1 b ; A40 χ0 (A;α) 11

b012

=

B1 b ; A20 12

b013

=

B1 χ3 (A;B;α) ; A30 χ0 (A;α)

b023

=

B1 1 b . A0 χ0 (A;α) 23

The proof is the similar to that of TLb5 . 6.7.1

Isomorphism classes in TLb6

Represent TLb6 as a union of the following subsets: U61 = {L(α) ∈ TLb6 : b23 , 0, b11 , 0}; U62 = {L(α) ∈ TLb6 : b23 , 0, b11 = 0, b01 , 0}; U63 = {L(α) ∈ TLb6 : b23 , 0, b11 = b01 = 0, b12 , 0, b00 , 0}; U64 = {L(α) ∈ TLb6 : b23 , 0, b11 = b01 = 0, b12 , 0, b00 = 0}; U65 = {L(α) ∈ TLb6 : b23 , 0, b11 = b01 = b12 = 0, b00 , 0}; U66 = {L(α) ∈ TLb6 : b23 , 0, b11 = b01 = b12 = b00 = 0}; U67 = {L(α) ∈ TLb6 : b23 = 0, b12 , 0, b11 , 0}; U68 = {L(α) ∈ TLb6 : b23 = 0, b12 , 0, b11 = 0, b01 , 0}; U69 = {L(α) ∈ TLb6 : b23 = 0, b12 , 0, b11 = b01 = 0, b00 , 0}; U610 = {L(α) ∈ TLb6 : b23 = 0, b12 , 0, b11 = b01 = b00 = 0}; U611 = {L(α) ∈ TLb6 : b23 = b12 = 0, b11 , 0, b13 , 0}; U612 = {L(α) ∈ TLb6 : b23 = b12 = 0, b11 , 0, b13 = 0, Π1 (α) , 0}; U613 = {L(α) ∈ TLb6 : b23 = b12 = 0, b11 , 0, b13 = Π1 (α) = 0}; U614 = {L(α) ∈ TLb6 : b23 = b12 = b11 = 0, b01 , 0, b13 , 0}; U615 = {L(α) ∈ TLb6 : b23 = b12 = b11 = 0, b01 , 0, b13 = 0}; U616 = {L(α) ∈ TLb6 : b23 = b12 = b11 = b01 = 0, b00 , 0, b13 , 0};

Classification of Filiform Leibniz Algebras in Low Dimensions  247

U617 = {L(α) ∈ TLb6 : b23 = b12 = b11 = b01 = 0, b00 , 0,

b13 = 0}; U618 = {L(α) ∈ TLb6 : b23 = b12 = b11 = b01 = b00 = 0, b13 , 0}; U619 = {L(α) ∈ TLb6 : b23 = b12 = b11 = b01 = b00 = b13 = 0}. Proposition 6.13. 1. Two algebras L(α0 ) and L(α) from U61 are isomorphic, if and only if b023 Π2 (α0 )

!2

b23 Π1 (α ) = Π2 (α)

(Π2 (α0 ))3 b03 12 04 b02 23 b11

!2

0

=

Π1 (α),

(Π2 (α))3 b31,2 b223 b411

.

2. For any λ1 , λ2 ∈ C, there exists L(α) ∈ U61 such that b23 Π2 (α)

!2 Π1 (α) = λ1 ,

(Π2 (α))3 b31,2 b223 b411

= λ2 .

Then orbits from the set U61 can be parameterized as L (λ1 , 0, 1, λ2 , 0, 1) λ1 , λ2 ∈ C. Proposition 6.14. 1. Two algebras L(α0 ) and L(α) from U62 are isomorphic, if and only if Π3 (α)4 b31,2 Π3 (α0 )4 b03 12 = . 05 b03 b323 b501 23 b01 2. For any λ ∈ C, there exists L(α) ∈ U62 such that

Π3 (α)4 b312 b323 b501

= λ.

Therefore orbits from U62 can be parameterized as L (0, 1, 0, λ, 0, 1) , λ ∈ C.

248  Leibniz Algebras

Proposition 6.15. 1. Two algebras L(α0 ) and L(α) from U67 are isomorphic, if and only if Π4 (α0 ) Π4 (α) = , b012 b02 b12 b211 11 (Π5 (α0 ))2 (Π5 (α))2 = . b012 b03 b12 b311 11 2. For any λ1 , λ2 ∈ C, there exists L(α) ∈ U67 such that (Π5 (α))2 Π4 (α) = λ2 . = λ , 1 b12 b211 b12 b311 The orbits from U67 are parameterized as L (λ1 , λ2 , 1, 1, 0, 0) , λ1 , λ2 ∈ C. Proposition 6.16. 1. Two algebras L(α0 ) and L(α) from U68 are isomorphic, if and only if (Π6 (α0 ))3 (Π6 (α))3 = 3 4 . 04 b03 b12 b01 12 b01 2. For any λ ∈ C, there exists L(α) ∈ U68 such that

(Π6 (α))3 b312 b401

= λ.

The orbits from the set U68 can be parameterized as L (λ, 1, 0, 1, 0, 0) , λ ∈ C. Proposition 6.17. 1. Two algebras L(α0 ) and L(α) from U611 are isomorphic, if and only if ! !6 b013 6 b13 0 Π1 (α ) = Π1 (α). b011 b11  6 2. For any λ ∈ C, there exists L(α) ∈ U611 such that bb13 Π1 (α) = λ. 11 The orbits from U611 can be parameterized as L (λ, 0, 1, 0, 1, 0) , λ ∈ C.

Classification of Filiform Leibniz Algebras in Low Dimensions  249

Proposition 6.18. The subsets U63 , U64 , U65 , U66 , U69 , U610 , U612 , U613 , U614 , U615 , U616 , U617 , 18 U6 and U619 are single orbits with representatives L(1, 0, 0, 1, 0, 1), L(0, 0, 0, 1, 0, 1), L(1, 0, 0, 0, 0, 1), L(0, 0, 0, 0, 0, 1), L(1, 0, 0, 1, 0, 0), L(0, 0, 0, 1, 0, 0), L(1, 0, 1, 0, 0, 0), L(0, 0, 1, 0, 0, 0), L(0, 1, 0, 0, 1, 0), L(0, 1, 0, 0, 0, 0), L(1, 0, 0, 0, 1, 0), L(1, 0, 0, 0, 0, 0), L(0, 0, 0, 0, 1, 0) and L(0, 0, 0, 0, 0, 0), respectively.

APPENDIX

A

Linear Algebra

A.1

VECTOR SPACES AND SUBSPACES

Definition A.1. A vector space over F is a set V together with a function V × V −→ V called addition, denoted (x, y) −→ x + y, and a function F × V −→ V called scalar multiplication and denoted (c, x) −→ cx, which satisfy the following axioms: 1. For each x, y ∈ V, x + y = y + x. 2. For each x, y, z ∈ V, (x + y) + z = x + (y + z). 3. There exists a zero vector in V, denoted 0, such that 0 + x = x for each x ∈ V. 4. For each x ∈ V, there exists −x ∈ V such that (−x) + x = 0. 5. For each a ∈ F and x, y ∈ V, a(x + y) = ax + ay. 6. For each a, b ∈ F and x ∈ V, (a + b)x = ax + bx. 7. For each a, b ∈ F and x ∈ V, (ab)x = a(bx). 8. For each x ∈ V, 1x = x. The elements of V are called vectors, whereas the elements of F are scalars. Facts: Let V be a vector space over a field F. • The vector 0 is the only additive identity in V. 251

252  Leibniz Algebras

• For each x ∈ V, the element −x is the only additive inverse for x in V. • For each x ∈ V, −x = (−1)x. • If a ∈ F and x ∈ V, then ax = 0 if and only if a = 0 or x = 0. • If x, y, z ∈ V and x + y = x + z, then y = z. Example A.1. • Any field is a vector space over itself and over its subfield. Particularly, the number fields Q, R, C are examples of vector spaces. • The set Fn (n ≥ 1) is a vector space over F. • The set of polynomials F[x1 , x2 , . . . , xn ] at variables x1 , x2 , . . . , xn with coefficients from a field F is a vector space over F. • The set of n × n matrices with entries in a field F is a vector space over F. • Let X be a non-empty set and F(X) = { f : X −→ F} be the set of functions on X with values in a field F. Define ( f ?g)(x) = f (x)+g(x) and (λ· f )(x) = λ f (x), for f, g ∈ F(X), λ ∈ F. The set (F(X), ?, ·) is a vector space over F. Let x1 , x2 , . . . , xk ∈ V and a1 , a2 , . . . , ak ∈ F. The vector x = a1 x1 + a2 x2 + · · · + ak xk ∈ V is called a linear combination of vectors x1 , x2 , . . . , xk with the coefficients a1 , a2 , . . . , ak . Definition A.2. A subset W ⊂ V of a vector space V is said to be a subspace if it is a vector space with respect to operations “ + ” and “ · ” defined in V. Note that this is equivalent to • For any x, y ∈ W one has x + y ∈ W; • For any a ∈ F and x ∈ W we have ax ∈ W.

Linear Algebra  253

Example A.2. 1. For any vector space V the subset {0} and V itself are subspaces of V. 2. The intersection of any nonempty collection of subspaces of V is a subspace of V. 3. Let W1 and W2 be subspaces of V and W1 + W2 = {x ∈ V| x1 ∈ W1 and x2 ∈ W2 }. Then W1 +W2 is a subspace of V containing W1 and W2 . It is the smallest subspace that contains them in the sense that any subspace that contains both W1 and W2 must contain W1 + W2 . 4. Let W1 and W2 be subspaces of V such that W1 ∩ W2 = {0}. Then the subspace W1 + W2 is denoted by W1 ⊕ W2 and it is called the direct sum of the subspaces W1 and W2 . If V = W1 ⊕ W2 then V is the direct sum of its subspaces W1 and W2 . Note that the sum and the direct sum of subspaces W1 , W2 , . . . , W s are subspaces and defined by induction. They are denoted by W1 + W2 + · · · + W s and W1 ⊕ W2 ⊕ · · · ⊕ W s , respectively. 5. For a set of vectors x1 , x2 , . . . , xk ∈ V one defines k X SpanF {x1 , x2 , . . . , xk } = {x ∈ V | x = ai xi , a1 , a2 , . . . , ak ∈ F}. i=1

The subset SpanF is a subspace of V. Definition A.3. A set of vectors x1 , x2 , . . . , xk ∈ V is said to be linearly independent if a1 x1 + a2 x2 + · · · + ak xk = 0 if and only if a1 = a2 = · · · = ak = 0. Definition A.4. A set of vectors x1 , x2 , . . . , xk ∈ V is said to be a basis of V if • The set {x1 , x2 , . . . , xk } is linearly independent; • V = SpanF {x1 , x2 , . . . , xk }. Definition A.5. If such a set of vectors x1 , x2 , . . . , xk ∈ V exists then the number of elements k of the set x1 , x2 , . . . , xk is said to be the dimension of the vector space V. The dimension of the vector space V is denoted by dimF V = k (sometimes just dim V = k if the main field is predefined).

254  Leibniz Algebras

Remark. Note that • dim(W1 + W2 ) = dim W1 + dim W2 − dim W1 ∩ W2 ; • dim(W1 ⊕ W2 ) = dim W1 + dim W2 . For any subspace W1 of a finite-dimensional space V there exists a subspace W2 such that V = W1 ⊕ W2 . The subspaces W1 and W2 are called the direct complements to each other. A.2

LINEAR TRANSFORMATIONS

Definition A.6. Let V1 and V2 be vector spaces over a field F. A function f : V1 −→ V2 is said to be a linear transformation if f (a1 x1 + a2 x2 ) = a1 f (x1 ) + a2 f (x2 ), for all x1 , x2 ∈ V1 , and a1 , a2 ∈ F. Let Hom(V1 , V2 ) denote the set of linear transformations from V1 to V2 . For f, g ∈ Hom(V1 , V2 ) and a ∈ F we define f ⊕ g and a f by the formulas ( f ⊕ g)(x) = f (x) + g(x) and (a f )(x) = a f (x), for all x ∈ V1 . The set Hom(V1 , V2 ) is a vector space over F with respect to the operations ⊕ and . Definition A.7. Vector spaces V1 and V2 are called isomorphic if there exists a bijective linear transformation f : V1 −→ V2 . Theorem A.1. Vector spaces V1 and V2 are isomorphic if and only if they have the same dimension. A.3 DUAL VECTOR SPACE

Let V be a vector space over a field F and F(V) be the vector space of functions defined on V. Consider a subset V ∗ of F(V) satisfying the conditions: • f (x1 + x2 ) = f (x1 ) + f (x2 ) for all x1 , x2 ∈ V; • f (λx) = λ f (x) for all x ∈ V, λ ∈ F.

Linear Algebra  255

The subset V ∗ is a subspace of the vector space F(V) and it is said to be a dual vector space to V. Let V be a finite-dimensional vector space. The dual space has the following properties: • dim V = dim V ∗ ; • The finite-dimensional vector spaces V, V ∗ and V ∗∗ are isomorphic; • Let {e1 , e2 , . . . , en } be a basis of V. Introduce the function ei (x) = xi , where xi is the coordinate of the vector in the basis {e1 , e2 , . . . , en }. An equivalent description of {ei } is as follows: ei (e j ) = δi j (the Kronecker delta symbol: 1 for i = j and 0 for i , j). The set of vectors {e1 , e2 , . . . , en } form a basis of V ∗ and it is said to be dual to {e1 , e2 , . . . , en } basis. A.4

TENSOR PRODUCTS

In this subsection we recall the notion of tensor product of vector spaces. Definition A.8. The tensor product V ⊗ W of vector spaces V and W over a field F is the quotient of the space V ∗ W whose basis is given by formal symbols v ⊗ w, v ∈ V, w ∈ W, by the subspace spanned by the elements (v1 + v2 ) ⊗ w − v1 ⊗ w − v2 ⊗ w, v ⊗ (w1 + w2 ) − v ⊗ w1 − v ⊗ w2 , av ⊗ w − a(v ⊗ w), v ⊗ aw − a(v ⊗ w), where v ∈ V, w ∈ W, a ∈ F. It is easy to see that V ⊗ W can be equivalently defined as the quotient of the free abelian group v · w generated by v ⊗ w, v ∈ V, w ∈ W by the subgroup generated by (v1 + v2 ) ⊗ w − v1 ⊗ w − v2 ⊗ w, v ⊗ (w1 + w2 ) − v ⊗ w1 − v ⊗ w2 , av ⊗ w − a(v ⊗ w), v ⊗ aw − a(v ⊗ w), where v ∈ V, w ∈ W, a ∈ F.

256  Leibniz Algebras

The elements v ⊗ w ∈ V ⊗ W for v ∈ V, w ∈ W are called pure tensors. Note that in general, there are elements of V ⊗ E which are not pure tensors. This allows one to define the tensor product of any number of vector spaces, V1 ⊗ · · · ⊗ Vn . Note that this tensor product is associative, in the sense that (V1 ⊗ V2 ) ⊗ V3 can be naturally identified with V1 ⊗ (V2 ⊗ V3 ). In particular, people often consider tensor products of the form ⊗n V = V ⊗ · · · ⊗ V (n times) for a given vector space V, and, more generally, E := V ⊗n ⊗ (V ∗ )⊗m . This space is called the space of tensors of type (n, m) on V. For instance, tensors of type (0, 1) are vectors, of type (1, 0)-linear functionals (covectors), of type (1, 1)-linear operators, of type (2, 0)-bilinear forms, of type (2, 1)-algebra structures, etc. If V is finite-dimensional with basis ei , i = 1, . . . , N, and ei is the dual basis of V ∗ , then a basis of E is the set of vectors ei1 ⊗ · · · ⊗ ein ⊗ e j1 ⊗ · · · ⊗ e jm , and a typical element of E is N P i1 ,...,in , j1 ,..., jm =1

j1 jm n T ij11,...,i ,..., jm ei1 ⊗ · · · ⊗ ein ⊗ e ⊗ · · · ⊗ e ,

where T is a multidimensional table of numbers. A.5 TENSOR ALGEBRA

The notion of tensor products allows us to give more conceptual (i.e., coordinate free) definitions of the free algebra, polynomial algebra, exterior algebra, and universal enveloping algebra of a Lie algebra. Namely, given a vector space V, define its tensor algebra T V over a field F to be T V = ⊕n≥0 V ⊗n , with multiplication defined by a · b := a⊗b, a ∈ V ⊗r , b ∈ V ⊗s . Observe that a choice of a basis {x1 , . . . , xm } in V defines an isomorphism of T V with the free algebra F hx1 , . . . , xm i. Also, one can make the following definition.

Linear Algebra  257

Definition A.9. (i) The symmetric algebra S V of V is the quotient of T V by the ideal generated by v ⊗ w − w ⊗ v, v, w ∈ V. (ii) The exterior algebra ∧V of V is the quotient of T V by the ideal generated by v ⊗ v, v ∈ V. (iii) If V is a Lie algebra, the universal enveloping algebra U(L) of V is the quotient of T V by the ideal generated by v ⊗ w − w ⊗ v − [v, w], v, w ∈ V. It is easy to see that a choice of a basis {x1 , . . . , xm } in V identifies S V with the polynomial algebra F[x1 , . . . , xm ], ∧V with the exterior algebra ∧k (x1 , . . . , xm ), and the universal enveloping algebra U(V) with one defined previously. Also, it is easy to see that we have decompositions M M SV = S n V, and ∧V = ∧n V. n≥0

A.6

n≥0

MATRIX OF A LINEAR TRANSFORMATION

Let V1 and V2 be vector spaces over a field F with the distinguished bases e = {e1 , e2 , . . . , en } and e0 = {e01 , e02 , . . . , e0m }, respectively. Consider an arbitrary linear transformation f : V1 −→ V2 and associate with it an m × n matrix A f with entries from F as follows (note that the sizes of A f are the same as the dimensions of V1 , V2 in reverse order). m P Represent the vector f (ek ) as linear combinations: f (ek ) = aik e0i . i=1

The matrix A f = (ai j )i=1,2,...,m, j=1,2,...,n is called the matrix of the linear transformation f with respect to bases e and e0 . The correspondence f 7−→ A f gives the isomorphism between the vector spaces Hom(V1 , V2 ) and Mm×n (F). Let now V1 = V2 = V then for f : V −→ V the matrix A f = (ai j )i, j=1,2,...,n and it is called the matrix of the linear transformation f of V on the basis e = {e1 , e2 , . . . , en }. If we take another basis e0 = {e01 , e02 , . . . , e0n } with the matrix A0f of f on e0 and the base change matrix M : e = {e1 , e2 , . . . , en } −→ e0 = {e01 , e02 , . . . , e0n } then one has A0f = M −1 A f M. One of the main problems of linear algebra is to choose M

258  Leibniz Algebras

so that M −1 AM is as nearly diagonal as possible. Let us consider the case when M −1 AM has the diagonal form. Definition A.10. A subspace W of V is said to be invariant with respect to a linear transformation f : V −→ V if f (W) ⊂ W. Definition A.11. A linear transformation f : V −→ V is diagonalizable if either one of the following two equivalent conditions holds: i) The vector space V decomposes into a direct sum of onedimensional invariant subspaces; ii) There exists a basis of V with respect to which the matrix of f is diagonal. Note that if there exists a k-dimensional invariant subspace W of an n-dimensional vector space V with respect to a linear transformation f : V −→ V then there exists a basis of V such that the matrix of f with respect to the basis has the form a11 a12 . . . a1k a1k+1 . . . a1n  a   21 a22 . . . a2k a2k+1 . . . a2n  ... ...   ak1 ak2 . . . akk akk+1 . . . akn  .  0 0 . . . 0 ak+1k . . . ak+1n    ... 0 0 . . . 0 ank+1 . . . ann Note that if there exist two such invariant subspaces W1 (dim W1 = k) and W2 with V = W1 ⊕ W2 then there exists a basis so that the matrix of f has the form a 0 ... 0   11 a12 . . . a1k a21 a22 . . . a2k 0 ... 0    ... ... a  . a . . . a 0 . . . 0 k1 k2 kk   0 . . . 0 ak+1k . . . ak+1n   0   ... 0 0 . . . 0 ank+1 . . . ann Moreover, if there exist ki -dimensional (i = 1, 2, . . . , s) invariant subspaces Wi (dim Wi = ki ) with V = W1 ⊕ W2 ⊕ · · · ⊕ W s then one can find a basis with respect to which the matrix of f has the form  A 0 0 0    1  0 A2 0 0  .  0 0 . . . 0  0 0 0 As

Linear Algebra  259

Definition A.12. i) A one-dimensional subspace V1 ⊂ V is said to be a proper subspace for the operator f : V −→ V if it is invariant, i.e., f (V1 ) ⊂ V1 . If V1 is such a subspace, then the effect of f on it is equivalent to multiplication by a scalar λ ∈ F. This scalar is called the eigenvalue of f (on V1 ). ii) The vector x ∈ V is said to be an eigenvector of f if x , 0 and the linear span Spanx is a proper subspace. In other words, f (x) = λx for an appropriate λ ∈ F. According to the definition above, diagonalizable operators f admit a decomposition of V into a direct sum of its proper subspaces. Let us now determine when f has at least one proper subspace. Definition A.13. Let V be a finite-dimensional vector space, f : V −→ V be a linear transformation and let A be its matrix in some basis. Denote by P(t) the polynomial det(tI − A) (I is the identity matrix) with coefficients in F and call it the characteristic polynomial of the operator f and of the matrix A. Theorem A.2. a) The characteristic polynomial of f does not depend on the choice of basis in which its matrix is represented; b) Any eigenvalue of f is a root of P(t) and any root of P(t) lying in F is an eigenvalue of f , corresponding to some (not necessarily the unique) proper subspace of V. Since the field F is algebraically closed any polynomial with leading coefficient equal to one is represented as a product of linear polyn Q nomials, therefore, P(t) = (t − λi )ri , where λi ∈ F, λi , λ j for i , j. i=1

The number r j is called the multiplicity of the root λi of the polynomial P(t). The set of all roots of the characteristic polynomial P(t) is called the spectrum of f . If all multiplicities are equal to one, the spectrum of f is said to be simple. Let us give the following important theorem on the characteristic polynomial of a linear transformation.

260  Leibniz Algebras

Theorem A.3. (The Cayley-Hamilton Theorem) If P(t) is the characteristic polynomial for a linear transformation f : V −→ V with a matrix A on some basis of V then P(A) = 0. Definition A.14. a) A matrix A of the form  λ   0  . . . 0

1 λ ... 0

0 1 ... 0

... 0 ... 0 ... ... ... λ

   

is called an r × r Jordan block Jr (λ) with the eigenvalue λ. b) A Jordan matrix is a matrix consisting of diagonal blocks  J (λ ) 0 0 0   r1 1  0 Jr2 (λ2 ) 0 0   0 0 ... 0  0 0 0 Jrs (λ s ) c) A Jordan basis for the operator f : V −→ V is a basis of the space V with respect to that the matrix of f is a Jordan matrix or, as it is usually said, that the matrix has the Jordan normal form. d) The solution of a matrix equation of the form X −1 AX = J, where A is a square matrix, X is an unknown non-singular matrix, and J is an unknown Jordan matrix, is called the reduction of A to Jordan normal form. Definition A.15. The vector x ∈ V is called a root vector of the operator f , corresponding to λ ∈ F, if there exists an r such that ( f − λI)r x = 0. Note that all eigenvectors are evidently root vectors. The set of all root vectors of f corresponding to λ is denoted by V(λ). Theorem A.4. a) The set V(λ) is an invariant subspace of V and V(λ) , {0} if and only if λ is an eigenvalue of f . b) V = ⊕V(λi ), where λi runs through all the eigenvalues of the operator f , i.e., λi , λ j , i, j = 1, 2, . . . , s.

Linear Algebra  261

A.7

JORDAN NORMAL FORM

The main goal of this section is to prove the following theorem on the existence and uniqueness of the Jordan normal form for matrices and linear operators. Theorem A.5. Let F be an algebraically closed field, V a finitedimensional linear space over F, and f : V → V be a linear operator. Then: a) A Jordan basis exists for the operator f , i.e., the matrix of the operator A in the original basis can be reduced by a change X of basis to the Jordan form X −1 AX = J. b) The matrix J is unique, apart from a permutation of its constituent Jordan blocks. The proof of the theorem is divided into a series of intermediate steps. We begin by constructing the direct composition V = ⊕ni=1 Vi , where the Vi , are invariant subspaces for f , which will later correspond to the set of Jordan blocks for f with the same number λ along the diagonal. In order to characterize these subspaces in an invariant manner, we recall that (Jr (λ) − λIr )n = 0, where Ir is a r × r identity matrix. If some power of an operator is zero then the operator is said to be nilpotent. Thus the operator f − λ is nilpotent on the subspace corresponding to the block Jr (λ). The same is true for its restriction to the sum of subspaces for fixed λ. This motivates the following definition. Definition A.16. The vector l ∈ V is called a root vector of the operator f , corresponding to λ ∈ F, if there exists an r such that ( f −λ)r l = 0 (here ( f − λ) denotes the operator f − λid). All eigenvectors are evidently root vectors. Proposition A.1. We denote by V(λ) the set of root vectors of the operator f in V corresponding to λ. Then V(λ) is a linear subspace in V and V(λ) , {0} if and only if λ is an eigenvalue of f . Proof. Let ( f − λ)r1 l1 = ( f − λ)r2 l2 = 0. Setting r = max(r1 , r2 ), we find that ( f − λ)r (l1 + l2 ) = 0 and ( f − λ)r1 (al1 ) = 0. Therefore, V(λ) is a subspace.

262  Leibniz Algebras

If λ is an eigenvalue of f , then there exists an eigenvector corresponding to λ such that V(λ) , {0}. Conversely, let l ∈ V(λ), l , 0. We select the smallest value of r for which ( f − λ)r l = 0. Obviously, r ≥ 1. The vector l0 = ( f − λ)r−1 l is an eigenvector of f with eigenvalue λ : l0 , 0 according to the choice of r and ( f − λ)l0 = 0, whence f (l0 ) = λl0 .  Proposition A.2. V = ⊕V(λi ), where λi runs through all the eigenvalues of the operator f , i.e., the different roots of the characteristic polynomial of f . Qs Proof. Let P(t) = i=1 (t − λi )ri be the characteristic polynomial of f , λi , λ j for i , j. Set Fi (t) = P(t)(t − λi )−ri , fi = Fi ( f ), Vi = im fi . We check the following series of assertions. a) f − λri i Vi = {0}, that is, Vi ⊂ V(λi ). Indeed, ( f − λi )ri fi = ( f − λi )ri Fi ( f ) = P( f ) = 0 according to the Cayley-Hamilton theorem. b) V = V1 + · · · + V s . Indeed, since the polynomials Fi (t) in aggregate are relatively prime, there exist polynomials Xi (t) such that s P Fi (t)Xi (t) = 1. Therefore, substituting f for t, we have

i=1

s X

Fi ( f )Xi ( f ) = id.

i=1

Applying this identity to any vector l ∈ V, we find l=

s X i=1

Fi (Xi )( f ), l ∈

s X

Vi .

i=1

c) P V = V1 ⊕ · · · ⊕ V s . Indeed, we choose 1 ≤ i ≤ s and verify that Vi ∩ ( V j ) = {0}. Let l be a vector from this intersection. Then j,i

( f − λi )ri l = 0, since l ∈ Vi ; Y X V j. Fi ( f )l = ( f − λ j )r j l = 0, since l ∈ j,i

j,i

Since (t − λi ) and Fi (t) are relatively prime polynomials, there exist polynomials X(t) and Y(t) such that X(t)(t − λi )ri + Y(t)Fi (t) = 1. ri

Linear Algebra  263

Substituting here f for t and applying the operator identity obtained to l, we obtain X( f )(0) + Y( f )(0) = l = 0. d) Vi = V(λi ). Indeed, we have already verified that Vi ⊂ V(λi ). To prove the converse we choose a vector l ∈ V(λi ) and represent it in the form l = l0 + l00 , l0 ∈ Vi , l00 ∈ ⊕ j,i Vi . There exists a number r0 such that 0 ( f − λi )r l00 = 0, because l00 = l − l0 ∈ V(λi ). In addition, Fi ( f )l00 = 0. 0 Writing the identity X(t)(t − λi )r + Y(t)Fi (t) = 1 and replacing t by f , we find that l00 = 0, so that l = l0 ∈ Vi .  Corollary A.1. If the spectrum of an operator f is simple, then f is diagonalizable. Proof. Indeed, the number of different eigenvalues of f then equals n = deg P(t) = dim V. Hence, in the decomposition V = ⊕ni=1 V(λi ) all spaces V(λi ) are one-dimensional and since each of them contains an eigenvector, the operator f becomes diagonal in a basis consisting of these vectors.  We now fix one of the eigenvalues λ and prove that the restriction of f to V(λ) has a Jordan basis, corresponding to this value of λ. To avoid introducing a new notation we shall assume up to the end of the proof of the proposition A.3 that f has only one eigenvalue λ and V = V(λ). Moreover, since any Jordan basis for the operator f is simultaneously a Jordan basis for the operator f +µ, where µ is any constant, we can even assume that λ = 0. Then, according to the Cayley-Hamilton theorem, the operator f is nilpotent: P(t) = tn , f n = 0. We shall now prove the following proposition. Proposition A.3. A nilpotent operator f on a finite-dimensional space V has a Jordan basis; the matrix of f in this basis is a combination of blocks of the form Jr (0). Proof. If we already have a Jordan basis in the space V, it is convenient to represent it by a diagram D, similar to the one shown here. · · ↓ ↓ · · · · ↓ ↓ ↓ ↓ · · · · In this diagram, the dots denote elements of the basis and the arrows describe the action of f (in the general case, the action of f − λ). The

264  Leibniz Algebras

operator f transforms to zero the elements in the lowest row, that is, the eigenvectors of f entering into the basis occur in this row. Each column thus stands for a basis of the invariant subspace, corresponding to one Jordan block, whose dimension equals the height of this column (the number of points in it): if f (eh ) = eh−1 , f (eh−1 ) = eh−2 , . . . , f (e1 ) = 0, then  0  f (e1 , . . . , eh ) = (e1 , . . . , eh )  . 0. . 0

1 0 ... 0

0 1 ... 0

... 0 ... 0 ... ... ... 0

   

Conversely, if we find a basis of V whose elements are transformed by f into other elements of the basis or into zero, so that the elements of this basis together with the action of f can be described by such a diagram, then it will be the Jordan basis for V. We shall prove existence by induction on the dimension of V. If dim V = 1, then the nilpotent operator f is a zero operator and any non-zero vector in V forms its Jordan basis. Now, let dim V = n > 1 and assume that the existence of a Jordan basis has already been proved for dimensions less than n. We denote by V0 ⊂ V the subspace of eigenvectors for f , that is, ker f . Since dim V0 > 0, we have dim V/V0 < n, while the operator f : V → V induces the operator f˜ : V/V0 → V/V0 : f˜(l + V0 ) = f (l) + V0 . (The correctness of the definition of f and its linearity are obvious.) According to the induction hypothesis, f˜ has a Jordan basis. We can assume that it is non-empty. Otherwise V = V0 and any basis of V0 will be a Jordan basis for f˜. Let us construct the diagram D˜ for elements of the Jordan basis of f˜. We take the uppermost vector e˜i , i = 1, . . . , m in each column, and set e˜i = ei + V0 , ei ∈ V. We shall now construct the diagram D of vectors of the space V as follows. For i = 1, . . . , m the ith column in the diagram D will consist (top to bottom) of the vectors ei , fei , . . . , f hi −1 (ei ), f hi (ei ), where hi is the height of the ith ˜ Since f˜hi (e˜i ) = 0, f hi (ei ) ∈ V0 and f hi +1 (ei ) = column in the diagram D. 0. We select a basis of the linear span of the vectors f h1 (e1 ), . . . , f hm (em ) in V0 , extend it to a basis of V0 , and insert the additional vectors as

Linear Algebra  265

additional columns (of unit height) in the bottom row of the diagram D; f transforms them into zero. Thus the diagram D consisting of vectors of the space V together with the action of f on its elements has exactly the form required for a Jordan basis. We have only to check that the elements of D actually form a basis of V. We shall first show that the linear span of D equals V. Let l ∈ V, l˜ = m hP i −1 P l + V0 . By assumption, l˜ = ai j f˜ j (e˜i ). Since V0 is invariant under i=1 j=0

f , it follows that l−

hi −1 m X X

ai j f j (ei ) ∈ V0 .

i=1 j=0

But all the vectors f j (ei ), j ≤ hi − 1 lie in the rows of the diagram D, beginning with the second from the bottom, and the subspace V0 is generated by the elements of the first row of D by construction. Therefore l can be represented as a linear combination of the elements of D. It remains to verify that the elements of D are linearly independent. First of all, the elements in the bottom row of D are linearly independent. Indeed, if some non-trivial linear combination of them equals m P zero, then it must have the form ai f hi (ei ) = 0, because the remaini=1

ing elements of the bottom row extend the basis of the linear span of { f h1 (e1 ), . . . , f hm (em )} up to a basis of V0 . But all the hi ≥ 1, therefore  m  X h −1  f  ai f i (ei ) = 0, i=1

so that

m X i=1

ai f hi −1 (ei ) ∈ V0

and

m X

ai f˜hi −1 (e˜i ) = 0.

i=1

It follows from the last relation that all the ai = 0, because the vectors f˜hi −1 (e˜i ) comprise the bottom row of the diagram D˜ and are part of a basis of V/V0 . Finally, we shall show that if there exists a non-trivial linear combination of the vectors of D equal to zero, then it is possible to obtain from it a non-trivial linear dependence between the vectors in the

266  Leibniz Algebras

bottom row of D. Indeed, consider the top row of D, which contains the non-zero coefficients of this imagined linear combination. Let the number of this row (counting from the bottom) be h. We apply to this combination the operator f h−1 . Evidently, the part of this combination corresponding to the hth row will transform into a non-trivial linear combination of elements of the bottom row, while the remaining terms will vanish. This completes the proof of the proposition.  Now we have only to verify the part of Theorem A.5 that refers to uniqueness. Let an arbitrary Jordan basis of the operator f be fixed. Any diagonal element of the matrix f in this basis is obviously one of the eigenvalues λ of this operator. Examine the part of the basis corresponding to all of the blocks of matrices with this value of λ and denote by Vλ its linear span. Since (Jr (λ) − λ)r = 0, we have Vλ ⊂ V(λ), where V(λ) is the root space of V0 . In addition, V = ⊕Vλi by definition of the Jordan basis and V = ⊕V(λi ) by Proposition A.2, where in both cases λi runs through all eigenvalues of f once. Therefore, dim Vλi = dim V(λi ) and Vλi = V(λi ). Hence, the sum of the dimensions of the Jordan blocks, corresponding to each λ − id, independent of the choice of Jordan basis and, moreover, the linear spans of the corresponding subsets of the basis Vλi are basis-independent. It is thus sufficient to check the uniqueness theorem for the case V = V(λ) or even for V = V(0). We construct the diagram D corresponding to a given Jordan basis of V = V(0). The dimensions of the Jordan blocks are the heights of its columns; if the columns in the diagram are arranged in decreasing order, these heights are uniquely determined if the lengths of the rows in the diagram are known, beginning with the bottom row, in decreasing order. We shall show that the length of the bottom row equals the dimension of V0 = ker f . Indeed, we take any eigenvector l for f and represent it as a linear combination of the elements of D. All vectors lying above the bottom row will appear in this linear combination with zero coefficients. Indeed, if the highest vectors with non-zero coefficients were to lie in a row with number h ≥ 2, then the vector f h−1 (l) = 0 would be a non-trivial linear combination of the elements of the bottom row of D (cf. the end of the proof of Proposition A.3), and this contradicts the linear independence of the elements of D. This means that the bottom row of D forms a basis of L0 , so that its length equals dim L0 ; hence, this length is the same for all Jordan bases. In

Linear Algebra  267

exactly the same way, the length of the second row does not depend on the choice of basis, so that, in the notation used in this section, it equals the dimension of ker f˜ in V/V0 . This completes the proof of uniqueness and of Theorem A.5. Remarks. a) Let the operator f be represented by a matrix A in some basis. Then the problem of reducing A to Jordan form can be solved as follows. Calculate the characteristic polynomial of A and its roots. Calculate the dimensions of the Jordan blocks, corresponding to the roots λ. For this, it is sufficient to calculate the lengths of the rows of the corresponding diagrams, that is, dim ker(A − λI), dim ker(A − λI)2 − dim ker(A − λI), dim ker(A − λI)3 − dim ker(A − λI)2 , . . . . Construct the Jordan form J of the matrix A and solve the matrix equation AX − XJ = 0. The space of solutions of this linear system of equations will, generally speaking, be multidimensional, and the solutions will also include singular matrices. But according to the existence theorem, non-singular solutions necessarily exist; any one can be chosen. b) One of the most important applications of the Jordan form is for the calculation of functions of a matrix (thus far we have considered only polynomial functions). Assume, for example, that we must find a large power AN of the matrix A. Since the degree of the Jordan matrix is easy to calculate, an efficient method is to use the formula AN = XJ N X −1 , where A = XJX −1 . The point is that the matrix X is calculated once and for all and does not depend on N. The same formula can be used to estimate the growth of the elements of the matrix AN . c) It is easy to calculate the minimal polynomial of a matrix A in terms of the Jordan form. Indeed, we shall restrict ourselves for simplicity to the case of a field with zero characteristic. Then the minimal polynomial of Jr (λ) equals (t − λ)r , the minimal polynomial of the block matrix (Jri (λ)) equals (t − λ)max(ri ) , and finally the minimal polynomial of the general Jordan matrix with diagonal elements Qs λ1 , . . . , λ s , (λi , λ j for i , j) equals j=1 (t − λ j )r j , where r j is the smallest dimension of the Jordan block corresponding to λ j .

APPENDIX

B

Elements of Representation Theory

Roughly speaking, representation theory studies symmetry in linear spaces. It is a beautiful mathematical subject which has many applications, ranging from number theory and combinatorics to geometry, probability theory, quantum mechanics and quantum field theory. Representation theory was born in 1896 in the work of the German mathematician F. G. Frobenius. This work was triggered by a letter to Frobenius by R. Dedekind. In this letter Dedekind made the following observation: take the multiplication table of a finite group G and turn it into a matrix XG by replacing every entry g of this table by a variable xg. Then the determinant of XG factors into a product of irreducible polynomials in {xg}, each of which occurs with multiplicity equal to its degree. Dedekind checked this surprising fact in a few special cases, but could not prove it in general. So he gave this problem to Frobenius. In order to find a solution of this problem, Frobenius created representation theory of finite groups. In this section we introduce the notion of representations of associative algebras. Definition B.1. A representation of an algebra A (also called a left Amodule) is a vector space V together with a homomorphism of algebras ρ : A −→ End(V). Similarly, a right A-module is a space V equipped with an antihomomorphism ρ : A −→ End(V) ; i.e., ρ satisfies ρ(ab) = ρ(b)ρ(a) and ρ(1) = 1. 269

270  Leibniz Algebras

The usual abbreviated notation for ρ(a)v is av for a left module and va for the right module. Then the property that ρ is an (anti)homomorphism can be written as a kind of associativity law: (ab)v = a(bv) for left modules, and (va)b = v(ab) for right modules. Here are some examples of representations. Example B.1. 1. V = {0}. 2. V = A, and ρ : A −→ End(A) is defined as follows: ρ(a) is the operator of left multiplication by a, so that ρ(a)b = ab (the usual product). This representation is called the regular representation of A. Similarly, one can equip A with a structure of a right Amodule by setting ρ(a)b := ba. 3. A = F. Then a representation of A is simply a vector space over F. 4. A = F[x1 , . . . , xn ]. Then a representation of A is just a vector space V over F with a collection of arbitrary linear operators ρ(x1 ), . . . , ρ(xn ) : V −→ V. Definition B.2. A subrepresentation of a representation V of an algebra A is a subspace W ⊂ V which is invariant under all the operators ρ(a) : V −→ V, a ∈ A. For instance, {0} and V are always subrepresentations. Definition B.3. A representation V , {0} of A is irreducible (or simple) if the only subrepresentations of V are {0} and V. Definition B.4. Let V1 , V2 be two representations of an algebra A. A homomorphism (or intertwining operator) ϕ : V1 −→ V2 is a linear operator which commutes with the action of A, i.e., ϕ(av) = aϕ(v) for any v ∈ V1 . A homomorphism ϕ is said to be an isomorphism of representations if it is an isomorphism of vector spaces. The set (space) of all homomorphisms of representations V1 −→ V2 is denoted by HomA (V1 , V2 ).

Elements of Representation Theory  271

Note that if a linear operator ϕ : V1 −→ V2 is an isomorphism of representations then so is the linear operator ϕ−1 : V2 −→ V1 . Two representations between which there exists an isomorphism are said to be isomorphic. For practical purposes, two isomorphic representations may be regarded as “the same”, although there could be subtleties related to the fact that an isomorphism between two representations, when it exists, is not unique. Definition B.5. Let V1 , V2 be representations of an algebra A. Then the space V1 ⊕ V2 has an obvious structure of a representation of A, given by a(v1 ⊕ v2 ) = av1 ⊕ av2 . Definition B.6. A nonzero representation V of an algebra A is said to be indecomposable if it is not isomorphic to a direct sum of two nonzero representations. It is obvious that an irreducible representation is indecomposable. On the other hand, the converse statement is false in general. One of the main problems of representation theory is to classify irreducible and indecomposable representations of a given algebra up to isomorphism. This problem is usually difficult and often can be solved only partially (say, for finite dimensional representations). Below we will give a number of examples in which this problem is partially or fully solved for specific algebras. We will now give Schur’s lemma. Although it is very easy to prove, it is fundamental in the whole subject of representation theory. Proposition B.1. (Schur’s lemma) Let V1 , V2 be representations of an algebra A over any field F (which need not be algebraically closed). Let ϕ : V1 −→ V2 be a nonzero homomorphism of representations. Then: (i). If V1 is irreducible then ϕ is injective; (ii). If V2 is irreducible ϕ is surjective. Thus, if both V1 and V2 are irreducible, ϕ is an isomorphism. Corollary B.1. (Schur’s lemma for algebraically closed fields) Let V be a finite dimensional irreducible representation of an algebra A over an algebraically closed field F, and ϕ : V −→ V be an intertwining operator. Then ϕ = λ · Id for some λ ∈ F (a scalar operator).

272  Leibniz Algebras

Remark. Note that this Corollary is false over the field of real numbers: it suffices to take A = C (regarded as an R-algebra), and V = A. Corollary B.2. Let A be a commutative algebra. Then every irreducible finite dimensional representation V of A is one-dimensional. Remark. Note that a one-dimensional representation of any algebra is automatically irreducible. Example B.2. 1. A = F. Since representations of A are simply vector spaces, V = A is the only irreducible and the only indecomposable representation. 2. A = F[x]. Since this algebra is commutative, the irreducible representations of A are its one-dimensional representations. They are defined by a single operator ρ(x). In the one-dimensional case, this is just a number from F. So all irreducible representations of A are Vλ = F, λ ∈ F, in which the action of A defined by ρ(x) = λ. Clearly, these representations are pairwise nonisomorphic.

APPENDIX

C

Zariski Topology

Let F be a fixed algebraically closed field. We define affine space over F denoted An , to be the set of all n−tuples of elements of F. An element P ∈ An is called a point, and if P = (a1 , . . . , an ) with ai ∈ F, then the ai are called the coordinates of P. Let A = F[x1 , . . . , xn ] be the polynomial ring in n variables over F. We will interpret the elements of A as functions from the affine n−space to F, by defining f (P) = f (a1 , . . . , an ), where f ∈ A and P ∈ An . Thus if f ∈ A is a polynomial, we can talk about the set of zeros of f , namely Z( f ) = {P ∈ An | f (P) = 0}. More generally, if T is any subset of A, we define the zero set of T to be the common zeros of all the elements of T : Z(T ) = {P ∈ An | f (P) = 0 for all f ∈ T }. Clearly, if J is the ideal of A generated by T , then Z(T ) = Z(J). Furthermore, since A is a noetherian ring, any ideal J has a finite set of generators f1 , . . . , fr we can regard Z(T ) as the common zeros of the finite set of polynomials f1 , . . . , fr . Definition C.1. A subset Y of An is an algebraic set if there exists a subset T ⊂ A such that Y = Z(T ). Example C.1. The space An is an algebraic set defined by f (x1 , x2 , . . . , xn ) ≡ 0. Example C.2. The set An \ {0} is not algebraic since any polynomial vanishing on all points An \ {0} is only zero polynomial. 273

274  Leibniz Algebras

Example C.3. If X ⊂ An and Y ⊂ Am are algebraic subsets then X × Y ⊂ An+m is also an algebraic subset. If both X, Y are irreducible then X × Y is also irreducible. Example C.4. Let us describe all algebraic sets X in A1 . Such a set is defined as a set of solutions of the system of equations f1 (x) = 0, f2 (x) = 0, . . . , fm (x) = 0, where f1 (x), f2 (x), . . . , fm (x) are polynomials of singe variable x. • If all of f1 (x), f2 (x), . . . , fm (x) are zero then X = A1 . • If f1 (x), f2 (x), . . . , fm (x) are relatively prime then there is no common root, hence X = ∅. • If gcd( f1 (x), f2 (x), . . . , fm (x)) = d(x) then d(x) = (x − α1 )(x − α2 ) . . . (x − αm ) and X = {α1 , α2 , . . . αm }, i.e., it consists of finite numbers of points. Proposition C.1. The union of two algebraic sets is an algebraic set. The intersection of any family of algebraic sets is an algebraic set. The empty set and the whole space are algebraic sets. We define the Zariski topology on An by taking the open subsets to be the complements of the algebraic sets. This is a topology, because according to the proposition, the intersection of two open sets is open, and the union of any family of open sets is open. Furthermore, the empty set and the whole space are both open. Example C.5. The Zariski topology in An is the smallest topology in An for which any polynomial f : An → A function is continuous. Note that a continuous function f : An → A need not be polynomial. For example, any one-to-one function f (x1 ) : A → A is continuous but it may be not polynomial. In general, a function f : A → A is continuous, if the coimage f −1 (y) = {x ∈ A | f (x) = y} is finite for any y ∈ A. Example C.6. Note that a polynomial map F defined by F = ( f1 (x), f2 (x), . . . , fm (x)) : An → Am is continuous, where fi ∈ A = F[x] and x = (x1 , x2 , ..., xn ). Indeed, if Y = Z(gi (y1 , y2 , . . . , ym ), i = 1, 2, . . . , k)

Zariski Topology  275

is a closed subset in Am then F −1 (Y) = {x = (x1 , x2 , . . . , xn ) ∈ An | F(x) ∈ Y} = {x = (x1 , x2 , . . . , xn ) ∈ An | gi ( f1 (x), . . . , fm (x)) = 0, i = 1, 2, . . . , k}. Definition C.2. A nonempty subset Y of a topological space X is irreducible if it cannot be expressed as a union Y = Y1 ∪ Y2 of its two proper closed subsets Y1 and Y2 . The empty set is not considered to be irreducible. Note that this definition is equivalent to each of the following: (i) Every nonempty open subset of Y is dense. (ii) Any pair of nonempty open subsets of Y intersect. Example C.7. A1 is irreducible, because its proper closed subsets are only finite sets, however A1 itself is infinite (because F is algebraically closed, hence infinite). Example C.8. Any nonempty open subset of an irreducible space is irreducible and dense. Proposition C.2. If Y is an irreducible subset of X, then its closure in X is also irreducible. Definition C.3. An affine algebraic variety or simply affine variety is an irreducible closed subset of An (with the induced topology). Now we need to explore the relationship between subsets of An and ideals in A more deeply. So for any subset Y ⊂ An , let us define the ideal of Y in A by I(Y) = { f ∈ A | f (P) = 0 for all P ∈ Y}. Now we have a function Z that maps subsets A to algebraic sets, and a function I which maps subsets of An to ideals. Their properties are summarized in the following proposition. Proposition C.3. a) If T 1 ⊆ T 2 are subsets of A, then Z(T 1 ) ⊇ Z(T 2 ). b) If Y1 ⊆ Y2 are subsets of An , then I(Y1 ) ⊇ I(Y2 ).

276  Leibniz Algebras

S n c) For any two subsets Y , Y of A , we have I(Y Y2 ) = 1 2 1 T I(Y1 ) I(Y2 ). √ √ d) For any ideal J ⊆ A one has I(Z(J)) = J, where J is the radical of J. e) For any subset Y ⊆ An we have Z(I(Y)) = Y, where Y the closure of Y. Theorem C.1. (Hilbert’s Nullstellensatz). Let F be an algebraically closed field, J be an ideal in A = F[x1 , . . . , xn ], and f ∈ A be a polynomial which vanishes at all points of Z(J). Then f r ∈ J for some integer r > 0. Proof. See Atiyah-Macdonald [22] or Zariski-Samuel [184].



Corollary C.1. There is a one-to-one inclusion-reversing correspondence between algebraic sets in An and radical ideals (i.e., ideals which are equal to their own radical) in A, given by Y 7−→ I(Y) and J 7−→ Z(J). Furthermore, an algebraic set is irreducible if and only if its ideal is prime. Example C.9. Let f be an irreducible polynomial in A = F[x, y]. Then f generates a prime ideal in A, since A is a unique factorization domain, so the zero set Y = Z( f ) is irreducible. It is called the affine curve defined by the equation f (x, y) = 0. If f has degree d, then Y is said to be a curve of degree d. Example C.10. More generally, if f is an irreducible polynomial in A = F[x1 , . . . , xn ], we obtain an affine variety Y = Z( f ), which is called a surface if n = 3, or a hypersurface if n > 3. Example C.11. A maximal ideal m of A = F[x1 , . . . , xn ] corresponds to a minimal irreducible closed subset of An , which must be a point, say P = (a1 , . . . , an ). This shows that every maximal ideal of A is of the form m = (x1 − a1 , . . . , xn − an ) , for some a1 , . . . , an ∈ F. Example C.12. If F is not algebraically closed, these results do not hold. For example, if F = R, the curve x2 + y2 + 1 = 0 in A2 has no points. Definition C.4. If Y ⊆ An is an affine algebraic set, we define the affine coordinate ring K[Y] (sometime A(Y)) of Y to be A/I(Y).

Zariski Topology  277

Remark. If Y is an affine variety, then A(Y) is an integral domain. Furthermore, A(Y) is a finitely generated F-algebra. Conversely, any finitely generated F-algebra B which is a domain is the affine coordinate ring of some affine variety. Indeed, write B as the quotient of a polynomial ring A = F[x1 , . . . , xn ] by an ideal J, and let Y = Z(J). Example C.13. • If X is a singleton then A(X) = F. • If X = An then I(X) = 0 and A(X) = F[x1 , . . . , xn ]. • Let X = {(x1 , x2 ) ∈ F × F | x1 x2 = 1}. Then



A(X) = F x1 , x1

−1



( =

) f (x1 ) m ≥ 0 and f (x1 ) ∈ F[x1 ] . x1 m

Remark. • The ring A(X) for an algebraic set X being an image of a noetherian ring is noetherian. • In A(X) the following version of Hilbert’s Nullstellensatz also holds, i.e., if a function f (x) ∈ A(X) vanishes on those points, where functions g1 (x), g2 (x), . . . , gm (x) vanish then f r ∈ (g1 , g2 , . . . , gm ) for some r > 0. Definition C.5. The function ϕ : X → Y from an affine variety X into an affine variety Y is said to be a morphism (or regular mapping) if f ◦ ϕ is a regular function on Y for any f ∈ A(X). Example C.14. Let F1 , F2 , . . . , Fm be polynomials in F[T 1 , . . . , T n ] then ϕ : An → Am defined by ϕ(x) = ϕ(x1 , . . . , xn ) = (F1 (x), F2 (x) . . . , Fm (x)) is a morphism.

278  Leibniz Algebras

Example C.15. Let X = Z(y2 − x3 + 1) be affine algebraic variety in A2 with the coordinate system (x, y) and Y = Z((t3 − s2 + 1), (r − s2 )) be affine algebraic variety in A3 with the coordinate system (s, t, r). Then the function ϕ defined by ϕ(x, y) = (x, y, x2 ) is a morphism from X into Y. Indeed, if f (s, t, r) = t3 − s2 + 1 then ( f ◦ ϕ)(x, y) = f (x, y, x2 ) = y3 − x2 +1 is identically zero on X. If f (s, t, r) = r− s2 then ( f ◦ϕ)(x, y) = f (x, y, x2 ) = x2 − x2 is identically zero on X. Example C.16. Let X = Z(y2 − x3 + 1) be affine algebraic variety in A2 with the coordinate system (x, y) and Y = Z((t3 − s2 + 1), (r − s2 )) be affine algebraic variety in A3 with the coordinate system (s, t, r). Then the function ϕ defined by ϕ(x, y) = (x, y, x2 ) is a morphism from X into Y. Definition C.6. The morphism ϕ : X → Y is called isomorphism if there exists a morphism from Y into X such that if f ◦ g = idY and g ◦ f = idX . Example C.17. The parabola X = {(x, y) ∈ F × F | y = xk }, where k is a fixed natural number, is isomorphic to the straight line Y = A, corresponding morphisms are f (x, y) = x and g(t) = (t, tk ). Indeed, ( f ◦ g)(t) = f (g(t)) = f (t, tk ) = t for any t ∈ Y and (g ◦ f )(x, y) = g( f (x, y)) = g(x) = (x, xk ) = (x, y) for any (x, y) ∈ X. Example C.18. The mapping f (t) = (t2 , t3 ) of the straight line Y = A to the curve X = {(x, y) ∈ F × F | x3 = y2 } is one to one. However, it is not isomorphism since the inverse mapping g(x, y) = yx is not regular at the origin. Example C.19. The set GL(n, F) = {g = (gi j )i, j=1,2,...,n | det(g) , 0} can be regarded as an affine variety as follows: GL(n, F) is identified with X = {(g, t) = ((gi j )i, j=1,2,...,n , t) ∈ Fn +1 | det(g)t − 1 = 0}. 2

Note that in this case the product (g1 , t1 )(g2 , t2 ) = (g1 g2 , t1 t2 ) map from X × X to X and the inverse (g, t)−1 = (g−1 , t−1 ) map from X to X × X are morphisms.

Zariski Topology  279

C.1

ACTION OF A GROUP

Let G be a group and X be a nonempty set. Definition C.7. An action of the group G on X is a function σ:G×X →X with: i) σ(e, x) = x, where e is the unit element of G and ii) σ(g, σ(h, x)) = σ(gh, x), for any g, x ∈ G and x ∈ X. We briefly write gx for σ(g, x), and call X a G-set. Let F be a field and F[X] = { f : X → F} be the set of all functions on X. It is an algebra over F with respect to point wise addition, multiplication and multiplication by scalar operations. Definition C.8. A function f : X → F is said to be invariant if f (gx) = f (x) for any g ∈ G and x ∈ X. The set of invariant functions on X, denoted by F[X]G , is a subalgebra of F[X]. The set Orb(x) = {y ∈ X | there exists g ∈ G such that y = gx} is called the orbit of the element x under the action of G. Theorem C.2. Let X be a G-set. Define a relation ∼ on X for x, y ∈ X as follows: x ∼ y if and only if σ(g, x) = y for some g ∈ G. Then ∼ is an equivalence relation on X. It is evident that the equivalence classes with respect to equivalence relation ∼ are the orbits under the action of the group G. The invariant functions are functions on X/ ∼. C.2 ALGEBRAIC GROUPS

Definition C.9. • An algebraic group is an affine variety G equipped with morphisms of varieties µ : G × G −→ G, ı : G −→ G that give G the structure of a group.

280  Leibniz Algebras

• A morphism f : G −→ H of algebraic groups is a morphism of varieties that is also a group homomorphism. Proposition C.4. • The kernel of a morphism f : G −→ H of algebraic groups is a closed subgroup of G, so it is an algebraic group in its own right. The same will turn out to be true about the images. • Translation by an element g ∈ G is an isomorphism of varieties, so all geometric properties at one point can be transferred to any other point. For example, as G has simple points, G is smooth. Example C.20. (i) The additive group Ga is the group (F, +), i.e., the affine variety A1 under addition; (ii) The multiplicative group Gm is the group (F∗ , ×), i.e., the principal open subset A1 \ {0} under multiplication; (iii) The group GLn = GLn (F) is the group of all invertible n × n matrices over F. As a variety, this is a principal open set in 2 Mn (F) = An corresponding to the determinant. Since the formulas for matrix multiplication and inversion involve only polynomials in the matrix entries and 1/det, the group structure maps are morphisms of varieties. Example C.21. (iv) Let V be an n-dimensional vector space over F. Then by fixing a basis we can define a structure of an algebraic group on GL(V) which is independent of the choice of basis. Of course, GL(V)  GLn (F); (v) The group S Ln = S Ln (F) is the closed subgroup of GLn defined by the zeros of f (x11 , x12 , . . . , xnn ) = det(A) − 1, where A ∈ GLn (F); (vi) The group Dn of invertible diagonal matrices is a closed subgroup of GLn . It is isomorphic to the direct product Gm × · · · × Gm (n copies). Here are more examples of algebraic groups.

Zariski Topology  281

Example C.22. (vii) The group Un of upper unitriangular matrices is another closed subgroup of GLn ; (viii) The orthogonal group On (F) = {A ∈ GLn | AA> = I}, with char(F) , 2; (ix) The special orthogonal group SOn = On ∩ SLn is a normal subgroup of On of index 2; (x) The group Sp2n = {A ∈ GL2n | A> JA = J}, where  symplectic  0 In In 0 is another closed subgroup.

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Index 0-filiform (null-filiform) Leibniz algebra, 91 1-filiform (filiform) Leibniz algebra, 91 D-map, 29 k-filiform Leibniz algebra, 91 Action of group, 212 Adapted basis, 170 Algebra of derivations, 30 Associative dialgebra, 34 Cartan subalgebra, 24, 67, 68 Cartan’s criterion of solvability, 74 Cartan-Killing Criterion, 20 Catalan number, 225 Center, 42 Characteristic sequence, 160 Chevalley-Eilenberg chain complex, 31 Classification of associative algebras, 10 Classification of Lie algebras, 12, 79 Classification procedure, 212 Conjugacy of Levi subalgebras, 48 Contraction, 128 Criterion on conjugacy of the Levi subalgebra, 58 Degeneration, 125–127, 147 Dendriform algebra, 30, 38

Derivation, 30 Derivation algebra, 41 Dialgebra, 36 Diassociative algebra, 30, 34 Engel’s theorem for Leibniz algebras, 44 Enveloping algebra, 36 Filiform Leibniz algebras, 195 Fitting component, 26 Fitting’s lemma, 25 Four-dimensional nilpotent Leibniz algebras, 95 Four-dimensional solvable Leibniz algebras, 104 Geometric classification, 127 Improper degeneration, 128 Inner automorphism, 48 Invariance argument, 147 Invariant form, 77 Invariant function, 212, 215 Irreducible component, 153, 157, 164 Isomorphism class, 246 Isomorphism classes of three-dimensional complex Leibniz algebras, 90 Isomorphism criterion, 210, 211, 213, 215, 223, 234, 246 Jacobi identity, 31 301

302  Index

Killing form, 77 Left annihilator, 42 Left Leibniz algebra identity, 27 Left normalizer, 67 Leibniz algebra, 27 Leibniz algebra identity, 31 Leibniz cohomology, 129 Leibniz module, 33 Levi subalgebra, 47 Levi’s theorem, 22 Levi’s theorem for Leibniz algebras, 47 Levi-Malcev theorem, 32 Lie algebra, 12 Liezator, 41 Linear algebra, 251 Loday algebras, 26 Loday diagram and Koszul duality, 40 Low-dimensional complex Leibniz algebra, 80 Low-dimensional nilpotent Leibniz algebras, 91 Maximal toral subalgebra, 25 Naturally graded algebra, 195 Naturally graded filiform Leibniz algebra, 164 Nil-independence, 105 Nilpotent Leibniz algebra, 43 Nilradical, 45, 108 Null-filiform Leibniz algebra, 92 Opposite Leibniz algebra, 27 Orbit’s closure, 125 Poincar´e-Birkhoff-Witt theorem, 36

Radical, 47 Rank of Leibniz algebra, 69 Reductive Lie algebra, 21 Regular element of Leibniz algebra, 69 Representation theory, 269 Richardson’s theorem, 135 Right annihilator, 42 Right Leibniz algebra identity, 27 Right multiplication operator, 41 Right normalizer, 67 Rigid algebra, 126 Rigid family of algebra, 126 Rigid nilpotent Leibniz algebras in dimension four, 153 Rigidity, 126, 147 Rigidity of Lie and Leibniz algebras, 124 Ring of polynomials, 1, 94 Roots, 23 Rota-Baxter algebra, 39 Schur’s Lemma, 22 Semisimple, 32 Semisimple associative algebra, 12 Semisimple Leibniz algebra, 60 Simple associative algebra, 12 Simple Leibniz algebra, 34, 60 Single orbit, 243 Solvable Leibniz algebra, 43, 105 Table of multiplication, 11, 43, 245 Three-dimensional Leibniz algebras, 81 Three-dimensional nilpotent Leibniz algebras, 106

Index  303

Three-dimensional rigid Leibniz algebras, 152 Toral Lie algebra, 25 Trivial degeneration, 128 Two-dimensional Leibniz algebras, 80

Variety of algebras, 126 Variety of Leibniz algebras, 150

Universal enveloping algebra, 37

Zariski topology, 273 Zinbiel algebra, 30, 37

Wedderburn-Artin theorem, 12 Weight, 23 Weyl’s theorem, 21