Classification of Leibniz algebras with a given nilradical and with some corresponding Lie algebra

Classification of Leibniz algebras with a given nilradical and with some corresponding Lie algebra IKBOLJON KARIMJANOV 2017 Classification of Leibniz algebras with a given nilradical and with some corresponding Lie algebra by IKBOLJON KARIMJANOV DOCTORAL DISSERTATION Submitted for the degree of DOCTOR EN MATEMÁTICAS en la UNIVERSIDAD DE SANTIAGO DE COMPOSTELA Santiago de Compostela, 2017 Classification of Leibniz algebras with a given nilradical and with some corresponding Lie algebra Fdo.: Ikboljon Karimjanov Memoria para optar al grado de Doctor realizada en el Departamento de Matemáticas en el Programa de Doctorado de Matemáticas, de la Facultad de Matemáticas de la Universidad de Santiago de Compostela, bajo la dirección de los Profesores Dr. Bakhrom Omirov y Dr. Manuel Ladra González. Santiago de Compostela, a 27 de abril de 2017. Fdo.: Bakhrom Omirov Fdo.: Manuel Ladra González Classification of Leibniz algebras with a given nilradical and with some corresponding Lie algebra AUTORIZACIÓN DEL DIRECTOR/TUTOR DE LA TESIS Dr. Bakhrom Omirov, Profesor del Institute of Mathematics, National University of Uzbekistan (Uzbekistán) y Dr. Manuel Ladra González, Profesor del Departamento de Matemáticas de la Universidad de Santiago de Compostela, como Directores de la Tesis Doctoral titulada Classification of Leibniz algebras with a given nilradical and with some corresponding Lie algebra, presentada por D. Ikboljon Karimjanov, alumno del programa de Doctorado Matemáticas, AUTORIZAMOS la presentación de la Tesis Doctoral indicada para optar al grado de Doctor por la Universidad de Santiago de Compostela, considerando que reúne los requisitos exigidos en el artículo 34 del reglamento de Estudos de Doutoramento, y que como Directores de la misma no incurre en causas de abstención establecidas en la ley 40/2015. Santiago de Compostela, a 27 de abril de 2017. Fdo.: Bakhrom Omirov Fdo.: Manuel Ladra González A maioría dos resultados presentados nesta memoria foron obtidos grazas ao financiamento da Consellería de Cultura, Educación e Ordenación Universitaria da Xunta de Galicia, na modalidade de Grupo de Referencia Competitiva, referencia GRC2013-045, incluído cofinanciamento do Fondo Europeo de Desenvolvemento Rexional (FEDER), (DOG 25/10/2013). Tamén queremos agradecer as axudas dos proxectos MTM2013-43687-P e MTM2016-79661-P (incluído cofinanciamento do FEDER) do Ministerio de Economía y Competitividad (España) e Agencia Estatal de Investigación. Unión Europea Fondo Europeo de Desarrollo Regional GRC2013-045 (Xunta de Galicia) con fondos FEDER Unha maneira de facer EUROPA la propiedad de derivación definida sobre las cadenas. Esto le motivó a introducir la noción de álgebra de Leibniz por la derecha (equivalentemente, por la izquierda), la cual es un álgebra no asociativa en la que el operador de la multiplicación por la derecha (equivalentemente, por la izquierda) es una derivación. Las álgebras de Leibniz generalizan álgebras de Lie de manera natural. Es bien sabido que existen tres tipos diferentes de álgebras de Lie: semisimples, solubles, y las que no son ni semisimples ni solubles. Por lo tanto, determinar la clasificación de las álgebras de Lie, en general, equivale a revelar la clasificación de cada uno de estos tres tipos. Sin embargo, se pueden condensar en dos por el Teorema de Levi-Malcev, que es una combinación de los resultados formulados en primer lugar por Levi [34] en 1905, y más tarde por Malcev [41] en 1945: cualquier álgebra de Lie de dimensión finita sobre un cuerpo de característica cero se puede expresar como una suma semidirecta (la descomposición de Levi-Malcev) de una subálgebra semisimple (llamada el factor de Levi) y su radical (su ideal soluble maximal). Esto reduce la tarea de clasificar todas las álgebras de Lie restringiendo la clasificación a las álgebras de Lie semisimples y solubles. La clasificación de las álgebras de Lie semisimples fue resuelta completamente por el bien conocido Teorema de Cartan: cualquier álgebra de Lie compleja o real semisimple se puede descomponer en una suma directa de ideales que son subálgebras simples las cuales son mutuamente ortogonales con respecto a la forma de Cartan-Killing. Así, el problema de clasificar las álgebras de Lie semisimples es equivalente a clasificar todas las álgebras de Lie simples no isomorfas; y la clasificación de las álgebras de Lie simples ya fue obtenida por Killing, Cartan y otros en la última década del siglo XIX (véase [25]). Por lo tanto, puede admitirse que el problema de la clasificación de las álgebras de Lie semisimples está totalmente resuelto en la actualidad. De hecho, principalmente Killing y Cartan, aunque otros autores también trabajaron en este tema, clasificaron las álgebras de Lie simples en cinco clases diferentes (las denominadas álgebras de Lie clásicas simples): las álgebras pertenecientes al grupo especial lineal, las álgebras ortogonales impares, las álgebras ortogonales pares, las álgebras simplécticas, más cinco álgebras de Lie que no tienen relación entre ellas y que no pertenecen a ninguna de las clases anteriores. Con respecto a la clasificación de las álgebras de Lie solubles, a pesar de los primeros intentos hecho por Lie [35,36] y por Bianchi [12], puede decirse que fue Dozias, en 1963, una de los primeros autores que se enfrentaron a este problema en serio: ella clasificó en su tesis doctoral las álgebras de Lie solubles xvi de dimensiones inferiores a 6 sobre el cuerpo de los números reales [28]. En ese mismo año Mubarakzjanov (véase [42–44] y [48]) también clasificó estas álgebras hasta dimensión 6 sobre el cuerpo de los números reales. Debido parcialmente a las dificultades para obtener una clasificación completa de las álgebras de Lie solubles, algunos autores consideraron la idea de una clasificación de extensiones solubles de ciertas clases de álgebras de Lie. En particular, las más relevantes fueron el análisis de todas las álgebras solubles, no nilpotentes, con un nilradical dado. Así, Rubin y Winternitz iniciaron una línea de investigación ( [4,5,31,45,50–55,57]) concerniente con la clasificación de las álgebras de Lie solubles con un nilradical dado, como las álgebras de Lie filiformes, las álgebras abelianas (también llamadas de 1-paso), las álgebras de Heisenberg, las álgebras de matrices triangulares estrictamente superiores, y así sucesivamente (para dimensiones arbitrariamente finitas). La investigación de las álgebras de Lie solubles con algunos tipos especiales de nilradical proviene de diferentes problemas de la física y fueron objeto de varios artículos [3,5,16,24,27], y muchas otras referencias dadas en esos trabajos. Malcev [41] ya había reducido en 1945 la clasificación de las álgebras de Lie complejas solubles a la clasificación de un subconjunto, las álgebras de Lie nilpotentes. Para ello, Malcev definió un tipo particular de álgebra, que llamó el álgebra escindida, cuya estructura está completamente determinada a partir de su ideal nilpotente maximal (llamado nilradical) y demostró que un álgebra de Lie soluble arbitraria está contenida en una única álgebra escindida minimal. La relación entre un álgebra y sus escisiones le llevó a la construcción de todas las álgebras de Lie solubles con una escisión dada. Mubarakzjanov demostró que la dimensión del álgebra escindida no excede del número de derivaciones nilindependientes del nilradical [42]. Así, de esta manera, la clasificación de todas las álgebras de Lie solubles se había reducido a la clasificación de las álgebras de Lie nilpotentes. Se han realizado muchos progresos en las clasificaciones de las álgebras de Lie nilpotentes. Con respecto a la clasificación de las álgebras de Lie nilpotentes, se han hecho muchos intentos en este tema, y se han publicado muchas listas de álgebras con mayor o menor fortuna. Durante los últimos 25 años se ha investigado activamente en la teoría de álgebras de Leibniz y se han dedicado numerosos trabajos al estudio de estas álgebras. Ayupov y Omirov clasificaron las álgebras de Leibniz complejas de dimensión 3 en 1999 [8]. Luego, comenzaron a investigar las álgebras de Leibniz nilpotentes. En la actualidad están clasificadas las álgebras de Leibniz nilpotentes cuya dimensión es menor que cinco. La clasificación de las álgebras xvii de Leibniz complejas nilpotentes de dimensión finita ya es un problema complicado. Debido a la falta de antisimetría, el problema de clasificar las álgebras de Leibniz complejas nilpotentes es más difícil. Se han realizado muchos progresos en el estudio de otras clasificaciones relativas a algunas propiedades particulares de las álgebras de Leibniz nilpotentes. Omirov consideró las álgebras de Leibniz nilpotentes graduadas. En [9], los autores clasificaron las álgebras de Leibniz nulo-filiformes y filiformes naturalmente graduadas. Después, muchos trabajos estuvieron dedicados al estudio de la sucesión característica de las álgebras de Leibniz nilpotentes [19–23]. Las álgebras de Leibniz simples y semisimples fueron definidas por Dzhumadil’daev [1]. En el trabajo [30], los autores investigaron las álgebras de Leibniz semisimples y demostraron que el teorema de escisión (Teorema de Cartan) para álgebras de Leibniz semisimples no es cierto en el caso general. De hecho, muchos resultados de la teoría de álgebras de Lie se han extendido al caso de álgebras de Leibniz. Por ejemplo, los resultados clásicos sobre subálgebras de Cartan [2, 46] y el teorema de Engel [7] se han determinado para el caso de álgebras de Leibniz. El análogo de la descomposición de Levi-Malcev para álgebras de Leibniz fue probado por D.W. Barnes [11], que afirma que cualquier álgebra de Leibniz se descompone en una suma semidirecta de su radical soluble y un álgebra de Lie semisimple. La parte semisimple se puede describir a partir de los ideales de Lie simples, por lo tanto, el problema principal de la descripción de las álgebras de Leibniz de dimensión finita consiste en el estudio de las álgebras de Leibniz solubles. Nuestro objetivo es probar el teorema de Mubarakzjanov para el caso de álgebras de Leibniz. En el siguiente teorema se extiende dicha afirmación para el caso de Leibniz. Teorema 1.2.2. Sea Run álgebra de Leibniz soluble y Nsu nilradical. Entonces la dimensión del espacio vectorial complementario a Nno es más grande que el número maximal de derivaciones nil-independientes de N. Usando este método, en el siguiente teorema se clasifican las álgebras de Leibniz solubles con nilradical nulo-filiforme. Teorema 1.2.6. Sea Run álgebra de Leibniz soluble cuyo nilradical es NFn. Entonces existe una base {e1, e2, . . . , en, x}del álgebra Rtal que la tabla de xviii multiplicación de Rcon respecto a esta base tiene la siguiente forma:      [ei, e1] = ei+1,1≤i≤n−1, [x, e1] = e1, [ei, x] = −iei,1≤i≤n. Este fue el primer paso para la investigación de las álgebras de Leibniz solubles con nilradical dado. Además, esta clasificación se extiende al caso en que el nilradical sea una suma directa de ideales nulo-filiformes y el espacio vectorial complementario del nilradical sea de dimensión uno. En la Sección 1.3 se investigan las álgebras de Leibniz solubles cuyo nilradical es el álgebra de Lie de matrices triangulares superiores. Dado que en el trabajo [55] se estudian las álgebras de Lie solubles con nilradical triangular, nosotros reducimos nuestro estudio a las álgebras de Leibniz que no son de Lie. En el siguiente corolario, se presentan algunas propiedades de la matriz de los operadores de las multiplicaciones por la derecha e izquierda para las álgebras de Leibniz solubles de dimensiones mínimas posibles con nilradical triangular. Corolario 1.3.3. Para un álgebra de Leibniz del conjunto L(n, 1), las matrices de los operadores de las multiplicaciones por la izquierda y por la derecha, A= (aij,pq)yB= (bij,pq), tienen las siguientes propiedades: (1) El número máximo de elementos fuera de la diagonal de la matriz Aes n−1; (2) El número máximo de elementos fuera de la diagonal de la matriz Bes n+ 1. En el Teorema 1.3.4 se demuestra que las álgebras de Leibniz solubles de dimensiones máximas posibles con nilradical triangular son álgebras de Lie. Además, se establece la clasificación de las álgebras de Leibniz solubles de dimensiones bajas con nilradicales triangulares. Teorema 1.3.4. Un álgebra de Leibniz soluble del conjunto L(n, n −1) es un álgebra de Lie. En el Capítulo 2 consideramos las álgebras de Leibniz solubles con nilradical filiforme naturalmente graduado. En los trabajos [4,52] se estudian las álgebras de Lie solubles con nilradicales filiformes naturalmente graduados. Se establece xix que la dimensión del espacio vectorial complementario es igual a 1 o 2. En los teoremas siguientes se clasifican las álgebras de Leibniz, que no son de Lie, con nilradicales nn,1yQ2n, y cuyo espacio vectorial complementario tiene dimensión 1. Teorema 2.1.5. Cualquier álgebra de Leibniz soluble de dimensión (n+1) con nilradical nn,1es isomorfa a una de las siguientes álgebras no isomorfas entre ellas: Rn+1,1(0,0,1), Rn+1,1(0,1,0), Rn+1,1(1,1,0), Rn+1,1(1,0,0). Teorema 2.1.6. Cualquier álgebra de Leibniz soluble de dimensión (2n+ 1) con nilradical Q2nes isomorfa a una de las siguientes álgebras no isomorfas entre ellas: R2n+1,1(0,0,1), R2n+1,1(0,1,0), R2n+1,1(1,1,0), R2n+1,1(1,0,0). En el caso de que el espacio vectorial complementario sea de dimensión 2 se establece que no existen álgebras de Leibniz, que no sean de Lie, con nilradicales nn,1yQ2n. Es bien conocido que existen dos clases de álgebras de Leibniz filiformes naturalmente graduadas, F1 nyF2 n. En los Teoremas 2.2.2, 2.2.3, 2.3.3 y 2.3.4 se clasifican las álgebras de Leibniz solubles con nilradicales F1 nyF2 n. Teorema 2.2.2. Cualquier álgebra de Leibniz soluble de dimensión (n+1) con nilradical F1 nes isomorfa a una de las siguientes álgebras no isomorfas entre ellas: R1, R2(α), R3, R4, R5(α4, . . . , αn). Además, el primer parámetro que no se anula {α4, . . . , αn}en las álgebras R5(α4, . . . , αn), puede ajustar la escala a 1. Teorema 2.2.3. Cualquier álgebra de Leibniz soluble de dimensión (n+2) con nilradical F1 nes isomorfa a un álgebra con la siguiente tabla de multiplicación: [ei, e1] = ei+1,2≤i≤n−1,[e1, x] = e1, [ei, y] = ei,2≤i≤n, [ei, x]=(i−1)ei,2≤i≤n, [x, e1] = −e1. xx Teorema 2.3.3. Cualquier álgebra de Leibniz soluble de dimensión (n+1) con nilradical F2 nes isomorfa a una de las siguientes álgebras no isomorfas entre ellas: L1(α), L2(α), L3, L4(α), L5(α), L6(α3, α4, . . . , αn, λ, δ). En el álgebra L6(α3, α4, . . . , αn, λ, δ)el primer parámetro que no se anula {α3, α4, . . . , αn, λ}puede ajustar la escala a 1. Teorema 2.3.4. Cualquier álgebra de Leibniz soluble de dimensión (n+2) con nilradical F2 nes isomorfa a una de las siguientes álgebras no isomorfas entre ellas: L1:     [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = e1,[x, e1] = −e1, [e2, y] = −[y, e2] = e2,[ei, x] = (i−1)ei,3≤i≤n, L2:     [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = e1,[x, e1] = −e1, [e2, y] = e2,[ei, x]=(i−1)ei,3≤i≤n. El último capítulo de la tesis se dedica a la investigación de las álgebras de Leibniz correspondientes a las álgebras de Lie de tipo diamante. Toda álgebra de Leibniz Lque no sea de Lie contiene un ideal no trivial (a partir de ahora denotado por I), que es el subespacio generado por los cuadrados de los elementos del álgebra L. Además, este ideal está contenido en el anulador por la derecha de L, esto es [L, I]=0. Obsérvese también que el ideal Ies el ideal minimal con la propiedad de que el álgebra cociente L/I es un álgebra de Lie; esta álgebra cociente se llama la liezación del álgebra de Leibniz Ly se dice que es el álgebra de Lie correspondiente al álgebra de Leibniz L. Un álgebra de Leibniz Lcon álgebra de Lie correspondiente de tipo diamante D=L/I se puede descomponer en una suma directa de espacios vectoriales L=D⊕I, donde Des la preimagen de Dpor el homomorfismo natural ϕ:L→D. Claramente, el Iideal puede considerarse como un D-módulo de Leibniz. Teniendo en cuenta que el ideal Iestá contenido en el anulador por la derecha del álgebra L, las multiplicaciones en Lse determinan a partir de los productos [D,D]y[I, D]. Dado que Ies un módulo de Leibniz sobre xxi el álgebra de Lie D, se tiene que el producto [I, D]corresponde a un elegido D-módulo de Lie a la derecha. Así, el principal problema de la descripción de las álgebras de Leibniz con álgebra de Lie correspondiente Dy con el ideal Ielegido por un D-módulo a la derecha específico consiste en identificar el producto [D,D]. Para un álgebra de Lie dada de tipo diamante Dde dimensión 4 construimos el así llamado módulo Fock sobre D, el espacio lineal F[x]de polinomios sobre x(Fdenota un cuerpo algebraicamente cerrado de característica cero) con la siguiente acción. Definición 3.2.1. El espacio lineal F[x]es llamado el D-módulo Fock si existe una acción (F[x],D)7→ F[x], la cual verifica lo siguiente: (p(x),1) 7→ p(x), (p(x), x)7→ xp(x), (p(x),∂ ∂x )7→ ∂ ∂x (p(x)), (p(x), e)7→ −x∂(p(x)) ∂x . para cualquier p(x)∈F[x]. En el siguiente teorema, se clasifican las álgebras de Leibniz de dimensión infinita correspondientes al módulo de Fock de las álgebras de Lie de tipo diamante. Teorema 3.2.2. El álgebra de Leibniz Lcon condiciones L/I ∼ =D, e Ies el DC-módulo de Fock, admite una base {1, x, ∂ ∂x, e, xt|t∈N∪{0}} tal que la tabla de multiplicaciones en esta base tiene la siguiente forma: [e, x] = x, [x, e] = −x, [e, ∂ ∂x ] = −∂ ∂x ,[∂ ∂x , e] = ∂ ∂x , [x, ∂ ∂x ] = 1,[∂ ∂x , x] = −1, [xt,1] = xt,[xt, x] = xt+1, [xt,∂ ∂x ] = txt−1,[xt, e] = −txt, donde los productos omitidos son iguales a cero. xxii En la Sección 3.3 estudiamos las álgebras de Leibniz cuya álgebra de Lie correspondiente es el álgebra de Lie de tipo diamante Dde dimensión cuatro y el ideal Ies una de sus representaciones de Leibniz indescomponibles de Dde dimensión finita, las cuales están descritas en [26]. En este artículo hay cuatro D-módulos indescomponibles: U1 n, U2 n, W1 nyW2 n. Teorema 3.3.1. Sea un álgebra de Leibniz compleja arbitraria con el álgebra de Lie correspondiente DC, y el ideal Iasociado definido como U1 nDC-módulo. Entonces admite una base {J, P+, P−, T, v0 0, v0 2k, v1 2k−1, v2 0, v2 2k}k=1,...,n/2, donde nes par, y la tabla de multiplicación [DC,DC]tiene la siguiente forma: •n= 4s          [J, P+] = −iP+,[J, P−] = iP−,[P+, P−] = −2iT, [P+, J] = iP+,[P−, J] = −iP−,[P−, P+] = 2iT + 2α1v2 2s, [J, T] = α1v2 2s[J, J] = α2v2 2s,[P+, P+] = α3v2 2s−2, [P−, P−] = α4v2 2s+2. •n= 4s−2                          [J, P+] = −iP+,[P+, J] = iP++ 2isβ1v2 2s−2, [J, P−] = iP−,[P−, J] = −iP−−2isβ1v2 2s, [P+, P−] = −2iT, [P−, P+] = 2iT + 2β2v1 2s−1, [J, J] = β1v1 2s−1,[J, T] = β2v1 2s−1, [P+, P+] = β3v1 2s−3,[P−, P−] = β4v1 2s+1, [P+, T] = 2isβ2v2 2s−2,[T, P+] = −i(2sβ2−(s−1)β3)v2 2s−2, [P−, T] = −2isβ2v2 2s,[T, P−] = i(4sβ2−(s−1)β4)v2 2s. donde αi, βi∈C,1≤i≤4. Teorema 3.3.2. Sea un álgebra de Leibniz compleja arbitraria con el álgebra de Lie correspondiente DC, y el ideal Iasociado definido como U2 nDC-módulo de Leibniz. Entonces admite una base {J, P+, P−, T, v0 2k−1, v1 0, v1 2k, v2 2k−1}k=1,...,n/2, donde nes par, y la tabla de multiplicación [DC,DC]tiene la siguiente forma: xxiii •n= 4s                                                                                 [J, P+] = −iP+, [P+, J] = iP++i(2s+ 1)γ1v2 2s−1, [J, P−] = iP−, [P−, J] = −iP−−i(2s+ 1)γ1v2 2s+1, [P+, P−] = −2iT, [P−, P+] = 2iT + 2γ2v1 2s, [J, J] = γ1v1 2s, [J, T] = γ2v1 2s, [P+, P+] = γ3v1 2s−2, [P−, P−] = γ4v1 2s+2, [P+, T] = i(2s+ 1)γ2v2 2s−1, [T, P+] = −i((2s+ 1)γ2−(2s−1)γ3 2)v2 2s−1, [P−, T] = −i(2s+ 1)γ2v2 2s+1, [T, P−] = i(2(2s+ 1)γ2−(2s−1)γ4 2)v2 2s+1, •n= 4s−2          [J, P+] = −iP+,[J, P−] = iP−,[P+, P−] = −2iT, [P+, J] = iP+,[P−, J] = −iP−,[P−, P+] = 2iT + 2δ1v2 2s−1, [J, T] = δ1v2 2s−1[J, J] = δ2v2 2s−1,[P+, P+] = δ3v2 2s−3, [P−, P−] = δ4v2 2s+1, donde γi, δi∈C,1≤i≤4. Teorema 3.3.4. Sea Lun álgebra de Leibniz compleja con el álgebra de Lie correspondiente de tipo diamante DC, y el ideal Iasociado definido como un DC-módulo de Leibniz por las representaciones indescomponibles W1 noW2 n. xxiv Entonces [DC,DC]tiene la siguiente forma: [J, P+] = −[P+, J] = −iP+, [J, P−] = −[P−, J] = iP−, [P+, P−] = −[P−, P+] = −2iT. Según el teorema de Ado, dada cualquier álgebra de Lie compleja gde dimensión finita, existe un álgebra de matrices isomorfa a g. De esta manera, toda álgebra de Lie compleja de dimensión finita puede representarse como una subálgebra de Lie del álgebra lineal general compleja gl(n, C), formada por todas las matrices complejas n×n, para algún n∈N. Nosotros consideramos el siguiente invariante valorado entero de g: µ(g) = min{dim(M)|Mes un g-módulo fiel}. Se deduce de la demostración del teorema de Ado que µ(g)puede acotarse por una función que depende solo de n. Este valor también es igual al valor minimal ntal que gl(C, n)contiene una subálgebra isomorfa a g: bµ(g) = min{n∈N| ∃ subálgebra de gl(C, n)isomorfa a g}. Dada un álgebra de Lie g, una representación de gin Cnes un homomorfismo de álgebras de Lie f:g→gl(Cn) = gl(C, n). El entero natural nse llama la dimensión de esta representación. Consideramos representaciones fieles porque tales representaciones nos permiten identificar un álgebra de Lie dada con su imagen por la representación, que es una subálgebra de Lie de gl(C, n). Las representaciones también se pueden definir usando espacios vectoriales arbitrarios Vde dimensión n(véase [29]). En tal caso, una representación sería un homomorfismo de álgebras de Lie de gal álgebra de Lie gl(V)de endomorfismos del espacio vectorial V, el cual es llamado un g-módulo. Sin embargo, basta con considerar representaciones en Cnporque siempre existe un n∈Ntal que Ves isomorfo a Cn. Muchos trabajos se dedican a encontrar el valor µ(g)para varias álgebras de Lie de dimensión finita. En [17], se encuentra el valor de µ(g)para álgebras de Lie abelianas y álgebras de Heisenberg, y además, se estima el valor de µ(g) para álgebras de Lie filiformes. xxv It is a first step for the investigation of solvable Leibniz algebras with given nilradical. Moreover, this classification is extended to the case when the nilradical is a direct sum of null-filiform ideals and the complementary vector space of the nilradical has one dimension. In Section 1.3 solvable Leibniz algebras whose nilradical is the Lie algebra of upper triangular matrices are investigated. Since in the work [55] solvable Lie algebras with triangular nilradical are studied, we reduce our study to non-Lie Leibniz algebras. In Corollary 1.3.3, some properties of the matrix of right and left multiplication operators for solvable Leibniz algebras of minimum possible dimensions with triangular nilradical are presented. In Theorem 1.3.4 it is proved that solvable Leibniz algebras of maximum possible dimensions with triangular nilradical are Lie algebras. Furthermore, the classification of the low-dimensional solvable Leibniz algebras with triangular nilradicals is established. In Chapter 2 we consider solvable Leibniz algebras with naturally graded filiform nilradical. In the works [4, 52] solvable Lie algebras with naturally graded filiform nilradical are described. In Theorems 2.1.5–2.1.6 the non-Lie Leibniz algebras with nn,1and Q2nnilradicals whose complementary vector space has dimension 1 are classified. In the case of complementary vector space of dimension two it is established that non-Lie Leibniz algebras with nn,1 and Q2nnilradicals do not exist. It is well known that there are two classes of naturally graded filiform Leibniz algebras F1 nand F2 n. In Theorems 2.2.2, 2.2.3, 2.3.3 and 2.3.4, solvable Leibniz algebras with nilradical F1 nand F2 nare classified. Every non-Lie Leibniz algebra Lcontains a non-trivial ideal (from now on denoted by I), which is the subspace spanned by the squares of elements of the algebra L. Moreover, this ideal is contained in the right annihilator of L, that is [L, I]=0. Note also that the ideal Iis the minimal ideal with the property that the quotient algebra L/I is a Lie algebra (the quotient algebra is called liezation of the Leibniz algebra L). One of the approaches to the investigation of Leibniz algebras is a description of such algebras whose quotient algebra with respect to the ideal Iis a given Lie algebra [6,18,47,49]. The map I×L/I →Idefined as (i, x)7→ [i, x]endows Iwith a structure of L/I-module. Considering the direct sum of vector spaces Q(L):=L/I ⊕I, xxxii the operation (−,−)defines a Leibniz algebra structure on Q(L)with multiplication [x, y] = [x, y],[x, i] = [x, i],[i, x] = 0,[i, j]=0, x, y ∈L, i, j ∈I. Therefore, for a given Lie algebra gand a g-module M, we can construct a Leibniz algebra L=g⊕Mby the above construction. The last chapter of the thesis is devoted to the investigation of Leibniz algebras corresponding to Diamond Lie algebras. Actually, for a Leibniz algebra Lcorresponding to the Diamond Lie algebra D=L/I we decompose it into direct sum of vector spaces L=D⊕I, where Dis the preimage of Dunder the natural homomorphism ϕ:L→D. Clearly, the ideal Ican be considered as a Leibniz D-module. Taking into account that the ideal Iis contained in the right annihilator of the algebra L, the multiplications in Lare determined by the products [D,D]and [I, D]. Since Iis a Leibniz module over the Lie algebra D, then the product [I, D]corresponds to a chosen right Lie D-module. Thus, the main problem of the description of Leibniz algebras corresponding to the Lie algebra Dand with the ideal Ichosen by specific right D-module consists of identifying the product [D,D]. For a given four-dimensional Diamond Lie algebra Dwe construct the socalled Fock module over D, the linear space F[x]of polynomials on x(Fdenotes an algebraically closed field of characteristic zero) with the action which is introduced in Section 3.2. In Theorem 3.2.2 infinite-dimensional Leibniz algebras corresponding to the Fock module of Diamond Lie algebras are classified. In Section 3.3 we study Leibniz algebras whose corresponding Lie algebra is the four-dimensional Diamond Lie algebra Dand the ideal Iis one of its finitedimensional indecomposable Leibniz representations of Dwhich are described in the work [26]. In this paper there are four indecomposable D-modules: U1 n, U2 n, W1 nand W2 n. According Ado’s Theorem, given any finite-dimensional complex Lie algebra g, there exists a matrix algebra isomorphic to g. In this way, every finite-dimensional complex Lie algebra can be represented as a Lie subalgebra of the complex general linear algebra gl(n, C), formed by all the complex n×n matrices, for some n∈N. We consider the following integer valued invariant of g: µ(g) = min{dim(M)|Mis a faithful g-module}. xxxiii It follows from the proof of Ado’s Theorem that µ(g)can be bounded by a function depending only on n. This value is also equal to the minimal value n such that gl(C, n)contains a subalgebra isomorphic to g: bµ(g) = min{n∈N| ∃ subalgebra of gl(C, n)isomorphic to g}. Given a Lie algebra g, a representation of gin Cnis a homomorphism of Lie algebra f:g→gl(Cn) = gl(C, n). The natural integer nis called the dimension of this representation. We consider faithful representations because such representations allow us to identify a given Lie algebra with its image under the representation, which is a Lie subalgebra of gl(C, n). Representations can be also defined by using arbitrary n-dimensional vector spaces V(see [29]). In this case, a representation would be a homomorphism of Lie algebras from gto the Lie algebra gl(V)of endomorphisms of the vector space V, which is called g-module. However, it is sufficient to consider representations on Cn because there always exists a unique n∈Nsuch that Vis isomorphic to Cn. Many works are devoted to finding the value µ(g)of several finitedimensional Lie algebras. In [17] the value of µ(g)for abelian Lie algebras and Heisenberg algebras is found, moreover, the value of µ(g)for filiform Lie algebras is estimated. In Section 3.4 we find a minimal faithful representation of the (2m+ 2)- dimensional complex general Diamond Lie algebra Dm(C), which is isomorphic to a subalgebra of the special linear Lie algebra sl(m+2,C). Then we construct Leibniz algebras with corresponding general Diamond Lie algebra and the ideal generated by the squares of elements in these faithful representations. Finally, in Section 3.5 we find a faithful representation of Dmwhich is isomorphic to a subalgebra of the symplectic Lie algebra sp(2m+2,R). We also investigate the Leibniz algebras constructed by this representation of general Diamond Lie algebras. xxxiv Chapter 1 Solvable Leibniz algebras with null-filiform and triangular nilradicals In this chapter we put the first steps to describing solvable Leibniz algebras with given nilradicals. It is known that any solvable Leibniz algebra can decomposed in sum of the nilradical and its complementary vector space. For the solvable Lie algebras Mubarakzjanov offered the method in which he said that the dimension of the complementary space is not greater than the maximal number of nil-independent derivations of the nilradical. Our goal is to show the validity of the Mubarakzjanov’s method for Leibniz algebras. Using this method the solvable Leibniz algebras with null-filiform nilradicals are classified. Moreover, we classify the minimal dimensional solvable Leibniz algebras whose nilradical is equal to the sum of null-filiform algebras. In the last section we describe solvable Leibniz algebras with triangular nilradicals. Furthermore, we establish that a solvable Leibniz algebra of maximal possible dimension with a given triangular nilradical is a Lie algebra. 1.1 Basic results from the theories of Lie and Leibniz algebras In this section we give necessary definitions and preliminary results. 1 2 1 Solvable Leibniz algebras with given nilradicals Definition 1.1.1 ( [15]).An algebra gover a field Kis called a Lie algebra if its multiplication (denoted by (x, y)7→ [x, y]) satisfies the identities: (1) [x, x] = 0, (2) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0, for all x, y, z in g. The product [x, y]is called the bracket of xand y. Identity (2) is called the Jacobi identity. Definition 1.1.2. An algebra Lover a field Kis called a Leibniz algebra if for any x, y, z ∈L, the Leibniz identity [[x, y], z] = [[x, z], y]+[x, [y, z]] is satisfied, where [−,−]is the multiplication in L. For the shortness, instead Leibniz identity [[x, y], z] = [[x, z], y]+[x, [y, z]] we will use below the notation {x, y, z}. For a Leibniz algebra Lwe consider the following derived and lower central series: (i) L(1) =L, L(n+1) = [L(n), L(n)], n > 1; (ii) L1=L, Ln+1 = [Ln, L], n > 1. Definition 1.1.3. An algebra Lis called solvable (nilpotent) if there exists s∈N(k∈N, respectively) such that L(s)= 0 (Lk= 0, respectively). The minimal number s(respectively, k) with such property is called index of solvability (respectively, of nilpotency) of the algebra L. Evidently, the index of nilpotency of an n-dimensional algebra is not greater than n+ 1. Definition 1.1.4. An n-dimensional Leibniz algebra is called null-filiform if dim Li=n+ 1 −i, 1≤i≤n+ 1. Evidently, any null-filiform Leibniz algebra has maximal index of nilpotency. Theorem 1.1.5 ( [7]).An arbitrary n-dimensional null-filiform Leibniz algebra is isomorphic to the algebra: NFn: [ei, e1] = ei+1,1≤i≤n−1, where {e1, e2, . . . , en}is a basis of the algebra. 1.1 Basic results from the theories of Lie and Leibniz algebras 3 Actually, a nilpotent Leibniz algebra is null-filiform if it is a one-generated algebra. Note, that this notion has no sense in Lie algebras case, because they are at least two-generated. Definition 1.1.6. A Leibniz algebra Lis said to be filiform if dim Li=n−i, for 2≤i≤n, where n= dim L. Definition 1.1.7. Given a filiform Leibniz algebra L, put Li=Li/Li+1,1≤ i≤n−1,and gr L=L1⊕L2⊕. . . Ln−1.Then [Li, Lj]⊆Li+jand we obtain the graded algebra gr L. If gr Land Lare isomorphic, denoted by gr L∼ =L, we say that the algebra Lis naturally graded. Thanks to [56] it is known two types of naturally graded filiform Lie algebras. Moreover the second class appears only in the case of even dimension. Theorem 1.1.8. Any complex naturally graded filiform Lie algebra is isomorphic to one of the following non-isomorphic algebras: nn,1: [ei, e1] = −[e1, ei] = ei+1,2≤i≤n−1. Q2n:[ei, e1] = −[e1, ei] = ei+1,2≤i≤2n−2, [ei, e2n+1−i] = −[e2n+1−i, ei] = (−1)ie2n,2≤i≤n. In the following theorem we present the classification of naturally graded filiform non-Lie Leibniz algebras. Theorem 1.1.9 ( [10]).Any complex n-dimensional naturally graded filiform non-Lie Leibniz algebra is isomorphic to one of the following non-isomorphic algebras: F1 n:[ei, e1] = ei+1,2≤i≤n−1,F2 n:([e1, e1] = e3, [ei, e1] = ei+1,3≤i≤n−1. Definition 1.1.10. The (unique) maximal nilpotent ideal of a Leibniz algebra is called the nilradical of the algebra. A derivation for Leibniz algebras is defined as usual. Definition 1.1.11. A linear map d:L→Lis called a derivation of Lif d([x, y]) = [d(x), y]+[x, d(y)] for any x, y ∈L. The space of all derivations is denoted by Der(L). 4 1 Solvable Leibniz algebras with given nilradicals For an arbitrary element x∈L, we consider the right multiplication operator Rx:L→Ldefined by Rx(z) = [z, x]. Right multiplication operators are derivations of the algebra Land are called inner derivations. The set R(L) = {Rx|x∈L}is a Lie algebra with respect to the commutator and the following identity holds: RxRy−RyRx=R[y,x]. The right annihilator of a Leibniz algebra L, denoted by Annr(L), is Annr(L) = {x∈L|[y, x] = 0 for all y∈L}. The left annihilator of a Leibniz algebra L, denoted by Annl(L), is Annl(L) = {x∈L|[x, y] = 0for all y∈L}. The center of a Leibniz algebra L, denoted by Center(L), is Center(L) = Annr(L)∩Annl(L) = {x∈L|[y, x] = [x, y] = 0 for all y∈L}. Definition 1.1.12 ( [42]).Let d1, d2, . . . , dnbe derivations of a Leibniz algebra L. The derivations d1, d2, . . . , dnare said to be nil-independent if α1d1+α2d2+···+αndn is not nilpotent for any scalars α1, α2, . . . , αn∈C, which are not all zero. In other words, if for any α1, α2, . . . , αn∈Cthere exists a natural number ksuch that (α1d1+α2d2+···+αnd2)k= 0, then α1=α2=··· =αn= 0. The classical Engel’s theorem for Lie algebras has the following analogue for Leibniz algebras. Theorem 1.1.13. A Leibniz algebra Lis nilpotent if and only if Rxis nilpotent for any x∈L. Similar to the case of finite-dimensional Lie algebras we have the following theorem. Theorem 1.1.14. A Leibniz algebra Lis solvable if and only if L2is a nilpotent Leibniz algebra. Further, we will use Lie’s theorem for proving the main result. Theorem 1.1.15 ( [32] Lie’s theorem).If Lis a solvable Lie algebra of linear transformations in a finite-dimensional vector space Vover an algebraically closed field of characteristic 0, then the matrices of Lcan be taken in simultaneous triangular form. 1.1 Basic results from the theories of Lie and Leibniz algebras 5 Let us consider the finite-dimensional Lie algebra T(n)of the uppertriangular n×nmatrices with n≥3over the field of the complex numbers. The products of the basis elements {Nij |1≤i < j ≤n}of T(n), where Nij is a matrix with the only non-zero entry at i-th row and j-th column equal to 1, can be computed by [Nij, Nkl] = δjkNil −δilNkj. For a natural number flet G(n, f)be a set of solvable Lie algebras of dimension 1 2n(n−1)+fwith nilradical T(n).Let Q=hX1, X2, . . . , Xfi, where Qis the complementary vector space of the nilradical T(n)to an algebra from G(n, f). Denote [Nij, Xα] = X 1≤q−p<n aα ij,pqNpq,[Xα, Xβ] = X 1≤q−p<n σαβ pq Npq,(1.1.1) where 1≤α, β ≤f, and aα ij,pq, σαβ pq ∈C, p < q ≤n. Let Nbe a vector column (N12, N23, . . . , N(n−1)n, N13, N24, . . . , N(n−2)n, . . . , N1n)T. Then we have RXα(N) = AαN, where Aα= (aα ij,pq),1≤i < j ≤n, 1≤p < q ≤n, are 1 2n(n−1) ×1 2n(n−1) complex matrices. The following lemma provides some information about the structure of the matrices above. Lemma 1.1.16 ( [55]).The structure matrices Aα= (aα ij,pq),1≤i<j≤ n, 1≤p<q≤n, have the following properties: (i) They are upper triangular; (ii) The only off-diagonal matrix elements that do not vanish identically and cannot be annulled by a redefinition of the elements Xαare: aα 12,2n, aα i(i+1),1n(2 ≤i≤n−2), aα (n−1)n,1(n−1); 6 1 Solvable Leibniz algebras with given nilradicals (iii) The diagonal elements aα i(i+1),i(i+1),1≤i≤n−1, are free to vary. The other diagonal elements satisfy aα ik,ik = k−1 X p=i aα p(p+1),p(p+1), k > i + 1. Lemma 1.1.17 ( [55]).The maximal number of non-nilpotent elements is fmax =n−1. 1.2 Solvable Leibniz algebras with null-filiform nilradical Let Rbe a solvable Leibniz algebra. Then it can be decomposed into the form R=N⊕Q, where Nis the nilradical and Qis the complementary vector space. Since the square of a solvable algebra is a nilpotent ideal and the finite sum of nilpotent ideals is a nilpotent ideal too, then the ideal R2is nilpotent, i.e. R2⊆Nand consequently, Q2⊆N. Lemma 1.2.1. Let x∈Qbe such that the operator Rx|Nis nilpotent. Then the subspace V=hx+Niis a nilpotent ideal of the algebra R. Proof. Since R2⊆N,Vis an ideal. We argue that it is nilpotent. If a∈N, then Ra|Nis a nilpotent operator. Let us suppose that there exists k∈N such that Ra|Nk= 0, then Ra|Vk+1 = 0. Hence Ra|Vis nilpotent. V is an ideal of the solvable Leibniz algebra R, then Inn(V)is a solvable Lie algebra of End(V), and so by Lie’s theorem there exists a basis such that Ra|V and Rx|Vare upper triangular; moreover, Ra|Vis nilpotent, which means that Ra|Vhas zero diagonal elements. On the other hand, by assumption, Rx|Nis nilpotent, then with a similar argument as the previous one, there exists s∈N such that Rx|Ns= 0, then Rx|Vs+1 = 0. Summarizing, Ra|Vand Rx|V are nilpotent and upper triangular, hence Ra|V+Rx|Vis nilpotent. Thus, by Engel’s theorem, Vis a nilpotent ideal. Theorem 1.2.2. Let Rbe a solvable Leibniz algebra and Nits nilradical. Then the dimension of the complementary vector space to Nis not greater than the maximal number of nil-independent derivations of N. 1.2 Solvable Leibniz algebras with null-filiform nilradical 7 Proof. We assert that every Rx|N,x∈Q, is a non-nilpotent outer derivation of N. Indeed, if there exists some x∈Qsuch that the operator Rx|Nis nilpotent, then the subspace V=hx+Niis a nilpotent ideal of the algebra R by Lemma 1.2.1, contradicting the maximality condition of N. Let {x1, . . . , xm}be a basis of Q. Then the operators Rx1|N, . . . , Rxm|N are nil-independent, since if for some scalars {α1, . . . , αm} ∈ C\{0}we have that m X i=1 αiRxi|Nk= 0, then Rk y|N, where y= m X i=1 αixi. Hence y= 0, and so αi= 0 for i= 1, . . . , m. Therefore, we see that the dimension of Qis 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 Rwe also have the inequality dim N≥dim R 2. From Theorem 1.2.2 we conclude the following properties of derivations of null-filiform Leibniz algebras. Proposition 1.2.3. Any derivation of the algebra NFnhas the following matrix form:        a1a2a3. . . an 0 2a1a2. . . an−1 0 0 3a1. . . an−2 . . .. . .. . .. . .. . . 0 0 0 . . . na1        . Proof. The proof is carried out by checking the derivation property on the algebra NFn. Corollary 1.2.4. The maximal number of nil-independent derivations of the n-dimensional null-filiform Leibniz algebra NFnis 1. Proof. Let Di=       ai 1ai 2ai 3. . . ai n 0 2ai 1ai 2. . . ai n−1 0 0 3ai 1. . . ai n−2 . . .. . .. . .. . .. . . 0 0 0 . . . nai 1        , i = 1,2, . . . , p, 14 1 Solvable Leibniz algebras with given nilradicals The equalities e1,[e1, x]=−e1,[x, e1]and f1,[f1, x]=−f1,[x, f1] imply that λ1=−α1,µ1=−γ1. From the equalities 0 = e1,[x, x]=ρ1e2and 0 = f1,[x, x]=ξ1f2, we get ρ1=ξ1= 0. In a similar way as in the proof of Theorem 1.2.6, the following equalities can be proved: [ei, x] = −iα1ei+ k X j=i+1 λj−i+1ej,2≤i≤k, [fi, x] = −iγ1fi+ s X j=i+1 µj−i+1fj,2≤i≤s. Summarizing, we have obtained the following multiplication table for the algebra R:                                                                                  [ei, e1] = ei+1,1≤i≤k−1, [fi, f1] = fi+1,1≤i≤s−1, [x, e1] = k X i=1 αiei+βsfs, [x, f1] = δkek+ s X i=1 γifi, [e1, x] = −α1e1+ k X i=2 λiei+ s X i=1 σifi, [f1, x] = k X i=1 τiei−γ1f1+ s X i=2 µifi, [ei, x] = −iα1ei+ k X j=i+1 λj−i+1ej,2≤i≤k, [fi, x] = −iγ1ei+ s X j=i+1 µj−i+1fj,2≤i≤s, [x, x] = k X i=2 ρiei+ s X i=2 ξifi. (1.2.3) 1.2 Solvable Leibniz algebras with null-filiform nilradical 15 Below, we analyze the different cases that can appear in terms of the possible values of α1and γ1. Case 1. Let α1=γ1= 0. Then the multiplication table (1.2.3) implies [ei, x]∈ h{ei+1, ei+2, . . . , ek}i,[fi, x]∈ h{fi+1, fi+2, . . . , fs}i,[e1, x]∈ h{e2, e3, . . . , ek, f1, f2, . . . , fs}i and [f1, x]∈ h{e1, e2, . . . , ek, f2, f3, . . . , fs}i. The above facts mean that the algebra Ris nilpotent, so we get a contradiction with the assumption of non-nilpotency of R. Therefore, this case is impossible. Case 2. Let α16= 0 and γ1= 0. Using the following change of basis: e0 1=1 α1k X i=1 αiei+βsfs, e0 i=1 α1 k X j=i αj−i+1ej,2≤i≤k, x0=1 α1 x, we assume that [x, e1] = e1. From the identity {x, x, e1}we have that e1= k X i=2 ρi[ei, e1]−[e1, x] = k X i=3 ρi−1ei+e1− k X i=2 λiei− s X i=1 σifi. Consequently, λ2=σi= 0 for 1≤i≤sand ρi=λi+1 for 2≤i≤k−1. From the identity {f1, x, e1}we conclude that 0 = [f1, x], e1= k P i=2 τi−1ei⇒τi= 0,1≤i≤k−1. From the identity {x, x, f1} 0 = s X i=3 ξi−1fi− s X i=2 γi[fi, x] + δk[ek, x] = s X i=3 ξi−1fi− s X i=2 γis X j=i+1 µj−i+1fj−kδkek = s X i=3 ξi−1fi− s X i=3i X j=3 γj−1µi−j+2fi−kδkek = s X i=3ξi−1− i X j=3 γj−1µi−j+2fi−kδkek. 16 1 Solvable Leibniz algebras with given nilradicals By comparison of coefficients at the elements of the basis we deduce that: ξi= i+1 X j=3 γj−1µi−j+3,2≤i≤s−1and δk= 0. Now we consider the following change of basis: f0 1=f1+τk kek, f0 i=fi,2≤i≤s. Then we obtain [f0 1, x] = [f1+τk kek, x] = s X i=2 µifi+τkek−τkek= s X i=2 µifi= s X i=2 µif0 i and [x, f0 1] = [x, f1+τk kek]=[x, f1] = s X i=2 γifi= s X i=2 γif0 i. 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, [x, f1] = s X i=2 γifi, [e1, x] = −e1+ k X i=2 λiei, [f1, x] = s X i=2 µifi, [ei, x] = −iei+ k X j=i+2 λj−i+1ej,2≤i≤k, [fi, x] = s X j=i+1 µj−i+1fj,2≤i≤s, [x, x] = k X i=2 ρiei+ s X i=2 ξifi. 1.2 Solvable Leibniz algebras with null-filiform nilradical 17 From the above multiplication table the following inclusions can be immediately derived: [x, NFk]⊆NFk,[NFk, x]⊆NFk,[x, NFs]⊆NFs,[NFs, x]⊆NFs. This completes the proof of the assertion established in the theorem for this case. Case 3. Let α1= 0 and γ16= 0. Due to symmetry of Cases 2 and 3, the proof of the assertion of the theorem follows by applying similar arguments as in Case 2. Case 4. Let α16= 0 and γ16= 0. Consider the following change of basis: e0 1=1 α1k X i=1 αiei+βsfs, e0 i=1 α1 k X j=i αj−i+1ej,2≤i≤k, f0 1=1 γ1s X i=1 γifi+δkek, f0 i=1 γ1 k X j=i γj−i+1fj,2≤i≤s, x0=1 α1 x. Then we derive [x0, e0 1] = 1 α1 x, 1 α1k X i=1 αiei+βsfs=1 α2 1 α1[x, e1] = 1 α1 [x, e1] = e0 1, [x0, f0 1] = 1 α1 x, 1 γ1s X i=1 γifi+δkek=1 α1γ1 γ1[x, f1] = γ1 α1 f0 1. From the identity {x, x, e1}we deduce: e1= k X i=2 ρi[ei, e1]−[e1, x] = k X i=3 ρi−1ei+α1e1− k X i=2 λiei− s X i=1 σifi. Therefore, α1= 1, λ1=−1, λ2=σi= 0,1≤i≤sand ρi=λi+1,2≤i≤ k−1. Expanding the identity {x, x, f1}we derive the equalities: γ1 α12f1= s X i=2 ξi[fi, f1]−γ1 α1 [f1, x] = s X i=3 ξi−1fi−γ1 α1 s X i=1 µifi−γ1 α1 k X i=1 τiei 18 1 Solvable Leibniz algebras with given nilradicals from which we have µ1=−γ1 α1, µ2=τi= 0,1≤i≤kand ξi=γ1 α1µi+1,2≤ i≤s−1. 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,[x, f1] = γ1 α1 f1, [e1, x] = −e1+ k X i=3 λiei,[f1, x] = −γ1 α1 f1+ s X i=3 µifi, [x, x] = k X i=2 ρiei+ s X i=2 ξifi. These products are sufficient in order to check the inclusions: [x, NFk]⊆NFk,[NFk, x]⊆NFk,[x, NFs]⊆NFs,[NFs, x]⊆NFs. Thus, the ideals NFkand NFsof the nilradical are also ideals of the algebra. Now we are going to describe solvable Leibniz algebras with nilradical NFk⊕NFsand with one-dimensional complementary vector space. Due to Theorem 1.2.7 we can assume that NFkand NFsare ideals of the algebra. Theorem 1.2.8. Let Rbe a solvable Leibniz algebra such that R=NFk⊕ NFs+Q, where NFk⊕NFsis the nilradical of Rand dim Q= 1. Let us assume that {e1, e2, . . . , ek}is a basis of NFk,{f1, f2, . . . , fs}is a basis of NFsand {x}is a basis of Q. Then the algebra Ris isomorphic to one of the following pairwise non-isomorphic algebras: R(α) :      [ei, e1] = ei+1,1≤i≤k−1,[fi, f1] = fi+1,1≤i≤s−1, [x, e1] = e1,[x, f1] = αf1, α 6= 0, [ei, x] = −iei,1≤i≤k, [fi, x] = −iαfi,1≤i≤s, R(β2, β3, . . . , βs, γ) :            [ei, e1] = ei+1,1≤i≤k−1,[fi, f1] = fi+1,1≤i≤s−1, [x, e1] = e1,[fi, x] = s X j=i+1 βj−i+1fj,1≤i≤s, [ei, x] = −iei,1≤i≤k, [x, x] = γfs. 1.2 Solvable Leibniz algebras with null-filiform nilradical 19 In the second family of algebras the first non-zero element of the vector (β2, β3, . . . , βs, γ)can be assumed to be equal to 1. Proof. Firstly, we note that the algebras NFk+Qand NFs+Qare not simultaneously nilpotent. Indeed, if they are both nilpotent, then we have: [ei, e1]∈ h{ei+1, . . . , ek}i,1≤i≤k−1, [x, e1]∈ h{e2, e3, . . . , ek}i, [ei, x]∈ h{ei+1, . . . , ek}i,1≤i≤k−1, [fi, f1]∈ h{fi+1, . . . , fs}i,1≤i≤s−1, [x, f1]∈ h{f2, f3, . . . , fs}i, [fj, x]∈ h{fj+1, . . . , fs}i,2≤i≤s−1 From the equalities 0 = e1,[x, x],0 = f1,[x, x]we conclude that: [x, x]∈ h{e2, e3, . . . , ek, f2, f3, . . . , fs}i. Therefore, R2⊆ {e2, e3, . . . , ek, f2, f3, . . . , fs}. Moreover, we have Ri⊆ {ei, ei+1, . . . , ek, fi, fi+1, . . . , fs}, which implies that Rmax{k,s}+1 ={0}. Thus, we have a contradiction to the assumption that Ris not nilpotent. Hence, the algebras NFk+Qand NFs+Qcannot be both nilpotent. Without loss of generality, we can assume that algebra NFk+Qis nonnilpotent. We take the quotient algebra by ideal NFs, then R/NFs∼ =NFk+Q. Thanks to Theorem 1.2.6, the structure of the algebra NFk+Qis known. Namely,      [ei, e1] = ei+1,1≤i≤k−1, [x, e1] = e1, [ei, x] = −iei,1≤i≤k. (1.2.4) Using the fact that NFkand NFsare ideals of Rand having in mind the 20 1 Solvable Leibniz algebras with given nilradicals multiplication table (1.2.4), we have that:                            [ei, e1] = ei+1,1≤i≤k−1,[fi, f1] = fi+1,1≤i≤s−1, [x, e1] = e1,[x, f1] = s X i=1 αifi, [ei, x] = −iei,1≤i≤k, [f1, x] = s X i=1 βifi, [x, x] = s X i=1 γifi. (1.2.5) If α16= 0, then in a similar way as the Case 1 of Theorem 1.2.6 we obtain the family of algebras R(α), where α6= 0. The fact that two algebras in the family R(α)with different values of parameter αare not isomorphic can be easily determined by a general change of basis and considering the expansion of the product [x0, f0 1]in both bases. Now consider α1= 0. Then by the change of basis x0=x−(α2f1+α3f2+···+αsfs−1), we can suppose [x, f1] = 0. From the identity {f1, f1, x}we get β1= 0. Similarly to the proof of Equations (1.2.1), we can prove that [fi, x] = s X m=i+1 βm−i+1fj,1≤i≤s. The identity {x, f1, x}implies the following chain of equalities: 0 = −[x, x], f1=− s X m=3 γm−1fm. Consequently, γi= 0,2≤i≤s−1. Thus, we obtain the products of the family R(β2, β3, . . . , βs, γ)            [fi, f1] = fi+1,1≤i≤s−1, [fi, x] = s X m=i+1 βm−i+1fm,1≤i≤s, [x, x] = γsfs. 1.2 Solvable Leibniz algebras with null-filiform nilradical 21 Now we are going to study the isomorphism inside the family R(β2, β3, . . . , βs, γ). Taking into account that, under general basis transformation, the products (1.2.5) should not be changed, we conclude that it is sufficient to take the following change of basis: f0 i=Ai−1 1 s X j=i Aj−i+1fj,(A16= 0),1≤i≤s, x0=x. Then we have [f0 1, x0] = s X i=1 Ai[fi, x] = s−1 X i=1 Ais X j=i+1 βj−i+1fj= s X i=2i−1 X j=1 AjBi−j+1fi. On the other hand [f0 1, x0] = s X i=2 β0 if0 i= s−1 X i=1 Ai 1β0 i+1s−i X j=1 Ajfi+j= s X i=2i−1 X j=1 Aj 1Ai−jβ0 j+1fi. Comparing coefficients at the elements of the basis we deduce that: k−1 X i=1 Aiβk−i+1 = k−1 X i=1 Ai 1Ak−iβ0 i+1, k = 2,3, . . . , s. From these systems of equations it follows: β0 i=βi Ai−1 1 ,2≤i≤s. If we consider γ0 sAs 1fs=γ0 sf0 s= [x0, x0] = [x, x] = γsfs, then we obtain γ0 s=γs As 1 . It is easy to see that by choosing an adequate value for the parameter A1, then the first non-zero element of the vector (β2, β3, . . . , βs, γ)can be assumed to be equal to 1. 22 1 Solvable Leibniz algebras with given nilradicals Therefore, two algebras R(β2, β3, . . . , βs, γ)and R(β0 2, β0 3, . . . , β0 s, γ0)with different set of parameters are not isomorphic. For given parameters αand β2, β3, . . . , βs, γ, the algebras R(α)and R(β2, β3, . . . , βs, γ)are not isomorphic because k+s= dim R(α)26= dim R(β2, β3, . . . , βs, γ)2=k+s−1. Remark 1.2.9. In the case when all the coefficients (β2, β3, . . . , βs, γ)are equal to zero we have the split algebra (NFk+Q)⊕NFs. Therefore, in the non-split case, we can always assume that (β2, β3, . . . , βs, γ)6= (0,0,0, . . . , 0). Now, by an induction process, we are going to generalize Theorem 1.2.8 to the case when the nilradical is a direct sum (greater than 2) of several copies of null-filiform ideals. Theorem 1.2.10. Let Rbe a solvable Leibniz algebra such that R=NFn1⊕ NFn2⊕ ···⊕NFns+Q, where NFn1⊕NFn2⊕··· ⊕ NFnsis the nilradical of Rand dim Q= 1. There exist p, q ∈Nwith p6= 0 and p+q=s, a basis {ei 1, ei 2, . . . , ei ni}of NFni, for 1≤i≤p, a basis {fk 1, fk 2, . . . , fk nk}of NFnp+k, for 1≤k≤q, and a basis {x}of Qsuch that the multiplication table of the algebra Ris given by: Rp,q :                                    [ej i, ej 1] = ej i+1,1≤i≤nj−1, [fk i, fk 1] = fk i+1,1≤i≤nk−1, [x, ej 1] = δjej 1, δj6= 0 [fk i, x] = nk X m=i+1 βk m−i+1fk m,1≤i≤nk, [ej i, x] = −iδjej i,1≤i≤nj, [x, x] = k X m=1 γmfnm, (1.2.6) where 1≤j≤p, 1≤k≤qand δ1= 1. Moreover, the first non-zero component of the vectors (βk 2, βk 3, . . . , βk nk, γk)can be assumed to be equal to 1. Moreover, the algebras are pairwise non-isomorphic. 1.2 Solvable Leibniz algebras with null-filiform nilradical 23 Proof. By induction on s: If s= 1, then p= 1, q = 0, so R1,0is the algebra given in Theorem 1.2.6. If s= 2, then we have two cases: either p= 2, q = 0 or p= 1, q = 1, which were considered in Theorem 1.2.7. Namely, we have two families of algebras: R(α), which corresponds to R2,0, and R(β2, β3, . . . , βs, γ), which corresponds to R1,1. Let us assume that the theorem is true for sand 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 ∼ =NFn1⊕NFn2⊕···⊕NnFs+Q. Then we get the multiplication table given in (1.2.6). Note that the multiplication table for the algebra Rcan be obtained from (1.2.6) by adding the products [es+1 i, es+1 1] = es+1 i+1 ,1≤i≤ns+1 −1, [x, es+1 1] = ns+1 X m=1 αs+1 mes+1 m, [es+1 1, x] = ns+1 X m=1 βs+1 mes+1 m, [x, x] = ns+1 X m=1 γs+1 mes+1 m. If αs+1 16= 0, then in an analogous way as in proof of Theorem 1.2.6, we deduce that [es+1 i, es+1 1] = es+1 i+1 ,1≤i≤ns+1 −1, [x, es+1 1] = αs+1 s+1es+1 1, [es+1 i, x] = −iαs+1es+1 i,1≤i≤ns+1. Therefore we get the algebra Rp+1,q. 30 1 Solvable Leibniz algebras with given nilradicals From Leibniz identity {Xα, Ni(i+1), Xα}, for 1≤i≤n−1, we obtain the restrictions: aα i(i+1),i(i+1)(aα i(i+1),1n+bα i(i+1),1n) = 0,2≤i≤n−2, aα 12,12bα 12,1n=aα (n−1)n,(n−1)nbα (n−1)n,1n= 0. Let us resume the obtained products of the basis elements. For 1≤α≤f we have                                                                          [N12, Xα] = aα 12,12N12 +aα 12,2nN2n, [Ni(i+1), Xα] = aα i(i+1),i(i+1)Ni(i+1) +aα i(i+1),1nN1n,2≤i≤n−2, [N(n−1)n, Xα] = aα (n−1)n,(n−1)nN(n−1)n+aα (n−1)n,1(n−1)N1(n−1), [Nij, Xα] = j−1 X p=i aα p(p+1),p(p+1)Nij, j > i + 1, [Xα, N12] = −aα 12,12N12 −aα 12,2nN2n+bα 12,1nN1n, [Xα, Ni(i+1)] = −aα i(i+1),i(i+1)Ni(i+1) +bα i(i+1),1nN1n,2≤i≤n−2, [Xα, N(n−1)n] = −aα (n−1)n,(n−1)nN(n−1)n−aα (n−1)n,1(n−1)N1(n−1) +bα (n−1)n,1nN1n, [Xα, Nij] = − j−1 X p=i aα p(p+1),p(p+1)Nij, j > i + 1, [Xα, Xβ] = σαβN1n, with restrictions on parameters: aα i(i+1),i(i+1)(aα i(i+1),1n+bα i(i+1),1n) = 0,2≤i≤n−2, aα 12,12bα 12,1n=aα (n−1)n,(n−1)nbα (n−1)n,1n= 0. Note that for solvable non-Lie Leibniz algebras of the set L(n, f)the following equality holds [Xγ, N1n] = [N1n, Xγ] = 0,1≤γ≤f. (1.3.2) 1.3 Solvable Leibniz algebras with triangular nilradical 31 Indeed, if we assume the contrary, then taking into account that [Xγ, N1n] = −[N1n, Xγ]we can assume [Xγ, N1n]6= 0 for some γ∈ {1, . . . , f}. Simplifying the following products using the Leibniz identity [Xγ,[N12, Xα]+[Xα, N12]],[Xγ,[Ni(i+1), Xα]+[Xα, Ni(i+1)]], [Xγ,[N(n−1)n, Xα]+[Xα, N(n−1)n]],[Xγ,[Xα, Xβ]+[Xβ, Xα]], [Xγ,[Xα, Xα]], we obtain bα 12,1n=bα (n−1)n,1n=σαα = 0, bα i(i+1),1n=−aα i(i+1),1n, σαβ =−σβα. Thus we get a Lie algebra, which is a contradiction. Corollary 1.3.3. For a Leibniz algebra of the set L(n, 1), the matrices of the left and right multiplication operators, A= (aij,pq)and B= (bij,pq), have the following properties: (1) The maximum number of off-diagonal elements of the matrix Ais n−1; (2) The maximum number of off-diagonal elements of the matrix Bis n+ 1. Theorem 1.3.4. A solvable Leibniz algebra of the set L(n, n −1) is a Lie algebra. Proof. Making suitable change of basis we can assume that operator RX1acts as follows [N12, X1] = N12 +a1 12,2nN2n, [Ni(i+1), X1] = a1 i(i+1),1nN1n,2≤i≤n−2, [N(n−1)n, X1] = a1 (n−1)n,1(n−1)N1(n−1), [N1j, X1] = N1j, j > 2. Since [N1n, X1] = N1n,then from Equation (1.3.2) it follows that the algebra is a Lie algebra. So, we present a description of solvable Leibniz algebras with nilradical T(n). Moreover, in the case of maximal possible dimension we show that this algebra is a Lie algebra. 32 1 Solvable Leibniz algebras with given nilradicals Now we give an illustration for low dimensions of solvable Leibniz algebras with nilradical T(n). Note that the Lie algebra T(3) is nothing else but the Heisenberg algebra h1. Solvable Leibniz algebras with Heisenberg nilradical were described in [14]. Therefore, we give the description of solvable Leibniz algebras with nilradical T(4). We know that the complementary vector space to the nilradical T(4) has dimension less than four. In case when dimension of the complementary space is equal to 3 we obtain a Lie algebra (see Theorem 1.3.4), which falls into the classification already obtained in [55]. So, we will consider the dimension of the complementary vector space to be equal to 1 and 2. The Leibniz algebras L(4,1). From previous section we have that the algebra L(4,1) admits a basis {N12, N23, N34, N13, N24, N14, X}in which the multiplication table has the following form:                                    [N12, X] = a12,12N12 +a12,24N24, [X, N12] = −a12,12N12 −a12,24N24 +b12,14N14, [N23, X] = a23,23N23 +a23,14N14, [X, N23] = −a23,23N23 +b23,14N14, [N34, X] = −(a12,12 +a23,23)N34 +a34,13N13, [X, N34] = (a12,12 +a23,23)N34 −a34,13N13 +b34,14N14, [N13, X] = −[X, N13] = (a12,12 +a23,23)N13, [N24, X] = −[X, N24] = −a12,12N24, [X, X] = σ14N14, (1.3.3) where a12,12b12,14 =a23,23(a23,14 +b23,14) = (a12,12 +a23,23)b34,14 = 0. Since L(4,1) is a non-nilpotent Leibniz algebra we conclude (a12,12, a23,23)6= (0,0). Case 1. Let a12,12 = 0. Then a23,23 6= 0, b23,14 =−a23,14 and b34,14 = 0. Taking the change of basis as follows: X0=1 a23,23 X, N0 23 =N23 +a23,14 a23,23 N14, N0 34 =N34 −a34,13 2a23,23 N13 1.3 Solvable Leibniz algebras with triangular nilradical 33 the multiplication (1.3.3) transforms into [N12, X] = a12,24N24,[X, N12] = −a12,24N24 +b12,14N14, [N23, X] = −[X, N23] = N23,[N34, X] = −[X, N34] = −N34, [N13, X] = −[X, N13] = N13,[X, X] = σ14N14, where (b12,14, σ14)6= (0,0). Case 2. If a12,12 6= 0,then b12,14 = 0.Taking by the scaling X0=1 a12,12 X, we can assume a12,12 = 1. Subcase 2.1. Let a23,23 = 0.Then b34,14 = 0. Applying the change of basis N0 12 =N12 +a12,24 2N24, N0 34 =N34 −a34,13 2N13 the products (1.3.3) simplify to the following: [N12, X] = −[X, N12] = N12,[N34, X] = −[X, N34] = −N34, [N13, X] = −[X, N13] = N13,[N24, X] = −[X, N24] = −N24, [N23, X] = a23,14N14,[X, N23] = b23,14N14, [X, X] = σ14N14, where (a23,14 +b23,14, σ14)6= (0,0). Subcase 2.2. Let a23,23 6= 0. Then b23,14 =−a23,14. Subcase 2.2.1. Let a23,23 =−1. Then substituting N0 23 =N23 −a23,14N14, N0 12 =N12 +a12,24 2N24 we derive to an algebra with the following multiplication table: [N12, X] = −[X, N12] = N12,[N23, X] = [X, N23] = −N23, [N34, X] = a34,13N13,[X, N34] = −a34,13N13 +b34,14N14, [N24, X] = −[X, N24] = −N24,[X, X] = σ14N14, where (b12,14, σ14)6= (0,0). Note that by permuting the indexes of the basis elements of the above algebra one obtains an algebra from Case 1. 34 1 Solvable Leibniz algebras with given nilradicals Subcase 2.2.2. Let a23,23 6=−1. Then b34,14 = 0. Setting N0 12 =N12 +a12,24 2N24, N0 23 =N23 +a23,14 a23,23 N14, N0 34 =σ14(N34 −a34,13 2(1 + a23,23)N13), N0 24 =σ14N24, N0 14 =σ14N14, we get an algebra with the following table of multiplications: [N23, X] = −[X, N23] = a23,23N23,[N12, X] = −[X, N12] = N12, [N34, X] = −[X, N34] = −(1 + a23,23)N34,[N24, X] = −[X, N24] = −N24, [N13, X] = −[X, N13] = (1 + a23,23)N13,[X, X] = N14, where (1 + a23,23)a23,23 6= 0. Non-isomorphisms of obtained algebras can be easily established by considering the dimensions of derived series of the algebras. Thus, the following theorem is proved. Theorem 1.3.5. An arbitrary non-Lie Leibniz algebra of the set L(4,1) is isomorphic to one of the following pairwise non-isomorphic algebras: L1: [N12, X] = a12,24N24,[X, N12] = −a12,24N24 +b12,14N14, [N23, X] = −[X, N23] = N23,[N34, X] = −[X, N34] = −N34, [N13, X] = −[X, N13] = N13,[X, X] = σ14N14, where (b12,14, σ14)6= (0,0). L2: [N12, X] = −[X, N12] = N12,[N34, X] = −[X, N34] = −N34 [N13, X] = −[X, N13] = N13,[N24, X] = −[X, N24] = −N24, [N23, X] = a23,14N14,[X, N23] = b23,14N14, [X, X] = σ14N14, where (a23,14 +b23,14, σ14)6= (0,0). L3: [N23, X] = −[X, N23] = a23,23N23, [N12, X] = −[X, N12] = N12, [N34, X] = −[X, N34] = −(1 + a23,23)N34, [N24, X] = −[X, N24] = −N24, [N13, X] = −[X, N13] = (1 + a23,23)N13, [X, X] = N14, 1.3 Solvable Leibniz algebras with triangular nilradical 35 where (1 + a23,23)a23,23 6= 0. The Leibniz algebras L(4,2). Classification of Leibniz algebras in this set is presented in the following theorem. Theorem 1.3.6. An arbitrary non-Lie Leibniz algebra of the set L(4,2) admits a basis {N12, N23, N34, N13, N24, N14, X1, X2}in which the multiplication table has the following form: [N12, X1] = −[X1, N12] = N12,[N34, X1] = −[X1, N34] = −N34, [N13, X1] = −[X1, N13] = N13,[N24, X1] = −[X1, N24] = −N24, [N23, X2] = −[X2, N23] = N23,[N34, X2] = −[X2, N34] = −N34, [N13, X2] = −[X2, N13] = N13,[X1, X1] = σ11N14, [X2, X2] = σ22N14,[X1, X2] = σ12N14, [X2, X1] = σ21N14. Proof. From Lemmas 1.3.1 and 1.3.2 we have [N12, X1] = a1 12,12N12 +a1 12,24N24, [X1, N12] = −a1 12,12N12 −a1 12,24N24 +b1 12,14N14, [N23, X1] = a1 23,23N23 +a1 23,14N14, [X1, N23] = −a1 23,23N23 +b1 23,14N14, [N34, X1] = −(a1 12,12 +a1 23,23)N34 +a1 34,13N13, [X1, N34] = (a1 12,12 +a1 23,23)N34 −a1 34,13N13 +b1 34,14N14, [N13, X1] = −[X1, N13] = (a1 12,12 +a1 23,23)N13, [N24, X1] = −[X1, N24] = −a1 12,12N24, [N12, X2] = a2 12,12N12 +a2 12,24N24, [X2, N12] = −a2 12,12N12 −a2 12,24N24 +b2 12,14N14, [N23, X2] = a2 23,23N23 +a2 23,14N14, [X2, N23] = −a2 23,23N23 +b2 23,14N14, [N34, X2] = −(a2 12,12 +a2 23,23)N34 +a2 34,13N13, [X2, N34] = (a2 12,12 +a2 23,23)N34 −a2 34,13N13 +b2 34,14N14, [N13, X2] = −[X2, N13] = (a2 12,12 +a2 23,23)N13, [N24, X2] = −[X2, N24] = −a2 12,12N24, 36 1 Solvable Leibniz algebras with given nilradicals with the restrictions a1 12,12b1 12,14 =a1 23,23(a1 23,14 +b1 23,14) = (a1 12,12 +a1 23,23)b1 34,14 = 0, a2 12,12b2 12,14 =a2 23,23(a2 23,14 +b2 23,14) = (a2 12,12 +a2 23,23)b2 34,14 = 0. Taking the change of basis X10=a2 23,23 a1 12,12a2 23,23 −a2 12,12a1 23,23 X1−a1 23,23 a1 12,12a2 23,23 −a2 12,12a1 23,23 X2, X20=−a2 12,12 a1 12,12a2 23,23 −a2 12,12a1 23,23 X1+a1 12,12 a1 12,12a2 23,23 −a2 12,12a1 23,23 X2, we deduce [N23, X1] = a1 23,14N14, [N12, X1] = −[X1, N12] = N12 +a1 12,24N24 [X1, N23] = b1 23,14N14, [N34, X1] = −[X1, N34] = −N34 +a1 34,13N13, [N12, X2] = a2 12,24N24, [N23, X2] = −[X2, N23] = N23 +a2 23,14N14, [N13, X1] = −[X1, N13] = N13, [X2, N12] = −a2 12,24N24 +b2 12,14N14, [N24, X1] = −[X1, N24] = −N24, [N34, X2] = −[X2, N34] = −N34 +a2 34,13N13, [N13, X2] = −[X2, N13] = N13. Applying the Leibniz identity to the following triples of elements: {N12, X1, X2},{N23, X1, X2},{N34, X1, X2},{X1, N23, X2},{X2, N12, X1} we get a2 12,24 =a1 23,14 =a1 34,13 =a2 34,13 =b1 23,14 =b2 12,14 = 0. Finally, taking the basis transformation: N0 12 =N12 +a1 12,24 2N24, N0 23 =N23 +a2 23,14N14, we obtain the multiplication table listed in the assertion of the theorem. Chapter 2 Solvable Leibniz algebras with naturally graded filiform nilradicals All solvable Lie algebras whose nilradical is naturally graded filiform Lie algebra nn,1are classified in [52]. Further, solvable Lie algebras whose nilradical is naturally graded filiform Lie algebra Q2nare classified in [4]. We give the classifications of solvable non-Lie Leibniz algebras whose nilradical are naturally graded filiform Lie and naturally graded filiform non-Lie Leibniz algebras, separately. 2.1 Solvable Leibniz algebras with naturally graded filiform Lie nilradicals It is proved that the dimension of a solvable Lie algebra whose nilradical is isomorphic to an n-dimensional naturally graded filiform Lie algebra is not greater than n+ 2. Below, we present its classification. In order to agree with the multiplication tables of algebras in Theorems 1.1.8 and 1.1.9, we make the following change of basis in the classification of [52]: e0 i=en+1−i,1≤i≤n, x =−f. We also use different notation to denote the algebras that appear in [52]. 37 38 2 Solvable Leibniz algebras with filiform nilradicals Theorem 2.1.1 ( [52]).There are three types of solvable Lie algebras of dimension (n+ 1) whose nilradical is isomorphic to nn,1(n≥4). The isomorphism classes in the basis {e1, . . . , en, x}are represented by the following algebras: Sn+1(α, β) :      [ei, e1] = −[e1, ei] = ei+1,2≤i≤n−1, [ei, x] = −[x, ei] = (i−2)α+βei,2≤i≤n, [e1, x] = −[x, e1] = αe1. The mutually non-isomorphic algebras of this type are Sn+1,1(β) = Sn+1(1, β) (depending on the value of β, in this case there are three different classes, β= 0,β=n−2and β /∈ {0, n −2}) and Sn+1,2=Sn+1(0,1). Sn+1,3:     [ei, e1] = −[e1, ei] = ei+1,2≤i≤n−1, [ei, x] = −[x, ei] = (i−1) ei,2≤i≤n, [e1, x] = −[x, e1] = e1+e2. Sn+1,4(α3, α4, . . . , αn−1) :        [ei, e1] = −[e1, ei] = ei+1,2≤i≤n−1, [ei, x] = −[x, ei] = ei+ n X l=i+2 αl+1−iel,2≤i≤n, where at least one αisatisfies αi6= 0 and the first non-vanishing parameter {α3, . . . , αn−1}can be assumed to be equal to 1. Theorem 2.1.2 ( [52]).There exists only one class of solvable Lie algebras of dimension n+ 2 with nilradical nn,1. It is represented by a basis {e1, e2, . . . , en, x, y}and the Lie brackets are Sn+2 :         [ei, e1] = −[e1, ei] = ei+1,2≤i≤n−1, [ei, x] = −[x, ei] = (i−2) ei,2≤i≤n, [e1, x] = −[x, e1] = e1, [ei, y] = −[y, ei] = ei,2≤i≤n. Now we recall the classification given in [4] after the following change of basis: e0 1=−e1, x0=−Y1, y0=−Y2. Concerning solvable Lie algebras with nilradical Q2nwe present the following proposition. 2.1 Solvable Leibniz algebras with filiform Lie nilradicals 39 Proposition 2.1.3 ( [4]).Any solvable Lie algebra of dimension 2n+1 with nilradical isomorphic to Q2nis isomorphic to one of the following algebras: Q2n+1,1(α) :                [ei, e1] = −[e1, ei] = ei+1,2≤i≤2n−2, [ei, e2n+1−i] = −[e2n+1−i, ei] = (−1)ie2n,2≤i≤n, [e1, x] = −[x, e1] = e1, [ei, x] = −[x, ei] = (i−2 + α)ei,2≤i≤2n−1, [e2n, x] = −[x, e2n] = (2n−3−2α)e2n. Q2n+1,2:                [ei, e1] = −[e1, ei] = ei+1,2≤i≤2n−2, [ei, e2n+1−i] = −[e2n+1−i, ei] = (−1)ie2n,2≤i≤n, [e1, x] = −[x, e1] = e1+ε e2n, ε = 0,1, [ei, x] = −[x, ei] = (i−n)ei,2≤i≤2n−1, [e2n, x] = −[x, e2n] = e2n. Q2n+1,3(α) :                              [ei, e1] = −[e1, ei] = ei+1,2≤i≤2n−2, [ei, e2n+1−i] = −[e2n+1−i, ei] = (−1)ie2n,2≤i≤n, [e2+i, x] = −[x, e2+i] = e2+i+b2n−3−i 2c X k=2 α2k+1 e2k+1+i, 0≤i≤2n−6, [e2n−i, x] = −[x, e2n−i] = e2n−i, i = 1,2,3, [e2n, x] = −[x, e2n] = 2 e2n. Proposition 2.1.4 ( [4]).For any n≥3there is unique (2n+ 2)-dimensional solvable Lie algebra having a nilradical isomorphic to Q2n:                      [ei, e1] = −[e1, ei] = ei+1,2≤i≤2n−2, [ei, e2n+1−i] = −[e2n+1−i, ei] = (−1)ie2n,2≤i≤n, [ei, x] = −[x, ei] = i ei,1≤i≤2n−1, [e2n, x] = −[x, e2n] = (2n+ 1) e2n, [ei, y] = −[y, ei] = ei,1≤i≤2n−1, [e2n, y] = −[y, e2n] = 2 e2n. 46 2 Solvable Leibniz algebras with filiform nilradicals From the equality 0 = d([e1, e2]) = [d(e1), e2]+[e1, d(e2)] = β1e3, we get β1= 0. Further, we have d(e3) = d([e1, e1]) = [d(e1), e1]+[e1, d(e1)] = (2α1+α2)e3+ n−1 X i=3 αiei+1. On the other hand, d(e3) = d([e2, e1]) = [d(e2), e1]+[e2, d(e1)] = (α1+β2)e3+ n−1 X i=3 βiei+1. Therefore, β2=α1+α2, βi=αi,3≤i≤n−1. Applying the property of derivation to the products [ei, e1] = ei+1 and by an induction on i, it is easy to get that the following equalities for 3≤i≤n: d(ei) = (i−1)α1+α2ei+ n X j=i+1 αj−i+2 ej,3≤i≤n. From Proposition 2.2.1 we conclude that the number of nil-independent outer derivations of the algebra F1 nis equal to two. Therefore, we have that any solvable Leibniz algebra whose nilradical is F1 nhas dimension either n+ 1 or n+ 2. Below, we present the description of such Leibniz algebras when their dimension is equal to n+ 1. Theorem 2.2.2. An arbitrary (n+1)-dimensional solvable Leibniz algebra with nilradical F1 nis isomorphic to one of the following pairwise non-isomorphic algebras: R1:         [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −e1−e2, [e1, x] = e1, [ei, x]=(i−1)ei,2≤i≤n, 2.2 Solvable Leibniz algebras with nilradical F1 n47 R2(α) :          [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −e1, [e1, x] = e1, [ei, x]=(i−1 + α)ei,2≤i≤n, R3:                [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −e1, [e1, x] = e1, [ei, x]=(i−n)ei,2≤i≤n, [x, x] = en. R4:                [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −e1, [e1, x] = e1+en, [ei, x] = (i+ 1 −n)ei,2≤i≤n, [x, x] = −en−1. R5(α4, . . . , αn) :                        [e1, e1] = e3, [ei, e1] = ei+1,2≤i≤n−1, [e1, x] = e2+ n−1 X i=4 αiei [ei, x] = ei+ n X j=i+2 αj−i+2 ej,2≤i≤n. Moreover, the first non-vanishing parameter {α4, . . . , αn}in the algebras R5(α4, . . . , αn), can be scaled to 1. Proof. From Theorem 1.1.9 and arguments after Proposition 2.2.1 we deduce that there exists a basis {e1, e2, . . . , en, x}such that the multiplication table of the algebra F1 nis completed with the products coming from Rx|F1 n (ei),1≤ 48 2 Solvable Leibniz algebras with filiform nilradicals i≤n, i.e. [e1, x] = n X i=1 αiei,[e2, x] = (α1+α2)e2+ n−1 X i=3 αiei+β en, [ei, x] = (i−1)α1+α2ei+ n X j=i+1 αj−i+2 ej,3≤i≤n. We denote the remaining products as follows: [x, e1] = n X i=1 βiei,[x, e2] = n X i=1 γiei,[x, x] = n X i=1 δiei. From the chain of equalities 0 = [x, e3] = x, [e2, e1]=[x, e2], e1−[x, e1], e2=[x, e2], e1 = (γ1+γ2)e3+ n X i=4 γi−1ei, we conclude that γ2=−γ1, γi= 0,3≤i≤n−1. Since γ1e3=e1,[x, e2]=[e1, x], e2−[e1, e2], x= 0, then γ1= 0. The identity {e1, x, e1}implies β1=−α1. Applying the Leibniz identity to the elements of the form {x, x, e2}and {x, e2, x}, we conclude: ((n−1)α1+α2γn= 0, (n−2)α1γn= 0. Note that γn= 0 otherwise α1=α2= 0 and then we get a contradiction with the non-nilpotency of the derivation D(see Proposition 2.2.1). Now we are going to consider the possible cases of the parameters α1and α2. Case 1. Let α16= 0. Case 1.1. Let α16=β2. Then taking the following change of basis: x0=−1 α1 x, e0 1=e1−1 α1 n X i=2 βiei, e0 i=−1 α1(−α1+β2)ei+ n X j=i+1 βj−i+2 ej,2≤i≤n, 2.2 Solvable Leibniz algebras with nilradical F1 n49 we obtain [e1, e1] = e3,[e1, x] = n X i=1 µiei,[ei, e1] = ei+1,2≤i≤n−1, [x, e1] = e1,[e2, x] = n X i=1 ηiei,[x, e2]=0,[x, x] = n X i=1 θiei. The equalities 0 = [e1, e2], x=e1,[e2, x]+[e1, x], e2=e1, n X i=1 ηiei=η1e3, imply η1= 0. Consider [e3, x] = [e1, e1], x=e1,[e1, x]+[e1, x], e1 =µ1e3+ (µ1+µ2)e3+ n−1 X i=3 µiei+1 = (2µ1+µ2)e3+ n−1 X i=3 µiei+1. On the other hand, [e3, x] = [e2, e1], x=e2,[e1, x]+[e2, x], e1=µ1e3+η2e3+ n−1 X i=3 ηiei+1 = (µ1+η2)e3+ n−1 X i=3 ηiei+1. The comparison of both right-hand sides implies: η2=µ1+µ2, ηi=µi,3≤i≤n−1, this means: [e2, x] = (µ1+µ2)e2+ n−1 X i=3 µiei+ηnenand [e3, x] = (2µ1+µ2)e3+ n−1 X i=3 µiei+1. Now we shall prove the following equalities by an induction on i: [ei, x] = (i−1)µ1+µ2ei+ n X j=i+1 µj−i+2 ej,3≤i≤n. (2.2.1) 50 2 Solvable Leibniz algebras with filiform nilradicals Obviously, the equality holds for i= 3. Let us assume that the equality holds for 3< i < n, and we shall prove it for i+ 1: [ei+1, x] = [ei, e1], x=ei,[e1, x]+[ei, x], e1 =µ1ei+1 +(i−1)µ1+µ2ei+1 + n X j=i+2 µj−i+1 ej = (iµ1+µ2)ei+1 + n X j=i+2 µj−i+1 ej; so the induction proves the equalities (2.2.1) for any i, 3≤i≤n. Applying the Leibniz identity to the triples of elements {e1, x, e1}, {e1, x, x},{x, e1, x}, we deduce that: µ1=−1, µ2=θ1= 0, θi=µi+1,2≤i≤n−1. Below, we summarize the multiplication table of the algebra                                              [e1, e1] = e3 [ei, e1] = ei+1,2≤i≤n−1, [e1, x] = −e1+ n X i=3 µiei, [e2, x] = −e2+ n−1 X i=3 µiei+ηnen, [ei, x] = −(i−1) ei+ n X j=i+1 µj−i+2 ej,3≤i≤n, [x, e1] = e1,[x, x] = n−1 X i=2 µi+1 ei+θnen. Let us take the change of basis in the following form: e0 1=e1+ n X i=3 Aiei, e0 2=e2+ n X i=3 Aiei, 2.2 Solvable Leibniz algebras with nilradical F1 n51 e0 i=ei+ n X j=i+1 Aj−i+2ej,3≤i≤n, x0= n−1 X i=2 Ai+1ei+Ben+x, where A3=µ3, Ai=1 (i−2)µi+ i−1 X j=3 Ajµi−j+2,4≤i≤n, and B=1 n−1θn+ n X j=3 Ajµn−j+3. Then [x0, e0 1] = hn−1 X i=2 Ai+1ei+Ben+x, e1i=e1+ n X i=3 Aiei=e0 1, [e0 1, x0]=[e1, x] + n X i=3 Ai[ei, x] =−e1+ n X i=3 µiei+ n X i=3 Ai−(i−1)ei+ n X j=i+1 µj−i+2ej =−e1− n X i=3 Aiei+ n X i=3 µiei− n X i=3 Ai(i−2)ei+ n X i=3 Ain X j=i+1 µj−i+2ej =−e1− n X i=3 Aiei+ n X i=3 µiei− n X i=3 Ai(i−2)ei+ n X i=4 i−1 X j=3 Ajµi−j+2ei =−e1− n X i=3 Aiei+ (µ3−A3)e3 + n X i=4 −Ai(i−2) + µi+ i−1 X j=3 Ajµi−j+2ei =−e1− n X i=3 Aiei=−e0 1, 52 2 Solvable Leibniz algebras with filiform nilradicals [e0 2, x0]=[e2, x] + n X i=3 Ai[ei, x] =−e2+ n−1 X i=3 µiei+ηnen+ n X i=3 Ai−(i−1)ei+ n X j=i+1 µj−i+2 ej =−e2− n X i=3 Aiei+ n−1 X i=3 µiei+ηnen− n X i=3 Ai(i−2)ei + n X i=3 Ain X j=i+1 µj−i+2 ej =−e2− n X i=3 Aiei+ n−1 X i=3 µiei+ηnen− n X i=3 Ai(i−2)ei + n X i=4 i−1 X j=3 Ajµi−j+2ei =−e2− n X i=3 Aiei+ (µ3−A3)e3 + n−1 X i=4 −Ai(i−2) + µi+ i−1 X j=3 Ajµi−j+2ei +ηn−(n−2)An+ n−1 X i=3 Aiµn−i+2en=−e0 2+η0e0 n, [x0, x0] = n−1 X i=2 Ai+1[ei, x] + B[en, x]+[x, x] = n−1 X i=2 Ai+1 −(i−1)ei+ n X j=i+1 µj−i+2ej −B(n−1)en+ n−1 X i=2 µi+1ei+θnen 2.2 Solvable Leibniz algebras with nilradical F1 n53 =− n−1 X i=2 Ai+1(i−1)ei+ n−1 X i=2 µi+1ei−B(n−1)en+θnen + n−1 X i=2 Ai+1 n X j=i+1 µj−i+2ej =− n−1 X i=2 Ai+1(i−1)ei+ n−1 X i=2 µi+1ei−B(n−1)en+θnen + n X i=3 i X j=3 Ajµi−j+3ei = (µ3−A3)e2+ n−1 X i=3 −Ai+1(i−1) + µi+1 + i X j=3 Ajµi−j+3ei +−B(n−1) + θn+ n X j=3 Ajµn−j+3en= 0. With a similar induction as the given for Equations (2.2.1), it is easy to check that the following equalities hold: [ei, x] = −(i−1) ei,3≤i≤n. Thus, we obtain the following multiplication table:      [e1, e1] = e3,[ei, e1] = ei+1,2≤i≤n−1, [x, e1] = e1,[e1, x] = −e1, [e2, x] = −e2+ηen,[ei, x] = −(i−1)ei,3≤i≤n. If η6= 0, then by taking the change of basis e0 2=e2+η n−2en, we get η0= 0. Finally, by applying the change of basis x0=−xand e0 1=e1−e2, we get the algebra R1. Case 1.2. Let α1=β2. Then by taking the following change of basis: e0 1=e1−e2, e0 i=ei,2≤i≤n, 54 2 Solvable Leibniz algebras with filiform nilradicals we can assume that the multiplication table is the following                                              [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −α1e1+ n X i=3 βiei, [e1, x] = α1e1+ (αn−β)en, [e2, x] = (α1+α2)e2+ n−1 X i=3 αiei+βen, [ei, x] = (i−1)α1+α2ei+ n X j=i+1 αj−i+2ej,3≤i≤n, [x, x] = n X i=1 δiei. Now, by taking x0=1 α1 x−1 α1 n−1 X i=2 βi+1ei, and renaming the parameters, we get F:                                        [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −e1, [e1, x] = e1+βen, [e2, x] = (1 + α2)e2+ n−1 X i=3 αiei+λen, [ei, x] = (i−1 + α2)ei+ n X j=i+1 αj−i+2ej,3≤i≤n, [x, x] = n X i=1 δiei. Making the change of basis x0=x, e0 1=e1, e0 i=ei+ n X j=i+1 Aj−i+2ej,2≤i≤n, 2.2 Solvable Leibniz algebras with nilradical F1 n55 where A3=−α3, Ai=−1 i−1αi+ i−1 X j=3 Ajαi−j+2,4≤i≤n−1, An=−1 n−2λ+ n−1 X j=3 Ajαn−j+2, and applying the Leibniz identity, we obtain the family of algebras F(α, β, γ) :                [ei, e1] = ei+1,2≤i≤n−1, [x, e1] = −e1, [e1, x] = e1+βen, [ei, x] = (i−1 + α)ei,2≤i≤n, [x, x] = −βen−1+γen. Below, we shall investigate the isomorphism inside the family. For this purpose we consider the general change of generator basis elements in the family F(α, β, γ), e0 1= n X i=1 Aiei, e0 2= n X i=1 Biei, x0=Cx + n X i=1 Piei. Then we obtain in the new basis {e0 1, e0 2, . . . , e0 n, x0}the behavior of the parameters with the following expressions: α0=α, β0=A1β+ (n−2 + α)An An−2 1B2 , γ0=γA1+ (n−1 + α)(PnA1−P1An) An−3 1B2 . Case 1.2.1. α6= 2 −n. Taking An=−A1β n−2 + α, we get β0= 0. Case 1.2.1.1. α6= 1 −n. Putting Pn=−γA1+ (n−1 + α)P1An (n−1 + α)A1 , 62 2 Solvable Leibniz algebras with filiform nilradicals Theorem 2.2.3 ( [33]).An arbitrary (n+ 2)-dimensional solvable Leibniz algebra with nilradical F1 nis isomorphic to an algebra with the following multiplication table: [ei, e1] = ei+1,2≤i≤n−1,[e1, x] = e1, [ei, y] = ei,2≤i≤n, [ei, x] = (i−1)ei,2≤i≤n, [x, e1] = −e1. 2.3 Solvable Leibniz algebras with nilradical F2 n In this section we describe solvable Leibniz algebras with nilradical F2 n. Proposition 2.3.1. An arbitrary derivation of the algebra F2 nhas the following matrix form: D=          α1α2α3α4. . . αn−1αn 0β0 0 . . . 0γ 0 0 2α1α3. . . αn−2αn−1 0 0 0 3α1. . . αn−3αn−2 . . .. . .. . .. . .. . . . . .. . . 0 0 0 0 . . . 0 (n−1)α1          . Proof. The proof follows by straightforward calculations in a similar way as the proof of Proposition 2.2.1. It is easy to check that the number of nil-independent derivations of the algebra F2 nis equal to 2. Corollary 2.3.2. The dimension of a solvable Leibniz algebra with nilradical F2 nis either n+ 1 or n+ 2. Theorem 2.3.3. An (n+1)-dimensional solvable Leibniz algebra with nilradical F2 nis isomorphic to one of the following pairwise non-isomorphic algebras: L1(α) :      [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = −e1,[ei, x] = −(i−1) ei,3≤i≤n, [x, e1] = e1,[x, x] = α e2, α ∈ {0,1}. 2.3 Solvable Leibniz algebras with nilradical F2 n63 L2(α) :      [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = −e1,[ei, x] = −(i−1) ei,3≤i≤n, [x, e1] = e1,[e2, x] = α e2, α 6= 0. L3:     [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = −e1,[ei, x] = −(i−1) ei,3≤i≤n, [x, e1] = e1,[e2, x] = (1 −n)e2+en. L4(α) :          [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = −e1,[ei, x] = −(i−1) ei,3≤i≤n, [x, e1] = e1,[e2, x] = −α e2, α 6= 1, [x, e2] = α e2. L5(α) :                [e1, e1] = e3, [e1, x] = −e1−α e2, α ∈ {0,1},[e2, x] = −e2, [ei, e1] = ei+1,3≤i≤n−1, [ei, x] = −(i−1) ei,3≤i≤n, [x, e1] = e1+α e2,[x, e2] = e2. L6(α3, α4, . . . , αn, λ, δ) :                              [e1, e1] = e3, [e1, x] = n X i=3 αiei,[e2, x] = e2, [ei, e1] = ei+1,3≤i≤n−1, [ei, x] = n X j=i+1 αj−i+2 ej,3≤i≤n−1, [x, x] = λ en,[x, e2] = δ e2, δ ∈ {0,−1}. In the algebra L6(α3, α4, . . . , αn, λ, δ)the first non-vanishing parameter {α3, α4, . . . , αn, λ}can be scaled to 1. Proof. Let Rbe a solvable Leibniz algebra satisfying the conditions of the theorem, then there exists a basis {e1, e2, . . . , en, x}, such that {e1, e2, . . . , en} is the standard basis of F2 n, and for non nilpotent outer derivations of the algebra F2 n, we have that [ei, x] = Rx|F2 n(ei),1≤i≤n. 64 2 Solvable Leibniz algebras with filiform nilradicals Due to Proposition 2.3.1 we can assume that [e1, x] = n X i=1 αiei,[e2, x] = β2e2+βnen, [ei, x] = (i−1)α1ei+ n X j=i+1 αj−i+2 ej,3≤i≤n. Let us introduce the following notations: [x, e1] = n X i=1 γiei,[x, e2] = n X i=1 δiei,[x, x] = n X i=1 λiei. Considering the Leibniz identity for the triples of elements {e1, x, x}, {e1, x, e1},{x, e2, e1}, we obtain λ1= 0, γ1=−α1and [x, e2] = δ2e2+δnen. By setting e0 2=δ2e2+δnen, we can assume that [x, e2] = δe2. Now we distinguish the following possible cases Case 1. Let α16= 0. Then the following change of basis e0 1=e1+1 γ1 n X j=3 γjej, e0 2=e2, e0 i=ei+1 γ1 n X j=i+1 γj−i+2 ej,3≤i≤n, x0=1 γ1 x, implies [x0, e0 1] = e0 1+γe0 2(where γ=γ2 γ1) and the rest of products remains unchanging. From the equalities: e1+γ(1 + δ)e2=x, [x, e1]=[x, x], e1−[x, e1], x = n X i=4 λi−1ei− n X i=1 αiei−γβ2e2−γβnen, we deduce α1=−1, α3= 0, α2=−γ(1 + δ+β2), λi=αi+1,3≤i≤n−2,and λn−1=αn+γβn. 2.3 Solvable Leibniz algebras with nilradical F2 n65 In addition, if we take the following change of basis: e0 1=e1+ n X i=4 Aiei, e0 2=e2, e0 i=ei+ n X j=i+2 Aj−i+2 ej,3≤i≤n, x0= n−1 X i=3 Ai+1 ei+B en+x, where Aj=1 j−2αj, j = 4,5, Ai=1 i−2αi+ i−2 P j=4 Ajαi−j+2,6≤i≤n and B=1 n−1λn+ n−1 P j=4 Ajαi−j+3, then we have [e0 1, x0] = −e0 1+α2e0 2,[e0 2, x0] = β2e0 2+βne0 n, [e0 i, x0] = −(i−1) e0 i,3≤i≤n, [x0, x0] = λ2e0 2+γβne0 n−1. Finally, we obtain the following multiplication table of the algebra R:      [e1, x] = −e1−γ(1 + δ+β2)e2,[e2, x] = β2e2+βnen, [x, e1] = e1+γ e2,[ei, x] = −(i−1) ei,3≤i≤n, [x, e2] = δ e2,[x, x] = λ2e2+γβnen−1. Considering the Leibniz identity for the triples of elements {x, x, e2}, {x, x, x},{x, e1, x}, we obtain: δβn=δ(δ+β2) = δλ2=γδ(δ+β2) = 0. Notice that if e2∈Annr(R), then dim Annr(R) = n−1and if e2/∈Annr(R), then dim Annr(R) = n−2. Now we analyze the following possible subcases: Case 1.1. Let e2∈Annr(R). Then δ= 0 and making the change e0 1= e1+γe2we can assume [x, e1] = e1. In this case, we must consider two additional subcases: Case 1.1.1. Let e2∈Center(R). Then dim Center(R) = 1 and β2=βn= 0. We have two options: if λ2= 0, then we get the split algebra L1(0); if λ26= 0, then we obtain the algebra L1(1) (by scaling the basis). Case 1.1.2. Let e2/∈Center(R). Then dim Center(R) = 0 and (β2, βn)6= (0,0). 66 2 Solvable Leibniz algebras with filiform nilradicals Let us take the following general change of basis: e0 1= n X i=1 Aiei, e0 2= n X i=1 Biei, e0 i=Ai−2 1A1ei+ n X j=i+1 Aj−i+2 ei,3≤i≤n, x0= n X i=1 Ciei+Cn+1 x, where (A1B2−A2B1)Cn+1 6= 0. From 0 = [e0 2, e0 1] = [e0 2, e0 2], we obtain that B1=Bi= 0,3≤i≤n−1, i.e. e0 2=B2e2+Bnenand A1B26= 0. The equalities e0 1= [x0, e0 1] = A1C1e3+ n X i=4 A1Ci−1ei+A1Cn+1 e1, imply that Cn+1 = 1, A2= 0, A3=A1C1, Ai=A1Ci−1,4≤i≤n. Similarly, from B2β0 2e2+ (Bnβ0 2+β0 nAn−1 1)en=β0 2e0 2+β0 ne0 n= [e0 2, x0] =B2β2e2+B2βn−(n−1)Bnen, and λ0 2B2e2+λ0 2Bnen=λ0 2e0 2= [x0, x0] = (λ2+C2β2)e2+ (C2 1−2C3)e3 + n−1 X i=4 C1Ci−1−(i−1)Ciei+C1Cn−1−(n−1)Cn+C2βnen, we obtain Ci=1 (i−1)!Ci−1 1,3≤i≤n−1and β0 2=β2, β0 n=B2βn−Bn(β2+n−1) An−1 1 , λ0 2=λ2+β2C2 B2 , λ0 2Bn=C1Cn−1−(n−1)Cn+C2βn. 2.3 Solvable Leibniz algebras with nilradical F2 n67 Now we need to distinguish two subcases: Case 1.1.2.1. Let β2= 1 −n. Putting C2=−λ2 1−n, Cn=C1Cn−1+C2βn n−1, we get λ0 2= 0 and β0 n=B2βn An−1 1 . If βn= 0, then we get the algebra L2(α)for α= 1 −n. If βn6= 0, then making A1=n−1 √βnB2, we obtain β0 n= 1 and the algebra L3. Case 1.1.2.2. Let β26= 1−n. Taking the change Bn=B2βn β2+n−1, we obtain βn= 0. Since β26= 0, we set C2=−λ2 β2, Cn=C1Cn−1+C2βn n−1and we get λ2= 0, i.e., the algebra L2(α)is obtained, for α /∈ {1−n, 0}. Case 1.2. Let e2/∈Annr(R). Then δ6= 0 and β2=−δ, βn=λ2= 0. Let us consider the general change of basis in the following form: e0 1= n X i=1 Aiei, e0 2= n X i=1 Biei, e0 i=Ai−2 1A1ei+ n X j=i+1 Aj−i+2 ej,3≤i≤n, x0= n X i=1 Ciei+Cn+1 x, where (A1B2−A2B1)Cn+1 6= 0. Then from 0 = [e0 2, e0 1] = [e0 2, e0 2], we derive that B1=Bi= 0,3≤i≤n−1, i.e. e0 2=B2e2+Bnenand A1B26= 0. Similarly, from the equations: e0 1+γ0e0 2= [x0, e0 1] = A1Cn+1 e1+Cn+1(A1γ+A2δ)e2+A1C1e3+ n X i=4 A1Ci−1ei, and δ0(B2e2+Bnen) = δ0e0 2= [x0, e0 2] = B2δ e2, we obtain Cn+1 = 1, A3=A1C1, Ai=A1Ci−1,4≤i≤n−1, γ0=A1γ+A2(δ−1) B2 , A1Cn−1=An+γ0Bn, δ0=δ, δ0Bn= 0. Now we distinguish the following possible subcases: Case 1.2.1. Let δ6= 1. Then by the substitution A2=−A1γ δ−1, An= A1Cn−1into the above conditions, we get γ0= 0 and the algebra L4(α). 68 2 Solvable Leibniz algebras with filiform nilradicals Case 1.2.2. Let δ= 1. Then Bn= 0. If γ= 0, then γ0= 0. If γ6= 0, then by putting B2=A1γand An=A1Cn−1−Bn, we get γ0= 1. Thus, the algebras L5(α), α ∈ {0,1}are obtained. Case 2. Let α1= 0. Then β26= 0 and by replacing xby x0=1 β2 x, we can assume [e2, x0] = e2+βnen. Under these conditions, the multiplication table of the algebra Rhas the form:                        [e1, x] = n X i=2 αiei,[e2, x] = e2+βnen, [x, e1] = n X i=2 γiei,[ei, x] = n X j=i+1 αj−i+2 ej,3≤i≤n−1, [x, e2] = δ e2,[x, x] = n X i=2 λiei. Making the transformation x0=x−γ3e1− n−1 P i=3 γi+1 ei, we can assume [x, e1] = γ e2. Similarly as above, we obtain the conditions: γ(δ+ 1) = α2δ−γ=βnδ=δ(δ+ 1) = λ2δ= 0. Now we distinguish the following subcases depending on the possible values of the parameter δ: Case 2.1. Let δ6= 0. Then dim Annr(R) = n−2and βn=λ2= 0, δ = −1, α2=−γ. By means of the change of the basis element e0 1=e1+γe2, we can suppose that [x0, e1] = 0. Taking the general change of basis as in the above considered cases, we derive the following conditions for the parameters α0 i=αi Ai−2 1 ,3≤i≤n, λ0 n=λn An−1 1 . Consequently, we deduce the algebra L6(α3, α4, . . . , αn, λ, −1). Case 2.2. Let δ= 0. Then dim Annr(R) = n−1and γ= 0. Taking the change of basis e0 2=e2+βnenand x0=x−λ2e2, we can assume that 2.3 Solvable Leibniz algebras with nilradical F2 n69 [e2, x] = e2,[x, x] = λnen. Therefore, we have the products [e1, x] = n X i=2 αiei,[e2, x] = e2,[ei, x] = n X j=i+1 αj−i+2 ej,3≤i≤n−1, [x, x] = λnen. Applying similar arguments to general transformation of bases, we have α0 2= 0, α0 i=αi Ai−2 1 ,3≤i≤n, λ0 n=λn An−1 1 . Thus, we obtain the algebra L6(α3, α4, . . . , αn, λ, 0). Theorem 2.3.4. An arbitrary (n+2)-dimensional solvable Leibniz algebra with nilradical F2 nis isomorphic to one of the following non-isomorphic algebras: L1:     [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = e1,[x, e1] = −e1, [e2, y] = −[y, e2] = e2,[ei, x] = (i−1)ei,3≤i≤n, L2:     [e1, e1] = e3,[ei, e1] = ei+1,3≤i≤n−1, [e1, x] = e1,[x, e1] = −e1, [e2, y] = e2,[ei, x]=(i−1)ei,3≤i≤n. Proof. Let Rx|F2 n=          α1α2α3α4. . . αn−1αn 0β0 0 . . . 0γ 0 0 2α1α3. . . αn−2αn−1 0 0 0 3α1. . . αn−3αn−2 . . .. . .. . .. . .. . . . . .. . . 0 0 0 0 . . . 0 (n−1)α1          and Ry|F2 n=          λ1λ2λ3λ4. . . λn−1λn 0µ0 0 . . . 0ν 0 0 2λ1λ3. . . λn−2λn−1 0 0 0 3λ1. . . λn−3λn−2 . . .. . .. . .. . .. . . . . .. . . 0 0 0 0 . . . 0 (n−1)λ1          70 2 Solvable Leibniz algebras with filiform nilradicals be two nil-independent outer derivations of the algebra F2 n. Taking the change of the basis elements x, y x0=µ α1µ−γ1βx−β α1µ−γ1βy, y0=−γ1 α1µ−γ1βx+α1 α1µ−γ1βy, we can assume that α1=µ= 1, λ1=β= 0. Thus, we have the products: [e1, x] = e1+ n X i=2 αiei,[e2, x] = γ en, [ei, x]=(i−1) ei+ n X j=i+1 αj−i+2 ej,3≤i≤n, [e1, y] = n X i=2 λiei,[e2, y] = e2+ν en, [ei, y] = n X j=i+1 λj−i+2 ej,3≤i≤n. Applying similar arguments and changes of bases which we have used in Theorem 2.3.3, we obtain isomorphism classes of algebras whose representative algebras are L1and L2. Remark 2.3.5. In fact, the algebra L1=I1⊕J1, where I1=NFn−1+hxiand J1=he2, yi, verifies that I1is a solvable Leibniz algebra with nilradical NFn−1 and J1is a two-dimensional solvable Lie algebra. The algebra L2=I2⊕J2, where I2=NFn−1+hxiand J2=he2, yi, verifies that J2is a two-dimensional solvable non-Lie Leibniz algebra. Thus, from Theorem 2.3.4, we conclude that any (n+ 2)-dimensional solvable Leibniz algebra with nilradical F2 nis split. Chapter 3 Leibniz algebras corresponding to the Diamond Lie algebras 3.1 Preliminary definitions and results The real Diamond Lie algebra DRis a four-dimensional Lie algebra with basis {J, P1, P2, T }and non-zero relations: [J, P1] = P2,[J, P2] = −P1,[P1, P2] = T. The complexification of the Diamond Lie algebra: DC=D⊗RCdisplays the following (complex) basis: {P+=P1−iP2, P−=P1+iP2, T, J},where iis the imaginary unit, whose nonzero commutators are [J, P+] = iP+,[J, P−] = −iP−,[P+, P−] = 2iT. If we change the basis J0=−iJ, P0 1=P+, P0 2=P−, T0= 2iT then we can assume that [J, P1] = P1,[J, P2] = −P2,[P1, P2] = T. (3.1.1) In the paper [26] the authors construct for any n∈Na(3n+3)-dimensional Lie module Vnover the algebra D, which is endowed with a basis {vj k}j=0,1,2 k=0,...,n. 71 78 3 Leibniz algebras corresponding to the Diamond Lie algebras Proof. We will consider two cases n= 4sand n= 4s−2. Let us introduce notation [J, J] = a0 0v0 0+ n/2 X k=1 a0 2kv0 2k+ n/2 X k=1 a1 2k−1v1 2k−1+a2 0v2 0+ n/2 X k=1 a2 2kv2 2k. Case 1. Let n= 4s. Taking the following change of basis: J0=J+ia0 0 2sv0 0+ s−1 X k=1 ia0 2k 2s−2kv0 2k+ 2s X k=s+1 ia0 2k 2s−2kv0 2k+ 2s X k=1 ia1 2k−1 2s−2k+ 1v1 2k−1 +ia2 0 2sv2 0+ s−1 X k=1 ia2 2k 2s−2kv2 2k+ 2s X k=s+1 ia2 2k 2s−2kv2 2k, we can assume that [J, J] = a0 2sv0 2s+a2 2sv2 2s. Lifting from the quotient Lie algebra Dto the Leibniz algebra Lwe have [J, P+] = −iP++b0 0v0 0+ 2s X k=1 b0 2kv0 2k+ 2s X k=1 b1 2k−1v1 2k−1+b2 0v2 0+ 2s X k=1 b2 2kv2 2k, [J, P−] = iP−+c0 0v0 0+ 2s X k=1 c0 2kv0 2k+ 2s X k=1 c1 2k−1v1 2k−1+c2 0v2 0+ 2s X k=1 c2 2kv2 2k, [P+, P−] = −2iT +d0 0v0 0+ 2s X k=1 d0 2kv0 2k+ 2s X k=1 d1 2k−1v1 2k−1+d2 0v2 0+ 2s X k=1 d2 2kv2 2k. Making the change of basis elements as follows: P0 +=P++ib0 0v0 0+ 2s X k=1 ib0 2kv0 2k+ 2s X k=1 ib1 2k−1v1 2k−1+ib2 0v2 0+ 2s X k=1 ib2 2kv2 2k, P0 −=P−−ic0 0v0 0− 2s X k=1 ic0 2kv0 2k− 2s X k=1 ic1 2k−1v1 2k−1−ic2 0v2 0− 2s X k=1 ic2 2kv2 2k, T0=T+i 2(d0 0v0 0+ 2s X k=1 d0 2kv0 2k+ (d1 1+ib0 0)v1 1 + 2s X k=2 (d1 2k−1+i(2k−1)b0 2k−2)v1 2k−1+d2 0v2 0+ 2s X k=1 (d2 2k+ 2ikb1 2k−1)v2 2k), 3.3.1 Leibniz algebras whose ideal Iis the DC-module U1 n79 we derive the products [J, P+] = −iP+,[J, P−] = iP−,[P+, P−] = −2iT. Considering the chain of equalities P+=i[J, P+] = [J, [P+, J]] = [[J, P+], J]−[[J, J], P+], P−=−i[J, P−] = [J, [P−, J]] = [[J, P−], J]−[[J, J], P−], we conclude [P+, J] = iP++ia0 2s(2s+ 1)v1 2s−1and [P−, J] = −iP−−ia0 2s(2s+ 1)v1 2s+1. We set [P+, P+] = q0 0v0 0+ 2s X k=1 q0 2kv0 2k+ 2s X k=1 q1 2k−1v1 2k−1+q2 0v2 0+ 2s X k=1 q2 2kv2 2k, [P−, P−] = l0 0v0 0+ 2s X k=1 l0 2kv0 2k+ 2s X k=1 l1 2k−1v1 2k−1+l2 0v2 0+ 2s X k=1 l2 2kv2 2k, [J, T] = r0 0v0 0+ 2s X k=1 r0 2kv0 2k+ 2s X k=1 r1 2k−1v1 2k−1+r2 0v2 0+ 2s X k=1 r2 2kv2 2k. Applying the Leibniz identity to the following triples of elements we get further constraints on the structure constants Leibniz identity Constraint {P+, J, P+}=⇒a0 2s=q0 0=q2 0=q1 2k−1= 0,1≤k≤2s, q0 2k=q2 2k= 0, k 6=s−1, s > 1 {P−, J, P−}=⇒l0 0=l2 0=l1 2k−1= 0,1≤k≤2s, l0 2k=l2 2k= 0 = 0, k 6=s+ 1. Thus, we obtain [J, J] = a2 2sv2 2s,[P+, J] = iP+,[P−, J] = −iP−, [P+, P+] = q0 2s−2v0 2s−2+q2 2s−2v2 2s−2,[P−, P−] = l0 2s+2v0 2s+2 +l2 2s+2v2 2s+2. 80 3 Leibniz algebras corresponding to the Diamond Lie algebras Note that for s= 1, we have [P+, P+] = q0 0v0 0+q2 0v2 0, which agrees with the case s > 1. Similarly, we derive the following constraints: Leibniz identity Constraint {J, T, J}=⇒[J, T] = r0 2sv0 2s+r2 2sv2 2s, {P+, J, P−}=⇒[T, J] = 0, {J, P+, T}=⇒[P+, T ] = i(2s+ 1)r0 2sv1 2s−1, {J, P−, T}=⇒[P−, T ] = −i(2s+ 1)r0 2sv1 2s+1. Taking into account the product [P+, P−] = −2iT in the following chain of equalities −2i[J, T] = [J, [P+, P−]] = [[J, P+], P−]−[[J, P−], P+] = −i[P+, P−]−i[P−, P+], we deduce [P−, P+] = 2iT + 2r0 2sv0 2s+ 2r2 2sv2 2s. In order to identify the products [T, P+]and [T, P−], we introduce the notations: [T, P+] = m0 0v0 0+ 2s X k=1 m0 2kv0 2k+ 2s X k=1 m1 2k−1v1 2k−1+m2 0v2 0+ 2s X k=1 m2 2kv2 2k, [T, P−] = t0 0v0 0+ 2s X k=1 t0 2kv0 2k+ 2s X k=1 t1 2k−1v1 2k−1+t2 0v2 0+ 2s X k=1 t2 2kv2 2k. In a similar way as above we obtain Leibniz identity Constraint {T, P+, J}=⇒[T, P+] = m1 2s−1v1 2s−1, {T, P−, J}=⇒[T, P−] = t1 2s+1v1 2s+1, {P+, P−, T}=⇒[T, T] = −s(2s+ 1)r0 2sv2 2s, {P+, P+, T}=⇒q0 2s−2=−(s+ 1)(2s+ 1)r0 2s, {P+, P+, P−}=⇒m1 2s−1=−i/2(2s+ 1)(2s2+s+ 1)r0 2s, {P−, P−, T}=⇒l0 2s+2 =−(2s+ 1)(s+ 1)r0 2s, {T, P+, P−}=⇒t1 2s+1 =−i/2(2s+ 1)(2s2+s+ 3)r0 2s. 3.3.1 Leibniz algebras whose ideal Iis the DC-module U1 n81 From these restrictions, we derive [P+, P+] = −(s+ 1)(2s+ 1)r0 2sv0 2s−2+q2 2s−2v2 2s−2, [T, P+] = −i/2(2s+ 1)(2s2+s+ 1)r0 2sv1 2s−1, [P−, P−] = −(2s+ 1)(s+ 1)r0 2sv0 2s+2 +l2 2s+2v2 2s+2, [T, P−] = −i/2(2s+ 1)(2s2+s+ 3)r0 2sv1 2s+1. Finally, if we apply the Leibniz identity to the triple of elements {P−, P+, P−}, we obtain r0 2s= 0. Thus, by assuming (r2 2s, a2 2s, q2 2s−2, l2 2s+2) = (α1, α2, α3, α4), we get the first family of the theorem. Case 2. Let n= 4s−2. Taking the change in the following form: J0=J+ia0 0 2s−1v0 0+ 2s−1 X k−1 ia0 2k 2s−2k−1v0 2k+ s−1 X k=1 ia1 2k−1 2s−2kv1 2k−1 + 2s−1 X k=s+1 ia1 2k−1 2s−2kv1 2k−1+ia2 0 2s−1v2 0+ 2s−1 X k=1 ia2 2k 2s−2k−1v2 2k, we can assume that [J, J] = a1 2s−1v1 2s−1. Analogously to the previous case, we can deduce the products [J, P+] = −iP+,[J, P−] = iP−,[P+, P−] = −2iT. Verifying the Leibniz identity on triples of elements we have the following restrictions: Leibniz identity Constraint {J, J, P+}=⇒[P+, J] = iP++ 2isa1 2s−1v2 2s−2, {J, J, P−}=⇒[P−, J] = −iP−−2isa1 2s−1v2 2s. 82 3 Leibniz algebras corresponding to the Diamond Lie algebras We put [P+, P+] = q0 0v0 0+ 2s−1 X k=1 q0 2kv0 2k+ 2s−1 X k=1 q1 2k−1v1 2k−1+q2 0v2 0+ 2s−1 X k=1 q2 2kv2 2k, [P−, P−] = l0 0v0 0+ 2s−1 X k=1 l0 2kv0 2k+ 2s−1 X k=1 l1 2k−1v1 2k−1+l2 0v2 0+ 2s−1 X k=1 l2 2kv2 2k, [J, T] = r0 0v0 0+ 2s−1 X k=1 r0 2kv0 2k+ 2s−1 X k=1 r1 2k−1v1 2k−1+r2 0v2 0+ 2s−1 X k=1 r2 2kv2 2k. From the Leibniz identity, we have Leibniz identity Constraint {P+, J, P+}=⇒[P+, P+] = q1 2s−3v1 2s−3, {P−, J, P−}=⇒[P−, P−] = l1 2s+1v1 2s+1, {J, J, T}=⇒[J, T] = r1 2s−1v1 2s−1, {P+, J, P−}=⇒[T, J] = 0, {J, P+, T}=⇒[P+, T ] = 2isr1 2s−1v2 2s−2, {J, P−, T}=⇒[P−, T ] = −2isr1 2s−1v2 2s, {J, P+, P−}=⇒[P−, P+] = 2iT + 2r1 2s−1v1 2s−1. Setting [T, P+] = m0 0v0 0+ 2s−1 X k=1 m0 2kv0 2k+ 2s−1 X k=1 m1 2k−1v1 2k−1+m2 0v2 0+ 2s−1 X k=1 m2 2kv2 2k, [T, P−] = t0 0v0 0+ 2s−1 X k=1 t0 2kv0 2k+ 2s−1 X k=1 t1 2k−1v1 2k−1+t2 0v2 0+ 2s−1 X k=1 t2 2kv2 2k, and applying the Leibniz identity to the following triples of elements: {T, P+, J},{T, P−, J},{P+, P−, T}, {P+, P+, P−},{T, P+, P−},{P−, P+, P−}, 3.3.2 Leibniz algebras whose ideal Iis the DC-module U2 n83 we derive [T, P+] = (i(s−1)q1 2s−3−2isr1 2s−1)v2 2s−2, [T, P−] = (4isr1 2s−1−i(s−1)l1 2s−2)v2 2s, [T, T ] = 0. Finally, by denoting (a1 2s−1, r1 2s−1, q1 2s−3, l1 2s+1) = (β1, β2, β3, β4), we obtain the second family. 3.3.2 Leibniz algebras whose ideal Iis the DC-module U2 n Suppose that the ideal Iis defined as a Leibniz DC-module by the irreducible representation U2 nand {v0 2k−1, v1 0, v1 2k, v2 2k−1}k=1,...,n/2for even nis the basis of Ichosen as in Proposition 3.1.1. Then the products [I, DC]have the following form:                              [v0 2k−1, J] = i 2(n−4k+ 2)v0 2k−1, k = 1, . . . , n 2, [v1 2k, J] = i 2(n−4k)v1 2k, k = 0,...,n 2, [v2 2k−1, J] = i 2(n−4k+ 2)v2 2k−1, k = 1, . . . , n 2, [v0 2k−1, P+] = (n−2k+ 2)v1 2k−2, k = 1,...,n 2, [v1 2k, P+]=(n−2k+ 1)v2 2k−1, k = 1, . . . , n 2, [v0 2k−1, P−] = 2kv1 2k, k = 1,...,n 2, [v1 2k, P−] = (2k+ 1)v2 2k+1, k = 0,...,n 2−1, [v0 2k−1, T] = −i/2(n−4k+ 2)v2 2k−1, k = 1,...,n 2. Theorem 3.3.2. An arbitrary complex Leibniz algebra with corresponding Lie algebra DCand with the ideal Idefined as a Leibniz DC-module U2 nadmits a basis {J, P+, P−, T, v0 2k−1, v1 0, v1 2k, v2 2k−1}k=1,...,n/2, where nis even, and such that the multiplication table [DC,DC]has the following form: 84 3 Leibniz algebras corresponding to the Diamond Lie algebras •n= 4s                                                                                 [J, P+] = −iP+, [P+, J] = iP++i(2s+ 1)γ1v2 2s−1, [J, P−] = iP−, [P−, J] = −iP−−i(2s+ 1)γ1v2 2s+1, [P+, P−] = −2iT, [P−, P+] = 2iT + 2γ2v1 2s, [J, J] = γ1v1 2s, [J, T] = γ2v1 2s, [P+, P+] = γ3v1 2s−2, [P−, P−] = γ4v1 2s+2, [P+, T] = i(2s+ 1)γ2v2 2s−1, [T, P+] = −i((2s+ 1)γ2−(2s−1)γ3 2)v2 2s−1, [P−, T] = −i(2s+ 1)γ2v2 2s+1, [T, P−] = i(2(2s+ 1)γ2−(2s−1)γ4 2)v2 2s+1, •n= 4s−2          [J, P+] = −iP+,[J, P−] = iP−,[P+, P−] = −2iT, [P+, J] = iP+,[P−, J] = −iP−,[P−, P+] = 2iT + 2δ1v2 2s−1, [J, T] = δ1v2 2s−1[J, J] = δ2v2 2s−1,[P+, P+] = δ3v2 2s−3, [P−, P−] = δ4v2 2s+1, where γi, δi∈C,1≤i≤4. Proof. Let us denote [J, J] = n/2 X k=1 a0 2k−1v0 2k−1+a1 0v1 0+ n/2 X k=1 a1 2kv1 2k+ n/2 X k=1 a2 2k−1v2 2k−1. 3.3.2 Leibniz algebras whose ideal Iis the DC-module U2 n85 In a similar way to the proof of Theorem 3.3.1 we will consider the cases n= 4sand n= 4s−2. Case 1. Let n= 4s. Taking the change of element Jas follows: J0=J+ 2s X k=1 ia0 2k−1 2s−2k+ 1v0 2k−1+ia1 0 2sv1 0+ s−1 X k=1 ia1 2k 2s−2kv1 2k+ 2s X k=s+1 ia1 2k 2s−2kv1 2k + 2s X k=1 ia2 2k−1 2s−2k+ 1v2 2k−1, we can assume that [J, J] = a1 2sv1 2s. Applying similar arguments as in the proof of Theorem 3.3.1, we derive [J, P+] = −iP+,[J, P−] = iP−,[P+, P−] = −2iT. We set [P+, P+] = 2s X k=1 q0 2k−1v0 2k−1+q1 0v1 0+ 2s X k=1 q1 2kv1 2k+ 2s X k=1 q2 2k−1v2 2k−1, [P−, P−] = 2s X k=1 l0 2k−1v0 2k−1+l1 0v1 0+ 2s X k=1 l1 2kv1 2k+ 2s X k=1 l2 2k−1v2 2k−1, [J, T] = 2s X k=1 r0 2k−1v0 2k−1+r1 0v1 0+ 2s X k=1 r1 2kv1 2k+ 2s X k=1 r2 2k−1v2 2k−1. Considering the Leibniz identity to the following triples of elements: {J, P+, J},{J, P−, J},{P+, J, P+},{P−, J, P−},{J, J, T},{P+, J, P−}, we deduce restrictions which imply the expressions for the products [P+, J] = iP++ia1 2s(2s+ 1)v2 2s−1,[P−, J] = −iP−−ia1 2s(2s+ 1)v2 2s+1, [P+, P+] = q1 2s−2v1 2s−2,[P−, P−] = l1 2s+2v1 2s+2, [J, T] = r1 2sv1 2s,[T, J] = 0. 86 3 Leibniz algebras corresponding to the Diamond Lie algebras Moreover, we have Leibniz identity Constraint {J, P+, T}=⇒[P+, T ] = ir1 2s(2s+ 1)v2 2s−1, {J, P−, T}=⇒[P−, T ] = −ir1 2s(2s+ 1)v2 2s+1, {J, P+, P−}=⇒[P−, P+] = 2iT + 2r1 2sv1 2s. We also denote [T, P+] = 2s X k=1 m0 2k−1v0 2k−1+m1 0v1 0+ 2s X k=1 m1 2kv1 2k+ 2s X k=1 m2 2k−1v2 2k−1, [T, P−] = 2s X k=1 t0 2k−1v0 2k−1+t1 0v1 0+ 2s X k=1 t1 2kv1 2k+ 2s X k=1 t2 2k−1v2 2k−1. Applying the Leibniz identity to the triples {T, P+, J},{T, P−, J}, {P+, P−, P−}, we obtain [T, P+] = m0 2s−1v0 2s−1+m2 2s−1v2 2s−1, [T, P−] = t0 2s+1v0 2s+1 +t2 2s+1v2 2s+1, [T, T ] = 0. Finally, we have Leibniz identity Constraint {P+, P+, P−}=⇒m0 2s−1= 0, m2 2s−1= 1/2i(2s−1)q1 2s−2−i(2s+ 1)r1 2s, {T, P+, P−}=⇒t0 2s+1 = 0, {P−, P+, P−}=⇒t2 2s+1 = 2i(2s+ 1)r1 2s−1/2i(2s−1)l1 2s+2. Denoting the parameters (a1 2s, r1 2s, q1 2s−2, l1 2s+2)by (γ1, γ2, γ3, γ4), we obtain the first family of the theorem. Case 2. Let n= 4s−2. The second family of the theorem is obtained by applying similar arguments as in the previous case. 3.3.3 Leibniz algebras whose ideal Iis either W1 nor W2 n87 3.3.3 Leibniz algebras whose ideal Iis the Leibniz DC-module either W1 nor W2 n Lemma 3.3.3. Let Lbe a Leibniz algebra such that L/I ∼ =DC, where DC is the Diamond Lie algebra and Iis its right DC-module. If there exists a basis {X1, X2, . . . , Xn}of Isuch that [Xk, J] = αkXk, αk/∈{−2i, 0,2i}, for 1≤k≤n, where iis the imaginary unit, then [DC,DC]has the following form: [J, P+] = −[P+, J] = −iP+, [J, P−] = −[P−, J] = iP−, [P+, P−] = −[P−, P+] = −2iT, Proof. Here we shall use the multiplication table (3.1.1) of the complex Diamond Lie algebra. Let us assume that [J, J] = n P k=1 mkXk.Then by setting J0:= J− n P k=1 mk αkXk,we can assume that [J, J] = 0. Let us denote [J, P+] = −iP++ n X k=1 qkXk,[J, P−] = −P−+ n X k=1 rkXk. Taking the following basis transformation: J0=J, P0 +=P+− n X k=1 qkXk, P0 −=P−+ n X k=1 rkXk, T0=i 2[P0 +, P0 −], we can assume that [J, P+] = −iP+,[J, P−] = −P−,[P+, P−] = −2iT. Applying the Leibniz identity for the triples {J, J, P+},{J, J, P−}we derive [P+, J] = −[J, P+],[P−, J] = −[J, P−]. We put [J, T] = n X k=1 tkXk. 94 3 Leibniz algebras corresponding to the Diamond Lie algebras Let us denote by V=R2m+2 the natural ψ(Dm)-module and endow it with aDm-module structure, V×Dm→V, given by (x, e):=xψ(e), where x∈Vand e∈Dm. Then the action of Dmon V=hX1, X2, . . . , X2m+2iis given below:                          (Xk, J) = −X2m+3−k,2≤k≤m+ 1, (Xk, J) = X2m+3−k, m + 2 ≤k≤2m+ 1, (X1, Pk) = Xk+1,1≤k≤m, (X2m+2−k, Pk) = −X2m+2,1≤k≤m, (X1, Qk) = X2m+2−k,1≤k≤m, (Xk+1, Qk) = X2m+2,1≤k≤m, (X1, T) = 2X2m+2,1≤k≤m, (3.5.1) and the remaining products in the action being zero. Theorem 3.5.2. An arbitrary real Leibniz algebra with corresponding Lie algebra Dm, and the Iideal associated as Dm-module defined by (3.5.1), admits a basis {J, P1, P2, . . . , Pm, Q1, Q2, . . . , Qm, T, X1, X2, . . . , X2m+2}such that the multiplication table [Dm,Dm]has the following form:      [J, J] = a1X2m+2,[J, Pk] = −[Pk, J] = Qk, [J, Qk] = −[Qk, J] = −Pk,[Pk, Qk] = −[Qk, Pk] = T, [Pk, Ps] = [Qk, Qs] = bk,sX2m+2,[Pk, Qs] = [Qk, Ps] = ck,sX2m+2, with the restrictions bk,s =−bs,k, ck,s =cs,k,1≤k, s ≤m, k 6=s. Proof. Let us assume that [J, J] = 2m+2 X k=1 δkXk,[J, T] = 2m+2 X k=1 ρkXk 3.5 Leibniz algebras constructed by a faithful representation 95 Then by setting J0=J−1 2ρ2m+2X1+ m+1 X k=2 δ2m+3−kXk− 2m+1 X k=m+2 δ2m+3−kXk, and considering the Leibniz identity for the triple {J, T, J}, we get [J, J] = δ2m+2X2m+2,[J, T] = ρ1X1. Let us suppose [J, Pk] = Qk+ 2m+2 X s=1 λk,sXs,[J, Qk] = −Pk+ 2m+2 X s=1 µk,sXs,1≤k≤m. Taking the following basis transformation: J0=J, P0 k=Pk− 2m+2 X s=1 µk,sXs, Q0 k=Qk+ 2m+2 X k=1 λk,sXs, T0= [P0 1, Q0 1],1≤k≤m, we can assume that [J, Pk] = Qk,[J, Qk] = −Pk,[P1, Q1] = T, 1≤k≤m. By applying the Leibniz identity to the triples {J, J, Pk},{J, J, Qk}we derive [Pk, J] = −[J, Pk],[Qk, J] = −[J, Qk],1≤k≤m. The verification of the Leibniz identity leads to the following restrictions. Leibniz identity Constraints {J, Pk, T}=⇒[Qk, T ] = ρ1Xk+1,1≤k≤m, {J, Qk, T}=⇒[Pk, T ] = −ρ1X2m+2−k,1≤k≤m, {P1, T, Q1}=⇒[T, T] = 0, 96 3 Leibniz algebras corresponding to the Diamond Lie algebras We set                [Pj, Qj] = T+ 2m+2 P t=1 βj,tXt,[Qk, Pk] = −T+ 2m+2 P t=1 γk,tXt, [Pk, Ps] = 2m+2 P t=1 ηk,s,tXt,[Qk, Qs] = 2m+2 P t=1 θk,s,tXt, [Pk, Qs] = 2m+2 P t=1 νk,s,tXt, k 6=s, [Qk, Ps] = 2m+2 P t=1 ξk,s,tXt, k 6=s, where 2≤j≤m, 1≤k, s ≤m. From the Leibniz identity for the triples {Pk, Ps, T},{Qk, Qs, T}and {J, Pk, Qk}we obtain ηk,k,1=θk,k,1=1 2ρ1, ηk,s,1=θk,s,1= 0, θk,k,t =−ηk,k,t, 1≤k, s ≤m, k 6=s, 2≤t≤2m+ 2. Analogously, from {J, Pk, Qs},{J, Qk, Qs}and {J, Pk, Ps}, we get [Pk, Ps] = −[Qs, Qk],[Pk, Qs]=[Ps, Qk],(3.5.2) [Qk, Ps]=[Qs, Pk],1≤k, s ≤m, k 6=s. By applying the Leibniz identity to the triples {P1, J, Q1},{P1, P1, Q1} and {Q1, P1, Q1}, we have [T, J] = 2m+2 X s=2 2η1,1,sXs, [T, P1] = 3 2ρ1X2m+1 +η1,1,2X2m+2, [T, Q1] = −3 2ρ1X2. By the next identities {Q1, J, P1}and {P1, J, P1}we deduce γ1,s =−4η1,1,2m+3−s, γ1,k = 4η1,1,2m+3−k, η1,1,2m+2 =γ1,1=γ1,2m+2 =η1,1,t = 0, with 2≤s≤m+ 1, m + 2 ≤k≤2m+ 1,2≤t≤2m+ 1. 3.5 Leibniz algebras constructed by a faithful representation 97 Hence, we have      [T, J]=0,[Q1, P1] = −T, [P1, P1] = 1 2ρ1X1,[Q1, Q1] = 1 2ρ1X1, [T, P1] = 3 2ρ1X2m+1,[T, Q1] = −3 2ρ1X2, By using the next Leibniz identity for {Pk, Pk, Qk},{Qk, Pk, Qk}we get [T, Pk] = 3 2ρ1X2m+2−k−βk,1Xk+1+(βk,2m+2−k+ηk,k,k+1)X2m+2,2≤k≤m. [T, Qk] = −3 2ρ1Xk+1 −βk,1X2m+2−k By applying the Leibniz identity to the triples of elements {Pk, J, Qk}and {Qk, J, Pk}, we get βk,s =γk,s =−2ηk,k,2m+3−s, βk,t =γk,t = 2ηk,k,2m+3−t, ηk,k,2m+2 = 0, where 2≤k≤m, 2≤s≤m+ 1, m + 2 ≤t≤2m+ 1. By the next Leibniz identity applied to {Pk, J, Pk}and {T, Pk, Qk}, we have γk,1=βk,1= 0, ηk,k,s = 0,2≤k≤m, 2≤s≤2m+ 1. So, we have          [Pk, Qk] = −[Qk, Pk] = T, [Pk, Pk] = [Qk, Qk] = 1 2ρ1X1, [T, Pk] = 3 2ρ1X2m+2−k, [T, Qk] = −3 2ρ1Xk+1, where 2≤k≤m. By verifying the Leibniz identity on elements, we obtain the following restrictions. Leibniz identity Constraints {Pk, Ps, T}=⇒ηk,s,1= 0,1≤k, s ≤m, k 6=s, {Qk, Qs, T}=⇒θk,s,1= 0,1≤k, s ≤m, k 6=s, {Pk, Qs, T}=⇒νk,s,1= 0,1≤k, s ≤m, k 6=s, {Qk, Ps, T}=⇒ξk,s,1= 0,1≤k, s ≤m, k 6=s. 98 3 Leibniz algebras corresponding to the Diamond Lie algebras By applying the Leibniz identity to the triples {Pk, Ps, J},{Qk, Qs, J}, we get [[Pk, Ps], J] = −[Qk, Ps]−[Pk, Qs],[[Qk, Qs], J] = [Qk, Ps]+[Pk, Qs], it follows that [[Pk, Ps], J] = −[[Qk, Qs], J], hence ξk,s,2m+2 =−νk,s,2m+2, θk,s,t =−ηk,s,t,1≤k, s ≤m, 2≤t≤2m+ 1, k 6=s. and νk,s,t +ξk,s,t =−ηk,s,2m+3−t2≤t≤m+ 1, νk,s,t +ξk,s,t =ηk,s,2m+3−tm+ 2 ≤t≤2m+ 1. Let us consider the identity [[Qk, Ps], J] = [Qk,[Ps, J]] + [[Qk, J], Ps] = −[Qk, Qs]+[Pk, Ps] We have that θk,s,2m+2 =ηk,s,2m+2 and ξk,s,t =−2ηk,s,2m+3−t,2≤t≤m+ 1, ξk,s,t = 2ηk,s,2m+3−t, m + 2 ≤t≤m+ 1, and νk,s,t =ηk,s,2m+3−t,2≤t≤m+ 1, νk,s,t =−ηk,s,2m+3−t, m + 2 ≤t≤2m+ 1. Analogously, by applying the Leibniz identity to the triple {Pk, Qs, J}, we get νk,s,t =−2ηk,s,2m+3−t,2≤t≤m+ 1, νk,s,t = 2ηk,s,2m+3−t, m + 2 ≤t≤2m+ 1. We get that νk,s,t = 0,1≤k, s ≤m, 2≤t≤2m+ 1, k 6=s. It implies that ηk,s,t =ξk,s,t =θk,s,t = 0 for 1≤k, s ≤m, 2≤t≤2m+ 1, k 6=s. Hence, we have [Pk, Ps] = [Qk, Qs] = ηk,s,2m+2X2m+2,1≤k, s ≤m, k 6=s, [Pk, Qs]=[Qk, Ps] = νk,s,2m+2X2m+2,1≤k, s ≤m, k 6=s. 3.5 Leibniz algebras constructed by a faithful representation 99 By equation (3.5.2) we have the following restrictions ηk,s,2m+2 =−ηs,k,2m+2, νk,s,2m+2 =νs,k,2m+2,1≤k, s ≤m, k 6=s. Finally, we apply the Leibniz identity to the triple {Pk, Pk, Ps}, with k6= s, and we obtain ρ1= 0. 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