Naturally graded Zinbiel algebras with nilindex n - 3

NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 J.Q. ADASHEV, L.M. CAMACHO, S. G ´ OMEZ-VIDAL, I.A. KARIMJANOV Abs ac . We p esen he classi ica ion o a subclass o n-dimensional na u ally g aded Zinbiel algeb as. This subclass has he nilindex n−3 and he cha ac e is ic sequence (n−3,2,1).In ac , his esul comple es he classi ica ion o na u ally g aded Zinbiel algeb as o nilindex n−3. Ma hema ics Subjec Classi ica ion 2010: 17A32. Key Wo ds and Ph ases: Zinbiel algeb a, Leibniz algeb a, nilpo ency, cha ac e is ic sequence 1. In oduc ion. In ensi e in es iga ion on Lie algeb as leads o he appea ance o a new algeb aic objec – Leibniz algeb as. The Leibniz algeb as in oduced by Loday in [7] a e a ”non commu a i e” algeb as analogue o Lie algeb as. I should be men ioned ha Leibniz algeb as inhe i an impo an Lie algeb a p ope y: he ope a o o igh mul iplica ion on an elemen o an algeb a is a de i a ion. Leibniz algeb as o m a Koszul ope ad in he sense o V. Ginzbu g and M. Kap ano [6]. Unde he Koszul duali y he ope ad o Lie algeb as is dual he ope ad o associa i e and commu a i e algeb as. The no ion o dual Leibniz algeb a de ined by J.-L. Loday [8] is p ecisely he dual ope ad o Leibniz algeb as in his sense. In his pape , we s udy algeb as which a e he dual o Leibniz algeb as in Koszul sense. J.-L. Loday s udied in [8] ca ego ical p ope ies o Leibniz algeb as and conside ed in his connec ion a new objec – Zinbiel algeb as (Leibniz is w i en in e e se o de ). Since he ca ego y o Zinbiel algeb as is Koszul dual o he ca ego y o Leibniz algeb as, some imes hey a e also called dual Leibniz algeb as. In [2, 5, 9] some c ucial p ope ies o Zinbiel algeb as we e ob ained. Pa icula ly, in [5], he au ho s p o e ha e e y ini e-dimensional Zinbiel algeb a o e complex numbe s is nilpo en . Howe e , he s udy o nilpo en algeb as is oo complex and should be ca ied ou wi h addi ional condi ions, such as condi ions on nilindex, a ious ypes o g ada ions, cha ac e is ic sequence and o he s. The aim o his wo k is o con inue he s udy o complex ini e-dimensional na u ally g aded Zinbiel algeb as. The n-dimensional Zinbiel algeb as o nilindex kwi h n−2≤k≤na e classi ied in [1, 2]. The classi ica ion o complex n-dimensional na u ally g aded Zinbiel algeb as o nilindex n−3 is a di icul p oblem and i should be di ided in o h ee cases. Namely, i is necessa y o conside he possibili ies o he cha ac e is ic sequence o such algeb as: (n−3,3), (n−3,1,1,1) and (n−3,2,1). The classi ica ion o complex na u ally g aded Zinbiel algeb as o nilindex n−3 wi h cha ac e is ic sequence equal o (n−3,3) and (n−3,1,1,1) has been done in [1]. The knowledge o na u ally g aded algeb as o a ce ain amily o e s signi ican in o ma ion abou hei s uc u al p ope ies. In his pape we ob ain he classi ica ion o na u ally g aded Zinbiel algeb as o nilindex n−3 wi h cha ac e is ic sequence (n−3,2,1).Thus, we comple e he s udy o he n−3 case. All he spaces and he algeb as a e conside ed o e he ield o complex numbe s. We omi he p oduc s which a e equal o ze o o con enience. Th oughou all he wo k we use he so wa e Ma hema ica (see [3]) o compu e he Zinbiel iden- i y in low dimensions and o o mula e he gene aliza ions o he calcula ions, which a e p o ed o a bi a y dimension. Mo eo e , he p og am allows us o cons uc new bases using some gene al ans o ma ion o he gene a o s o he algeb a. Since he di ec sum o nilpo en Zinbiel algeb as is nilpo en , we shall conside only non spli algeb as. 2. P elimina ies In his sec ion we in oduce some de ini ions, no a ions and esul s, which a e necessa y o he unde s anding o g aded Zinbiel algeb as. 1 2 J.Q. ADASHEV, L.M. CAMACHO, S. G´ OMEZ-VIDAL, I.A. KARIMJANOV De ini ion 2.1. A ec o space Zo e a ield Kwi h a bilinea ope a ion “◦” is called Zinbiel algeb a i o any x, y, z ∈ Z he ollowing iden i y (2.1) (x◦y)◦z=x◦(y◦z) + x◦(z◦y) holds. Examples o Zinbiel algeb as can be ound in [2, 5, 8]. Z(a, b, c) deno es he ollowing polynomial: Z(a, b, c) = (a◦b)◦c−a◦(b◦c)−a◦(c◦b). Zinbiel algeb as a e de ined by he iden i y Z(a, b, c) = 0. Fo a gi en Zinbiel algeb a Z he sequence o wo-sided ideals de ined ecu si ely as ollow: Z1=Z,Zk+1 =Z ◦ Zk, k ≥1. is said o be he lowe cen al se ies. De ini ion 2.2. A Zinbiel algeb a Zis called nilpo en i he e exis s s∈Nsuch ha Zs6= 0 and Zs+1 = 0.The minimal numbe ssa is ying his p ope y is called he index o nilpo ency o nilindex o he algeb a Z. Fo a gi en Zinbiel algeb a Zwe in oduce deno a ions: R(Z) = {x∈ Z | y◦x= 0 o any y∈ Z} − − he igh annihila o o Z, L(Z) = {x∈ Z | x◦y= 0 o any y∈ Z} − − he le annihila o o Z, Cen (Z) = {x, y ∈ Z | x◦y=y◦x= 0 o any y∈ Z} − − he cen e o Z. I is easy o see ha he cen e and he igh annihila o o Za e wo-sided ideals. Le us deno e by Lx he ope a o o le mul iplica ion on elemen x, i.e. Lx:Z −→ Z such ha Lx(y) = x◦y o any y∈ Z. Le Zbe a complex n-dimensional Zinbiel algeb a and xbe an elemen o he se Z Z2. Fo he ope a o Lxwe de ine a descending sequence C(x) = (n1, n2,...,nk) wi h n1+···+nk=n, which consis s o he dimensions o he Jo dan blocks o he ope a o Lx. In he se o such sequences we conside he lexicog aphic o de , ha is, C(x) = (n1, n2,...,nk)< C(y) = (m1, m2,...,ms) i he e exis s isuch ha ni< miand nj=mj o j < i. Taking in o accoun he equali y n1+···+nk= m1+···+mssuch compa ison is always applicable. De ini ion 2.3. The sequence C(Z) = max{C(x) : x∈ Z Z2}is called he cha ac e is ic sequence o he algeb a Z. In [5], he au ho s p o e ha Zinbiel algeb as o ini e dimension a e nilpo en . Since we ocused ou a en ion on ini e dimension complex nilpo en Zinbiel algeb as. Le Zbe a ini e-dimensional nilpo en Zinbiel algeb a wi h nilindex equal o s. Fo i(1 ≤i≤s) we pu Zi=Zi/Zi+1 and we ob ain he g aded Zinbiel algeb a g (Z) = Z1⊕ Z2⊕...⊕ Zs,whe e Zi◦ Zj⊆ Zi+j. An algeb a Zi called na u ally g aded i Z∼ =g (Z).I is no di icul o see ha Zi+1 =Z1◦ Zi in he na u ally g aded algeb a Z. Le Zbe a na u ally g aded Zinbiel algeb a wi h cha ac e is ic sequence (n−3,2,1).By de ini ion o cha ac e is ic sequence he e exis s a basis {e1, e2,...,en}in he algeb a Zsuch ha he ope a o Le1has one block Jn−3o size (n−3),one block J2o size 2 and one block J1o size one. No e ha he e will be six possibili ies o he ope a o s Le1.By a change o basis i is easy o p o e ha he six cases can be educed o he ollowing h ee cases: I.   Jn−30 0 0J20 0 0 J1  , II.   J20 0 0Jn−30 0 0 J1  , III.  J10 0 0Jn−30 0 0 J2  . De ini ion 2.4. A Zinbiel algeb a Zis called ei he o i s ype ( ype I), second ype ( ype II) o hi d ype ( ype III) i he ope a o Le1has he o m: I.   Jn−30 0 0J20 0 0 J1  , II.   J20 0 0Jn−30 0 0 J1  , III.  J10 0 0Jn−30 0 0 J2   NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 3 espec i ely. F om now on we deno e by Cj i he combina o ial numbe s Cj i=i j. The ollowing esul holds: Lemma 2.5. [4] Le Zbe a Zinbiel algeb a such ha e1◦ei=ei+1 o 1≤i≤k−1,wi h espec o he adap ed basis {e1,...,ek, ek+1,...,en}.Then ei◦ej=Cj i+j−1ei+j, o 2≤i+j≤k 3. Main Resul 3.1. Type I. Algeb as o ype I wi h n≥8.So, we ha e he ollowing b acke s:            e1◦ei=ei+1,1≤i≤n−4, e1◦en−3= 0, e1◦en−2=en−1, e1◦en−1= 0, e1◦en= 0. I is easy o see ha Zi⊇ heiiwhe e 1 ≤i≤n−3. I is e iden ha dim(Z1)>1.In ac , i dim(Z1) = 1, hen he algeb a Zis one-degene a ed and he e o e i is a ze o- ili o m algeb a, bu i is no an algeb a o nilindex n−3.Le us assume ha en−2∈ Z 1and en∈ Z 2, hen en−1∈ Z 1+1. We can dis inguish he ollowing cases: Case I. I 1= 2= 1. Then we ha e ha Z1=< e1, en−2, en>, Z2=< e2, en−1>, Z3=< e3>, . . . , Zn−3=< en−3> and he ollowing p oduc s: e1◦e1=e2, e1◦en−2=en−1, en−2◦e1=α1e2+α2en−1, en−2◦en−2=α3e2+α4en−1, en−2◦en=α5e2+α6en−1, en◦e1=β1e2+β2en−1, en◦en−2=β3e2+β4en−1, en◦en=β5e2+β6en−1, e1◦e2=e3, en−2◦e2=γ1e3, en−2◦en−1=γ2e3, en◦e2=γ3e3, en◦en−1=γ4e3. F om he equali y Z(e1, en, e1) = Z(e1, en, en) = 0 we ha e β1=β5= 0. Le us conside he equali ies Z(e1, en−2, e1) = Z(e1, en−1, e1) = 0 hen i ollows α1= 0. F om he equali ies Z(e1, e1, en−2) = Z(en−2, e1, e1) = Z(en−2, en−1, e1) = Z(e1, e1, en) = 0 Z(e1, en, e2) = Z(en, en−1, e1) = Z(e1, en−2, en−2) = Z(e1, en−1, en−2) = 0 Z(e1, en−2, en) = Z(e1, en, en−1) = Z(e1, e1, en−1) = Z(e1, en−2, e2) = 0 Z(en−2, en, e1) = Z(en−2, en−2, e1) = 0 we ob ain γ1=γ2=γ3=γ4=α3=α5=β3= 0, and e2◦en−2=en−2◦e2=e2◦en−1=en−1◦e2=e2◦en=en◦e2= 0. Now, by ma hema ical induc ion me hod, we p o e ha en−1◦ek= 0 and ek◦en−1= 0 wi h 2≤k≤n−3. •I k= 2, hen we ha e en−1◦e2=e2◦en−1= 0. •Le us suppose ha o some k he equali ies en−1◦ek= 0 and ek◦en−1= 0 a e ue. We p o e i o k+ 1. en−1◦ek+1 =en−1◦(e1◦ek) = (en−1◦e1)◦ek−en−1◦(ek◦e1) = =−C1 ken−1◦ek+1 =−ken−1◦ek+1, en−1◦ek+1 = 0. ek+1 ◦en−1= (e1◦ek)◦en−1=e1◦(ek◦en−1) + e1◦(en−1◦ek) = = 0 4 J.Q. ADASHEV, L.M. CAMACHO, S. G´ OMEZ-VIDAL, I.A. KARIMJANOV As in p e ious cases, i easy o see ha ek◦en−2=en−2◦ek= 0 and ek◦en=en◦ek= 0 o 2≤k≤n−3. Thus, we ha e ob ained he ollowing amily o algeb as: Z(a1, a2, a3, a4, a5, a6) :                        ei◦ej=Cj i+j−1ei+j,2≤i+j≤n−3, e1◦en−2=en−1, en−2◦e1=a1en−1, en−2◦en−2=a2en−1, en−2◦en=a3en−1, en◦e1=a4en−1, en◦en−2=a5en−1, en◦en=a6en−1, whe e we omi he p oduc s ha a e equal o ze o. Theo em 3.1. An a bi a y Zinbiel algeb a o he amily Z(a1, a2, a3, a4, a5, a6)is isomo phic o one o he ollowing pai wise non-isomo phic algeb as: Z1(1,0,0,0,1,0), Z2(0,0,0,0,1,0), Z3(0,1,0,1,0,0), Z4(0,0,0,1,0,0), Z5(0,1,0,0,0,0), Z6(1,1,0,0,0,0), Z7(λ, 0,0,0,0,0), λ ∈C, Z8(0, λ, 1,0,0,1), λ ∈C {0}, Z9(α, −α (α−1)2,1,0,0,1), α ∈C {0,1}, Z10(0,0,1,0,1,1), Z11(1,0,1,0,1,1), Z12(0,0,1,1,0,0), Z13(0,0,1,0,0,0), Z14(λ, 1,1,0,1,1), λ ∈C, Z15(0,1,1,−1,1,1), Z16(1,1,1,0,1,1). P oo . Le Zbe sa is ying o he hypo hesis o he heo em. Due o he p ope y o na u al g ada ion o he algeb a i is enough o conside he ollowing change o gene a o s: e′ 1=P1e1+Pn−2en−2+Pnen, e′ n−2=Q1e1+Qn−2en−2+Qnen, e′ n=R1e1+Rn−2en−2+Rnen. Making he gene al change o basis in he amily Z(a1, a2, a3, a4, a5, a6),we de i e he exp essions o he new pa ame e s in he new basis (1): a′ 1=a1P1Qn−2+a2Pn−2Qn−2+a3PnQn−2+a4P1Qn+a5Pn−2Qn+a6PnQn P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn , a′ 2=a2Q2 n−2+a3Qn−2Qn+a5Qn−2Qn+a6Q2 n P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn , a′ 3=a2Qn−2Rn−2+a3Qn−2Rn+a5QnRn−2+a6QnRn P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn , a′ 4=a1P1Rn−2+a2Pn−2Rn−2+a3PnRn−2+a4P1Rn+a5Pn−2Rn+a6PnRn P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn , a′ 5=a2Qn−2Rn−2+a3QnRn−2+a5Qn−2Rn+a6QnRn P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn , a′ 6=a2R2 n−2+a3Rn−2Rn+a5Rn−2Rn+a6R2 n P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn , and he ollowing es ic ions: (2)          Q1=R1= 0, P1Rn−2+a2Pn−2Rn−2+a3Pn−2Rn+a5PnRn−2+a6PnRn= 0, P1Qn−2+a2Pn−2Qn−2+a3Pn−2Qn+a5PnQn−2+a6PnQn6= 0, P1(Qn−2Rn−QnRn−2)6= 0. We can dis inguish wo cases: Case 1. Le en∈R(Z) be, hen a3=a6= 0. NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 5 F om he es ic ions, (P1+a2Pn−2+a5Pn)Rn−2= 0, (P1+a2Pn−2+a5Pn)Qn−26= 0, P1(Qn−2Rn−QnRn−2)6= 0.    ⇒Rn−2= 0. i ollows ha P1Qn−2Rn6= 0.Thus, he new pa ame e s a e: a′ 1=a1P1Qn−2+a2Pn−2Qn−2+a4P1Qn+a5Pn−2Qn Qn−2(P1+a2Pn−2+a5Pn), a′ 2=a2Qn−2+a5Qn P1+a2Pn−2+a5Pn , a′ 4=Rn(a4P1+a5Pn−2) Qn−2(P1+a2Pn−2+a5Pn), a′ 5=a5Rn P1+a2Pn−2+a5Pn , We obse e ha he nulli y o a5is in a ian . Mo eo e , i is easy o check ha he nulli y o he ollowing exp ession a′ 2a′ 4−a′ 1a′ 5=(a2a4−a1a5)P1Rn (P1+a2Pn−2+a5Pn)2 is in a ian . Thus, we can dis inguish he ollowing non-isomo phic cases: Case 1.1. Le a56= 0 be. Then choosing Rn=P1+a2Pn−2+a5Pn a5 , Pn−2=−a4P1 a5 , Qn=−a2Qn−2 a5 we ha e a′ 5= 1, a′ 4= 0, a′ 2= 0, a′ 1=(a2a4−a1a5)P1 (a2a4−a5)P1−a2 5Pn and he de e minan is o med by he po encies o he ollowing non-ze o ac o s: P1Qn−2a5((a2a4− a5)P1−a2 5Pn). a) I a2a4−a1a56= 0, choosing Pn=P1(a1−1) a5we ecei e a′ 1= 1.I ollows he algeb a Z1(1,0,0,0,1,0). b) I a2a4−a1a5= 0, hen we ob ain a′ 1= 0 and we ha e he algeb a Z2(0,0,0,0,1,0). Case 1.2. Le a5= 0 be. Then, a′ 5= 0 and we ha e a′ 1=a1P1Qn−2+a2Pn−2Qn−2+a4P1Qn Qn−2(P1+a2Pn−2), a′ 2=a2Qn−2 P1+a2Pn−2 , a′ 4=a4P1Rn Qn−2(P1+a2Pn−2). wi h P1Qn−2Rn(P1+a2Pn−2)6= 0. We obse e ha he nulli ies o a2and a4a e in a ian , so we can dis inguish he ollowing cases: a) Le a46= 0 be. Then, choosing Rn=Qn−2(P1+a2Pn−2) a4P1 , Qn=−Qn−2(a1P1+a2Pn−2) a4P1 , we ge a′ 4= 1 and a′ 1= 0. a.1) I a26= 0, hen choosing Qn−2=P1+a2Pn−2 a2 ,we ob ain a′ 2= 1 and he algeb a Z3(0,1,0,1,0,0). a.2) I a2= 0, hen we ha e a′ 2= 0 and he algeb a Z4(0,0,0,1,0,0). 6 J.Q. ADASHEV, L.M. CAMACHO, S. G´ OMEZ-VIDAL, I.A. KARIMJANOV b) Le a4= 0 be. Then a′ 4= 0 and we ha e a′ 1=a1P1+a2Pn−2 P1+a2Pn−2 , a′ 2=a2Qn−2 P1+a2Pn−2 . We ha e ha he nulli y o he ollowing exp ession: a′ 1−1 = P1(a1−1) P1+a2Pn−2 . is in a ian . b.1) Le a26= 0 be. Then, choosing Qn−2=P1+a2Pn−2 a2 ,we ob ain a′ 2= 1. •I a1−16= 0, hen pu ing Pn−2=−a1P1 a2,we ha e a′ 1= 0 and he algeb a Z5(0,1,0,0,0,0). The de e minan o change o basis consis s o he po encies o he ollowing non-ze o ac o s a2(a1−1)P1Rn. •I a1−1 = 0, hen a′ 1= 1 and we ob ain Z6(1,1,0,0,0,0). b.2) Le a2= 0 be. Then, we ha e a′ 2= 0, a′ 1=a1=λ∈Cand he amily Z7(λ, 0,0,0,0,0),wi h λ∈C. Case 2. Le en/∈R(Z) be, hen (a3, a6)6= (0,0).We can suppose ha a36= 0,in ano he case, a3= 0 and a66= 0 we make he ollowing change o basis ′ 1= 1+ 3.Thus, a36= 0.Taking in o accoun he exp essions gi en in (1), he es ic ions (2) and he ollowing exp ession: ∆ = a3 3a4+a2 3a4a5−a1a2 3a4a5+a2a3a2 4a5−a1a3a4a2 5− −a1a2 3a6−3a2a3a4a6+a1a2a3a4a6−a2 2a2 4a6+ +a3a5a6+a2 1a3a5a6+a2a4a5a6+a1a2a4a5a6− −a1a2 5a6−a2a2 6+ 2a1a2a2 6−a2 1a2a2 6, he nulli y o he ollowing exp essions a e in a ian a′2 3−a′ 3a′ 5+a′2 5−a′ 2a′ 6=(a2 3−a3a5+a2 5−a2a6)(Qn−2Rn−QnRn−2)2 P1Qn−2+a2Pn−2Qn−2+a5PnQn−2+a3Pn−2Qn+a6PnQn , a′ 3a′ 5−a′ 2a′ 6=(a3a5−a2a6)(Qn−2Rn−QnRn−2)2 P1Qn−2+a2Pn−2Qn−2+a5PnQn−2+a3Pn−2Qn+a6PnQn , ∆′=∆P2 1(Qn−2Rn−QnRn−2)4 (a2Qn−2Rn−2+a5QnRn−2+a3Qn−2Rn+a6QnRn)2, a′ 3−a′ 5=(a3−a5)(Qn−2Rn−QnRn−2) P1Qn−2+a2Pn−2Qn−2+a5PnQn−2+a3Pn−2Qn+a6PnQn . We can dis inguish he ollowing non isomo phic cases: Case 2.1. Le a3a5−a2a66= 0 be. Then, choosing Pn−2=−(a6P1QnRn−2+a2a5Qn−2R2 n−2+a2 5QnR2 n−2−a6P1Qn−2Rn+ +a3a5Qn−2Rn−2Rn+a2a6Qn−2Rn−2Rn+ 2a5a6QnRn−2Rn+a3a6Qn−2R2 n+ +a6QnR2 n)1 (a3a5−a2a6)(Qn−2Rn−QnRn−2) Pn=−(−a3P1QnRn−2−a2 2Qn−2Rn−2−a2a5QnR2 n−2+a3P1Qn−2Rn− −2a2a3Qn−2Rn−2Rn−a3a5QnRn−2Rn−a2a6QnRn−2Rn−a2 3Qn−2R2 n− −a3a6QnR2 n)1 (a3a5−a2a6)(Qn−2Rn−QnRn−2) and using he es ic ion (2), we ob ain a′ 3= 1. NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 7 a) Le a3−a56= 0 be. Then, choosing Qn=−a2Qn−2Rn−2+a5Qn−2Rn a3Rn−2+a6Rn , Rn−2=a3Qn−2−a5Qn−2−a6Rn a3 we ge a′ 5= 0, a′ 6= 1 and a′ 2=λ∈C {0}.The de e minan o he change o basis is o med by he po encies o he ollowing non-ze o ac o s: (a3Rn−2+a6Rn)P1Qn−2(a2R2 n−2+a3Rn−2Rn+a5Rn−2Rn+a6R2 n). a.1) Le ∆ 6= 0 be. Then, we choose Rn=−(a2a3a4−a3a5−a2a4a5+a1a2 5+a2a6−a1a2a6)P1 (a3−a5)(a3a5−a2a6), Qn−2= (a2 3a4−a1a2 3a6−2a2a3a4a6+a3a5a6+a1a3a5a6+ +a2a4a5a6−a1a2 5a6−a2a2 6+a1a2a2 6)P1 (a3−a5)2(a3a5−a2a6), and we ge a′ 1=a′ 4= 0 and he amily Z8(0, λ, 1,0,0,1), λ ∈C {0}.The de e minan o change o basis is o med by he non-ze o po encies o he ollowing ac o s (a3−a5)(a3a5−a2a6)∆P1. a.2) Le ∆ = 0 be. Then, we ha e ∆′=a′ 5= 0, a′ 3=a′ 6= 1, a′ 4−a′ 1−3a′ 2a′ 4+a′ 2a′ 1a′ 4−a′2 2a′2 4−a′ 2(a′ 1−1)2= 0. Thus, we ob ain he ollowing amily ei◦ej=Cj i+j−1ei+j,2≤i+j≤n−3, e1◦en−2=en−1, en−2◦e1=αen−1, en−2◦en=en−1, en−2◦en−2=βen−1,wi h β6= 0 en◦en=en−1, en◦e1=γen−1,wi h γ−α−3βγ +αβγ −β2γ2−β(1 −α)2= 0 Now, we make he gene ic change o basis e′ 1=P1e1+Pn−2en−2+Pnen, e′ n−2=Q1e1+Qn−2en−2+Qnen, e′ n=R1e1+Rn−2en−2+Rnen. and we ha e he exp essions o he new pa ame e s and he new es ic ions: α′=αP1Qn−2+βPn−2Qn−2+γP1Qn+PnQn−2+Pn−2Qn+PnQn P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn , β′=βQ2 n−2+ 2Qn−2Qn+Q2 n P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn , γ′=αP1Rn−2+βPn−2Rn−2+γP1Rn+PnRn−2+Pn−2Rn+PnRn P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn , 1 = βQn−2Rn−2+Qn−2Rn+QnRn−2+QnRn P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn , 1 = βR2 n−2+ 2Rn−2Rn+R2 n P1Qn−2+βPn−2Qn−2+Pn−2Qn+PnQn−2+PnQn , 0 = P1Rn−2+βPn−2Rn−2+Pn−2Rn+PnRn−2+PnRn= 0.(∗∗) 8 J.Q. ADASHEV, L.M. CAMACHO, S. G´ OMEZ-VIDAL, I.A. KARIMJANOV Pu ing Pn−2=−P1QnRn−2−P1Qn−2Rn+βQn−2Rn−2Rn+Qn−2R2 n+QnR2 n β(QnRn−2−Qn−2Rn), Pn= (P1QnRn−2+β2Qn−2R2 n−2−P1Qn−2Rn+ 2βQn−2Rn−2Rn+ +βQnRn−2Rn+Qn−2R2 n+QnR2 n)1 β(QnRn−2−Qn−2Rn), Qn−2=Rn−2+Rn, Qn=−βQn−2Rn−2 Rn−2+Rn =−βRn−2, we ge α′=−(−P1Rn−2+ 2βP1Rn−2−αβP1Rn−2+β2γP1Rn−2+ +βR2 n−2−β2R2 n−2−P1Rn+βP1Rn−αβP1Rn+ +Rn−2Rn−βRn−2Rn+R2 n−βR2 n)1 β(βR2 n−2+Rn−2Rn+R2 n), γ′= (P1Rn−2−βP1Rn−2+αβP1Rn−2−βR2 n−2+P1Rn+ +βγP1Rn−Rn−2Rn−R2 n)1 β(βR2 n−2+Rn−2Rn+R2 n), β′=β6= 0 wi h P1(βR2 n−2+Rn−2Rn+R2 n)6= 0.I is easy o p o e ha : γ′−α′−3β′γ′+α′β′γ′−β′2γ′2−β′(α′−1)2=γ−α−3βγ +αβγ −β2γ2−β(α−1)2 βR2 n−2+Rn−2Rn+R2 n P2 1= 0. Now, i we choose Rn=1 2(P1+βγP1−Rn−2)− −q(P1+βγP1−Rn−2)2+ 4(P1Rn−2−βP1Rn−2+αβP1Rn−2−βR2 n−2) 2 we ge γ′= 0, β′=β6= 0, α′+β′(α′−1)2= 0 wi h α′∈C {0,1}, hus β′=−α′ (α′−1)2 and we ha e he amily Z9(α, −α′ (α′−1)2,1,0,0,1),wi h α∈C {0,1}. b) Le a3−a5= 0 be. Then, we ha e a′ 3=a′ 5= 1, a2 3−a2a66= 0 and a′ 1=((a1−1)Qn−2+a4Qn)P1 a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn + 1, a′ 2=a2Q2 n−2+ 2a3Qn−2Qn+a6Q2 n a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn , a′ 4=((a1−1)Rn−2+a4Rn)P1 a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn , a′ 6=a2R2 n−2+ 2a3Rn−2Rn+a6R2 n a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn , wi h P1(QnRn−2−Qn−2Rn)(a2Qn−2Rn−2+a3QnRn−2+a3Qn−2Rn+a6QnRn)6= 0, P1Rn−2+a2Pn−2Rn−2+a3Pn−2Rn+a5PnRn−2+a6PnRn= 0. We can suppose a′ 2= 0 , •a26= 0,i we choose Qn−2=−a3±pa2 3−a2a6 a2 Qn,we ge a′ 2= 0, •a2= 0,i we choose Qn−2=−a6Qn 2a3 we ha e a′ 2= 0. NATURALLY GRADED ZINBIEL ALGEBRAS WITH NILINDEX n−3 9 Analogously, we can suppose ha a′ 4= 0 , using Rn. Now, we ha e he new amily: ei◦ej=Cj i+j−1ei+j,2≤i+j≤n−3, e1◦en−2=en−1, en−2◦e1=a′ 1en−1, en−2◦en=en−1, en◦en−2=en−1, en◦en=a′ 6en−1. and we make a gene ic change o basis. Choosing Pnand Pn−2(as in p e ious cases) we ge a′′ 3=a′′ 5= 1. Pu ing Rn−2= 0, Qn−2=−a6Qn 2we ha e a′′ 4=a′′ 2= 0 and a′′ 1=(1 −a′ 1)P1+Rn Rn , a′′ 6=2Rn Qn I is easy o check ha he nulli y o a′′ 1−1 is in a ian because a′′ 1−1 = (a′ 1−1)P1Qn−2 (Qn−2+a6Qn)Rn Mo eo e , choosing Qn= 2Rnwe ob ain a′ 6= 1.The de e minan o change o basis is o med by he non-ze o po encies o he ollowing ac o s a6P1QnRn. Now, we can dis inguish wo cases: b.1) I a′ 1−16= 0,choosing Rn= (a′ 1−1)P1we ge a′′ 1= 0 and he algeb a Z10(0,0,1,0,1,1). b.2) I a′ 1−1 = 0, hen a′′ 1= 1 and we ha e Z11(1,0,1,0,1,1). Case 2.2. Le a3a5−a2a6= 0 be. As a36= 0 ⇒a5=a2a6 a3.We subs i u e in (1) and in (2) and we choose Pn−2=−a3P1Rn−2+a2a6PnRn−2+a3a6PnRn a3(a2Rn−2+a3Rn). Now, aking in o accoun he new pa ame e s (1), we ge o: a′ 2=(a2Qn−2+a3Qn)(a3Qn−2+a6Qn)(a2Rn−2+a3Rn) a2 3P1(Qn−2Rn−QnRn−2), a′ 3=(a3Qn−2+a6Qn)(a2Rn−2+a3Rn)2 a2 3P1(Qn−2Rn−QnRn−2)6= 0. wi h a3P1(QnRn−2−Qn−2Rn)(a2R1+a3Rn)6= 0 and ha he nulli y o he ollowing exp essions: a′ 1a′ 6−a′ 3a′ 4=a3(a1a6−a3a4)(a2Rn−2+a3Rn)(Qn−2Rn−QnRn−2)P1 (a3a6PnQn+a3P1Qn−2+a2a3Pn−2Qn−2+a2a6PnQn−2+a2 3Pn−2Qn)2, a′2 3−a′ 2a′ 6=(a2 3−a2a6)(a3Qn−2+a6Qn)(a2Rn−2+a3Rn)(Qn−2Rn−QnRn−2) (a3a6PnQn+a3P1Qn−2+a2a3Pn−2Qn−2+a2a6PnQn−2+a2 3Pn−2Qn)2, a e in a ian . Thus, we can dis inguish he non isomo phic cases: