Right and left solvable extensions of an associative Leibniz algebra

Transcription

1 Communictions in Algebr ISSN: Print) Online) Journl homepge: Right nd left solvble extensions of n ssocitive Leibniz lgebr A Shbnsky To cite this rticle: A Shbnsky 2017) Right nd left solvble extensions of n ssocitive Leibniz lgebr, Communictions in Algebr, 45:6, , DOI: / To link to this rticle: Accepted uthor version posted online: 07 Oct 2016 Published online: 07 Oct 2016 Submit your rticle to this journl Article views: 87 View relted rticles View Crossmrk dt Citing rticles: 1 View citing rticles Full Terms & Conditions of ccess nd use cn be found t Downlod by: [The UC Irvine Librries] Dte: 29 September 2017, At: 11:55

2 COMMUNICATIONS IN ALGEBRA 2017, VOL 45, NO 6, Right nd left solvble extensions of n ssocitive Leibniz lgebr A Shbnsky Deprtment of Mthemtics nd Sttistics, The University of Toledo, Toledo, USA ABSTRACT A sequence of nilpotent Leibniz lgebrs denoted by N n,18 is introduced Here n denotes the dimension of the lgebr defined for n 4; the first term in the sequence is R 18 in the list of four-dimensionl nilpotent Leibniz lgebrs introduced by Albeverio et l [4] Then ll possible right nd left solvble indecomposble extensions over the field R re constructed so tht N n,18 servessthenilrdiclofthecorrespondingsolvblelgebrstheconstruction continues Winternitz nd collegues progrm estblished to clssify solvble Lie lgebrs using specil properties rther thn trying to extend one dimension t time 1 Introduction ARTICLE HISTORY Received 2 April 2016 Revised 30 June 2016 Communicted by A Elduque KEYWORDS Derivtion; Leibniz lgebr; nil-independence; nilpotency; nilrdicl; solvbility MATHEMATICS SUBJECT CLASSIFICATION 17A30; 17A32; 17A36; 17A60; 17B30 LeibnizlgebrswerediscoveredbyBlochin1965[21]whoclledthemD lgebrslterontheywere considered by Lody nd Cuvier[43, 57 59] s non-ntisymmetric nlogue of Lie lgebrs It mkes everylielgebrbeleibnizlgebr,buttheconverseisnottrueexctlylodynmedthemleibniz lgebrs fter Gottfried Wilhelm Leibniz Since then mny reserchers re working on them nd s one of the results they found nlogs of importnt theorems in Lie theory such s the nlogue of Levi s theorem which ws proved by Brnes [15] He showed tht ny finite-dimensionl complex Leibniz lgebr is decomposed into semidirect sum of the solvble rdicl nd semisimple Lie lgebr Therefore the biggest chllenge in the clssifiction problem of finite-dimensionl complex Leibniz lgebrs is to study the solvble prt And to clssify solvble Leibniz lgebrs, we need nilpotent Leibniz lgebrs used s the nilrdicls sme sinthecseoflielgebrs[61] There re lso other importnt nlogous theorems such s Engel s theorem hs been generlized by severl reserchers like Ayupov nd Omirov [8] in stronger form by Ptsourkos [68], nd finlly Brnes [14] gve simple proof of this theorem Mny other uthors work on proving nlogous properties djusted to Leibniz lgebrs nd their work could be found in the cittions [2, 5, 12, 16, 18, 19,22,39,44,46,51,60,64,65] Every Leibniz lgebr stisfies generlized version of the Jcobi identity clled the Leibniz identity There re two Leibniz identities: the left nd right Leibniz identity We cll Leibniz lgebrs right Leibniz lgebrs if they stisfy the right Leibniz identity, nd left if they stisfy the left Leibniz identity A left Leibniz lgebr is not necessrily right Leibniz lgebr[44] In this pper we construct both right nd leftsolvbleleibnizlgebrs 1 Leibniz lgebrs inherit n importnt property of Lie lgebrs which is tht the rightleft) multipliction opertor of right left) Leibniz lgebr is derivtion [40] Besides the lgebr of right left) multipliction opertors is endowed with structure of Lie lgebr by mens of the commuttor[40] CONTACT A Shbnsky shbn@rocketsutoledoedu Deprtment of Mthemtics nd Sttistics, The University of Toledo, 2801WBncroftSt,ToledoOH43606,USA 1 WhenwesyLeibnizlgebr,wemenrightLeibnizlgebr,unlessthechoiceisunderstoodfromthecontext 2017Tylor&Frncis

3 2634 A SHABANSKAYA Another importnt property is tht the quotient lgebr by the two-sided idel generted by the squre elements of Leibniz lgebr is Lie lgebr [67], where such idel is the miniml, belin nd in the cse of non-lie Leibniz lgebrs it is lwys nontrivil The clssifiction of Leibniz lgebrs strted in 1993 by Lody[58] In tht pper he gve in prticulr the exmples of Leibniz lgebrs in dimensions one nd two over the field tht does not hve ny zero divisors, such tht the Leibniz lgebr in dimension one is belin nd coincides with Lie lgebr In dimension two there re two non Lie Leibniz lgebrs: one is nilpotentnull-filiform) nd one is solvble Leibniz lgebrs in dimension three over the complex numbers were clssified by Omirov nd Ayupov in1998[8,9]ndwsreviewedin2012bycssetl[38],wheretheynoticedthtoneisomorphism clss does not hve Leibniz lgebr structure Therefore there re four nilpotent one is null-filiform, three re filiform) non Lie Leibniz lgebrs nd seven solvble, such tht one nilpotent nd two solvble lgebrs contin prmeter Two-dimensionl nd three-dimensionl left Leibniz lgebrs were clssified by Demir et l[44] over the field of chrcteristic 0 In dimension two they found two non Lie Leibniz lgebrs: one is nilpotent, sme s the right Leibniz lgebr [58], nd one is solvble left Leibniz lgebr In dimension three there re five nilpotent Leibniz lgebrsone of them depends on prmeter) isomorphic to the right Leibniz lgebrs[38] Also there re seven solvble non Lie left Leibniz lgebrs, where two of them re one prmetric fmilies Four-dimensionl nilpotent complex Leibniz lgebrs, one of them is R 18, which is the first term in the sequence of the nilpotent Leibniz lgebrs N n,18, n 4), were clssified by Albeverio et l [4] Altogether they found 21 nilpotent complex non Lie Leibniz lgebrs, where the first 10 of them re either null-filiform or filiform[10] nd remining 11 re ssocitive lgebrs In one of them the prmeter just tkestwovlues0or1,sowethinkthtsuchlgebrdoesnotcontinprmeter,thereforetherere three nilpotent one prmetric fmilies mong the ssocitive lgebrs in[4] Four-dimensionl solvble Leibniz lgebrs in dimension four over the field of complex numbers were clssified by Cñete nd Khudoyberdiyev[37] They found 38 non Lie Leibniz lgebrs, where 16 of them contin up to two prmeters Besides in dimension four Omirov et l [67] proved tht there only exists one non solvble 4- dimensionlleibnizlgebrsl 2 C = {e,f,h,x}definedsfollows sl 2 C : [e,h] = [h,e] = 2e, [h,f] = [f,h] = 2f, [e,f] = [f,e] = h, whichisdecomposblelielgebrtthesmetime In dimension five Khudoyberdiyev et l [56] studied solvble Leibniz lgebrs with 3-dimensionl nilrdicls They showed tht five-dimensionl solvble Leibniz lgebr with three-dimensionl Heisenberg nilrdicl is Lie lgebr Also they found 22 non isomorphic clsses of solvble Leibniz lgebrs including one with the Heisenberg nilrdicl, where 12 of them re prmetric fmilies depending on up to four prmeters Few words bout the clssifiction of semisimple nd simple Leibniz lgebrs The definition of simple Leibniz lgebrs ws suggested by Dzhumdil dev nd Abdykssymov [45] According to them L is clledsimpleleibnizlgebriflltwo-sidedidelsoflre0, L nn ndlsemisimpleleibnizlgebrs were defined by Demir et l [44] s well s n nlogue of the Killing form for Leibniz lgebrs They lso showed if the Leibniz lgebr is semisimple then the Killing form is nondegenerte, but the converse is not true Further work on semisimple nd simple complex Leibniz lgebrs with the clssifiction of some prticulr cses ws performed by Cmcho et l[35], Gómez-Vidl et l[51], Omirov et l[67], Rkhimov et l[74] In other dimensions strting from dimension 5) the clssifiction of nilpotent Leibniz lgebrs hs ttempts to move to ny finite dimension over the field of chrcteristic zero, where there re the following directions of the reserch bsed on the clsses of nilpotent lgebrs: Ayupov nd Omirov[10] proved tht in every finite dimension n there is the only null-filiform non Lie Leibniz lgebr defined s [x i,x 1 ] = x i+1, 1 i n 1)

4 COMMUNICATIONS IN ALGEBRA 2635 Two kinds of filiform Leibniz lgebrs were clssified in ny finite dimension such s nturlly grded filiform Leibniz lgebrs studied by Vergne[90] nd Ayupov nd Omirov[10] nd non chrcteristiclly nilpotent filiform Leibniz lgebrs, which were clssified by Ayupov nd Omirov [10] nd Khudoyberdiyev nd Ldr[54] The filiform Leibniz lgebrs whose nturl grdtion is filiform Lie lgebr were considered by Omirov nd Rkhimov [66] This clss is denoted by TLeib n+1 They presented n effective lgorithm to control the behvior of the structure constnts under dpted trnsformtions of bsis in the rbitrry fixed dimension Also in one prticulr cse, they gve the formuls in dimensions less thn 10 Rkhimov nd Hssn [71] considered subtype of TLeib n+1 denoted by Cedµ n ) nd provided the isomorphism criteri for specific 5 nd 6-dimensionl cses Their other work on the clssifictionof5nd6-dimensionlfiliformlgebrsfromthefmilytleib n+1 couldbefoundin[72] The filiform Leibniz lgebrs which re one-dimensionl centrl extensions of filiform Lie lgebrs up to dimension nine were clssified by Rkhimov nd Hssn [73] If the nturl grdtion is non-lie filiform Leibniz lgebr, then such cse ws studied by Ayupov[10] nd Omirov nd Gómez[50], where Omirov nd Gómez[50] noticed tht becuse of the type of their nturl grdtion such filiform Leibniz lgebrs belong to two clsses either FLeib n+1 or SLeib n+1 Rkhimov nd Bekbev [70] introduced method to clssify such Leibniz lgebrs in ny finite fixed dimension bsed on lgebric invrints nd gve the ppliction of the developed method from dimension 5 up to dimension 9 Rkhimov nd Sid Husin [75, 76] clssified up to isomorphism the subclss FLeib n+1 for n = 4, 5, 6 nd delt withtheclssifictioninlowdimensionsofthereminingsubclsssleib n+1 AltogetherfiliformLeibniz lgebrs were only clssified up to dimension less thn 10 with some other work not mentioned bove given in the cittions[1, 3, 25, 49, 63, 69, 77 79] Nturlly grded qusi-filiform Leibniz lgebrs were studied by Cmcho et l [31] The study of nturlly grded Leibniz lgebrs of the certin nilindex nd different chrcteristic sequences could be found in the following references [24, 29] There is n ongoing study presented by Cmcho et l[26, 28, 32 34] on p-filiform nd nturlly grded p-filiform Leibniz lgebrs[25] ThesubjectofthispperistoddtotheprogrmofclssifyingsolvbleLeibnizlgebrsoverthefield of rel numbers in ny finite dimension pplying Winternitz nd his collegues Snobl, Rubin, Krsek, Trembly, Ndogmo [62, 80, 85 89] pproch, which ws estblished to clssify solvble Lie lgebrs, whereby they strt with prticulr nilpotent Lie lgebr to find ll its possible solvble extensions Some other work on solvble Lie lgebrs following the sme pproch could be found in the following references[82, 84, 91] Moreover, tht pproch is bsed on wht ws shown by Mubrkzynov in[61]: the dimension of the complimentry vector spce to the nilrdicl does not exceed the number of nil-independent derivtions of the nilrdicl Tht result ws extended to Leibniz lgebrs by Css, Ldr, Omirov nd Krimjnov in [41] with the help of the pper [8], where they clssified solvble Leibniz lgebrs with null-filiform nilrdicl Similrly to Lie lgebrs, the inequlity dimnill) 1 2 diml[61]islsotrueforsolvbleleibnizlgebrl There is lredy lot of work performed on such pproch to clssify solvble Leibniz lgebrs over the field of chrcteristic zero: Omirov nd his collegues Css, Khudoyberdiyev, Ldr, Krimjnov, Cmcho nd Msutov clssified solvble Leibniz lgebrs whose nilrdicls re direct sum of nullfiliform lgebrs [55], nturlly grded filiform [40], tringulr [53] nd finlly filiform [36] Bosko- Dunbr, Dunbr, Hird nd Stgg ttempted to clssify left solvble Leibniz lgebrs with Heisenberg nilrdicl[23] SomeotherworkndkindsofLeibnizlgebrsnotdiscussedreshowninthecittions[6,7,11,13, 17,20,27,30,42,47,48,52,92] The strting point of the present rticle is the four-dimensionl nilpotent non Lie right Leibniz lgebrs[4]introducedbyalbeverioetl[4]ofwhichtherere21suchlgebrsoverthefieldofcomplex numbersaswehvesid,thefirst10ofthemreeithernull-filiformorfiliform[10]ndremining11 re ssocitive lgebrs, nd there re three nilpotent one prmetric fmilies mong the ssocitive lgebrs Our choice is to work with one ssocitive Leibniz lgebr from the remining 11 R 18 An interesting observtion is tht it is left nd right Leibniz lgebr t the sme time, which is true for ll those ssocitive Leibniz lgebrs Also it does not depend on the prmeter nd over the field of rel

5 2636 A SHABANSKAYA numbers it remins the sme Other nilpotent Leibniz lgebrs in dimensions two, three nd four re either null-filiform or filiform nd solvble extensions of such lgebrs were studied by Omirov nd his collegues[36, 40, 41] WecretesequenceofnilpotentLeibnizlgebrsN n,18, n 4),wherethefirstterminthesequence isthefour-dimensionllgebrn 4,18,whichisexctly R 18 usingthenottionofthepper[4] ForsequenceN n,18 wefindllpossiblesolvbleindecomposbleextensionssleftndrightleibniz lgebrs in every finite dimension Such extensions of dimensions one nd, in one prticulr cse two, re possiblerightsolvbleextensionswithcodimensiononenilrdicln n,18 refoundfollowingthesteps intheorems51,52nd53withtheminresultsummrizedintheorem53,whereitisshownthere re four such solvble lgebrs The mjor result for the right solvble extensions with codimension twonilrdicln n,18 issttedintheorem54,whereitisprovedthereisonlyonesuchleibnizlgebr We follow the steps in Theorems 61, 62 nd 63 to find one-dimensionl left solvble extensions We noticeintheorem63thtthereisthesmenumberofthem,butonelgebrisleftndrighttthesme time We find the only two-dimensionl solvble extension s well stted in Theorem 64 As regrds nottion, we use e 1,e 2,,e r to denote the r-dimensionl subspce generted by e 1,e 2,,e r, where r N, LS is the lower centrl series nd DS is the derived series nd refer to them collectively s the chrcteristic series Besides g is used to denote solvble right Leibniz lgebrs nd l is solvble left Leibniz lgebrs Throughoutthepperllthelgebrsrefinitedimensionloverthefieldofrelnumbersndifthe brcket is not given, then it is ssumed to be zero, except the brckets for the nilrdicl, which re not giventllseeremrk51)tosvespce We use Mple softwre to compute the Leibniz identity, the bsorption see[81, 83] nd Section 32), the chnge of bsis for solvble Leibniz lgebrs in some prticulr dimensions, which re generlized nd proved in n rbitrry finite dimension 2 Preliminries In this section we give some bsic definitions encountered working with Leibniz lgebrs Definition 21 1) AvectorspceLoverfieldFwithbilineropertion [, ] : L LisclledLeibnizlgebr iffornyx, y, z LthesoclledLeibnizidentity [[x,y],z] = [[x,z],y] + [x,[y,z]] holds This Leibniz identity is known s the right Leibniz identity nd we cll L in this cse right Leibniz lgebr 2) There exists the version corresponding to the left Leibniz identity [z,[x,y]] = [[z,x],y] + [x,[z,y]], ndleibnizlgebrlisclledleftleibnizlgebr Remrk21 Inddition,ifLstisfies [x,x] = 0foreveryx L,thenitisLielgebrThereforeevery LielgebrisLeibnizlgebr,buttheconverseisnottrue Definition22 Thetwo-sidedidelCL) = {x L : [x,y] = [y,x] = 0}issidtobethecenterofL Definition23 Alinermpd : L LofLeibnizlgebrLisderivtionifforllx, y L d[x,y]) = [dx),y] + [x,dy)]

6 COMMUNICATIONS IN ALGEBRA 2637 In terms of this definition for x L, where L is right Leibniz lgebr, the right multipliction opertor R x : L L defined s R x y) = [y,x], y L is derivtion for left Leibniz lgebr L, the leftmultiplictionopertorl x : L L, L x y) = [x,y], y Lisderivtion) AnyrightLeibnizlgebrLisssocitedwiththelgebrofrightmultiplictionsRL) = {R x x L} endowedwiththestructureoflielgebrbymensofthecommuttor [R x,r y ] = R x R y R y R x = R [y,x],whichdefinesnntihomomorphismbetweenlndrl) ForleftLeibnizlgebrL,thecorrespondinglgebrofleftmultiplictionsLL) = {L x x L}is endowed with the structure of Lie lgebr by mens of the commuttor s well [L x,l y ] = L x L y L y L x = L [x,y] InthiscsewehvehomomorphismbetweenLndLL) Definition 24 For given Leibniz lgebr L, we define the sequence of two-sided idels s follows: L 0 = L, L k+1 = [L k,l], k 0) L 0) = L, L k+1) = [L k),l k) ], k 0), residtobethelowercentrlseriesndthederivedseriesofl,respectively ALeibnizlgebrLissidtobenilpotentsolvble)ifthereexistsn NsuchthtL n = 0L n) = 0) Theminimlsuchnumbernissidtobetheindexofnilpotencysolvbility) ItfollowsthtifLisLeibnizlgebrwithL 2 = 0,thenitisssocitive 3 Constructing solvble Leibniz lgebrs with given nilrdicl Every solvble Leibniz lgebr L contins unique mximl nilpotent idel clled the nilrdicl nd denoted nill) such tht dimnill) 1 2 diml) [61] Let us consider the problem of constructing solvble Leibniz lgebrs L with given nilrdicl N = nill) Suppose {e 1,e 2,e 3,,e n } is bsis forthenilrdiclnd {e n+1,,e p }isbsisforsubspcecomplementrytothenilrdicl IfLissolvbleLeibnizlgebr[8],then nd we hve the following structure equtions [L,L] N 31) [e i,e j ] = C k ij e k,[e,e i ] = A k i e k,[e i,e ] = A k i e k,[e,e b ] = B k b e k, 32) where1 i,j,k,m nndn+1,b p 31 Solvble right Leibniz lgebrs Clcultion shows tht the right Leibniz identity is equivlent to the following conditions: A k i Cm kj = A k j Cm ki +Ck ij Am k, Ak i Cm kj = Ck ij Am k +Ak j Cm ik, Ck ij Am k = Ak i Cm kj +Ak j Cm ik, 33) B k b Cm ki = Ak i Am kb +Ak bi Am k, Ak i Am kb = Bk b Cm ki +Ak ib Am k, Bk b Cm ik = Am kb Ak i Am k Ak ib 34) ThentheentriesofthemtricesA,whichre A k i ),muststisfytheequtions33)whichcomefrom ll the possible Leibniz identities between the triples {e,e i,e j } In 34) we hve reltions on structure constntsobtinedfromllleibnizidentitiesbetweenthetriples {e,e b,e i } Since N is the nilrdicl of L, no nontrivil liner combintion of the mtrices A, n + 1 p) is nilpotent which mens tht the mtrices A must be nil-independent, tht is, no non trivil combintion of them is nilpotent[41, 61] LetusnowconsidertherightmultiplictionopertorR e ndrestrictitton, n+1 p)notice weshllgetouterderivtionsofthenilrdicln = nill)[41]thenfindingthemtricesa isthesme sfindingouterderivtionsr e ofnfurtherthecommuttors[r eb,r e ] = R [e,e b ], n+1,b p) dueto31)consistofinnerderivtionsofnsothosecommuttorsgivethestructureconstntsb k b s

7 2638 A SHABANSKAYA showninthelstequtionof34)butonlyuptotheelementsinthecenterofthenilrdicln,becuse if e i, 1 i n) is in the center of N then ) R ei = 0, where R ei is n inner derivtion of the N ) N nilrdicl N Notice tht outer derivtions cn be nilpotent wheres inner derivtions of N) must be nilpotent 32 Solvble left Leibniz lgebrs The left Leibniz identity is equivlent to the following conditions: A k i Cm jk = Ak j Cm ki +Ck ji Am k, Ak i Cm jk = Ck ji Am k +Ak j Cm ik, Ck ij Am k = Ak i Cm kj +Ak j Cm ik, 35) B k b Cm ik = Ak i Am kb +Ak ib Am k, Ak ib Am k = Bk b Cm ik +Ak i Am kb, Bk b Cm ki = Am k Ak bi Am bk Ak i 36) ThentheentriesofthemtricesA,whichreA ) k i,muststisfytheequtions35)whichcomefromll thepossibleleibnizidentitiesbetweenthetriples {e,e i,e j }InthiscseA = L e ) N, n+1 p) reouterderivtionsndthecommuttors[l e,l eb ] = L [e,e b ]givethestructureconstntsb k b sshown inthelstequtionof36)butonlyuptotheelementsinthecenterofthenilrdicln Once the left or right Leibniz identities re stisfied in the most generl possible wy nd the outer derivtions re found: i) We cn crry out the technique of bsorption [81, 83], which mens we cn simplify solvble Leibniz lgebr without ffecting the nilrdicl in32) pplying the trnsformtion n e i = e i, 1 i n), e = e + d k e k, n +1 p) ii) Wecnchngebsissuchthtthebrcketsforthenilrdiclin32)reunchngedtoremovell the possible prmeters 4 ThenilpotentsequenceN n,18 41 N n,18 InN n,18, n 4)thepositiveintegerndenotesthedimensionofthelgebrThecenterofthislgebris CN n,18 ) = e n 1, e n N n,18 cnbedescribedexplicitlysfollows:inthebsis {e 1,e 2,e 3,e 4,e 5,,e n } it hs only the following non-zero brckets [e 1,e i ] = e n 1, [e i,e 1 ] = e n 1, [e i,e i ] = e n, 2 i n 2, n 4) 41) The dimensions of the idels in the chrcteristic series re k=1 LS = DS = [n, 2, 0], n 4) ItgivesthtN n,18 isnssocitiveleibnizlgebrndstisfiestheleftndrightleibnizidentities,soit istheleftndrightleibnizlgebrtthesmetime ThenottionforN n,18 couldbeexplinedsfollows:subscriptn, n 4)denotesthedimensionof thelgebrnd18isthenumberingbsedonthefour-dimensionlnilpotentleibnizlgebr R 18 using thenottionofthepper[4],whichisexctlythislgebrforn = 4BesidesN emphsizesthefcttht this nilpotent Leibniz lgebr is t the sme time the nilrdicl of the solvble indecomposble left nd right Leibniz lgebrs we construct in this pper Itisshownbelowthtsolvbleindecomposblerightleft)LeibnizlgebrswiththenilrdiclN n,18 onlyexistfordimg = n +1nddimg = n +2diml = n +1nddiml = n +2)

8 COMMUNICATIONS IN ALGEBRA 2639 For the solvble indecomposble right Leibniz lgebrs with codimension one nilrdicl, we use the nottiong n+1,i wheren+1indictesthedimensionofthelgebrgndiitsnumberingwithinthelist of lgebrs There re four types of such lgebrs up to isomorphism so 1 i 4 There is the only solvble indecomposble right Leibniz lgebr up to isomorphism with codimension two nilrdicl g n+2,1 We lso hve four solvble indecomposble left Leibniz lgebrs up to isomorphism with codimension one nilrdicl N n,18 denoted l n+1,1,g n+1,2,l n+1,3 nd l n+1,4, where g n+1,2 is left nd right Leibniz lgebr t the sme time There is the only solvble indecomposble left Leibniz with codimension two nilrdiclswelldenotedl n+2,1 Therereonlytworightleft)lgebrsthtcontinprmeters,suchsg n+1,2,g n+1,3 g n+1,2,l n+1,3 ) Altogether right left) lgebrs depend on t most n, n 4) prmeters, where g n+1,2 depends on n 1prmetersndg n+1,3,l n+1,3 dependononlyoneinthiscsewedonotconsider ǫ = 0,1tobe prmeter s it is discrete vlue Therefore those lgebrs define continuous fmily of lgebrs We hvelsorestrictedthevluesoftheprmeters,whereneeded,smuchspossiblesostomkeech lgebr unique within its clss 5 Clssifiction of solvble indecomposble right Leibniz lgebrs with nilrdicl N n,18 Our gol in this section nd Section 6 is to find ll the possible right nd left solvble indecomposble extensionsofthenilpotentleibnizlgebrn n,18,whichservessthenilrdicloftheextendedlgebr Remrk 51 It is ssumed throughout this section nd next section tht the solvble right Leibniz lgebrs g nd the solvble left Leibniz lgebrs l hve the nilrdicl N n,18 ; however, for the ske of simplicity the brckets of the nilrdicl will be mostly omitted 51 SolvbleindecomposblerightLeibnizlgebrswithcodimensiononenilrdiclN n,18 ThenilpotentLeibnizlgebrN n,18, n 4)isdefinedin41)Suppose {e n+1 }isinthecomplementry subspcetothenilrdicln n,18 ndgisthecorrespondingrightsolvbleleibnizlgebrsince [g,g] N n,18,thereforewehve [e 1,e i ] = e n 1,[e i,e 1 ] = e n 1,[e i,e i ] = e n, 2 i n 2), n n [e r,e n+1 ] = j,r e j,[e n+1,e k ] = b j,k e j, 1 k n,1 r n +1) j=1 j=1 Theorem51 Set 2,2 := in51)tostisfytherightleibnizidentity,thererethefollowingcsesbsed on the conditions involving prmeters, ech gives continuous fmily of solvble Leibniz lgebrs 1) If = 0, 1,1 + = 0, 2 1,1 = 0,then n [e 1,e n+1 ] = 1,1 e 1 + n 1,1 e n 1 + n,1 e n,[e i,e n+1 ] = 1,2 e 1 +e i + k,i e k, k=n 1 [e n 1,e n+1 ] = 1,1 +)e n 1,[e n,e n+1 ] = 2e n,[e n+1,e n+1 ] = n,n+1 e n, [e n+1,e 1 ] = 1,1 e 1 n 1,1 e n 1 + 1,1 n,1 e n,[e n+1,e i ] = 1,2 e 1 e i 2 1,1 n 1,i e n 1 + n,i + 2 ) 1,2 n,1 e n, 2 i n 2), [e n+1,e n 1 ] = 1,1 )e n 1 2 1,1 51)

9 2640 A SHABANSKAYA 2) If 1,1 :=, = 0,thenthebrcketsforthelgebrre [e 1,e n+1 ] = e 1 + n 1,1 e n 1 3b n,1 e n,[e i,e n+1 ] = 1,2 e 1 +e i + n k=n 1 k,i e k, [e n,e n+1 ] = 2e n,[e n+1,e n+1 ] = n 1,n+1 e n 1 + n,n+1 e n,[e n+1,e 1 ] = e 1 n 1,1 e n 1 +b n,1 e n,[e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + n,i 2 ) 1,2b n,1 e n, 2 i n 2) 3) If = 0, 1,1 = 0,thenwehve [e 1,e n+1 ] = 1,1 e 1 + n 1,1 e n 1 + n,1 e n,[e i,e n+1 ] = 1,2 e 1 + [e n 1,e n+1 ] = 1,1 e n 1,[e n+1,e n+1 ] = n,n+1 e n,[e n+1,e 1 ] = 1,1 e 1 n 1,1 e n 1 n,1 e n, n k=n 1 k,i e k, [e n+1,e i ] = 1,2 e 1 n 1,i e n 1 +b n,i e n, 2 i n 2), [e n+1,e n 1 ] = 1,1 e n 1 4) If 1,1 := 2, = 0,then n [e 1,e n+1 ] = 2e 1 + n 1,1 e n 1,[e i,e n+1 ] = 1,2 e 1 +e i + k,i e k, k=n 1 [e n 1,e n+1 ] = 3e n 1,[e n,e n+1 ] = 2e n,[e n+1,e n+1 ] = n,n+1 e n,[e n+1,e 1 ] = 2e 1 n 1,1 e n 1 +b n,1 e n,[e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + n,i + ) 1,2b n,1 e n, 2 i n 2), [e n+1,e n 1 ] = 3e n 1 Proof LetusdenoteR en+1 byrtostisfytherightleibnizidentityin51),wehvetoconsiderinr 2 sthe, 1,1 ) plnethefollowingthreelinerequtions: 1,1 =, = 0, 1,1 = 2Theyhveone point of the intersection 0, 0), which is excluded in order to void nilpotent Leibniz lgebrs As result, weconsiderbelowfourregionsofr 2 ;collectivelytheirunionisr 2 \0,0)correspondingtollpossible orderedpirs, 1,1 )excludingtheoriginthefirstregiongivenin1)istheopensubsetofr 2 forwhich llofthethreeequtionsrenotstisfiedndisgenericcsetheeqution 1,1 = correspondsto region 2),theeqution = 0correspondstoregion 3)ndfinllytheeqution 1,1 = 2corresponds to region 4) 1) Suppose = 0, 1,1 + = 0, 2 1,1 = 0Theproofisoff-lodedtoTble1 2) Suppose 1,1 :=, = 0 Following similr computtions s in steps 1) 17) of the generic cse nd fter tht considering the right Leibniz identity R[e n+1,e i ]) = [Re n+1 ),e i ] + [e n+1,re i )], 2 i n 2),wederivetheresult 3) Suppose = 0, 1,1 = 0 We pply steps 1) 18) of cse 1) Then for steps 19) nd 20), we considertheidentitiesr e1 [e n+1,e n+1 ]) = [R e1 e n+1 ),e n+1 ] + [e n+1,r e1 e n+1 )],R[e n+1,e i ]) = [Re n+1 ),e i ] + [e n+1,re i )], 2 i n 2)ndprovetheresult 4) Suppose 1,1 := 2, = 0 We repet steps 1) 18) of the generic cse Then considering the identityr[e n+1,e i ]) = [Re n+1 ),e i ] + [e n+1,re i )], 2 i n 2),wededucetheresult

10 COMMUNICATIONS IN ALGEBRA 2641 Tble 1 Right Leibniz identities in the generic cse in Theorem 51 Steps Ordered triple Result 1 R[e 1,e 2 ] [e n 1,e n+1 ] = 1,1 ++ ) n 2 k=3 k,2 e n 1 + 2,1 e n k,n 1 = 0, 1 k n 2), n,n 1 := 2,1, n 1,n 1 := 1,1 ++ n 2 k=3 k,2 2 R[e 2,e 2 ] [e n,e n+1 ] = 2e n k,n = 0, 1 k n 1), n,n := 2 3 R[e 2,e n+1 ] 1,n+1 = 2,n+1 = 0 [e n+1,e n+1 ] = n k=3 k,n+1 e k 4 R[e i,e n+1 ] i,n+1 = 0, 3 i n 2)Combiningwith3, 5 R[e 2,e 1 ] [e n+1,e n+1 ] = n 1,n+1 e n 1 + n,n+1 e n 2,1 = 0 [e n 1,e n+1 ] = 1,1 ++ ) n 2 k=3 k,2 e n 1, [e 1,e n+1 ] = 1,1 e 1 + n k=3 k,1 e k 6 R[e 1,e i ] i,1 = 0, 3 i n 2) [e 1,e n+1 ] = 1,1 e 1 + n 1,1 e n 1 + n,1 e n 7 R[e i,e i ] i,i :=, 3 i n 2) [e j,e n+1 ] = j 1 k=1 k,je k +e j + n k=j+1 k,j e k, 2 j n 2) 8 R[e i k,e i ] 1,i := 1,i k, i k,i := i,i k, 2 +k i n 2, 1 k n 4),where kisfixed [e j,e n+1 ] = 1,2 e 1 j 1 m=2 j,me m +e j + nm=j+1 m,j e m, 2 j n 2) 9 R e2 [en+1,e 1 ] ) b k,n 1 = 0, 1 k n 2),b n,n 1 := b 2,1,b n 1,n 1 := b 1,1 + n 2 k=2 b k,2 [e n+1,e n 1 ] = b 1,1 + ) n 2 k=2 b k,2 e n 1 +b 2,1 e n 10 R e2 [en+1,e 2 ] ) [e n+1,e n ] = 0 b k,n = 0, 1 k n) 11 R e1 [e2,e n+1 ] ) b 2,1 = 0, b 1,1 := 1,1 b n 1,n 1 := 1,1 + n 2 k=2 b k,2 nd [e n+1,e n 1 ] = 1,1 + ) n 2 k=2 b k,2 e n 1, [e n+1,e 1 ] = 1,1 e 1 + n k=3 b k,1 e k 12 R e1 [ei,e n+1 ] ) b i,1 = 0, 3 i n 2) [e n+1,e 1 ] = 1,1 e 1 +b n 1,1 e n 1 +b n,1 e n 13 R ei [ei,e n+1 ] ) b 1,i := 1,2, b i,i :=, 2 i n 2) 14 R ek [en+1,e i ] ) b i,k := b k,i, 2 i n 3, 1 +i k n 2),whereiisfixed [e n+1,e j ] = 1,2 e 1 + j 1 m=2 b j,me m e j + n m=j+1 b m,j e m, 2 j n 2) 15 R ek [ei,e n+1 ] ) b k,i := k,i, 2 i n 3, 1 +i k n 2),whereiisfixed b n 1,n 1 := 1,1 + n 2 m=3 m,2, [e n+1,e j ] = 1,2 e 1 + j 1 m=2 j,me m e j + n 2 m=j+1 m,je m + nm=n 1 b m,j e m, 2 j n 2),nd [e n+1,e n 1 ] = 1,1 + ) n 2 m=3 m,2 e n 1 16 R ei [ek,e n+1 ] ) k,i = 0, 2 i n 3, 1 +i k n 2),whereiisfixed [e j,e n+1 ] = 1,2 e 1 +e j + n m=n 1 m,j e m, [e n+1,e j ] = 1,2 e 1 e j + n m=n 1 b m,j e m, 2 j n 2), [e n 1,e n+1 ] = 1,1 + ) e n 1, [e n+1,e n 1 ] = ) 1,1 en 1 17 R[e n+1,e 1 ] b n 1,1 := n 1,1,b n,1 := 1,1 n,1 2 1,1 [e n+1,e 1 ] = 1,1 e 1 n 1,1 e n 1 + 1,1 n,1 2 e 1,1 n 18 R[e n+1,e n+1 ] n 1,n+1 = 0 [e n+1,e n+1 ] = n,n+1 e n 19 R ei [en+1,e n+1 ] ) b n 1,i := n 1,i, b n,i := n,i + 2 1,2 n,1 2 [e 1,1 n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + n,i + 2 1,2 n,1 2 )e 1,1 n, 2 i n 2) Remrk 52 In the tbles throughout the pper n ordered triple is shorthnd nottion for derivtion property of the multipliction opertors, which re either R z [x,y] ) = [Rz x),y] + [x,r z y)] or L z [x,y] ) = [Lz x),y] + [x,l z y)]wessign R en+1 := R nd L en+1 := L

11 2642 A SHABANSKAYA Theorem 52 Applying the technique of bsorption see Section 32), we cn further simplify the lgebrs inechofthefourcsesintheorem51sfollows: 1) Suppose = 0, 1,1 + = 0, 2 1,1 = 0Wehve [e 1,e n+1 ] = 1,1 e 1 + n,1 e n,[e 2,e n+1 ] = 1,2 e 1 +e 2 + n 1,2 e n 1 + n,2 e n, [e i,e n+1 ] = 1,2 e 1 +e i + n 1,i e n 1,[e n 2,e n+1 ] = 1,2 e 1 +e n 2, [e n 1,e n+1 ] = + 1,1 )e n 1,[e n,e n+1 ] = 2e n,[e n+1,e 1 ] = 1,1 e 1 + 1,1 n,1 [e n+1,e 2 ] = 1,2 e 1 e 2 n 1,2 e n 1 + n, ,2 n,1 2 1,1 ) e n, 2 1,1 e n, [e n+1,e i ] = 1,2 e 1 e i n 1,i e n ,2 n,1 e n,3 i n 3), 2 1,1 [e n+1,e n 2 ] = 1,2 e 1 e n ,2 n,1 e n,[e n+1,e n 1 ] = 1,1 )e n 1 2 1,1 2) Suppose 1,1 :=, = 0Thenthebrcketsforthelgebrre [e 1,e n+1 ] = e 1 3b n,1 e n,[e 2,e n+1 ] = 1,2 e 1 +e 2 + n 1,2 e n 1 + n,2 e n, [e i,e n+1 ] = 1,2 e 1 +e i + n 1,i e n 1,[e n 2,e n+1 ] = 1,2 e 1 +e n 2,[e n,e n+1 ] = 2e n, [e n+1,e n+1 ] = n 1,n+1 e n 1,[e n+1,e 1 ] = e 1 +b n,1 e n,[e n+1,e 2 ] = 1,2 e 1 e 2 n 1,2 e n 1 + n,2 2 ) 1,2b n,1 e n, [e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 2 1,2b n,1 e n, 3 i n 3), [e n+1,e n 2 ] = 1,2 e 1 e n 2 2 1,2b n,1 e n 3) If = 0, 1,1 = 0,then [e 1,e n+1 ] = 1,1 e 1 + n,1 e n,[e 2,e n+1 ] = 1,2 e 1 + n 1,2 e n 1 + n,2 e n,[e i,e n+1 ] = 1,2 e 1 + n 1,i e n 1, 3 i n 3), [e n 2,e n+1 ] = 1,2 e 1,[e n 1,e n+1 ] = 1,1 e n 1, [e n+1,e n+1 ] = n,n+1 e n,[e n+1,e 1 ] = 1,1 e 1 n,1 e n,[e n+1,e j ] = 1,2 e 1 n 1,j e n 1 +b n,j e n,2 j n 3),[e n+1,e n 2 ] = 1,2 e 1 +b n,n 2 e n, [e n+1,e n 1 ] = 1,1 e n 1 4) If 1,1 := 2, = 0,thenthebrcketsforthesolvbleLeibnizlgebrre [e 1,e n+1 ] = 2e 1,[e 2,e n+1 ] = 1,2 e 1 +e 2 + n 1,2 e n 1 + n,2 e n,[e i,e n+1 ] = 1,2 e 1 +e i + n 1,i e n 1,[e n 2,e n+1 ] = 1,2 e 1 +e n 2,[e n 1,e n+1 ] = 3e n 1,[e n,e n+1 ] = 2e n, [e n+1,e 1 ] = 2e 1 +b n,1 e n,[e n+1,e 2 ] = 1,2 e 1 e 2 n 1,2 e n 1 + n,2 + ) 1,2b n,1 e n, [e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + 1,2b n,1 e n, 3 i n 3), [e n+1,e n 2 ] = 1,2 e 1 e n 2 + 1,2b n,1 e n,[e n+1,e n 1 ] = 3e n 1 Proof We show the generic cse corresponding to region 1) Other cses re proved pplying pplicble trnsformtions nd properly renming the entries

12 COMMUNICATIONS IN ALGEBRA ) Suppose = 0, 1,1 + = 0, 2 1,1 = 0 The right derivtion) nd left not derivtion) multipliction opertors restricted to the nilrdicl re 1,1 1,2 1,2 1,2 1,2 1, R en+1 = , n 1,1 n 1,2 n 1,3 n 1,4 n 1,5 n 1,n 2 1,1 + 0 n,1 n,2 n,3 n,4 n,5 n,n ,1 1,2 1,2 1,2 1,2 1, L en+1 = n 1,1 n 1,2 n 1,3 n 1,4 n 1,5 n 1,n 2 1,1 0 1,1 n,1 2 1,1 n,2 n,3 n,4 n,5 n,n ,2 n,1 2 1, ,2 n,1 2 1, ,2 n,1 2 1, ,2 n,1 2 1, ,2 n,1 2 1,1 The trnsformtion e i = e i, 1 i n), e n+1 = e n+1 + n 1,n 2 e 1 removes n 1,n 2 in R en+1 nd n 1,n 2 in L en+1, but it ffects other entries s well, such s the entries in the n 1,2), n 1,3),,n 1,n 3) positions in L en+1 nd R en+1, which we renme by n 1,2, n 1,3,, n 1,n 3 nd n 1,2, n 1,3,, n 1,n 3, respectively Consecutively werenmethecoefficientinfrontofe n inthebrcket [e n+1,e n+1 ]bckby n,n+1 Wefixjndpplythetrnsformtione i = e i, 1 i n), e n+1 = e n+1 n,j e j, 3 j n 2), whichremoves n,j inr en+1 nd n,j fromthe n,j) th entriesinl en+1 Thenwechngetheentryin the n 1,1)positioninR en+1 ndl en+1 to n 1,1 nd n 1,1,respectively,ndthecoefficient infrontofe n inthebrcket [e n+1,e n+1 ]bckto n,n+1 We pply the trnsformtion e i = e i, 1 i n), e n+1 = e n+1 n 1,1 e 2 to remove n 1,1 in R en+1 nd n 1,1 in L en+1 As consequence, we renme the entry in the n,2) position in R en+1 nd L en+1 bck by n,2 nd n, ,2 n,1 2 1,1, respectively We lso chnge the coefficient in frontofe n inthebrcket [e n+1,e n+1 ]bckto n,n+1 Finlly pplying the trnsformtion e i = e i, 1 i n), e n+1 = e n+1 n,n+1 2 e n, we remove n,n+1 inthebrcket [e n+1,e n+1 ]ndprovetheresult

13 2644 A SHABANSKAYA Theorem 53 There re four solvble indecomposble right Leibniz lgebrs up to isomorphism with codimensiononenilrdicln n,18, n 4),whichregivenbelow: i) g n+1,1 : [e n+1,e 1 ] = e 1,[e 1,e n+1 ] = e 1,[e n+1,e i ] = e i,[e i,e n+1 ] = e i, 2 i n 2),[e n,e n+1 ] = 2e n,[e n+1,e n+1 ] = e n 1, DS = [n +1, n, 2, 0],LS = [n +1, n, n,], ii) g n+1,2 : [e n+1,e i ] = e i, [e i,e n+1 ] = e i,i = 1, n 1), [e 2,e n+1 ] = ǫe n,ǫ = 0,1), [e n+1,e j ] = j 1 e n,2 j n 2), [e k,e n+1 ] = be n,3 k n 2),[e n+1,e n+1 ] = ce n, m 1 m, 3 m n 3), DS = [n +1, 3, 0], LS = [n +1, 3, 2, 2,] Remrk53 Togurnteethelgebrsreuniquewithinitsclss,weshouldlwyshvetht m 1 m,3 m n 3), otherwise permuting e 3,e 4,,e n 2, we obtin isomorphic lgebrs In cse when ǫ = 0 nd b = 0 by permuting e 2 nd e k, we my ssume tht 1 k 1,3 k n 2) If ǫ = 1,b = 1,n 6), then we should hve s well tht 1 k 1,3 k n 2), becuse permuting e 2 nd e k gives isomorphic lgebrs Finlly if n = 5 nd ǫ = 1,b = 0, then pplying the trnsformtione 1 = e 1,e i = be i,2 i 4),e 5 = b2 e 5,e 6 = e 6,wemyssumethteither b 1or 1 2 iii) g n+1,3 : [e n+1,e 1 ] = e 1, [e 1,e n+1 ] = e 1, [e n+1,e i ] = e i,[e i,e n+1 ] = e i, 2 i n 2), [e n+1,e n 1 ] = 1)e n 1,[e n 1,e n+1 ] = +1)e n 1, [e n,e n+1 ] = 2e n, DS = [n +1, n, 2, 0],LS = [n +1, n, n,], iv) g n+1,4 : [e n+1,e 1 ] = e 1,[e 1,e n+1 ] = e 1, [e n+1,e i ] = e 1 e i,[e i,e n+1 ] = e 1 +e i, 2 i n 2), [e n+1,e n 1 ] = 2e n 1, [e j,e n+1 ] = 2e j, n 1 j n), DS = [n +1, n, 2, 0],LS = [n +1, n, n,] Proof One pplies the chnge of bsis trnsformtions keeping the nilrdicl N n,18 given in 41) unchnged We show the cse corresponding to region 1) nd cse 4) Other cses re proved pplying pplicble trnsformtions of cse 1) 1 which is generic, nd crefully observing how they ffect other entries, so the detils for those cses will be mostly omitted

14 COMMUNICATIONS IN ALGEBRA ) Suppose = 0, 1,1 + = 0, 2 1,1 = 0 The right derivtion) nd left not derivtion) multipliction opertors re 1,1 1,2 1,2 1,2 1,2 1,2 1, R en+1 =, n 1,2 n 1,3 n 1,4 n 1,5 n 1,n ,1 0 n,1 n, ,1 1,2 1,2 1,2 1,2 1,2 1, L en+1 = n 1,2 n 1,3 n 1,4 n 1,5 n 1,n 3 0 1,1 0 1,1 n,1 2 1,1 n, ,2 n,1 2 1,2 n,1 2 1,2 n,1 2 1,2 n, ,1 2 1,1 2 1,1 2 1,1 1,2 n,1 2 1,2 n,1 2 1,1 2 1, ) Let 1,1 = The trnsformtion e 1 = e 1, e i = e i + 1,2 1,1 e 1, 2 i n 2), e j = e j, n 1 j n +1)removes 1,2 inr en+1 nd 1,2 inl en+1 fromtheentriesinthe 1,2), 1,3),, 1, n 2) positions, respectively This trnsformtion ffects other entries s well As the result wechngethe n,2) nd entryinr en+1 ndl en+1 to n,2 + 1,2 n,1 1,1 ndnmetheentriesinthe n,3), n,4),,n,n 2)positionsinR en+1 ndl en+1 by 1,2 n,1 1,1 Applying the trnsformtion e 1 = e 1 n,1 2 1,1 e n, e i = e i n 1,i 1,1 e n 1, 2 i n 3), e j = e j, n 2 j n + 1), we remove n 1,2, n 1,3,, n 1,n 3 in R en+1 nd n 1,2, n 1,3,, n 1,n 3 in L en+1 ; n,1 nd 1,1 n,1 2 1,1 from the n,1) st entry in R en+1 ndl en+1,respectively,keepingotherentriesunchnged ) n,2 + 1,2 n,1 1,1 ) e n, e i = e i We pply the trnsformtion e 1 = e 1, e 2 = e 2 1,2 n,1 1,1 ) e n, 3 i n 2), e k = e k, n 1 k n + 1) to remove n,2 + 1,2 n,1 1,1 fromthe n,2) nd entryinr en+1 ndl en+1 ; 1,2 n,1 1,1 fromthe n,i) th entriesinr en+1 ndl en+1

15 2646 A SHABANSKAYA Finlly pplying the trnsformtion e i = e i, 1 i n), e n+1 = e n+1, we scle to unity Thenwerenme 1,1 by 1,1 ndobtinleibnizlgebr [e 1,e n+1 ] = 1,1 e 1,[e i,e n+1 ] = e i,[e n 1,e n+1 ] = 1,1 +1)e n 1,[e n,e n+1 ] = 2e n, [e n+1,e 1 ] = 1,1 e 1,[e n+1,e i ] = e i,[e n+1,e n 1 ] = 1,1 1)e n 1, 2 i n 2, 1,1 = 1, 1,1 = 1, 1,1 = 2) 52) 2) Suppose 1,1 =, = 0Westrtwithpplyingthesecondtrnsformtionofthepreviouscse, crefullywtchinghowitffectstheentriesthenweconsiderthetrnsformtione 1 = e 1, e 2 = e 2 n,2+ n,1 1,2 2 e n, e i = e i 1,2 n,1 2 e n, 3 i n 2), e k = e k, n 1 k n + 1) to remove n,2 + 1,2 n,1 fromthe n,2) nd entryinr en+1 ndl en+1 ; 1,2 n,1 fromthe n,i) th entries inr en+1 ndl en+1 Furtherpplyingthetrnsformtione j = e j, 1 j n), e n+1 = e n+1,we scletounityndrenming 1,2 by 1,2,weobtinLeibnizlgebr { [e1,e n+1 ] = e 1,[e i,e n+1 ] = 1,2 e 1 +e i,[e j,e n+1 ] = 2e j,n 1 j n), [e n+1,e 1 ] = e 1,[e n+1,e i ] = 1,2 e 1 e i,2 i n 2),[e n+1,e n 1 ] = 2e n 1 If 1,2 = 0 then we hve limiting cse of 52) with 1,1 = 1 If 1,2 = 0 then pplying the trnsformtione 1 = 1,2e 1, e i = e i, 2 i n 2), e n 1 = 1,2e n 1, e j = e j, n j n+1), wederiveleibnizlgebrg n+1,4 2) Suppose 1,1 :=, = 0Inthiscsewehve { [e1,e n+1 ] = e 1,[e i,e n+1 ] = e i,[e n,e n+1 ] = 2e n,[e n+1,e n+1 ] = n 1,n+1 e n 1, [e n+1,e 1 ] = e 1,[e n+1,e i ] = e i, 2 i n 2) If n 1,n+1 = 0thenweobtinlimitingcseof52)with 1,1 = 1If n 1,n+1 = 0thenpplying thetrnsformtione 1 = n 1,n+1e 1, e i = e i, 2 i n 2), e n 1 = n 1,n+1e n 1, e j = e j, n j n +1),wehveLeibnizlgebrg n+1,1 3) Suppose = 0nd 1,1 = 0Wehve [e 1,e n+1 ] = e 1,[e 2,e n+1 ] = n,2 e n,[e k,e n+1 ] = be n,3 k n 2), [e n 1,e n+1 ] = e n 1,[e n+1,e n+1 ] = ce n,[e n+1,e 1 ] = e 1,[e n+1,e j ] = j 1 e n, 53) 2 j n 2),[e n+1,e n 1 ] = e n 1 If n,2 = 0 then pplying the trnsformtion e 1 = e 1, e i = n,2 e i, 2 i n 1), e n = n,2 ) 2 e n, e n+1 = e n+1,wescle n,2 tounitychngingtheotherentriestowhttheywereoriginlly, we obtin Leibniz lgebr [e 1,e n+1 ] = e 1,[e 2,e n+1 ] = e n,[e k,e n+1 ] = be n, 3 k n 2), [e n 1,e n+1 ] = e n 1,[e n+1,e n+1 ] = ce n,[e n+1,e 1 ] = e 1,[e n+1,e j ] = j 1 e n, 54) 2 j n 2),[e n+1,e n 1 ] = e n 1 Altogether53)with n,2 = 0nd54)giveusLeibnizlgebrg n+1,2 4) Suppose 1,1 := 2, = 0Wepplythefollowingtrnsformtions Thetrnsformtione 1 = e 1, e i = e i 1,2 e 1, 2 i n 2), e j = e j, n 1 j n+1)removes 1,2 in R en+1 nd 1,2 in L en+1 from the entries in the 1,2), 1,3),,1,n 2) positions, respectively In L en+1 this trnsformtion lso removes the prt 1,2b n,1 from the n,2) nd entry n,2 + 1,2b n,1 nd 1,2b n,1 fromtheentriesinthe n,3), n,4),,n,n 2)positions Applying the trnsformtion e 1 = e 1 b n,1 2 e n, e i = e i n 1,i 2 e n 1, 2 i n 3), e j = e j, n 2 j n + 1), we remove n 1,2, n 1,3,, n 1,n 3 in R en+1 nd

16 COMMUNICATIONS IN ALGEBRA 2647 n 1,2, n 1,3,, n 1,n 3 inl en+1 ;b n,1 fromthe n,1) st entryinl en+1 withoutffecting other entries The trnsformtion e 1 = e 1, e 2 = e 2 n,2 e n, e i = e i, 3 i n + 1) removes n,2 from the n,2) nd entryinr en+1 ndl en+1 Finlly we pply the trnsformtion e i = e i, 1 i n), e n+1 = e n+1 to scle to unity nd obtinlimitingcseof52)with 1,1 = 2 Altogether52)ndllitslimitingcsesfterreplcing 1,1 withgiveusleibnizlgebrg n+1,3 52 SolvbleindecomposblerightLeibnizlgebrswithcodimensiontwonilrdiclN n,18 Thenon-zeroinnerderivtionsofN n,18, n 4)regivenby n 2 R e1 = E n 1,i =, i= R ei = E n 1,1 +E n,i, 2 i n 2), wheree n 1,i isthemtrixwhichhs1inthe n 1,i) th entryndllotherentriesrezero,e n 1,1 isthe mtrixwith1inthe n 1,1) st positionndzerosinllotherplcesnde n,i isthemtrixthths1in the n,i) th positionndzeroseverywhereelse The nilrdicl N n,18 in the bsis {e 1,e 2,e 3,,e n } is defined in 41), where e n 1 nd e n re in the centerofn n,18 Thevectorscomplementrytothenilrdiclree n+1 nde n+2 OnecouldnoticethttwoouterderivtionsR en+1 ndr en+2 re nil-independent [41,61]onlyin cse 1)inTheorem52 Remrk 54 If we hve three or more outer derivtions, then they re nil-dependent mens liner combintion of them is nilpotent, which cnnot hppen s we construct solvble Leibniz lgebrs Therefore the solvble lgebrs we re constructing re of codimension t most two By tking liner combintion of R en+1 nd R en+2 nd keeping in mind tht no nontrivil liner combintion of the mtrices R en+1 nd R en+2 cn be nilpotent mtrix, one could set 1,1 ) ) 1 b1,1 ) ) 0 = nd =, becuse t the sme time we hve tht = 0, 1,1 + 1 b 1 = 0, 2 1,1 = 0 the sme is true for b nd b 1,1 ) Therefore the vector spce of outer derivtions s the n n mtrices re R en+1 = 1 1,2 1,2 1,2 1,2 1,2 1, , n 1,2 n 1,3 n 1,4 n 1,5 n 1,n n,1 n,

17 2648 A SHABANSKAYA R en+2 = 0 b 1,2 b 1,2 b 1,2 b 1,2 b 1,2 b 1, b n 1,2 b n 1,3 b n 1,4 b n 1,5 b n 1,n b n,1 b n, Generl pproch to find solvble right Leibniz lgebrs with codimension two nilrdicln n,18 2 i) We consider R [er,e s ] = [R es,r er ] = R es R er R er R es, n + 1 r n + 2, 1 s n + 2) ndcomprewith n 2 k=1 c kr ek,becusee n 1 nde n reinthecenterofn n,18, n 4)definedin 41) ii) We write down [e r,e s ], n + 1 r n + 2, 1 s n + 2) including liner combintion of e n 1 nde n swellweddthebrcketsfromthenilrdicln n,18, n 4)ndouterderivtions R en+1 ndr en+2 iii) One stisfies the right Leibniz identity: [[e r,e s ],e t ] = [[e r,e t ],e s ] + [e r,[e s,e t ]] or, equivlently, R et [e r,e s ]) = [R et e r ),e s ] + [e r,r et e s )], 1 r,s,t n +2) for ll the brckets obtinedin step ii) iv) We pply the chnge of bsis trnsformtions without ffecting the nilrdicl to remove s mny prmeters s possible i) One could find tht R [en+1,e i ] = R ei,1 i n 2),R [en+1,e k ] = 0, n 1 k n +1), n 2 R [en+1,e n+2 ] = n 4) n,1 b 1,2 R e2 n,1 b 1,2 R ej, R [en+2,e 1 ] = 0, R [en+2,e m ] = b 1,2 R e1 R em, 2 m n 2), R [en+2,e n 1 ] = R [en+2,e n ] = 0, n 2 R [en+2,e n+1 ] = 4 n) n,1 b 1,2 R e2 + n,1 b 1,2 R ej, R [en+2,e n+2 ] = 0 Atthesmetime,wededucetht 1,2 = 0, b n,1 := 2 n,1, b n,2 := n,2 n 3) n,1 b 1,2, b n 1,i = 0, 2 i n 3)ndwemkesuchchngesintheouterderivtionsR en+1 ndr en+2 ii) Weincludelinercombintionofe n 1 nde n ndobtin [e n+1,e i ] = e i +c n 1,i e n 1 +c n,i e n, 1 i n 2),[e n+1,e k ] = c n 1,k e n 1 +c n,k e n, n 2 n 1 k n +1), [e n+1,e n+2 ] = n 4) n,1 b 1,2 e 2 n,1 b 1,2 e j +c n 1,n+2 e n 1 j=3 +c n,n+2 e n,[e n+2,e 1 ] = d n 1,1 e n 1 +d n,1 e n, [e n+2,e m ] = b 1,2 e 1 e m +d n 1,m e n 1 +d n,m e n, 2 m n 2), [e n+2,e n 1 ] = d n 1,n 1 e n 1 +d n,n 1 e n, n 2 [e n+2,e n ] = d n 1,n e n 1 +d n,n e n, [e n+2,e n+1 ] = 4 n) n,1 b 1,2 e 2 + n,1 b 1,2 e j j=3 +d n 1,n+1 e n 1 +d n,n+1 e n,[e n+2,e n+2 ] = d n 1,n+2 e n 1 +d n,n+2 e n j=3 j=3 2 WhenweworkwiththeleftLeibnizlgebrs,weinterchngesndrinstep i)nd ii),therightmultiplictionopertorto theleft,ndtherightleibnizidentitytotheleftleibnizidentityinstep iii)

18 COMMUNICATIONS IN ALGEBRA 2649 BesideswehvethebrcketsfromouterderivtionsR en+1 ndr en+2 ndn n,18, n 4) iii) We stisfy the right Leibniz identity by checking the derivtion property or right Leibniz identities forllpossibletriplesasconsequence,weobtinthtr en+1 ndr en+2 restrictedtothenilrdicl sn nmtricesdonotchngendotherbrcketsremodifiedto [e n+1,e 1 ] = e 1 + n,1 e n, [e n+1,e 2 ] = e 2 n 1,2 e n 1 + n,2 e n, [e n+1,e i ] = e i n 1,i e n 1, 3 i n 3), [e n+1,e n 2 ] = e n 2, [e n+1,e n 1 ] = 2e n 1, n 2 [e n+1,e n+1 ] = c n,n+1 e n, [e n+1,e n+2 ] = n 4) n,1 b 1,2 e 2 n,1 b 1,2 e j +c n 1,n+2 e n 1 + ) c n,n+1 n 4) n,1 b 1,2 n,2 en, [e n+2,e 2 ] = b 1,2 e 1 e 2 + ) n,2 n 5) n,1 b 1,2 en, [e n+2,e i ] = b 1,2 e 1 e i +2 n,1 b 1,2 e n, 3 i n 2), [e n+2,e n 1 ] = e n 1, n 2 [e n+2,e n+1 ] = 4 n) n,1 b 1,2 e 2 + n,1 b 1,2 e j c n 1,n+2 e n 1 +d n,n+1 e n, j=3 [e n+2,e n+2 ] = d n,n+1 + n 3)n 4) ) ) 2 n,1 b 1,2 n 4)n,1 b 1,2 n,2 e n AltogetherthenilrdiclN n,18, n 4),theouterderivtionsndthereminingbrcketsgiveus continuos fmily of solvble right Leibniz lgebrs depending on the prmeters iv) Finlly we pply the chnge of bsis trnsformtions given below to remove ll prmeters keeping thenilrdicln n,18 givenin41)unchnged Thetrnsformtione 1 = e 1, e i = e i +b 1,2 e 1, 2 i n 2), e j = e j, n 1 j n +2) removes b 1,2 in R en+2 nd b 1,2 in L en+2 from the entries in the 1,2), 1,3),,1,n 2) positions, respectively Then we chnge the n,2) nd entry in R en+2 to n,2 n 5) n,1 b 1,2, in R en+1 nd L en+1 to n,2 + n,1 b 1,2 We nme the entries in the n,3), n,4),,n,n 2) positionsinr en+2 by2 n,1 b 1,2,inR en+1 ndl en+1 by n,1 b 1,2 Applying the trnsformtion e 1 = e 1, e i = e i n 1,i e n 1, 2 i n 3), e j = e j, n 2 j n + 2), we remove n 1,2, n 1,3,, n 1,n 3 in R en+1 nd n 1,2, n 1,3,, n 1,n 3 inl en+1 Thistrnsformtionffectsthecoefficientsinfrontofe n 1 in [e n+2,e n+1 ] nd [e n+1,e n+2 ]Wechngethembckto c n 1,n+2 ndc n 1,n+2,respectively The trnsformtion e 1 = e 1 n,1 e n, e 2 = e 2 + n 5) n,1 b 1,2 n,2 ) en, e i = e i 2 n,1 b 1,2 e n, 3 i n 2), e k = e k, n 1 k n+2)removestheelementfromthe n,1) positioninr en+1, R en+2 ndl en+1,theelementfromthe n,2) nd positioninr en+2 ndl en+2, but it ffects the n,2) nd entries in R en+1 nd L en+1, so we chnge them to n 4) n,1 b 1,2 It lso removes the elements from the n,i) th positions in R en+2 nd L en+2, but chnges them in R en+1 nd L en+1 to n,1 b 1,2 As the finl ffect of this trnsformtion, it chnges the coefficientsinfrontofe n in [e n+1,e n+2 ]nd [e n+2,e n+1 ]toc n,n+1 n 3)n 4) n,1 b 1,2 ) 2 ndd n,n+1 + n 3)n 4) n,1 b 1,2 ) 2 n 4)n,1 b 1,2 n,2,respectively We finish the chnge of bsis with the technique of bsorption Applying the trnsformtion e i = e i, 1 i n), e n+1 = e n n) n,1 b 1,2 e 2 + n 2 j=3 n,1b 1,2 e j, e n+2 = e n+2, we remove the elements from the entries in the n,2), n,3), n,4),,n,n 2) positions in R en+1 nd L en+1 Then we chnge the coefficient in front of e n in [e n+1,e n+1 ] to c n,n+1 n 3)n 4) n,1 b 1,2 ) 2 This trnsformtion lso removesthecoefficientsinfrontofe 2, e 3, e 4,,e n 2 in [e n+2,e n+1 ]nd [e n+1,e n+2 ] Wepplytrnsformtione i = e i, 1 i n), e n+1 = e n+1 c n 1,n+2 e n 1, e n+2 = e n+2 to removethecoefficientsc n 1,n+2 nd c n 1,n+2 infrontofe n 1 in [e n+1,e n+2 ]nd[e n+2,e n+1 ], respectively j=3

19 2650 A SHABANSKAYA n 3)n 4) n,1 b 1,2) 2 c n,n+1 ) e n Thetrnsformtione i = e i, 1 i n), e n+1 = e n+1 + 2, e n+2 = e n d n,n+1 + n 3)n 4) ) ) 2 n,1 b 1,2 n 4)n,1 b 1,2 n,2 e n removes the remining coefficient c n,n+1 n 3)n 4) n,1 b 1,2 ) 2 in front of en in [e n+1,e n+1 ], [e n+1,e n+2 ], nd d n,n+1 + n 3)n 4) n,1 b 1,2 ) 2 n 4)n,1 b 1,2 n,2 in front of e n in [e n+2,e n+1 ], [e n+2,e n+2 ] We obtin the lgebr g n+2,1 To summrize, we formulte the following theorem: Theorem 54 There is one solvble indecomposble right Leibniz lgebr up to isomorphism with codimensiontwonilrdicln n,18, n 4),whichisgivenbelow: g n+2,1 : [e i,e n+1 ] = e i, [e n+1,e i ] = e i, [e j,e n+1 ] = 2e j, n 1 j n), 1 i n 2), [e n+1,e n 1 ] = 2e n 1, [e k,e n+2 ] = e k, [e n+2,e k ] = e k, 2 k n 1), [e n,e n+2 ] = 2e n, DS = [n +2, n, 2, 0], LS = [n +2, n, n,] 6 ClssifictionofsolvbleindecomposbleleftLeibnizlgebrswithnilrdiclN n,18 61 SolvbleindecomposbleleftLeibnizlgebrswithcodimensiononenilrdiclN n,18 Clssifiction of one-dimensionl left solvble extensions of N n,18, n 4) follows the sme steps s we used for the right solvble Leibniz lgebrs We consider the sme eqution 51) nd we hve the following theorem: Theorem61 Set 2,2 := in51)tostisfytheleftleibnizidentity,thererethefollowingcsesbsed on the conditions involving prmeters, ech gives continuous fmily of solvble Leibniz lgebrs 1) If = 0, 1,1 + = 0, 2 1,1 = 0,then [e 1,e n+1 ] = 1,1 e 1 + n 1,1 e n 1 + 1,1b n,1 n e n, [e i,e n+1 ] = 1,2 e 1 +e i + k,i e k, 2 1,1 k=n 1 [e n 1,e n+1 ] = + 1,1 )e n 1, [e n+1,e n+1 ] = n,n+1 e n, [e n+1,e 1 ] = 1,1 e 1 n 1,1 e n 1 n,i 2 ) 1,2b n,1 e n, 2 i n 2), 2 1,1 +b n,1 e n,[e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + [e n+1,e n 1 ] = 1,1 )e n 1,[e n+1,e n ] = 2e n 2) If 1,1 :=, = 0,thenthebrcketsforthelgebrre n [e 1,e n+1 ] = e 1 + n 1,1 e n 1 + n,1 e n, [e i,e n+1 ] = 1,2 e 1 +e i + k,i e k, k=n 1 [e n+1,e n+1 ] = n 1,n+1 e n 1 + n,n+1 e n, [e n+1,e 1 ] = e 1 n 1,1 e n 1 3 n,1 e n, [e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + n,i + 2 ) 1,2 n,1 e n, 2 i n 2), [e n+1,e n ] = 2e n 3) If = 0, 1,1 = 0,then [e 1,e n+1 ] = 1,1 e 1 + n 1,1 e n 1 + n,1 e n, [e i,e n+1 ] = 1,2 e 1 + n 1,i e n 1 + n,i e n, [e n 1,e n+1 ] = 1,1 e n 1, [e n+1,e n+1 ] = n,n+1 e n, [e n+1,e 1 ] = 1,1 e 1 n 1,1 e n 1 n,1 e n, [e n+1,e i ] = 1,2 e 1 n 1,i e n 1 +b n,i e n, 2 i n 2), [e n+1,e n 1 ] = 1,1 e n 1

20 COMMUNICATIONS IN ALGEBRA 2651 Remrk61 ThislgebristhesmestherightLeibnizlgebrincse 3)inTheorem51 4) If 1,1 := 2, = 0,then n [e 1,e n+1 ] = 2e 1 + n 1,1 e n 1 + n,1 e n, [e i,e n+1 ] = 1,2 e 1 +e i + k,i e k, k=n 1 [e n 1,e n+1 ] = 3e n 1, [e n+1,e n+1 ] = n,n+1 e n, [e n+1,e 1 ] = 2e 1 n 1,1 e n 1, [e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 + n,i ) 1,2 n,1 e n, 2 i n 2), [e n+1,e n 1 ] = 3e n 1, [e n+1,e n ] = 2e n Proof 1) Suppose = 0, 1,1 + = 0, 2 1,1 = 0TheproofofthiscsecouldbefoundinTble2 2) Suppose 1,1 :=, = 0ConsideringthesmeleftLeibnizidentitiessinthegenericcse,where theidentityinstep 19)doesnotgivenyextrconditionsontheprmeters,weprovetheresult 3) Suppose = 0, 1,1 = 0Wepplysteps 1) 19)ofcse 1)ndforthelsttwosteps 20)nd 21), weconsidertheidentities:l e1 [e n+1,e n+1 ]) = [L e1 e n+1 ),e n+1 ]+[e n+1,l e1 e n+1 )], L[e i,e n+1 ] = [Le i ),e n+1 ] + [e i,le n+1 )], 2 i n 2) 4) Suppose 1,1 := 2, = 0Werepetsteps 1) 20)ofthegenericcsendderivetheresult Theorem 62 Applying the technique of bsorption see Section 32), we cn further simplify the lgebrs given in Theorem 61 s follows: 1) Suppose = 0, 1,1 + = 0, 2 1,1 = 0Wehve [e 1,e n+1 ] = 1,1 e 1 + 1,1b n,1 e n,[e 2,e n+1 ] = 1,2 e 1 +e 2 + n 1,2 e n 1 + n,2 e n, 2 1,1 [e i,e n+1 ] = 1,2 e 1 +e i + n 1,i e n 1,[e n 2,e n+1 ] = 1,2 e 1 +e n 2,[e n 1,e n+1 ] = + 1,1 )e n 1,[e n+1,e 1 ] = 1,1 e 1 +b n,1 e n,[e n+1,e 2 ] = 1,2 e 1 e 2 n 1,2 e n 1 + n,2 2 ) 1,2b n,1 e n,[e n+1,e i ] = 1,2 e 1 e i n 1,i e n 1 2 1,2b n,1 e n, 2 1,1 2 1,1 3 i n 3),[e n+1,e n 2 ] = 1,2 e 1 e n 2 2 1,2b n,1 e n, 2 1,1 [e n+1,e n 1 ] = 1,1 )e n 1,[e n+1,e n ] = 2e n 2) Suppose 1,1 :=, = 0Thenthebrcketsforthelgebrre [e 1,e n+1 ] = e 1 + n,1 e n,[e 2,e n+1 ] = 1,2 e 1 +e 2 + n 1,2 e n 1 + n,2 e n, [e i,e n+1 ] = 1,2 e 1 +e i + n 1,i e n 1,[e n 2,e n+1 ] = 1,2 e 1 +e n 2, [e n+1,e n+1 ] = n 1,n+1 e n 1,[e n+1,e 1 ] = e 1 3 n,1 e n,[e n+1,e 2 ] = 1,2 e 1 e 2 n 1,2 e n 1 + n,2 + 2 ) 1,2 n,1 e n,[e n+1,e i ] = 1,2 e 1 e i n 1,i e n ,2 n,1 e n, 3 i n 3),[e n+1,e n 2 ] = 1,2 e 1 e n ,2 n,1 e n,[e n+1,e n ] = 2e n