ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS
OF NILINDEX N+M
L.M. CAMACHO, J. R. G´
OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV
Abs ac . We p esen he desc ip ion up o isomo phism o Leibniz supe al-
geb as wi h cha ac e is ic sequence (n|m1, . . . , mk) and nilindex n+m, whe e
m=m1+· · · +mk, n and m(m6= 0) a e dimensions o e en and odd pa s,
espec i ely.
Ma hema ics Subjec Classi ica ion 2000: 17A32, 17B30.
Key Wo ds and Ph ases: Lie supe algeb as, Leibniz supe algeb as, nilindex,
cha ac e is ic sequence.
1. In oduc ion
Ex ensi e in es iga ions o Lie algeb as heo y ha e lead o appea ance o mo e
gene al algeb aic objec s, such as Mal’ce algeb as, Lie supe algeb as, Leibniz al-
geb as, and o he s.
The well-known Lie supe algeb as a e gene aliza ions o Lie algeb as and o
many yea s hey a ac he a en ion o bo h he ma hema icians and physicis s.
The sys ema ical exposi ion o basic Lie supe algeb as heo y can be ound in he
monog aph [12] and he pape s ela ed wi h he nilpo en Lie supe algeb as a e [5],
[8], [10], and [11].
Leibniz supe algeb as a e he gene aliza ion o Leibniz algeb as and, on he o he
hand, hey na u ally gene alize Lie supe algeb as.
Recall ha Leibniz algeb as a e a ”non an isymme ic” gene aliza ion o Lie
algeb as [13]. The s udy o nilpo en Leibniz algeb as [1]–[3] shows ha many
nilpo en p ope ies o Lie algeb as can be ex ended o nilpo en Leibniz algeb as.
The esul s o he nilpo en Leibniz algeb as may help us o in es iga e he nilpo en
Leibniz supe algeb as.
In he desc ip ion o Leibniz supe algeb as s uc u e he c ucial ask is o p o e
he exis ence o a sui able basis (so-called he adap ed basis) in which he mul i-
plica ion o he supe algeb a has he mos con enien o m.
In con as o Lie supe algeb as o which p oblem o he desc ip ion o supe -
algeb as wi h he maximal nilindex is di icul [10], o nilpo en Leibniz supe al-
geb as i u ns ou o be compa a i ely easy and was sol ed in [1]. The dis inc i e
p ope y o such Leibniz supe algeb as is ha hey a e single-gene a ed. The nex
s ep - he desc ip ion o Leibniz supe algeb as wi h he dimensions o e en and
odd pa s, espec i ely equal o nand m, and wi h nilindex n+ma his momen
seems o be e y complica ed. The e o e, such Leibniz supe algeb as can be s ud-
ied by applying es ic ions on hei cha ac e is ic sequences [4]–[9]. Following his
Pa ially suppo ed by he PAICYT, FQM143 o he Jun a de Andaluc´ıa (Spain). The hi d
au ho was suppo ed by g an NATO-Rein eg a ion e . CBP.EAP.RIG.983169.
1
2 L.M. CAMACHO, J. R. G ´
OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV
app oach we in es iga e such Leibniz supe algeb as wi h cha ac e is ic sequence
C(L) = (n|m1, . . . , mk), whe e m1+···+mk=m.
Taking in o accoun he esul s o he pape s [6] and [9], whe e some cases o
nilpo en Leibniz supe algeb as we e desc ibed, in his wo k we in es iga e he es
cases. Thus, we comple e he desc ip ion o Leibniz supe algeb as wi h cha ac e -
is ic sequence C(L) = (n|m1,...,mk) and nilindex n+m.
All o e he wo k we conside spaces and algeb as o e he ield o complex num-
be s. By as e isks (∗) we deno e he app op ia e coe icien s a he basic elemen s
o supe algeb a.
2. P elimina ies
AZ2-g aded ec o space G=G0⊕G1is called a Lie supe algeb a i i is equipped
wi h a p oduc [−,−] which sa is ies he ollowing condi ions:
1. [Gα,Gβ]⊆ Gα+β(mod 2),
2. [x, y] = −(−1)αβ[y, x],
3. (−1)αγ [x, [y, z]]+(−1)αβ[y, [z, x]]+(−1)βγ [z, [x, y]] = 0 – Jacobi supe iden i y
o any x∈ Gα, y ∈ Gβ, z ∈ Gγand α, β, γ ∈Z2.
AZ2-g aded ec o space L=L0⊕ L1is called a Leibniz supe algeb a i i is
equipped wi h a p oduc [−,−] which sa is ies he ollowing condi ions:
1. [Lα,Lβ]⊆ Lα+β(mod 2),
2. [x, [y, z]] = [[x, y], z]−(−1)αβ[[x, z], y]−Leibniz supe iden i y,
o any x∈ L, y ∈ Lα, z ∈ Lβand α, β ∈Z2.
The ec o spaces L0and L1a e said o be e en and odd pa s o he supe algeb a
L, espec i ely.
No e ha i in L he g aded iden i y [x, y] = −(−1)αβ[y, x] holds, hen he Leib-
niz supe iden i y and Jacobi supe iden i y coincide. Thus, Leibniz supe algeb as
a e a gene aliza ion o Lie supe algeb as.
Fo examples o Leibniz supe algeb as we e e o [1].
Deno e by Leibn,m he se o Leibniz supe algeb as wi h dimensions o he e en
pa and he odd pa equal o nand m, espec i ely.
Le V=V0⊕V1, W =W0⊕W1be wo Z2-g aded spaces. A linea map
:V→Wis called o deg ee α(deno ed as deg( ) = α), i (Vβ)⊆Wα+β o all
β∈Z2.
Le Land L′be Leibniz supe algeb as. A linea map :L → L′is called a
homomo phism o Leibniz supe algeb as i
1. (L0)⊆ L′
0and (L1)⊆ L′
1,i.e. deg( ) = 0;
2. ([x, y]) = [ (x), (y)] o all x, y ∈ L.
Mo eo e , i is bijec ion hen i is called an isomo phism o Leibniz supe algeb as
Land L′.
Fo a gi en Leibniz supe algeb a Lwe de ine a descending cen al sequence in
he ollowing way:
L1=L,Lk+1 = [Lk,L1], k ≥1.
De ini ion 2.1. A Leibniz supe algeb a Lis called nilpo en , i he e exis s s∈N
such ha Ls= 0.The minimal numbe swi h his p ope y is called nilindex o he
supe algeb a L.
The se s
R(L) = {z∈ L | [L, z] = 0}and Z(L) = {z∈ L | [L, z] = [z, L] = 0}
ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 3
a e called he igh annihila o and he cen e o a supe algeb a L, espec i ely.
Using he Leibniz supe iden i y i is no di icul o see ha R(L) is an ideal
o he supe algeb a L. Mo eo e , he elemen s o he o m [a, b] + (−1)αβ[b, a],
(a∈ Lα, b ∈ Lβ) belong o R(L).
The ollowing heo em desc ibes he nilpo en Leibniz supe algeb as wi h maxi-
mal nilindex.
Theo em 2.2. [1] Le Lbe a Leibniz supe algeb a o he a ie y Leibn,m wi h
nilindex equal o n+m+ 1.Then Lis isomo phic o one o he ollowing non-
isomo phic supe algeb as:
[ei, e1] = ei+1,1≤i≤n−1; [ei, e1] = ei+1,1≤i≤n+m−1,
[ei, e2] = 2ei+2,1≤i≤n+m−2,
(omi ed p oduc s a e equal o ze o).
I should be no ed ha o he second supe algeb a we ha e m=nwhen n+m
is e en and m=n+ 1 i n+mis odd. Mo eo e , i is clea ha he Leibniz
supe algeb a has he maximal nilindex i and only i i is single-gene a ed.
Le L=L0⊕ L1be a nilpo en Leibniz supe algeb a. Fo an a bi a y elemen
x∈L0, he ope a o o igh mul iplica ion Rxis a nilpo en endomo phism o he
space Li,whe e i∈ {0,1}.Deno e by Ci(x) (i∈ {0,1}) he descending sequence o
he dimensions o Jo dan blocks o he ope a o Rx.Conside he lexicog aphical
o de on he se Ci(L0).
De ini ion 2.3. A sequence
C(L) = max
x∈L0 [L0,L0]C0(x)max
ex∈L0 [L0,L0]C1(ex)
is said o be he cha ac e is ic sequence o he Leibniz supe algeb a L.
Simila as in case o Lie supe algeb as [7] (Co olla y 3.0.1) i can be p o ed ha
he cha ac e is ic sequence is an in a ian unde isomo phisms.
Fo Leibniz supe algeb as we in oduce he analogue o he ze o- ili o m Leibniz
algeb as.
De ini ion 2.4. A Leibniz supe algeb a L ∈ Leibn,m is called ze o- ili o m i C(L) =
(n|m)
Deno e by ZFn,m he se o all ze o- ili o m Leibniz supe algeb as om Leibn,m.
F om [2] i can be concluded ha he e en pa o a ze o- ili o m Leibniz supe -
algeb a is a ze o- ili o m Leibniz algeb a, he e o e a ze o- ili o m supe algeb a is
no a Lie supe algeb a.
Fu he , we need he esul on exis ence o an adap ed basis o ze o- ili o m
Leibniz supe algeb as.
Theo em 2.5. [9] In an a bi a y supe algeb a om ZFn,m he e exis s a basis
{x1, x2,...,xn, y1, y2,...,ym}which sa is ies he ollowing condi ions:















[xi, x1] = xi+1,1≤i≤n−1,
[xn, x1] = 0,
[xi, xk] = 0,1≤i≤n, 2≤k≤n,
[yj, x1] = yj+1,1≤j≤m−1,
[ym, x1] = 0,
[yj, xk] = 0,1≤j≤m, 2≤k≤n.
4 L.M. CAMACHO, J. R. G ´
OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV
3. Ze o- ili o m Leibniz supe algeb a wi h nilindex equal o n+m
This sec ion is de o ed o he desc ip ion o ze o- ili o m Leibniz supe algeb as
wi h nilindex equal o n+m.
Le L ∈ ZFn,m wi h nilindex equal o n+m. E iden ly, Lhas wo gene a o s.
Mo eo e , om Theo em 2.5 i ollows ha one gene a o lies in L0and he second
gene a o lies in L1. Wi hou loss o gene ali y i can be assumed ha in an adap ed
basis he gene a o s a e x1and y1.
In he adap ed basis o Lwe in oduce he no a ions.
[xi, y1] =
m
X
j=2
αi,jyj,1≤i≤n, [yi, y1] =
n
X
j=2
βi,jxj,1≤i≤m.
In he abo e no a ion he ollowing lemma holds.
Lemma 3.1.
[yi, yj] =
min{i+j−1,m}−i
X
s=0
(−1)sCs
j−1
n−j+s+1
X
=2
βi+s, x +j−s−1,(1)
whe e 1≤i, j ≤m.
P oo . The p oo is deduced by induc ion on ja any alue o i.
Since in he wo k [9] he se ZFn,2was al eady desc ibed, we conside he se
ZFn,m(m≥3).
Case ZF2,m (m≥3).
Theo em 3.2. Le Lbe a Leibniz supe algeb a wi h he nilindex m+ 2 om
ZF2,m (m≥3).Then mis odd and Lis isomo phic o he ollowing supe algeb a:
[x1, x1] = x2,
[yi, x1] = yi+1,1≤i≤m−1,
[x1, yi] = −yi+1,1≤i≤m−1,
[yi, ym+1−i] = (−1)j+1x2,1≤i≤m−1.
P oo . F om (1) we easily ob ain
[yi, yj] = (−1)j−1βi+j−1,2x2,2≤i+j≤m+ 1,
[yi, yj] = 0, m + 2 ≤i+j≤2m. (2)
I should be no ed ha βm,26= 0.Indeed, i βm,2= 0, hen Lm−1={x2, ym−1, ym},
Lm={x2, ym}and Lm+1 ={ym}which imply ha [ym−1, y1] = ax2and [x2, y1] =
bym,whe e ab 6= 0.
The chain o he equali ies
abym= [ax2, y1] = [[ym−1, y1], y1] = 1
2[ym−1,[y1, y1]] = 0
implies a con adic ion o he p ope y ab 6= 0.The e o e, βm,26= 0.
The simple analysis o he p oduc s leads o x2∈ Z(L) (since x2∈ Lm+1 ⊆
Z(L)).
Using he Leibniz supe iden i y we ha e
[x1, yi] = α1,2yi+1 +···+α1,m−i+1ym,1≤i≤m−1.
ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 5
The exp ession [y1, x1] + [x1, y1] lies in R(L).Hence
(1 + α1,2)y2+α1,3y3+···+α1,mym(3)
belongs o R(L),as well.
I ei he α1,26=−1 o he e exis s i(3 ≤i≤m) such ha α1,i 6= 0, hen
mul iplying he linea combina ion (3) om he igh side equi ed imes o x1
we deduce ym∈ R(L).Howe e , by (2) we ha e [y1, ym] = (−1)m−1βm,2x2which
implies ha βm,2= 0 and we ge a con adic ion wi h condi ion βm,26= 0.
I α1,2=−1 and α1,i = 0 (3 ≤i≤m), hen by applying he Leibniz supe iden i y
o he basic elemen s {x1, yi, yi}we ob ain β2i,2= 0 o 1 ≤i≤[m
2].
No e ha in case mis e en we ob ain βm,2= 0 which is a con adic ion. The e-
o e, mis odd.
Le us in oduce new no a ions
γs=β2s−1,2,1≤s≤m+ 1
2.
Then we ob ain he amily L(γ1, γ2,...,γm+1
2) :







[x1, x1] = x2,
[yi, x1] = yi+1,1≤i≤m−1,
[x1, yi] = −yi+1,1≤i≤m−1,
[yi, yj] = (−1)j−1γi+j
2,2x2, i +jis e en,2≤i+j≤m+ 1, m is odd.
Make he ollowing gene al ans o ma ion o he gene a o basic elemen s:
x′
1=b1x1, y′
1=
m+1
2
X
s=1
a2s−1y2s−1.
Then x′
2=b2
1x2and
y′
2i−1=b2(i−1)
1
m−2(i−1)+1
2
X
s=1
a2s−1y2s+2i−3,1≤i≤m+ 1
2,
y′
2i=b2i−1
1
m−2(i−1)−1
2
X
s=1
a2s−1y2s+2i−2,1≤i≤m−1
2.
Choosing he pa ame e s aias ollows
a1=s1
bm−3
1γm+1
2
, a3=−a1γm−1
2
2γm+1
2
,
ai=−a2
1γm−i+2
2+ 2a1a3γm−i+4
2+···+ (2a1ai−2+···+ 2ai−3
2ai+1
2)γm−1
2
2a1γm+1
2
−
−
(2a3ai−2+···+ 2ai−3
2ai+5
2+a2
i+1
2
)γm+1
2
2a1γm+1
2
, o i+ 1
2odd.
ai=−
a2
1γm−i+2
2+ 2a1a3γm−i+4
2+···+ (2a1ai−2+···+ 2ai−5
2ai+3
2+a2
i−1
2
)γm−1
2
2a1γm+1
2
−
−(2a3ai−2+···+ 2ai−1
2ai+3
2)γm+1
2
2a1γm+1
2
, o i+ 1
2e en.

6 L.M. CAMACHO, J. R. G ´
OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV
when 4 ≤i≤m, we ob ain [y′
m, y′
1] = x′
2,[y′
i, y′
1] = 0 o 1 ≤i≤m−1.
Then applying Leibniz supe iden i y ge he es b acke s
[y′
i, y′
j] = 0,1≤i, j ≤m, i +j6=m+ 1,
[y′
i, y′
j] = (−1)j−1x′
2,1≤i, j ≤m, i +j=m+ 1..
Thus, we ob ain he supe algeb a o he heo em. 
Case ZFn,m (n≥3, m ≥3).
Lemma 3.3. Any Leibniz supe algeb a om ZFn,m (n≥3, m ≥3) has nilindex
less ha n+m.
P oo . Le us assume he con a y, i.e. Lis a Leibniz supe algeb a om ZFn,m (n≥
3, m ≥3) and Lhas he nilindex equal o n+m. Then in he adap ed basis we
ha e
L={x1, x2,...,xn, y1, y2,...,ym},
L2={x2,...,xn, y2,...,ym},
L3⊃ {x3,...,xn, y3,...,ym}.
Le us suppose ha L3={x3,...,xn, y2,...,ym},i.e. x2/∈ L3and y2∈ L3.
Then he e exi s i0(2 ≤i0≤n) such ha [xi0, y1] = αi0,2y2+···+αi0,mymwi h
αi0,26= 0.Since xi∈ R(L) o 2 ≤i≤nand R(L) is an ideal, hen αi0,2y2+···+
αi−0,mym∈ R(L).
Mul iplying he p oduc [xi0, y1] on he igh side consequen ly o he basic
elemen x1(m−1)− imes we easily ob ain ha y2, y3,...,ym∈ R(L), ha is
L2=R(L).
By induc ion one can p o e he ollowing
[xi, y1] = 




m+1−i
P
j=2
α1,jyj+i−1,i i+ 1 ≤m,
0,i i+ 1 > m.
(4)
Since L2=R(L) hen y2can appea only in he p oduc s [xi, y1] o 2 ≤i≤n)
o [yj, x1] o 2 ≤j≤m−1).Howe e , in he i s case om (4) we con-
clude ha y2does no lie in L3and in he second case he elemen y2can no
be ob ained, i.e. in bo h cases we ha e a con adic ion wi h he assump ion
L3={x3,...,xn, y2,...,ym}.
Thus, L3={x2,...,xn, y3,...,ym}.Le sbe a na u al numbe such ha x2∈
Ls Ls+1.
Suppose s≤m. Then we ha e
Li={x2,...,xn, yi,...,ym},2≤i≤s,
Ls+1 ={x3,...,xn, ys,...,ym}
and in he equali y [ys−1, y1] =
n
P
j=2
βs−1,jxj he coe icien βs−1,2is no ze o.
F om Lemma 3.1 we ha e
[y1, ys] =
s−1
X
i=0
(−1)iCi
s−1
n−s+i+1
X
=2
β1+i, x +s−i−1,
ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 7
in which he coe icien βs−1,2occu s. Taking in o accoun he equali y [ys, y1] =
n
X
j=2
βs,jxjwe conclude ha x3∈lin < [y1, ys],[ys, y1], x4,...,xn>. The e o e
Ls+2 =< x3,...,xn, ys+1,...,ym>, i.e. ys∈ Ls+1 Ls+2 and α2,s 6= 0.
Conside he equali ies
[ys−1,[y1, y1]] = 2[[ys−1, y1], y1] = 2 "n
X
=2
βs−1, [x , y1]#= 2βs−1,2[x2, y1]+ X
i≥s+1
(∗)yi.
On he o he hand [ys−1,[y1, y1]] = 0,because [y1, y1]∈ R(L).
The basic elemen ysappea s only in he p oduc [x2, y1].Hence we ha e ha
βs−1,2α2,sys+P
i≥s+1
(∗)yi= 0 which implies βs−1,2α2,s = 0.This con adic s o he
assump ion s≤m.
Le us conside now he case s=m+ 1.Then we ha e
L={x1, x2,...,xn, y1, y2,...,ym},
Li={x2, x3,...,xn, yi, yi+1 ...,ym},2≤i≤m,
Lm+i−1={xi, xi+1 ...,xn},2≤i≤n.
Since x2∈ Lm+1 we ha e [ym, y1] =
n
P
i=2
βm,ixiwi h βm,26= 0.
The sum [y1, x1] + [x1, y1] lies in R(L) since [y1, x1] + [x1, y1] = (1 + α1,2)y2+
α1,3y3+···+α1,mym∈ R(L).
I [y1, x1] + [x1, y1] = 0, hen using he Leibniz supe iden i y we ha e
[x1,[ym, y1]] = [[x1, ym], y1] + [[x1, y1], ym] = −[y2, ym] = βm,2x3+X
i≥4
(∗)xi.
On he o he hand
[x1,[ym, y1]] =
n
X
i=2
βm,i[x1, xi] = 0.
Hence, βm,2= 0 which is a con adic ion.
Thus, [y1, x1] + [x1, y1]6= 0.Con inuing he same a gumen a ion as in he p oo
o Theo em 3.2 we ob ain ym∈ R(L). The e o e
[y1, ym] =
m−1
X
i=0
(−1)iCi
m−1
n−m+i+1
X
=2
β1+i, x +m−i−1= 0.
The minimal alue o he exp ession +m−i−1 is eached when i=m−1
and = 2.Thus, [y1, ym] = (−1)m−1Cm−1
m−1βm,2x2+P
i≥4
(∗)xiwhich implies ha
βm,2= 0.Tha is a con adic ion wi h he assump ion ha nilindex o Lis equal
o n+m. 
4. Leibniz supe algeb as wi h he cha ac e is ic sequence
(n|m1, m2,...,mk)and nilindex n+m
Leibniz supe algeb as wi h he cha ac e is ic sequence equal o (n|m−1,1) and
wi h he nilindex n+mwe e examined in [6]. The e o e, in his sec ion we shall
conside he Leibniz supe algeb as Lo nilindex n+mwi h he cha ac e is ic
sequence equal o (n|m1, m2,...,mk) wi h condi ions m1≤m−2.
8 L.M. CAMACHO, J. R. G ´
OMEZ, B.A. OMIROV AND A.KH. KHUDOYBERDIYEV
F om he de ini ion o cha ac e is ic sequence he e exi s a basis {x1, x2,...xn, y1,
y2,...ym}in which he ope a o Rx1|L1has he ollowing o m:
Rx|L1=



Jmj10··· 0
0Jmj2··· 0
··· ··· ··· ···
0 0 ··· Jmjk



,
whe e (mj1, mj2, . . . , mjk) is a pe mu a ion o (m1, m2,...,mk).Wi hou loss o
gene ali y, by a shi ing o he basic elemen s we can assume ha ope a o Rx1|L1
has he ollowing o m
Rx1|L1=



Jm10··· 0
0Jm2··· 0
··· ··· ··· ···
0 0 ··· Jmk



.
I means ha he basis {x1, x2,...xn, y1, y2,...ym}sa is ies he ollowing condi-
ions:
[xi, x1] = xi+1,1≤i≤n−1,
[yj, x1] = yj+1, o j /∈ {m1, m1+m2,...,m1+m2+···+mk},
[yj, x1] = 0, o j∈ {m1, m1+m2,...,m1+m2+···+mk}.
(5)
I is clea ha wo gene a o s can no lie in L0.In ac , in [2] he able o
mul iplica ion o he Leibniz algeb a L0is p esen ed and i has only one gene a o .
Theo em 4.1. Le Lbe a Leibniz supe algeb a o nilindex n+mwi h cha ac e is ic
sequence (n|m1, m2,...,mk),whe e m1≤m−2.Then bo h gene a o s can no
belong o L1a he same ime.
P oo . Le Leibniz supe algeb a L=L0⊕L1has nilindex n+mand le {x1, x2,...,xn}
be a basis o L0and {y1, y2,...,ym}a basis o L1. Suppose ha wo gene a o s lie
in L1.Then hey should be om he se
{y1, ym1+1, ym1+m2+1,...,ym1+m2+···+mk−2+1, ym1+m2+···+mk−1+1}
Wi hou loss o gene ali y, he gene a o s can be chosen as {y1, ym1+1}.
Conside he ollowing cases:
Case 1. Le [y1, y1]∈ L0 L2
0.Then conside he Leibniz supe algeb a gene a ed by
he elemen < y1> . Since [y1, y1]∈ L0 L2
0we can assume [y1, y1] = x1.Then om
he p oduc s in (5) we deduce {x1, x2,...,xn, y2, y3,...,ym1} ⊆< y1> . I i easy o
see ha ym1+1 /∈< y1> . Indeed, i ym1+1 ∈< y1>, hen {ym1+2,...,ym1+m2} ⊆<
y1>which implies C(L)≥(n|m1+m2, m3,...,mk).Tha is a con adic ion o
he condi ion o cha ac e is ic sequence o L, because C(L) = (n|m1, m2,...,mk).
Thus, he Leibniz supe algeb a gene a ed by he basic elemen y1consis o
{x1, x2,...,xn, y1, y2,...,ym1}.
Since he supe algeb a < y1>is single-gene a ed hen om Theo em 2.2 we ha e
ha ei he m1=no m1=n+1 and he mul iplica ion in < y1>has he ollowing
o m: [xi, x1] = xi+1,1≤i≤n−1,
[yj, x1] = yj+1,1≤j≤m1−1,
[xi, y1] = 1
2yi+1,1≤i≤m1−1,
[yj, y1] = xj,1≤j≤n.
ON COMPLEX NILPOTENT LEIBNIZ SUPERALGEBRAS OF NILINDEX N+M 9
Case m1=n. Since y1and yn+1 a e gene a o s we ha e
L={x1, x2,...,xn, y1,...,yn, yn+1,...,ym},
L2={x1, x2,...,xn, y2,...,yn, yn+2,...,ym}.
Besides, x1/∈ L3.O he wise, i x1∈ L3, hen he e exis s z∈ L1such ha
z∈ L2/L3.The eby z∈lin < [y1, y1],[y1, yn+1],[yn+1, y1],[yn+1, yn+1]>and
aking in o accoun ha [yi, yj]∈ L0we ob ain z∈ L0which is a con adic ion.
Thus,
L3={x2,...,xn, y2,...,yn, yn+2,...,ym}.
I L2k={xi, xi+1,...,xn, yj, . . . , yn, yn+2,...,ym}, hen by a simila way one
can p o e ha L2k+1 ={xi+1,...,xn, yj,...,yn, yn+2,...,ym}. In ac , i z∈
L2k/L2k+1, hen zhas o be gene a ed by 2kp oduc s o he gene a o s (bu hey
a e om L1). Hence his p oduc s belong o L0,and we ha e z∈ L0.
Applying he simila a gumen a ion we ge
L2k+2 ={xi+1,...,xn, yj+1,...,yn, yn+2,...,ym}.
Con inuing wi h he p ocess, we ob ain ha L2n+1 ={yi1, yi2,...,yik}and
L2n+2 = 0.Since dim(L2n+1/L2n+2) = 1 hen L2n+1 ={yn+2}and nilindex should
be equal o 2n+ 2.Thus, m=n+ 2 and we ha e
L={x1, x2,...,xn, y1,...,yn, yn+1, yn+2},
L2k={xk,...,xn, yk+1,...,yn, yn+2},1≤k≤n−1,
L2k+1 ={xk+1,...,xn, yk+1,...,yn, yn+2},1≤k≤n−1,
L2n={xn, yn+2},L2n+1 ={yn+2},L2n+2 ={0}.
Fu he mo e, L2n= [L2n−1,L] =<[xn, y1],[xn, yn+1],[yn, y1],[yn, yn+1],[yn+2, y1],
[yn+2, yn+1]> . No e ha he elemen yn+2 can be ob ained only om p oduc
[xn, yn+1] (because [xn, y1] = 0, o he wise we ge a con adic ion wi h he p ope y
o cha ac e is ic sequence). Howe e ,
[xn, yn+1] = [[xn−1, x1], yn+1] = [xn−1,[x1, yn+1]] + [[xn−1, yn+1], x1] = 0,
which deduce L2n={xn}, ha is a con adic ion wi h he condi ion o nilindex.
Case m1=n+ 1. In his case, simila o he p e ious case, we ge a con adic-
ion.
Case 2. Le [y1, y1]/∈ L0 L2
0and [ym1+1, ym1+1]∈ L0 L2
0.Then applying he
same a gumen s o ym1+1 as o y1in Case 1, we ob ain a con adic ion wi h he
ac ha bo h gene a o s lie in L1,as well.
Case 3. Le [y1, y1]/∈ L0 L2
0and [ym1+1, ym1+1]/∈ L0 L2
0.Then, wi hou loss o
gene ali y, we can assume ha
[y1, ym1+1] = x1,
[ym1+1, y1] =
n
P
i=1
bixi.
I b1= 1, hen making he change o basis y′
1=y1+ym1+1 we ob ain [y′
1, y′
1]∈
L0 L2
0.The e o e his case can be educed o Case 1.
I b16= 1, hen [y1, ym1+1]−[ym1+1, y1] = (1 −b1)x1+b2x2+···+bnxn∈ R(L)
and since xi∈ R(L) (2 ≤i≤n) we ge x1∈ R(L). F om he Leibniz supe iden i y