Some results on the Schur multiplier of nilpotent Lie algebras

0.19 MB·

by  Peyman Niroomand

SOME RESULTS ON THE SCHUR MULTIPLIER OF
NILPOTENT LIE ALGEBRAS
7
1
PEYMANNIROOMANDANDFARANGISJOHARI
0
2
n Abstract. For a non-abelian LiealgebraL of dimension n withthe derived
a subalgebra of dimension m , the author earlier proved that the dimension of
J itsSchurmultiplierisboundedby 21(n+m−2)(n−m−1)+1. Inthecurrent
work,weobtaintheclassofallnilpotentLiealgebraswhichattainstheabove
4
bound. Furthermore,wealsoimprovethisboundasmuchaspossible.
1

]
A
R 1. Introduction, Motivation and Preliminaries
.
h Analogous to the Schur multiplier of a group, the Schur multiplier of a Lie
t algebra,M(L), canbe defined asM(L)∼=R∩F2/[R,F]where L∼=F/R andF is
a
m a free Lie algebra (see [5, 13, 14] for more information).
Thereareseveralworkstoshowthatthe resultsonthe Schurmultiplier offinite
[
p-groups (p a prime) have analogues on the Schur multiplier M(L) of a nilpotent

1 LiealgebraLofdimensionn. Forinstance,in[13,Theorem3.1],theauthorproved
v
that for a non-abelian nilpotent Lie algebra of dimension n, we have
6
5 1
(1.1) dimM(L)≤ (n+m−2)(n−m−1)+1
9 2
3 and the equality holds when L ∼= H(1) ⊕ A(n − 3), in where H(m) and A(n)
0
denote the Heisenberg Lie algebra of dimension 2m+1 (a Lie algebra such that
.
1 L2 =Z(L)anddimL2 =1)andabelianLiealgebraofdimensionn. Thisimproves
0
the earlier results obtained by the same author in [15, Main Theorem] for Lie
7
algebras. Recently, the structure of all p-groups of class two for which |M(G)|
1
: attainsthe bound[13,Theorem3.1]isclassifiedin[17],andthen[13,Theorem3.1]
v
has been improved by Hatui in [7].
i
X In the present article, we give some new inequalities on the exterior square and
r the Schur multiplier of Lie algebras. Then we classify all nilpotent Lie algebras
a
that attain the bound 1.1. More precisely, they are exactly nilpotent Lie algebras
of class two. Moreover, for nilpotent Lie algebras of class at last 3, we improve
1.1 as much as possible. It develops some key results of [7, 17] for the class of Lie
algebras by a different way.
Fortheconvenienceofthereader,wegivesomeresultswithoutproofswhichwill
be used in the next section. For a Lie algebra L, we use notation L(ab) instead of
L/L2.
Lemma 1.1. [2, Proposition 3] Let A and B be two Lie algebras. Then
M(A⊕B)∼=M(A)⊕M(B)⊕(A(ab)⊗modB(ab)),
Date:January17,2017.
Keywords andphrases. Tensorsquare,exteriorsquare,capability,Schurmultiplier,p-groups,
relativeSchurmultiplier,locallyfinitegroups.
1
2 P.NIROOMANDANDF.JOHARI
in where A(ab)⊗ B(ab) is the standard tensor product A and B.
mod
Schurmultipliers ofabelianandHeisenbergLie algebrasarewellknown. See for
instance [11, Lemma 2.6].
Lemma 1.2. We have
1
(i) dimM(A(n))= n(n−1).
2
(ii) dimM(H(1))=2.
(iii) dimM(H(m))=2m2−m−1 for all m≥2.
Our next aim is to exhibit a close relation between the Lie algebra M(L) and
M(L/K) where K is an ideal of L.
Lemma 1.3. [13, Corollary 2.3] Let L be a finite dimensional Lie algebra, K an
ideal of L and H =L/K. Then
dimM(L)+dim(L2∩K)≤dimM(H)+dimM(K)+dim(H(ab)⊗ K(ab)).
mod
ThenextlemmagivesanupperboundfortheSchurmultiplierofann-dimensional
nilpotent Lie algebra with the derived subalgebra of maximum dimension.
Lemma 1.4. [14,Theorem3.1] An n-dimensional nilpotent Lie algebra L in which
dimL2 =n−2 and n≥4 has dimM(L)≤dimL2.
The following theorem improves the earlier bound on the dimension of Schur
multiplier.
Theorem 1.5. [13, Theorem3.1] Let L be an n-dimensional non-abelian nilpotent
Lie algebra with the derived subalgebra of dimension m. Then
1
dimM(L)≤ (n+m−2)(n−m−1)+1.
2
In particular, when m=1 the bound is attained if and only if L=H(1)⊕A(n−3).
From [5] L∧L and L⊗L denote the exterior square and the tensor square of
a Lie algebra L, respectively. The authors assume that the reader is familiar with
these concepts.
Proposition 1.6. [5, Proposition 1.1] Let L be a Lie algebra such that I and K
are ideals in L and K ⊆I. Then the sequence K ∧L→I ∧L→L/K∧I/K →0
is exact.
Lemma 1.7. [5, Theorem 35 (iii)] Let L be a Lie algebra. Then 0 → M(L) →
L∧L−κ→′ L2 →0 is exact, in where κ′ :L∧L→L2 is given by l∧l 7→[l,l ].
1 1
The following lemmas are useful in our main results.
Lemma 1.8. [9, Lemma 2.14] Let I be a central ideal of L. Then
I ∧L∼=(cid:0)I⊗modL/L2(cid:1)/hx⊗(x+L2)|x∈Ii.
Moreover, if I ⊆L2, then I∧L∼=I ⊗modL/L2.
The nextlemma illustratesthe Lie algebraL∧Lis isomorphicto afactorofthe
free Lie algebra F2.
SCHUR MULTIPLIER OF NILPOTENT LIE ALGEBRAS 3
π
Lemma 1.9. [12, Theorem 2.10] Let 0→R→F −→L→0 be a free presentation
of a Lie algebra L. Then
δ :L∧L→F2/[R,F]
x∧y 7→[x˜,y˜]+[R,F]
is an isomorphism, in where π(x˜+R)=x and π(y˜+R)=y.
The next result is extract from the works of Batten, Moneyhun and Stitzinger
(1996).
Lemma 1.10. [2, Lemma 1] Let L be a Lie algebra such that dimL/Z(L) = n.
1
Then dimL2 ≤ n(n−1).
2
2. Main result
In this section, after examining certain upper bounds for L∧L and M(L), we
investigate the numerical inequality on the dimension M(L). Then we classify all
nilpotentLiealgebrasthatattainstheupperboundTheorem1.5. Theyareexactly
nilpotent Lie algebras of class two. Moreover,for nilpotent Lie algebras of class at
least 3, we also improve the bound Theorem 1.5.
First, we begin with the following result for a Lie algebra, similar to the result
of Blackburn for the group theory case [3].
π
Let 0→R→F −→L→0 be a free presentation of a Lie algebra L. Then
Theorem 2.1. Let L be a finite dimensional nilpotent non-abelian Lie algebra of
class two. Then
0→kerg →L2⊗ L(ab) −→g M(L)→M(L(ab))→L2 →0
mod
is exact, in where
g :x⊗(z+L2)∈L2⊗ L(ab) 7→[x˜,z˜]+[R,F]∈M(L)=R∩F2/[R,F],
mod
π(x˜+R)=x and π(z˜+R)=z. Moreover, K =h[x,y]⊗z+L2+[z,x]⊗y+L2+
[y,z]⊗x+L2|x,y,z ∈Li⊆kerg.
Proof. By using [1, Lemma 1.2] for c=1, we have the following exact sequence
0→kerg →L2⊗ L(ab) −→g M(L)→M(L(ab))→L2 →0
mod
in where
g :x⊗(z+L2)∈L2⊗ L(ab) →[x˜,z˜]+[R,F]∈M(L)=R∩F2/[R,F],
mod
π(x˜+R) = x and π(z˜+R) = z. Putting K = h[x,y]⊗z+L2+[z,x]⊗y+L2+
[y,z]⊗x+L2|x,y,z ∈Li. By using the Jacobi identities,
g([x,y]⊗z+[z,x]⊗y+[y,z]⊗x)=[[x˜,y˜],z˜]+[[z˜,x˜],y˜]+[[y˜,z˜],x˜]+[R,F]=0.
Thus K ⊆kerg. (cid:3)
The following two theorems are similar to the results of Ellis in [6] and Hauti in
[7] for the case of group theory.
Theorem 2.2. Let L be a Lie algebra. Then
(i) Li/Li+1∧L/Li+1 ∼=Li/Li+1⊗modL/L2 for all i≥2.
(ii) The natural sequence Li+1∧L −α−i−+→1 Li∧L −η→i Li/Li+1⊗ L/L2 → 0
mod
is exact for all i≥2.
4 P.NIROOMANDANDF.JOHARI
Proof. (i) SinceLi/Li+1 ⊆Z(L/Li+1)∩L2/Li+1,Lemma1.8impliesLi/Li+1∧
L/Li+1 ∼=Li/Li+1⊗modL/L2 for all i≥2.
(ii) The result follows from Proposition 1.6 and part (i).
(cid:3)
Proposition 2.3. Let L be a Lie algebra. Then
(i) the map γ :L(ab)⊗ L(ab)⊗ L(ab) →L2/L3⊗ L/L2 given by
L mod mod mod
(x+L2)⊗(y+L2)⊗(z+L2)7→([x,y]+L3⊗z+L2)+([z,x]+L3⊗y+L2)+([y,z]+L3⊗x+L2)
is a Lie homomorphism. If any two element of the set {x,y,z} are linearly
dependent. Then γ (x+L2⊗y+L2⊗z+L2)=0.
L
(ii) Define the map
γ′ :(L/Z(L))(ab)⊗ (L/Z(L))(ab)⊗ (L/Z(L))(ab) →L2/L3⊗ (L/Z(L))(ab)
2 mod mod mod
x+(L2+Z(L)) ⊗ y+(L2+Z(L)) ⊗ z+(L2+Z(L)) 7→
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
[x,y]+L3⊗z+(L2+Z(L)) + [z,x]+L3⊗y+(L2+Z(L)) +
(cid:0) (cid:1) (cid:0) (cid:1)
[y,z]+L3⊗x+(L2+Z(L)) .
(cid:0) (cid:1)
Then γ′ is a Lie homomorphism. Moreover, if any two element of the set
2
{x,y,z} are linearly dependent, then
γ′(x+(L2+Z(L))⊗y+(L2+Z(L))⊗z+(L2+Z(L)))=0.
2
(iii) The map
γ′ :(L/Z(L))(ab)⊗ (L/Z(L))(ab)⊗ (L/Z(L))(ab)⊗ (L/Z(L))(ab)
3 mod mod mod
→L3⊗ (L/Z(L))(ab) given by
mod
x+(L2+Z(L)) ⊗ y+(L2+Z(L)) ⊗ z+(L2+Z(L)) + w+(L2+Z(L))
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
7→ [[x,y],z]⊗w+(L2+Z(L)) + [w,[x,y]]⊗z+(L2+Z(L)) +
(cid:0) (cid:1) (cid:0) (cid:1)
[[z,w],x]⊗y+(L2+Z(L)) + [y,[z,w]]⊗x+(L2+Z(L))
(cid:0) (cid:1) (cid:0) (cid:1)
is a Lie homomorphism.
Proof. Clearly, γ is a Lie homomorphism. Let x = βy, for a scalar β. We claim
L
that γ (x+L2⊗y+L2⊗z+L2)=0. Since [x,y]=0, we have
L
γ (x+L2⊗y+L2⊗z+L2)=
L
([z,x]+L3⊗y+L2)+([y,z]+L3⊗x+L2)=
([z,βy]+L3⊗y+L2)+([y,z]+L3⊗βy+L2)=
β ([z,y]+L3⊗y+L2)+([y,z]+L3⊗y+L2) =
(cid:0) (cid:1)
β ([z,y]+L3⊗y+L2)−([z,y]+L3⊗y+L2) =0.
(cid:0) (cid:1)
Thus γ (x+L2 ⊗y +L2 ⊗z +L2) = 0. The cases (ii) and (iii) obtained by a
L
similar way. (cid:3)
The following preliminary result will also play an important role in the next.
Lemma 2.4. The following natural sequence of abelian Lie algebras
Li/Li+1⊗ (Z(L)+L2)/L2 −τ→i′ Li/Li+1⊗ L/L2 −δ→i Li/Li+1⊗ L/(Z(L)+L2) →0
mod(cid:0) (cid:1) mod mod(cid:0) (cid:1)
is exact for all i≥2.
SCHUR MULTIPLIER OF NILPOTENT LIE ALGEBRAS 5
Proof. The proof is straightforward. (cid:3)
We now make an observation to the Proposition 1.6 and Theorem 2.2 (ii), we
can easily check that the following maps are homomorphisms
η :l∧l ∈L∧L7→(l+L2∧l +L2)∈L/L2∧L/L2,
1 1 1
η :x∧y ∈Li∧L7→(x+Li+1⊗y+L2)∈Li/Li+1⊗ L/L2,
i mod
α :x∧z ∈Li∧L7→(x∧z)∈Li−1∧L,andκ′ :(x ∧z )∈L∧Li−1 7→[x ,z ]∈Li
i i 1 1 1 1
for all i≥2. Put K =h[x,y]∧z+[z,x]∧y+[y,z]∧x|x,y,z ∈Li⊆L2∧L and
K =h[[x,y],z]∧w+[w,[x,y]]∧z+[[z,w],x]∧y+[y,[z,w]]∧x|w,x,y,z ∈Li⊆L3∧L.
1
By remembering the last homomorphisms, we are now ready to prove
Proposition 2.5. Consider the canonical homomorphisms
τ :Li∧Z(L)→Li∧L,
i
τ′ :Li/Li+1⊗ (Z(L)+L2)/L2 →Li/Li+1⊗ L/L2
i mod(cid:0) (cid:1) mod
and
δ :Li/Li+1⊗ L/L2։Li/Li+1⊗ L/(Z(L)+L2)
i mod mod(cid:0) (cid:1)
for all i ≥ 2. Then αi(Imτi) = 0Li−1∧L, ηi (Imτi) = Imτi′ = kerδi, Imτ2 ⊆
(cid:12)
K ⊆ kerα , dimτ′ ≤ dimImγ ≤ dimK ≤(cid:12)dImiτmi kerα , Im(δ ) = Imγ′ and
2 2 L 2 2 2
(cid:12)
ker(δ )=Imτ′ for alli≥2.Moreover, K ⊆kerα anddim(cid:12)ImImγLγ′ ≤dimK ≤
2 2 1 3 3 1
(cid:12)
(cid:12)ImγL
dimkerα .
3
Proof. We claim that αi(x ∧z) = 0Li−1∧L for all z ∈ Z(L) and x ∈ Li for all
i ≥ 2. There exists x¯ ∈ L ∧ Li−1 such that κ′(x¯) = x. We may assume that
i
x¯= r β (l′ ∧l ), where l ∈Li−1,l′ ∈L and β is scalar. Then
Pt=1 t t t t t t
r r
′ ′ ′ ′
x=κi(x¯)=κi(Xβt(lt∧lt))=Xβt([lt,lt]).
t=1 t=1
Therefore
r r
′ ′
αi(x∧z)=x∧z =(Xβt[lt,lt])∧z =Xβt([lt,lt]∧z)
t=1 t=1
r
′ ′
=Xβt(lt∧[lt,z]−lt∧[lt,z])=0Li−1∧L.
t=1
Thus αi(Imτi)=0Li−1∧L. Consider the restriction of homomorphism ηi to Imτi as
follows
η :x∧z ∈Imτ 7→(x+Li+1⊗z+L2)∈Li/Li+1⊗ L/L2.
i i mod
(cid:12)
(cid:12)Imτi
Obviously, Im(η ) = Imτ′. By invoking Lemma 1.9 , δ : L∧L → F2/[R,F]
i i
(cid:12)
(cid:12)Imτi
is an isomorphism and δ([x,y]∧z+[z,x]∧y+[y,z]∧x) =[[x˜,y˜],z˜]+[[z˜,x˜],y˜]+
[[y˜,z˜],x˜]+[R,F] = 0. Using the Jacobi identities, we have [[x˜,y˜],z˜]+[[z˜,x˜],y˜]+
[[y˜,z˜],x˜]+[R,F]=0 for x,y,z ∈L. Thus
[x,y]∧z+[z,x]∧y+[y,z]∧x=0L∧L for all x,y,z ∈L.
6 P.NIROOMANDANDF.JOHARI
Henceα (K)=0andsoK ⊆kerα .Bythe restrictionofη toK andProposition
2 2 2
2.3 (i), we have η (K) = Imγ . Thus dimImγ ≤ dimK ≤ dimkerα . Now we
2 L L 2
show that Imτ′ ⊆Imγ . Let z ∈Z(L) and d= t α [x ,y ] for x ,y ∈L. Since
2 L Pi=1 i i i i i
[x ,y ]+L3⊗z+L2 + [z,x ]+L3⊗y +L2 + [y ,z]+L3⊗x +L2 =
(cid:0) i i (cid:1) (cid:0) i i (cid:1) (cid:0) i i (cid:1)
[x ,y ]+L3⊗z+L2 for all 1≤i≤t,
(cid:0) i i (cid:1)
we obtain [x ,y ]+L3⊗z+L2 ∈Imγ . Thus
(cid:0) i i (cid:1) L
t
d+L3⊗z+L2 =(Xαi[xi,yi]+L3)⊗z+L2 =
i=1
t
X(αi[xi,yi]+L3⊗z+L2)∈ImγL.
i=1
Therefore Imτ′ ⊆ Imγ . Similarly Imτ ⊆ K ⊆ kerα . By the restriction of δ to
2 L 2 2 2
Imγ , we have
L
δ :Imγ →L2/L3⊗ L/(L2+Z(L)) given by
2 L mod
(cid:12)
(cid:12)ImγL
([x,y]+L3⊗z+L2)+([z,x]+L3⊗y+L2)+([y,z]+L3⊗x+L2)7→
[x,y]+L3⊗z+(L2+Z(L)) + [z,x]+L3⊗y+(L2+Z(L)) +
(cid:0) (cid:1) (cid:0) (cid:1)
[y,z]+L3⊗x+(L2+Z(L)) .
(cid:0) (cid:1)
Obviously,Im(δ )=Imγ′ andker(δ )=Imτ′.WeshowthatK ⊆kerα
2 2 2 2 1 3
(cid:12) (cid:12)
and dimImγ′ ≤d(cid:12)IimmγLK ≤dimkerα . Sin(cid:12)cIemγL
3 1 3
α ([[x,y],z]∧w+[w,[x,y]]∧z+[[z,w],x]∧y+[y,[z,w]]∧x)=
3
[[x,y],z]∧w+[w,[x,y]]∧z+[[z,w],x]∧y+[y,[z,w]]∧x=
[x,y]∧[z,w]−z∧[[x,y],w]+[w,[x,y]]∧z+[z,w]∧[x,y]
−x∧[[z,w],y]+[y,[z,w]]∧x=
[x,y]∧[z,w]+z∧[w,[x,y]]+[w,[x,y]]∧z+[z,w]∧[x,y]
+x∧[y,[z,w]]+[y,[z,w]]∧x=0L2∧L,
wehave K ⊆kerα . Similarly, Proposition2.3 (iii)implies η (K )=Imγ′ andso
1 3 3 1 3
dimImγ′ ≤dimK ≤dimkerα , as required. (cid:3)
3 1 3
Theorem 2.6. Let L be a finite dimensional nilpotent non-abelian Lie algebra of
class c. Then
c
dimL∧L+dimImγL ≤dimL∧L+Xdimkerαi
i=2
c
=dimL/L2∧L/L2+Xdim(Li/Li+1⊗modL/L2).
i=2
Proof. By using Proposition1.6 andTheorem 2.2 (ii), the following two sequences
(2.1) L2∧L −α→2 L∧L −η→1 L/L2∧L/L2 →0
and
(2.2) Li+1∧L −α−i−+→1 Li∧L −η→i Li/Li+1⊗ L/L2 →0,
mod
SCHUR MULTIPLIER OF NILPOTENT LIE ALGEBRAS 7
are exact for all i≥2.
Using 2.1 and 2.2, we have
(2.3)
c c
dimL∧L+Xdimkerαi =dim(L/L2∧L/L2)+X(dimLi/Li+1⊗modL/L2).
i=2 i=2
Now Proposition 2.5 implies dimImγ ≤dimkerα . Hence 2.3 deduces that
L 2
dimL∧L+dimImγ ≤dimL∧L+dimkerα
L 2
c
≤dimL∧L+Xdimkerαi =
i=2
c
dim(L/L2∧L/L2)+Xdim(Li/Li+1⊗modL/L2),
i=2
as required. (cid:3)
Theorem 2.7. Let L be a finite dimensional nilpotent non-abelian Lie algebra of
class c. Then
c
′
dimL∧L+dimImγ2 ≤dimL∧L+Xdimkerαi
i=2
c
=dim(L/L2∧L/L2)+Xdim(Li/Li+1⊗mod(L/Z(L))(ab)).
i=2
Proof. By Proposition 2.5, we have dimImγ ≤dimkerα . Thus
L 2
′
dimL∧L+dimImγ −dimImτ ≤
L 2
′
dimL∧L+dimkerα −dimImτ .
2 2
Proposition 2.5 implies dimImτ′ ≤dimImτ ≤dimkerα for all i≥2. Hence
i i i
′
dimL∧L+dimkerα −dimImτ
2 2
c
′
≤dimL∧L+X(dimkerαi−dimImτi).
i=2
Hence
c
′ ′
dimL∧L+dimImγL−dimImτ2 ≤dimL∧L+X(dimkerαi−dimImτi).
i=2
On the other hand, with the aid of Theorem 2.6, we have
c
′
dimL∧L+X(dimkerαi−dimImτi)
i=2
c
=dim(L/L2∧L/L2)+X(cid:0)dim(Li/Li+1⊗modL/L2)−dimImτi′(cid:1).
i=2
Now Lemma 2.4 implies
dim(Li/Li+1⊗ L/L2)=dim(Li/Li+1⊗ (L/Z(L))(ab))+dimImτ′,
mod mod i
8 P.NIROOMANDANDF.JOHARI
for i≥2. Thus
c
dim(L/L2∧L/L2)+X(cid:0)dim(Li/Li+1⊗modL/L2)−dimImτi′(cid:1)=
i=2
c
dim(L/L2∧L/L2)+Xdim(Li/Li+1⊗mod(L/Z(L))(ab)).
i=2
′ ′
By the fact that dimImγ −dimImτ =dimImγ in the proof of Proposition 2.5,
L 2 2
we have
c
dimL∧L+dimImγ2′ ≤dimL/L2∧L/L2+Xdim(Li/Li+1⊗mod(L/Z(L))(ab)),
i=2
as required. (cid:3)
Recall that a Lie algebra L is called stem provided that Z(L)⊆L2.
Theorem 2.8. Let L be a finite dimensional nilpotent stem Lie algebra of class 3.
Then, we have
′ ′
dimL∧L+dimImγ +dimImγ
2 3
≤dim(L/L2∧L/L2)+dim(L2/L3⊗ L(ab))+dim(L3⊗ L(ab)).
mod mod
Proof. Proposition 2.5 implies dimImγ′ ≤ dimK ≤dimkerα . But dimkerα ≤
3 1 3 3
dimL3∧L and Lemma 1.8 implies that dimL3∧L=dimL3⊗ L(ab). Now the
mod
result directly obtained from Theorem 2.7. (cid:3)
Looking the proof of [13, Theorem 3.1], we have
Proposition 2.9. Let L be a nilpotent Lie algebra of dimension n such that
dimL2 =1. Then L∼=H(m)⊕A(n−2m−1).
1
(i) If m=1, then dimM(L)= (n−1)(n−2)+1.
2
1
(ii) If m≥2, then dimM(L)= (n−1)(n−2)−1.
2
Lemma 2.10. Let L be a finite dimensional nilpotent Lie algebra and dimL/L2 =
1. Then dimL=1.
Proof. Since dimL/L2 = 1 and the Frattini subalgebra L is equal to L2, we have
dimL=1, as required. (cid:3)
The following theorem improves Theorem 1.5.
Theorem 2.11. Let L be an n-dimensional non-abelian nilpotent Lie algebra of
class c with the derived subalgebra of dimension m and t=dim(Z(L)/Z(L)∩L2).
Then
1
dimM(L)≤ (n+m−2)(n−m−1)−t(m−1)+1.
2
In particular, when m=1 the bound is attained if and only if L=H(1)⊕A(n−3).
SCHUR MULTIPLIER OF NILPOTENT LIE ALGEBRAS 9
Proof. If m = 1, then the result holds by Proposition 2.9. Thus we may assume
thatm>1.ByinvokingLemmas1.2and1.7,dimL/L2∧L/L2 =dimM(L/L2)=
1
(n−m)(n−m−1). Therefore Theorem 2.7 implies
2
c
′
dimL∧L+dimImγ2 ≤dimL∧L+Xdimkerαi
i=2
c
=dim(L/L2∧L/L2)+Xdim(Li/Li+1⊗mod(L/Z(L))(ab))
i=2
c
1
= (n−m)(n−m−1)+Xdim(Li/Li+1⊗mod(L/Z(L))(ab)).
2
i=2
On the other hand,
c c
Xdim(Li/Li+1⊗mod(L/Z(L))(ab))=dim(cid:0)(MLi/Li+1)⊗mod(L/Z(L))(ab)(cid:1)
i=2 i=2
=dimL2dim(L/Z(L))(ab) =m(n−m−t).
Thus
1
′
dimL∧L+dimImγ ≤ (n−m)(n−m−1)+m(n−m−t).
2 2
′
Now we are going to obtain a lower bound for the dimension of Imγ .
2
Ifdim(L/Z(L))(ab) =1,byusingLemma2.10,wehavedim(L/Z(L))=1,andso
Lisabelianthatisimpossible. Hencedim(L/Z(L))(ab) ≥2.Setd=dim(L/(Z(L)+
L2)). We claim that d−2 ≤ dimImγ′ for d ≥ 2. If d = dim(L/(Z(L)+L2)) =
2
dim(L/Z(L))(ab) = 2, then Imγ′ = 0, by Proposition 2.3 (ii). Thus d−2 = 0 =
2
dimImγ′. Suppose that d=dim(L/(Z(L)+L2))≥3. We can choose a basis
2
{x +Z(L)+L2,...,x +Z(L)+L2}
1 d
forL/(Z(L)+L2) suchthat [x ,x ]+L3 is non-trivialin L2/L3. We claimthat all
1 2
elements of the set
B ={γ′ (x +(Z(L)+L2))⊗(x +(Z(L)+L2))⊗(x +(Z(L)+L2)) |3≤i≤d}
2(cid:0) 1 2 i (cid:1)
are linearly independent. Since
d
L2/L3⊗mod(L/Z(L))(ab) ∼=M(cid:0)L2/L3⊗modhxi+(Z(L)+L2)i(cid:1)
i=1
and for i≥3
γ′ x +(Z(L)+L2)⊗x +(Z(L)+L2)⊗x +(Z(L)+L2)
2(cid:0) 1 2 i (cid:1)
=[x ,x ]+L3⊗x +(L2+Z(L))+[x ,x ]+L3⊗x +(L2+Z(L))
1 2 i i 1 2
+[x ,x ]+L3⊗x +(L2+Z(L)),
2 i 1
we have
γ′ x +(Z(L)+L2)⊗x +(Z(L)+L2)⊗x +(Z(L)+L2) ∈
2(cid:0) 1 2 i (cid:1)
h[x ,x ]+L3⊗x +(L2+Z(L))i⊕h[x ,x ]+L3⊗x +(L2+Z(L))i⊕
1 2 i i 1 2
h[x ,x ]+L3⊗x +(L2+Z(L))i.
2 i 1
10 P.NIROOMANDANDF.JOHARI
Since[x ,x ]∈/ L3 andx ∈/ L2+Z(L),[x ,x ]+L3⊗x +(L2+Z(L))isnon-trivial
1 2 i 1 2 i
element in L2/L3⊗ hx +(Z(L)+L2)i, γ′ x +(Z(L)+L2)⊗x +(Z(L)+
mod i 2(cid:0) 1 2
L2)⊗x +(Z(L)+L2) 6=0. Hence all elements of
i (cid:1)
γ′(B)={[x ,x ]+L3⊗x +(L2+Z(L))+[x ,x ]+L3⊗x +(L2+Z(L))
2 1 2 i i 1 2
+[x ,x ]+L3⊗x +(L2+Z(L))|3≤i≤d}
2 i 1
are linearly independent and so d−2≤dimImγ′. By Lemma 1.7, we have
2
1
dimM(L)+dimL2+d−2≤ (n−m)(n−m−1)+m(n−m−t).
2
Since d=n−m−t, we have
1
dimM(L)+m+(n−m−t−2)≤ (n−m)(n−m−1)+m(n−m−t)=
2
1
(n−m)(n−m−1)+(m−1)(n−m−1)−(m−1)(n−m−1)+m(n−m−t)=
2
1
(n+m−2)(n−m−1)−m(n−m)+(n−m−1)+m+m(n−m−t)=
2
1
(n+m−2)(n−m−1)+(n−m−1)+m−mt=
2
1
(n+m−2)(n−m−1)+n−1−mt.
2
Thus
1
dimM(L)≤ (n+m−2)(n−m−1)+n−1−mt−m−(n−m−t−2)
2
1
(n+m−2)(n−m−1)−mt−1+t+2=
2
1
(n+m−2)(n−m−1)−t(m−1)+1,
2
as claimed. If m=1, then the converse holds by Theorem 1.5. (cid:3)
According to the notation and terminology of the classification of nilpotent Lie
algebras of dimension at most 6 in [4], let
L =hx ,...,x |[x ,x ]=x ,[x ,x ]=x i
5,8 1 5 1 2 4 1 3 5
and
L =hx ,...,x |[x ,x ]=x ,[x ,x ]=x ,[x ,x ]=x i.
6,26 1 6 1 2 4 1 3 5 2 3 6
Note that from the notation of [8, 16], L is also denoted by L(4,5,2,4).
5,8
The following example shows that the upper bound of Theorem 2.11 can be
obtained.
Example 2.12. Let L=L ⊕A(1). By Lemmas 1.1 and 2.13, we have
5,8
dimM(L)=dimM(L )+dimL /(L )2 =6+3=9.
5,8 5,8 5,8
Since n=6,m=2 and t=1,
1 1
dimM(L)= (n+m−2)(n−m−1)−t(m−1)+1= 6(6−3)−1+1=9.
2 2
The following lemma gives the structure of all n-dimensionalnilpotent Lie alge-
bra with the derived subalgebra of dimension two when the bound of Theorem 1.5
is attained.