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The Leibniz algebras whose subalgebras are ideals

  • Leonid A. Kurdachenko EMAIL logo , Nikolai N. Semko and Igor Ya. Subbotin
Published/Copyright: February 22, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

In this paper we obtain the description of the Leibniz algebras whose subalgebras are ideals.

Keywords: Leibniz algebra; Lie algebra; Cyclic subalgebra; Left center; Right center; Center of a Leibniz algebra; Nilpotent subalgebras; Abelian subalgebras; Extraspecial subalgebras; Bilinear form
MSC 2010: 17A32; 17A60; 17B30

Introduction

An algebra L over a field F is said to be a Leibniz algebra (more precisely a let Leibniz algebra) if it satisfies the Leibniz identity

[[a,b],c]=[a,[b,c]]−[b,[a,c]]  for  all  a,  b,  c∈L. (LI)

Leibniz algebras are generalizations of Lie algebras. Indeed, a Leibniz algebra L is a Lie algebra if and only if [a, a] = 0 for every element a ∈ L. For this reason, we may consider Leibniz algebras as "non-anticommutative" analogs of Lie algebras.

Leibniz algebra appeared first in the papers of A.M. Bloh [1–3], in which he called them the D-algebras. However, that time, the real study of these algebras has not begun. Only two decades later, a real interest in these algebras raised. The impetus for this was the work of J.L. Loday [11], who introduced the term Leibniz algebras since it was Leibniz who discovered and proved the "Leibniz rule" for differentiation of functions. The Leibniz algebras appeared to be naturally related to several topics such as differential geometry, homological algebra, classical algebraic topology, algebraic K-theory, loop spaces, noncommutative geometry, and so on. They found some applications in physics (see, for example, [4, 7, 8]). Some papers concerning Leibniz algebras are devoted to the study of homological problems [5, 9, 12, 13]. The theory of Leibniz algebras has been developing quite intensively but very uneven. On one hand, some deep structural theorems were obtained as analogues of the corresponding results about Lie algebras. On the other hand, the study of Leibniz algebras does not look consistent and systematic. For any algebraic structure there are some natural questions that were not previously considered for Leibniz algebras. Thus, for example, a natural question on the structure of the cyclic subalgebras in Leibniz algebras has just recently been considered [6]. Another natural question is a question about the structure of Leibniz algebras, whose subalgebras are ideals. We note that it is not hard to prove that a Lie algebra, whose subalgebras are ideals, is abelian. But it is not true for Leibniz algebras. The following simple example justifies it. Let L be a vector space of dimension 2 over a field F,{a, b} be a basis of L. Define the operation [,] by the following rule: [a, a] = b, [b, b] = [b, a] = [a, b] = 0. A direct check shows that L becomes a Leibniz algebra. If λa + μ b is an arbitrary element of L and λ ≠ 0, then [λ a + μ b, λ a + μ b] = λ2b. By λ2 ≠ 0 we obtain that the subalgebra generated by λ a + μ b includes Fb. Since L/Fb is abelian, 〈λ a + μ b〉 is an ideal. Hence every cyclic subalgebra of L is an ideal. It follows that every subalgebra of L is an ideal. As we shall see later, any non-abelian Leibniz algebra, whose subalgebras are ideals, is built of these algebras as the bricks. Now consider more details.

If L is a Leibniz algebra and M is a subset of L, then by 〈M〉 we denote the subalgebra generated by M.

As usual, a Leibniz algebra L is called abelian if [x, y] = 0 for all elements x, y ∈ L. In an abelian Leibniz algebra every subspace is a subalgebra and an ideal.

A Leibniz algebra L is called an extraspecial algebra, if it satisfies the following condition:

“(G) is non-trivial and has dimension 1,

L / ζ(L) is abelian.

As it turned out, not in any extraspecial Leibniz algebra every subalgebra is an ideal. Now, we give an example of an extraspecial Leibniz algebra showing it. Moreover, the presence of subalgebras, which are not ideals, depend on the choice of the field.

Let F be a field, put L = Fa ⊕ Fb ⊕ Fc. Define on L an operation [, ] by the following rule: c = [a, a] = [b, b] = [a, b], [c, c] = [c, a] = [c, b] = [a, c] = [b, c] = [b, a] = 0. From this definition we can see that [L, L] ≤ Fc, c ∈ ζ (L), 〈c〉 = Fc. Since [x, y], [y, z], [x, z] ∈ ζ(L), the equality

[[x,y],  z]=[x,[y,  z]]−[y,[x,z]]

occurs automatically. Thus, L is a Leibniz algebra. Let x be an arbitrary element of L. Then x = λ a + μ b + vc for some λ, μ, v ∈ F. We have

                                            [x,x]=[λa+μb+vc;λa+μb+vc]=λ2[a,a]+λμ[a,b]+λv[a,c]+λμ[b,a]+μ2[b,b]+μv[b,c]+λv[c,a]+μv[c,b]+                                    v2[c,c]=λ2c+λμc+μ2c=(λ2+λμ+μ2)c.

Let F = 𝔽2. If (λ, μ) ≠ (0, 0), then λ2 + λμ + μ2 = 1, that is [x, x] = c whenever x ∉ Fc. It follows that ζ(L) = Fc and 〈x〉 = Fx ⊕ Fc. It follows that 〈x〉 is an ideal of L. Since Fc is an ideal, we obtain that every subalgebra of L is an ideal.

Let F = 𝔽5. Suppose that λ2 + λμ + μ2 = 0. It follows that (λ + 1/2μ)2 = μ2(1/4 - 1). In field 𝔽5 a solution of an equation 4x = 1 is 4, so that 1/4 - 1 = 3. But the equation x2 = 3 has no solutions in 𝔽5. This shows that the equality λ2 + λμ + μ2 = 0 is possible only if λ = μ = 0. Thus if (λ, μ) ≠ (0, 0), then [x, x] ≠ 0 and [x, x] ∈ Fc. Hence in this case, every subalgebra of L is an ideal.

If F = ℚ, then using the similar arguments we obtain again that every subalgebra of L is an ideal and the center of L is Fc.

Now consider the case when F = 𝔽3. For element x = a + b we have [a + b, a + b] = 3c = 0. It follows that 〈x〉 = Fx. But [x, a] = [a + b, a] = c ∉ Fx, which shows that a cyclic subalgebra 〈x〉 is not an ideal.

The following theorem describes the structure of Leibniz algebras, in which every subalgebra is an ideal. But first, it will be useful to recall the following definitions.

A Leibniz algebra L has one specific ideal. Denote by Leib(L) the subspace generated by the elements [a, a], a ∈ L. It is not hard to prove that Leib(L) is an ideal of L. From its choice it follows that L/ Leib(L) is a Lie algebra. And conversely, if H is an ideal of L such that L/H is a Lie algebra, then Leib(L) ≤ H.

The ideal Leib(L) is called the Leibniz kernel of algebra L.

We also note the following important property of the Leibniz kernel:

[[a,a],x]=0  for  arbitrary  elements  a,x∈  L.

The left (respectively right) center ζleft(L) (respectively ζ right(L)) of a Leibniz algebra L is defined by the rule

ζleft(L)={x  ∈  L|[x,y]=0  for  each  element  y  ∈  L}.

(respectively

ζright(L)={x  ∈  L|[y,x]=0  for  each  element  y  ∈  L}).

It is not hard to prove that the left center of L is an ideal. Moreover, Leib(L) ≤ ζleft(L), so that L/ζleft(L) is a Lie algebra. In general, the left and the right centers are different, moreover, the left center is an ideal, but it is not true for the right center. One can find the corresponding counterexample in [10].

The center ζ(L) of L is defined by the rule

ζ(L)={x∈L|[x,y]=0=[y,x]  for  each  element  y∈L}.

The center is an ideal of L. In particular, we can consider the factor-algebra L/ζ(L).

Theorem A

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. If L is non-abelian, then L = E ⊕ Z where Z ≤ ζ(L) and E is an extraspecial subalgebra such that [a, a] ≠ 0 for every element a ∉ ζ(E).

In Section 2 we establish a relation of extraspecial Leibniz algebras L such that [a, a] ≠ 0 for every non-zero element a and the bilinear forms with no non-zero isotropic elements, which reduces the description of extraspecial Leibniz algebras to the description of these forms, and got some results on the structure of such forms. Here we present the main result of Section 2.

Let V be a vector space over a field F having countable dimension and Φ be a bilinear form on V. A basis {aj|j ∈ ℕ} is called left orthogonal, if Φ(aj, ak) = 0 whenever j > k.

Theorem B

Let V be a vector space of countable dimension over a field F and Φ be a bilinear form on V. If Φ(a, a) ≠ 0 for each element 0 ≠ a ∈ V, then V has a let orthogonal basis.

It should be noted that for the spaces of uncountable dimension a bilinear forms structure may be rather complicated. Even in the case of alternating forms there are quite exotic examples (see, for example, [14], Chapter 3).

1 On the structure of Leibniz algebras whose subalgebras are ideals

Let L be a Leibniz algebra over a field F. If B is a subspace of L and if [L, B], [B, L] ≤ B, then we will say that B is L-invariant.

Lemma 1.1

Let L be a Leibniz algebra over a field F, a ∈ L. Suppose that every subalgebra of L is an ideal of L. If A is an abelian subalgebra of L, then every subspace of A is L-invariant. Moreover for every element x ∈ L there are elements ax, βx ∈ F, which depend only of x, such that [a, x] = αxa, [x, a] = βxa for any element a ∈ A.

Proof

Let B be a subspace of A, b ∈ B. Since A is abelian, [b, b] = 0, so that subspace Fb is a subalgebra. Then it is an ideal. It follows that [x, b], [b, x] ∈ Fb ≤ B, so that B is L-invariant.

Let a, c ∈ A, x ∈ L, then by above proved [a, x] = αa, [c, x] = γ c for some elements α, γ ∈ F. We have [a - c, x] = [a, x] - [c, x] = αa - γc. On the other hand, a - c ∈ A, so that F (a - c) must be L-invariant, i.e. [a - c, x] = η(a - c) = η a - η c. It follows that α a - γ c = η a - η c, and hence α = η = γ. For [x, a] the arguments are similar. □

Lemma 1.2

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. If a is an element of L such that [a, a] ≠ 0, then 〈a〉 = Fa ⊕ F[a, a] and [a, [a, a]] = [[a, a], a] = 0.

Proof

Put b = [a, a]. If b ∈ Fa, then [a, a] = γa for some 0 ≠ γ ∈ F. In this case we have 0 = [[a, a], a] = [γa, a] = γ[a, a], which implies that [a, a] = 0, and we obtain a contradiction. This contradiction shows that the elements a, b are linearly independent. We have b = [a, a] ∈ Leib(L) ≤ ζleft(L). Since the left center is an abelian ideal, Lemma 1.2 shows that [a, b] = αb for some element α ∈ F. Suppose that α ∈ 0 and consider an element d = a - α−1b. By such a choice d ∉ Fb, so that Fd ∩ Fb = 〈0〉. We have

[d,d]=[a−α−1b,a−α−1b]=[a,a]−[a,α−1b]=b−α−1[a,b]=b−b=0.

It follows that 〈d〉 = Fd. On the other hand, [d, a] = [a, a] = b ∉ 〈d〉. It shows that a subalgebra 〈d〉 is not an ideal, and we obtain a contradiction. This contradiction shows that α = 0, that is [a, b] = 0. □

Lemma 1.3

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. Then factor-algebra L/Leib(L) is abelian.

Proof

Indeed, L/Leib(L) is a Lie algebra, and, moreover, it is a Lie algebra, whose subalgebras are ideals. So it is abelian. □

Corollary 1.4

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. Then a factor-algebra L/ζleft(L) is abelian.

Lemma 1.5

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals, C = ζleft(L). If A is a maximal abelian ideal including C, then A = C.

Proof

Let a ∈ A and x be an arbitrary element of L. If c is a non-zero element of C, then [c, x] = 0 = 0c. Lemma 1.1 shows that [a, x] = 0a = 0. It follows that a ∈ C, that is A = C. □

Let L be a Leibniz algebra over a field F, M be a non-empty subset of L and H be a subalgebra of L. Put

AnnHleft(M)={a∈H|[a, M]=0},AnnHright(M)={a∈H|[M, a]=0}.

The subset AnnHleft(M) is called the left annihilator or let centralizer of M in subalgebra H; the subset AnnHright(M) is called the right annihilator or right centralizer of M in subalgebra H. The intersection

AnnH(M)=AnnHleft(M)∩AnnHright(M)={a∈H|[a, M]=〈 0 〉=[M,a]}

is called annihilator or centralizer of M in subalgebra H.

It is not hard to see that all these subsets are subalgebras of L. Moreover, if M is a left ideal of L, then AnnLleft(M) is an ideal of L. Indeed, let x be an arbitrary element of L, a ∈ AnnHleft(M), b ∈ M. Then

[[a,x],b]=[a,[x,b]]−[x,[a,b]]=0−[x,0]=0  and   [[x,a],b]=[x,[a,b]]−[a,[x,b]]=[x,0]−0=0.

If M is an ideal of L, then AnnL(M) is an ideal of L. Indeed, let x be an arbitrary element of L, a ∈ AnnH(M), b ∈ M. Using the above arguments, we obtain that [[a, x], b] = [[x, a], b] = 0. Further

[b,[a,x]]=[[b,a],x]]+[a,[b,x]]]=[0,x]+0=0  and     [b,[x,a]]=[[b,x],a]+[x,[b,a]]]=0+[x,0]=0.

Let L be a Lie algebra. Define the lower central series of L

L=γ1(L)γ2(L)≥…≥γα(L)≥γα+1(L)≥…γδ(L)

by the following rule: γ1(L) = L, γ2(L) = [L, L], and recursively γα+1(L) = [L, γα(L)] for all ordinals α, and γλ(L) = ∩μ<λ γμ(L) for the limit ordinals λ. The last term γδ(L) is called the lower hypocenter of L. We have γδ(L) = [L, γδ(L)].

If α = k is a positive integer, then γk(L) = [L, [L, [L, …, L] … L] is aleft normed product of k copies of L.

Consider the factor γk(L)/γk+1(L), k ∈ ℕ. By definition, [L, γk(L)] = γk+1(L), and also [γk(L), L] = [γk(L), γ1(L)] ≤ γk+1(L) (see, for example, [10], Proposition 2.2).

As usual, we say that a Leibniz algebra L is nilpotent, if there exists a positive integer k such that γk(L) = 〈0〉. More precisely, L is said to be nilpotent of nilpotency class c if γc+1(L) = 〈0〉, but γc(L) ≠ 〈0〉. We denote by ncl(L) the nilpotency class of L.

Lemma 1.6

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals, C = ζleft(L). Then A = AnnLleft(C) = AnnL(C) is a nilpotent ideal, ncl(A) ≤ 2, codimF(A) ≤ 1.

Proof

As we noted above, C is an ideal of L, so that A is an ideal of L. Let a be an arbitrary element of A, then [a, c] = 0 for each element c ∈ C. On the other hand, [c, a] = 0 by the choice of c. It follows that C ≤ ζ(A). By Corollary 1.4, a factor-algebra L/C is abelian, so that [A, A] ≤ C and A is nilpotent of nilpotency class at most 2.

Let x ∈ L. Lemma 1.1 shows that there exists an element αx ∈ F such that [x, c] = αx c for every element c ∈ C. Consider now the mapping f : L → F, defined by the rule f(x) = αx for each element x ∈ L. If y ∈ L, then

αx+yc=[x+y,c]=[x,c]+[y,c]=αxc+αyc=(αx+αy)c  for  every  c∈C.

This shows that f(x + y) = αx+y = αx + αy = f(x) + f(y). Furthermore, if β ∈ F, then

αβxc=[βx,c]=β[x,c]=β(αxc)=(βαx)c  for  every  c∈C,

which shows that f(βx) = βαx = βf(x). It follows that mapping f is linear. Finally, Ker(f) = {x ∈ L|αx = 0}. In other words, x ∈ Ker(f) is equivalent to [x, c] = 0 for each c ∈ C. This proves that Ker(f) = AnnLleft(C) = A. Hence if L ≠ A, then dimF(L/A) = 1. □

Define now the upper central series

0=ζ0(L)≤ζ1(L)≤ζ2(L)≤…≤ζα(L)≤ζα+1(L)≤…ζγ(L)=ζ∞(L)

of a Leibniz algebra L by the following rule: ζ1(L) = ζ(L) is the center of L, and recursively ζα+1(L)/ζα(L)= ζ(L/ζα(L)) for all ordinals α, and ζλ(L) = ∪μ<λ ζμ(L) for the limit ordinals λ. By the definition each term of this series is an ideal of L. The last term ζ∞(L) of this series is called the upper hypercenter of L. Denote by zl(L) the length of upper central series of L.

If L = ζ∞(L), then L is said to be hypercentral Leibniz algebra.

Lemma 1.7

Let L be a Leibniz algebra over a field F and A be a non-zero ideal of L. If A ≤ ζ∞(L), then A ∩ ζ(L) ≠ 〈0〉.

Proof

Suppose that A ∩ ζ(L) = 〈0〉. Since A ≤ ζ∞(L), there is an ordinal α such that A ∩ ζα(L) ≠ 〈0〉. We choose the least ordinal λ with this property. Clearly λ is a non-limit ordinal, so that C = A ∩ ζλ(L) ≠ 〈0〉, but 〈0〉 = A ∩ ζλ−1(L). The inclusion C ≤ ζλ(L) implies that [L, C], [C, L] ≤ ζλ−1(L). On the other hand, A is an ideal of L, and therefore [L, C], [C, L] ≤ A. We obtain the inclusion [L, C], [C, L] ≤ A ∩ ζλ−1(L) = 〈0〉. It follows that 〈0〉 ≠ C ≤ ζ(L) and we obtain a contradiction. This contradiction proves a result. □

Corollary 1.8

Let L be a hypercentral Leibniz algebra over a field F and A be a non-zero ideal of L. Then A ∩ ζ(L) ≠ 〈0〉.

Lemma 1.9

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. Suppose that L is nilpotent and ncl(L) = 2. If a is an element of L such that [a, a] = 0, then a ∈ ζ(L).

Proof

The equality [a, a] = 0 implies that 〈a〉 = Fa, so that dimF(〈a〉) = 1. Since 〈a〉 is an ideal, Corollary 1.8 implies that 〈0〉 ≠ 〈a〉 ∩ ζ(L). It follows that 〈a〉 = 〈a〉 ∩ ζ(L), i.e. a ∈ ζ(L). □

Lemma 1.10

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. Suppose that L is nilpotent and ncl(L) = 2. If a ∉ ζ(L), then codimF(AnnLleft(〈a〉)) = 1 = codimF(AnnLright(〈a〉)).

Proof

Put c = [a, a]. Since a ∉ ζ(L), Lemma 1.9 shows that c ≠ 0. Then by Lemma 1.2 we obtain that A = 〈a〉 = Fa ⊕ Fc. Furthermore, c ∈ ζ(L), so that [a, c] = [c, a] = 0.

Let x ∈ L. Since L/ζ(L) is abelian, [x, a] ∈ ζ(L). On the other hand, A is an ideal, so that [x, a] ∈ A, that is [x, a] ∈ A ∩ ζ(L) = Fc. It follows that [x, a] = αxc for some element αx ∈ F. Consider the mapping f : L → F defined by the rule f(x) = αx for each element x ∈ L. If y ∈ L, then

αx+yc=[x+y,a]=[x,a]+[y,a]=αxc+αyc=(αx+αy)c.

This shows that f(x + y) = αx + y = αx + αy = f(x)+f(y). Furthermore, if β ∈ F, then

αβxc=[βx,a]=β[x,a]=β(αxc)=(βαx)c,

which shows that f(β x) = βαx = β f(x). It follows that mapping fis linear. Finally, Ker(f) = {x ∈ L|αx = 0}. In other words, x ∈ Ker(f) is equivalent to [x, a] = 0. This proves that Ker(f) = AnnLleft(a). It is not hard to see that AnnLleft(a) = AnnLleft(A). The choice of a shows that AnnLleft(a) ≠ L. It follows that dimF(L/AnnLleft(A)) = 1. □

Lemma 1.11

Let L be an extraspecial Leibniz algebra overafield F. Then every subalgebra of L is an ideal if and only if [a, a] ≠ 0 for every element a ≠ ζ(L).

Proof

Suppose that every subalgebra of L is an ideal and a is a non-zero element of L such that [a, a] = 0. Then Lemma 1.9 shows that a ∈ ζ(L).

Conversely, assume that L is an extraspecial algebra such that [a, a] ≠ 0 for every element a ∉ ζ(L). Let x be an arbitrary element of L. If x ∈ ζ(L), then 〈x〉 = Fx, and a cyclic subalgebra 〈x〉 is an ideal. If x ∉ ζ(L), then y = [x, x] ∈ ζ(L). It follows that [x, [x, x]] = [[x, x], x] = 0, and 〈x〉 = Fx ⊕ F[x, x]. Since [x, x] ≠ 0, F[x, x] = ζ(L). Hence every cyclic subalgebra includes the center of L. Since L/ζ(L) is abelian, every cyclic subalgebra of L is an ideal. It follows that every subalgebra of L is an ideal. □

Lemma 1.12

Let L be a Leibniz algebra over a field F and suppose that L is nilpotent and ncl(L) = 2. If a, b ∉ ζ(L) and 〈a〉 ∩ 〈b〉 = 〈0〉, then L includes a subalgebra, which is not an ideal.

Proof

Suppose the contrary. Let every subalgebra of L be an ideal. If we assume that [a, a] = 0, then as in Lemma 1.9 we obtain that a ∈ ζ(L) . But this contradicts the choice of a. This contradiction proves that [a, a] = c ≠ 0. Since a is nilpotent of class 2, c ∈ ζ(L), which implies that [a, c] = [c, a] = 0. In a similar way, [b, b] = d ≠ 0 and [b, d] = [d, b] = 0. The subalgebras 〈a〉 and 〈b〉 are ideals, and therefore [a, b], [b, a] ∈ 〈a〉 ∩ 〈b〉 = 〈0〉. It follows that [a, b] = [b, a] = 0. Let x = a + b, then

[x,x]=[a+b,a+b]=[a,a]+[b,b]=c+d.

As above 〈x〉 = Fx ⊕ F[x, x] and 〈x〉 ∩ ζ(L) = F(c + d). On the other hand, [x, a] = c, [x, b] = d. It follows that an ideal, generated by x, must contain Fc ⊕ Fd, and we obtain a contradiction. This contradiction proves a result. □

Proposition 1.13

Let L be a Leibniz algebra over a field F, whose subalgebras are ideals. Suppose that L is nilpotent and ncl(L) = 2. Then L = E ⊕ Z where E is an extraspecial subalgebra, Z ≤ ζ(L).

Proof

If we suppose that [a, a] = 0 for every element a ∈ L, then Lemma 1.9 shows that L = ζ(L), and we obtain a contradiction. Hence there exists an element a such that [a, a] = c ≠ 0. Lemma 1.2 shows that 〈a〉 = Fa ⊕ Fc. Let B be a basis of L containing a, c. Then B = {aα|α < γ} for some ordinal α. We can put a1 = a, a2 = c. Consider the subalgebra 〈a1, a2, a3〉. If [a3, a3] = 0, then Lemma 1.9 shows that a3 ∈ ζ(L). In this case, 〈a1, a2, a3〉 = 〈a1〉 ⊕ Fa3, where subalgebra 〈a1〉 is extraspecial, Fa3 ≤ ζ(L).

Assume now that [a3, a3] = b ≠ 0. Lemma 1.2 shows that 〈a3〉 = Fa3 ⊕ Fb. In this case, dimF(〈a1, a3〉)≤ dimF(〈a1〉) + 2. If dimF(〈a1, a3〉) = dimF(〈a1〉) + 2, then 〈a1〉 ∩ 〈a3〉 = < 0 >, and Lemma 1.12 implies that in this case, L includes a subalgebra, which is not an ideal. This contradiction proves that dimF(〈a1, a3〉) = dimF(〈a1〉) + 1. If 〈a1, a3〉 ∩ ζ(L) ≠ Fc, then 〈a1, a3〉 ∩ ζ(L) = Fc ⊕ Fd for some element d ∈ ζ(L). In this case, 〈a1, a3〉 = 〈a1〉 ⊕ Fd. If 〈a1, a3〉 ∩ ζ(L) = Fc, then 〈a1, a3〉 is extraspecial.

Suppose that we have already constructed the subalgebras Lβ for all β < α satisfying the following conditions: av∈ Lβ for all v ≤ β, Lβ = Eβ ⊕ Zβ where Eβ is an extraspecial subalgebra, ζ(Eβ) = Fc, Zβ ≤ ζ(L), Ev≤Eβ, Zv ≤ Zβ whenever v ≤ β. If α is a not limit ordinal, then α – 1 exists. Let η be aleast ordinal such that aη ∉ Lα – 1. Consider a subalgebra Lα = 〈Lα – 1, aη〉. If [aη, aη] = 0, then Lemma 1.9 shows that aη∈ ζ(L). In this case,

Lα=Lα−1⊕Faη=Eα−1⊕Zα−1⊕Faη=Eα⊕Zα

where Eα = Eα – 1, Zα = Zα – 1 ⊕ Faη.

If [aη, aη] = u ≠ 0. Lemma 1.2 shows that 〈aη〉 = Faη ⊕ Fu. In this case, dimF(Lα) ≤ dimF(Lα – 1) + 2. If dimF(Lα) = dimF(Lα – 1) + 2, then Lα – 1 ∩ 〈aη〉 = 〈0〉. In particular, 〈a1〉 ∩ 〈aη〉 = 〈0〉, and Lemma 1.12 implies that in this case L includes a subalgebra, which is not an ideal. This contradiction proves that dimF(Lα) = dimF(Lα – 1) + 1. If Lα ∩ ζ(L) ≠ Fc, then Lα ∩ ζ(L) = Fc ⊕ Fv for some element v ∈ ζ(L). In this case,

Lα=Lα−1⊕Fv=Eα−1⊕Zα−1⊕Fv=Eα⊕Zα

where Eα = Eα – 1, Zα = Zα – 1 ⊕ Fv.

If Lα ∩ ζ(L) = Fc, then Lα/Fc = Lα – 1/Fc ⊕〈aη〉/Fc. In this case, Eα = 〈Eα – 1, aη〉 is extraspecial and Lα = Eα ⊕ Zα where Zα = Zα – 1.

Assume that α is a limit ordinal. Put Eα = ∪β < α Eβ, Zα = ∪β < α Zβ, Lα = ∪β < α Lβ. Then clearly Lα = Eα ⊕ Zα, Eα is extraspecial, ζ(Eα) = Fc, Zα≤ζ(L).

For α = γ we proved the result. □

Proof of Theorem A. Let C be a left center of L. By Lemma 1.5 C is a maximal abelian ideal of L and Corollary 1.4 shows that L/C is abelian. Since [C, L] = 〈0〉, AnnLleft(C) = AnnL(C) . Then Lemma 1.6 shows that a is a nilpotent ideal, ncl(A) ≤ 2, and dimF(L/A) ≤ 1. Suppose that L = A. By Proposition 1.13, L = E ⊕ Z where E is an extraspecial subalgebra, Z ≤ ζ(L). Lemma 1.11 shows that [a, a] ≠ 0 for every element a ∈ ζ(E) = E ∩ C.

Suppose now that L ≠ A and choose an element d ∉ A. Put e = [d, d]. By Lemma 1.2 〈d〉 = Fd ⊕ Fe and [d, e] = [e, d] = 0. Since C is abelian, Lemma 1.1 shows that there are the elements λ, ρ ∈ F such that [d, c] = λ c, [c, d] = ρ c for each element c ∈ C. Since e ∈ Leib(L) ≤ ζleft(L), λ = ρ = 0. In other words, [d, c] = 0 for each element c ∈ C. This means that d ∈ AnnLleft(C) = A. But this contradicts the choice of d. This contradiction shows that L = A, which proves the result. □

2 Extraspecial Leibniz algebras and bilinear forms

We can connect a bilinear form to an extraspecial algebra in the following way. Let Z = ζ(L), V = L/Z and c be a fixed non-zero element of Z. Define the mapping Φ : V × V → F by the following rule: If x, y ∈ L, then [x, y] ∈ Z, so that [x, y] = αc for some element α ∈ F. Put Φ(x + Z, y + Z) = α. This mapping is definitely correct. Indeed, let x1, y1 be elements of L such that x1 + Z = x + Z, y1 + Z = y + Z. Then x1 = x + c1, y1 = y + c2 for some elements c1, c2 ∉ Z. Then

[x1,y1]=[x+c1,y+c2]=[x,y]+[x,c2]+[c1,y]+[c1,c2]=[x,y].

The mapping Φ is bilinear. In fact, let x, y, u ∉ L, [x, u] = λc, [y, u] = μ c. Then [x + y, u] = [x, u] + [y, u] = λc + μc = (λ + μ)c, so that

Φ(x+Z+y+Z,u+Z)=Φ(x+y+Z,u+Z)=(λ+μ)c=λc+μc=Φ(x+Z,u+Z)+Φ(y+Z,u+Z).

Similarly we can show that

Φ(x+Z,y+Z+u+Z)=Φ(x+Z,u+Z)+Φ(x+Z,y+Z) .

Let β ∈ F, then [β x, y] = β[x, y] = β(α c) = (βα)c. It follows that

Φ(β(x+Z),y+Z)=Φ(βx+Z,y+Z)=(βα)c=β(αc)=βΦ(x+Z,y+Z) 

Similarly we can show that

Φ(x+Z),β(y+Z)=βΦ(x+Z,y+Z) .

By the definition of an extraspecial algebra we obtain that a bilinear form Φ is non-degenerate. Moreover, Lemma 1.9 shows that Φ(x, x) ≠ 0 for every non-zero element x.

Conversely, let V be a vector space over a field F and Φ be a bilinear form on V such that Φ(x, x) ≠ 0 for every non-zero element x. Put L = V ⊕ F. Define the operation [,] on L by the following rule: if a, b ∈ V, α, β ∈ F, then [(a, α), (b, β)] = (0, Φ(a, b)). Put C = {(0, α)|α∈ F}. Then dimF(C) = 1. By this definition [L, L] = [L, C] = ]C, L] = [C, C] = 〈0〉. It follows that the constructed algebra is a Leibniz algebra. Furthermore, C ≤ ζ(L). Moreover, C = ζ(L). Indeed, let (z, γ) ∈ ζ(L) and suppose that z ≠ 0. Then [(z, γ), (a, α)] = [(a, α), (z, γ)] = (0, 0), in particular, [(z, γ), (z, γ)] = (0, 0). But [(z, γ), (z, γ)] = (Φ(z, z), 0). Since z ≠ 0, (Φ(z, z) ≠ 0, and we obtain a contradiction. This contradiction proves the equality C = ζ(L).

Let V be a vector space over afield F, U a subspace of V and Φ be a bilinear form on V. Put

⊥U={x∈V|Φ(x;u)=0  for all elements u∈U},U⊥={x∈V|Φ(u,x)=0 for  all  elements  u∈U}.

Clearly ⊥U and U⊥ are the subspaces of V, ⊥U is called left orthogonal complement of U in V, U⊥ is called right orthogonal complement of U in V.

Using usual methods of linear algebra the following proposition can be proved.

Proposition 2.1

Let V be a finite dimensional vector space over a field F, U a subspace of V and Φ be a non-degenerate bilinear form on V. If the restriction of Φ on U is non-degenerate, then dimF(⊥U) = dimF(U⊥) = dimF(V)–dimF(U).

Corollary 2.2

Let V be a finite dimensional vector space over a field F and Φ be a bilinear form on V. If Φ(a, a) ≠ 0 for each element 0 ≠ a ∈ V, then V has a let orthogonal basis.

Indeed, let U be an arbitrary non-zero subspace of V. If we suppose that a restriction of Φ on U is degenerate, then ⊥U ∩ U ≠ 〈0〉. Let 0 ≠ a ∈ ⊥ U ∩ U, then Φ(a, u) = 0 for all elements u ∈ U. In particular, Φ(a, a) = 0, and we obtain a contradiction. Hence the restriction of Φ on every non-zero subspace is non-degenerate, and we can apply Proposition 2.1.

We note that if V is a finite dimensional vector space and Φ is a bilinear form on V, such that V has a left orthogonal basis, then the matrix of a form Φ in this basis is triangular.

Proof of Theorem B.

Theorem B

Let V be a vector space over a field F, having countable dimension, and Φ be a bilinear form on V. If ϕ(a, a) ≠ 0 for each element 0 ≠ a ∈ V, then V has a let orthogonal basis.

Choose in space V an arbitrary basis aj|j ∈ ℕ}. Put Vj = Fa1 + … + Faj, j ∈ N. Since dimF(V2) = 2 is finite, Corollary 2.2 shows that V2 has a basis {v1, v2} such that [v2, v1] = 0. Since dimF(V3) is finite, Proposition 2.1 implies that dimF(V3 ∩ ⊥V2) = dimF(VdimF(V2) = 1. By the above remarks, the restriction of Φ on every non-zero subspace is non-degenerate, which implies that (V3 ∩ ⊥V2) ∩ V2 = 〈0〉. Let 0 ≠ v3∈ V3 ∩ ⊥V2, then {v1, v2, v3} is a basis of V3 such that [v2, v1] = [v3, v1] = [v3, v2] = 0. Using similar arguments, ordinary induction and the fact that V = ∩n∈ ℕ Vn, we proved the above statement. □

Corollary 2.3

Let L be an extraspecial Leibniz algebra over a field F, having countable dimension. If [a, a] ≠ 0 for every element a ∉ ζ(L), then L has a basis {en|n ∈ ℕ} such that [e1, en] = [en, e1] = 0, [ej, en] ∈ Fe1 for all j, n ∈ ℕ, [ej, en] = 0 whenever j > n.

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Received: 2016-10-17
Accepted: 2017-1-17
Published Online: 2017-2-22

© 2017 Kurdachenko et al.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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https://doi.org/10.1515/math-2017-0010
Keywords for this article
Leibniz algebra; Lie algebra; Cyclic subalgebra; Left center; Right center; Center of a Leibniz algebra; Nilpotent subalgebras; Abelian subalgebras; Extraspecial subalgebras; Bilinear form
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