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arXiv:1106.0393v1 [math.RA] 2 Jun 2011

A kind of infinite-dimensional Novikov

algebras and its realization∗

Liangyun Chen, Yao Ma, Haijun Yu

Department of Mathematics, Northeast Normal University,

Changchun 130024, CHINA

Abstract

In this paper, we construct a kind of infinite-dimensional Novikov algebras and

give its realization by hyperbolic sine functions and hyperbolic cosine functions.

Key words: Novikov algebras, left symmetric algebras, adjoining Lie algebras.

AMS Subject Classification: 17A30.

§1 Introduction

The Hamilton operator is an important operator of the calculus of variations. When

I. M. Gel’fand and I. Ya. Dorfman [5-7] studied the following operator:

Hij=

?

k

cijku(1)

k+ dijku(0)

k

d

dx,

cijk∈ C,dijk= cijk+ cjik,

they gave the definition of Novikov algebras. Concretely, let cijkbe the structural

coefficients, a product of L = L(e0,e1,···) be ◦ such that

ei◦ ej=

?

cijkek.

Then the product is Hamilton operator if and only if ◦ satisfies

(a ◦ b) ◦ c = (a ◦ c) ◦ b

∗Supported by NNSF of China (No.10871057),

province (No.201115006), Scientific Research Foundation for Returned Scholars Ministry of

Education of China and the Fundamental Research Funds for the Central Universities.

Corresponding author(L. Chen): chenly640@nenu.edu.cn

Natural Science Foundation of Jilin

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(a ◦ b) ◦ c + c ◦ (a ◦ b) = (c ◦ b) ◦ a + a ◦ (c ◦ b).

In 1987, E. I. Zel’manov[15] began to study Novikov algebras and proved that

the dimension of finite simple Novikov algebras over a field of characteristic zero is

one. In algebras, what are paid attention to by mathematician are classifications

and structures, but so far we haven’t got the systematic theory for general Novikov

algebras. In 1992, J. M. Osborn [9-10] had finished the classification of infinite simple

Novikov algebras with nilpotent elements over a field of characteristic zero and finite

simple Novikov algebras with nilpotent elements over a field of characteristic p > 0.

In 1995, X. P. Xu [11-14] developed his theory and got the classification of simple

Novikov algebras over an algebraically closed field of characteristic zero. C. M. Bai

and D. J. Meng [1-3] has serial work on low dimensional Novikov algebras, such as

the structure and classification. We construct two kinds of Novikov algebras [4].

Recently, people obtain some properties in Novikov superalgebras [8, 16].

In this paper, we construct a new infinite-dimensional Novikov algebras and give

its realization by hyperbolic sine functions and hyperbolic cosine functions.

Definition 1.1Let (A,◦) be an algebra over F such that:

(a,b,c) = (b,a,c), (1.1)

(a ◦ b) ◦ c = (a ◦ c) ◦ b,∀a,b,c ∈ A, (1.2)

then A is called a Novikov algebra over F.

Remark 1.2(1) Condition (1.1) is usually written by

a ◦ (b ◦ c) − (a ◦ b) ◦ c = b ◦ (a ◦ c) − (b ◦ a) ◦ c. (1.3)

(2) An algebra A is called a left symmetric algebra if it only satisfies (1.1). It is

clear that left symmetric algebras contain Novikov algebras.

Remark 1.3 (1) If (A,◦) is a left symmetric algebra satisfying

[a,b] = a ◦ b − b ◦ a,∀a,b ∈ A, (1.4)

then (A,[,]) is a Lie algebra. Usually, it is called an adjoining Lie algebra.

(2) Let (A,·) be a commutative algebra, then (A,d0,◦) is a Novikov algebra if

d0is a derivation of A with a bilinear operator ◦ such that

a ◦ b = a · d0(b),∀a,b ∈ A. (1.5)

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§2 Main results

Lemma 2.1 Let { b0, a1, b1, a2, b2, ··· an, bn, ··· } be a basis of the linear space

A over a field F of characteristic p ?= 2 satisfying

aman=1

2(bm+n− bm−n),

bmbn=1

2(bm+n+ bm−n),

ambn= bnam=1

2(am+n+ am−n),

(2.1)

where b−m= bm, a−m= −am. Then A is a commutative and associative algebra.

Proof. It is clear that A is a commutative algebra over F.

(ak,an,am) = ak(anam) − (akan)am

1

2ak(bm+n− bn−m) −1

1

4(ak+m+n+ ak−m−n− ak+n−m− ak−n+m

−am+k+n− am−k−n+ am+k−n+ am−k+n)

=

2(bk+n− bk−n)am

=

= 0.

Similarly, we have that (bk,bn,bm) = (ak,an,bm) = (ak,bn,am) = (bk,an,am) =

(bk,bn,am) = (bk,an,bm) = (ak,bn,bm) = 0. Then (a,b,c) = 0,∀a,b,c ∈ A. The

result follows.

?

Corollary 2.2b0of Lemma 2.1 is a unity of A.

Lemma 2.3 Let A be a commutative and associative algebra satisfying Lemma

2.1. Then the following statements hold:

1) If D0is a linear transformation of A such that

?

D0(an) = nbn, n = 1,2,3,···,

D0(bn) = nan, n = 0,1,2,···,

(2.2)

then D0is a derivation of A.

2)If aD0is a linear transformation of A such that

(aD0)(b) = aD0(b),∀a,b ∈ A, (2.3)

then aD0is a derivation of A.

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3)D1= {aD0|a ∈ A} is a subalgebra of Lie algebra DerA.

Proof. (1) We have

D0(anam) = D0

?1

2(bn+m− bn−m)

?

=1

2((m + n)an+m− (n − m)an−m)

and

D0(an)am+ anD0(am) = nbnam+ manbm

=

n

2(an+m− an−m) +m

1

2((m + n)am+n− (n − m)an−m).

2(an+m− am−n)

=

So D0is a derivation of A.

2)For ∀a,b,c ∈ A, we have

(aD0)(bc) = aD0(bc) = aD0(b)c + abD0(c) = (aD0)(b)c + b(aD0)(c),

so aD0is a derivation of A.

3) For ∀a,b,c ∈ A, we have

[aD0,bD0](c) = (aD0)(bD0)(c) − (bD0)(aD0)(c)

= aD0(b)D0(c) − bD0(a)D0(c)

= (aD0(b) − bD0(a))D0(c).

Then [aD0,bD0] = (aD0(b) − bD0(a))D0∈ D1, and so 3) holds.

?

Theorem 2.4Let A be a commutative and associative algebra satisfying Lemma

2.1, and let a be an element of A. If D0satisfies Lemma 2.3 and ◦ satisfies

b ◦ c = baD0(c),∀b,c ∈ A, (2.4)

then the following statements hold:

1)(A,aD0,◦) is a Novikov algebra.

2)(A,aD0,[,]) is an adjoining Lie algebra of (A,aD0,◦) and [ , ] such that

[b,c] = a(bD0(c) − cD0(b)),∀b,c ∈ A. (2.5)

Proof. 1) By Lemma 2.3, aD0is a derivation of the commutative algebra A. So

(A,aD0,◦) is a Novikov algebra by Remark 1.3 (2).

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2)(A,aD0,[,]) is an adjoining Lie algebra of (A,aD0,◦) by Remark 1.3 (1).

For ∀b,c ∈ A,∃a ∈ A, we have

[b,c] = b ◦ c − c ◦ b = baD0(c) − caD0(b) = a(bD0(c) − cD0(b))

since A is commutative. Hence we obtain the desired result.

?

Let b0 be a unity of A. If we set a = b0 in Theorem 2.4, then an◦ am =

anb0D0(am) = an(mbm) =

m

2(am+n+ an−m). Similarly, we obtain the following

corollary:

Corollary 2.5 Let A be a commutative and associative algebra satisfying Lemma

2.1. Then the following statements hold:

an◦ am=m

bn◦ bm=m

an◦ bm=m

bn◦ am=m

2(an+m+ an−m)

2(an+m+ am−n)

2(bn+m− bn−m)

2(bn+m+ bn−m)

(2.6)

and

[an,am] =1

[bn,bm] =1

[an,bm] =1

[bn,am] =1

2(m − n)an+m+1

2(m − n)an+m−1

2(m − n)bn+m−1

2(m − n)bn+m+1

2(m + n)an−m

2(m + n)an−m

2(n + m)bn−m

2(m + n)bn−m.

(2.7)

The following, let sinhx =ex−e−x

2

, coshx =ex+e−x

2

the field F be assumed R or

C. We will construct Novikov algebras over the linear space which is generated by

sinhx and coshx.

First, let T be a linear space generated by {sinhmx,coshnx|m,n ∈ N} over F.

Lemma 2.6T satisfying the above product is a commutative associative algebra.

Proof. Since the above product is commutative and associative, we only need

that T is closed for the product. In fact,

sinhmxsinhnx =1

2[cosh(m + n)x − cosh(m − n)x]

coshmxcoshnx =1

2[cosh(m + n)x + cosh(m − n)x]

sinhmxcoshnx =1

2[sinh(m + n)x + sinh(m − n)x].

(2.8)

So T is a commutative and associative algebra.

?

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Lemma 2.7 Let T be a linear space generated by {sinhmx,coshnx|m,n ∈ N}

over F, then {1,sinhmx,coshnx|m,n ∈ N0} is a basis of T .

Proof. For ∀n ∈ N0, suppose that there are c0,ai,bj∈ F,i,j ∈ N0such that

c0+ a1sinhx + b1coshx + ··· + ansinhnx + bncoshnx = 0. (2.9)

We take derivative for (2.9) such that its derivative order is 2k − 1 (k ∈ N0), and

put x = 0. Then we have

a1+ 22k−1a2+ ··· + n2k−1an= 0.

Let k = 1,2,···,n, then we obtain the following system of n linear equations:

a1+ 2a2+ ··· + nan= 0

a1+ 23a2+ ··· + n3an= 0

···························

a1+ 22n−1a2+ ··· + n2n−1an= 0.

(2.10)

If a1,···,an are seen to be unknown, then the coefficient matrix of (2.10) is the

Vandermonde matrix whose determinant is not 0, so ai= 0,i = 1,···,n.

We take derivative for (2.9) such that its derivative order is 2k (k ∈ N0), and

put x = 0. Then we have

b1+ 22kb2+ ··· + n2kbn= 0.

Let k = 1,2,···,n, then we obtain the following system of n linear equations:

b1+ 22b2+ ··· + n2bn= 0

b1+ 24b2+ ··· + n4bn= 0

···························

b1+ 22nb2+ ··· + n2nbn= 0.

(2.11)

If b1,···,bn are seen to be unknown, then the coefficient matrix of (2.11) is the

Vandermonde matrix whose determinant is not 0, so bi = 0,i = 1,···,n. Since

for ∀i ∈ N0, ai = 0 and bi = 0 satisfy (2.9), we have c0 = 0. Hence {1,

sinhx, coshx, ···, sinhnx, coshnx} are linearly independent for ∀n ∈ N0, then

{1,sinhnx,coshmx|n,m ∈ N0} are linearly independent and so they form a basis

of T as desired.

?

Theorem 2.8Let A1, A2be commutative and associative algebras over F. If ϕ:

A1−→ A2is an isomorphism and D1∈ DerA1, then the following statements hold:

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1)D2:= ϕD1ϕ−1∈ DerA2.

2)ϕ: (A1,D1,◦) −→ (A2,D2,◦) is also an isomorphism of Novikov algebras.

Proof. 1) For ∀a,b ∈ A1, we have

(ϕD1ϕ−1)(ϕ(a)ϕ(b)) = (ϕD1ϕ−1)(ϕ(ab))

= ϕD1(ab) = ϕ(D1(a)b + aD1(b)) = ϕ(D1(a))ϕ(b) + ϕ(a)ϕ(D1(b))

= (ϕD1ϕ−1)(ϕ(a))ϕ(b) + ϕ(a)(ϕD1ϕ−1)(ϕ(b)).

So 1) holds.

2) For ∀a,b ∈ A1, we have

ϕ(a ◦ b) = ϕ(aD1(b)) = ϕ(a)ϕ(D1(b))

= ϕ(a)(ϕD1ϕ−1)(ϕ(b)) = ϕ(a)D2(ϕ(b))

= ϕ(a) ◦ ϕ(b).

So 2) holds.

?

Theorem 2.9Let A be a commutative and associative algebra over F satisfy-

ing Lemma 2.1, D0be its derivation satisfying (2.2) and T be a commutative and

associative algebra over F satisfying Lemmas 2.6 and 2.7. If ϕ : A −→ T satisfies

ϕ(bm) = coshmx, m = 0,1,2,···,ϕ(an) = sinhnx, n = 1,2,···, (2.12)

then the following statements hold:

1)ϕ is an isomorphism of commutative and associative algebras.

2)ϕD0ϕ−1=

d

dx.

3)ϕ : (A,aD0,◦) −→ (T ,ϕ(a)d

dx,◦) is an isomorphism of Novikov algebras.

Proof. It is clear by Lemma 2.7, (2.1) and (2.8).

2) By (2.2) and (2.12), we have

ϕD0ϕ−1(sinhnx) = ϕD0(an)

= ϕ(nbn) = ncoshnx

=dsinhnx

dx

ϕD0ϕ−1(coshnx) = ϕD0(bn)

,

= ϕ(nan) = nsinhnx

=dcoshnx

dx

.

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So 2) holds.

3) It is clear that ϕ(aD0)ϕ−1= ϕ(a)d/dx. By (2.12) and (2.2), we have

ϕ(aD0)ϕ−1(sinhnx) = ϕ(aD0)(an)

= ϕ(aD0(an)) = ϕ(anbn)

= ϕ(a)ϕ(nbn) = ϕ(a)ncoshnx

= ϕ(a)d(sinhnx)/dx.

Similarly, we have ϕ(aD0)ϕ−1(coshnx) = ϕ(a)d(coshnx)/dx. So ϕ(aD0)ϕ−1=

ϕ(a)d/dx.

By Theorems 2.4, 2.8 and Remark 1.3 (2), we have

ϕ(b ◦ c) = ϕ(baD0(c))

= ϕ(b)ϕ(aD0(c))

= ϕ(b)[ϕ(aD0)ϕ−1(ϕ(c))]

= ϕ(b)ϕ(a)d/dx(ϕ(c))

= ϕ(b) ◦ ϕ(c),∀b,c ∈ A.

So ϕ : (A0,aD0,◦) −→ (T ,ϕ(a)d

dx,◦) is an isomorphism of Novikov algebras.

?

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