On split regular BiHom-Poisson color algebras

Yaling Tao; Yan Cao

doi:10.1515/math-2021-0039

Article Open Access

On split regular BiHom-Poisson color algebras

Yaling Tao and Yan Cao

Published/Copyright: July 22, 2021

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$Open Mathematics$

From the journal Open Mathematics Volume 19 Issue 1

Abstract

The purpose of this paper is to introduce the class of split regular BiHom-Poisson color algebras, which can be considered as the natural extension of split regular BiHom-Poisson algebras and of split regular Poisson color algebras. Using the property of connections of roots for this kind of algebras, we prove that such a split regular BiHom-Poisson color algebra L is of the form L = ⊕ [ α ] ∈ Λ / ∼ I [ α ] with I [ α ] a well described (graded) ideal of L , satisfying [ I [ α ] , I [ β ] ] + I [ α ] I [ β ] = 0 if [ α ] ≠ [ β ] . In particular, a necessary and sufficient condition for the simplicity of this algebra is determined, and it is shown that L is the direct sum of the family of its simple (graded) ideals.

Keywords: BiHom-Lie color algebra; BiHom-Poisson algebra; root space; root system

MSC 2010: 17B75; 17A60; 17B22; 17B65

1 Introduction

The interest in Poisson algebras has grown in the last few years, motivated especially by their applications in geometry and mathematical physics. For example, Poisson algebras play a fundamental role in deformation of commutative associative algebras [1]. Moreover, the cohomology group, deformation, tensor product and Γ -graded of Poisson algebras have been studied by many authors in [2,3, 4,5]. A Hom-algebra is an algebra such that a linear homomorphism appears in the identities satisfied by its multiplication. This class of algebras appeared in the study of quasi-deformations of vector fields, in particular quasi-deformations of Witt and Virasoro algebras in [6]. So far, many authors have studied Hom-type algebras [7,8, 9,10,11, 12,13]. In particular, the notion of Hom-Lie color algebra was introduced in [8] and presented the methods to construct this color algebra, which can be viewed as an extension of a Hom-Lie algebra to a Γ -graded algebra, where Γ is any abelian group. Furthermore, a BiHom-algebra is an algebra in such a way that the identities defining the structure are twisted by two homomorphisms ϕ , ψ . A BiHom-Poisson color algebra has simultaneously a BiHom-Lie algebra structure and a BiHom-associative algebra structure, satisfying the BiHom-Leibniz identity.

The class of the split algebras is especially related to addition quantum numbers, graded contractions and deformations. For instance, for a physical system which displays a symmetry of L , it is interesting to know in detail the structure of the split decomposition because its roots can be seen as certain eigenvalues which are the additive quantum numbers characterizing the state of such system. Recently, the structure of different classes of split algebras such as split regular Hom-Poisson algebras, split regular Hom-Poisson color algebras, split regular BiHom-Lie superalgebras, split BiHom-Leibniz superalgebras and split Leibniz triple systems have been studied by using techniques of connections of roots (see for instance [14,15, 16,17,18, 19,20,21, 22,23,24, 25,26]). Later, these techniques of connections become powerful to study not only split algebras but also graded algebras and algebras having multiplicative bases [27,28, 29,30]. The purpose of this paper is to consider the decomposition and simplicity of split regular BiHom-Poisson color algebras by the techniques of connections of roots.

2 Preliminaries

First we recall the definitions of Lie color algebra, Poisson color algebra, Hom-Lie color algebra and Hom-Poisson color algebra. The following definition is well known from the theory of graded algebra.

Definition 2.1

[8] Let Γ be an abelian group. A bi-character on Γ is a map ε : Γ × Γ → K ⧹ { 0 } satisfying

ε ( α , β ) ε ( β , α ) = 1 ,
ε ( α , β + γ ) = ε ( α , β ) ε ( α , γ ) ,
ε ( α + β , γ ) = ε ( α , γ ) ε ( β , γ ) ,

for all α , β , γ ∈ Γ .

It is clear that ε ( α , 0 ) = ε ( 0 , α ) = 1 for any α ∈ Γ , where 0 denotes the identity element of Γ .

Definition 2.2

[14] Let L = ⊕ g ∈ Γ L g be a Γ -graded K -vector space. For a nonzero homogeneous element v ∈ L , denote by v ¯ the unique group element in Γ such that v ∈ L v ¯ , which will be called the homogeneous degree of v . We shall say that L is a Lie color algebra if it is endowed with a K -bilinear map [ ⋅ , ⋅ ] : L × L → L satisfying

[ v , w ] = − ε ( v ¯ , w ¯ ) [ w , v ] , (Skew-symmetry),

[ v , [ w , t ] ] = [ [ v , w ] , t ] + ε ( v ¯ , w ¯ ) [ w , [ v , t ] ] (Jacobi identity),

for all homogeneous elements v , w , t ∈ L .

Lie superalgebras are examples of Lie color algebras with Γ = Z 2 and ε ( i , j ) = ( − 1 ) i j , for any i , j ∈ Z 2 . We also note that L 0 is a Lie algebra.

Definition 2.3

[5] A Poisson color algebra is a Γ -graded vector space L = ⊕ g ∈ Γ L g , an even bilinear mapping [ ⋅ , ⋅ ] : L × L → L , and a bi-character ε on Γ satisfying the following conditions:

( L , ε ) is an associative color algebra,
( L , [ ⋅ , ⋅ ] ) is a Lie color algebra,
Leibniz color identity [ x y , z ] = x [ y , z ] + ε ( y ¯ , z ¯ ) [ x , z ] y ,

for any x , y , z ∈ L , y ¯ , z ¯ denote the homogeneous degree of y , z .

Definition 2.4

[8] A Hom-Lie color algebra L is a quadruple ( L , [ ⋅ , ⋅ ] , ϕ , ε ) consisting of a Γ -graded space L , an even bilinear mapping [ ⋅ , ⋅ ] : L × L → L , a homomorphism ϕ : L → L and a bi-character ε on Γ satisfying

[ x , y ] = − ε ( x ¯ , y ¯ ) [ y , x ] ,
ε ( z ¯ , x ¯ ) [ ϕ ( x ) , [ y , z ] ] + ε ( x ¯ , y ¯ ) [ ϕ ( y ) , [ z , x ] ] + ε ( y ¯ , z ¯ ) [ ϕ ( z ) , [ x , y ] ] = 0 ,

for any x , y , z ∈ L , y ¯ , z ¯ denote the homogeneous degree of y , z . When ϕ is an algebra automorphism it is said that L is a regular Hom-Lie color algebra.

Definition 2.5

[17] A Hom-Poisson color algebra is a Hom-Lie color algebra ( L , [ ⋅ , ⋅ ] , ϕ , ε ) endowed with a Hom-associative color product, that is, a bilinear product denoted by juxtaposition such that

ϕ ( x ) ( y z ) = ( x y ) ϕ ( z ) ,

for all x , y , z ∈ L , and such that the Hom-Leibniz color identity

[ x y , ϕ ( z ) ] = ϕ ( x ) [ y , z ] + ε ( y ¯ , z ¯ ) [ x , z ] ϕ ( y )

holds for any x , y , z ∈ L , y ¯ , z ¯ denote the homogeneous degree of y , z .

If ϕ is furthermore a Poisson automorphism, that is, a linear bijective on such that ϕ ( [ x , y ] ) = [ ϕ ( x ) , ϕ ( y ) ] and ϕ ( x y ) = ϕ ( x ) ϕ ( y ) for any x , y ∈ L , then L is called a regular Hom-Poisson color algebra.

Definition 2.6

A BiHom-Lie color algebra L is a quintuple ( L , [ ⋅ , ⋅ ] , ϕ , ψ , ε ) consisting of a Γ -graded space L , an even bilinear mapping [ ⋅ , ⋅ ] : L × L → L , two homomorphisms ϕ , ψ and a bi-character ε on Γ satisfying

ϕ ∘ ψ = ψ ∘ ϕ ,
[ ψ ( x ) , ϕ ( y ) ] = − ε ( x ¯ , y ¯ ) [ ψ ( y ) , ϕ ( x ) ] (BiHom-skew-symmetry),
ε ( z ¯ , x ¯ ) [ ψ 2 ( x ) , [ ψ ( y ) , ϕ ( z ) ] ] + ε ( x ¯ , y ¯ ) [ ψ 2 ( y ) , [ ψ ( z ) , ϕ ( x ) ] ] + ε ( y ¯ , z ¯ ) [ ψ 2 ( z ) , [ ψ ( x ) , ϕ ( y ) ] ] = 0 (BiHom-Jacobi identity),

for any x , y , z ∈ L , x ¯ , y ¯ , z ¯ denote the homogeneous degree of x , y , z .

When ϕ , ψ furthermore are algebra automorphisms, it is said that L is a regular BiHom-Lie color algebra.

Definition 2.7

A BiHom-Poisson color algebra is a BiHom-Lie color algebra endowed with a BiHom-associative color product, that is, a bilinear product denoted by juxtaposition such that

ϕ ( x ) ( y z ) = ( x y ) ψ ( z ) ,

for all x , y , z ∈ L , and such that the BiHom-Leibniz color identity

[ x y , ϕ ψ ( z ) ] = ϕ ( x ) [ y , ϕ ( z ) ] + ε ( y ¯ , z ¯ ) [ x , ψ ( z ) ] ϕ ( y )

holds for any x , y , z ∈ L , x ¯ , y ¯ , z ¯ denote the homogeneous degree of x , y , z .

If ϕ , ψ is furthermore a Poisson automorphism, then L is called a regular BiHom-Poisson color algebra.

Poisson color algebras are examples of BiHom-Poisson color algebras by taking ϕ = ψ = Id . Hom-Poisson color algebras are also examples of BiHom-Poisson color algebras by considering ψ = ϕ .

Example 2.8

Let ( L , [ ⋅ , ⋅ ] , ε ) be a Poisson color algebra, ϕ , ψ : L → L two automorphisms and ϕ ∘ ψ = ψ ∘ ϕ . If we endow the underlying linear space L with a new product [ ⋅ , ⋅ ] ′ : L × L → L defined by [ x , y ] ′ ≔ [ ϕ ( x ) , ψ ( y ) ] , ( x y ) ′ = ϕ ( x ) ψ ( y ) for any x , y ∈ L , we have that ( L , [ ⋅ , ⋅ ] ′ , ϕ , ψ , ε ) becomes a regular BiHom-Poisson color algebra.

Proof

First we check that ( ϕ ( x ) ( y z ) ′ ) ′ = ( ( x y ) ′ ψ ( z ) ) ′ .

( ϕ ( x ) ( y z ) ′ ) ′ = ( ϕ ( x ) ( ϕ ( y ) ψ ( z ) ) ) ′ = ϕ 2 ( x ) ψ ( ϕ ( y ) ψ ( z ) ) = ϕ 2 ( x ) ψ ϕ ( y ) ψ 2 ( z )

and

( ( x y ) ′ ψ ( z ) ) ′ = ( ( ϕ ( x ) ψ ( y ) ) ψ ( z ) ) ′ = ϕ ( ϕ ( x ) ψ ( y ) ) ψ 2 ( z ) = ϕ 2 ( x ) ϕ ψ ( y ) ψ 2 ( z ) = ϕ 2 ( x ) ψ ϕ ( y ) ψ 2 ( z ) .

That is, ( ϕ ( x ) ( y z ) ′ ) ′ = ( ( x y ) ′ ψ ( z ) ) ′ .

Next check that [ ( x y ) ′ , ϕ ψ ( z ) ] ′ = ( ϕ ( x ) [ y , ϕ ( z ) ] ′ ) ′ + ε ( y ¯ , z ¯ ) ( [ x , ψ ( z ) ] ′ ϕ ( y ) ) ′ , by Leibniz color identity and [ x , y ] ′ ≔ [ ϕ ( x ) , ψ ( y ) ] , we get

[ ( x y ) ′ , ϕ ψ ( z ) ] ′ = [ ϕ ( x y ) ′ , ψ ( ϕ ψ ( z ) ) ] = [ ϕ ( ϕ ( x ) ψ ( y ) ) , ψ ( ϕ ψ ( z ) ) ] = [ ϕ 2 ( x ) ψ ϕ ( y ) , ψ 2 ϕ ( z ) ] = ϕ 2 ( x ) [ ψ ϕ ( y ) , ψ 2 ϕ ( z ) ] + ε ( y ¯ , z ¯ ) [ ϕ 2 ( x ) , ψ 2 ϕ ( z ) ] ψ ϕ ( y ) ,

( ϕ ( x ) [ y , ϕ ( z ) ] ′ ) ′ = ϕ 2 ( x ) [ ψ ( y ) , ψ ϕ ( z ) ] ′ = ϕ 2 ( x ) [ ψ ϕ ( y ) , ψ 2 ϕ ( z ) ] ,

and

ε ( y ¯ , z ¯ ) ( [ x , ψ ( z ) ] ′ ϕ ( y ) ) ′ = ε ( y ¯ , z ¯ ) [ ϕ ( x ) , ψ ϕ ( z ) ] ′ ψ ϕ ( y ) = ε ( y ¯ , z ¯ ) [ ϕ 2 , ψ 2 ϕ ( z ) ] ψ ϕ ( y ) .

Hence, [ x y , ϕ ψ ( z ) ] = ϕ ( x ) [ y , ϕ ( z ) ] + ε ( y ¯ , z ¯ ) [ x , ψ ( z ) ] ϕ ( y ) . This gives the conclusion.□

Throughout this paper we will consider a regular BiHom-Poisson color algebra L being of arbitrary dimension and over an arbitrary base field K . N denotes the set of all non-negative integers and Z denotes the set of all integers. A subalgebra A of L is a graded subspace such that [ A , A ] + A A ⊂ A and ϕ ( A ) = ψ ( A ) = A . A subalgebra I of L is called an ideal if [ I , L ] + I L + L I ⊂ I and ϕ ( I ) = ψ ( I ) = I . A BiHom-Poisson color algebra L will be called simple if [ L , L ] + L L ≠ 0 and its only ideals are {0} and L .

Let us introduce the class of split algebras in the framework of regular Hom-Poisson color algebras L . First, we recall that a Hom-Poisson color algebra ( L , [ ⋅ , ⋅ ] , ϕ , ε ) , over a base field K , is called split with respect to a maximal Abelian subalgebra H of L , if L can be written as the direct sum

L = H ⊕ ( ⊕ α ∈ Γ L α ) ,

where

L α = { v α ∈ L : [ h 0 , ϕ ( v α ) ] = α ( h 0 ) ϕ ( v α ) for any h 0 ∈ H 0 } ,

being any α : H 0 → K , α ∈ Γ , a nonzero linear functional on H 0 such that L α ≠ 0 .

Definition 2.9

Denote by H = ⊕ g ∈ Γ H g a maximal Abelian (graded) subalgebra, of a regular BiHom-Poisson color algebra L . For a linear functional α : H 0 → K , we define the root space of L (with respect to H ) associated with α as the subspace

L α = { v α ∈ L : [ h 0 , ϕ ( v α ) ] = α ( h 0 ) ϕ ψ ( v α ) for any h 0 ∈ H 0 } .

The elements α : H 0 → K satisfying L α ≠ 0 are called roots of L with respect to H . We denote Λ ≔ { α ∈ ( H 0 ) ∗ ⧹ { 0 } : L α ≠ 0 } . We say that L is a split regular BiHom-Poisson color algebra, with respect to H , if

L = H ⊕ ( ⊕ α ∈ Λ L α ) .

We also say that Λ is the root system of L .

Note that when ϕ = ψ = Id , the split Poisson color algebras become examples of split regular BiHom-Poisson color algebras and when ϕ = ψ , the split regular Hom-Poisson color algebra become examples of split regular BiHom-Poisson color algebras. Hence, the present paper extends the results in [15].

From now on L = H ⊕ ( ⊕ α ∈ Λ L α ) denotes a split regular BiHom-Poisson color algebras. Also, and for an easier notation, the mappings ϕ ∣ H , ψ ∣ H , ϕ ∣ H − 1 , ψ ∣ H − 1 : H → H will be denoted by ϕ , ψ , ϕ − 1 , ψ − 1 , respectively.

It is clear that the root space associated with the zero root L 0 satisfies H ⊂ L 0 . Conversely, given any v 0 ∈ L 0 we can write v 0 = h ⊕ ( ⊕ i = 1 n v α i ) , where h ∈ H and v α i ∈ L α i for i = 1 , … , n , with α i ≠ α j if i ≠ j . Since for any h 0 ∈ H 0 we have [ h 0 , v 0 ] = 0 , then 0 = [ h 0 , h ⊕ ( ⊕ i = 1 n ϕ ϕ − 1 ( v α i ) ) ] = ⊕ i = 1 n α i ϕ ( h 0 ) ψ ( v α i ) . From here, taking into account the direct character of the sum with the fact α i ≠ 0 gives us that any v α i = 0 . Hence, v 0 = h ∈ H . Consequently,

(2.1) H = L 0 .

Lemma 2.10

Let L be a split regular BiHom-Poisson color algebra. Then, for any α , β ∈ Λ ∪ { 0 } , the following assertions hold.

ϕ ( L α ) = L α ϕ − 1 and ϕ − 1 ( L α ) = L α ϕ .
ψ ( L α ) = L α ψ − 1 and ψ − 1 ( L α ) = L α ψ .
[ L α , L β ] ⊂ L α ϕ − 1 + β ψ − 1 .
L α L β ⊂ L α ϕ − 1 + β ψ − 1 .
If α ∈ Λ , then α ϕ − z 1 ψ − z 2 ∈ Λ for any z 1 , z 2 ∈ Z .

Proof

1. For any h 0 ∈ H 0 and v α ∈ L α , since

(2.2) [ h 0 , ϕ ( v α ) ] = α ( h 0 ) ϕ ψ ( v α ) ,

we have that by writing h 0 ′ = ϕ ( h 0 ) then

[ h 0 ′ , ϕ 2 ( v α ) ] = ϕ ( [ h 0 , ϕ ( v α ) ] ) = α ( h 0 ) ϕ 2 ψ ( v α ) = α ϕ − 1 ( h 0 ′ ) ϕ 2 ψ ( v α ) = α ϕ − 1 ( h 0 ′ ) ϕ ψ ( ϕ ( v α ) ) .

Therefore, we get ϕ ( v α ) ∈ L α ϕ − 1 and so

(2.3) ϕ ( L α ) ⊂ L α ϕ − 1 .

Now, let us show

L α ϕ − 1 ⊂ ϕ ( L α ) .

Indeed, for any h 0 ∈ H 0 and v α ∈ L α , equation (2.2) shows [ ϕ − 1 ( h 0 ) , v α ] = α ( h 0 ) ψ ( v α ) . From here, we get [ ϕ ( h 0 ) , v α ] = α ϕ 2 ( h 0 ) ψ ( v α ) and conclude

(2.4) ϕ − 1 ( L α ) ⊂ L α ϕ .

Hence, since for any x ∈ L α ϕ − 1 , we can write x = ϕ ( ϕ − 1 ( x ) ) and by equation (2.4) we have ϕ − 1 ( x ) ∈ L α , we conclude L α ϕ − 1 ⊂ ϕ ( L α ) . This fact together with (2.3) show ϕ ( L α ) = L α ϕ − 1 .

To show

ϕ − 1 ( L α ) = L α ϕ .

By equation (2.4), while the fact L α ϕ ⊂ ϕ − 1 ( L α ) is a consequence of writing any element x ∈ L α ϕ of the form x = ϕ − 1 ( ϕ ( x ) ) and apply equation (2.3).

2. To verify

(2.5) ψ ( L α ) ⊂ L α ψ − 1 ,

observe that equation (2.2) gives us [ ψ ( h 0 ) , ψ ϕ ( v α ) ] = α ( h 0 ) ψ ϕ ψ ( v α ) , and so [ ψ ( h 0 ) , ϕ ψ ( v α ) ] = α ψ − 1 ( ψ ( h 0 ) ) ϕ ψ ( ψ ( v α ) ) . Since equation (2.2) and the identity ψ − 1 ϕ = ϕ ψ − 1 also give us

(2.6) ψ − 1 ( L α ) ⊂ L α ψ ,

we conclude as above that ψ ( L α ) = L α ψ − 1 . We can argue similarly with equations (2.5) and (2.6) to get ψ − 1 ( L α ) = L α ψ .

3. For each h 0 ∈ H 0 , v α ∈ L α , i , v β ∈ L β , j , we can write

[ h 0 , ϕ ( [ v α , v β ] ) ] = [ ψ 2 ψ − 2 ( h 0 ) , ϕ ( [ v α , v β ] ) ] .

So, by denoting h 0 ′ = ψ − 2 ( h 0 ) , we can apply BiHom-Jacobi identity and BiHom-skew-symmetry to get

[ ψ 2 ( h 0 ′ ) , ϕ ( [ v α , v β ] ) ] = [ ψ 2 ( h 0 ′ ) , [ ψ ψ − 1 ϕ ( v α ) , ϕ ( v β ) ] ] = − ε ( h 0 ′ ¯ , α ¯ + β ¯ ) [ ψ ϕ ( v α ) , [ ψ ( v β ) , ϕ ( h 0 ′ ) ] ] − ε ( α ¯ + h 0 ′ ¯ , β ¯ ) [ ψ 2 ( v β ) , [ ψ ( h 0 ′ ) , ϕ ψ − 1 ϕ ( v α ) ] ] = ε ( h 0 ′ ¯ , α ¯ ) [ ψ ϕ ( v α ) , [ ψ ( h 0 ′ ) , ϕ ( v β ) ] ] − ε ( α ¯ + h 0 ′ ¯ , β ¯ ) [ ψ 2 ( v β ) , ϕ [ ϕ − 1 ψ ( h 0 ′ ) , ψ − 1 ϕ ( v α ) ] ] = ε ( h 0 ′ ¯ , α ¯ ) [ ψ ϕ ( v α ) , [ ψ ( h 0 ′ ) , ϕ ( v β ) ] ] + [ [ ψ 2 ϕ − 1 ( h 0 ′ ) , ϕ ( v α ) ] , ϕ ψ ( v β ) ] = ε ( h 0 ′ ¯ , α ¯ ) β ψ ( h 0 ′ ) [ ψ ϕ ( v α ) , ϕ ψ ( v β ) ] + α ( ψ 2 ϕ − 1 ( h 0 ′ ) ) [ ϕ φ ( v α ) , ϕ ψ ( v β ) ] = ε ( h 0 ′ ¯ , α ¯ ) ( β ψ + α ψ 2 ϕ − 1 ) ( h 0 ′ ) [ ψ ϕ ( v α ) , ϕ ψ ( v β ) ] = ε ( h 0 ′ ¯ , α ¯ ) ( β ψ + α ψ 2 ϕ − 1 ) ( h 0 ′ ) [ ϕ ψ ( v α ) , ϕ ψ ( v β ) ] = ε ( h 0 ′ ¯ , α ¯ ) ( β ψ + α ψ 2 ϕ − 1 ) ( h 0 ′ ) ϕ ψ ( [ v α , v β ] ) = ( β ψ + α ψ 2 ϕ − 1 ) ( h 0 ′ ) ϕ ψ ( [ v α , v β ] ) .

Taking into account h 0 ′ = ψ − 2 ( h 0 ) we have shown

[ h 0 , ϕ ( [ v α , v β ] ) ] = ( β ψ − 1 + α ϕ − 1 ) ( h 0 ) ϕ ψ ( [ v α , v β ] ) .

From here, [ L α , L β ] ⊂ L α ϕ − 1 + β ψ − 1 .

From Lemma 2.10-3, we can assert that

[ L α , g 1 , L β , g 2 ] ⊂ L α ϕ − 1 + β ψ − 1 , g 1 + g 2 ,

for any g 1 , g 2 ∈ Γ .

4. Let h 0 ∈ H 0 , v α ∈ L α , i , v β ∈ L β , j , we can write

[ h 0 , ϕ ( v α v β ) ] = [ ψ ψ − 1 ( h 0 ) , ϕ ( v α v β ) ] ,

and denote h 0 ′ = ψ − 1 ( h 0 ) . By applying the BiHom-Leibniz color identity and BiHom-skew-symmetry, we get

[ ψ ( h 0 ′ ) , ϕ ( v α v β ) ] = − ε ( h 0 ′ ¯ , ( v α v β ¯ ) ) [ ψ ( v α v β ) , ϕ ( h 0 ′ ) ] = − ε ( h 0 ′ ¯ , ( v α v β ¯ ) ) [ ψ ( v α ) ψ ( v β ) , ϕ ψ ( ψ − 1 ( h 0 ′ ) ) ] = − ε ( h 0 ′ ¯ , ( v α v β ¯ ) ) ε ( v β ¯ , h 0 ′ ¯ ) [ ψ ( v α ) , h 0 ′ ] ϕ ψ ( v β ) − ε ( h 0 ′ ¯ , ( v α v β ¯ ) ) ϕ ψ ( v α ) [ ψ ( v β ) , ϕ ψ − 1 ( h 0 ′ ) ] = − [ ψ ( v α ) , ϕ ϕ − 1 ( h 0 ′ ) ] ϕ ψ ( v β ) − ϕ ψ ( v α ) [ ψ ( v β ) , ϕ ψ − 1 ( h 0 ′ ) ] = − ( − ε ( v α ¯ , h 0 ′ ¯ ) ) [ ψ ϕ − 1 ( h 0 ′ ) , ϕ ( v α ) ] ϕ ψ ( v β ) − ( − ε ( v β ¯ , h 0 ′ ¯ ) ) ϕ ψ ( v α ) [ h 0 ′ , ϕ ( v β ) ] = [ ψ ϕ − 1 ( h 0 ′ ) , ϕ ( v α ) ] ϕ ψ ( v β ) + ϕ ψ ( v α ) [ h 0 ′ , ϕ ( v β ) ] = [ ψ ϕ − 1 ( ψ − 1 ( h 0 ) ) , ϕ ( v α ) ] ϕ ψ ( v β ) + ϕ ψ ( v α ) [ ψ − 1 ( h 0 ) , ϕ ( v β ) ] = [ ϕ − 1 ( h 0 ) , ϕ ( v α ) ] ϕ ψ ( v β ) + ϕ ψ ( v α ) [ ψ − 1 ( h 0 ) , ϕ ( v β ) ] = α ϕ − 1 ( h 0 ) ϕ ψ ( v α ) ϕ ψ ( v β ) + ϕ ψ ( v α ) β ψ − 1 ( h 0 ) ϕ ψ ( v β ) = ( α ϕ − 1 + β ψ − 1 ) ( h 0 ) ϕ ψ ( v α ) ϕ ψ ( v β ) = ( α ϕ − 1 + β ψ − 1 ) ( h 0 ) ϕ ψ ( v α v β ) .

From here, L α L β ⊂ L α ϕ − 1 + β ψ − 1 .

From Lemma 2.10-4 we can assert that

L α , g 1 L β , g 2 ⊂ L α ϕ − 1 + β ψ − 1 , g 1 + g 2 ,

for any g 1 , g 2 ∈ Γ .

5. This is a consequence of Lemma 2.10-1,2.□

Definition 2.11

A root system Λ of a split BiHom-Poisson color algebra is called symmetric if it satisfies that α ∈ Λ implies − α ∈ Λ .

3 Decompositions

In the following, L denotes a split regular BiHom-Poisson color algebra with a symmetric root system Λ and L = H ⊕ ( ⊕ α ∈ Λ L α ) the corresponding root decomposition. We begin by developing the techniques of connections of roots in this section.

Definition 3.1

Let α and β be two nonzero roots. We shall say that α is connected to β if there exists α 1 , … , α k ∈ Λ such that

If k = 1 , then

α 1 ∈ { α ϕ − n ψ − r : n , r ∈ N } ∩ { ± β ϕ − m ψ − s : m , s ∈ N } .

If k ≥ 2 , then

α 1 ∈ { α ϕ − n ψ − r : n , r ∈ N } .
α 1 ϕ − 1 + α 2 ψ − 1 ∈ Λ ,
α 1 ϕ − 2 + α 2 ϕ − 1 ψ − 1 + α 3 ψ − 1 ∈ Λ , ⋯

α 1 ϕ − i + α 2 ϕ − i + 1 ψ − 1 + α 3 ϕ − i + 2 ψ − 1 + ⋯ + α i ϕ − 1 ψ − 1 + α i + 1 ψ − 1 ∈ Λ , ⋯ α 1 ϕ − k + 2 + α 2 ϕ − k + 3 ψ − 1 + α 3 ϕ − k + 4 ψ − 1 + ⋯ + α k − 2 ϕ − 1 ψ − 1 + α k − 1 ψ − 1 ∈ Λ .
α 1 ϕ − k + 1 + α 2 ϕ − k + 2 ψ − 1 + α 3 ϕ − k + 3 ψ − 1 + ⋯ + α i ϕ − k + i ψ − 1 + ⋯ + α k − 1 ϕ − 1 ψ − 1 + α k ψ − 1 ∈ { ± β ϕ − m ψ − s : m , s ∈ N } .

We shall also say that { α 1 , … , α k } is a connection from α to β .

Our next goal is to show that the connection is an equivalence relation on Λ .

Proposition 3.2

The relation ∼ in Λ , defined by α ∼ β if and only if α is connected to β , is of equivalence.

Proof

This can be proved completely analogously to [15, Corollary 2.1].□

For any α ∈ Λ , we denote by

Λ α ≔ { β ∈ Λ : β ∼ α } .

Clearly if β ∈ Λ α , then − β ∈ Λ α and, by Proposition 3.2, if γ ∉ Λ α , then Λ α ∩ Λ γ = ∅ .

Our next goal is to associate an adequate ideal L Λ α of L with any Λ α . For Λ α , α ∈ Λ , we define

H Λ α ≔ span K { [ L β ψ − 1 , L − β ϕ − 1 ] + L β ψ − 1 L − β ϕ − 1 : β ∈ Λ α }

and

V Λ α ≔ ⊕ β ∈ Λ α L β = ⊕ β ∈ Λ α , g ∈ Γ L β , g .

We denote by L Λ α the following graded subspace of L ,

L Λ α ≔ H Λ α ⊕ V Λ α .

Proposition 3.3

For any α ∈ Λ , we have [ L Λ α , L Λ α ] + L Λ α L Λ α ⊂ L Λ α .

Proof

First we have to check that L Λ α satisfies [ L Λ α , L Λ α ] ⊂ L Λ α . Taking into account H = L 0 , then [ H Λ α , H Λ α ] = 0 and we have

(3.7) [ L Λ α , L Λ α ] = [ H Λ α ⊕ V Λ α , H Λ α ⊕ V Λ α ] ⊂ [ H Λ α , V Λ α ] + [ V Λ α , H Λ α ] + Σ β , γ ∈ Λ α [ L β , L γ ] .

Let us consider the first summand [ H Λ α , V Λ α ] in equation (3.7). Given β ∈ Λ α , we have [ H Λ α , L β ] ⊂ [ L 0 , L β ] ⊂ L β ψ − 1 , being β ψ − 1 ∈ Λ α by Lemma 2.10-4. Hence,

(3.8) [ H Λ α , V Λ α ] ⊂ V Λ α .

Similarly, we can also get

(3.9) [ V Λ α , H Λ α ] ⊂ V Λ α .

Next consider the third summand Σ β , γ ∈ Λ α [ L β , L γ ] . Given β , γ ∈ Λ α such that [ L β , L γ ] ≠ 0 , if β ϕ − 1 + γ ψ − 1 = 0 , then clearly [ L β , L γ ] ⊂ H Λ α . Suppose that β ϕ − 1 + γ ψ − 1 ≠ 0 , since [ L β , L γ ] ≠ 0 together with Lemma 2.10 ensures that β ϕ − 1 + γ ψ − 1 ∈ Λ , we have that { β , γ } is a connection from β to β ϕ − 1 + γ ψ − 1 . The transitivity of ∼ gives now that β ϕ − 1 + γ ψ − 1 ∈ Λ α and so

(3.10) [ L β , L γ ] ⊂ L β ϕ − 1 + γ ψ − 1 ⊂ V Λ α .

From equations (3.7)–(3.10), we conclude that [ L Λ α , L Λ α ] ⊂ L Λ α .

Second, we will verify that L Λ α L Λ α ⊂ L Λ α . We have

L Λ α L Λ α = ( H Λ α ⊕ V Λ α ) ( H Λ α ⊕ V Λ α ) ⊂ H Λ α H Λ α + H Λ α V Λ α + V Λ α H Λ α + Σ β , γ ∈ Λ α ( L β L γ ) .

By arguing as above, we have

H Λ α V Λ α + V Λ α H Λ α + Σ β , γ ∈ Λ α ( L β L γ ) ⊂ H Λ α .

Hence, it just remains to check that H Λ α H Λ α , observe that

H Λ α H Λ α ⊂ ( Σ β ∈ Λ α ( [ L β ψ − 1 , L − β ϕ − 1 ] + L β ψ − 1 L − β ϕ − 1 ) ) H ⊂ ( Σ β ∈ Λ α [ L β ψ − 1 , L − β ϕ − 1 ] ) H + ( Σ β ∈ Λ α L β ψ − 1 L − β ϕ − 1 ) H .

From the above, by BiHom-Leibniz color identity, we have

[ L β ψ − 1 , L − β ϕ − 1 ] H = [ L β ψ − 1 , L − β ϕ − 1 ] ϕ ( ϕ − 1 ( H ) ) ⊂ [ L β ψ − 1 ϕ − 1 ( H ) , ϕ ψ ( ψ − 1 ( L − β ϕ − 1 ) ) ] + ϕ ( L β ψ − 1 ) [ ϕ − 1 ( H ) , ϕ ( ψ − 1 ( L − β ϕ − 1 ) ) ] ⊂ [ L β ψ − 1 ϕ − 1 , L − β ϕ − 2 ] + L β ψ − 1 ϕ − 1 L − β ϕ − 2 ⊂ H Λ α .

By BiHom-associativity, we have

( L β ψ − 1 L − β ϕ − 1 ) H = ( L β ψ − 1 L − β ϕ − 1 ) ψ ( ψ − 1 ( H ) ) ⊂ ϕ ( L β ψ − 1 ) ( L − β ϕ − 1 ψ − 1 ( H ) ) ⊂ L β ψ − 1 ϕ − 1 L − β ϕ − 2 ⊂ H Λ α . □

Proposition 3.4

For any α ∈ Λ , we have ϕ ( L Λ α ) = L Λ α and ψ ( L Λ α ) = L Λ α .

Proof

This is a direct consequence of Lemma 2.10-1,2.□

Proposition 3.5

If γ ∉ Λ α , then [ L Λ α , L Λ γ ] + L Λ α L Λ γ = 0 .

Proof

We have

(3.11) [ L Λ α , L Λ γ ] = [ H Λ α ⊕ V Λ α , H Λ γ ⊕ V Λ γ ] ⊂ [ H Λ α , V Λ γ ] + [ V Λ α , H Λ γ ] + [ V Λ α , V Λ γ ]

and

(3.12) L Λ α L Λ γ = ( H Λ α ⊕ V Λ α ) ( H Λ γ ⊕ V Λ γ ) ⊂ H Λ α H Λ γ + H Λ α V Λ γ + V Λ α H Λ γ + V Λ α V Λ γ .

First, we consider [ V Λ α , V Λ γ ] + V Λ α V Λ γ and suppose that there exist β ∈ Λ α and η ∈ Λ γ such that [ L β , L η ] + L β L η ≠ 0 . As necessarily β ϕ − 1 ≠ − η ψ − 1 , then β ϕ − 1 + η ψ − 1 ∈ Λ . So { β , η , − β ϕ − 1 } is a connection between β and η . By the transitivity of the connection relation we have γ ∈ Λ α , a contradiction. Hence, [ L β , L η ] + L β L η = 0 and so

(3.13) [ V Λ α , V Λ γ ] + V Λ α V Λ γ = 0 .

Second, we consider [ H Λ α , V Λ γ ] + H Λ α V Λ γ , and suppose there exist β ∈ Λ α and η ∈ Λ γ such that

[ [ L β , L − β ] , L η ] + [ L β L − β , L η ] + [ L β , L − β ] L η + ( L β L − β ) L η ≠ 0 .

The following is divided into four situations to discuss.

Case 1:

[ [ L β , L − β ] , L η ] ≠ 0 .

BiHom-skew-symmetry and BiHom-Jacobi identity give

0 ≠ [ [ ψ ( ψ − 1 ( L β ) ) , ϕ ( ϕ − 1 ( L − β ) ) ] , ψ 2 ( ψ − 2 ( L η ) ) ] ⊂ [ ψ 2 ( ψ − 1 ( L β ) ) , [ ψ ( ϕ − 1 ( L − β ) ) , ϕ ( ψ − 2 ( L η ) ) ] ] + [ ψ 2 ( ϕ − 1 ( L − β ) ) , [ ψ ( ψ − 2 ( L η ) ) , ϕ ( ψ − 1 ( L β ) ) ] ] ⊂ [ ψ ( L β ) , [ ψ ( ϕ − 1 ( L − β ) ) , ϕ ( ψ − 2 ( L η ) ) ] ] + [ ψ 2 ( ϕ − 1 ( L − β ) ) , [ ψ − 1 ( L η ) , ϕ ( ψ − 1 ( L β ) ) ] ] .

We get either [ ψ ( ϕ − 1 ( L − β ) ) , ϕ ( ψ − 2 ( L η ) ) ] ≠ 0 or [ ψ − 1 ( L η ) , ϕ ( ψ − 1 ( L β ) ) ] ≠ 0 . In any case, which contradicts equation (3.13). Hence, [ [ L β , L − β ] , L η ] = 0 .

Case 2:

[ L β L − β , L η ] ≠ 0 .

BiHom-Leibniz color identity gives

0 ≠ [ L β L − β , ϕ ψ ( ϕ − 1 ψ − 1 ( L η ) ) ] ⊂ ϕ ( L β ) [ L − β , ϕ ( ϕ − 1 ψ − 1 ( L η ) ) ] + [ L β , ψ ( ϕ − 1 ψ − 1 ( L η ) ) ] ϕ ( L − β ) = ϕ ( L β ) [ L − β , ψ − 1 ( L η ) ] + [ L β , ϕ − 1 ( L η ) ] ϕ ( L − β ) .

We get either [ L − β , ψ − 1 ( L η ) ] ≠ 0 or [ L β , ϕ − 1 ( L η ) ] ≠ 0 . In any case, which contradicts equation (3.13). Hence, [ L β L − β , L η ] = 0 .

Case 3:

[ L β , L − β ] L η ≠ 0 .

BiHom-Leibniz color identity gives

0 ≠ [ L β , ψ ( ψ − 1 ( L − β ) ) ] ϕ ( ϕ − 1 ( L η ) ) ⊂ [ L β ( ϕ − 1 ( L η ) ) , ϕ ψ ( ψ − 1 ( L − β ) ) ] + ϕ ( L β ) [ ϕ − 1 ( L η ) , ϕ ( ψ − 1 ( L − β ) ) ] = [ L β ( ϕ − 1 ( L η ) ) , ϕ ( L − β ) ] + ϕ ( L β ) [ ϕ − 1 ( L η ) , ϕ ψ − 1 ( L − β ) ] .

We get either [ L β ϕ − 1 ( L η ) , ϕ ( ( L − β ) ) ] ≠ 0 or [ ϕ − 1 ( L η ) , ϕ ψ − 1 ( L − β ) ] ≠ 0 . In any case, which contradicts equation (3.13). Hence, [ L β , L − β ] L η = 0 .

Case 4:

( L β L − β ) L η ≠ 0 .

BiHom-associativity gives

0 ≠ ( L β L − β ) ψ ( ψ − 1 ( L η ) ) ⊂ ϕ ( L β ) ( L − β ( ψ − 1 ( L η ) ) ) ,

which contradicts equation (3.13). Hence, ( L β L − β ) L η = 0 . Consequently,

(3.14) [ H Λ α , V Λ γ ] + H Λ α V Λ γ = 0 .

In a similar way, we get

(3.15) [ V Λ α , H Λ γ ] + V Λ α H Λ γ = 0 .

Finally, we consider H Λ α H Λ γ , suppose there exist β ∈ Λ α and η ∈ Λ γ such that

[ L β , L − β ] [ L η , L − η ] + [ L β , L − β ] ( L η L − η ) + ( L β L − β ) [ L η , L − η ] + ( L β L − β ) ( L η L − η ) ≠ 0 .

As above, BiHom-Leibniz color identity or BiHom-associativity identity gives us that this fact implies

[ H Λ α , V Λ γ ] + H Λ α V Λ γ + [ V Λ α , H Λ γ ] + V Λ α H Λ γ ≠ 0 ,

a contradiction either with equation (3.14) or equation (3.15). From here,

H Λ α H Λ γ = 0 .

Hence, [ L Λ α , L Λ γ ] + L Λ α L Λ γ = 0 . The proof is completed.□

Proposition 3.6

For any α ∈ Λ , we have H Λ α H + H H Λ α ⊂ H Λ α .

Proof

Fix any β ∈ Λ α . On one hand, by BiHom-Leibniz color identity, we get

[ L β , L − β ] H + H [ L β , L − β ] = [ L β , ψ ( ψ − 1 ( L − β ) ) ] ϕ ( ϕ − 1 ( H ) ) + ϕ ( ϕ − 1 ( H ) ) [ L β , ϕ ( ϕ − 1 ( L − β ) ) ] ⊂ [ L β ( ϕ − 1 ( H ) ) , ϕ ψ ( ψ − 1 ( L − β ) ) ] + ϕ ( L β ) [ ϕ − 1 ( H ) , ϕ ( ψ − 1 ( L − β ) ) ] + [ ϕ − 1 ( H ) L β , ϕ ψ ( ϕ − 1 ( L − β ) ) ] + [ ϕ − 1 ( H ) , ψ ( ϕ − 1 ( L − β ) ) ] ϕ ( L β ) ⊂ [ L β ϕ − 1 , L − β ϕ − 1 ] + L β ϕ − 1 L − β ϕ − 1 + [ L β ϕ − 1 , L − β ψ − 1 ] + L − β ψ − 1 L β ϕ − 1 ⊂ H Λ α .

On the other hand, by BiHom-associativity, we get

( L β L − β ) H + H ( L β L − β ) = ( L β L − β ) ψ ( ψ − 1 ( H ) ) + ϕ ( ϕ − 1 ( H ) ) ( L β L − β ) ⊂ ϕ ( L β ) ( L − β ψ − 1 ( H ) ) + ( ϕ − 1 ( H ) L β ) ψ ( L − β ) ⊂ L β ϕ − 1 L − β ψ − 1 ⊂ H Λ α .

The proof is completed.□

Theorem 3.7

The following assertions hold.

For any α ∈ Λ , the subalgebra
L Λ α = H Λ α ⊕ V Λ α
of L associated with Λ α is an ideal of L .
If L is simple, then there exists a connection from α to β for any α , β ∈ Λ and H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) .

Proof

Since [ L Λ α , H ] = [ L Λ α , L 0 ] ⊂ V Λ α , taking into account Propositions 3.3 and 3.5, we have
[ L Λ α , L ] = [ L Λ α , H ⊕ ( ⊕ β ∈ Λ α L β ) ⊕ ( ⊕ γ ∉ Λ α L γ ) ] ⊂ L Λ α .
By Propositions 3.3 and 3.6, we get
L Λ α L + L L Λ α = L Λ α ( H ⊕ ( ⊕ β ∈ Λ α L β ) ⊕ ( ⊕ γ ∉ Λ α L γ ) ) + ( H ⊕ ( ⊕ β ∈ Λ α L β ) ⊕ ( ⊕ γ ∉ Λ α L γ ) ) L Λ α ⊂ L Λ α .
And by Proposition 3.4 also have ϕ ( L Λ α ) = L Λ α , ψ ( L Λ α ) = L Λ α . So we conclude that L Λ α is an ideal of L .
The simplicity of L implies L Λ α = L . From here, it is clear that Λ α = Λ and H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) .□

Theorem 3.8

For a vector space complement U of

span K { [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 : α ∈ Λ }

in H, we have

L = U + ∑ [ α ] ∈ Λ / ∼ I [ α ] ,

where any I [ α ] is one of the ideals L Λ α of L described in Theorem 3.7-1, satisfying [ I [ α ] , I [ β ] ] + I [ α ] I [ β ] = 0 , whenever [ α ] ≠ [ β ] .

Proof

By Proposition 3.2, we can consider the quotient set Λ / ∼ ≔ { [ α ] : α ∈ Λ } . Let us denote by I [ α ] ≔ L Λ α . We have I [ α ] is well defined and by Theorem 3.7-1, an ideal of L . Therefore,

L = U + ∑ [ α ] ∈ Λ / ∼ I [ α ] .

By applying Proposition 3.5 we also obtain [ I [ α ] , I [ β ] ] + I [ α ] I [ β ] = 0 if [ α ] ≠ [ β ] .□

Definition 3.9

Let Z ( L ) be the center of L satisfying

Z ( L ) ≔ { x ∈ L : [ x , L ] + x L + L x = 0 } .

Theorem 3.10

If Z ( L ) = 0 and H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) , then L is the direct sum of the ideals given in Theorem 3.7,

L = ⊕ [ α ] ∈ Λ / ∼ I [ α ] .

Furthermore, [ I [ α ] , I [ β ] ] + I [ α ] I [ β ] = 0 , whenever [ α ] ≠ [ β ] .

Proof

Since H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) , we get L = ⊕ [ α ] ∈ Λ / ∼ I [ α ] . To finish, we show the direct character of the sum. Given x ∈ I [ α ] ∩ ∑ [ β ] ∈ Λ / ∼ [ β ] ≠ [ α ] I [ β ] . Since x ∈ I [ α ] , then using again the equation [ I [ α ] , I [ β ] ] + I [ α ] I [ β ] = 0 , for [ α ] ≠ [ β ] , we obtain

x , ∑ [ β ] ∈ Λ / ∼ [ β ] ≠ [ α ] I [ β ] + x ∑ [ β ] ∈ Λ / ∼ [ β ] ≠ [ α ] I [ β ] + ∑ [ β ] ∈ Λ / ∼ [ β ] ≠ [ α ] I [ β ] x = 0 .

In a similar way, since x ∈ ∑ [ β ] ∈ Λ / ∼ [ β ] ≠ [ α ] I [ β ] , we have

[ x , I [ α ] ] + x I [ α ] + I [ α ] x = 0 .

It implies [ x , L ] + x L + L x = 0 , that is, x ∈ Z ( L ) . Thus x = 0 , as desired.□

4 The simple components

In this section, we study if any of the components in the decomposition given in Corollary 3.10 is simple. Under certain conditions we give an affirmative answer.

Observe the grading of I , we have

(4.16) I = ⊕ g ∈ Γ I g = ⊕ g ∈ Γ ( ( I g ∩ H g ) ⊕ ( ⊕ α ∈ Λ ( I g ∩ L α , g ) ) ) .

Lemma 4.1

Let L be a split regular BiHom-Poisson color algebra, suppose

H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) .

If I is an ideal of L such that I ⊂ H , then I ⊂ Z ( L ) .

Proof

Observe that [ I , H ] ⊂ [ H , H ] = 0 and

[ I , ⊕ α ∈ Λ L α ] + I ( ⊕ α ∈ Λ L α ) + ( ⊕ α ∈ Λ L α ) I ⊂ I ∩ ( ⊕ α ∈ Λ L α ) ⊂ H ∩ ( ⊕ α ∈ Λ L α ) = 0 .

Since H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) , by the BiHom-Leibniz color identity and the above observation, that H I + I H = 0 . So I ⊂ Z ( L ) .□

Let us introduce the concepts of root-multiplicativity and maximal length in the framework of split BiHom-Poisson color algebras, in a similar way to the ones for split BiHom-Lie algebras (see [15]). For each g ∈ Γ , we denote by Λ g ≔ { α ∈ Λ : L α , g ≠ 0 } .

Definition 4.2

A split regular BiHom-Poisson color algebra L is root-multiplicative if given α ∈ Λ g i and β ∈ Λ g j , with g i , g j ∈ Γ , such that α + β ∈ Λ , then

[ L α , g i , L β , g j ] + L α , g i L β , g j ≠ 0 .

Definition 4.3

A split regular BiHom-Poisson color algebra L is of maximal length if for any α ∈ Λ g , g ∈ Γ , we have dim L κ α , κ g = 1 for κ ∈ { ± 1 } .

Observe that if L is of maximal length, then equation (4.16) let us assert that given any nonzero ideal I of L then

(4.17) I = ⊕ g ∈ Γ ( ( I g ∩ H g ) ⊕ ( ⊕ α ∈ Λ g I L α , g ) ) ,

where Λ g I ≔ { α ∈ Λ : I g ∩ L α , g ≠ 0 } for each g ∈ Γ .

Theorem 4.4

Let L be a split regular BiHom-Poisson color algebra of maximal length, root multiplicative and with Z ( L ) = 0 . Then L is simple if and only if it has all its nonzero roots connected and H = ∑ α ∈ Λ ( [ L α ψ − 1 , L − α ϕ − 1 ] + L α ψ − 1 L − α ϕ − 1 ) .

Proof

The first implication is Theorem 3.7-2. To prove the converse, consider I a nonzero ideal of L . By Lemma 4.1 and equation (4.17), we can write

I = ⊕ g ∈ Γ ( ( I g ∩ H g ) ⊕ ( ⊕ α ∈ Λ g I L α , g ) ) ,

with Λ g I ⊂ Λ g for any g ∈ Γ and some Λ g I ≠ ∅ . Hence, we may choose α 0 ∈ Λ g I being so

(4.18) 0 ≠ L α 0 , g ⊂ I .

Since ϕ ( I ) = I and ψ ( I ) = I and by making use of Lemma 2.10 we can assert that

(4.19) if α ∈ Λ I , then { α ϕ z 1 ψ z 2 : z 1 , z 2 ∈ Z } ⊂ Λ I .

In particular,

(4.20) { L α 0 ϕ z 1 ψ z 2 , g : z 1 , z 2 ∈ Z } ⊂ I .

Now, let us take any β ∈ Λ satisfying β ∉ { ± α 0 ϕ z 1 ψ z 2 : z 1 , z 2 ∈ Z } . Since α 0 and β are connected, we have a connection { α 1 , … , α k } , k ≥ 2 , from α 0 to β satisfying:

α 1 = a 0 ϕ − n ψ − r , for some n , r ∈ N , α 1 ϕ − 1 + α 2 ψ − 1 ∈ Λ , α 1 ϕ − 2 + α 2 ϕ − 1 ψ − 1 + α 3 ψ − 1 ∈ Λ , ⋯ α 1 ϕ − i + α 2 ϕ − i + 1 ψ − 1 + α 3 ϕ − i + 2 ψ − 1 + ⋯ + α i ϕ − 1 ψ − 1 + α i + 1 ψ − 1 ∈ Λ , ⋯ α 1 ϕ − k + 2 + α 2 ϕ − k + 3 ψ − 1 + α 3 ϕ − k + 4 ψ − 1 + ⋯ + α k − 2 ϕ − 1 ψ − 1 + α k − 1 ψ − 1 ∈ Λ , α 1 ϕ − k + 1 + α 2 ϕ − k + 2 ψ − 1 + α 3 ϕ − k + 3 ψ − 1 + ⋯ + α i ϕ − k + i ψ − 1 + ⋯ + α k − 1 ϕ − 1 ψ − 1 + α k ψ − 1 = ε β ϕ − m ψ − s for some m , s ∈ N and ε ∈ { ± 1 } .

Consider α 1 , α 2 and α 1 ϕ − 1 + α 2 ψ − 1 . Since α 2 ∈ Λ , there exists g 1 ∈ Γ such that L α 2 , g 1 ≠ 0 and so α 2 ∈ Λ g 1 . From here, we have α 1 ∈ Λ g and α 2 ∈ Λ g 1 , such that α 1 ϕ − 1 + α 2 ψ − 1 ∈ Λ g + g 1 . The root-multiplicativity and maximal length of L show 0 ≠ [ L α 1 , g , L α 2 , g 1 ] = L α 1 ϕ − 1 + α 2 ψ − 1 , g + g 1 or 0 ≠ L α 1 , g L α 2 , g 1 + L α 2 , g 1 L α 1 , g = L α 1 ϕ − 1 + α 2 ψ − 1 , g + g 1 . Since 0 ≠ L α 1 , g ⊂ I as a consequence of equation (4.18) we get

0 ≠ L α 1 ϕ − 1 + α 2 ψ − 1 , g + g 1 ⊂ I .

We can argue in a similar way from α 1 ϕ − 1 + α 2 ψ − 1 , α 3 and α 1 ϕ − 2 + α 2 ϕ − 1 ψ − 1 + α 3 ψ − 1 to get

0 ≠ L α 1 ϕ − 2 + α 2 ϕ − 1 ψ − 1 + α 3 ψ − 1 , g 2 ⊂ I ,

for some g 2 ∈ Γ . Following this process with the connection { α 1 , … , α k } , we obtain that

0 ≠ L α 1 ϕ − k + 1 + α 2 ϕ − k + 2 ψ − 1 + ⋯ + α k ψ − 1 , g 3 ⊂ I ,

and so either 0 ≠ L β ϕ − m ψ − s , g 3 ⊂ I or 0 ≠ L − β ϕ − m ψ − s , g 3 ⊂ I for some g 3 ∈ Γ . That is,

0 ≠ L ε β ϕ − m ψ − s , g 3 ⊂ I ,

for some ε ∈ { ± 1 } , some g 3 ∈ Γ .

From equations (4.19)–(4.20), we get either 0 ≠ L α ϕ − z 1 ψ − z 2 , g 3 ⊂ I or 0 ≠ L − α ϕ − z 1 ψ − z 2 , g 3 ⊂ I for some g 3 ∈ Γ . That is,

0 ≠ L ε α ϕ − z 1 ψ − z 2 , g 3 ⊂ I ,

for any α ∈ Λ , some ε ∈ { ± 1 } , some g 3 ∈ Γ .

This can be reformulated by saying that for any α ∈ Λ either { α ϕ − z 1 ψ − z 2 } or { − α ϕ − z 1 ψ − z 2 } is contained in Λ g I .

Taking into account H = ∑ β ∈ Λ ( [ L β ψ − 1 , L − β ϕ − 1 ] + L β ψ − 1 L − β ϕ − 1 ) , we have

(4.21) H ⊂ I .

Now for any α ∈ Λ , since L α = [ H , L α ψ ] by the maximal length of L , equation (4.21) gives L α ⊂ I , and so I = L . Consequently, L is simple.□

Acknowledgement

The authors would like to thank the referee for valuable comments and suggestions proposed in this paper.

Funding information: This research was supported by NNSF of China (No. 11801121), NSF of Heilongjiang province (No. QC2018006) and the Fundamental Research Foundation for Universities of Heilongjiang Province (No. LGYC2018JC002).
Conflict of interest: Authors state no conflict of interest.

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Received: 2020-11-25

Revised: 2021-02-20

Accepted: 2021-02-24

Published Online: 2021-07-22

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2021-0039

Keywords for this article

BiHom-Lie color algebra; BiHom-Poisson algebra; root space; root system

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