_{1}

^{*}

The present paper is a continuation of [1], where we considered braided infinitesimal Hopf algebras (
*i.e.*, infinitesimal Hopf algebras in the Yetter-Drin feld category
for any Hopf algebra
*H*), and constructed their Drinfeld double as a generalization of Aguiar’s result. In this paper we mainly investigate the necessary and sufficient condition for a braided infinitesimal bialgebra to be a braided Lie bialgebra (
*i.e.*, a Lie bialgebra in the category
).

An infinitesimal bialgebra is a triple ( A , m , Δ ) , where ( A , m ) is an associative algebra (possibly without unit), ( A , Δ ) is a coassociative coalgebra (possibly without counit) such that

Δ ( x y ) = x y 1 ⊗ y 2 + x 1 ⊗ x 2 y , x , y ∈ A .

Infinitesimal bialgebras were introduced by Joni and Rota in [

One of the motivations of studying infinitesimal bialgebras is that they are closely related to Drinfeld’s Lie bialgebras (see [

Motivated by [

To give a positive answer to the question above, we organize this paper as follows.

In Section 1, we recall some basic definitions about Yetter-Drinfeld modules and braided Lie bialgerbas. In Section 2, we introduce the notion of the balanceator of a braided infinitesimal bialgerba and show that a braided infinitesimal bialgerba gives rise to a braided Lie bialgerba if and only if the balanceator is symmetric (see Theorem 2.3).

In this paper, k always denotes a fixed field, often omitted from the notation. We use Sweedler’s ( [

A left-left Yetter-Drinfeld module M is both a left H-module and a left H- comodule satisfying the compatibility condition

h 1 m ( − 1 ) ⊗ h 2 ⋅ m 0 = ( h 1 ⋅ m ) ( − 1 ) h 2 ⊗ ( h 1 ⋅ m ) 0 (2.1)

for all h ∈ H and m ∈ M . The equation (1.1) is equivalent to

ρ ( h ⋅ m ) = h 1 m ( − 1 ) S ( h 3 ) ⊗ ( h 2 ⋅ m 0 ) . (2.2)

By [

C M , N ( m ⊗ n ) = m ( − 1 ) ⋅ n ⊗ m ( 0 ) ,

for all m ∈ M ∈ Y H H D and n ∈ N ∈ Y H H D .

Let A be an object in Y H H D , the braiding τ is called symmetric on A if the following condition holds:

( ( a ( − 1 ) ⋅ b ) ( − 1 ) ⋅ a 0 ) ⊗ ( a ( − 1 ) ⋅ b ) 0 = a ⊗ b , (2.3)

which is equivalent to the following condition:

a ( − 1 ) ⋅ b ⊗ a 0 = b 0 ⊗ S − 1 ( b ( − 1 ) ) ⋅ a , (2.4)

for any a , b ∈ A .

In the category Y H H D , we call an (co)algebra simply if it is both a left H- module (co)algebra and a left H-comodule (co)algebra. For more details about (co)module-(co)algebras, the reader can refer to [

A braided Lie algebra ( [

(1) H-anti-commutativity: [ l , l ′ ] = − [ l ( − 1 ) ⋅ l ′ , l 0 ] , l , l ′ ∈ L ,

(2) H-Jacobi identity:

{ l ⊗ l ′ ⊗ l ″ } + { ( τ ⊗ 1 ) ( 1 ⊗ τ ) ( l ⊗ l ′ ⊗ l ″ ) } + { ( 1 ⊗ τ ) ( τ ⊗ 1 ) ( l ⊗ l ′ ⊗ l ″ ) } = 0 ,

for all l , l ′ , l ″ ∈ L , where { l ⊗ l ′ ⊗ l ″ } denotes [ l , [ l ′ , l ″ ] ] and τ the braiding for L.

Let A be an associative algebra in Y H H D . Assume that the braiding is symmetric on A. Define

[ a , b ] = a b − ( a ( − 1 ) ⋅ b ) a 0 , a , b ∈ A .

Then ( A , [ , ] ) is a braided Lie algebra (see [

A braided Lie coalgebra ( [

(1) H-anti-cocommutativity: δ = − τ δ ,

(2) H-coJacobi identity:

( i d + ( i d ⊗ τ ) ( τ ⊗ i d ) + ( τ ⊗ i d ) ( i d ⊗ τ ) ) ( i d ⊗ δ ) δ = 0,

where τ denotes the braiding for L.

Dually, let ( C , Δ ) be a coassociative coalgebra in Y H H D . Assume that the braiding on C is symmetric. Define δ : C → C ⊗ C , by

c ↦ c 1 ⊗ c 2 − c 1 ( − 1 ) ⋅ c 2 ⊗ c 10 , c ∈ C .

Then ( C , δ ) is a braided Lie coalgebras in Y H H D (see [

A braided Lie bialgebra ( [

δ [ x , y ] = ( ( [ , ] ⊗ i d ) ( i d ⊗ δ ) + ( i d ⊗ [ , ] ) ( τ ⊗ i d ) ( i d ⊗ δ ) ) ( i d ⊗ i d − τ ) ( x ⊗ y ) , x , y ∈ L ,

where τ denotes the braiding for L.

In this section, we will study the relation between braided infinitesimal bialge- bras and braided Lie bialgebras as a generalization of Aguiar’s result in [

Let ( A , m , Δ ) be a braided e-bialgebra in Y H H D . For any x , y , z ∈ A , define an action of A on A ⊗ A by

x → ( y ⊗ z ) = x y ⊗ z − x ( − 1 ) ⋅ y ⊗ ( x 0 ( − 1 ) ⋅ z ) x 00 .

Then the action ® is a morphism in Y H H D . In fact, for any x , y , z ∈ A and h ∈ H , we have

h 1 ⋅ x → h 2 ⋅ ( y ⊗ z ) = h 1 ⋅ x → ( h 2 ⋅ y ⊗ h 3 ⋅ z ) = ( h 1 ⋅ x ) ( h 2 ⋅ y ) ⊗ ( h 3 ⋅ z ) − ( h 1 ⋅ x ) ( − 1 ) ⋅ h 2 ⋅ y ⊗ ( ( h 1 ⋅ x ) 0 ( − 1 ) ⋅ h 3 ⋅ z ) ( h 1 ⋅ x ) 00 = ( h 1 ⋅ x ) ( h 2 ⋅ y ) ⊗ ( h 3 ⋅ z ) − h 11 x ( − 1 ) S ( h 13 ) ⋅ h 2 ⋅ y ⊗ ( ( h 12 ⋅ x 0 ) ( − 1 ) ⋅ h 3 ⋅ z ) ( h 12 ⋅ x 0 ) 0 = ( h 1 ⋅ x ) ( h 2 ⋅ y ) ⊗ ( h 3 ⋅ z ) − h 1 x ( − 1 ) ⋅ y ⊗ ( ( h 2 ⋅ x 0 ) ( − 1 ) ⋅ h 3 ⋅ z ) ( h 2 ⋅ x 0 ) 0 = ( h 1 ⋅ x ) ( h 2 ⋅ y ) ⊗ ( h 3 ⋅ z ) − h 1 x ( − 1 ) ⋅ y ⊗ ( h 21 x 0 ( − 1 ) S ( h 23 ) ⋅ h 3 ⋅ z ) ( h 22 ⋅ x 00 ) = ( h 1 ⋅ x ) ( h 2 ⋅ y ) ⊗ ( h 3 ⋅ z ) − h 1 x ( − 1 ) ⋅ y ⊗ ( h 2 x 0 ( − 1 ) ⋅ z ) ( h 3 ⋅ x 00 ) = h 1 ⋅ ( x y ) ⊗ ( h 2 ⋅ z ) − h 1 x ( − 1 ) ⋅ y ⊗ h 2 ⋅ ( ( x 0 ( − 1 ) ⋅ z ) x 00 ) = h ⋅ ( x y ⊗ z − x ( − 1 ) ⋅ y ⊗ ( x 0 ( − 1 ) ⋅ z ) x 00 ) .

So ® is left H-linear. To show the left H-colinearity of the action ®, we compute

ρ ( x → ( y ⊗ z ) ) = ρ ( x y ⊗ z − x ( − 1 ) ⋅ y ⊗ ( x 0 ( − 1 ) ⋅ z ) x 00 ) = x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 y 0 ⊗ z 0 − ( x ( − 1 ) ⋅ y ) ( − 1 ) ( x 0 ( − 1 ) ⋅ z ) ( − 1 ) x 00 ( − 1 ) ⊗ ( x ( − 1 ) ⋅ y ) 0 ⊗ ( x 0 ( − 1 ) ⋅ z ) 0 x 000 = x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 y 0 ⊗ z 0 − x ( − 11 ) y ( − 1 ) S ( x ( − 13 ) ) x ( − 1 ) 4 z ( − 1 ) S ( x ( − 1 ) 6 ) x ( − 1 ) 7 ⊗ x ( − 1 ) 2 ⋅ y 0 ⊗ ( x ( − 1 ) 5 ⋅ z 0 ) x 0 = x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 y 0 ⊗ z 0 − x ( − 11 ) y ( − 1 ) z ( − 1 ) ⊗ x ( − 1 ) 2 ⋅ y 0 ⊗ ( x ( − 1 ) 3 ⋅ z 0 ) x 0 ,

and

( i d ⊗ → ) ρ ( x ⊗ y ⊗ z ) = ( i d ⊗ → ) ( x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 ⊗ y 0 ⊗ z 0 ) = x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 y 0 ⊗ z 0 − x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 ( − 1 ) ⋅ y 0 ⊗ ( x 00 ( − 1 ) ⋅ z 0 ) x 000 = x ( − 1 ) y ( − 1 ) z ( − 1 ) ⊗ x 0 y 0 ⊗ z 0 − x ( − 1 ) 1 y ( − 1 ) z ( − 1 ) ⊗ x ( − 1 ) 2 ⋅ y 0 ⊗ ( x ( − 1 ) 3 ⋅ z 0 ) x 0 ,

as desired.

Definition 2.1. Let ( A , m , Δ ) be a braided infinitesimal bialgebra and τ the braiding of A. The map B : A ⊗ A → A ⊗ A defined by

B ( x , y ) = x → τ Δ ( y ) + τ ( y → τ Δ ( x ) ) , x , y ∈ A , (3.1)

is called the balanceator of A. The balanceator B is called symmetric if B = B ∘ τ . The braided infinitesimal bialgebra A is called balanced if B ≡ 0 on A.

Condition (2.1) can be written as follows:

B ( x , y ) = x ( y 1 ( − 1 ) ⋅ y 2 ) ⊗ y 10 − x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ⊗ ( x 0 ( − 1 ) ⋅ y 10 ) x 00 + ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ⊗ ( x ( − 1 ) ⋅ y ) 0 x 02 − ( ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 02

Obviously,

B ( x ( − 1 ) ⋅ y , x 0 ) = ( x ( − 1 ) ⋅ y ) ( x 0 ( − 1 ) ⋅ x 02 ) ⊗ x 010 − ( x ( − 1 ) ⋅ y 1 ) x 0 ⊗ y 2 + x ( − 1 ) ⋅ y 1 ⊗ x 0 y 2 − ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 010 ) ( x ( − 1 ) ⋅ y ) 00 .

Lemma 2.2. Let ( A , m , Δ ) be a braided infinitesimal bialgebra and x , y ∈ A . Assume that the braiding τ on A is symmetric. Then the following equations hold:

(1) ( ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ) ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 00 = x 1 ⊗ x 2 y ,

(2) ( ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 010 = x 1 ( − 1 ) ⋅ ( x 2 y ) ⊗ x 10 ,

(3) ( x ( − 1 ) 1 ⋅ y 1 ) ( − 1 ) ⋅ ( ( x ( − 1 ) 2 ⋅ y 2 ) x 0 ) ⊗ ( x ( − 1 ) 1 ⋅ y 1 ) 0 = ( x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ) x 0 ⊗ y 10 .

Proof. (1) Since the braiding τ on A is symmetric, for all x , y ∈ A , we have ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 0 ⊗ ( x ( − 1 ) ⋅ y ) 0 = x ⊗ y , then

( i d ⊗ m ) ( Δ ⊗ i d ) ( ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 0 ⊗ ( x ( − 1 ) ⋅ y ) 0 ) = ( i d ⊗ m ) ( Δ ⊗ i d ) ( x ⊗ y )

that is,

( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 00 = x 1 ⊗ x 2 y .

So (1) holds.

(2) To show the Equation (2.2), we need the following computation:

( ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 010 = ( ( x 1 ( − 1 ) x 2 ( − 1 ) ⋅ y ) ( − 1 ) x 10 ( − 1 ) ⋅ x 20 ) ( x 1 ( − 1 ) x 2 ( − 1 ) ⋅ y ) 0 ⊗ x 100 = ( x 1 ( − 1 ) 1 x 2 ( − 1 ) 1 y ( − 1 ) S ( x 2 ( − 1 ) 3 ) S ( x 12 ( − 1 ) 3 ) x 10 ( − 1 ) ⋅ x 20 ) ( x 1 ( − 1 ) 2 x 2 ( − 1 ) 2 ⋅ y 0 ) ⊗ x 100 = ( x 1 ( − 1 ) 1 x 2 ( − 1 ) 1 y ( − 1 ) S ( x 2 ( − 1 ) 3 ) ⋅ x 20 ) ( x 1 ( − 1 ) 2 x 2 ( − 1 ) 2 ⋅ y 0 ) ⊗ x 10 = ( x 1 ( − 1 ) 1 ( x 2 ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 20 ) ( x 1 ( − 1 ) 2 ( x 2 ( − 1 ) ⋅ y ) 0 ) ⊗ x 10 = x 1 ( − 1 ) ⋅ ( ( ( x 2 ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 20 ) ( x 2 ( − 1 ) ⋅ y ) 0 ) ⊗ x 10 = x 1 ( − 1 ) ⋅ ( x 2 y ) ⊗ x 10 .

The last equality holds since τ is symmetric on A. Hence (2) holds.

(3) Finally, we check the Equation (2.3) as follows:

( x ( − 1 ) 1 ⋅ y 1 ) ( − 1 ) ( ( x ( − 1 ) 2 ⋅ y 2 ) x 0 ) ⊗ ( x ( − 1 ) 1 ⋅ y 1 ) 0 = ( x ( − 1 ) 11 y 1 ( − 1 ) S ( x ( − 1 ) 13 ) ) ⋅ ( ( x ( − 1 ) 2 ⋅ y 2 ) x 0 ) ⊗ x ( − 1 ) 12 ⋅ y 10 = ( x ( − 1 ) 1 y 1 ( − 1 ) S ( x ( − 1 ) 3 ) ) ⋅ ( ( x ( − 1 ) 4 ⋅ y 2 ) x 0 ) ⊗ x ( − 1 ) 2 ⋅ y 10 = ( x ( − 1 ) 11 y 1 ( − 1 ) 1 S ( x ( − 1 ) 32 ) x ( − 1 ) 4 ⋅ y 2 ) ( x ( − 1 ) 12 y 1 ( − 1 ) 2 S ( x ( − 1 ) 31 ) ⋅ x 0 ) ⊗ x ( − 1 ) 2 ⋅ y 10 = ( x ( − 1 ) 11 y 1 ( − 1 ) 1 ⋅ y 2 ) ( x ( − 1 ) 12 y 1 ( − 1 ) 2 S ( x ( − 1 ) 3 ) ⋅ x 0 ) ⊗ x ( − 1 ) 2 ⋅ y 10 = ( x ( − 1 ) 1 y 1 ( − 1 ) ⋅ y 2 ) ( ( x ( − 1 ) 2 ⋅ y 10 ) ( − 1 ) ⋅ x 0 ) ⊗ ( x ( − 1 ) 2 ⋅ y 10 ) 0 = ( x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ) ( ( x 0 ( − 1 ) ⋅ y 10 ) ( − 1 ) ⋅ x 00 ) ⊗ ( x 0 ( − 1 ) ⋅ y 10 ) 0 = ( x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ) x 0 ⊗ y 10 .

The last equality holds since τ is symmetric on A. Hence (3) holds as required. ,

Therem 2.3. Let ( A , m , Δ ) be a braided infinitesimal bialgebra. Assume that the braiding τ on A is symmetric. Then ( A , [ , ] = m − m τ , δ = Δ − τ Δ ) is a braided Lie bialgebra if and only if B = B ∘ τ .

Proof. Since ( A , m ) is an associative algebra and ( A , Δ ) is a coassociative coalgebra in Y H H D , ( A , [ , ] = m − m τ ) is a braided Lie algebra and ( A , δ = Δ − τ Δ ) is a braided Lie coalgebra. Therefore it remains to check the compatible condition:

δ [ x , y ] = ( ( [ , ] ⊗ i d ) ( i d ⊗ δ ) + ( i d ⊗ [ , ] ) ( τ ⊗ i d ) ( i d ⊗ δ ) ) ( i d ⊗ i d − τ ) ( x ⊗ y ) ,

for all x , y ∈ A . In fact, on the one hand, we have

δ [ x , y ] = δ ( x y − ( x ( − 1 ) ⋅ y ) x 0 ) = ( 1 − τ ) Δ ( x y ) − ( 1 − τ ) Δ ( ( x ( − 1 ) ⋅ y ) x 0 ) = ( 1 − τ ) ( x 1 ⊗ x 2 y + x y 1 ⊗ y 2 ) − ( 1 − τ ) ( ( x ( − 1 ) 1 ⋅ y 1 ) ⊗ ( x ( − 1 ) 2 ⋅ y 2 ) x 0 + ( x ( − 1 ) ⋅ y ) x 01 ⊗ x 02 ) = x 1 ⊗ x 2 y + x y 1 ⊗ y 2 − x 1 ( − 1 ) ⋅ ( x 2 y ) ⊗ x 10 − ( x y 1 ) ( − 1 ) ⋅ y 2 ⊗ ( x y 1 ) 0 − ( x ( − 1 ) ⋅ y ) x 01 ⊗ x 02 − ( x ( − 1 ) 1 ⋅ y 1 ) ⊗ ( x ( − 1 ) 2 ⋅ y 2 ) x 0 + ( x ( − 1 ) 1 ⋅ y 1 ) ( − 1 ) ⋅ ( ( x ( − 1 ) 2 ⋅ y 2 ) x 0 ) ⊗ ( x ( − 1 ) 1 ⋅ y 1 ) 0 + ( ( x ( − 1 ) ⋅ y ) x 01 ) ( − 1 ) ⋅ x 02 ⊗ ( ( x ( − 1 ) ⋅ y ) x 01 ) 0 .

On the other hand, we have

( ( [ , ] ⊗ i d ) ( i d ⊗ δ ) + ( i d ⊗ [ , ] ) ( τ ⊗ i d ) ( i d ⊗ δ ) ) ( i d ⊗ i d − τ ) ( x ⊗ y ) = ( ( [ , ] ⊗ i d ) ( i d ⊗ δ ) + ( i d ⊗ [ , ] ) ( τ ⊗ i d ) ( i d ⊗ δ ) ) ( x y − ( x ( − 1 ) ⋅ y ) x 0 ) = x y 1 ⊗ y 2 − ( x ( − 1 ) ⋅ y 1 ) x 0 ⊗ y 2 − x ( y 1 ( − 1 ) ⋅ y 2 ) ⊗ y 10 + ( x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ) x 0 ⊗ y 10 − ( x ( − 1 ) ⋅ y ) x 01 ⊗ x 02 + ( ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 02 + ( x ( − 1 ) ⋅ y ) ( x 01 ( − 1 ) ⋅ x 02 ) ⊗ x 010 + x ( − 1 ) ⋅ y 1 ⊗ x 0 y 2 − ( ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 010 − x ( − 1 ) ⋅ y 1 ⊗ ( x 0 ( − 1 ) ⋅ y 2 ) x 00 − x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ⊗ x 0 y 10 + x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ⊗ ( x 0 ( − 1 ) ⋅ y 10 ) x 00 − ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ⊗ ( x ( − 1 ) ⋅ y ) 0 x 02 + ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 00 + ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ⊗ ( x ( − 1 ) ⋅ y ) 0 x 010 − ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 010 ) ( x ( − 1 ) ⋅ y ) 00 .

According to Lemma 2.2, we have

x y 1 ⊗ y 2 + ( x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ) x 0 ⊗ y 10 − ( x ( − 1 ) ⋅ y ) x 01 ⊗ x 02 − ( ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 010 − x ( − 1 ) ⋅ y 1 ⊗ ( x 0 ( − 1 ) ⋅ y 2 ) x 00 − x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ⊗ x 0 y 10 + ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 02 ) ( x ( − 1 ) ⋅ y ) 00 + ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ⊗ ( x ( − 1 ) ⋅ y ) 0 x 010 = x y 1 ⊗ y 2 + ( x ( − 1 ) 1 ⋅ y 1 ) ( − 1 ) ⋅ ( ( x ( − 1 ) 2 ⋅ y 2 ) x 0 ) ⊗ ( x ( − 1 ) 1 ⋅ y 1 ) 0 − ( x ( − 1 ) ⋅ y ) x 01 ⊗ x 02 − x 1 ( − 1 ) ⋅ ( x 2 y ) ⊗ x 10 − x ( − 1 ) 1 ⋅ y 1 ⊗ ( x ( − 1 ) 2 ⋅ y 2 ) x 0 − ( x y 1 ) ( − 1 ) ⋅ y 2 ⊗ ( x y 1 ) 0 + x 1 ⊗ x 2 y + ( ( x ( − 1 ) ⋅ y ) x 01 ) ( − 1 ) ⋅ x 02 ⊗ ( ( x ( − 1 ) ⋅ y ) x 01 ) 0 = δ [ x , y ] .

Therefore,

( ( [ , ] ⊗ i d ) ( i d ⊗ δ ) + ( i d ⊗ [ , ] ) ( τ ⊗ i d ) ( i d ⊗ δ ) ) ( i d ⊗ i d − τ ) ( x ⊗ y ) = δ [ x , y ] − x ( y 1 ( − 1 ) ⋅ y 2 ) ⊗ y 10 + ( ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ) ( x ( − 1 ) ⋅ y ) 0 ⊗ x 02 + x ( − 1 ) y 1 ( − 1 ) ⋅ y 2 ⊗ ( x 0 ( − 1 ) ⋅ y 10 ) x 00 − ( x ( − 1 ) ⋅ y ) ( − 1 ) ⋅ x 01 ⊗ ( x ( − 1 ) ⋅ y ) 0 x 02 + ( x ( − 1 ) ⋅ y ) ( x 01 ( − 1 ) ⋅ x 02 ) ⊗ x 010 − ( x ( − 1 ) ⋅ y 1 ) x 0 ⊗ y 2 + x ( − 1 ) ⋅ y 1 ⊗ x 0 y 2 − ( x ( − 1 ) ⋅ y ) ( − 1 ) x 01 ( − 1 ) ⋅ x 02 ⊗ ( ( x ( − 1 ) ⋅ y ) 0 ( − 1 ) ⋅ x 010 ) ( x ( − 1 ) ⋅ y ) 00 = δ [ x , y ] − B ( x , y ) + B ( x ( − 1 ) ⋅ y , x 0 ) = δ [ x , y ] − B ( x , y ) + B ∘ τ ( x , y ) ,

as desired. We complete the proof. ,

Corollary 2.4. Let ( A , m , Δ ) be a braided infinitesimal bialgebra. Assume that the braiding τ on A is symmetric and the balanceator B = 0 . Then ( A , [ , ] = m − m τ , δ = Δ − τ Δ ) is a braided Lie bialgebra.

Proof. Straightforward from Theorem 2.3. ,

Example 2.5. Let q be an 2th root of unit of k and G the cyclic group of order 2 generated by g, H = k G be the group algebra in the usual way. We consider the algebra A 4 = k [ x ] / ( x 4 ) . By [

Δ ( 1 ) = 0 , Δ ( x ) = x ⊗ x 2 − 1 ⊗ x 3 , Δ ( x 2 ) = x 2 ⊗ x 2 , Δ ( x 3 ) = x 3 ⊗ x 2 .

Define the left-H-module action and the left-H-comodule coaction of A by

g i ⋅ x j = q i j x j , ρ ( x j ) = g j ⊗ x j , i = 0 , 1 , j = 0 , 1 , 2 , 3.

It is not hard to check that the multiplication and the comultiplicaition are both H-linear and H-colinear, therefore A 4 is a braided infinitesimal bialgebra. Since B ( x , x ) = 2 x 2 ⊗ x 2 − q x ⊗ x 2 − q x 2 ⊗ x − q x 3 ⊗ x − x ⊗ x 3 and τ ( x ⊗ x ) = ( x ( − 1 ) ⋅ x ) x 0 = ( g ⋅ x ) x = q x ⊗ x , it is clear that B ( x , x ) = B τ ( x , x ) if and only if q = 1 . If q = 1 , it is not hard to check that the balanceator is symmetric on A 4 . By Theorem 2.3, ( A 4 , [ , ] = m − m τ , δ = Δ − τ Δ ) is a braided Lie bialgebra.

Example 2.6. Let q be a 4th root of unit of k. Consider the Hopf algebra H = k G , where G is a cyclic group of order 4 generated by g. Recall from [

Δ ( a b c d ) = ( 0 a 0 c ) ⊗ ( 0 1 0 0 ) − ( 0 1 0 0 ) ⊗ ( c d 0 0 )

and the H-module action, the H-comodule coaction:

g k ⋅ E i j = q 2 k ( i + j ) E i j , ρ ( E i j ) = g 2 ( i + j ) ⊗ E i j , k = 0 , 1 , 2 , 3 , i , j = 1 , 2.

Since

B ( E 11 , E 21 ) = 2 ( E 12 ⊗ E 22 − E 11 ⊗ E 12 ) ,

B ( E 11 ( − 1 ) ⋅ E 21 , E 11 0 ) = B ( E 21 , E 11 ) = 2 ( E 22 ⊗ E 12 − E 11 ⊗ E 11 ) ,

we claim that the balanceator is not symmetric. By Theorem 2.3, ( M 2 ( k ) , [ , ] = m − m τ , δ = Δ − τ Δ ) is not a braided Lie bialgebra, where m is the multiplication of A.

Let A 1 = { ( a b 0 a ) | a , b ∈ k } ⊂ M 2 ( k ) . It is clear that A 1 is both H-stable and

H-costable, hence A 1 is also a braided infinitesimal bialgebra contained in A. One can check easily that the balanceator B = 0 on A 1 . By Corollary 2.4, ( A 1 , [ , ] = m − m τ , δ = Δ − τ Δ ) is a braided Lie bialgebra.

The paper is partially supported by the China Postdoctoral Science Foundation (No. 2015M571725), the Key University Science Research Project of Anhui Province (Nos. KJ2015A294 and KJ2016A545), the outstanding top-notch talent cultivation project of Anhui Province (No. gxfx2017123) and the NSF of Chuzhou University (No. 2015qd01).

Wang, S.X. (2017) From Braided Infinitesimal Bialgebras to Braided Lie Bialgebras. Advances in Pure Mathematics, 7, 366-374. https://doi.org/10.4236/apm.2017.77023