## Lie algebras with triangular decomposition

1.1 Let $I$ be a finite set. Let $Q=ℤ\left[I\right]$, ${Q}^{+}=ℕ\left[I\right]$, and ${Q}^{-}=-{Q}^{+}$.
Let $𝔤$ be a Lie algebra that satisfies the following:

1. $𝔤$ is $Q$-graded, i.e., $𝔤$ is a direct sum of spaces ${𝔤}_{\alpha },\alpha \in Q$, and $𝔤α , 𝔤β ⊆ 𝔤α+β$ for all $\alpha ,\beta \in Q$.
2. The Lie algebra $𝔤$ has a triangular decomposition with respect to $Q$, i.e., $𝔤= 𝔫-⊕𝔥⊕𝔫+$ where $𝔫+= ⨁ α∈Q+∖{0} 𝔤α, 𝔫-= ⨁ α∈Q-∖{0} 𝔤-α, 𝔥=𝔤0.$
3. For each $\alpha \in Q$ the vector space ${𝔤}_{\alpha }$ is finite dimensional.
4. For each $i\in I$, $dim\left({𝔤}_{i}\right)=1$.
5. The subspaces ${𝔤}_{i}$, $i\in I$ generate $𝔤$ as a lie algebra.
6. There is a nondegenerate invariant symmetric bilinear form $⟨,⟩:𝔤 × 𝔤\to k$ on $𝔤$ such that
1. The restriciton $⟨,⟩:{𝔤}_{\alpha } × {𝔤}_{-\alpha }\to k$ is nondegenerate for each $\alpha \in Q$ such that ${𝔤}_{\alpha }\ne 0$. In particular $⟨,⟩$ is a nondegenerate form on $𝔥$.
2. If $\alpha ,\beta \in Q$, $\alpha \ne \beta$ and $x\in {𝔤}_{\alpha }$ and $y\in {𝔤}_{\beta }$ then $⟨x,y⟩=0$.
7. There is a linear map $\theta :𝔤\to 𝔤$ called the Chevalley involution, such that
1. $\theta$ is a Lie algebra automorphism.
2. $\theta \left(h\right)=-h$ for all $h\in 𝔥$.
3. $\theta \left({𝔤}_{\alpha }\right)=\theta \left({𝔤}_{-\alpha }\right)$ for all $\alpha \in Q$.
4. $⟨\theta \left(x\right),\theta \left(y\right)⟩=⟨x,y⟩$ for all $x,y\in 𝔤$.

1.2 For each $i\in i$ let us fix an element ${X}_{i}\in {g}_{i}$. Let ${Y}_{i}\in {𝔤}_{-i}$ be dual to ${X}_{i}$ with respect to the form $⟨,⟩$ and let ${H}_{i}=⟨{X}_{i},{Y}_{i}⟩\in 𝔥$.

1.3 We shall let ${𝔤}^{+}$ and ${𝔟}^{-}$ be the Lie subalgebras of $𝔤$ given by $𝔟+=𝔥⊕𝔫+, 𝔟-=𝔥⊕𝔫-.$

1.4 Given an element $x\in 𝔤$ we write $x=x-+x𝔥+x+,$ where ${x}^{-}\in {𝔫}^{-},{x}_{𝔥}\in 𝔥,{x}^{+}\in {𝔫}^{+}$.

1.5 The finite dimensional simple Lie algebras over $ℂ$ and the Kac-Moody Lie algebras ${𝔤}^{\prime }\left(A\right)$ in [Kc] are Lie algebras that satisfy the conditions in (1.1). See Theorems 1.2 and 2.2 of [Kc].

## Lie bialgebra structure on ${𝔟}^{+}$ and $D\left({𝔟}^{+}\right)$

Let $𝔤$ be a Lie algebra with invariang form ${⟨,⟩}_{𝔤}$ satisfying the conditions of Section 1. Let ${⟨,⟩}_{𝔥}$ denote the restriction of the form ${⟨,⟩}_{𝔤}$ to $𝔥$. Given an element $x\in 𝔤$ we write we write $x=x-+x𝔥+x+,$ where ${x}^{-}\in {𝔫}^{-},{x}_{𝔥}\in 𝔥,{x}^{+}\in {𝔫}^{+}$. The triple $𝔤 × 𝔥,𝔭1,𝔭2 ,where 𝔭1= x,x𝔥∣x∈ 𝔟+ and 𝔭2= y,-y𝔥∣ y∈𝔟-$ with the scalar product and bracket on $𝔤 × 𝔥$ given by $⟨ (x,x𝔥) , (y,y𝔥) ⟩ = ⟨ x , y ⟩𝔤- ⟨ x𝔥 , y𝔥 ⟩ 𝔥, and [ ( x,x𝔥 ) , ( y,y𝔥 ) ] = ( [ x , y ], [ h𝔥,y𝔥 ] ) = ( [ x,y ], 0 )$ respectively, is a Manin triple.

 Proof: We must show that ${𝔭}_{1}$ and ${𝔭}_{2}$ are isotropic Lie subalgebras of $𝔤\otimes 𝔥$. $𝔤\otimes 𝔥={𝔭}_{1}\oplus {𝔭}_{2}$. The form $⟨,⟩$ on $𝔤 × 𝔥$ is invariant. The form $⟨,⟩$ on $𝔤 × 𝔥$ is nondegenerate. The proofs of these facts follow in 2.2-2.5. $\square$

2.2 If $x,y\in {𝔟}^{+}$ then $\left[x,y\right]\in {𝔫}^{+}$ and $[ (x,x𝔥),(y,y𝔥) ] = ( [x,y],0 )∈𝔭1$ So ${𝔭}_{1}$ is a Lie subalgebra of $𝔤\otimes 𝔥$. Using the fact that $x,y\in {𝔟}^{+}$, so that $⟨x,y⟩=⟨{x}_{𝔥},{y}_{𝔥}⟩$, We have that $⟨ (x,x𝔥) , (y,y𝔥) ⟩= 0.$ Thus ${𝔭}_{1}$ is isotropic.
Similarly one shows taht ${𝔭}_{2}$ is an isotropic Lie algebra of $𝔤 × 𝔥$.

2.3 If $\left(x,y\right)\in 𝔤 × 𝔥$ then $(a,h)= a++ a𝔥+h 2 , a𝔥+h 2 + a-+ a𝔥-h 2 , -a𝔥+h 2 ∈𝔭1+𝔭2.$ So ${𝔭}_{1}+{𝔭}_{2}=𝔤 × 𝔥$.
If $\left(a,h\right)\in {𝔭}_{1}\cap {𝔭}_{2}$ then $a={a}_{𝔥}$ and $h={a}_{𝔥}$ and $h=-{a}_{𝔥}$. It follows that $a={a}_{𝔥}=h=0$. Thus ${𝔭}_{1}\cap {𝔭}_{2}=0$ and we have that $𝔤 × 𝔥={𝔭}_{1}\oplus {𝔭}_{2}$.

2.4 Recall that for $\left(x,h\right),\left(y,{h}^{\prime }\right)\in 𝔤 × 𝔥$, $⟨\left(x,h\right),\left(y,{h}^{\prime }\right)⟩={⟨x,y⟩}_{𝔤}-⟨h,{h}^{\prime }{⟩}_{𝔥}$. It follows that $⟨ [ (x,h) , (y,h′) ] , (z,h′′) ⟩ = ⟨ [ x , y ] , z ⟩𝔤 - ⟨ [ h , h′ ] , h′′ ⟩ = ⟨ y , [ z , x ] ⟩𝔤 - ⟨ h′ , [ h′′ , h] ⟩𝔥 = ⟨ (y,h′) , ( [ z , x ] , [ h′′ , h ] ) ⟩ = ⟨ (y,h′) , [ ( z , h′′ ) , (x,h) ] ⟩.$ It follows that the form $⟨,⟩$ is an invariant form on $𝔤 × 𝔥$.

2.5 Let $x,h\right)\in 𝔤 × 𝔥$ such that $x,h\right)\ne \left(0,0\right)$. Since $⟨,{⟩}_{𝔥}$ is nondegenerate, if $h\ne 0$, there exists some ${h}^{\prime }\in 𝔥$ such that $⟨h,{h}^{\prime }{⟩}_{𝔥}\ne 0$. Thus if $h\ne 0$, $⟨ (x,h) , (0,h′) ⟩ = - ⟨ h , h′ ⟩𝔥 ≠0.$ If $h=0$ then $x\ne 0$, and by the nondegeneracy of $⟨,{⟩}_{𝔤}$ there exists some $y\in 𝔤$ such that $⟨x,y{⟩}_{𝔤}\ne 0$ . Thus, if $h=0$, $⟨ (x,0) , (y,0) ⟩ = ⟨ x , y ⟩𝔤 ≠0.$ Thus $⟨,⟩$ is a nondegenerate form on $𝔤 × 𝔥$.

2.6 For each $i\in I$ let us fix an element $X\in {𝔤}_{i}$. Let ${Y}_{i}\in {𝔤}_{-i}$ be dual to ${X}_{i}\in {𝔤}_{i}$ with respect to the form $⟨,⟩$ and let ${H}_{i}=\left[{X}_{i},{Y}_{i}\right]\in 𝔥$. The elements ${X}_{i},{Y}_{i},{H}_{i},i\in I$ generate $𝔤$. The manin triple in (2.1) determines a Lie algebra structure on ${𝔟}^{+}$ with cocommututator $\phi :{𝔤}_{A}\to {\bigwedge }^{2}{𝔤}_{A}$ given by $φHi=0, φXi= 12 Xi∧Hi.$

 Proof: Thus the cobracket on ${𝔭}_{1}$ is determined by the equation $⟨ φ((z,z𝔥)) , (y,-y𝔥) ⊗ (f,-f𝔥) ⟩ = ⟨ (z,z𝔥) , [ (y,-y𝔥) , (f,-f𝔥) ] ⟩$ where $y,f\in {𝔟}^{-}$ and $z\in {𝔟}^{+}$. We have $⟨ φ((z,z𝔥)) , (y,-y𝔥) ⊗ (f,-h𝔥) ⟩ = ⟨ (z,z𝔥) , [ (y,-y𝔥) , (f,-f𝔥) ] ⟩ = ⟨ (z,z𝔥) , ( [ y , f ], [ -y𝔥 , -f𝔥 ] ⟩ = ⟨ z , [ y , f ] ⟩ - ⟨ , z𝔥 [ -y𝔥 , -f𝔥 ]⟩.$ Let ${H}_{i}=\left[{X}_{i},{Y}_{i}\right]\in 𝔥$. It follows that $⟨\phi \left(\left({H}_{i},{H}_{i}\right)\right),\left(y,x\right)\otimes \left(f,e\right)⟩=⟨{H}_{i},\left[y,f\right]⟩-⟨{X}_{i},\left[x,e\right]⟩=0-0=0$, since $\left[y,f\right]\in {𝔫}^{-}$ and $\left[x,e\right]\in {𝔫}^{+}$. Thus $φHi,Hi.$ $⟨ φXi,0 , y,-y𝔥⊗ f,-f𝔥 ⟩ = ⟨ Xi , [ y , f ] ⟩- ⟨ 0 , [ -y𝔥 , -f𝔥 ] ⟩ = ⟨ Xi , [ y , f ] ⟩ - ⟨ 0 , [ -y𝔥 , -f𝔥 ] ⟩ = ⟨ Xi , [ y , f ] ⟩ = ⟨ [ Xi , y ] , f ⟩ = ⟨ [ Xi , ⟨ y , Xi ⟩ Yi+y𝔥 ] , f ⟩ = ⟨ y , Xi ⟩ ⟨ [ Xi , Yi ] , f ⟩+ ⟨ [ Xi , y𝔥 ] , f ⟩ = ⟨ y , Xi ⟩ ⟨ Hi , f ⟩ - ⟨ [ Xi , f ] , y𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f ⟩ - ⟨ [ Xi , ⟨ f , Xi ⟩ Yi+f𝔥 ] , y𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f ⟩ - ⟨ f , Xi ⟩ ⟨ [ X , Yi ] , y𝔥 ⟩+ ⟨ [ Xi , f𝔥 ] , y𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f ⟩- ⟨ f , Xi ⟩ ⟨ Hi , y ⟩+0.$ On the other hand, $⟨ Xi,0∧ Hi,Hi , y,-y𝔥 ⟩ = ⟨ Xi,0⊗ Hi,Hi- Hi,Hi⊗ Xi,0 , y,-y𝔥⊗ f,-f𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f𝔥+f𝔥 ⟩ - ⟨ Hi , y+y𝔥 ⟩ ⟨ Xi , f ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f𝔥+f𝔥 ⟩ - ⟨ Hi , y𝔥+y𝔥 ⟩ ⟨ Xi , f ⟩ = 2 ⟨ Xi , y ⟩ ⟨ Hi , f𝔥 ⟩ - ⟨ Hi , y𝔥 ⟩ ⟨ Xi , f ⟩ .$ It follows that $φ Xi,0 = 12 Xi,0∧ Hi,Hi .$ $\square$

The double $D\left({𝔟}^{+}\right)\cong 𝔤 × 𝔥$.

 Proof: This is clear from the form of the Manin triple in (2.1). $\square$

## Quasitriangular Lie bialgebra structure on 𝔤

Let $𝔤$ be a Lie algebra with invariant form $⟨,⟩$ satisfying the conditions of Section 1.

1. There is a Lie bialgebras on $𝔤$ determined by the Manin triple $\left({𝔭}_{1},{𝔭}_{1},{𝔭}_{2}\right)$ given by $𝔭 = 𝔤 × 𝔤, 𝔭1 = x,y∈ 𝔤 × 𝔤A∣ x=y ≅𝔤 𝔭2 = x,y∈ 𝔟- × 𝔟+ ∣ x𝔥+y𝔥=0 ,$ and Lie bracket and invariant scalar product $[ x1,y1 , x2,y2 ] = [ x1 , x2 ] , [ y1 , y2 ] , ⟨ x1,y1 , x2,y2 ⟩ = ⟨ x1 , x2 ⟩- ⟨ y1 , y2 ⟩,$ for all ${x}_{1},{x}_{2},{y}_{1},{y}_{2}\in {𝔤}_{A}$ respectively.
2. The Lie bialgebra structure on $𝔤$ determined by the Manin triple in 1) is given by the cocommutator $\phi :{𝔤}_{A}\to {\bigwedge }^{2}{𝔤}_{A}$ determined by $φHi = 0, φXi = 12 Xi∧Hi, φYi = 12 Yi∧Hi,$ for all $i\in I$.
3. For each $\alpha \in {Q}^{+}\setminus \left\{0\right\}$, let $Xαj∣ 1≤j≤dim𝔤α X-αj∣ 1≤j≤dim𝔤α$ be a basis of ${𝔤}_{\alpha }$ and a dual basis in ${𝔤}_{-\alpha }$. Let $\left\{{\stackrel{˜}{H}}_{i}\right\}$ be an orthonormal basis of $𝔥$. The Lie algebra $𝔤$ is a quasitriangular Lie bialgebra with $r$-matrix given by $r= ∑ α∈Q+∖ 0 ∑ j=1 dim𝔤α Xαj⊗ X-αj + 12 ∑ H ˜ i ⊗ H ˜ i .$

 Proof: We must show that ${𝔭}_{1}$ is an isotropic Lie subalgebra of $𝔤$. ${𝔭}_{2}$ is an isotropic Lie subalgebra of $𝔤$. $𝔭=𝔤 × 𝔤={𝔭}_{1}\oplus {𝔭}_{2}$. The bilinear form $⟨,⟩$ is invariant. The bilinear form $⟨,⟩$ is nondegenerate. These are proved in (3.2)-(3.6). This is proved in (3.7). This follows in (3.8). $\square$

3.3 ${𝔭}_{1}$ is an isotropic Lie subalgebra of $𝔤 × 𝔤$. THe fact that ${𝔭}_{1}$ is an isotropic Lie subalgebra of $𝔤 × 𝔤$ follows from the following computations. $⟨ x,x , y,y ⟩ ⟨ x , y ⟩ - ⟨ x , y ⟩ =0 [ x,x , y,y ]= [ x , y ], [ x , y ] ∈𝔭1.$

3.3 Let $\left(y,x\right),\left(f,e\right)\in {𝔭}_{2}$ so that $y=y-+y𝔥, f=f-+f𝔥, and x=x++x𝔥, e=e++e𝔥,$ and ${x}_{𝔥}=-{y}_{𝔥}$ and ${f}_{𝔥}={e}_{𝔥}$. Then $⟨ y,x , f,e ⟩= ⟨ y , f ⟩ - ⟨ x , e ⟩ = ⟨ y𝔥 , f𝔥 ⟩ - ⟨ x𝔥 , e𝔥 ⟩ = ⟨ y𝔥 , f𝔥 ⟩ - ⟨ -y𝔥 , -f𝔥 ⟩=0.$ Thus ${𝔭}_{2}$ is isotropic. Note that since $\left[y,f\right]\in {𝔫}^{-}$, and $\left[x,e\right]\in {𝔫}^{+}$, it follows that $\left[y,f{\right]}_{𝔥}=\left[x,e{\right]}_{𝔥}=0$, and thus $[ y,x , f,e ]= [ y , f ], [ x , e ] ∈𝔭2.$ Thus ${𝔭}_{2}$ is a Lie subalgebra.

3.4 Let $\left(y,x\right)\in {𝔭}_{1}\cap {𝔭}_{2}$. It follows that $y\in {𝔟}^{-},x\in {𝔟}^{+},y=x$ and ${x}_{𝔥}={y}_{𝔥}$ and ${x}_{𝔥}+{y}_{𝔥}=0$. Thus ${x}^{+}=0$ and ${y}^{-}=0$ giving also that ${x}_{𝔥}={y}_{𝔥}$. Thus if $char\left(k\right)\ne 2$ we have that ${x}_{𝔥}={y}_{𝔥}=x=y=0$. So ${𝔭}_{1}\cap {𝔭}_{2}=0$. Given two elements $x,y\in 𝔤$ we may write $x,y= x+-y++ x𝔥-y𝔥 2 , -x-+ y-- x𝔥-y𝔥 2 + x-+y++ x𝔥+y𝔥 2 , x-+y+- x𝔥+y𝔥 2$ to see that $\left(x,y\right)\in {𝔭}_{1}+{𝔭}_{2}$. It follows that $𝔤×𝔤={𝔭}_{1}+{𝔭}_{2}$.

3.5 Recall that $⟨\left({x}_{1},{x}_{2}\right),\left({x}_{1},{y}_{2}\right)⟩=⟨{x}_{1},{y}_{1}⟩-⟨{x}_{2},{y}_{2}⟩$. It follows that $⟨ [ x1,y1 , x2,y2 ] , x3,y3 ⟩ = ⟨ [ x1 , x2 ] , x3 ⟩- ⟨ [ y1 , y2 ] , y3 ⟩ = ⟨ x2 , [ x3 , x1 ] ⟩- ⟨ y2 , [ y3 , y1 ] ⟩ = ⟨ x2,y2 , [ x3 , x1 ], [ y3 , y1 ] ⟩ = ⟨ x2,y2 , [ x3,y3 , x1,y1 ] ⟩.$ Thus, the form $⟨,⟩$ is invariant.

3.6 Let $\left(x,y\right)\in 𝔤 × 𝔤$ such that $\left(x,y\right)\ne \left(0,0\right)$. If $x\ne 0$ then, since the form on $𝔤$ is nondegenerate there exists some $z\in 𝔤$ such that $\left(x,z\right)\ne 0$. Thus $⟨ x,y , z,0 ⟩= ⟨ x , 0 ⟩- ⟨ y , 0 ⟩= ⟨ x , z ⟩≠0.$ If $x=0$ then $y\ne 0$ and there exists $w\in 𝔤$ such that $⟨y,w⟩\ne 0$. It follows that $⟨ x,y,0,w ⟩= ⟨ x , 0 ⟩ - ⟨ y , w ⟩= -⟨ y , w ⟩ ≠0.$ Thus, the form on $𝔤 × 𝔤$ is nondegenerate.

3.7 The cobracket on ${𝔭}_{1}$ is determined by $⟨ φz,z , y,x⊗ f,e ⟩ = ⟨ z,z , [ y,x , f,e ] ⟩$ where $y,f\in {𝔟}^{-}$ and $x,e\in {𝔟}^{+}$. We have $⟨ φz,z , y,x⊗ f,e ⟩= ⟨ z,z , [ y,x , f,e ] ⟩= ⟨ z,z , [ y , f ], [ x , e ] ⟩= ⟨ z , [ y , f ] ⟩ - ⟨ z , [ x , e ] ⟩.$ Let ${H}_{i}\in 𝔥$. Then, since $\left[y,f\right]\in {𝔫}^{-}$ and $\left[x,e\right]\in {𝔫}^{+}$, it follows that $⟨\phi \left(\left({H}_{i},{H}_{i}\right)\right),\left(y,x\right)\otimes \left(f,e\right)⟩=⟨{H}_{i},\left[y,f\right]⟩-⟨{H}_{i},\left[x,e\right]⟩=0-0=0$. Thus, $φHi,Hi =0.$ The computations to determine $\phi \left({X}_{i}\right)$ are as follows. $⟨ φXi,Xi , y,x⊗ f,e ⟩ = ⟨ Xi , [ y , f ] ⟩- ⟨ Xi , [ x , e ] ⟩ = ⟨ Xi , [ y , f ] ⟩ -0 = ⟨ [ Xi , y ] , f ⟩ = ⟨ [ Xi , ⟨ y , Xi ⟩ Yi ] +y𝔥 ⟩ = ⟨ y , Xi ⟩ ⟨ [ Xi , Yi ] , f ⟩+ ⟨ [ Xi , y𝔥 ] , f ⟩ = ⟨ y , Xi ⟩ ⟨ Hi , f ⟩- ⟨ [ Xi , f ] , y𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f ⟩ - ⟨ [ Xi , ⟨ f , Xi ⟩ Yi+f𝔥 ] , y𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f ⟩- ⟨ f , Xi ⟩ ⟨ [ Xi , Yi ] , y𝔥 ⟩+ ⟨ [ Xi , f𝔥 ] , y𝔥 ⟩ = ⟨ Xi , y ⟩ ⟨ Hi , f ⟩ - ⟨ f , Xi ⟩⟨ Hi , y ⟩+0.$ On the other hand, since ${y}_{𝔥}=-{x}_{𝔥}$ and ${f}_{𝔥}=-{e}_{𝔥}$ it follows that $⟨ Xi,Xi ∧ Hi,Hi , y,x⊗ f,e ⟩ = ⟨ Xi,Xi⊗ Hi,Hi- Hi,Hi⊗ Xi,Xi⊗ , y,x⊗ f,e ⟩ = ⟨ Xi , y-x ⟩ ⟨ Hi , f-e ⟩ - ⟨ Hi , y-x ⟩ ⟨ Xi , f-e ⟩ = ⟨ Xi , y-x ⟩ ⟨ Hi , f𝔥+f𝔥 ⟩- ⟨ Hi , y𝔥+y𝔥 ⟩ ⟨ Xi , f-e ⟩ = 2 ⟨ Xi , y-x ⟩ ⟨ Hi , f ⟩- ⟨ Hi , y ⟩ ⟨ Xi , f-e ⟩ .$ Since ${X}_{i}\in {𝔫}^{+}$ and $x\in {𝔟}^{+}$ it follows that $⟨{X}_{i},x⟩=0$. Similarly, $⟨{X}_{i},e⟩=0$. Thus $⟨ Xi,Xi∧ Hi,Hi , y,x⊗ f,e ⟩= 2 ⟨ Xi , y ⟩ ⟨ Hi , f ⟩- ⟨ Hi , y ⟩ ⟨ Xi , f ⟩ .$ It follows that $φXi,Xi = 12 Xi,Xi∧ Hi,Hi .$ The formula $φYi,Yi = 12 Yi,Yi∧ Hi,Hi .$ is proved similarly.

3.8 We know that $D\left({𝔟}^{+}\right)\cong 𝔤 × 𝔥$ from Corollary (2.7). The set $Xαj,0 ∣ α∈Q+ ∖ 0 , 1≤j≤dim𝔤α ∪ { 1 2 H ˜ i , H ˜ i }$ is a basis of ${𝔭}_{1}$ with dual basis (with respect to the form on $𝔤 × 𝔥$) $X-αj,0 ∣ α∈Q+ ∖ 0 , 1≤j≤dim𝔤α ∪ { 1 2 H ˜ i , - H ˜ i }$ in ${𝔭}_{2}$. The $r$-matrix $r ˆ = ∑ α∈Q+∖ 0 ∑ j=1 dim𝔤α Xαj,0 ⊗ X-αj,0 + 12 ∑ H ˜ i ⊗ H ˜ i$ gives a quasitriangular Lie bialgebra structure on $𝔤 × 𝔥$. The subspace $0 × 𝔥$ of $𝔤 × 𝔥$ is an ideal of $𝔤 × 𝔥$ and $𝔤\cong \left(𝔤 × 𝔥\right)/\left(0 × 𝔥\right)$. Thus the projection $r∈ 𝔤 × 𝔥 / 0 × 𝔥 ⊗ 𝔤 × 𝔥 / 0 × 𝔥 r= = ∑ α∈Q+∖ 0 ∑ j=1 dim𝔤α Xαj,0 ⊗ X-αj,0 + 12 ∑ H i ,0 ⊗ H i ,0$ of the $r$-matrix $\stackrel{ˆ}{r}$ gives a quasitriangular Lie bialgebra structure on $𝔤$.
It remains to check that htis bialgebra structure is the same as that given by the Manin triple in 1). By (2.6) we know that the cocommutator determined by $r$ satisfies $φHi=0 φXi= 12 Xi∧Hi,$ on the subalgebra ${𝔟}^{+}$. We can calculate $\phi \left({Y}_{i}\right)$ by using the Cartan involution $\theta$. $ad⊗2Yi θ⊗θ r = θ⊗θ ad⊗2Xi r = θ × θ 12 Xi∧Hi = - 12 Yi∧Hi.$ Since $θ⊗θr = ∑ α∈Q+ ∑ j=1 dim𝔤α θXαj⊗ θX-αj + 12 ∑ θHi ⊗ θHi = ∑ α∈Q+∖ 0 ∑ j=1 dim𝔤α Xαj⊗ X-αj + 12 ∑ H i ⊗ H i = r21,$ and ${r}^{21}+{r}^{12}$ is invariant, it follows that $ad⊗2 Yi r= - ad⊗2 Yi r21= 12 Yi∧Hi.$ Thus we have shown that the bialgbera structure on $𝔤$ determined by the $r$-matrix and the bialgebra structure on $𝔤$ determined by the Manin triple given in 1) are identical.

## References

The definitions and proofs of the fact that Kac-Moody Lie algebras satisfy the properties given in Section 1 are contained in the first few pages of the following standard book.

[Kc] V. Kac, Infinite dimensional Lie algebras, Third Ed., Cambridge University Press 1990. MR1104219

The examples of bialgebra structures geven here appear in example 3.2 of the following paper by Drinfel'd.

[D] V.G. Drinfeld, Quantum Groups, Vol. 1 of Proccedings of the International Congress of Mathematicians (Berkeley, Calif., 1986). Amer. Math. Soc., Providence, RI, 1987, pp. 198–820. MR0934283

[DHL] H.-D. Doebner, Hennig, J. D. and W. Lücke, Mathematical guide to quantum groups, Quantum groups (Clausthal, 1989), Lecture Notes in Phys., 370, Springer, Berlin, 1990, pp. 29–63. MR1201823

[J] N. Jacobson, Lie algebras, Interscience Publishers, New York, 1962.