## Basic properties

[D1 Thm. 1] Let $𝔤$ be a Lie algebra and let $\delta :𝔘𝔤\to 𝔘𝔤\otimes 𝔘𝔤$ be a map that turns $𝔘𝔤$ into a co-Poisson Hpf algebra with the standard multiplication and comultiplication. Then $\delta \left(𝔤\right)\subseteq 𝔤\otimes 𝔤$, and the map $\phi :𝔤\to 𝔤\otimes 𝔤$ indused by $\delta$ turns $𝔤$ into a Lie bialgebra. Conversely, if $\left(𝔤,\phi \right)$ is a Lie bialgebra, then there is a uniqe map $\delta :𝔘𝔤\to 𝔘𝔤\otimes 𝔘𝔤$ which is an extension of $\phi$ and which turns $𝔘𝔤$ into a co-Poisson Hopf algebra.

 Proof: Suppose that $\left(𝔤,\phi \right)$ is a Lie bialgebra. Let $x,y\in 𝔘𝔤$ and let $δx= ∑ x x(1)⊗x(2), δy= ∑ y y(1)⊗y(2), Δx= ∑ x x(1)⊗x(2), Δy= ∑ y y(1)⊗y(2).$ Computation 1. The co-Poisson homomorphism condition gives us that $δxy = δ∘mx⊗y = m⊗m δ⊗x⊗y = m⊗m id⊗σ⊗id Δx⊗δy+ δx⊗Δy = Δxδy +δxΔy = ∑x(1)y(1) ⊗x(2)y(2)+ x(1)y(1)⊗ x(2)y(2)$ Computation 2. Now let us analyse the condition that $\delta$ is a co-Poisson map. $id⊗Δ δx = δ∘mx⊗y = ∑x(1)⊗ x(2)(1) ⊗ x(2)(2).$ On the other hand this must also be equal to $δ12+ δ13∘ Δx = δ12+ δ13 ∑ x(1)⊗x(2) = ∑ x(1)1 ⊗ x(1)2 ⊗x(2)+ x(2)1 ⊗x(1)⊗ x(2)2.$ Computation 3. The following fact will also be useful: $Δxy = Δ∘mx⊗y = m⊗ Δ⊗Δ x⊗y = m⊗m Δ⊗x⊗y.$ $⇒$: Suppose that $\left(𝔤,\phi \right)$ is a Lie bialgebra. We must show that there is a unique extension $\phi :𝔤\to 𝔤\otimes 𝔤$ to a map $\delta :𝔘𝔤\to 𝔘𝔤\otimes 𝔘𝔤$ such that $\delta \circ m=\left(m\otimes m\right)\circ {\delta }^{\otimes }$, $\left(id\otimes \Delta \right)\circ \delta =\left({\delta }^{12}+{\delta }^{13}\right)\circ \Delta$, $\delta$ is a Lie cobracket on $𝔘𝔤$. Computation 1 shows that if $\delta$ satisfies 1) then $\delta \left(xy\right)$ is completely determined by $\Delta \left(x\right)$, $\Delta \left(y\right)$, $\delta \left(x\right)$, $\delta \left(y\right)$ and the multipliation in $𝔘𝔤$. Since the elements of $𝔤$ generate $𝔘𝔤$ (under multiplication) it follows that there is a unique map $\delta :𝔘𝔤\to 𝔘𝔤\otimes 𝔘𝔤$ such that $\delta \left(x\right)=\phi \left(x\right)$ for all $x\in 𝔤$ and such that $\delta$ satisfies 1). 2) Let $x,y=𝔘𝔤$ and let us assume that we know that $id⊗Δδx = δ12+δ13 Δx,and id⊗Δδy = δ12+δ13 Δy,$ Consider the following $id⊗ΔΔxy = id⊗Δ m⊗mδ⊗ x⊗y = m⊗Δ∘m δ⊗xy = m⊗m⊗mΔ⊗ δ⊗x⊗y = m⊗m⊗m id⊗Δ⊗ δ⊗x⊗y.$ On the other hand we have that $δ12+ δ13 Δxy = δ⊗id+ σ⊗id id⊗δ Δ∘mx⊗y = δ⊗id+ σ⊗id id⊗δ m⊗mΔ⊗ x⊗y = δ∘m⊗m + σ⊗id m⊗δ∘m Δ⊗x⊗y = m⊗m∘δ⊗ ⊗m+ σ⊗id m⊗m⊗m∘δ⊗ Δ⊗x⊗y = m⊗m⊗m δ⊗⊗id⊗+ σ⊗⊗id⊗ id⊗∘δ⊗ Δ⊗x⊗y = m⊗m⊗m δ⊗12+ δ⊗13 Δ⊗x⊗y.$ Thus we need only show that $id⊗Δ⊗ δ⊗x⊗y= δ⊗12+ δ⊗13 Δ⊗x⊗y.$ The computation $a⊗b,c⊗d e⊗f = a⊗b,ce⊗df = ace⊗ b,df + a,ce ⊗bdf = ace⊗ b,d f+ace⊗d b,f + a,c e⊗bdf+ c a,e ⊗bdf = ac⊗ b,d e⊗f + c⊗d ae⊗ b,f + a,c ⊗bd e⊗f+ c⊗d a,e ⊗bf$ shows that ${\left\{,\right\}}^{\otimes }$ is a Poisson bracket. Dualizing this computation will show that ${\delta }^{\otimes }\left(x\otimes y\right)$ satisifies the appropriate condition if $\delta \left(x\right)$ and $\delta \left(y\right)$ do. 3) It remains to show that $\delta$ is a Lie cobracket. Since ${\phi }^{\ast }:{𝔤}^{\ast }\otimes {𝔤}^{\ast }\to {𝔤}^{\ast }$ is a Lie bracket on ${𝔤}^{\ast }$, the skew symmetry condition implies that of $\phi \left(𝔤\right)\subseteq 𝔤\wedge 𝔤$. Now assume that $x,y\in 𝔘𝔤$ and that we know that $δx= ∑ x x(1)⊗x(2) = - ∑ x x(2)⊗x(1) ∈𝔘𝔤⊗𝔘𝔤, δy= ∑ y y(1)⊗y(2) = - ∑ y y(2)⊗y(1) ∈𝔘𝔤⊗𝔘𝔤.$ Since $𝔘𝔤$ is cocommutative it follows that $δx= ∑ x x(1)⊗x(2) = ∑ x x(2)⊗x(1), δx= ∑ y y(1)⊗y(2) = ∑ y y(2)⊗y(1).$ Now it follows from computation 1 that $\delta \left(xy\right)\in 𝔘𝔤\otimes 𝔘𝔤$ satisfies the skew-symmetry condition. We must check that $\delta$ satisfies the co-Jacobi identity. Let $x,y\in 𝔘𝔤$ and let us assume that we know that $1+ζ+ζ2 δ⊗idδx=0, and 1+ζ+ζ2 δ⊗idδy=0.$ We need to show that $\left(1+\zeta +{\zeta }^{2}\right)\left(\delta \otimes id\right)\delta \left(xy\right)=0$. It follows from $δ⊗idδ∘m = δ⊗idm⊗m δ⊗ = δ∘m⊗m δ⊗ = m⊗m⊗m δ⊗⊗id⊗δ⊗ δ⊗ = m⊗m⊗m δ⊗⊗id⊗ δ⊗,$ that $id+ζ+ζ2 δ∘m= m⊗m⊗m id⊗+ζ⊗+ ζ⊗2 δ⊗⊗id⊗ δ⊗$ Thus it is sufficient to show that ${\delta }^{\otimes }$ satisfies the co-Jacobi identity. This follows (dually) from the fact that ${\left\{,\right\}}^{\otimes }$ satisfies the Jacobi identity. $⇐$: Suppose that $𝔘𝔤$ is a co-Poisson Hopf algebra with the usual multiplication and comultiplication. We need to show two things: $\delta \left(𝔤\right)\subseteq 𝔤\otimes 𝔤$ $\delta$, restricted to $𝔤$, satisfies the cocycle condition for a bialgebra. 1) First let us show that $\delta \left(1\right)=0$. Since $\Delta \left(1\right)=1\otimes 1$, the co-Poisson homomorphism condition gives that $δ1 = δ1⋅1 = ∑1(1)⊗1(2)+ 1(1)⊗1(2) = 2δ1.$ It follows that $\delta \left(1\right)=0$ if $char\left(k\right)\ne 2$. Now suppose that $x\in 𝔤$. Then $\Delta \left(x\right)=1\otimes x+x\otimes 1$, and computation 2 becomes $δ12+δ13 Δx = δ12+δ13 x⊗1+1⊗x = ∑ x(1)⊗x(2) ⊗1+0+0+ x(1)⊗1⊗x(2) = id⊗Δδx = ∑ x(1)⊗ δx(2)$ It follows that ${x}^{\left(2\right)}$ is a primitive element of $𝔘𝔤$ and thus is an element of $𝔤$. The result follows by noting that $\delta \left(x\right)\in 𝔘𝔤\otimes 𝔘𝔤$ since ${\delta }^{\ast }$ must satisfy the skew symmetry condition for a Lie algebra. 2) Suppose that $x,y\in 𝔤$. Then $\Delta \left(x\right)=x\otimes 1+1\otimes x$ and $\Delta \left(y\right)=y\otimes 1+1\otimes y$. It follows from computation 1 above that $δ [ x , y ] = δxy-yx = ∑x(1)y(1)⊗ x(2)y(2)+ x(1)y(1)⊗ x(2)y(2)- y(1)x(1)⊗ y(2)x(2)- y(1)x(1)⊗ y(2)x(2) = ∑xy(1)⊗y(2) +y(1)⊗xy(2) +x(1)y⊗x(2) +x(1)⊗x(2)y -yx(1)⊗x(2) -x(1)⊗yx(2) -y(1)x⊗y(2) -y(1)⊗y(2)x = ∑ [ x , y(1) ] ⊗y(2) +y(1)⊗ [ x , y(2) ] - [ y , x(1) ]⊗x(2) -x(1)⊗ [ y , x(2) ] = ∑ ad⊗2x y(1)⊗y(2) - ad⊗2y x(1)⊗x(2) = ad⊗2x δy- ad⊗2yδ x.$ So $\delta$ satisfies the $1$-cocycle condition. Thus $\left(𝔤,\delta {\mid }^{𝔤}\right)$ is a bialgebra. $\square$

## References

[Bou] N. Bourbaki, Lie groups and Lie algebras, Chapters I-III, Springer-Verlag, 1989.

[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

The theorem that a Lie bialgebra structure determinse a co-Poisson Hopf algebra structure on its enveloping algebra is due to Drinfel'd and appears as Theorem 1 in the following article.

[D1] V.G. Drinfel'd, Hopf algebras and the quantum Yang-Baxter equation, Soviet Math. Dokl. 32 (1985), 254–258. MR0802128

[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

The following book is a standard introductory book on Lie algebras; a book which remains a standard reference. This book has been released for a very reasonable price by Dover publishers.

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