Lie bialgebra

In mathematics, a Lie bialgebra is the Lie-theoretic case of a bialgebra: it is a set with a Lie algebra and a Lie coalgebra structure which are compatible.

It is a bialgebra where the multiplication is skew-symmetric and satisfies a dual Jacobi identity, so that the dual vector space is a Lie algebra, whereas the comultiplication is a 1-cocycle, so that the multiplication and comultiplication are compatible. The cocycle condition implies that, in practice, one studies only classes of bialgebras that are cohomologous to a Lie bialgebra on a coboundary.

They are also called Poisson-Hopf algebras, and are the Lie algebra of a Poisson–Lie group.

Lie bialgebras occur naturally in the study of the Yang–Baxter equations.

Definition
A vector space $$\mathfrak{g}$$ is a Lie bialgebra if it is a Lie algebra, and there is the structure of Lie algebra also on the dual vector space $$\mathfrak{g}^*$$ which is compatible. More precisely the Lie algebra structure on $$\mathfrak{g}$$ is given by a Lie bracket $$[\ ,\ ]:\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$$ and the Lie algebra structure on $$\mathfrak{g}^*$$ is given by a Lie bracket $$\delta^*:\mathfrak{g}^* \otimes \mathfrak{g}^* \to \mathfrak{g}^*$$. Then the map dual to $$\delta^*$$ is called the cocommutator, $$\delta:\mathfrak{g} \to \mathfrak{g} \otimes \mathfrak{g}$$ and the compatibility condition is the following cocycle relation:


 * $$\delta([X,Y]) = \left(

\operatorname{ad}_X \otimes 1 + 1 \otimes \operatorname{ad}_X \right) \delta(Y) - \left( \operatorname{ad}_Y \otimes 1 + 1 \otimes \operatorname{ad}_Y \right) \delta(X) $$

where $$\operatorname{ad}_XY=[X,Y]$$ is the adjoint. Note that this definition is symmetric and $$\mathfrak{g}^*$$ is also a Lie bialgebra, the dual Lie bialgebra.

Example
Let $$\mathfrak{g}$$ be any semisimple Lie algebra. To specify a Lie bialgebra structure we thus need to specify a compatible Lie algebra structure on the dual vector space. Choose a Cartan subalgebra $$\mathfrak{t}\subset \mathfrak{g}$$ and a choice of positive roots. Let $$\mathfrak{b}_\pm\subset \mathfrak{g}$$ be the corresponding opposite Borel subalgebras, so that $$\mathfrak{t} = \mathfrak{b}_-\cap\mathfrak{b}_+$$ and there is a natural projection $$\pi:\mathfrak{b}_\pm \to \mathfrak{t}$$. Then define a Lie algebra
 * $$\mathfrak{g'}:=\{ (X_-,X_+)\in \mathfrak{b}_-\times\mathfrak{b}_+\ \bigl\vert\  \pi(X_-)+\pi(X_+)=0\}$$

which is a subalgebra of the product $$\mathfrak{b}_-\times\mathfrak{b}_+$$, and has the same dimension as $$\mathfrak{g}$$. Now identify $$\mathfrak{g'}$$ with dual of $$\mathfrak{g}$$ via the pairing
 * $$ \langle (X_-,X_+), Y \rangle := K(X_+-X_-,Y)$$

where $$Y\in \mathfrak{g}$$ and $$ K $$ is the Killing form. This defines a Lie bialgebra structure on $$\mathfrak{g}$$, and is the "standard" example: it underlies the Drinfeld-Jimbo quantum group. Note that $$\mathfrak{g'}$$ is solvable, whereas $$\mathfrak{g}$$ is semisimple.

Relation to Poisson–Lie groups
The Lie algebra $$\mathfrak{g}$$ of a Poisson–Lie group G has a natural structure of Lie bialgebra. In brief the Lie group structure gives the Lie bracket on $$\mathfrak{g}$$ as usual, and the linearisation of the Poisson structure on G gives the Lie bracket on $$\mathfrak{g^*}$$ (recalling that a linear Poisson structure on a vector space is the same thing as a Lie bracket on the dual vector space). In more detail, let G be a Poisson–Lie group, with $$f_1,f_2 \in C^\infty(G)$$ being two smooth functions on the group manifold. Let $$\xi= (df)_e$$ be the differential at the identity element. Clearly, $$\xi \in \mathfrak{g}^*$$. The Poisson structure on the group then induces a bracket on $$\mathfrak{g}^*$$, as


 * $$[\xi_1,\xi_2]=(d\{f_1,f_2\})_e\,$$

where $$\{,\}$$ is the Poisson bracket. Given $$\eta$$ be the Poisson bivector on the manifold, define $$\eta^R$$ to be the right-translate of the bivector to the identity element in G. Then one has that


 * $$\eta^R:G\to \mathfrak{g} \otimes \mathfrak{g}$$

The cocommutator is then the tangent map:


 * $$\delta = T_e \eta^R\,$$

so that


 * $$[\xi_1,\xi_2]= \delta^*(\xi_1 \otimes \xi_2)$$

is the dual of the cocommutator.