# Lie bialgebroid

In differential geometry, a field in mathematics, a Lie bialgebroid consists of two compatible Lie algebroids defined on dual vector bundles. Lie bialgebroids are the vector bundle version of Lie bialgebras.

## Definition
### Preliminary notions

A Lie algebroid consists of a bilinear skew-symmetric operation $[\cdot ,\cdot ]$ on the sections $\Gamma (A)$ of a vector bundle $A\to M$ over a smooth manifold $M$, together with a vector bundle morphism $\rho :A\to TM$ subject to the Leibniz rule
    $[\phi ,f\cdot \psi ]=\rho (\phi )[f]\cdot \psi +f\cdot [\phi ,\psi ],$
and Jacobi identity
    $[\phi ,[\psi _{1},\psi _{2}]]=[[\phi ,\psi _{1}],\psi _{2}]+[\psi _{1},[\phi ,\psi _{2}]]$
where $\phi ,\psi _{k}$ are sections of $A$ and $f$ is a smooth function on $M$.
The Lie bracket $[\cdot ,\cdot ]_{A}$ can be extended to multivector fields $\Gamma (\wedge A)$ graded symmetric via the Leibniz rule
    $[\Phi \wedge \Psi ,\mathrm {X} ]_{A}=\Phi \wedge [\Psi ,\mathrm {X} ]_{A}+(-1)^{|\Psi |(|\mathrm {X} |-1)}[\Phi ,\mathrm {X} ]_{A}\wedge \Psi$
for homogeneous multivector fields $\phi ,\psi ,X$.
The Lie algebroid differential is an $\mathbb {R}$-linear operator $d_{A}$ on the $A$-forms $\Omega _{A}(M)=\Gamma (\wedge A^{*})$ of degree 1 subject to the Leibniz rule
    $d_{A}(\alpha \wedge \beta )=(d_{A}\alpha )\wedge \beta +(-1)^{|\alpha |}\alpha \wedge d_{A}\beta$
for $A$-forms $\alpha$ and $\beta$. It is uniquely characterized by the conditions
    $(d_{A}f)(\phi )=\rho (\phi )[f]$
and
    $(d_{A}\alpha )[\phi ,\psi ]=\rho (\phi )[\alpha (\psi )]-\rho (\psi )[\alpha (\phi )]-\alpha [\phi ,\psi ]$
for functions $f$ on $M$, $A$-1-forms $\alpha \in \Gamma (A^{*})$ and $\phi ,\psi$ sections of $A$.

### The definition

A Lie bialgebroid consists of two Lie algebroids $(A,\rho _{A},[\cdot ,\cdot ]_{A})$ and $(A^{*},\rho _{*},[\cdot ,\cdot ]_{*})$ on the dual vector bundles $A\to M$ and $A^{*}\to M$, subject to the compatibility
    $d_{*}[\phi ,\psi ]_{A}=[d_{*}\phi ,\psi ]_{A}+[\phi ,d_{*}\psi ]_{A}$
for all sections $\phi ,\psi$ of $A$. Here $d_{*}$ denotes the Lie algebroid differential of $A^{*}$ which also operates on the multivector fields $\Gamma (\wedge A)$.

### Symmetry of the definition

It can be shown that the definition is symmetric in $A$ and $A^{*}$, i.e. $(A,A^{*})$ is a Lie bialgebroid if and only if $(A^{*},A)$ is.

## Examples

* A Lie bialgebra consists of two Lie algebras $({\mathfrak {g}},[\cdot ,\cdot ]_{\mathfrak {g}})$ and $({\mathfrak {g}}^{*},[\cdot ,\cdot ]_{*})$ on dual vector spaces ${\mathfrak {g}}$ and ${\mathfrak {g}}^{*}$ such that the Chevalley–Eilenberg differential $\delta _{*}$ is a derivation of the ${\mathfrak {g}}$-bracket.
* A Poisson manifold $(M,\pi )$ gives naturally rise to a Lie bialgebroid on $TM$ (with the commutator bracket of tangent vector fields) and $T^{*}M$ (with the Lie bracket induced by the Poisson structure). The $T^{*}M$-differential is $d_{*}=[\pi ,\cdot ]$ and the compatibility follows then from the Jacobi identity of the Schouten bracket.

## Infinitesimal version of a Poisson groupoid

It is well known that the infinitesimal version of a Lie groupoid is a Lie algebroid (as a special case, the infinitesimal version of a Lie group is a Lie algebra). Therefore, one can ask which structures need to be differentiated in order to obtain a Lie bialgebroid.

### Definition of Poisson groupoid

A Poisson groupoid is a Lie groupoid $G\rightrightarrows M$ together with a Poisson structure $\pi$ on $G$ such that the graph $m\subset G\times G\times (G,-\pi )$ of the multiplication map is coisotropic. An example of a Poisson-Lie groupoid is a Poisson-Lie group (where $M$ is a point). Another example is a symplectic groupoid (where the Poisson structure is non-degenerate on $TG$).

### Differentiation of the structure

Remember the construction of a Lie algebroid from a Lie groupoid. We take the $t$-tangent fibers (or equivalently the $s$-tangent fibers) and consider their vector bundle pulled back to the base manifold $M$. A section of this vector bundle can be identified with a $G$-invariant $t$-vector field on $G$ which form a Lie algebra with respect to the commutator bracket on $TG$.
We thus take the Lie algebroid $A\to M$ of the Poisson groupoid. It can be shown that the Poisson structure induces a fiber-linear Poisson structure on $A$. Analogous to the construction of the cotangent Lie algebroid of a Poisson manifold there is a Lie algebroid structure on $A^{*}$ induced by this Poisson structure. Analogous to the Poisson manifold case one can show that $A$ and $A^{*}$ form a Lie bialgebroid.

## Double of a Lie bialgebroid and superlanguage of Lie bialgebroids

For Lie bialgebras $({\mathfrak {g}},{\mathfrak {g}}^{*})$ there is the notion of Manin triples, i.e. $c={\mathfrak {g}}+{\mathfrak {g}}^{*}$ can be endowed with the structure of a Lie algebra such that ${\mathfrak {g}}$ and ${\mathfrak {g}}^{*}$ are subalgebras and $c$ contains the representation of ${\mathfrak {g}}$ on ${\mathfrak {g}}^{*}$, vice versa. The sum structure is just
    $[X+\alpha ,Y+\beta ]=[X,Y]_{g}+\mathrm {ad} _{\alpha }Y-\mathrm {ad} _{\beta }X+[\alpha ,\beta ]_{*}+\mathrm {ad} _{X}^{*}\beta -\mathrm {ad} _{Y}^{*}\alpha$.

### Courant algebroids

It turns out that the naive generalization to Lie algebroids does not give a Lie algebroid any more. Instead one has to modify either the Jacobi identity or violate the skew-symmetry and is thus lead to Courant algebroids.

### Superlanguage

The appropriate superlanguage of a Lie algebroid $A$ is $\Pi A$, the supermanifold whose space of (super)functions are the $A$-forms. On this space the Lie algebroid can be encoded via its Lie algebroid differential, which is just an odd vector field.
As a first guess the super-realization of a Lie bialgebroid $(A,A^{*})$ should be $\Pi A+\Pi A^{*}$. But unfortunately $d_{A}+d_{*}|\Pi A+\Pi A^{*}$ is not a differential, basically because $A+A^{*}$ is not a Lie algebroid. Instead using the larger N-graded manifold $T^{*}[2]A[1]=T^{*}[2]A^{*}[1]$ to which we can lift $d_{A}$ and $d_{*}$ as odd Hamiltonian vector fields, then their sum squares to $0$ iff $(A,A^{*})$ is a Lie bialgebroid.