Lie bialgebroid


A Lie bialgebroid is a mathematical structure in the area of non-Riemannian differential geometry. In brief a Lie bialgebroid are two compatible Lie algebroids defined on dual vector bundles. They form the vector bundle version of a Lie bialgebra.

Definition

Preliminary notions

Remember that a Lie algebroid is defined as a skew-symmetric operation on the sections Γ of a vector bundle A→M over a smooth manifold M together with a vector bundle morphism ρ: A→TM subject to the Leibniz rule
and Jacobi identity
where Φ, ψk are sections of A and f is a smooth function on M.
The Lie bracket A can be extended to multivector fields Γ graded symmetric via the Leibniz rule
for homogeneous multivector fields Φ, Ψ, Χ.
The Lie algebroid differential is an R-linear operator dA on the A-forms ΩA = Γ of degree 1 subject to the Leibniz-rule
for A-forms α and β. It is uniquely characterized by the conditions
and
for functions f on M, A-1-forms α∈Γ and Φ, ψ sections of A.

The definition

A Lie bialgebroid are two Lie algebroids and on dual vector bundles A→M and A*M subject to the compatibility
for all sections Φ, ψ of A.
Here d* denotes the Lie algebroid differential of A* which also operates on the multivector fields Γ.

Symmetry of the definition

It can be shown that the definition is symmetric in A and A*, i.e. is a Lie bialgebroid iff is.

Examples

1. A Lie bialgebra are two Lie algebras and on dual vector spaces g and g* such that the Chevalley–Eilenberg differential δ* is a derivation of the g-bracket.
2. A Poisson manifold gives naturally rise to a Lie bialgebroid on TM and T*M with the Lie bracket induced by the Poisson structure. The T*M-differential is d*= 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. 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 together with a Poisson structure π on G such that the multiplication graph mG×G× is coisotropic. An example of a Poisson Lie groupoid is a Poisson Lie group. Another example is a symplectic groupoid.

Differentiation of the structure

Remember the construction of a Lie algebroid from a Lie groupoid. We take the t-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→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 bialgebroids there is the notion of Manin triples, i.e. c=g+g* can be endowed with the structure of a Lie algebra such that g and g* are subalgebras and c contains the representation of g on g*, vice versa. The sum structure is just

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 ΠA, the supermanifold whose space of 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 should be ΠA+ΠA*. But unfortunately dA +d*|ΠA+ΠA* is not a differential, basically because A+A* is not a Lie algebroid. Instead using the larger N-graded manifold T*A = T*A* to which we can lift dA and d* as odd Hamiltonian vector fields, then their sum squares to 0 iff is a Lie bialgebroid.