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

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Preliminary notions

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A Lie algebroid consists of a bilinear skew-symmetric operation   on the sections   of a vector bundle   over a smooth manifold  , together with a vector bundle morphism   subject to the Leibniz rule

 

and Jacobi identity

 

where   are sections of   and   is a smooth function on  .

The Lie bracket   can be extended to multivector fields   graded symmetric via the Leibniz rule

 

for homogeneous multivector fields  .

The Lie algebroid differential is an  -linear operator   on the  -forms   of degree 1 subject to the Leibniz rule

 

for  -forms   and  . It is uniquely characterized by the conditions

 

and

 

for functions   on  ,  -1-forms   and   sections of  .

The definition

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A Lie bialgebroid consists of two Lie algebroids   and   on the dual vector bundles   and  , subject to the compatibility

 

for all sections   of  . Here   denotes the Lie algebroid differential of   which also operates on the multivector fields  .

Symmetry of the definition

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It can be shown that the definition is symmetric in   and  , i.e.   is a Lie bialgebroid if and only if   is.

Examples

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  1. A Lie bialgebra consists of two Lie algebras   and   on dual vector spaces   and   such that the Chevalley–Eilenberg differential   is a derivation of the  -bracket.
  2. A Poisson manifold   gives naturally rise to a Lie bialgebroid on   (with the commutator bracket of tangent vector fields) and   (with the Lie bracket induced by the Poisson structure). The  -differential is   and the compatibility follows then from the Jacobi identity of the Schouten bracket.

Infinitesimal version of a Poisson groupoid

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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

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A Poisson groupoid is a Lie groupoid   together with a Poisson structure   on   such that the graph   of the multiplication map is coisotropic. An example of a Poisson-Lie groupoid is a Poisson-Lie group (where   is a point). Another example is a symplectic groupoid (where the Poisson structure is non-degenerate on  ).

Differentiation of the structure

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Remember the construction of a Lie algebroid from a Lie groupoid. We take the  -tangent fibers (or equivalently the  -tangent fibers) and consider their vector bundle pulled back to the base manifold  . A section of this vector bundle can be identified with a  -invariant  -vector field on   which form a Lie algebra with respect to the commutator bracket on  .

We thus take the Lie algebroid   of the Poisson groupoid. It can be shown that the Poisson structure induces a fiber-linear Poisson structure on  . Analogous to the construction of the cotangent Lie algebroid of a Poisson manifold there is a Lie algebroid structure on   induced by this Poisson structure. Analogous to the Poisson manifold case one can show that   and   form a Lie bialgebroid.

Double of a Lie bialgebroid and superlanguage of Lie bialgebroids

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For Lie bialgebras   there is the notion of Manin triples, i.e.   can be endowed with the structure of a Lie algebra such that   and   are subalgebras and   contains the representation of   on  , vice versa. The sum structure is just

 .

Courant algebroids

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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.[1]

Superlanguage

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The appropriate superlanguage of a Lie algebroid   is  , the supermanifold whose space of (super)functions are the  -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  . But unfortunately   is not a differential, basically because   is not a Lie algebroid. Instead using the larger N-graded manifold   to which we can lift   and   as odd Hamiltonian vector fields, then their sum squares to   iff   is a Lie bialgebroid.

References

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  1. ^ Z.-J. Liu, A. Weinstein and P. Xu: Manin triples for Lie bialgebroids, Journ. of diff. geom. vol. 45, pp. 547–574 (1997)
  • C. Albert and P. Dazord: Théorie des groupoïdes symplectiques: Chapitre II, Groupoïdes symplectiques. (in Publications du Département de Mathématiques de l’Université Claude Bernard, Lyon I, nouvelle série, pp. 27–99, 1990)
  • Y. Kosmann-Schwarzbach: The Lie bialgebroid of a Poisson–Nijenhuis manifold. (Lett. Math. Phys., 38:421–428, 1996)
  • K. Mackenzie, P. Xu: Integration of Lie bialgebroids (1997),
  • K. Mackenzie, P. Xu: Lie bialgebroids and Poisson groupoids (Duke J. Math, 1994)
  • A. Weinstein: Symplectic groupoids and Poisson manifolds (AMS Bull, 1987),