Manin triple

Manin triples and Lie bialgebras

There is an equivalence of categories between finite-dimensional Manin triples and finite-dimensional Lie bialgebras.

More precisely, if (\mathfrak{g}, \mathfrak{p}, \mathfrak{q}) is a finite-dimensional Manin triple, then \mathfrak{p} can be made into a Lie bialgebra by letting the cocommutator map \mathfrak{p} \to \mathfrak{p} \otimes \mathfrak{p} be the dual of the Lie bracket \mathfrak{q} \otimes \mathfrak{q} \to \mathfrak{q} (using the fact that the symmetric bilinear form on \mathfrak{g} identifies \mathfrak{q} with the dual of \mathfrak{p}).

Conversely if \mathfrak{p} is a Lie bialgebra then one can construct a Manin triple (\mathfrak{p} \oplus \mathfrak{p}^, \mathfrak{p}, \mathfrak{p}^) by letting \mathfrak{q} be the dual of \mathfrak{p} and defining the commutator of \mathfrak{p} and \mathfrak{q} to make the bilinear form on \mathfrak{g} = \mathfrak{p} \oplus \mathfrak{q} invariant.

Examples

• Suppose that \mathfrak{a} is a complex semisimple Lie algebra with invariant symmetric bilinear form (\cdot,\cdot). Then there is a Manin triple (\mathfrak{g}, \mathfrak{p}, \mathfrak{q}) with \mathfrak{g} = \mathfrak{a} \oplus \mathfrak{a}, with the scalar product on \mathfrak{g} given by ( (w,x),(y,z) ) = (w,y) - (x,z). The subalgebra \mathfrak{p} is the space of diagonal elements (x,x), and the subalgebra \mathfrak{q} is the space of elements (x,y) with x in a fixed Borel subalgebra containing a Cartan subalgebra \mathfrak{h}, y in the opposite Borel subalgebra, and where x and y have the same component in \mathfrak{h}.

References

References

1. (1987). "Quantum groups". [[American Mathematical Society]].
2. Delorme, Patrick. (2001-12-01). "Classification des triples de Manin pour les algèbres de Lie réductives complexes: Avec un appendice de Guillaume Macey". [[Journal of Algebra]].