Just as the exterior algebra (and tensor algebra) of vector fields on a manifold form a Lie algebra (over the base field K), the de Rham complex of differential forms on a manifold form a Lie coalgebra (over the base field K). Further, there is a pairing between vector fields and differential forms.
However, the situation is subtler: the Lie bracket is not linear over the algebra of smooth functions (the error is the Lie derivative), nor is the exterior derivative: (it is a derivation, not linear over functions): they are not tensors. They are not linear over functions, but they behave in a consistent way, which is not captured simply by the notion of Lie algebra and Lie coalgebra.
Further, in the de Rham complex, the derivation is not only defined for , but is also defined for .
A Lie algebra structure on a vector space is a map which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map that satisfies the Jacobi identity.
Dually, a Lie coalgebra structure on a vector space E is a linear map which is antisymmetric (this means that it satisfies , where is the canonical flip ) and satisfies the so-called cocycle condition (also known as the co-Leibniz rule)
Due to the antisymmetry condition, the map can be also written as a map .
The dual of the Lie bracket of a Lie algebra yields a map (the cocommutator)
where the isomorphism holds in finite dimension; dually for the dual of Lie comultiplication. In this context, the Jacobi identity corresponds to the cocycle condition.
More explicitly, let E be a Lie coalgebra over a field of characteristic neither 2 nor 3. The dual space E* carries the structure of a bracket defined by
α([x, y]) = dα(x∧y), for all α ∈ E and x,y ∈ E*.
We show that this endows E* with a Lie bracket. It suffices to check the Jacobi identity. For any x, y, z ∈ E* and α ∈ E,
where the latter step follows from the standard identification of the dual of a wedge product with the wedge product of the duals. Finally, this gives
Since d2 = 0, it follows that
, for any α, x, y, and z.
Thus, by the double-duality isomorphism (more precisely, by the double-duality monomorphism, since the vector space needs not be finite-dimensional), the Jacobi identity is satisfied.
In particular, note that this proof demonstrates that the cocycle condition d2 = 0 is in a sense dual to the Jacobi identity.