Lie coalgebra

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In mathematics a Lie coalgebra is the dual structure to a Lie algebra.

In finite dimensions, these are dual objects: the dual vector space to a Lie algebra naturally has the structure of a Lie coalgebra, and conversely.

1 Definition
- 1.1 Relation to de Rham complex
2 The Lie algebra on the dual
3 Notes

[edit] Definition

Let E be a vector space over a field k equipped with a linear mapping $d\colon E \to E \wedge E$ from E to the exterior product of E with itself. It is possible to extend d uniquely to a graded derivation^[1] of degree 1 on the exterior algebra of E:

$d\colon \bigwedge^\bullet E\rightarrow \bigwedge^{\bullet+1} E.$

Then the pair (E, d) is said to be a Lie coalgebra if d² = 0, i.e., if the graded components of the exterior algebra with derivation $(\bigwedge^* E, d)$ form a cochain complex:

$E\ \rightarrow^{\!\!\!\!\!\!d}\ E\wedge E\ \rightarrow^{\!\!\!\!\!\!d}\ \bigwedge^3 E\rightarrow^{\!\!\!\!\!\!d}\ \dots$

[edit] Relation to de Rham complex

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 $C^\infty(M)$ (the error is the Lie derivative), nor is the exterior derivative: $d(fg) = (df)g + f(dg) \neq f(dg)$ (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 $\Omega^1 \to \Omega^2$ , but is also defined for $C^\infty(M) \to \Omega^1(M)$ .

[edit] The Lie algebra on the dual

A Lie algebra structure on a vector space is a map $[\cdot,\cdot]\colon \mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$ which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map $[\cdot,\cdot]\colon \mathfrak{g} \wedge \mathfrak{g} \to \mathfrak{g}$ that satisfies the Jacobi identity.

Dually, a Lie coalgebra structure on a vector space is a map $d\colon E \to E \wedge E$ which satisfies the cocycle condition. The dual of the Lie bracket yields a map (the cocommutator)

$[\cdot,\cdot]^*\colon \mathfrak{g}^* \to (\mathfrak{g} \wedge \mathfrak{g})^* \cong \mathfrak{g}^* \wedge \mathfrak{g}^*$

where the isomorphism $\cong$ 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. 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,

$d^2\alpha (x\wedge y\wedge z) = \frac{1}{3} d^2\alpha(x\wedge y\wedge z + y\wedge z\wedge x + z\wedge x\wedge y) = \frac{1}{3} \left(d\alpha([x, y]\wedge z) + d\alpha([y, z]\wedge x) +d\alpha([z, x]\wedge y)\right),$

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

$d^2\alpha (x\wedge y\wedge z) = \frac{1}{3} \left(\alpha([[x, y], z]) + \alpha([[y, z], x])+\alpha([[z, x], y])\right).$