Lie coalgebra

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.

## Contents

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

## Definition

Let E be a vector space over a field 𝕜 equipped with a linear mapping d:E->E∧E from E to the exterior product of E with itself. It is possible to extend d uniquely to a graded derivation (this means that, for any a,b∈E which are homogeneous elements, d(a∧b)=(da)∧b+(−1)deg⁡aa∧(db)) of degree 1 on the exterior algebra of E:

d:⋀∙E->⋀∙+1E.

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 (⋀*E,d) form a cochain complex:

E →d E∧E →d ⋀3E→d ⋯

### 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∞(M) (the error is the Lie derivative), nor is the exterior derivative: d(fg)=(df)g+f(dg)≠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 Ω1->Ω2, but is also defined for C∞(M)->Ω1(M).

## The Lie algebra on the dual

A Lie algebra structure on a vector space is a map [⋅,⋅]:𝔤×𝔤->𝔤 which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map [⋅,⋅]:𝔤∧𝔤->𝔤 that satisfies the Jacobi identity.

Dually, a Lie coalgebra structure on a vector space E is a linear map d:E->E⊗E which is antisymmetric (this means that it satisfies τ∘d=−d, where τ is the canonical flip E⊗E->E⊗E) and satisfies the so-called cocycle condition (also known as the co-Leibniz rule)

(d⊗id)∘d=(id⊗d)∘d+(id⊗τ)∘(d⊗id)∘d.

Due to the antisymmetry condition, the map d:E->E⊗E can be also written as a map d:E->E∧E.

The dual of the Lie bracket of a Lie algebra 𝔤 yields a map (the cocommutator)

[⋅,⋅]*:𝔤*->(𝔤∧𝔤)*≅𝔤*∧𝔤*

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,

d2α(x∧y∧z)=13d2α(x∧y∧z+y∧z∧x+z∧x∧y)=13(dα([x,y]∧z)+dα([y,z]∧x)+dα([z,x]∧y)),

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

d2α(x∧y∧z)=13(α([[x,y],z])+α([[y,z],x])+α([[z,x],y])).

Since d2=0, it follows that

α([[x,y],z]+[[y,z],x]+[[z,x],y])=0, 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.

## References

* Michaelis, Walter (1980), "Lie coalgebras", Advances in Mathematics 38 (1): 1–54, doi:10.1016/0001-8708(80)90056-0, ISSN 0001-8708

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