# 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.

## Definition

Let $E$ be a vector space over a field $\mathbb {k}$ 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 (this means that, for any $a,b\in E$ which are homogeneous elements, $d(a\wedge b)=(da)\wedge b+(-1)^{\deg a}a\wedge (db)$) 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 $d^{2}=0$, i.e., if the graded components of the exterior algebra with derivation (⋀$^∗$E, d){\textstyle (\bigwedge ^{*}E, d)} form a cochain complex:
    $E\ \xrightarrow {d} \ E\wedge E\ \xrightarrow {d} \ \bigwedge ^{3}E\xrightarrow {d} \ \cdots$

### 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 $\mathbb {k}$), the de Rham complex of differential forms on a manifold form a Lie coalgebra (over the base field $\mathbb {k}$). 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 $C^{\infty }(M)\to \Omega ^{1}(M)$.

## 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 E is a linear map $d\colon E\to E\otimes E$ which is antisymmetric (this means that it satisfies $\tau \circ d=-d$, where $\tau$ is the canonical flip $E\otimes E\to E\otimes E$) and satisfies the so-called cocycle condition (also known as the co-Leibniz rule)
    $\left(d\otimes \mathrm {id} \right)\circ d=\left(\mathrm {id} \otimes d\right)\circ d+\left(\mathrm {id} \otimes \tau \right)\circ \left(d\otimes \mathrm {id} \right)\circ d$.
Due to the antisymmetry condition, the map $d\colon E\to E\otimes E$ can be also written as a map $d\colon E\to E\wedge E$.
The dual of the Lie bracket of a Lie algebra ${\mathfrak {g}}$ 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 over a field of characteristic neither 2 nor 3. The dual space $E^{*}$ carries the structure of a bracket defined by
$\alpha ([x,y])=d\alpha (x\wedge y)$, for all $\alpha \in E$ and $x,y\in E^{*}$.
We show that this endows $E^{*}$ with a Lie bracket. It suffices to check the Jacobi identity. For any $x,y,z\in E^{*}$ and $\alpha \in E$,
    ${\begin{aligned}d^{2}\alpha (x\wedge y\wedge z)&={\frac {1}{3}}d^{2}\alpha (x\wedge y\wedge z+y\wedge z\wedge x+z\wedge x\wedge y)\\&={\frac {1}{3}}\left(d\alpha ([x,y]\wedge z)+d\alpha ([y,z]\wedge x)+d\alpha ([z,x]\wedge y)\right),\end{aligned}}$
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
    $d^{2}\alpha (x\wedge y\wedge z)={\frac {1}{3}}\left(\alpha ([[x,y],z])+\alpha ([[y,z],x])+\alpha ([[z,x],y])\right).$
Since $d^{2}=0$, it follows that
    $\alpha ([[x,y],z]+[[y,z],x]+[[z,x],y])=0$, for any $\alpha$, $x$, $y$, 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 $d^{2}=0$ is in a sense dual to the Jacobi identity.