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 $d:E\to E\wedge E$ 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, $d(a\wedge b)=(da)\wedge b+(-1{)}^{\mathrm{deg}a}a\wedge (db)$) of degree 1 on the exterior algebra of $E$:

$d:{\bigwedge }^{\bullet }E\to {\bigwedge }^{\bullet +1}E.$

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 $({\bigwedge }^{*}E,d)$ form a cochain complex:

$E\stackrel{d}{\to }E\wedge E\stackrel{d}{\to }{\bigwedge }^{3}E\stackrel{d}{\to }\cdots$

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

The Lie algebra on the dual

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

Dually, a Lie coalgebra structure on a vector space E is a linear map $d:E\to E\otimes E$ 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)

$(d\otimes \mathrm{i}\mathrm{d})\circ d=(\mathrm{i}\mathrm{d}\otimes d)\circ d+(\mathrm{i}\mathrm{d}\otimes \tau )\circ (d\otimes \mathrm{i}\mathrm{d})\circ d$.

Due to the antisymmetry condition, the map $d:E\to E\otimes E$ can be also written as a map $d:E\to E\wedge E$.

The dual of the Lie bracket of a Lie algebra $\mathfrak{𝔤}$ yields a map (the cocommutator)

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

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 $\alpha \in E$,

$\begin{array}{rl}{d}^2\alpha (x\wedge y\wedge z)& =\frac{1}{3}{d}^2\alpha (x\wedge y\wedge z+y\wedge z\wedge x+z\wedge x\wedge y)\\ & =\frac{1}{3}(d\alpha ([x,y]\wedge z)+d\alpha ([y,z]\wedge x)+d\alpha ([z,x]\wedge y)),\end{array}$

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

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

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.