Lie coalgebra explained

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 k equipped with a linear mapping

d\colonE\toE\wedgeE

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\wedgeb)=(da)\wedgeb+(-1)^\dega\wedge(db)

) of degree 1 on the exterior algebra of E:

d\colonwedge^\bulletE → wedge^\bullet+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

(\bigwedge^* E, d)

form a cochain complex:

E \xrightarrow{d} E\wedgeE \xrightarrow{d} wedge³E\xrightarrow{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^infty(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

\Omega¹\to\Omega²

, 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

[ ⋅ , ⋅ ]\colonak{g} x ak{g}\toak{g}

which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map

[ ⋅ , ⋅ ]\colon ak{g}\wedgeak{g}\toak{g}

that satisfies the Jacobi identity.

Dually, a Lie coalgebra structure on a vector space E is a linear map

d\colonE\toE ⊗ E

which is antisymmetric (this means that it satisfies

\tau\circd=-d

, where

\tau

is the canonical flip

E ⊗ E\toE ⊗ E

) and satisfies the so-called cocycle condition (also known as the co-Leibniz rule)

\left(d ⊗ id\right)\circd=\left(id ⊗ d\right)\circd+\left(id ⊗ \tau\right)\circ\left(d ⊗ id\right)\circd

Due to the antisymmetry condition, the map

d\colonE\toE ⊗ E

can be also written as a map

d\colonE\toE\wedgeE

The dual of the Lie bracket of a Lie algebra

akg

yields a map (the cocommutator)

[ ⋅ , ⋅ ]^*\colonak{g}^*\to(ak{g}\wedgeak{g})^*\congak{g}^*\wedgeak{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

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,

\begin{align} d^2\alpha(x\wedgey\wedgez)&=

	1
	3

d^{2\alpha(x\wedge}y\wedgez+y\wedgez\wedgex+z\wedgex\wedgey)\\ &=

	1
	3

\left(d\alpha([x,y]\wedgez)+d\alpha([y,z]\wedgex)+d\alpha([z,x]\wedgey)\right), \end{align}

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\wedgey\wedgez)=

	1
	3

\left(\alpha([[x,y],z])+\alpha([[y,z],x])+\alpha([[z,x],y])\right).

Since d² = 0, it follows that

\alpha([[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.