Extensible automorphism

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 $d:E\to E\wedge E$ 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 ∈ 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 d² = 0, i.e., if the graded components of the exterior algebra with derivation $({\bigwedge }^{*}E,d)$ form a cochain complex:

$E{\to }^{d}E\wedge E{\to }^d{\bigwedge }^{3}E{\to }^d\dots$

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 $C^{\mathrm{\infty }}(M)$ (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

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,

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

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

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

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.

References

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