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 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 &isin; 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 d2 = 0, i.e., if the graded components of the exterior algebra with derivation $(\bigwedge^* E, d)$ form a cochain complex:
 * $$E\ \xrightarrow{d}\ E\wedge E\ \xrightarrow{d}\ \bigwedge^3 E\xrightarrow{d}\ \cdots$$

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 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: $$d(fg) = (df)g + f(dg) \neq f(dg)$$ (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&and;y), for all &alpha; &isin; E and x,y &isin; 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\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{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\wedge y\wedge z) = \frac{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 &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 d2 = 0 is in a sense dual to the Jacobi identity.