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 $dcolon\; E\; o\; E\; wedge\; E$ 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, $d(a\; wedge\; b)\; =\; (da)wedge\; b\; +\; (-1)^\{operatorname\{deg\}\; a\}\; a\; wedge(db)$.] of degree 1 on the exterior algebra of "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 form a cochain complex::

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: $d(fg)\; =\; (df)g\; +\; f(dg)\; eq\; f(dg)$ (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 $Omega^1\; o\; Omega^2$, but is also defined for $C^infty(M)\; o\; Omega^1(M)$.

The Lie algebra on the dual

A Lie algebra structure on a vector space is a map $[cdot,cdot]\; colon\; mathfrak\{g\}\; imesmathfrak\{g\}\; omathfrak\{g\}$ which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map $[cdot,cdot]\; colonmathfrak\{g\}\; wedge\; mathfrak\{g\}\; o\; mathfrak\{g\}$ that satisfies the Jacobi identity.

Dually, a Lie coalgebra structure on a vector space is a map $dcolon\; E\; o\; E\; wedge\; E$ which satisfies the cocycle condition. The dual of the Lie bracket yields a map (the cocommutator):$[cdot,cdot]\; ^*colon\; mathfrak\{g\}^*\; o\; (mathfrak\{g\}\; wedge\; mathfrak\{g\})^*\; cong\; mathfrak\{g\}^*\; wedge\; mathfrak\{g\}^*$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. The dual space "E"^* carries the structure of a bracket defined by:α( ["x", "y"] ) = "d"α("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", :$d^2alpha\; (xwedge\; ywedge\; z)\; =\; frac\{1\}\{3\}\; d^2alpha(xwedge\; ywedge\; z\; +\; ywedge\; zwedge\; x\; +\; zwedge\; xwedge\; y)\; =\; frac\{1\}\{3\}\; left(dalpha(\; [x,\; y]\; wedge\; z)\; +\; dalpha(\; [y,\; z]\; wedge\; x)\; +dalpha(\; [z,\; x]\; wedge\; y)\; ight),$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:$d^2alpha\; (xwedge\; ywedge\; z)\; =\; frac\{1\}\{3\}\; left(alpha(\; [x,\; y]\; ,\; z]\; )\; +\; alpha(\; [y,\; z]\; ,\; x]\; )+alpha(\; [z,\; x]\; ,\; y]\; )\; ight).$Since "d"² = 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 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.

Notes