- Lie coalgebra
In
mathematics a Lie coalgebra is the dual structure to aLie algebra .In finite dimensions, these are dual objects: the
dual vector space to aLie 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 theexterior product of "E" with itself. It is possible to extend "d" uniquely to agraded derivation [This means that, for any "a", "b" ∈ "E" which arehomogeneous element s, d(a wedge b) = (da)wedge b + (-1)^{operatorname{deg} a} a wedge(db).] of degree 1 on theexterior algebra of "E"::dcolon igwedge^ullet E ightarrow igwedge^{ullet+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 theexterior algebra with derivation igwedge^* E, d)form acochain complex ::E ightarrow^{!!!!!!d} Ewedge E ightarrow^{!!!!!!d} igwedge^3 E ightarrow^{!!!!!!d} dotsRelation to de Rham complex
Just as the exterior algebra (and tensor algebra) of
vector field s on a manifold form a Lie algebra (over the base field "K"), thede 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 theexterior derivative : d(fg) = (df)g + f(dg) eq f(dg) (it is a derivation, not linear over functions): they are nottensor s. 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
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