Cocommutator Map

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 ''$\mathbb$'' 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 ''$a,b \in E$'' which are homogeneous elements, $d(a \wedge b) = (da)\wedge b + (-1)^ a \wedge(db)$) of degree 1 on the exterior algebra of ''$E$'':
:$d\colon \bigwedge^\bullet E\rightarrow \bigwedge^ E.$
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 $(\bigwedge^* E, d)$ form a cochain complex:
:$E\ \xrightarrow\ E\wedge E\ \xrightarrow\ \bigwedge^3 E\xrightarrow\ \cdots$

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 ''$\mathbb$''), the de Rham complex of differential forms on a manifold form a Lie coalgebra (over the base field ''$\mathbb$''). 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,\cdotcolon \mathfrak\times\mathfrak\to\mathfrak which is skew-symmetric, and satisfies the Jacobi identity. Equivalently, a map cdot,\cdotcolon
\mathfrak \wedge \mathfrak \to \mathfrak 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'')

:$\left(d\otimes \mathrm\right)\circ d = \left(\mathrm\otimes d\right)\circ d+\left(\mathrm \otimes \tau\right)\circ\left(d\otimes \mathrm\right)\circ d$.

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^* \to (\mathfrak \wedge \mathfrak)^* \cong \mathfrak^* \wedge \mathfrak^*
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

$\alpha($, y = d\alpha(x \wedge y), for all ''$\alpha \in E$'' and ''$x, y \in E^*$''.

We show that this endows ''$E^*$'' with a Lie bracket. It suffices to check the Jacobi identity. For any ''$x,y,z \in E^*$'' and ''$\alpha \in E$'',
:$\begin d^2\alpha (x\wedge y\wedge z) &= \frac d^2\alpha(x\wedge y\wedge z + y\wedge z\wedge x + z\wedge x\wedge y) \\ &= \frac \left(d\alpha($, ywedge z) + d\alpha(, zwedge x) +d\alpha(, xwedge y)\right),
\end
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 \left(\alpha( x, y z]) + \alpha( y, z x])+\alpha( z, x y])\right).
Since ''$d^2=0$'', 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 (algebraic topology), cocycle condition ''$d^2=0$'' is in a sense dual to the Jacobi identity.

References

*{{Citation , last1=Michaelis , first1=Walter , title=Lie coalgebras , doi=10.1016/0001-8708(80)90056-0 , doi-access=free , mr=594993 , year=1980 , journal=Advances in Mathematics , issn=0001-8708 , volume=38 , issue=1 , pages=1–54

Coalgebras
Lie algebras