Lie coalgebra

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In mathematics a Lie coalgebra is a way of endowing a vector space with an algebraic structure in such a way that its dual vector space naturally carries the structure of a Lie algebra. Thus, informally, a Lie coalgebra is the dual of a Lie algebra.

It is possible to give a direct definition of a Lie coalgebra without referring to Lie algebras. Let E be a vector space over a field k equipped with a linear mapping d : E → E ∧ E from E to the exterior product of E with itself. It is possible to extend d uniquely to a graded derivation[1] of degree 1 on the exterior algebra of E:

d:\bigwedge^\cdot E\rightarrow \bigwedge^{\cdot+1} E.

Then the pair (E, d) is said to be a Lie coalgebra if d2 = 0.

As an alternative view, regarding the homogeneous components of the mapping d as a sequence,

E\ \rightarrow^{\!\!\!\!\!\!d}\ E\wedge E\ \rightarrow^{\!\!\!\!\!\!d}\ \bigwedge^3 E\rightarrow^{\!\!\!\!\!\!d}\ \dots\qquad\hbox{ } (1)

E is a Lie coalgebra if, and only if, (1) is a cochain complex of vector spaces.

Thus a Lie coalgebra bears a certain formal similarity to the de Rham complex, and to the duality between differential forms and vector fields.[2]

[edit] The Lie algebra on the dual

Let E be a Lie coalgebra. 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} \left(d\alpha([x, y]\wedge z) + d\alpha([y, z]\wedge x) +d\alpha([z, x]\wedge y)\right),

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

α([[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 d2 = 0 is in a sense dual to the Jacobi identity.

[edit] Notes

  1. ^ This means that, for any a, b ∈ E which are homogeneous elements, d(a∧b) = (da)∧b + (-1)deg aa ∧ (db).
  2. ^ Warning: the de Rham complex uses a more subtle definition of d in order to couple the derivational structure of vector fields on the ring of functions with the fact that vector fields are a module under multiplication by functions.
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