Dual of The Lie Bracket Given a smooth manifold U, we have a map $\wedge^2\Gamma(U,TU)\to \Gamma(U,TU)$ given by $X\wedge Y\mapsto [X,Y]$, where $TU$ denotes the tangent bundle. Is it possible to describe the map $\Gamma(U,T^*U)\to \Gamma(U,\wedge^2 T^*U)$ corresponding to this map. 
 A: Yes. The dual of the Lie bracket is the exterior differential that maps 1-forms into 2-forms. See any good textbook on differential geometry.
A: To expand on Leonid's comment, if $\omega$ is a 1-form and $X,Y$ are vector fields, then
$$ d\omega(X \wedge Y) = X \omega(Y) - Y \omega(X) - \omega([X,Y]). $$
If the first two terms were not there, then one could say, as in Victor's answer, that the exterior derivative is (minus) the transpose of the Lie bracket of vector fields.  The fact that the first two terms are there is symptomatic of Leonid's observation that the Lie bracket is not really a tensorial map.
A: The dual of a Lie bracket is the differential in the corresponding Chevalley-Eilenberg algebra.
Background, formulas, details and examples are at
nLab: Chevalley-Eilenberg algebra.
This makes sense for "Lie bracket" understood in the general sense of Lie algebroids and $L_\infty$-algebras and fully generally for $\infty$-Lie algebroids.
In the case at hand, when regarding $T X$ as a Lie algebroid (instead of regarding $\Gamma(T X)$ as just a Lie algebra) the corresponding CE-algebra is the de Rham complex 
$$
  CE(T X) = (\Omega^\bullet(X), d_{dR})
$$
and the general formula for the dual of a Lie bracket on a Lie algebroid reproduces the familiar formula for the de Rham differential.
