coboundary definition of exterior derivative
Let be a smooth manifold, and
let denote the Lie-algebra of smooth vector fields;
where is a list of vector fields. Recall also that is a module. The action is given by a directional derivative, and takes the form
With these preliminaries out of the way, we have the following description of the exterior derivative operator . For , we have
where indicates the omission of the argument .
The above expression (1) of can be taken as the definition of the exterior derivative. Letting the arguments be coordinate vector fields, it is not hard to show that the above definition is equivalent to the usual definition of as a derivation of the exterior algebra of differential forms, or the local coordinate definition of . The nice feature of (1) is that it is equivalent to the definition of the coboundary operator for Lie algebra cohomology. Thus, we see that de Rham cohomology, which is the cohomology of the cochain complex , is just zeroth-order Lie algebra cohomology of with coefficients in . The bit about “zeroth order” means that we are considering cochains that are zeroth order differential operators of their arguments — in other words, differential forms.
|Title||coboundary definition of exterior derivative|
|Date of creation||2013-03-22 15:38:06|
|Last modified on||2013-03-22 15:38:06|
|Last modified by||rmilson (146)|