From Cartan to base forms

Consider a principal -bundle . A differential form on of rank can be described as a local map:

where “local” means that the value of the function at point only depends on the value of at point . In this language, the de Rham differential is:

where is the Lie derivative along of a function . Similarly, a Cartan cocycle provides us with a -invariant (i.e. respecting the action of ) local map:

The connection is a :

where denotes -invariant vector fields on . Generally speaking does not respect the operation of commutator. But the difference between and is necessarily a -invariant vertical vector field. Since we have assumed that acts freely, we can present it in the following form:

where is the curvature of :

and

is the action of . Notice that for all , and in this sense is a form of rank 2. It is essentially a 2-form on with values in the vector bundle . We observe the following:

For every Cartan we can construct the base as follows:

It follows from -invariance of and Eq. (13) that is a -invariant function on , i.e. can be considered as a function on .

The proof that so defined is a closed form goes as follows. We start with the definition of de Rham differential on :

In the formula above we consider a function on . Let us now consider it a -invariant function on , and also use Eq. (12) to present as . We get:

The first line can be presented as follows:

The first term vanishes because we assumed that is a Cartan cocycle; the second term cancels with the second line of Eq. (15) using Eq. (14). This completes the proof of being a closed form.