Case of coset space 
Here we will consider the case of a coset space as an example
Remember that means that we identify 
and 
for 
.
Let
be a subspace of
which is transversal to
and satisfies:
1 The Cartan complex of 
Consider
—
the the right-invariant
-valued 1-form on
with values
in
. Notice that the contraction with
acts as follows:
The action of the Cartan differential on functions
is:
In order for
to be nilpotent, we require the
-invariance of
:
2 Connection in the principle bundle
2.1 Definition of a connection
We can construct a connection in
in the following way. Let us consider
horizonthal those vectors
which are orthogonal to the fibers:
The so defined distribution of horizonthal planes is
-invariant if Eq. (
18) is satisfied. The
connection form is a
-valued form on
.
It computes for every vector its vertical component:
2.2 Computation of curvature
Since the curvature is a tensor, it is enough to compute it on left-invariant
vector fields. For
and
we have the corresponding left-invariant
vector fields
and
, which are defined as follows:
The commutator of the two left-invariant vector fields is given by:
Let us project and then lift:
This can be also written as follows:
We will now compute the curvature using the definition Eq. (
10), in the
following way. Notice that for any two vector fields
and
we have:
and therefore:
Let us define the right-invariant vector fields by the formula:
This implies:
Therefore the curvature on the left-invariant vector fields is determined by
the following formula:
Since the curvature is a tensor, we can now compute it for arbitrary vector
fields. Given two vector fields
and
, we have:
In other words:
where
is defined in Eq. (
22).
3 From Cartan model to base forms
Let us consider a class of the Cartan model represented by
satisfying the conditions described in
the description of Cartan complex of . We will construct
as
follows (with
is defined in Eq. (
22)):
We claim that this is a base form. The horizonthality is obvious, because
the differentials only enter through
. The invariance is also
straightforward using Eq. (
19). It remains to verify that
is closed. This should follow from the general theory using
Eq. (
17). Let us verify it directly. It is useful to interpret
as a differentiation over an odd parameter
:
Now consider this expression:
(the only difference with the RHS of Eq. (
23) being
It can be interpreted as the result of application of the de Rham
to the
equivariant form
, following by the substitution
.
Because
, we know that (
24) is equal to
, with the same substitution for
.
This substitution has the effect of turning
into
and therefore
.
is not a trick, but actually a definition of the de Rham operator
. Indeed, pseudo-differential forms on
are defined as functions on
. Notice that
. Then our
