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:
Now, as a check of Eq. (21), it implies for the right-invariant fields:
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 :
The differentiation over 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 is the coordinate on
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 .