On this page:
1 The Cartan complex of
2 Connection in the principle bundle
2.1 Definition of a connection
2.2 Computation of curvature
3 From Cartan model to base forms

Case of coset space

Here we will consider the case of a coset space as an example

    1 The Cartan complex of

    2 Connection in the principle bundle

      2.1 Definition of a connection

      2.2 Computation of curvature

    3 From Cartan model to base forms

Remember that means that we identify and for .

Let be a subspace of which is transversal to and satisfies:

(18)

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 :

(19)

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:

(20)

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:

(21)

Let us define the right-invariant vector fields by the formula:

(22)

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 :

(23)

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:

(24)

(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 .