For preliminary discussion, see Cartan connection applications.

In Riemannian geometry, we can introduce a coordinate system over the Riemannian manifold (at least, over a chart), giving n coordinates

$x_{i}\;\text{,}\qquad i = 1, \dots, n$

for an n-dimensional manifold. Locally, at least, this gives a basis for the 1-forms, dxi where d is the exterior derivative. The dual basis for the tangent space T is ei.

Now, let's choose an orthonormal basis for the fibers of T. The rest is index manipulation.

## Example

Take a 3-sphere with the radius R and give it polar coordinates α, θ, φ.

e(eα)/R,
e(eθ)/R sin(α) and
e(eφ)/R sin(α) sin(θ)

form an orthonormal basis of T.

Call these e1, e2 and e3. Given the metric η, we can ignore the covariant and contravariant distinction for T.

$e_1 = R\, d\alpha$
$e_2 = R\, \sin{(\alpha)} d\theta$
$e_3 = R\, \sin{(\alpha)} \sin{(\theta)} d\phi$.

So,

$de_1=0$
$de_2=R \cos{(\alpha)} d\alpha \wedge d\theta$
$de_3=R (\cos{(\alpha)} \sin{(\theta)} d\alpha \wedge d\phi + \sin{(\alpha)} \cos{(\theta)} d\theta \wedge d\phi)$.

from the relation

$d_\mathbf{A} e = de + A \wedge e = 0$,

we get

$A_{12} = -\cos{(\alpha)} \, d\theta$
$A_{13} = -\cos{(\alpha)} \, \sin{(\theta)} d\phi$
$A_{23} = -\cos{(\theta)} \, d\phi$.

(dAη=0 tells us A is antisymmetric)

So, $\mathbf{F} = d\mathbf{A} + \mathbf{A} \wedge \mathbf{A}$,

$F_{12}=\sin{(\alpha)} d\alpha\wedge d\theta$
$F_{13}=\sin{(\alpha)} \sin{(\theta)} d\alpha\wedge d\phi$
$F_{23}=\sin^2{(\alpha)} \sin{(\theta)} d\theta\wedge d\phi$