Curvature form

In differential geometry, the curvature form describes curvature of a connection on a principal bundle. It can be considered as an alternative to or generalization of curvature tensor in Riemannian geometry.

1 Definition
- 1.1 Curvature form in a vector bundle
2 Bianchi identities
3 References
4 See also

Definition

Let G be a Lie group with Lie algebra $\mathfrak g$ , and P → B be a principal G-bundle. Let ω be an Ehresmann connection on P (which is a $\mathfrak g$ -valued one-form on P).

Then the curvature form is the $\mathfrak g$ -valued 2-form on P defined by

$\Omega=d\omega %2B{1\over 2}[\omega,\omega]=D\omega.$

Here $d$ stands for exterior derivative, $[\cdot,\cdot]$ is defined by $[\alpha \otimes X, \beta \otimes Y]�:= \alpha \wedge \beta \otimes [X, Y]_\mathfrak{g}$ and D denotes the exterior covariant derivative. In other terms,

$\,\Omega(X,Y)=d\omega(X,Y) %2B [\omega(X),\omega(Y)].$

Curvature form in a vector bundle

If E → B is a vector bundle. then one can also think of ω as a matrix of 1-forms and the above formula becomes the structure equation:

$\,\Omega=d\omega %2B\omega\wedge \omega,$

where $\wedge$ is the wedge product. More precisely, if $\omega^i_{\ j}$ and $\Omega^i_{\ j}$ denote components of ω and Ω correspondingly, (so each $\omega^i_{\ j}$ is a usual 1-form and each $\Omega^i_{\ j}$ is a usual 2-form) then

$\Omega^i_{\ j}=d\omega^i_{\ j} %2B\sum_k \omega^i_{\ k}\wedge\omega^k_{\ j}.$

For example, for the tangent bundle of a Riemannian manifold, the structure group is O(n) and Ω is a 2-form with values in o(n), the antisymmetric matrices. In this case the form Ω is an alternative description of the curvature tensor, i.e.

$\,R(X,Y)=\Omega(X,Y),$

using the standard notation for the Riemannian curvature tensor,

Bianchi identities

If $\theta$ is the canonical vector-valued 1-form on the frame bundle, the torsion $\Theta$ of the connection form $\omega$ is the vector-valued 2-form defined by the structure equation