# Curvature form

(Redirected from Differential Bianchi identity)

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

## Definition

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

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

${\displaystyle \Omega =d\omega +{1 \over 2}[\omega \wedge \omega ]=D\omega .}$

Here ${\displaystyle d}$ stands for exterior derivative, ${\displaystyle [\cdot \wedge \cdot ]}$ is defined in the article "Lie algebra-valued form" and D denotes the exterior covariant derivative. In other terms,[1]

${\displaystyle \,\Omega (X,Y)=d\omega (X,Y)+{1 \over 2}[\omega (X),\omega (Y)]}$

where X, Y are tangent vectors to P.

There is also another expression for Ω: if X, Y are horizontal vector fields on P, then[2]

${\displaystyle 2\Omega (X,Y)=-\omega ([X,Y])=-[X,Y]+h[X,Y]}$

where hZ means the horizontal component of Z and on the right we identified a vertical vector field and a Lie algebra element generating it (fundamental vector field).

A connection is said to be flat if its curvature vanishes: Ω = 0. Equivalently, a connection is flat if the structure group can be reduced to the same underlying group but with the discrete topology. See also: flat vector bundle.

### Curvature form in a vector bundle

If EB 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 of Élie Cartan:

${\displaystyle \,\Omega =d\omega +\omega \wedge \omega ,}$

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

${\displaystyle \Omega _{\ j}^{i}=d\omega _{\ j}^{i}+\sum _{k}\omega _{\ k}^{i}\wedge \omega _{\ j}^{k}.}$

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 the Lie algebra of O(n), i.e. the antisymmetric matrices. In this case the form Ω is an alternative description of the curvature tensor, i.e.

${\displaystyle \,R(X,Y)=\Omega (X,Y),}$

using the standard notation for the Riemannian curvature tensor.

## Bianchi identities

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

${\displaystyle \Theta =d\theta +\omega \wedge \theta =D\theta ,}$

where as above D denotes the exterior covariant derivative.

The first Bianchi identity takes the form

${\displaystyle D\Theta =\Omega \wedge \theta .}$

The second Bianchi identity takes the form

${\displaystyle \,D\Omega =0}$

and is valid more generally for any connection in a principal bundle.

1. ^ since ${\displaystyle [\omega \wedge \omega ](X,Y)={1/2}([\omega (X),\omega (Y)]-[\omega (Y),\omega (X)])}$
2. ^ Proof: ${\displaystyle 2\Omega (X,Y)=2d\omega (X,Y)=X\omega (Y)-Y\omega (X)-\omega ([X,Y])=-\omega ([X,Y]).}$