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Curvature form

## Summary

In differential geometry, the curvature form describes curvature of a connection on a principal bundle. The Riemann curvature tensor in Riemannian geometry can be considered as a special case.

## 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

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

(In another convention, 1/2 does not appear.) 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]

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

where hZ means the horizontal component of Z, on the right we identified a vertical vector field and a Lie algebra element generating it (fundamental vector field), and ${\displaystyle \sigma \in \{1,2\}}$  is the inverse of the normalization factor used by convention in the formula for the exterior derivative.

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.

### 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 E. Cartan:

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

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

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

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 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, 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.

The Bianchi identities can be written in tensor notation as: ${\displaystyle R_{abmn;\ell }+R_{ab\ell m;n}+R_{abn\ell ;m}=0.}$

The contracted Bianchi identities are used to derive the Einstein tensor in the Einstein field equations, the bulk of general theory of relativity.[clarification needed]

## Notes

1. ^ since ${\displaystyle [\omega \wedge \omega ](X,Y)={\frac {1}{2}}([\omega (X),\omega (Y)]-[\omega (Y),\omega (X)])}$ . Here we use also the ${\displaystyle \sigma =2}$  Kobayashi convention for the exterior derivative of a one form which is then ${\displaystyle d\omega (X,Y)={\frac {1}{2}}(X\omega (Y)-Y\omega (X)-\omega ([X,Y]))}$
2. ^ Proof: ${\displaystyle \sigma \Omega (X,Y)=\sigma d\omega (X,Y)=X\omega (Y)-Y\omega (X)-\omega ([X,Y])=-\omega ([X,Y]).}$