Curvature of a connection: Difference between revisions

From Diffgeom
 
(3 intermediate revisions by the same user not shown)
Line 19: Line 19:


<math>R(X,Y)Z = \nabla_X (\nabla_Y Z) - \nabla_Y (\nabla_X Z) - \nabla_{[X,Y]}Z</math>
<math>R(X,Y)Z = \nabla_X (\nabla_Y Z) - \nabla_Y (\nabla_X Z) - \nabla_{[X,Y]}Z</math>
===In local coordinates===
{{further|[[curvature matrix of a connection]]}}
Consider a system of local coordinate charts for <math>M</math> such that the vector bundle <math>E</math> is trivial on each chart. For any connection <math>\nabla</math>, we can write a matrix that, in local coordinates, describes the curvature of <math>\nabla</math>. This matrix is sometimes denoted as <math>\Omega</math>, and is defined by:
<math>\Omega := d\omega + \omega \wedge \omega</math>
Here, <math>\omega</math> is a [[matrix of connection forms]].


===In the linear case===
===In the linear case===

Latest revision as of 12:24, 22 May 2008

This curvature is also sometimes known as the Riemann curvature tensor. However, the latter term is usually reserved for situations where we have a linear connection, in particular, the Riemann curvature tensor arising from the Levi-Civita connection for a Riemannian or pseudo-Riemannian manifold

Definition

Given data

Definition part

The curvature of is defined as the map:

where

Note that itself outputs a linear map . We can thus write this as:

In local coordinates

Further information: curvature matrix of a connection

Consider a system of local coordinate charts for such that the vector bundle is trivial on each chart. For any connection , we can write a matrix that, in local coordinates, describes the curvature of . This matrix is sometimes denoted as , and is defined by:

Here, is a matrix of connection forms.

In the linear case

In the special case where (the case of a linear connection) we get that . We can thus think of this map as a (1,3)-tensor because it takes as input three vector fields and outputs one vector field.

Properties

Tensoriality

Further information: Curvature is tensorial

The curvature is tensorial in all three arguments. This is best proved by proving -linearity in all arguments, via a computation.

Antisymmetry

Further information: Curvature is antisymmetric in first two variables

We have the following identity: