Curvature of a connection

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

A connected differential manifold $M$
A vector bundle $E$ over $M$
A connection $\nabla$ for $E$

Definition part

The curvature of $\nabla$ is defined as the map:

$R (X, Y) = \nabla_{X} \circ \nabla_{Y} - \nabla_{Y} \circ \nabla_{X} - \nabla_{[X, Y]}$

where $X, Y \in Γ$

Note that $R (X, Y)$ itself outputs a linear map $Γ (E) \to Γ (E)$ . We can thus write this as:

$R (X, Y) Z = \nabla_{X} (\nabla_{Y} Z) - \nabla_{Y} (\nabla_{X} Z) - \nabla_{[X, Y]} Z$

In local coordinates

Further information: curvature matrix of a connection

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

$Ω : = d ω + ω \land ω$

Here, $Ω$ is a matrix of connection forms.

In the linear case

In the special case where $E = T M$ (the case of a linear connection) we get that $X, Y, Z \in Γ (T M)$ . 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 $C^{\infty}$ -linearity in all arguments, via a computation.

Antisymmetry

Further information: Curvature is antisymmetric in first two variables

We have the following identity:

$R (X, Y) = - R (Y, X)$