Curvature of a connection

(Redirected from Riemann curvature tensor)
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 $X, Y \in \Gamma$

Note that $R(X,Y)$ itself outputs a linear map $\Gamma(E) \to \Gamma(E)$. 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 $\Omega$, and is defined by:

$\Omega := d\omega + \omega \wedge \omega$

Here, $\omega$ is a matrix of connection forms.

In the linear case

In the special case where $E = TM$ (the case of a linear connection) we get that $X,Y, Z \in \Gamma(TM)$. 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:

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