Connection vs Curvature

Question

Why is twice a connection usually referred to the curvature: $\overline{\nabla}\circ\nabla=F^\nabla$
Is there an axiomatic definition of curvature, e.g. it is module-linear operator etc?

Answer 1

In broad terms, the curvature is a measure of non-flatness of a connection.

The definition of the curvature depends on the context.

In vector bundles, a connection gives us a way to identify nearby fibers, so we can differentiate sections. Technically, it is easier to define a connection as a way to differentiate sections of a vector bundle.

Definition. A connection $\nabla$ in a vector bundle $E$ over manifold $M$ is a (globally defined) map $$ \nabla : \Gamma(E) \to \Gamma(T^* M) \otimes \Gamma(E) \tag{1} $$ that satisfies the Leibniz rule: $$ \nabla(f \, s) = \mathrm{d}f \otimes s + f \nabla s \tag{2} $$

Asking for the Leibniz rule on tensor products $$ \nabla (s \otimes t) = (\nabla s) \otimes t + s \otimes (\nabla t) \tag{3} $$ and commuting with the contraction $C : T^*M \times TM \to \mathbb{R}$, that is, $$ \nabla \circ C = C \circ \nabla \tag{4} $$ we extend the notion of the connection on all (finite) tensor products of the bundle $E$ and its dual $E^* := \mathrm{Hom}(E,\mathbb{R})$.

To work in more familiar terms we use covariant derivatives.

The covariant derivative $\nabla_X s$ of a section $s \in \Gamma(E)$ along a vector field $X$ is defined as $$ \nabla_X s := C(X, \nabla s) \tag{5} $$

When we try to give a meaning to $\nabla \circ \nabla$ we run into a delicate issue, namely, the above construction is not yet applicable to sections of $\Gamma(T^* M) \otimes \Gamma(E)$ because we also need to specify a way to differentiate covectors.

The notation $(1)$ really means that the connection has values in the bundle of $E$-valued $1$-forms, and $\nabla s$ can be also seen as the exterior covariant derivative $\mathrm{d}^{\nabla} s$ of a $0$-form $s \in \Gamma(E)$. It is instructive to think of $\nabla_X s$ as of a $1$-form $(\nabla s)(X) \equiv (\mathrm{d}^{\nabla} s)(X)$, that is a $E$-valued linear operator on $T M$.

Definition. The exterior covariant derivative of a $1$-form $\omega \in \Gamma(T^* M) \otimes \Gamma(E)$ is defined by $$ \mathrm{d}^{\nabla} \omega (X,Y) = \nabla_X \omega (Y) - \nabla_Y \omega (X) - \omega \tag{6}([X,Y]) $$

This definition makes sense without knowing a connection in tensor bundles (that is, $TM$, $T^* M$ and finite tensor products thereof)!

Definition. The curvature $K^{\nabla}$ of connection $\nabla$ in a vector bundle $E$ is defined on each section $s \in \Gamma(E)$ as the covariant exterior derivative of the $1$-form $\nabla s$, that is $$ K^{\nabla} = \mathrm{d}^{\nabla} \circ \nabla \tag{7} $$

Exercise. Verify that $K^{\nabla}$ is an $\mathrm{End}(E,E)$-valued $2$-form on $M$.

For more details I recommend I. Madsen, J. Tornehave, From Calculus to Cohomology.