Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 3

In a local frame of $E$, a connection on a vector bundle has the formwhere $A$ is a matrix of one-forms. Under a frame change $g$, its matrix transforms asThe covariant exterior derivative on an $E$-valued $r$-form $\sigma$ isThe curvature form of a connection isUsing the graded Leibniz rule,

Regard an $E^{\vee }$-valued $r$-form $\lambda$ as a row vector and an $\mathrm{End}(E)$-valued $r$-form $\mu$ as a matrix. The dual connection and endomorphism connection areFor the curvature two-form,because substituting $F=dA+A\wedge A$ makes all terms cancel. This is the Bianchi identity.

For an endomorphism-valued $r$-form $\mu$ and an $E$-valued form $\sigma$, direct expansion gives the compatible Leibniz ruleThe cancellation of the two middle $A$-terms proves the identity.

For a path $\gamma :[0,1]\to B$, a section $s(t)$ of ${\gamma }^{\ast }E$ is parallel whenExistence and uniqueness for this linear ordinary differential equation define the parallel transportTransport along the reversed path solves the inverse initial-value problem, so

If $P(t)$ denotes transport from $0$ to $t$, thensatisfies the horizontal equation for the induced endomorphism connection. Uniqueness therefore gives

Let $b^{\prime }\in B$. Since $B$ is path-connected, choose a path $\gamma$ from $b$ to $b^{\prime }$. Horizontality of $\mu$ and part c implyThus $\mu (b^{\prime })$ is conjugate to the isomorphism $\mu (b)$ and is itself an isomorphism. Since $b^{\prime }$ was arbitrary, $\mu$ is fiberwise invertible everywhere.