Regard an $E^\vee$-valued $r$-form $\lambda$ as a row vector and an $\operatorname{End}(E)$-valued $r$-form $\mu$ as a matrix. The <dual connection> and endomorphism connection are
$$
d^{\mathcal A^\vee}\lambda
=d\lambda-(-1)^r\lambda\wedge A,
$$
$$
d^{\operatorname{End}(\mathcal A)}\mu
=d\mu+A\wedge\mu-(-1)^r\mu\wedge A.
$$
For the curvature two-form,
$$
d^{\operatorname{End}(\mathcal A)}F=dF+A\wedge F-F\wedge A=0,
$$
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 rule
$$
d^{\mathcal A}(\mu\wedge\sigma)
=d^{\operatorname{End}(\mathcal A)}\mu\wedge\sigma
+(-1)^r\mu\wedge d^{\mathcal A}\sigma.
$$
The cancellation of the two middle $A$-terms proves the identity.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-115/3/b/solution