Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 17 / 5 / b

Solution

Fix the coefficient-column convention: write the local frame as a row

e = (e_{1}, \dots, e_{r})

and a section as

s = e u

for a column

u

θ = (θ_{ij})

\nabla e_{j} = \sum_{i} e_{i} θ_{ij}, \nabla (e u) = e (d u + θ u) .

R^{E} (X, Y) s = \nabla_{X} \nabla_{Y} s - \nabla_{Y} \nabla_{X} s - \nabla_{[X, Y]} s .

The same Leibniz cancellations as for tangent-bundle curvature show that this is linear over smooth functions in

X, Y, s

. Thus it is a two-form with values in the endomorphisms of

E

Θ = (Θ_{ij})

is specified by

R^{E} (X, Y) e_{j} = \sum_{i} e_{i} Θ_{ij} (X, Y)

d + θ \land

. Squaring it on

u

gives

(d + θ \land)^{2} u = d θ u - θ \land d u + θ \land d u + θ \land θ u .

Θ = d θ + θ \land θ, Θ_{ij} = d θ_{ij} + k \sum θ_{ik} \land θ_{kj} .

Using

d^{2} = 0

d Θ = d θ \land θ - θ \land d θ .

On the other hand, associativity of the endomorphism-valued exterior product gives

Θ \land θ - θ \land Θ = d θ \land θ + θ \land θ \land θ - θ \land d θ - θ \land θ \land θ = d θ \land θ - θ \land d θ .

The cubic terms cancel, proving the requested Bianchi identity

d Θ = Θ \land θ - θ \land Θ.

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