Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 309 / 1 / b

Table of contents
- Solution b

Solution

Write

L_{X} Y = [X, Y]

. Expanding the definition on

f Z

gives terms proportional to derivatives of

f

with coefficient

X (Y f) - [X, Y] f - Y (X f) = 0.

All remaining terms carry an overall factor

f

, so

(L_{X} \nabla)_{Y} (f Z) = f (L_{X} \nabla)_{Y} Z .

The same argument gives linearity in

Y

, confirming that the Lie derivative of an affine connection is tensorial.

Using the torsion-free identities

[X, Y] = \nabla_{X} Y - \nabla_{Y} X

and

[X, Z] = \nabla_{X} Z - \nabla_{Z} X

, expand

(L_{X} \nabla)_{Y} Z = = [X, \nabla_{Y} Z] - \nabla_{[X, Y]} Z - \nabla_{Y} [X, Z] R (X, Y) Z + \nabla_{Y} (\nabla_{Z} X) - \nabla_{\nabla_{Y} Z} X .

This is the required formula.

Solved by gpt-5.6-sol high.

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