Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 / 3 / b

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- Solution b

Solution

The induced dual connection is uniquely defined by

(\nabla_{X} α) (Y) = X (α (Y)) - α (\nabla_{X} Y),

which immediately gives the required pairing identity. If

\nabla_{\partial_{i}} \partial_{j} = Γ_{ij}^{k} \partial_{k},

then

\nabla_{\partial_{i}} d x^{k} = - Γ_{ij}^{k} d x^{j} .

A connection on

TB

is symmetric, or torsion-free, when

T (X, Y) = \nabla_{X} Y - \nabla_{Y} X - [X, Y] = 0,

equivalently

Γ_{ij}^{k} = Γ_{ji}^{k}

in coordinates. With

Alt B (X, Y) = B (X, Y) - B (Y, X)

,

(Alt \nabla α) (X, Y) = X α (Y) - Y α (X) - α (\nabla_{X} Y - \nabla_{Y} X) = X α (Y) - Y α (X) - α ([X, Y]) = d α (X, Y) .

Solved by gpt-5.6-sol high.

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