The induced <dual connection> is uniquely defined by
$$
(\nabla_X\alpha)(Y)=X(\alpha(Y))-\alpha(\nabla_XY),
$$
which immediately gives the required pairing identity. If
$$
\nabla_{\partial_i}\partial_j=\Gamma^k_{ij}\partial_k,
$$
then
$$
\nabla_{\partial_i}dx^k=-\Gamma^k_{ij}dx^j.
$$

A connection on $TB$ is symmetric, or <torsion-free connection>[torsion-free], when
$$
T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]=0,
$$
equivalently $\Gamma^k_{ij}=\Gamma^k_{ji}$ in coordinates. With $\operatorname{Alt}B(X,Y)=B(X,Y)-B(Y,X)$,
$$
\begin{aligned}
(\operatorname{Alt}\nabla\alpha)(X,Y)
&=X\alpha(Y)-Y\alpha(X)-\alpha(\nabla_XY-\nabla_YX)\\
&=X\alpha(Y)-Y\alpha(X)-\alpha([X,Y])\\
&=d\alpha(X,Y).
\end{aligned}
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-115/3/b/solution