The <solder form> on $X$ is the $TX$-valued one-form
$$
\theta(V)=V.
$$
In a coordinate frame it is $\theta=dx^i\otimes\partial_i$. The torsion form of $\nabla$ is
$$
T=d^\nabla\theta.
$$
Equivalently, for vector fields $U,V$,
$$
T(U,V)=\nabla_UV-\nabla_VU-[U,V].
$$
Writing
$$
\nabla_{\partial_i}\partial_j=\Gamma^k{}_{ji}\partial_k,
$$
and using $[\partial_i,\partial_j]=0$ gives
$$
T(\partial_i,\partial_j)
=(\Gamma^k{}_{ji}-\Gamma^k{}_{ij})\partial_k.
$$
Hence $\nabla$ is <torsion-free connection>[torsion-free] exactly when
$$
\Gamma^k{}_{ji}=\Gamma^k{}_{ij}
$$
for every $i,j,k$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-115/4/a/solution