The connection is orthogonal, or metric-compatible, when $\nabla g=0$; equivalently, its <parallel transport> preserves the <Riemannian metric>. Compatibility of the induced connections with tensor contraction gives
$$
\partial_i(g(U,V))
=g(\nabla_{\partial_i}U,V)+g(U,\nabla_{\partial_i}V)
$$
for all vector fields $U,V$. Taking $U=\partial_j$ and $V=\partial_k$ yields
$$
\frac{\partial g_{jk}}{\partial x^i}
=\Gamma^\ell{}_{ji}g_{\ell k}
+\Gamma^\ell{}_{ki}g_{j\ell}
=\Gamma_{jki}+\Gamma_{kji}.
$$
Conversely, this coordinate identity makes every component of $\nabla g$ vanish, so it is equivalent to orthogonality.

Solution

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

 Solution

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