Treating $x^\lambda$ as a scalar during differentiation,
$$
\nabla_\mu\nabla_\nu x^\lambda
=\partial_\mu\partial_\nu x^\lambda
-\Gamma^\rho_{\mu\nu}\partial_\rho x^\lambda
=-\Gamma^\lambda_{\mu\nu}.
$$
Hence
$$
\boxed{\Box_gx^\lambda=0
\iff g^{\mu\nu}\Gamma^\lambda_{\mu\nu}=0}.
$$
Metric compatibility and the determinant identity give the contracted-Christoffel formula
$$
g^{\mu\nu}\Gamma^\lambda_{\mu\nu}
=-\frac1{\sqrt{-g}}\partial_\mu
(\sqrt{-g}\,g^{\lambda\mu}).
$$
Therefore
$$
\boxed{g^{\mu\nu}\Gamma^\lambda_{\mu\nu}=0
\iff\partial_\mu(\sqrt{-g}\,g^{\lambda\mu})=0}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/1/b/solution