Parallel propagation means $\nabla_Te_\alpha=0$, where $T=\dot\lambda$. By <metric compatibility>,
$$
\frac d{dt}g(e_\alpha,e_\beta)
=g(\nabla_Te_\alpha,e_\beta)+g(e_\alpha,\nabla_Te_\beta)=0.
$$
The inner products therefore retain their initial values $\eta_{\alpha\beta}$, so the <parallel-propagated orthonormal frame> remains orthonormal.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 1 / d / i / Solution

Along a curve with tangent $T$, a frame is parallel propagated when $\nabla_Te_a=0$. Metric compatibility makes every inner product $g(e_a,e_b)$ constant along the curve, so an initially orthonormal frame remains orthonormal.


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