differential geometry - Is there a way to see that $nabla_mu g_{nu rho} = 0$ without explicit computation, where $nabla_mu$ refers to the covariant derivative?

In books, it is usually said that this is a consequence of the fact that parallel transport preserves dot product. How ?

Answer

This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here.

We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions and that it reduce to the well-known partial derivative on scalars. No one really debates these properties, but as it turns out they don't uniquely define a derivative. For that we add on the properties of being torsion-free and metric compatible. Summarizing Carroll, we have

$$\begin{align} \nabla(T + S) & = \nabla T + \nabla S && \text{(linearity)} \\ \nabla(T \otimes S) & = (\nabla T) \otimes S + T \otimes \nabla S && \text{(product rule)} \\ \nabla_\mu ({T^\lambda}_{\lambda\rho}) & = {{(\nabla T)_\mu}^\lambda}_{\lambda\rho} && \text{(example of commuting with contractions)} \\ \nabla_\mu \phi & = \partial_\mu \phi && \text{(reduction to partial derivative)} \\ \Gamma^\lambda_{\mu\nu} & = \Gamma^\lambda_{\nu\mu} && \text{(torsion-free connection)} \\ \nabla_\rho g_{\mu\nu} & = 0 && \text{(metric compatibility).} \end{align}$$

You can have covariant derivatives that are not metric compatible. This just means the differential structure of spacetime provided by the derivative doesn't play nicely with the structure induced by the metric, and such constructions turn out to be not very useful for general relativity.

When people "derive" metric compatibility from parallel transport, they are taking as an axiom "parallel transport [which depends only on the covariant derivative] is compatible with inner products [which depend only on the metric]." It is the same as just saying $\nabla_\rho g_{\mu\nu} = 0$, but more complicated. Better would be to assume $\nabla_\rho g_{\mu\nu} = 0$ from the start and prove that this means parallel transport preserves inner products.