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

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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.

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As one way, consider the parallel transport of two vectors $p^\mu$ and $q^\nu$ along the same curve $C$ such that the same angle between the two, $$ \cos\phi=\frac{p^\alpha q_\alpha}{\sqrt{p^\beta q_\beta}\sqrt{p^\gamma q_\gamma}} $$ is preserved. Parallel transport tells us that for a general vector $f^\alpha$, $$ \frac{Df^\alpha}{D\tau}={f^\alpha}_{\,;\gamma}\dot{x}^\gamma=0 $$ Thus, we have $Dp^\alpha/D\tau=Dq^\alpha/D\tau=0$ and to preserve the angle, we require that $$ \dot{x}^\gamma\left(g_{\alpha\beta}p^\alpha q^\beta\right)_{;\gamma}=0\tag{1} $$ Expanding the covariant derivative, $$ \dot{x}^\gamma\left(g_{\alpha\beta;\gamma}\,p^\alpha\,q^\beta+g_{\alpha\beta}\,p^\alpha_{\,\,;\gamma}\,q^\beta+g_{\alpha\beta}\,p^\alpha\,q^\beta_{\,\,;\gamma}\right)=0 $$ By definition the last two terms are zero (they are parallel transported), so we must have that $$ g_{\alpha\beta;\gamma}=0 $$ for (1) to be true.

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  • $\begingroup$ I cannot quite connect the fact that the vectors are parallel transported and the last two terms being zero. BTW $\frac{Df^\alpha}{D\tau}$ stands for $\frac{df^\alpha}{d\tau}$ right? $\endgroup$
    – Gonenc
    Jun 14 '15 at 9:25
  • $\begingroup$ Look at the line above (1), expand the line you question, then compare. I think I made my notation clear enough, the use of D indicates parallel transport, if you are familiar with another definition that means parallel transport, then yes, otherwise, no. $\endgroup$
    – Kyle Kanos
    Jun 14 '15 at 11:21
  • $\begingroup$ Oh I forgot about the factor of $\dot x$ multiplying the brackets. Thanks $\endgroup$
    – Gonenc
    Jun 14 '15 at 11:31

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