Is there any intuitive explanation for why the absolute gradient of the metric tensor $\nabla_{\alpha} g_{\mu \nu} = 0$ in every coordinate system?
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1$\begingroup$ Metric connection. $\endgroup$– AccidentalFourierTransformCommented Jun 13, 2018 at 18:13
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2$\begingroup$ Possible duplicates: physics.stackexchange.com/q/47919/2451 , physics.stackexchange.com/q/189374/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jun 13, 2018 at 19:48
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2$\begingroup$ Possible duplicate of Why is the covariant derivative of the metric tensor zero? $\endgroup$– user4552Commented Jun 13, 2018 at 20:04
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1 Answer
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By the fundamental theorem of Riemannian geometry, on a manifold $M$ with metric $g$, it is always possible to choose a connection $\nabla$ such that,
$$\partial_X \langle Y,Z\rangle = \langle \nabla_X Y,Z\rangle + \langle Y, \nabla_X Z \rangle$$
where $X,Y$ and $Z$ are vector fields. Converting to explicit index notation, it is possible to show that this condition implies we can always choose a connection such that $\nabla_a g_{bc} = 0$.
Thus, to answer your question directly, we can always choose a connection, i.e. choose a means to parallel transport tangent vectors, in such a way that the metric compatibility conditions is true.
It should be stressed, this choice of connection, the Levi-Civita connection (which has the added condition of being torsion free) is only one choice of connection, for the tangent bundle on $M$, and there are of course other choices and other bundles to consider, for which it is not true.
