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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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marked as duplicate by AccidentalFourierTransform, Ben Crowell, sammy gerbil, Kyle Kanos, Qmechanic Jun 14 '18 at 11:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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