General Relativity – Why Is the Covariant Derivative of the Determinant of the Metric Zero?

differential-geometrydifferentiationgeneral-relativitymetric-tensor

This question, metric determinant and its partial and covariant derivative,
seems to indicate $$\nabla_a \sqrt{g}=0.$$ Why is this the case? I've always learned that $$\nabla_a f= \partial_a f,$$ hence surely $$\nabla_a \sqrt{g}= \partial_a\sqrt{g} \neq 0. $$

Where's the hole in my logic?

Best Answer

Comments to the post (v2):

  1. Note that $\sqrt{|g|}$ transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of $\sqrt{|g|}$ does not necessarily coincide with the partial derivative of $\sqrt{|g|}$.

  2. Here is a heuristic explanation using local coordinates. The Levi-Civita connection is compatible with the metric $g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$. That a connection $\nabla$ is compatible with a metric means that $\nabla_{\lambda}g_{\mu\nu}=0$. Using linearity and Leibniz rule, the covariant derivative $\nabla_{\lambda}$ then annihilates any sufficiently nice function $f(g_{00},g_{01}, \ldots)$ of the metric. In particular, the square root of the determinant $\sqrt{|g|}$, so $\nabla_{\lambda}\sqrt{|g|}=0$.

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