I learn GR now, and there is a strange thing that I discovered.

It is well-known, that the condition $\nabla_{\mu}g_{\alpha \beta}=0$ is specified, when we choose specific metric-compatible Levi-Civita connection on our manifold.

Otherwise, there is a proof, which does not use properties of Levi-Civita connection at all, but only uses the fact, that covariant derivative has Leibniz rule and respect index raising rules:

$\nabla_{\alpha}A_{\mu}=\nabla_{\alpha}(g_{\mu\nu}A^{\nu})=g_{\mu\nu}\nabla_{\alpha}A^{\nu}+A^{\nu}\nabla_{\alpha}g_{\mu \nu}$

Now we say, that for any vector $A$ the object $\nabla_{\alpha}A_{\mu}$ must be a tensor, namely $\nabla_{\alpha}A_{\mu} \equiv D_{\alpha \mu}$, and it must respect index raising rules, so by definition:

$g^{\mu \nu}\nabla_{\alpha}A_{\mu}=g^{\mu\nu}D_{\alpha\nu}=D_{\alpha}^{\mu}=\nabla_{\alpha}A^{\mu}\implies\nabla_{\alpha}A_{\mu}=g_{\mu\nu}\nabla_{\alpha}A^{\nu}$

Now we immediatly conclude, that in first equation $A^{\nu}\nabla_{\alpha}g_{\mu \nu}=0$, and hence $\nabla_{\alpha}g_{\mu \nu}=0$.

There must be some mistake, but I can't see it.

  • 3
    $\begingroup$ It is wrong that from the definition of $D$ and without assuming that the connection is not that of Levi-Civita, $\Delta^\mu_\alpha =\nabla_\alpha( A^\mu)$ as you instead assume. This is exactly what you want prove. $\endgroup$ Oct 23, 2019 at 9:13

1 Answer 1


You made a mistake when you wanted to raise one of the indices of $D_{\alpha \nu}$, i.e., $$ g^{\mu \nu} D_{\alpha \nu} \neq D^\mu_\alpha \,, $$ since the connection is not Levi-Civita and you are not allowed to "drag" the metric in the covatiant derivative. When you write $g^{\mu \nu} D_{\alpha \nu} = D^\mu_\alpha$, you implicitly assume that the connection is Levi-Civita.


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