# Why metric tensor can be not covariantly constant?

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.

• 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. – Valter Moretti Oct 23 '19 at 9:13

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.