Metricity Equation: ∇_{μ}g_{αβ}=0, ∇_{μ}g^αβ ≠ 0?

[kuecken]

∇_{\mu}g_{\alpha\beta}=0 is the metricity equation (∇ the covariant derivative).
However, I wondered whether also ∇_{\mu}g^{\alpha\beta}=0 holds?
When I checked it, it did not. But I really want it to hold, so I was wondering whether I made a mistake.
Thank you.

 

[WannabeNewton]

Hi kuecken. In fact $\nabla_{\mu}g_{\alpha\beta} = 0 \Leftrightarrow \nabla_{\mu}g^{\alpha\beta} = 0$.

This is actually very easy to see:

$0 = \nabla_{\mu}\delta^{\alpha}{}{}_{\beta} = \nabla_{\mu}(g^{\alpha \gamma}g_{\gamma \beta}) = g^{\alpha\gamma}\nabla_{\mu}g_{\gamma \beta} + g_{\gamma\beta}\nabla_{\mu}g^{\alpha \gamma}$ holds identically hence if $\nabla_{\mu}g_{\alpha\beta} = 0$ then $\nabla_{\mu}g^{\alpha\beta} = 0$ and similarly for the converse.

 

[kuecken]

This is the expression I arrived at. However, can I argue that the g_{\gamma\beta} in front of ∇ is arbitrary? Or how do I do that last step

 

[WannabeNewton]

So we have $g_{\gamma\beta}\nabla_{\mu}g^{\alpha\gamma} = 0$. Contract both sides with $g^{\nu\beta}$ to get $\delta^{\nu}{}{}_{\gamma}\nabla_{\mu}g^{\alpha\gamma} = \nabla_{\mu}g^{\alpha\nu} = 0$.

 

[dextercioby]

Because of the metricity condition, one can define the so-called contravariant derivatives:

\nabla^{\mu}T_{\alpha} = g^{\mu\nu} \nabla_{\nu}T_{\alpha}

= \nabla_{\nu}\left(g^{\mu\nu}T_{\alpha}\right)

that is freely take out and put back the metric tensor under the covariant/contravariant differentiation sign.