The gothicized metric and the Palatini formalism In the Palatini formalism of GR, we had two results treating the metric $g_{\mu\nu}$ and the connection $\Gamma^\alpha_{\mu\nu}$ separately as dynamical variables, which are 


*

*The vaccum field equations of GR.


And


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*That the connection is necessarily the metric connection.


The later was obtained from varying the action w.r.t. the symmetric connection and we thus obtain:
$\nabla_\alpha (\sqrt{-g}g^{\mu\nu})=0$ , where we deduced the last result.
Now my question is how to obtain $\nabla_\alpha g_{\mu\nu}=0$ to be able to reach to such a result?
p.s. Ray D'Inverno called the tensor density $\sqrt{-g}g^{\mu\nu}$ a gothicized metric.
 A: The main idea answering this question was from this and here I'd like to answer in detail:
We only have this:  $\nabla_\lambda \sqrt{-g} g^{\mu\nu}=0$  ------(1
I will need some formulas which I will deduce firstly.
$\delta^\mu_\nu=g^{\mu\alpha} g_{\alpha \nu} \implies \nabla_\lambda \delta^\mu_\nu=g_{\alpha \nu}\nabla_\lambda g^{\mu\alpha} +g^{\mu\alpha} \nabla_\lambda g_{\alpha \nu}$
$\implies g_{\alpha \nu}\nabla_\lambda g^{\mu\alpha} =-g^{\mu\alpha} \nabla_\lambda g_{\alpha \nu}$
$\implies \nabla_\lambda g^{\mu\beta}=-g^{\mu \alpha} g^{\beta\nu}\nabla_\lambda g_{\alpha \nu}$       -------(2
We take as a priliminary (From the calculus of matrices) $\partial_\lambda \sqrt{-g}=\frac{1}{2}\sqrt{-g}g^{\mu\nu}\partial_\lambda g_{\mu\nu}$------(3
Now the tricks begin:
$\nabla_\lambda \sqrt{-g}=\partial_\lambda \sqrt{-g}- \Gamma^\zeta_{\zeta\lambda} \sqrt{-g}$ $~~~~~~~~~~as$ $\sqrt{-g}$ is a scalar density of weight +1.
$~~~~~~~~~~~~~= \frac{1}{2}\sqrt{-g}g^{\mu\nu}\partial_\lambda g_{\mu\nu}- \Gamma^\zeta_{\zeta\lambda} \sqrt{-g}$ $~~~~~~~~~$Using (3
$~~~~~~~~~~~~=\frac{1}{2}\sqrt{-g}g^{\mu\nu}\partial_\lambda g_{\mu\nu}-\sqrt{-g}g^{\mu\nu} \left(\frac{1}{2}\Gamma^\zeta_{\mu\lambda}g_{\zeta\nu}+\frac{1}{2}\Gamma^\zeta_{\nu\lambda}g_{\mu\zeta}\right)$
$~~~~~~~~~~~~=\frac{1}{2}\sqrt{-g}g^{\mu\nu}\nabla_\lambda g_{\mu\nu}$------(4
Now,
$\nabla_\lambda (\sqrt{-g}g^{\alpha\beta})=\sqrt{-g}\nabla_{\lambda}g^{\alpha\beta}+g^{\alpha\beta}\nabla_\lambda\sqrt{-g}$ 
Now use (1 and (3 to obtain:
$\nabla_\lambda (\sqrt{-g}g^{\alpha\beta})=-\sqrt{-g}g^{ \alpha\mu} g^{\beta\nu}\nabla_\lambda g_{\mu \nu}+\frac{1}{2}\sqrt{-g}g^{\alpha\beta}g^{\mu\nu}\nabla_\lambda g_{\mu\nu}$
$~~~~~~~~~~~~~~~~~~~~~~~=\sqrt{-g}\left(-g^{ \alpha\mu} g^{\beta\nu}+\frac{1}{2}g^{\alpha\beta}g^{\mu\nu}\right)\nabla_\lambda g_{\mu\nu}$
$~~~~~~~~~~~~~~~~~~~~~~~=0$
Contracting with $g_{\alpha\beta}$ we get
$g^{\mu\nu}\nabla_\lambda g_{\mu\nu}=0$------(5
Now here another trick:
$4 \nabla_\lambda \sqrt{-g}=\nabla_\lambda (g_{\mu\nu}g^{\mu\nu}\sqrt{-g})$
$~~~~~~~~~~~~~~~~=g_{\mu\nu}\nabla_\lambda (g^{\mu\nu}\sqrt{-g}) + \sqrt{-g} (g^{\mu\nu} \nabla_\lambda g_{\mu\nu})$
Now using (1 and (5 we get:
$\nabla_\lambda \sqrt{-g}=0$ 
And from this using leibniz in (1 we can get easily: $\nabla_\lambda g^{\mu\nu}=0$ and hence using a simple calculation we get $\nabla_\lambda g_{\mu\nu}=0$. And the results of the Palatini formalism then follows easily.
