Why is the covariant derivative of the metric tensor with UPPER indices equal to zero? I've shown that $\nabla_{\lambda} g_{\mu\nu} = 0 $ rigorously by the following method:
$ \nabla_{\lambda} g_{\mu\nu} = \partial_{\lambda}g_{\mu\nu} - \Gamma^{\rho}_{\lambda\mu} g_{\rho\nu} - \Gamma^{\rho}_{\lambda\nu} g_{\mu\rho} $
$ = \partial_{\lambda}g_{\mu\nu} - \frac{1}{2}g_{\rho\nu}g^{\rho\sigma}(\partial_{\lambda}g_{\mu\sigma} + \partial_{\mu}g_{\sigma\lambda} - \partial_{\sigma}g_{\lambda\mu}) - \frac{1}{2}g_{\mu\rho}g^{\rho\sigma} (\partial_{\lambda}g_{\nu\sigma} + \partial_{\nu}g_{\sigma\lambda} - \partial_{\sigma}g_{\lambda\nu}) $
We have that $ g_{\rho\nu}g^{\rho\sigma} = \delta^{\sigma}_{\nu}$ and $ g_{\mu\rho}g^{\rho\sigma} = \delta^{\sigma}_{\mu} $ so,
$ = \partial_{\lambda}g_{\mu\nu} - \frac{1}{2}\delta^{\sigma}_{\nu} ( \partial_{\lambda}g_{\mu\sigma} + \partial_{\mu}g_{\sigma\lambda} - \partial_{\sigma}g_{\lambda\mu}) - \frac{1}{2} \delta^{\sigma}_{\mu} (\partial_{\lambda}g_{\nu\sigma} + \partial_{\nu}g_{\sigma\lambda} - \partial_{\sigma}g_{\lambda\nu}) $
$ = \partial_{\lambda}g_{\mu\nu} - \frac{1}{2}(\partial_{\lambda}g_{\mu\nu} + \partial_{\mu}g_{\nu\lambda} - \partial_{\nu}g_{\lambda\mu}) - \frac{1}{2} ( \partial_{\lambda}g_{\nu\mu} + \partial_{\nu}g_{\mu\lambda} - \partial_{\mu}g_{\lambda\nu}) $
$ = \partial_{\lambda}g_{\mu\nu} - \frac{1}{2} \partial_{\lambda}g_{\mu\nu} - \frac{1}{2}\partial_{\lambda}g_{\nu\mu}$
and with $g_{\mu\nu} = g_{\nu\mu} $ we have that $\nabla_{\lambda}g_{\mu\nu} = 0 $
Great. Now I'm trying to show that $\nabla_{\lambda}g^{\mu\nu} = 0$ and I'm having trouble. I've been advised to "cleverly" use the result from $\nabla_{\lambda}g_{\mu\nu} = 0$ in proving the second case, but I'm not seeing it and am getting caught up in index gymnastics -- or missing something carelessly. We are working under the condition that $g_{\mu\nu} \neq g^{\mu\nu}$. Can someone please help to show me the proof for the case of the metric with upper indices under this regime?
 A: A hint: the condition which defines the inverse metric is $g_{\mu\nu} g^{\nu
\rho} = \delta_\mu{}^\rho$, and we can differentiate this equality: one side is a constant and the other can be expanded with the product rule.
If you do not trust the instinct that $\nabla_\mu \delta_\nu{}^\rho = 0$, you can show that it is true in a few different ways, I'd do it like this:
$$ \nabla_\mu \delta_\nu{}^\rho = \partial_\mu \delta_\nu{}^\rho 
+ \Gamma_{\mu \alpha}{}^\rho \delta_\nu{}^\alpha
- \Gamma_{\mu \nu}{}^\alpha \delta_\alpha{}^\rho = 
\Gamma_{\mu \nu}{}^\rho  - \Gamma_{\mu \nu}{}^\rho = 0\,.
$$
A: If you don't trust that the covariant derivative of the Kroneker delta is zero (which you shouldn't assume), I'd start with figuring out what:
$$\nabla_{a}g^{bc} = \nabla_{a}\left(g^{bd}g^{ce}g_{de}\right)$$
reduces to.
A: The covariant derivative is independent of the chart used, if the value at a selected point $p$ is zero in one chart, it is zero in every chart.
Using geodetic curves from $p$ allows to project the tangent space at $p$ to the manifold, this is called the exponential chart. Parallel transport along these geodetic curves gives constant coefficients in the exponential chart and derived charts for the tangent and cotangent bundles. This has as consequence that the metric coefficients are constant for this chart, both in lower and upper indices, as metric on the tangent and on the cotangent space.
Now the derivative of a constant is zero.
