Why isn't $g^{\mu \nu}\partial_{\nu}$ equal to $\partial^{\mu}$ in General Relativity? The question is pretty short. I have been told that unlike for flat spacetime, in curved spacetime we cannot write
$g^{\mu \nu}\partial_{\nu}f=\partial^{\mu}f$
With $f$ an scalar function, but I don't really get why.
 A: Well I'll answer but I would be interested to know if my answer is wrong. I thought we defined $\partial^a$ such that
$$
\partial^a f \equiv g^{a\mu} \partial_\mu f .
$$
The thing to be careful about is then the fact that
$$
\partial_\mu (g^{a\mu} f) = (\partial_\mu g^{a\mu}) f + g^{a\mu} \partial_\mu f 
$$
so usually
$$
\partial_\mu (g^{a\mu} f) \ne \partial^a f .
$$
It also follows that you cannot assume that $\partial^a$ operators commute, nor even that $\partial^\mu \partial_\mu$ is equal to $\partial_\mu \partial^\mu$ unless you make some other definition of $\partial^a$.
A: For a scalar function; $\nabla_{\mu}f=\partial_{\mu}f$, its a definition, we know $\nabla_{\mu}f$ is a $(0,1)$ tensor so $g^{\mu\nu}\nabla_{\mu}f=\nabla^{\nu}f$ but there doesn't exist any similar definition where we identify contravariant version of $\nabla^{\mu}f$ with $\partial^{\mu}f$. So ultimately it boils down to the definition of $\partial^{\mu}$.
A: The definition
$$
\partial^{\mu} = g^{\mu \nu} \, \partial_{\nu} \tag{1}
$$
doesn't have more sense than
$$
x_{\mu} = g_{\mu \nu} \, x^{\nu}. \tag{2}
$$
Using (1) may lead to many troubles that you want to avoid at all cost.  Don't use it!
However, you may use the same for the covariant derivative:
$$
\nabla^{\mu} = g^{\mu \nu} \, \nabla_{\nu}, \tag{3}
$$
which may be usefull since $\nabla_{\lambda} \, g^{\mu \nu} = 0$.
