TITLE: Does $d x^\mu=g^{\mu\nu} \partial_\nu$, or $\partial_\nu=g_{\nu \mu}dx^\mu$?
QUESTION [5 upvotes]: As the title suggests: does $d x^\mu=g^{\mu\nu} \partial_\nu$, or $\partial_\nu=g_{\nu \mu}dx^\mu$?
This is merely my own guess, from the material that I'm reading; so I'm not sure... The partial derivative operator is still an differential operator in my point of view.
This may be too basic for you, but thank you very much.

REPLY [2 votes]: SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $\delta_{\mu \nu}dx^\mu\otimes dx^\nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.

My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $\mu_0\in \{1, 2, \ldots n\}$ be fixed. The 1-form $dx^{\mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, \ldots, dx^n$ are 
$$
(\delta^{\mu_0}{}_{\nu}\ :\ \nu=1, 2, \ldots, n), $$
and we denote them by $\delta^{\mu_0}{}_{\nu}$. (This $\delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important). 
Dually, the vector field $\frac{\partial}{\partial x^{\mu_0}}$ is the tensor whose coordinates are $\delta_{\mu_0}{}^\nu$. And this is the end of the story unless we introduce a Riemannian metric. 
If we introduce a such metric $g_{\mu\nu}$ (or $g_{\mu\nu}dx^\mu\otimes dx^\nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{\mu\nu}=\delta_{\mu\nu}$, then raising the second index in the first tensor we obtain the second; 
$$
\delta^{\mu_0}{}_\nu \delta^{\nu \rho}=\delta^{\mu_0\rho}. $$ 

WARNING! If the metric is not $\delta_{\mu \nu}dx^\mu\otimes dx^\nu$, then the above fails, and (in the language of Ivo's answer) it is not true that 
$$
\left(\frac{\partial}{\partial x^\mu}\right)_b =dx^\mu, \quad \frac{\partial}{\partial x^\mu}=(dx^\mu)^{\#}.$$ 
Let me make an explicit example: parametrize $\mathbb S^2$ (minus a semicircle) as 
$$
(\cos \theta, \sin \theta\cos \phi), \qquad \text{where }\theta\in(0, \pi),\phi\in(0, 2\pi).$$
Then the metric tensor of $\mathbb S^2$ is 
$$
g = d\theta^2 + \sin^2\phi d\phi^2,$$
hence the matrix $g_{\mu \nu}$ is 
$$
\begin{bmatrix} 1 & 0 \\ 0 & \sin^2\phi\end{bmatrix}. $$
In particular, $g_{\mu \nu}$ is not the Kronecker symbol. 
Now, let us compute $(d\phi)_b$. The components of this tensor, in the basis $d\theta, d\phi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute 
$$
(0, 1)\begin{bmatrix} 1 & 0 \\ 0 & \sin^2\phi\end{bmatrix} = (0, \sin^2\phi).$$ 
Hence 
$$
(d\phi)_b= \sin^2\phi \frac{\partial}{\partial \phi}.$$