Inverse metric tensor as a map? I understand that the metric tensor is a natural map between vectors and 1-forms. For instance, this is shown below:
$g(v,.) $
That is a 1-form since eats a vector and gives you a real number. However, it's not clear to me how to pass from 1-form to a vector using the metric. Is there something as in the case of the start? For instance, I have to go from the 1 form.
$du=dt-dx$
to the vector:
$\nabla u = -\partial_{ t}-\partial_{x}$
The result is intuitive but the procedure it is not clear to me.
 A: In components the metric is given by $g_{\mu\nu}dx^\mu\otimes dx^\nu$ which eats vectors whose component forms are $v^\mu\partial_\mu$. Since the metric is non-degenerate, it has an inverse $g^{-1}=g^{\mu\nu}\partial_\mu\otimes\partial_\nu$ which satisfies $\delta^\mu_\nu=g^{\mu\lambda}g_{\lambda\nu}$. In this way $g$ defines a map from $TM$ to $T^*M$ as you say and $g^{-1}$ defines a map from $T^*M$ to $TM$ in the same way, and these maps are indeed inverses of each other.
These ideas are usually formalized in a coordinate-independent language by the so-called musical isomorphisms.
A: You take the inverse matrix of $g_{ab}$ to be $g^{ab}$, which you can then treat as an operator that eats a pair of one forms and spits out reals.  Then, it should be pretty clear that mapping a vector to a one form and back to a vector is the identity operation, and that the inner product of two vectors is equal to the inner product of the two one-forms created from those vectors.
In the case of Minkowski, the operation is kind of confusing, because the Minkowski metric in $(t, x, y, z)$ coordinates is its own inverse.
A: I wanted to offer a short comment on Richard Myers' answer, which however contained to much math to fit into the comment section, therefore i'm adding it here as an Answer.
Let $p$ be a point in $M$, and $g$ a metric on $M$, and $\alpha$ is a one-form on $M$. One-forms are linear functionals on vectors, i.e. $\alpha_p : T_pM \rightarrow \mathbb{R}$. The space of linear functionals on a vector space is called the (algebraic) dual space and denoted by a star, i.e. one writes $\alpha \in (T_pM)^* = T^*_pM$. As was explained in another answer, a metric $g$ on $M$ allows to define an inverse metric $g^{-1}$ on $T^*M$. Then for each one form $\alpha$ we have that $g^{-1}_p(\alpha_p,\cdot)$ maps linear functionals at $p$ into the reals, or in other words, $g^{-1}_p(\alpha_p,\cdot) \in (T_pM)^{**}$.
We can now use that for any finite-dimensional vector space $V$ we have $V \cong V^{**}$ canonically. Indeed: let $\varphi \in V^*$, then for each $v\in V$ we get the evaluation map $x_v\in V^{**}$ given as
$$ x_v(\varphi) = \varphi(v) \ . $$
This gives an embedding $V \subset V^{**}$.
Using this backwards, we get a vector.
