Confusion about 1-forms as introduced in "Gravitation" (Kip S. Thorne,...) In the book Gravitation in chapter 2, paragraph 5, they introduce the concept of 1-forms by thinking about the momentum 4-vector differently.
They first introduce the de Broglie 1-form as follows (I don't understand the following really well):

suppose we have particle in spacetime. Consider its associated de Broglie wave. If we diffract this, we get a pattern and this gives surfaces of the same integral phases.

Then they introduce the 1-form $\tilde{\mathbf{k}}$ which has input a vector $\mathbf{v}$ in spacetime and outputs a number $\langle \tilde{\mathbf{k}}, \mathbf{v}\rangle$ which denotes the number of pierces through these surfaces by the vector $\mathbf{v}$.
After this they define the momentum 1-form $\tilde{\mathbf{p}}$ where we define the surface to be the surfaces of $\tilde{\mathbf{k}}$, but multiplied by $\hbar$. Then they claim that if $\mathbf{p}$ is the ordinary momentum 4-vector then $\mathbf{p}\cdot\mathbf{v}=\langle \tilde{\mathbf{p}}, \mathbf{v}\rangle$.
I have to prove this (excercise 2.1) using the quantum-mechanical properties of the de Broglie wave $\psi=\exp(i(\mathbf{k}\cdot \mathbf{x}-\omega t))$.
Can someone explain the above a little bit more in detail to me? I find this motivation for a defintion of 1-forms not so clear. If I understand it I think I can solve the excercise.
 A: Notation
$\renewcommand{braket}[2]{\langle #1 | #2 \rangle}$
First let us define the term "1-form":

A 1-form is a linear function which takes a single vector as it's argument.

That is it.
It is no more confusing or complicated than that single statement.
Consider a vector $v$.
Given any other vector $w$ we can form the inner product $\braket{v}{w}$.
Note that
$$\braket{v}{a w + b x} = a \braket{v}{w} + b \braket{v}{x} \, .$$
Therefore, the function $f_v$ defined by the equation
$$f_v(w) = \braket{v}{w}$$
is a function which takes a single vector and is linear in its argument.
In other words, $f_v$ is a 1-form.
From this discussion you can see that any vector $v$ is directly associated to a 1-form $f_v$ defined by the equation $f_v(x) = \braket{v}{x}$.
Note that 1-forms are also called covectors, and the 1-form $f_v$ associated to vector $v$ is called the dual vector or just dual of $v$.
We now switch to using bold type for vectors to keep consistent with the original post.
Consider a plane wave expressed as $\exp[i \mathbf{k} \cdot \mathbf{x}]$.
The thing in the exponent is the inner product of a wave vector $\mathbf{k}$ with the position vector $\mathbf{x}$.
In fact we can write the wave in a suggestive form
\begin{align}
\exp \left[ i \mathbf{k} \cdot \mathbf{x} \right]
&= \exp \left[ i \braket{\mathbf{k}}{\mathbf{x}} \right] \\
&= \exp \left[ i f_\mathbf{k}(\mathbf{x)} \right]
\end{align}
where $f_\mathbf{k}$ is a 1-form as described above.
It seems that to clarify the notation your book puts a $\tilde{}$ on top of the symbols to indicate that something is a 1-form instead of a vector.
In other words, where I write $f_\mathbf{k}$ your book writes $\tilde{\mathbf{k}}$.
In fact I would say the notation $\braket{\tilde{\mathbf{k}}}{\mathbf{x}}$ used by your book is not so good because normally the symbol $\braket{\cdot}{\cdot}$ is used to denote the inner product of two vectors.
It's a bit strange to write it with a 1-form on the left.
However, we can just re-interpret the symbol $\braket{\tilde{\mathbf{k}}}{\mathbf{x}}$ as $\braket{\mathbf{k}}{\mathbf{x}}$ or $f_\mathbf{k}(\mathbf{x})$.
Physical meaning
We now discuss the physical meaning of $\tilde{\mathbf{k}}$.
As your book says, when you feed this 1-form a vector $\mathbf{v}$, it tells you the number of cycles that the wave goes through as you move by the displacement $\mathbf{x}$.
It's a 1-form for the simple reason that it eats one vector and produces a number, and is linear in the argument.
That's it, it's just a definition.
As your book says, you can picture the 1-form $\tilde{\mathbf{k}}$ as a set of planes because counting the number of planes crossed by a $\mathbf{x}$ gives the same result as $\mathbf{k} \cdot \mathbf{x}$, as shown in the diagram.

Summary


*

*A 1-form is just a linear function of a vector.

*Each vector $v$ is uniquely associated to a 1-form $f_v$ defined by the equation $f_v(x) = \braket{v}{x} = v \cdot x$.

*A geometric picture of the 1-form $f_k$ is a series of parallel planes spaced by distance $2 \pi / |k|$, and the number $f_k(x)$ is the number of planes crossed by vector $x$.
A: Suppose you have a skewed coordinate system, with vectors that have components $v^i$ defined relative to some basis vectors $\hat e_i$ such that $\vec v = v^i \hat e_i$ in the Einstein summation convention (we sum over any index which is repeated once above, once below).
The meaning of a skewed coordinate system is that $\hat e_i \cdot \hat e_j \ne \delta_{ij}$ for the Kronecker delta $\delta_{ij} = \{1\text{ if }i = k,\text{ else } 0\}.$ This messes with our nice dot product formula; no more can we say with great ease$$\vec u\cdot\vec v = \sum_i u^i~v^i$$because that is no longer true. Instead, we have:$$\vec u\cdot\vec v = u^i~v^j~(\hat e_i \cdot \hat e_j)$$ which means there is a symmetric matrix $g_{ij} = \hat e_i \cdot \hat e_j$ which governs such inner products. This matrix is called the metric.
The metric is symmetric and has a matrix inverse we can write as $g^{ij}$ defined by $g^{ij} g_{jk} = \delta^i_k$, where we now have a "correct" Kronecker $\delta$ expression. This can be used to define the dual basis$$\hat e^k = g^{ki}~\hat e_i,$$having the defining property that $\hat e^i \cdot \hat e_j = \delta^i_j.$In other words, to find the vector dual to $\hat e_1$, we find a vector $\vec u^1$ which is perpendicular to $\hat e_2, \hat e_3, \dots$ and then scale it to $\hat e^1 = k~\vec u^1$ by choosing $k$ such that $\hat e^1 \cdot \hat e_1 = 1$. [If you do not like matrix inversion or you are doing all of this in a nonflat space, if that space has an orientation (an antisymmetric tensor which is a linear map from $n$ vectors to a scalar, where $n$ is the dimension of the tangent spaces), you can use that to create a one-form which is maps those other basis vectors to zero.]
Then the vector $\vec u = u^i ~\hat e_i$ can also be written as $u_i~\hat e^i$ where $u_i = g_{ij} u^j$ and the dot product's nice, simple form can be restored as $$\vec u \cdot \vec v = u_i v^i = u^i v_i.$$
Now when you start to write familiar expressions in a crystal lattice, $$\psi = e^{i(~\vec k~\cdot~\vec x ~-~ \omega~t~)}$$With our new formalism we'd express your position vector $\vec x = x^i ~\hat e_i$ and so the natural "space" that your wavenumbers live in is the dual space. This is also the natural space for derivatives, it turns out: the differential of a function is still $$df = f(\vec x + d\vec x) - f(\vec x) \approx \sum_i \frac{\partial f}{\partial x^i} ~ dx^i $$ which means that the components $\left(\frac{\partial f}{\partial x^i}\right)_{\{x^j,j\ne i\}}$ (partial derivatives with respect to $x^i$ holding all of the other $x^j$ constant) must transform like the covectors transform, so that this can be written simply as $$df = (\partial_i f)~dx^i.$$The typical quantum momentum operator is $\hat p_i = -i~\hbar\partial_i$. Applying this as $p_a = -i \hbar \partial_a \exp\left[i~\big(k_b x^b - \omega t\big)\right]$ gives a covector momentum which acts on $v$ like $\langle p, v\rangle$.
I'm not 100% sure if that helps but hopefully that gives you an idea of how the idea of dual spaces connects to physics. In addition, if you're interested in using Einstein summations without explicit reference to coordinates, there is something called "abstract index notation" which you should look up.
Caution: many texts in solid state physics define $\hat e^i \cdot \hat e_j = 2\pi~\delta^i_j$, to save a $2\pi$ in some exponential expressions. Be careful when you see these things. 
A: The name "1-form" is only used because there are 2-forms and 3-forms and so on. An equivalent but probably better name for 1-forms in this case is that of "dual vector." Any vector space will have associated with it the dual space consisting of all linear functions that map vectors into scalars.
This is all a dual vector is: a thing that linearly maps vectors into scalars.
But people like to have visual aids. Let's stick to 2D. If you think of vectors as arrows, there is an analogous picture of dual vectors as contour lines. Or rather, if the collection of many arrows represents of vector field, the collection of many contour lines represents a dual vector field.
In particular, for finite-dimensional vector spaces there is a natural bijection between the space itself and the dual space: Every vector is uniquely associated with a dual vector and vice versa. The dual vector field associated with a vector field is represented by the contours orthogonal to the vector field's arrows. Conversely, if you start with a contour plot representing a dual vector field and draw arrows orthogonal to it with length proportional to contour density, you have yourself the arrow representation of the associated vector field.
In this picture of arrows and contours, the action of a dual vector field on a vector field is the scalar field that measures how many contours are pierced by each arrow. If the contours are closer (the dual vectors are larger), or the arrows are longer (the vectors are larger), the result will be a larger number. If the arrows point largely along contours, then the fields are misaligned and the result is smaller (recall this means the arrows associated with the contours, which are perpendicular to those contours, would be perpendicular to the original arrows).
The machinery may seem a bit overbearing, since in Euclidean geometry there really is no benefit to using contours instead of their associated arrows. But with non-flat metrics the connection between vectors and dual vectors becomes less trivial. Of course then our pictures tend not to work so well either.
