Geometric Understanding of 1-Forms
$\Begingroup$ in a Geometric Approach to Differential Forms by David Bachman You Can Find the Passage: .. . a 1-Form Is a Linear Function Which Acts on Vectors...
In A Geometric Approach to Differential Forms by David Bachman you can find the passage:
... a 1-form is a linear function which acts on vectors and returns numbers. For the moment let's just look at 1-forms on $T_p\mathbb R^2$ for some fixed point, $p.$ Recall that a linear function $\omega,$ is just one whose graph is a plane through the origin. Hence, we want to write down an equation of a plane through the origin in $T_p\mathbb R^2 \times \mathbb R,$ where one axis is labelled $dx,$ another $dy,$ and the third $\omega.$ This is easy: $\omega=a\; dx + b \;dy.$ Hence, to specify a 1-form on $T_p\mathbb R^2$ we only need to know two numbers: $a$ and $b.$
I am aiming at an intuitive idea of what he is referring to. So far this would be my summary:
1-forms are elements of $V^*$ that "eat" a vector to produce a scalar. In general I picture them as row vectors, but in the particular case of differentiable equations, it may work like this: the equations are elements of a vector space. In this case, then, the $dx$ in the integral $\int f(x) dx$ would be the 1-form. This becomes more intuitive when including a Jacobian transformation.
With this sketchy idea as background, I don't understand what $\omega,$ the plane through the origin, is; or why it has to go through the origin.
Perhaps I have problems understanding $\mathbb R^2$ in $T_p \mathbb R^2$ among other concepts. I see that this is the tangent plane at point $p.$ Here is a relevant passage:
Let’s look at the tangent line to the graph of $y = x^2$ at the point $(1, 1).$ We are no longer thinking of this tangent line as lying in the same plane that the graph does. Rather, it lies in $T_{(1,1)}\mathbb R^2.$ The horizontal axis for $T_{(1,1)}\mathbb R^2$ is the “dx” axis and the vertical axis is the “dy” axis. Hence, we can write the equation of the tangent line as $dy = 2dx.$
2 Answers
Maybe the story from a generic curved space perspective will give you a better understanding of the notation (the $\mathbb R^n$ case is somewhat special).
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1-forms on a Sphere
Suppose you are in $\mathbb R^3$. The two dimensional sphere $S$ in $\mathbb R^3$ is a two dimensional manifold (manifold = generic curved space). At any point $p\in S$ the sphere has a tangent plane which we denote with $T_pS$. The subset $T_pS\subset \mathbb R^3$ is a two dimensional affine subspace of $\mathbb R^3$ and can be given the structure of a vector space in a canonical way by $q\mapsto q-p$ (i.e. identifying it with the two dimensional vector subspace of $\mathbb R^3$ that is parallel to $T_pS$). The disjoint union of the $T_pS$ is the tangent bundle of $S$, i.e. $TS := \{(p, v):\ p\in S, v\in T_pS\}$.
Now that you have a vector space for every point of $S$ you can go crazy with your linear algebra and build all sorts of other vector spaces attached to every point of $S$. One such is to consider the dual $T_pS^*$ for every $p\in S$. Once you have applied your construction at every point you take the disjoint union and get the "bundle", for example the cotangent bundle $TS^* = \{(p, a):\ p\in S, a\in T_pS^*\}$.
Every time you have a bundle like $TS^*$ you also have its projection $\pi: TS^*\to S$ that projects the elements of the bundle to the base point $p$ on $S$ that they are attached to, i.e. $\pi: (p, a)\mapsto p$. The counter image $\pi^{-1}(p) = \{p\}\times T_pS^*$ is also called the fiber of $TS^*$ at $p$.
A $1$-form on $S$ is a so called section of $\pi:TS^*\to S$, meaning a function $\alpha: S\to TS^*$ such that $\alpha(p) = (p, a_p)$ for some $a_p\in T_pS^*$, or in other words a function $S\to TS^*$ that attaches to every point $p\in S$ an element of the fiber at $p$.
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