# Integration of differential forms - how to extend (and fix?) this intuition?

For context, I'm a first-year undergrad in a linear algebra / multivariable calculus course. I've developed some intuition about $k$-vectors and $k$-forms, and I want to know:

• Are there any problems with this intuition (mistakes or areas in which it is counterproductive)?
• How can this be extended to things like closed/exact forms, the exterior derivative, and integration of differential forms? $\newcommand{\reals}{\mathbb{R}}$

Oh, and I've heard that all of this can be extended to "fields" instead of just $\reals$, and the complex numbers are an example of such a field. I haven't worked with them, though, so I'm just going to use $\reals$ for now.

Okay, so I'm familiar with vectors (both as elements of an abstract vector space and as elements of $\reals^n$ - in my class, we're mainly working with $\reals^n$, though).

From what I understand, a bivector (in Euclidean space) can be thought of as an "oriented parallelogram" similarly to how vectors can be thought of as oriented line segments. Similarly, a trivector can be thought of as an oriented parallelepiped, and this is extended to higher dimensions that are harder to visualize.

We can construct a $k$-vector from $k$ regular vectors by taking the wedge product of those vectors. The wedge product is multilinear and alternating. (This defines an "orientation" - the equivalent of forward/back or clockwise/counterclockwise - for each $k$-vector that switches if you switch two vectors in the wedge product.)

Geometrically, if two oriented parallelepipeds have the same orientation, they "span the same subspace of $\reals^n$", and one's "component vectors" can be rotated and scaled to "meet" the other's without leaving that subspace or using reflections. (If they "share the same subspace" but cannot be rotated/scaled to "meet" each other, then one is a negative scalar multiple of the other.)

(I'd never state it formally like this, but this is just about getting across my intuition, so hopefully it makes sense.)

A covector is an element of the dual space of a vector space, $V^*$, and is a linear transformation from $V$ to $\reals$. $V^*$ is also a vector space. Covectors are also called $1$-forms. They can also be wedged together to create $k$-forms, which are alternating multilinear transformations from $V^n$ to $\reals$. (We write the set of all $k$-forms as $\Lambda^k(\reals^n)$.)

(I'm vaguely familiar with this visualization of $k$-forms, but the exterior derivative part loses me.)

A smooth vector field is a $\mathcal C^\infty$ function that maps every point in $\reals^n$ to a vector. (Though I've never heard the terms, I assume there are also "bivector fields", "trivector fields", and so on.)

A differential $k$-form is a smooth $k$-covector field, which is the same as a smooth vector field but replacing "vector" with "$k$-covector" or "$k$-form".

Is this intuition "correct"? Can it be extended to describe closed/exact $k$-forms, the exterior derivative, and/or integration of differential forms?

• (Also, I apologize if this should've been split into multiple questions! I wasn't quite sure, but I saw some other intuition-confirming questions asking multiple things.) – Deusovi Apr 9 '17 at 7:02

I apologize, but nothing in the above goes outside of the literal definition of these objects. If you wish to understand the construction of differential forms, you will need to go through some notes on multilinear algebra. For this you can consult Darling Differntial forms and Connections or Loring Tu's Introduction to Manifolds.

The first text starts off with the exterior algebra and the second begins the work in the second chapter I believe. The geometry of differential forms will literally be a direct consequence of their actions on tangent vectors. For instance, one thinks of $du \wedge dv$ as an infinitesimal area form on $P:=\{ \alpha \textbf{u} + \beta \textbf{v}: 0 \leq \alpha, \beta \leq 1\}$ since,

$$(du \wedge dv) \ (\textbf{v}, \textbf{w}) = \textbf{Area}(P)$$

There are a ton of things like this that will become clear later once you read about the multilinear alegebra. Things there will be done on an arbitrary finite dimensional vector spaces so you still have to translate things to make sense for $V = \mathbb{R}^n$. You are also welcome to read Grassman's original account on bi-vectors, wedge product here, etc.

• That seems backwards; I think that $(\mathbb v,\mathbb w)$ or $\mathbb v\wedge\mathbb w$ should be the parallelogram, not $(du\wedge dv)$. The 2-form is a function that applies to a parallelogram to measure its area. – mr_e_man Aug 17 '18 at 23:48
• @mr_e_man: You are exactly right. I made an edit. – Faraad Armwood Aug 19 '18 at 15:53
• You still have $du\wedge dv$ as the parallelogram, and $P$ is undefined now. What I'm suggesting is that the parallelogram is $P=\mathbb v\wedge\mathbb w$, so that $\text{Area}(\mathbb v\wedge\mathbb w)$ is being measured by $du\wedge dv$. And $du\wedge dv$ can be visualized as a different parallelogram, or as a lattice of points as in the OP's link. – mr_e_man Aug 21 '18 at 2:31
• @mr_e_man: P was still defined as the span. I've made my last edit. – Faraad Armwood Aug 22 '18 at 15:14
• The span of two vectors is the entire plane; its area is infinite. Anyway, I don't think anyone else is bothered by it; you don't have to edit. – mr_e_man Aug 24 '18 at 2:19