Exterior derivative for 1-forms -- derivation of the formula

Question

I would like a help with understanding the formula for exterior derivative. Suppose that we have a differential 1-form on $\mathbb{R}^2$ given by $\omega(x)(v) = (a(x)dx + b(x)dy)(v)$. In other words $\omega$ is a mapping of the type $\mathbb{R}^2 \to \text{Hom}(\mathbb{R}^2, \mathbb{R})$, to every point we assign a linear mapping of the type $\mathbb{R}^2 \to \mathbb{R}$.

Now, I would like to derive the formula for $d\omega$. This means that I do not want to start with the axioms for the $d$ and I do not want to start with the formula and prove it. I want to naturally get to the fact that

$$d\omega(x)(u)(v) = \left (\frac{\partial b}{\partial x}(x) - \frac{\partial a}{\partial y}(x)\right )dx \wedge dy(v, u).$$

For example, for a 0-form $f\colon \mathbb{R}^2 \to \mathbb{R}$, we naturally want to approximate it every point by a linear mapping. So, we naturally get to the fact that the limit

$$\lim_{\|h\|\to 0}\frac{f(x+h)-f(x) - L(h)}{\|h\|}$$

must be $0$, for some linear mapping $L$ at a particular point $x$. The next step is to prove that the mapping $L$ has to be given by the $1\times 2$ matrix

$$\left (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right ).$$

Is it possible to do a similar analogy of linear approximation for the 1-form $\omega$, or in general for a $k$-form on $\mathbb{R}^n$?

---

EDIT:

In particular I would like to know why

$$d\omega(x)(v)(u) = (d\omega(x)(v))(u) - (d\omega(x)(u))(v).$$

This is in some textbooks a definition and it can be simplified to the formula for $d\omega(x)(v)(u)$ stated above.

Answer 1

If you want a limit-based definition for higher forms, I think basing it on the generalized Stoke's theorem is the way to go:

$$ \int_{B(P, r)} \mathrm{d}\omega = \oint_{S(P, r)} \omega $$

where $B(P,r)$ is the disk around $P$ with radius $r$, and $S$ is its boundary: the circle around $P$ with radius $r$.

If $\mathrm{d}\omega = f \mathrm{d}x \mathrm{d}y$, then you can obtain the values of $P$ by integrating over small disks:

$$ f(P) = \lim_{r \to 0} \frac{1}{\pi r^2} \oint_{S(P, r)} \omega $$

You could replace disks and circles with other suitable shapes, such as squares.

---

If I understand correctly, I believe the way you want to conceive the derivative doesn't actually relate to the exterior derivative — what you are conceiving, I think, is the covariant derivative, or maybe of connections.

Answer 2

Here is an extremely abstract way to extract the exterior derivative $d \omega$ of a $1$-form $\omega$ from the derivative in the ordinary sense. View $\omega$ as a smooth map $T M \to \mathbb{R}$; we can then take its ordinary derivative $D\omega : TTM \to T\mathbb{R}$, and the question is how to extract a $2$-form from this data. Now, we can canonically make the identification $T \mathbb{R} = \mathbb{R} \times \mathbb{R}$, so that $D \omega$ viewed as a map $TTM \to \mathbb{R} \times \mathbb{R}$ just does $\omega \circ \pi$ on the first factor. In other words, the diagram $$ \require{AMScd} \begin{CD} TTM @>{D \omega}>> \mathbb{R} \times \mathbb{R}\\ @V \pi VV @VV\operatorname{pr}_1 V\\ TM @>{\omega}>> \mathbb{R} \end{CD} $$ is commutative. Let $k : T\mathbb{R} \to \mathbb{R}$ be the projection onto the second factor. (If you want to be extremely fancy, $k$ is the connector of the canonical affine connection on the vector space $\mathbb{R}$.)

At any rate, we wind up with a map $k \circ D \omega : TTM \to \mathbb{R}$. The double tangent bundle $TTM$ comes equipped with a canonical flip map $\operatorname{flip} : TTM \to TTM$. Note that the canonical flip is not an endomorphism of the double tangent bundle given its usual vector bundle strucutre; it is actually an involution which exchanges two different vector bundle structures (see here for more info). It turns out that the difference $$ \delta := k \circ D \omega - k \circ D \omega \circ \operatorname{flip} : TTM \to \mathbb{R} $$ annihilates the so-called vertical subbundle of $TTM$. (The general fact is that $\alpha - \alpha \circ \operatorname{flip}$ always has this property when $\alpha$ is linear with respect to both possible vector bundle structures on $TTM$, and this is true for $D\omega$ since $\omega : TM \to \mathbb{R}$ is itself fiberwise linear.) This means that $\delta$ factors through the projection $(\pi, D\pi) : TTM \to TM \times_M TM$, where $D\pi$ denotes the derivative of the projection $TM \to M$. That is, we obtain a map $$ d\omega : TM \times_M TM \to \mathbb{R}, $$ linear in both $TM$ arguments and satisfying $d \omega \circ (\pi, D\pi) = \delta$. Moreover, since $(\pi, D\pi) \circ \operatorname{flip} = (D\pi, \pi)$ and $\delta \circ \operatorname{flip} = -\delta$, the map $d \omega$ is alternating.

---

So, there you go. The exterior derivative $d \omega$ arises by taking in some sense the "alternating part" of the full derivative $D \omega$, and this alternating part is able to descend from a map $TTM \to \mathbb{R}$ down to a map $TM \times_M TM \to \mathbb{R}$. This process is actually quite similar to taking the Lie bracket of vector fields $X$ and $Y$ viewed as maps $M \to TM$; it turns out that the difference $DY \circ X - \operatorname{flip} \circ DX \circ Y : M \to TTM$ has image contained within the vertical subbundle of $TTM$, and therefore canonically projects down to a map $M \to TM$; this is exactly $[X, Y]$.

A similar process allows you to Lie differentiate arbitrary tensor fields by a fixed vector field; the correct modification is to replace "$\operatorname{flip} \circ DX$" by the "prolongation" of $X$ into the relevant tensor bundle. (The prolongation of $X : M \to TM$ into $TTM$ is $\operatorname{flip} \circ DX$, which also goes by the name the complete lift of $X$.)