I am trying to understand differential forms and their exterior derivative. I find the "algebraic" definition of the exterior derivative by some universal properties very unsatisfying and want a more geometric picture. This is how I picture differential forms, please correct me if this is wrong:
Let's consider the following intuition for differential forms:
A differential $k$-form measures flow through an infinitesimal $k$-parallelotope.
So for a $k$-form $\omega$ and tangent vectors $(v_i)_{i=1}^k$ at a point $p$ we interpret the real number $\omega_p(v_1,...,v_n)$ as the "flow through the infinitesimal parallelotope localised at the point $p$ spanned by the vectors $(v_i)_{i=1}^k$.
Following Stoke's theorem I think the right intuition for the exterior derivative should then be:
The $(k+1)$-form $d \omega$ measures the flow through an infinitesimal $(k+1)$-parallelotope by measuring the flow through the faces of the parallelotope individually by $\omega$ (since these are themselves $k$-parallelotopes).
Because then Stokes' theorem simply states informally
The total flow through a manifold (with boundary) is equal to the total flow through its boundary.
Using this I now want to "derive" a formula for the exterior derivative. Let's consider the flat manifold $\mathbb{R^2}$ together with a differential $1$-form $\omega$ on it. In the following I'll identify the tangent space $T_p\mathbb{R^2} \cong\mathbb{R^2}$, so $\omega$ simply assigns every vector in $\mathbb{R^2}$ a linear functional on $\mathbb{R^2}$. By using the intuition from above the exterior derivative can be approximated by looking at a non-infinitesimal parallelogram and individually calculating the flow through the sides and adding them (for small $h$):
$$d\omega_p(v_1, v_2) \ \approx \underbrace { \omega_{p}(v_1) - \omega_{p+hv_2}(v_1) }_\text{flow through the sides parallel to $v_1$} + \underbrace { \omega_p(v_2) - \omega_{p+hv_1}(v_2) }_\text{flow through the sides parallel to $v_2$}. $$
Here opposite sides have opposite sign because the flow "is entering through one side and exitting through the other side".
We now write $\omega$ in standard planar coordinates: $\omega = f^1 dx + f^2 dy$ for smooth functions $f_1,f_2$ on $\mathbb{R^2}.$ Then in the limit $h \to 0$ of the approximate expression divided by $h$ ("infinitesimalising" the parallelogram) we get
$$ d\omega_p(v_1, v_2) = \ \ \frac{\partial f^1}{\partial v_2} dx(v_1) + \frac{\partial f^2}{\partial v_2}dy(v_1) \ \ \ \ + \ \ \ \ \frac{\partial f^1}{\partial v_1} dx(v_2) + \frac{\partial f^2}{\partial v_1}dy(v_2).$$
This is a symmetric bilinear form in $v_1,v_2$. However, the form should be anti-symmetric such that $d\omega$ represents a differential $2$-form. In particular, an infinitesimal parallelogram of vanishing area needn't have vanishing flow, which clearly shouldn't be. But by replacing the "$+$" between the terms for sides parallel to $v_1$ and sides parallel to $v_2$ by a "$-$" one gets an anti-symmetric bilinear form and recovers the correct definition of the exterior derivative.
I however don't see why the flow through $v_2$ should be counted as negative and the flow through it's approximate counterpart $v_2+hv_1$ as positive.
I have now two questions:
- Is this a good way of thinking about differential forms? How could one intuitively think about the identity $d \circ d = 0$ in this picture?
- If yes, how does one explain the wrong sign?