# Calculus of Exterior and Differential Forms: Motivation

$$\def\pd{\partial}\def\Real{\mathbb{R}}$$

1. Let $$\Omega \subset \mathbb{R}^n$$ be some Euclidean domain, and consider
a curve $$\gamma:\underbrace{[0,1]}_{I} \to \Omega$$ given by
$$\gamma(t) = \begin{pmatrix} \gamma_1(t), & \ldots &, \gamma_n(t) \end{pmatrix}.$$Given a real-valued multivariate function $$f:\Omega\to \mathbb{R}$$ we’ll try to reconstruct the Newton-Leibnitz formula.
Suppose that we have a partition
$$0=t_0 < t_1 <\ldots < t_K=1$$, then
$$f(\gamma(1)) – f(\gamma(0)) = \sum_{k=0}^K \big(f(\gamma(t_{k+1})-f(\gamma(t_{k})\big)$$
If $$f$$ is continuously differentiable, then using the first Taylor approximation we get
$$f(\gamma(t_{k+1})) – f(\gamma(t_k)) = \Delta_k\cdot \frac{d}{dt} f(\gamma(t))\Big|_{t=t_k} + \text{o}(\Delta_k) ~~~~~~(**)$$
here $$\Delta_k = t_{k+1}-t_k$$.
The composed derivative is given by the so-called total differential formula we know from calculus:
$$\frac{d}{dt} f(\gamma(t)) = \sum_{l=1}^n \frac{\partial f}{\partial x_l} \frac{\partial \gamma_l}{\partial t}$$

Of course, this is a special case of the composition of two differentiable functions
$$F:R^k\to R^l,~~~ G:R^l \to R^m$$. The Jacobian matrix of the composed function is
the product of their respective Jacobians:
$$J_{F\circ G} = J_F\cdot J_G.$$

2. If we have the row vector
$$\frac{\pd f}{\pd x}(\gamma):=\begin{pmatrix} \frac{\partial f(\gamma)}{\partial x_1},&\ldots&,\frac{\partial f(\gamma)}{\partial x_n} \end{pmatrix}$$
and the differential approximation $$(**)$$:
$$f(\gamma(1))-f(\gamma(0)) \approx \sum_{k} \left(\frac{\pd f}{\pd x}\cdot \dot \gamma\right)\Big|_{t_k} \Delta_k$$
then in the limit we obtain
$$f(\gamma(1))-f(\gamma(0)) = \int_{0}^1 \frac{df}{dt}\cdot \dot \gamma dt = \int_{0}^1 df\cdot \dot \gamma$$
This integral holds for any (differentiable) re-parametrization of $$\gamma$$ that
maintains the same trajectory.
3. Example: let $$f(x,y) = x^2-y^2$$, and let $$\gamma(t) = (2t,t)$$.
Clearly, $$f(\gamma(1)) -f(\gamma(0)) = 2^2-1 = 3$$. We will verify this by integration.
The differential of $$f$$ is:
$$df = 2x dx + 2y dy$$
and the tangent is:
$$\dot \gamma(t) = 2 \partial_x + \partial_y$$
(where we use the fancy notation $$\partial_x,\partial_y$$ for the standard basis in $$\Real^2$$, and $$dx,dy$$ for the standard dual basis.Hence
$$\int_0^1 (df \cdot \dot \gamma) = \int_0^1 \begin{pmatrix} 2x & 2y \end{pmatrix} \begin{pmatrix} 2 \\1 \end{pmatrix} = = \int_0^1 (4x(t)-2y(t)) dt = \int_0^1 8t-2t dt = 6 \int_0^1 t dt = 3$$
4. Differential 1-forms If we drop the requirements that the components of the covector (linear functional)
coupled with $$\dot\gamma$$ are partial derivatives of a function, we arrive at the notion of 1-form
$$\omega = \sum a_k(x) dx_k,$$
with coefficients depending on $$x$$. We can integrate a 1-form along a curve $$\gamma$$ using exactly the same definition as above:
$\int_\gamma \omega:=\int_0^1 \sum a_k(\gamma(t))\frac{\partial \gamma_l}{\partial t}dt.$It is easy to check that this integral of 1-form along the curve does not change if we reparameterize the curve (that is if we represent $$t=t(s)$$ as a function of another variable $$s\in I$$).
5. Differential 1-forms: given the path integral
$$\int_\gamma df = \int_\gamma \sum a_k(x) dx_k$$
a differential 1-form (at point $$x$$) is the expression
$$\omega = \sum a_k(x) dx_k.$$
The components $$a_k(x)$$ are continuous (or smooth) functions, $$dx_k$$ are basis vectors
of linear functionals on tangent vectors. This is an arbitrary form, not necessarily
originated in a differential of a function.
6. Integration over the 1-form $$\omega$$ along a path $$\gamma$$ is defined as
$$\int_\gamma \omega = \int_0^1 \sum_k a_k(\gamma(t)) \frac{\partial \gamma_k}{\partial t} dt$$
7. Example: let $$\omega = xdx + ydy$$, and let $$\gamma(t):=(x_* t, y_* t)$$.
Then
$$\int_\gamma \omega = (x_*t,y_*t)\begin{pmatrix} x_* \\ y_* \end{pmatrix} dt = (x_*^2+y_*^2) \int_0^1 t dt = \frac{1}{2}(x_*^2+y_*^2)$$For another path $$\gamma’$$ (not to be confused with derivative prime), defined by
$$\gamma'(t) = \begin{cases} (2x_*t,0) & t\leq \frac{1}{2} \\ (2x_*,2y_*(t-\frac{1}{2})) & t\geq \frac{1}{2} \end{cases}$$
the curve would have
$$\int_\gamma \omega = \int_0^{0.5} …. + \int_{0.5}^1 …$$
but the end result will not change whatsoever.
8. Example: integrate the form $$\omega = y dx – xdy$$, along the curve
$$\gamma(t) = (x_*t, y_*t)$$
$$\int_\gamma \omega = \int (y,-x) \begin{pmatrix} x_* \\ y_* \end{pmatrix} dt = 0$$
Now, do the same along the arc
$$\gamma_a(t) = (\cos(\phi t),~~\sin (\phi t))$$
we get
$$\int_{\gamma_a} (\sin(\phi t), ~\cos(\phi t) \begin{pmatrix} -\sin(\phi t) \\ \cos(\phi t) \end{pmatrix} dt = \phi \int_0^1 (-\sin^2(\phi t) – \cos^2 (\phi t)) dt = -\phi$$
This is an example of a form whose path integral depends on the integration path.
9. Conserving/exact forms
consider a differential form, $$\sum a_k(x) dx_k,$$ and its integral along some curve.If that integral depends only on the end points of the curve, we call the
the form conservative or exact .
Incidentally, a form $$\omega$$ is a exact iff $$\omega = df$$ for some $$f$$.
We can explicitly construct the function $$f$$ by
$$f(x):=\int_\gamma \omega$$
10. A necessary condition for a form $$\omega = a_1(x) dx_1 + a_2(x) dx_2$$
to be exact is that
$$\frac{\partial a_1}{\partial x_2} = \frac{\partial a_2}{\partial x_1}.$$
This condition is not sufficient, for example the form
$$\omega = \frac{y dx }{x^2+y^2} – \frac{x dy}{x^2+y^2}$$
is not exact, but satisfies the necessary condition nonetheless, expect for at $$x=y=0$$ –
11. Exercise: Integrate $$dx/y$$ over the circle (oriented counterclockwise) $$\{x^2+y^2=R^2\}$$.