Calculus of Exterior and Differential Forms: Motivation


  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 \(\gamma(t):=(x_* t, y_* t) \).
    $$ \int_\gamma \omega
    = (x_*t,y_*t)\begin{pmatrix} x_* \\ y_* \end{pmatrix} dt
    = (x_*^2+y_*^2) \int_0^1 t dt = \frac{1}{2}(x_*^2+y_*^2) $$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 \(x=y=0\) –
    which is a singular point (we will learn more about that in the chapter about complex analysis).
  11. Exercise: Integrate \(dx/y\) over the circle (oriented counterclockwise) \(\{x^2+y^2=R^2\}\).
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