Exterior Differential forms

\(\def\pd{\partial}
\def\Real{\mathbb{R}}
\)

  • We define a differential \(k\)-form as
    $$\omega(x) = \sum_{I=\{i_1<i_2<\ldots<i_k\}} c_I(x) dx_{i_1}\wedge \cdots\wedge dx_{i_k},$$
    an exterior form with coefficients depending on the positions.

    (We will be using sometimes simplifying notation \(\xi_i = dx_i\) or \(\eta_i=dy_i\).)

  • Pullback maps: Let \(\Omega_1,\Omega_2\) be two Euclidean domains
    (not necessarily embedded in the same space) such that
    \(\psi:\Omega_1 \to \Omega_2\) is a smooth (continuously differentiable enough times) mapping between them.Let \(f\) be a real function on \(\Omega_2\). Then one can pull it back, by defining
    \[
    (\psi^*f)(x):=f(\psi(x)).
    \]There is no mystery here – this is just a way to substitute.

    Example: If \(\psi:\Real^2\to\Real^3\) is given by
    \[
    \psi(x_1,x_2)=(x_1^2, x_1x_2, x_2^2),
    \]
    and \(f:\Real^3\to\Real\) is a linear function \(ay_1+by_2+cy_3+d\), then
    \[
    \psi^*f(x_1,x_2)=ax_1^2+bx_1x_2+cx_2^2+d.
    \]

  • Similarly one can define the pullback of a differential form. Again, the pullback of a differential form is just a substitution.

    If \(\omega\) is a \(k\)-form on \(\Omega_2\), given by
    $$\omega(x) = \sum_{I=\{i_1<i_2<\ldots<i_k\}} c_I(x) dx_{i_1}\wedge \cdots\wedge dx_{i_k},$$
    and \(\psi:\Omega_1\to\Omega_2\) is given, in coordinates, as
    \[
    x_i=\psi_i(y_1,\ldots,y_m),
    \]
    then
    \[
    (\psi^*\omega)(y)=\sum_{I=\{i_1<i_2<\ldots<i_k\}} c_I(\psi(y)) dx_{i_1}(y)\wedge \cdots\wedge dx_{i_k}(y)=
    \sum_{I=\{i_1<i_2<\ldots<i_k\}} c_I(\psi(y))(\sum_l\frac{\pd x_{i_1}}{\pd y_l}dy_l)(y)\wedge \cdots\wedge (\sum_l\frac{\pd x_{i_1}}{\pd y_l}(y)dy_l),
    \]

    Example: if \(\psi(r,\phi)=(r\cos(\phi),r\sin(\phi))=(x,y)\),
    then
    \[
    \psi^*(dx\wedge dy)=(\cos(\phi)dr-r\sin(\phi)d\phi)\wedge(\sin(\phi)dr+r\cos(\phi)d\phi)=r dr\wedge d\phi.
    \]

  • If \(\gamma\) is a curve in
    \(\Omega_1\), it is mapped by \(\psi\) into a curve on \(\Omega_2\), again, by composition.
  • One can verify immediately that for 1-forms
    the composed path integrals of \(\psi^*\omega(\gamma)\) and its image in \(\Omega_2\)
    are the same:
    $$ \underbrace{\int_{\psi(\gamma)}df}_{\text{Integration in } \Omega_2} =
    \underbrace{\int_\gamma d \psi^*(f)}_{\text{Integration in } \Omega_1}.$$

  • Examples
    1. Define the 1-form \(\omega_1 = xdx + ydy = rdr \) – this is, clearly, just the differential \(dr^2/2\).
    2. Another 1-form \(\omega_2 = xdy-ydx\). Applied on a vector \(f=(s,t)\) at a point \(r=(x,y)\), it evaluates to
      $$ \omega_2(v) = \frac{xt – ys}.$$
      The physical meaning is therefore the angular momentum of the force \(v\) applied at \(r\).
    3. One more form: \(\omega_3=\omega_2/|r|^2\) – defined everywhere outside of the origin.
      In the polar coordinates \(d\phi := \frac{\mathbf{r}\times v}{|\mathbf{r}|^2} \),
      and we get the well-known identity:
      $$ \omega_1 \wedge \omega_2 = (rdr)\wedge d\phi = \frac{1}{2} (xdx+ydy)\wedge \frac{xdy+ydx}{x^2+y^2}
      = \frac{x^2dx\wedge dy – y^2 dy\wedge dx}{x^2+y^2} = dx\wedge dy $$

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