Differential Forms 09: Worked Computations and Pitfalls
Differential forms / 09
Worked computations and pitfalls
Most mistakes with forms are sign mistakes, orientation mistakes, or attempts to treat coordinate expressions as coordinate-free objects.
Calculus root: every pitfall is already visible in first-year calculus
The common mistakes in differential forms are higher-dimensional versions of familiar calculus mistakes: dropping the sign when reversing limits, forgetting a Jacobian in substitution, confusing a derivative with an antiderivative, or ignoring a boundary term.
Example: sign from reversed limits
In one variable,
\[\int_b^a f(x),円dx=-\int_a^b f(x),円dx.\]The identity \(dy\wedge dx=-dx\wedge dy\) is the same orientation sign in two variables.
Example: missing Jacobian
The polar-coordinate mistake
\[dx,円dy=dr,円d\theta\]is corrected by the form identity
\[dx\wedge dy=r,円dr\wedge d\theta.\]The wedge product forces the substitution factor to appear before the integral is evaluated.
This article collects the computations that make Part III usable. Each item should be checked by expanding into ordered coordinate bases.
1. Wedge signs
Let
\[\alpha=x,円dx+y,円dy,\qquad \beta=dx+2dy.\]Then
\[\begin{aligned} \alpha\wedge\beta &=x,円dx\wedge dx+2x,円dx\wedge dy \\ &\quad + y,円dy\wedge dx+2y,円dy\wedge dy \\ &=(2x-y),円dx\wedge dy. \end{aligned}\]The only surviving terms are those with distinct differentials.
Second sign check
In \(\mathbb R^3\),
\[dy\wedge dz\wedge dx=dx\wedge dy\wedge dz\]because the cyclic permutation \((x,y,z)\mapsto(y,z,x)\) is even. But
\(dy\wedge dx\wedge dz=-dx\wedge dy\wedge dz.\)
2. Exterior derivative
For
\[\eta=(x^2+y),円dx+(xy+\sin y),円dy,\]one has
\[d\eta=(Q_x-P_y),円dx\wedge dy=(y-1),円dx\wedge dy.\]For
\[\alpha=x,円dy\wedge dz+y^2,円dz\wedge dx+e^z,円dx\wedge dy,\]one obtains
\[d\alpha=(1+2y+e^z),円dx\wedge dy\wedge dz.\]3. Pullback under polar coordinates
Let \(F(r,\theta)=(r\cos\theta,r\sin\theta)\). Then
\[F^*dx=\cos\theta,円dr-r\sin\theta,円d\theta,\] \[F^*dy=\sin\theta,円dr+r\cos\theta,円d\theta.\]Therefore
\[F^*(dx\wedge dy)=r,円dr\wedge d\theta.\]The Jacobian appears because wedge products discard the repeated terms.
Pullback of a 1-form
For \(\eta=-y,円dx+x,円dy\),
\(\begin{aligned} F^*\eta &=-r\sin\theta(\cos\theta,円dr-r\sin\theta,円d\theta) \\ &\quad +r\cos\theta(\sin\theta,円dr+r\cos\theta,円d\theta) \\ &=r^2,円d\theta. \end{aligned}\)
4. Verifying Stokes on a rectangle
Let \(R=[0,a]\times[0,b]\) and \(\eta=x,円dy\). Since \(d\eta=dx\wedge dy\),
\[\int_R d\eta=ab.\]On the boundary, the horizontal sides have \(dy=0\). The left vertical side has \(x=0\). The right vertical side is parametrized by \((a,y)\), \(0\le y\le b\), so
\[\int_{\partial R}x,円dy=\int_0^b a,円dy=ab.\]Equivalently, using \(\eta=-y,円dx\) also gives \(d\eta=dx\wedge dy\). Then the top edge has orientation right-to-left and contributes
\[\int_{\text{top}}-b,円dx=ab.\]This example isolates the boundary-orientation trap: the nonzero contribution depends on which primitive of \(dx\wedge dy\) is used.
5. Cauchy-Green bookkeeping
For \(\omega=f(w)(w-z)^{-1}dw\),
\[d\omega=d\left({f(w)\over w-z}\right)\wedge dw.\]The derivative of \((w-z)^{-1}\) produces a multiple of \(dw\wedge dw\), hence vanishes. Only the \(\bar\partial\) term survives:
\[d\omega={f_{\bar w}(w)\over w-z},円d\bar w\wedge dw.\]This is the entire algebraic reason the Cauchy-Green formula measures failure of holomorphicity.
6. Closed non-exact forms
The form
\[\eta=\operatorname{Im}{dz\over z}\]is closed on \(\mathbb C^\times\), but
\[\int_{|z|=1}\eta=2\pi.\]The form
\[d\log|z|=\operatorname{Re}{dz\over z}\]is also closed on \(\mathbb C^\times\), but it is exact:
\[d\log|z|=d\left({1\over2}\log(x^2+y^2)\right).\]The imaginary part detects angle and winding; the real part detects radius and has a global potential on the punctured plane.
7. Surface area
For the graph \(\sigma(u,v)=(u,v,u^2+v^2)\),
\[\sigma_u=(1,0,2u),\qquad \sigma_v=(0,1,2v).\]Thus
\[\begin{aligned} E&=1+4u^2,\qquad F=4uv,\\ G&=1+4v^2. \end{aligned}\]and
\[\sqrt{EG-F^2}=\sqrt{1+4u^2+4v^2}.\]So the area over a region \(R\) is
\[\iint_R \sqrt{1+4u^2+4v^2},円du,円dv.\]8. Hodge-Weyl compatibility
Before solving a compact Poisson problem, do not try to solve \(\Delta u=f\) before checking
\[\int_M f,円\omega=0.\]For example, on \(S^1\) the equation \(-u''=1\) has no periodic solution because
\[\int_0^{2\pi}1,円d\theta\ne0.\]But \(-u''=\cos(5\theta)\) has the normalized solution
\[u={1\over25}\cos(5\theta).\]