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,

\[\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx.\]

The identity \(dy\wedge dx=-dx\wedge dy\) is the same orientation sign in two variables.

Example: missing Jacobian

The polar-coordinate mistake

\[dx\,dy=dr\,d\theta\]

is corrected by the form identity

\[dx\wedge dy=r\,dr\wedge d\theta.\]

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

\[\alpha=x\,dx+y\,dy,\qquad \beta=dx+2dy.\]

Then

\[\begin{aligned} \alpha\wedge\beta &=x\,dx\wedge dx+2x\,dx\wedge dy \\ &\quad + y\,dy\wedge dx+2y\,dy\wedge dy \\ &=(2x-y)\,dx\wedge dy. \end{aligned}\]

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

\[\eta=(x^2+y)\,dx+(xy+\sin y)\,dy,\]

one has

\[d\eta=(Q_x-P_y)\,dx\wedge dy=(y-1)\,dx\wedge dy.\]

For

\[\alpha=x\,dy\wedge dz+y^2\,dz\wedge dx+e^z\,dx\wedge dy,\]

one obtains

\[d\alpha=(1+2y+e^z)\,dx\wedge dy\wedge dz.\]

3. Pullback under polar coordinates

Let \(F(r,\theta)=(r\cos\theta,r\sin\theta)\). Then

\[F^*dx=\cos\theta\,dr-r\sin\theta\,d\theta,\] \[F^*dy=\sin\theta\,dr+r\cos\theta\,d\theta.\]

Therefore

\[F^*(dx\wedge dy)=r\,dr\wedge d\theta.\]

The Jacobian appears because wedge products discard the repeated terms.

Pullback of a 1-form

For \(\eta=-y\,dx+x\,dy\),

\(\begin{aligned} F^*\eta &=-r\sin\theta(\cos\theta\,dr-r\sin\theta\,d\theta) \\ &\quad +r\cos\theta(\sin\theta\,dr+r\cos\theta\,d\theta) \\ &=r^2\,d\theta. \end{aligned}\)

4. Verifying Stokes on a rectangle

Let \(R=[0,a]\times[0,b]\) and \(\eta=x\,dy\). 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

\[\int_{\partial R}x\,dy=\int_0^b a\,dy=ab.\]

Equivalently, using \(\eta=-y\,dx\) also gives \(d\eta=dx\wedge dy\). Then the top edge has orientation right-to-left and contributes

\[\int_{\text{top}}-b\,dx=ab.\]

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:

\[d\omega={f_{\bar w}(w)\over w-z}\,d\bar w\wedge dw.\]

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

\[\iint_R \sqrt{1+4u^2+4v^2}\,du\,dv.\]

8. Hodge-Weyl compatibility

Before solving a compact Poisson problem, do not try to solve \(\Delta u=f\) before checking

\[\int_M f\,\omega=0.\]

For example, on \(S^1\) the equation \(-u''=1\) has no periodic solution because

\[\int_0^{2\pi}1\,d\theta\ne0.\]

But \(-u''=\cos(5\theta)\) has the normalized solution

\[u={1\over25}\cos(5\theta).\]