So I know Stokes theorem for integrating exact differential forms over a chain:
$$\int_cd\omega=\int_{\partial c}\omega.$$
But what if there are holes? More precisely, consider the following famous example:
Consider the punctured plane $U=\Bbb R^2\setminus\{(0,0)\}$ and the differential form $$\omega=\frac{1}{x^2+y^2}(-ydx+xdy)$$ on $U$. It is known that line integral of $\omega$ along a circle $c:[0,1]\to U, c(t)=(\cos 2\pi t,\sin 2\pi t)$, is $2\pi$.
Stokes theorem does not really apply here because neither is $\omega$ exact nor is $c$ the boundary of another chain. But it is possible to "define" the following (I saw similar things from physics textbook and notes):$$d\omega=2\pi\delta(x,y)dx\wedge dy$$where $\delta$ is the Dirac delta function, giving infinity when $x=y=0$, and $0$ otherwise, such that the integral of $\delta(x,y)dx\wedge dy$ over a subset $S$ of $\Bbb R^2$ is $1$ if the origin is an interior point of $S$. Regard $c$ as the boundary of the unit disk $D$ at the origin and Stokes theorem can be "applied" just fine, by$$\int_Dd\omega=\int_{c}\omega.$$
Is there a way to formalise these ideas? If possible, I would like to see some reference to formal definitions of such generalised exterior derivative and proofs of such generalised Stokes theorem.
I know some complex analysis (Cauchy's integral formula), the definition of a distribution, and the definition of de Rham cohomology if it will help.
Best Answer
The usual way to arrive at this is by removing an $\varepsilon$-disk centered at the origin and taking the limit, but let me give a sketch of how you might do this with the Dirac delta function approach. I'm going to treat $\delta_0$ as the distribution with the property that for any smooth $f\colon\Bbb R^2\to\Bbb R$ with compact support, we have $$\delta_0(f) = \int_{\Bbb R^2}\delta_0(x,y) f(x,y)\,dx\wedge dy = 2\pi f(0).$$
You start with $\omega = \dfrac{-y\,dx+x\,dy}{x^2+y^2} = d\theta$ and we want to see that $d\omega = \delta_0 \,dx\wedge dy$ as a current. In particular, this means we want to see that for any smooth $f$ with compact support, we have $$\int_{\Bbb R^2} f\,d\omega = 2\pi f(0).$$ As usual, we start with the equation [think "integration by parts"] $$d(f\omega) = f\,d\omega + df\wedge\omega$$ and integrate over a large closed ball $B(0,R)$ with the property that $f=0$ on $\partial B(0,R)$. It will be convenient to use polar coordinates, of course, and then we see that $df\wedge\omega = \left(\dfrac{\partial f}{\partial r}dr + \dfrac{\partial f}{\partial\theta}d\theta\right)\wedge d\theta = \dfrac{\partial f}{\partial r}dr\wedge d\theta$. So \begin{align*} \int_{\Bbb R^2} f\,d\omega &= \int_{\Bbb R^2} d(f\omega) - \int_{\Bbb R^2} df\wedge\omega \\ &= \int_{B(0,R)} d(f\omega) - \int_{B(0,R)} df\wedge\omega \\ &= \int_{\partial B(0,R)} f\omega - \int_0^{2\pi}\int_0^R \dfrac{\partial f}{\partial r}dr\,d\theta \\ &= 0 - \int_0^{2\pi} \big(f(R,\theta)-f(0,\theta)\big)d\theta = 2\pi f(0), \end{align*} as needed.