3
$\begingroup$

I learned the Stokes theorem in the context of differential topology, it says that if I have a $n-1$ form $w$ with compact support on an oriented smooth n-manifold, then $ \int_{\partial M} w = \int_M dw$. However I was asked to prove the following fact using the stokes theorem: given a compact set $K \subset \mathbb C$, and a holomorphic function on $\mathbb C$, $\int_{ \partial K} f =0 $, how should I use the stokes theorem since $K$ here is necessarily a manifold? May be it involves arguing that $\partial K$ can be realized as the boundary of some manifold?

$\endgroup$
2
  • 1
    $\begingroup$ It isn't true: let $K$ be the closed real interval $[0, 1]$ and let $f(z) = z$ be the identity function. $K$ is a compact subset of $\mathbb{C}$, and it also is a oriented smooth 1-manifold, but the integral equals 1. $\endgroup$ Commented Jun 14, 2017 at 22:44
  • $\begingroup$ Hmm. Is it possible that the asker of the question meant to assume that $K$ is a nice set? $\endgroup$ Commented Jun 14, 2017 at 22:44

1 Answer 1

1
$\begingroup$

The result is true if we suppose that a compact set $K$ is a smooth oriented 2-manifold over $\mathbb{C}$, and that it's boundary, $\partial K$, is a smooth 1-manifold. Then we are integrating the 1-form $w = f(z)dz$, whose exterior derivative is known to be $0$ (because $f$ is holomorphic).

We can, without loss of generality, suppose that $K$ is connected. Being compact, $K$ is also closed and limited. Now, if we can prove that the boundary points of $K$ (topological boundary points, not the same thing as the boundary of a manifold) form a connected subset (which I believe to be true), then the theorem should become true under the following two hypothesis:

  1. $K$ doesn't have an empty interior
  2. Every open ball centered on a boundary point of $K$ intersects an interior point of $K$

Actually, (2) implies (1), but I'm leaving it there to emphasize it. I think that using this hypothesis, we can prove that $K$ is actually a (sufficiently smooth) 2-manifold. (1) rules out the example $K = \{x + 0i \in \mathbb{C}| x \in [0,1]\}$ that I gave on the comments, and (2) rules out other weird compact sets such as:

$K = \{z \in \mathbb{C} | |z + 1| \leq 1\} \cup \{x + 0i \in \mathbb{C}| x \in [0,1]\}$

The set above is the union of the closed disk centered on $z = -1$ and radius $1$ with the real line segment $[0, 1]$. The first part can be realized as a smooth oriented 2-manifold and satisfies property (2), while the second one can't. This is still informal though, I'm not sure how laborious or technical it is to prove

Theorem Every non-empty compact set $K$ satisfying (1) and (2) can be realized as a smooth oriented 2-manifold over $\mathbb{C}$.

EDIT: Maybe even this won't be enough - try to imagine the unitary disk $|z| \leq 1$, but create a perturbation on it's boundary using something like the Weierstrass function, a function continuous everywhere but differentiable nowhere. It may even be a continuous manifold, but I don't think it will be differentiable, let alone smooth enough.

EDIT2: It is true that the interior points of $K$ form a differentiable oriented 2-manifold (they are simply a submanifold of $\mathbb{C}$). It's boundary is precisely the closure of it's interior, which is contained in $K$ (and if $K$ satisfies (2), it is all of $K$). It doesn't need however to be a smooth nor even a differentiable manifold - google Koch Snowflake , center it on the origin and let it be your $K$. Stronger hypothesis are needed then, and I think this can't be done with nothing less than "supposing $K$ satisfies (2) and it's boundary points are a smooth 1-manifold".

$\endgroup$
5
  • $\begingroup$ Thank you very much that is a very detailed answer!!!!! Is there a stokes theorem more of a computational essence? I am pretty sure people do not have to deal with smooth manifold in calc 3. $\endgroup$
    – z.z
    Commented Jun 15, 2017 at 0:49
  • $\begingroup$ @z.z while the machinery involved in the generalized Stokes' Theorem is pretty abstract (differential forms and smooth manifolds are, at first, counter-intuitive notions and formalism-heavy), the work done internalizing those ideas pays off on the long run. It generalizes all the big multi-variable calculus theorems such as the Gauss-Ostrogradsky's, the Kelvin-Stokes', the Green's, the Gradient's, and it also gives a unified language for vector fields, scalar fields, parametrization, the $dx$, $dy$, $dA$, $dS$, orientation, etc. It is, really, the most modern way to learn it. But... $\endgroup$ Commented Jun 15, 2017 at 0:58
  • $\begingroup$ @z.z not all calculus courses teach it that way and it isn't necessarily a bad thing. Most references in applications (engineering, physics, social sciences, etc) still do multi-variable calculus the old way, using more intuitive notions but broken down to several specific cases. You could see the four theorems cited above for alternatives to the generalized Stokes' Theorem, as they cover basically all the most usual applications of the theorem. $\endgroup$ Commented Jun 15, 2017 at 1:01
  • $\begingroup$ @z.z computationally speaking, differential forms and manifolds aren't very complicated either. You take a manifold, which is essentially a parametrized set, you use this parametrization to pullback your differential form, which is (or should be) a purely mechanical process, and then you have iterated single-variable integrals which you can then solve trough your favorite single-variable techniques. You're welcome to ask around the forum with a specific example or calculation, should you have questions regarding computational aspects of Stokes' Theorem! $\endgroup$ Commented Jun 15, 2017 at 1:04
  • $\begingroup$ really really thank you!!!! That is really very helpful!!! $\endgroup$
    – z.z
    Commented Jun 15, 2017 at 1:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .