2
$\begingroup$

I'm sorry if this is an elementary question, but the book I'm using does not really shed any light on the following issue: for $S$ a compact, oriented 2-surface in $\Bbb R^3$ with boundary $\partial S$ and a smooth 1-form $\omega$ on $S$, we have Stokes' Theorem: $$ \int_S d\omega = \int_{\partial S} \omega $$ Now, one way we would want to use this theorem is to calculate integrals of the following form: $$ \int_S \alpha $$ Where $\alpha = \sum_{i,j} f_{i,j} \ dx_i\wedge dx_j$ is a 2-form and again $S$ is our 2-surface with boundary. In this case we would like to say $\alpha = d\omega$ for some 1-form $\omega$, so that we can reduce our integral over $S$ to an integral over $\partial S$. But how do we find this $\omega$? Is there a general procedure? I looked through the book I'm using and it doesn't really talk about this. Any help is appreciated, thanks.

$\endgroup$
  • 1
    $\begingroup$ See my answer to math.stackexchange.com/q/1683631/265466 for a general method that works on star-shaped domains and (with some modification) on any region diffeomorphic to one. $\endgroup$ – amd Jan 22 '18 at 1:55
1
$\begingroup$

A 2-form $\alpha$ is called exact if there exists a 1-form $\omega$ such that $\alpha=d\omega$. If such an $\alpha$ exists, then $\alpha$ is called a closed form (I.e., $d\alpha=0$, since $d\alpha=d(d\omega))=0$.

Every exact form is closed. But not every closed form is exact! See an example at the end of this answer. The Poincaré lemma in that link provides some additional conditions under which a closed form would be exact (and shows you how to compute $\omega$ from $\alpha$, i.e., luck not needed).

So you cannot always find such an $\omega$. However, if you start out with a $\omega$ that you want to integrate over $\partial M$, then you can use Stokes’ theorem to convert it into $\int_{ M} \alpha$.

Example: This is a fairly standard example. Consider the 1-form $$\frac{xdy-ydx}{x^2+y^2}$$ defined on $D=\Bbb R^2 \setminus \{0\}$. This form is well-defined and you May recognise it as “$d\theta$”. However, it is not the “d” of any 0-form (there doesn’t exist such a “$\theta$”!

$\endgroup$
1
$\begingroup$

I don't see any straightforward way to apply Stoke's Theorem without knowin any more information (and in general such a form you are looking for may not exist).

Maybe, since you have a surface in $\mathbb{R^3}$ you could see if this surface is the boundary of a solid in $\mathbb{R^3}$ and if you are lucky $dω$ will be much simpler than $ω$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.