3
$\begingroup$

Accepting Stokes' theorem (101) as a premise, how do I prove the fundamental theorem of calculus (102)? I know that the FTC is a straightforward specialization of Stokes' theorem and so proving $\text{(101)} \implies \text{(102)}$ should be trivial. I'm using the triviality of this implication to try to improve my understanding of Stokes' theorem. I vaguely remember Stokes' theorem from school, but didn't really understand it then and don't really understand it now. I wrote both theorems with the boundary-related part on the left and the interior-related part on the right.

Basically, what values do I pick for $\Omega$ and $\omega$ to turn (101) into (102)?

$$ \int_{\partial \Omega} \omega = \int_{\Omega} d \omega \tag{101} $$

$$ F(b) - F(a) =\int_{a}^{b} f(x)dx \tag{102} $$

Explaining the notation, $\omega$ is a differential form and and $\Omega$ is an orientable manifold. I think this means that $\Omega$ is not inherently equipped with an orientation but is capable of receiving one. So, $\Omega$ cannot be something like a Möbius strip.

I'm trying to get the left sides to match first, but I'm stuck.

So it seems like the most straightforward way to prove the implication is to have $\Omega$ be a closed interval on the real line $[a, b]$.

However, the boundary of $[a,b]$ is a set of two points $\{a, b\}$. I'm trying to understand what $\omega$ would have to be to make $\int_{\{a, b\}} \omega$ equal to $F(b)-F(a)$ rather than $F(b)+F(a)$.

What values to pick for $\Omega$ and $\omega$?

$\endgroup$
2
$\begingroup$

You should take $\omega = F$ (which is a $0$-form), and you should take $\Omega = [a,b]$ as you say, with the natural orientation (given by choosing the identity charts to be positively oriented, or to say it a different way, by choosing the nonvanishing $1$-form $dx$ on $[a,b]$, where $x$ is the identity coordinate function). This induces an orientation on $\partial[a,b] = \{a,b\}$, giving $\{a\}$ negative local orientation and $\{b\}$ positive local orientation, so that integrating $F$ on it gives $F(b)-F(a)$ instead of $F(b)+F(a)$. The statement of Stokes' theorem includes the assumption that $\partial \Omega$ is also oriented, in this specific way induced by the orientation on $\Omega$. Look here for more details in the 1-dimensional case.

$\endgroup$

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.