2
$\begingroup$

Let $M$ be a compact $m$ dimensional manifold with boundary $\partial M$.

Assume that $I_{1}, I_{2}$ are two linear functionals on $\Omega^{m}(M), \Omega^{m-1}(\partial M)$, respectively. Assume that we have $I_{1}(d\alpha )=I_{2} ( \alpha )$ for every $m-1$ differential form $\alpha$ on $M$.

Are $I_{1},I_{2}$ necessarily equal to a constant multiple of the usual integral?

$\endgroup$
4
  • 2
    $\begingroup$ See en.wikipedia.org/wiki/Current_(mathematics) especially the notion of a "boundary operator" $\endgroup$ Nov 28 '16 at 20:50
  • $\begingroup$ @WillieWong thank you for your very helpful comment. $\endgroup$ Nov 28 '16 at 21:24
  • $\begingroup$ @WillieWong but id I am not mistaken, I think that it does not give an explicit example since a form on the boundary is not necessarily a restricted from(From M to its boundary). $\endgroup$ Nov 29 '16 at 9:01
  • $\begingroup$ That's why it isn't an answer. I was uncertain how strictly you want to interpret that bit about the restrictions. $\endgroup$ Nov 29 '16 at 14:23
5
$\begingroup$

If you assume that $M$ is oriented, then up to a multiple $I_1$ and $I_2$ are the usual integral. In this case $\partial M$ is oriented, and since this is a manifold without boundary, the integral induces a linear isomorphism $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M))\to\mathbb R$. Now for any $\beta\in\Omega^{m-2}(\partial M)$ there exists an extension $\tilde\beta\in\Omega^{m-2}(M)$ and by naturality of $d$, we get $d\tilde\beta|_{\partial M}=d\beta$. Since $0=d(d\tilde\beta)$ the defining equation tells you that $0=I_1(d(d\tilde\beta))=I_2(d\beta)$. Thus you see that $I_2$ factorizes to the quotient $\Omega^{m-1}(\partial M)/d(\Omega^{m-2}(\partial M)$ and hence that is a number $a\in\mathbb R$ such that $I_2(\alpha)=a\int_{\partial M}\alpha$ for all $\alpha\in \Omega^{m-1}(\partial M)$.

Now consider $\alpha\in\Omega^{m-1}(M)$ and form $I_1(d\alpha)-a\int_M(d\alpha)$. By your defining equation the first term gives $I_2(\alpha)=a\int_{\partial M}\alpha$, so by Stokes $I_1-a\int_M$ vanishes on any exact form. But on a manifold with boundary any top degree form is exact, so $I_1=a\int_M$.

$\endgroup$
4
  • $\begingroup$ but your argument shows that $I_{2}$ is zero on the restriction of every exact form to the boundary. But this does not imply that $I_{21}$ vanishs at exact forms on the boundary which are not necessarily restriction of an exact form from $M$ to $\partial M$? Am I mistaken? $\endgroup$ Nov 29 '16 at 8:34
  • 1
    $\begingroup$ Any form on the boundary extends to a form on $M$, and the exterior derivative commutes with restriction to the boundary (which is a pullback). Hence any exact form on the boundary is the restriction of an exact form on $M$. $\endgroup$ Nov 29 '16 at 9:20
  • $\begingroup$ May I am missing some thing but assume that B1 and B2 are two forms on M which agree on the boundary this does not imply that dB1 and dB2 agree on the boundary. So if I am not mistaken, some thing is missing in your answer. $\endgroup$ Nov 29 '16 at 21:22
  • 1
    $\begingroup$ I have added more details to the answer. $\endgroup$ Nov 30 '16 at 9:02

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.