3
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ See en.wikipedia.org/wiki/Current_(mathematics) especially the notion of a "boundary operator" $\endgroup$
    – Willie Wong
    Commented Nov 28, 2016 at 20:50
  • $\begingroup$ @WillieWong thank you for your very helpful comment. $\endgroup$
    – Ali Taghavi
    Commented Nov 28, 2016 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). $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2016 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. $\endgroup$
    – Willie Wong
    Commented Nov 29, 2016 at 14:23

1 Answer 1

Reset to default
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 $I_1=a\int_M$.

Improve this answer
$\endgroup$
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? $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2016 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 $M$. $\endgroup$
    – Andreas Cap
    Commented Nov 29, 2016 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. $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2016 at 21:22
  • 2
    $\begingroup$ I have added more details to the answer. $\endgroup$
    – Andreas Cap
    Commented Nov 30, 2016 at 9:02

You must log in to answer this question.

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