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Can somebody help me understand this. Let $\omega$ be a closed two-form on $\mathbb{R}^3$ and $\eta$ a one-form such that $\omega=d\eta$. $M$ is an orientable manifold with boundary $\partial M$. $i:M\to\mathbb{R}^3$ and $j:\partial M\to M$ are inclusion maps.

What I don't understand is, why is it that you can claim that

$$\int_M i^*\omega =\int_M d(j^*(i^*\eta)) $$

I just forgot all the algebra rules involved.

I believe you can exchange the operators $d$ and $i^*$ like this $$i^*(d\eta) = d(i^*\eta)$$ right?

But how does the $j^*$ factor into all of this? As I said, I would love to know the algebraic rules on operations involving pullbacks because I used to have a list but I lost it and I cannot find much info online.

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  • $\begingroup$ I don't think the claim you've made even makes sense: $d(j^* i^* \eta)$ is a form on $\partial M$, but you're trying to integrate it over $M$? $\endgroup$ – Anthony Carapetis Jan 27 '18 at 2:17
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Maybe you have to use Stoke's Theorem:

$\int_Μd{(j^*(i^*(η))}=\int_{\partial M}j^*(i^*(η))=\int_{\partial M}(ij)^*(η))$

which makes sense since $ij$ is a function from $\partial M$ to $\mathbb{R^3}$.

The "algebraic" rule is that the pullback is a contravariant functor, meaning $(fg)^*=g^*f^*$

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  • $\begingroup$ Watch out. It's contravariant, not covariant. $\endgroup$ – Ted Shifrin Jan 26 '18 at 23:38
  • $\begingroup$ @nick-a Sorry maybe my question wasn't clear. I know how to continue from $\int_M d(j^*(i^*(\eta))$ using Stokes Theorem. I want to know how to get to $\int_M d(j^*(i^*(\eta))$ from $\int_M i^* \omega$. Maybe a bit of a silly question and it seems obvious to you but I am confused about how and why he introduced $j^*$ in that step. $\endgroup$ – berrygreen Jan 26 '18 at 23:53
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    $\begingroup$ As I think of it , it confuses me too... Can you tell us the book or sets of notes where you found this so I can have look and maybe find out from the context . $\endgroup$ – Nick A. Jan 27 '18 at 9:17

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