Is $\int_X d(F^{*}\omega) = 0$ because of a corollary of Stokes' Theorem?

Question

My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave.

The last part of Proposition 11.11 goes $$\int_X d(F^{*}\omega) = \int_X F^{*}(d\omega) = 0$$

Can we just skip commutativity of pullback and exterior derivative and directly say that $\int_X d(F^{*}\omega) = 0$ by Corollary 10.9 of Stokes' Theorem (Theorem 10.8)?

If we cannot skip, then why is it that $\int_X F^{*}(d\omega) = 0$?

By the way, if anyone wants to know the definition of domain with smooth boundary: I think this is the book's term for a subset, of a manifold, that is manifold with boundary. See here: Why is there a form with compact support on a connected oriented manifold with integral one but with support contained in a given open proper subset?

I realized my error: $X$ is not a manifold (without boundary)! We'll have to use commutativity and the fact that for any smooth $n$-dimensional manifold (without boundary) $M$, $\Omega^nM = Z^nM$, as pointed out below.

score 3 · Accepted Answer · answered May 07 '19 at 17:29

3

The point is that $\omega$ is an $n$-form on an $n$-dimensional manifold, so $d\omega = 0$. (The only $(n+1)$-form on an $n$-dimensional manifold is $0$.) On the other hand, $F^*\omega$ is an $n$-form on an $(n+1)$-dimensional manifold, so, as such, its exterior derivative would not necessarily have to be $0$.

answered May 07 '19 at 17:29

Ted Shifrin

115,160

I actually make the specific mistake of assuming that $X$ is a manifold (without boundary) right? (See edit) Thanks! – May 07 '19 at 17:40
Of course, the essential ingredient is that $X$ is a domain with boundary and the boundary has the same dimension as $M$. – Ted Shifrin May 07 '19 at 18:00
Why isn't it that $\dim(\partial X)=\dim(X)-1=\dim(M)-1$? – May 09 '19 at 07:41
Because $X$ started out $(n+1)$-dimensional? – Ted Shifrin May 09 '19 at 16:21

Is $\int_X d(F^{*}\omega) = 0$ because of a corollary of Stokes' Theorem?

1 Answers1