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In this problem, it is assumed that $M$ is orientable. My question is: why is this needed? Is this a mistake in the book, or is my solution wrong somehow?

18-1 (paraphrased): Suppose that $M$ is an oriented smooth manifold and $\omega$ is a closed $p$-form on $M$.

(a) Show that $\omega$ is exact if and only if the integral of $\omega$ over every smooth $p$-cycle is zero.

(b) Suppose that the $p$-th singular homology $H_p(M)$ is finitely generated by the smooth $p$-cycles $\{c_1,\ldots,c_m\}$. Show that $\omega$ is exact if and only if for each $i$, $\int_{c_i}\omega = 0$.

My solution:

(a) If $\omega = d\eta$ is exact, then by Stokes's theorem for chains, $\int_c\omega = \int_c d\eta = \int_{\partial c}\eta = \int_0 \eta = 0$, since $\partial c = 0$ for any $p$-cycle $c$. Conversely if $\int_c \omega = 0$ for any cycle $c$, then $c \mapsto \int_c\omega$ represents $0$ in singular cohomology. By de Rham's theorem, $\omega$ therefore is zero in de Rham cohomology, so $\omega$ is exact.

(b) If $\omega$ is exact, then the integral of $\omega$ over any $c_i$ is zero by the same proof in part (a). Conversely if for all $i$: $\int_{c_i}\omega = 0$, then by linearity $\int_c \omega = 0$ for any $p$-cycle $c$, since $c$ can be written as a linear combination of the generators $\{c_1,\ldots,c_m\}$. Hence by the proof in part (a), $\omega$ is exact. $\square$

Remark: Theorem 18.12 ("Stokes's Theorem for Chains") on p.481 of Lee's book does not assume orientability of $M$.

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  • $\begingroup$ @Baloown As far as I can tell, the integral of a form over a chain is well-defined independently of orientability. If $c:\Delta_p\to M$ is a chain, then by definition $\int_{c}\omega:=\int_{\Delta_p}c^*\omega$, where $\Delta_p\subset \mathbb{R}^p$ is the standard $p$-simplex. $\endgroup$ – Matthew Kvalheim Sep 12 '17 at 16:34
  • $\begingroup$ You need orientability in order to apply Stokes. $\endgroup$ – Randall Sep 12 '17 at 16:41
  • $\begingroup$ @Randall I am only integrating over a chain $c:\Delta_p\to M$, and $\Delta_p$ inherits an orientation as it is a submanifold (with corners) of $\mathbb{R}^p$. On p. 481 of Lee, Thm 18.12 is "Stokes's Theorem for Chains" which does not assume orientability of $M$. Or am I making a mistake? $\endgroup$ – Matthew Kvalheim Sep 12 '17 at 17:59
  • $\begingroup$ I see. Hmm.......... $\endgroup$ – Randall Sep 12 '17 at 18:55
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You're right -- there's no need for the "oriented" hypothesis. Good catch. I've added this to my correction list.

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  • $\begingroup$ Thank you very much for your answer! :-) $\endgroup$ – Matthew Kvalheim Sep 12 '17 at 22:08

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