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I am working through this problem:

Let $M$ be a compact oriented 3-manifold with boundary, where the boundary is $\partial M=S^1\times S^1$. Let $\theta_i\in\Omega^1(\partial M)$, $i = 1, 2$ be the 1-forms obtained by pulling back the standard volume form $\theta\in\Omega^1(S^1)$ by the first and second projections $\pi_i:S^1\times S^1\to S^1$, $i = 1, 2$. Prove that it is not possible to extend both $\theta_1$ and $\theta_2$ to closed 1-forms on M (i.e. one of $\theta_1$ must fail to extend in this way).

My progress so far: Assume both $\theta_i$ could be extended to closed forms, then their wedge product is also closed. Then I would like to use Stokes to reach a contradiction - the volume of the torus is obviously non zero so I hope I can write $vol(nonzero)=\int_M d\theta_1\wedge d\theta_2=\int_{\partial M} d(d\theta_1\wedge d\theta_2)=0$ Am I correct? Any help and/or correction would be appreciated.

Android Netizen
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  • $M$ is a 3 manifold, but $d\theta_1\wedge d\theta_2$ is a $4$-form. – Moya Aug 23 '16 at 01:58
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    Try to integrate $\theta_1\wedge \theta_2$ on $\partial M$ instead. –  Aug 23 '16 at 02:00
  • To add slightly to @ArcticChar's comment: Note that you have an error in Stokes's Theorem: It should be $\displaystyle\int_M d\phi = \int_{\partial M}\phi$. – Ted Shifrin Aug 24 '16 at 00:14

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You're on the right track. Elaborating on the hints by Arctic Char and Ted Shifrin ...

Suppose both $\theta_1$ and $\theta_2$ extend to closed one-forms $\omega_1$, $\omega_2$ on $M$. Notice this means that the three-form $d(\omega_1\wedge\omega_2) = d\omega_1\wedge\omega_2 \pm \omega_1\wedge (d\omega_2) = 0$ is identically zero. Now compute: $$ 0 = \int_M d(\omega_1\wedge\omega_2) = \int_{\partial M} \omega_1\wedge \omega_2 = \int_{\partial M} \theta_1\wedge\theta_2 \neq 0 $$

Neal
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  • It all makes sense now. I made a rookie mistake. Many thanks to you for the detailed solution and also many thanks to both people who provided hints. – Android Netizen Aug 24 '16 at 22:41