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Let $M$ be a closed smooth manifold and $Q\subset M$ a closed embedded submanifold. Furthermore, let $\omega$ be an exact differential form $\omega\in\Omega^k(M)$ and vanishing identically on $Q$ (i.e. $\omega_q=0$ for any $q\in\ Q$).

Can we always find a primitive $\alpha\in\Omega^{k-1}(M)$ (i.e. such that $d\alpha=\omega$), whose restriction to $Q$ also vanishes?

This question came up during class and maybe it is obvious but I can't even seem to convince myself whether it is true or not, so any help is greatly appreciated.

Edit: counterexamples are given in the comments for $k=\dim Q+1$ and for the case of $\omega$ being $1$-form with $Q$ disconnected. In the context of the class, we were specifically considering $\omega$ to be a $2$-form, but I am also interested in the general case.

Edit2: There is an answer dealing with the condition of the pullback of $\omega$ to $Q$ being $0$. However, I meant that $\omega$ itself vanishes identically in points that belong to the submanifold $Q$.

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  • $\begingroup$ Please elaborate on your phrase "with vanishing deRham cohomology." Are you assuming that $\omega$ is closed and that $H^k(M) = 0$? I assume you are, so then you do get $\alpha$ with $\omega = d\alpha$. I can certainly give counterexamples to your wish with $Q$ disconnected. $\endgroup$ Mar 16, 2021 at 17:49
  • $\begingroup$ Of course, it's trivially false if $k=\dim Q+1$, for example. $\endgroup$ Mar 16, 2021 at 17:51
  • $\begingroup$ No, I didnt mean vanishing $H^k(M)$, I'm sorry, just the cohomology class of $\omega$, I have edited the question. $\endgroup$ Mar 16, 2021 at 17:53
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    $\begingroup$ For the first, take a function $f\colon M\to\Bbb R$ and take $Q$ to be the union of the preimages of two regular values. For the second, I can take $\alpha$ to be the volume form of $Q$ (extended arbitrarily to $M$) and let $\omega = d\alpha$. Of course $\omega|_Q$ is identically $0$. $\endgroup$ Mar 16, 2021 at 18:02
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    $\begingroup$ I will leave the question as it is for now. The point was that, if an exact form vanishing identically on $Q$ could be given a primitive that would also vanish on $Q$, one of the steps in one of our proofs would be unnecessary. If I provide too much context, we'll end up just falling back to the proof that we were doing in class which is not what I was interested in. I hope I am making sense. I have also edited the question again to make some things clearer. $\endgroup$ Mar 16, 2021 at 19:21

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Here is an honest example of the failure. Take $\omega = dx_1\wedge dx_2 + dx_3\wedge dx_4$ to be the standard symplectic form on $M=\Bbb R^4=\Bbb R^2\times (\Bbb R^2)'$. Let $Q$ be the torus $\{x_1^2+x_2^2=x_3^2+x_4^2=1\}$. Then of course the restriction of $\omega$ to $Q$ is identically $0$. Let \begin{align*} C&=\{x_1^2+x_2^2=1, x_3=x_4=0\}=\partial D\subset\Bbb R^2 \quad\text{and} \\ C'&=\{x_1=x_2=0, x_3^2+x_4^2=1\}=\partial D'\subset (\Bbb R^2)'. \end{align*} Then note that if $\omega=d\alpha$, we have $\int_C \alpha = \int_D d\alpha = \int_D \omega\ne 0$ (and similarly for $C'$). Thus, the restriction of $\alpha$ to $Q$ cannot be identically $0$.

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  • $\begingroup$ Thank you very much for your answer. But I dont understand why would the symplectic form vanish on $Q$. I understand that its pullback maybe would, but the symplectic form is necessarily nondegenerate. What am I missing? $\endgroup$ Mar 16, 2021 at 18:43
  • $\begingroup$ Oh, by restriction I mean the pullback by the inclusion. That's standard notation, I'm afraid. $\endgroup$ Mar 16, 2021 at 18:44
  • $\begingroup$ Oh I see. I tried to write it in a way that would not cause confusion but I should have made it specifically clear. I will edit the question, I apologize. And thank you once again. $\endgroup$ Mar 16, 2021 at 18:46
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    $\begingroup$ Oh, wait a minute. You're saying that $\omega$ is identically $0$ at every point $q\in Q$? As you yourself said, this contradicts nondegeneracy of $\omega$. I interpreted your entire question to be about restriction, i.e., pullback to the submanifold. So the question totally does not apply to the symplectic form? $\endgroup$ Mar 16, 2021 at 18:51
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Proposition 6.8 of Cannas da Silva's "Lectures on Symplectic Geometry" says:

Proposition: Let $U$ be a tubular neighborhood of a compact submanifold $Q \subset M$, and $i: Q\to U$ the inclusion. If $\tau$ is a closed $k$-form on $U$ such that $i^*\tau = 0$, then $\tau$ is exact. Moreover, there is a $k-1$ form $\alpha$ with $d\alpha = \tau$ on $U$, such that $\alpha|_{T_QM} = 0$.

In the context of Moser's trick, this is applied to $\tau = \omega_1 - \omega_0 $, the difference of two symplectic forms that have been assumed or arranged to agree on $T_QM$, the tangent bundle of $M$ restricted over $Q$.

By the example in Ted Shifrin's answer, it is necessary to restrict to a tubular neighborhood: even if $\tau$ is globally exact, there need not exist a global primitive $\alpha$ that also vanishes on $Q$.

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  • $\begingroup$ The method from Proposition 6.8 can be generalized as follows: Let $Q\subset M$ be a subset such that there is a smooth map $\rho: [0,1]\times M \to M$ with $\rho(0,x)=x$ and $\rho(1,x)\in Q$ for all $x\in M$ and $\rho(t,x) \in Q$ for all $t\in[0,1]$ and $x\in Q$ (for instance, $\rho$ is a strong deformation retraction). Let $\omega$ be a closed form on $M$ which vanishes on $Q$. Then $\eta = \int_{[0,1]} \rho^* \omega$ is a primitive of $\omega$ which vanishes on $Q$. $\endgroup$
    – Pavel
    Jun 10, 2021 at 13:14

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