# If an exact form vanishes on a submanifold, can I find a primitive that also does?

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 $$Q$$.

• 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. Mar 16, 2021 at 17:49
• Of course, it's trivially false if $k=\dim Q+1$, for example. Mar 16, 2021 at 17:51
• No, I didnt mean vanishing $H^k(M)$, I'm sorry, just the cohomology class of $\omega$, I have edited the question. Mar 16, 2021 at 17:53
• 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 $0$. Mar 16, 2021 at 18:02
• 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. Mar 16, 2021 at 19:21

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 $$0$$.

• 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? Mar 16, 2021 at 18:43
• Oh, by restriction I mean the pullback by the inclusion. That's standard notation, I'm afraid. Mar 16, 2021 at 18:44
• 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. Mar 16, 2021 at 18:46
• 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? Mar 16, 2021 at 18:51

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 $$Q$$.

• 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 $Q$. Jun 10, 2021 at 13:14