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

Question

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 $Q$.

Answer 1

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 $0$.

Answer 2

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 $Q$.