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Let $M$ be an oriented, compact smooth $d$-dimensional manifold with boundary, and let $\omega_t$ be a smooth family of $d$-forms. Since $\omega_t$ are closed, they are exact, so there is a corresponding family of $d-1$ forms $\eta_t$ such that $d\eta_t=\omega_t$.

Can we always choose $\eta_t$ to be smooth in $t$ (or at least $C^1$ in $t$)?

For my application, we can assume that $\omega_t=\omega_0$ for all $t$, in some open neighbourhood of $\partial M$.

I don't think it matters but for the record, I am not interested in "long-time behaviour", i.e we can restrict $t \ge 0$ to be small- and work with variation $(\omega_t)_{t \in [0,\epsilon)}, \omega_0=\omega$.

Asaf Shachar
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  • This is not a proof, but I will be really surprised if it turns out to be impossible, as this tells about the existence of smooth section of an affine bundle over a vector space, though the two relevant vector space is infinite-dimensional. – cjackal Sep 22 '17 at 08:22
  • And it might be worth notice that a little Hodge theory says a path of exact forms on a closed manifold is always represented by the differential of a path of forms of degree one lower. (a forklore to symplectic geometers; google Moser's argument if you haven't heard of this.) The problem is in case of manifolds with boundary the codifferential is not the adjoint anymore. – cjackal Sep 22 '17 at 08:47

1 Answers1

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After some thought, this just follows from a little Hodge theory, and actually the argument that I will give from now on is quite familiar to symplectic geometers.

First notice that we can assume WLOG that $\omega_t$ vanishes near the boundary (by subtracting $\omega_0$ from $\omega_t$). Now, you can take the double of your manifold, and by our assumption, the family $\omega_t$ extends to the double. Thus we can use the corresponding argument applicable to closed manifolds. The argument is as follows:

First choose any Riemannian metric on your closed manifold so that you obtain a Hodge structure. We have assumed that the family $\omega_t$ is exact, which implies it lies on the orthogonal complement of the kernel of Hodge Laplacian $\Delta$. Thus we obtain a smooth family of $d$-forms $\Delta^{-1}(\omega_t)$, and you can now see that $\delta\Delta^{-1}(\omega_t)$ is a desired form (where $\delta$ is the codifferential):

By definition, $\omega_t=\Delta(\Delta^{-1}(\omega_t))=d\delta\Delta^{-1}(\omega_t)+\delta d\Delta^{-1}(\omega_t)$ holds, meaning it suffices to show that $\delta d\Delta^{-1}(\omega_t)$ always vanishes. Indeed,

$\left<\delta d\Delta^{-1}(\omega_t),\delta d\Delta^{-1}(\omega_t)\right>=\left<\omega_t-d\delta \Delta^{-1}(\omega_t),\delta d\Delta^{-1}(\omega_t)\right>=\left<d\omega_t-d^2\delta \Delta^{-1}(\omega_t),d\Delta^{-1}(\omega_t)\right>=\left<0,\delta d\Delta^{-1}(\omega_t)\right>=0$.

Edit. First let me state some well-known facts about the Hodge Laplacian $\Delta$. You may find proofs in introductory textbooks about the Hodge theory.

The Hodge Laplacian is defined as $\Delta=d\delta+\delta d$, which allows a local expression in terms of standard basis $\left\{e_1\wedge \cdots \wedge e_k\right\}$. This is a self-adjoint elliptic operator. Its kernel is the space of harmonic forms, which is finite-dimensional, and $\Delta$ is invertible (in sense of continuus operators) on the orthogonal complement of $\mathrm{ker}\Delta$, which is called Hodge theorem.

Now $\Delta^{-1}\omega_t$ is smooth as $\Delta^{-1}$ is a continuous linear operator. Of course the principle that the OP mentioned in the comment plays a role under the water, but it is a bit hard to explicate where it was used. If you know the elliptic regularity, then it would be automatic as $\Delta$ is elliptic. Next, $\Delta^{-1}$ is local as $\Delta$ is.

cjackal
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  • Thanks. Can you please elaborate on why $\Delta^{-1}\omega_t$ is smooth in $t$? Intuitively I guess this is because $\Delta^{-1}$ is a differential operator in "space", and derivatives w.r.t space commute with derivatives w.r.t "time". BTW, is there any formula for $\Delta^{-1}$ in terms of $d,\delta$? Is it a local operator? – Asaf Shachar Sep 22 '17 at 15:24