$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
This is a follow-up of this question.
Let $\M,\N$ be closed $d$-dimensional oriented Riemannian manifolds ($d \ge 2$). Let $f:\M \to \N$ be smooth, and let $\delta=d^*$ be the adjoint of the exterior derivative. Recall a form $\omega$ is called co-closed if $\delta \omega=0$.
Let $1 \le k \le d-1$ be fixed. Consider the following two properties $f$ can have: For every open $U \subseteq \N$,
- $\omega\in \Omega^k(U)$ is co-closed $\Rightarrow f^*\omega$ is co-closed.
$\,\,\,\,$ 2. $\omega\in \Omega^k(U)$ is closed and co-closed $\Rightarrow f^*\omega$ is closed and co-closed.
Question: Does property 2 implies property 1?
(Property 1 implies property 2, since closed forms are automatically preserved by all maps).
The point is that locally, the space of closed and co-closed forms is infinite-dimensional. Thus property $2$ gives us a "infinite-dimensional amount" of information about $f$, so maybe there is a chance the answer is positive.
Edit:
Here are some examples for non-isometric maps which satisfy property $(1)$. Let $d$ be even, and set $k=\frac{d}{2}$. Then, every conformal map satisfies property $(1)$. Indeed, if $f$ is conformal, then for every $k$-form $\omega$ on $\N$,
$$\delta \omega=0 \Rightarrow \delta(|\det df|^{1-2\frac{k}{d}} f^*\omega)=0.$$
This essentially follows from the fact $f^*(\star \omega)= |\det df|^{1-2\frac{k}{d}} \star f^*\omega$, which implies $$ d(|\det df|^{1-2\frac{k}{d}} \star f^*\omega)=df^*(\star \omega)=f^* d(\star \omega)=0. $$
Putting $k=\frac{d}{2}$ we get the desired property.
Of course, another natural question is whether or not all maps which satisfy $(1)$ are conformal.
