$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$
I have now posted also on MO.
Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds. Let $f:\M \to \N$ be smooth, and let $\delta=d^*$ be the adjoint of the exterior derivative.
Let $1 \le k \le d$. Consider the following two properties $f$ can have:
- $\delta^{\N} \omega=0 \Rightarrow \delta^{\M}(f^*\omega)=0$ for every $k$-form $ \omega \in \Omega^k(\N)$.
$\,\,\,\,$ 2.$\,\delta_{\nabla^{\Lambda_k(f^*{\TN})}} (\bigwedge^k df)=0$.
Question: Does property 1 implies property 2?
Here $df \in \Omega^1(\M,f^*{\TN})$, and $\bigwedge^k df\in \Omega^k\big(\M,\Lambda_k(f^*{\TN})\big)$ is the induced map. $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} $ is the adjoint of $d_{\nabla^{\Lambda_k(f^*{\TN})}}$. (where $\nabla^{\Lambda_k(f^*{\TN})}$ is the induced connection on $\Lambda_k(f^*{\TN})$ by the Levi-Civita connection on $\N$).
Note that for one-forms ($k=1$), the second property is exactly harmonicity of $f$.
Edit 1: In the Euclidean case, the answer is positive:
Let $\M=\N=\R^d$ (endowed with the standard metrics), property $(1)$ implies property $(2)$. Indeed, let $\omega=dx^{i_1} \wedge dx^{i_2} \wedge \dots \wedge dx^{i_k}$. Then property $(1)$ implies
$$ \delta(df^{i_1} \wedge df^{i_2} \wedge \dots \wedge df^{i_k})=0. \tag{3}$$
For a given increasing multi-index $I=(i_1,\dots,i_k)$, write $df^I=df^{i_1} \wedge df^{i_2} \wedge \dots \wedge df^{i_k}$, and $e_I=e_{i_1} \wedge e_{i_2} \wedge \dots \wedge e_{i_k}$. Then a direct calculation (which is valid only in the Euclidean context) shows that $$ \delta_{\nabla^{\Lambda_k(f^*T\R^d)}} (\bigwedge^k df)=\sum_I \delta (df^I) \otimes e_I $$
This is an equality of elements in $\Omega^{k-1}(\R^d,\Lambda_k(\R^d))$.
Edit 2: property $(2)$ does not imply property $(1)$.
Let $\M=\N=\R^2$ (with the standard metrics). Define $f(x,y)=(x,x)$. $f$ is affine, so $\delta(df)=0$.
Let $k=1$ and set $\omega=ydx$. Then $$ \delta(\omega)=\star d \star ydx=\star d ( y dy)=\star 0=0,$$ but
$$ \delta(f^*\omega)=\delta(x dx) =\star d \star xdx=\star d ( x dy)=\star (dx \wedge dy)=1 \neq 0.$$