Let $(M,g)$ be a smooth $d$-dimensional Riemannian manifold, $d$ even. Are there obstructions (I guess in terms of curvature) for $g$ to have the following property:

For every $p \in M$ there exist a coordinate system around $p$, such that the co-frame associated with it satisfy:

$$ \delta(dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{\frac{d}{2}}})=0 \, \, \text{for every choice of indices } 1 \le i_1 < i_2 < \dots < i_{\frac{d}{2}} \le d.$$

Here $\delta=d^*$ is the adjoint of the exterior derivative.

In fact, for my purposes it suffices that $dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{\frac{d}{2}}}$ would be co-closed only for two complementing sets of indices, but I am not sure this problem is easier to analyse in practice. (though any ideas on that would be welcomed).

For the full problem, we have $\binom{d}{\frac{d}{2}} \cdot \binom{d}{\frac{d}{2}-1}$ equations, while the metric has $\frac{d(d+1)}{2}$ degress of freedom, so this problem is probably overdetermined. (*Can we prove that a generic metric admits no solutions?*).

Given a coordinate system, we can write $dx^i=a^i_je^j$ where $e^j$ is some (positive) orthonormal coframe. Writing $A=a^i_j$, we get that $A^TA=G^{-1}$, where $G=g_{ij}$ is the coordinate representation of the metric. This means we can assume that $A= \sqrt{G^{-1}}$. Then

$$ dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{k}}=a^{i_1}_{j_1}\dots a^{i_k}_{j_k} e^{j_1} \wedge \dots \wedge e^{j_k}=$$ $$\sum_{1 \le j_1 < j_2 < \dots < j_k \le d} \sum_{\sigma \in S^k} a^{i_1}_{j_{\sigma(1)}}\dots a^{i_k}_{j_{\sigma(k)}} \text{sgn}(\sigma) e^{j_1} \wedge \dots \wedge e^{j_k}= \sum_J A^I_Je^J,$$ where $A^I_J$ is the $k$-minor of the matrix $A= \sqrt{G^{-1}}$ corresponding to columns $I=(i_1,\dots,i_k)$ and rows $J=(j_1,\dots,j_k)$. Here $e^J:=e^{j_1} \wedge \dots \wedge e^{j_k}$. Taking the Hodge dual, we obtain

$$\star dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{k}}=\sum_J A^I_Je^{J^c},$$

so the final equation is

$$d \star dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{k}}=\sum_J dA^I_J \wedge e^{J^c}+A^I_J \wedge de^{J^c}.$$

I am not sure how to proceed from here. My idea was to expand $dA^I_J ,de^{J^c}$ and get some first order equation on $G$ and its minors. Then, I guess that second differentiation might give us something which is related to the curvature. However, it doesn't seem easy to do so.

Finally, we might simplify things a bit by assuming $e^{J^c} $ is closed. This raises the question what are the obstructions for such an orthonormal co-frame $e^j$ to exist. Note that if we want the $e^i$ themselves to be closed (not just their wedge product) then this forces the metric to be flat. I am not sure what is the obstruction when $|J|=k>1$.