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 $|J|=k>1$.



Obstructions for the wedge of coordinate differentials to be harmonic
**Update (1 June 2018):** I have now figured out the 'little linear algebra lemma' in all dimensions $d = 2n$ and can give a complete answer to the OP's question: A metric $(M^{2n},g)$ possesses coordinate charts of the kind that the OP desires if and only if it is 'locally conformally unimodular Hessian', i.e., every point $p\in M$ has an open neighborhood $V$ on which there exist coordinates $x^1,\ldots,x^{2n}$ and a function $u$ with positive definite $x$-Hessian satisfying the Monge-Ampère equation $\det(\mathrm{Hess}\_x(u)) = 1$ so that $g$ is a multiple of the Hessian metric
$$
h = \frac{\partial^2u}{\partial x^i\partial x^j}\,\mathrm{d}x^i\mathrm{d}x^j\,.
\tag0
$$
When $d = 2n > 2$, the generic metric $g$ is not locally conformally unimodular Hessian, so such coordinate charts do not exist.


Because the proof is a bit clearer in the case $d=4$, I'm going to leave that in and simply indicate how the fundamental lemma (equation (2) below) changes in higher dimensions after the $d=4$ argument. What follows up until then is what I had written before:


I now have a complete answer in the case $d=4$ (the case $d=2$ being trivial). (I suspect that the answer holds for all (even) $d$, but that would require proving a little linear algebra lemma that I don't see an immediate proof of, but maybe it will come to me.)


Here is the answer: When $d=2n>2$ there are definitely obstructions (though, what they are explicitly in terms of curvature, I haven't a clue at this point), as my original argument (retained below) shows. 


Meanwhile, in the case $d=4$, one has the following necessary and sufficient condition: $(M^4,g)$ has the desired local coordinate systems if and only if $(M^4,g)$ is locally conformally *unimodular Hessian*, i.e., every point $p\in M$ has a neighborhood $V$ on which there exist coordinates $x^1,x^2,x^3,x^4$ and a function $u$ with positive definite $x$-Hessian satisfying the Monge-Ampère equation $\det(\mathrm{Hess}\_x(u)) = 1$ so that $g$ is a multiple of the Hessian metric
$$
h = \frac{\partial^2u}{\partial x^i\partial x^j}\,\mathrm{d}x^i\mathrm{d}x^j\,.
\tag1
$$
In fact, for such coordinates, we have $\mathrm{d}\bigl(\ast\_g(\mathrm{d}x^i\wedge\mathrm{d}x^j)\bigr) = 0$ 
for all $1\le i<j\le 4$.


*Note:* The set of such metrics *properly* contains the locally conformally flat metrics in dimension $4$.


The argument: Assume that $x^1,x^2,x^3,x^4$ are local coordinates on a simply connected set $V\subset M$ that satisfy the condition $\mathrm{d}\bigl(\ast\_g(\mathrm{d}x^i\wedge\mathrm{d}x^j)\bigr) = 0$ for all $1\le i<j\le 4$. Replacing $g$ by a scalar multiple $h$, we can assume that the volume form of $h$ is $\mathrm{d}x^1\wedge\mathrm{d}x^2\wedge\mathrm{d}x^3\wedge\mathrm{d}x^4$. By linear algebra (special to dimension $4$), there exists a basis $\alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ of $1$-forms on $V$ such that the following identities hold:
$$
\begin{aligned}
\ast\_h(\mathrm{d}x^1\wedge\mathrm{d}x^2)&= \alpha\_3\wedge\alpha\_4\\
\ast\_h(\mathrm{d}x^1\wedge\mathrm{d}x^3)&= \alpha\_4\wedge\alpha\_2\\
\ast\_h(\mathrm{d}x^1\wedge\mathrm{d}x^4)&= \alpha\_2\wedge\alpha\_3\\
\end{aligned}\qquad
\begin{aligned}
\ast\_h(\mathrm{d}x^3\wedge\mathrm{d}x^4)&= \alpha\_1\wedge\alpha\_2\\
\ast\_h(\mathrm{d}x^4\wedge\mathrm{d}x^2)&= \alpha\_1\wedge\alpha\_3\\
\ast\_h(\mathrm{d}x^2\wedge\mathrm{d}x^3)&= \alpha\_1\wedge\alpha\_4\\
\end{aligned}
\tag2
$$
In fact, the solution of these equations (unique up to replacing each $\alpha\_i$ by $-\alpha\_i$) is given by 
$$
\alpha\_i = h\_{ij}\,\mathrm{d}x^j\qquad\qquad\text{where}\quad 
h = h\_{ij}\,\mathrm{d}x^i\mathrm{d}x^j.
$$
Now, the closure assumption is equivalent to assuming that $\mathrm{d}(\alpha\_i\wedge\alpha\_j) = 0$ for $1\le i<j\le 4$, and, as remarked above, 
this implies that the $\alpha\_i$ themselves are closed. In particular, there exist functions $p\_i$ such that $\alpha\_i = \mathrm{d}p\_i$ where, because of the symmetry of $h\_{ij}=h\_{ji}$, it follows that
$$
\frac{\partial p\_i}{\partial x^j} = h\_{ij} = \frac{\partial p\_j}{\partial x^i}.
$$
Hence there exists a function $u$ such that $\mathrm{d}u = p\_j\,\mathrm{d}x^j$.
Whence we have  

$$
h\_{ij} = \frac{\partial^2u}{\partial x^i\partial x^j}\,,
$$ 
so that $h$ has the Hessian form above, where, because of the volume form normalization, $u$ satisfies the specified Monge-Ampère equation in the $x$-coordinate system.


**Higher dimensions:** For general $d=2n$ the version of $(2)$ that has to be proved is the following one: If $\xi^1,\ldots,\xi^{2n}$ is any positively oriented basis of $1$-forms on $V$ such that $h = h\_{ij}\xi^i\xi^j$ with $\det(h\_{ij})= 1$, then, for any multi-index $I = i\_1i\_2\cdots i\_n$ with $1\le i\_1<i\_2<\cdots<i\_n\le 2n$, if we set $\xi^I = \xi^{i\_1}{\wedge}\cdots\wedge\xi^{i\_n}$, then we have
$$
\ast\_h(\xi^I) = \sigma^{IJ}\alpha\_{J}
$$
where $J=j\_1j\_2\cdots j\_n$ with $1\le j\_1<j\_2<\cdots<j\_n\le 2n$ is the multi-index of length $n$ that is disjoint from $I$, $\sigma^{IJ}=\pm1$ is the sign of the permutation that $IJ$ represents of $12\cdots(2n)$, and $\alpha\_J = \alpha\_{j\_1}{\wedge}\cdots\wedge\alpha\_{j\_n}$ where 
$$
\alpha\_i = h\_{ij}\,\xi^j.
$$
(This lemma about the Hodge star operator can be proved using techniques very similar to the case $n=4$.) 


Once you have this lemma, taking $\xi^i = \mathrm{d}x^i$, the proof proceeds as in the case $d=4$: We show that $\mathrm{d}\alpha\_j = 0$, so that $\alpha\_j = \mathrm{d}p\_j$ for some function $p\_j$, etc.


**Original answer (of the OP's side question):**


Since the OP asked, I can say something about the simpler case of being able to do this for two complementary pairs of $\tfrac12d$-sets of indices. (This has some bearing on the case of all co-closure for *all* $\tfrac12d$-sets of indices, since that is asking for much more.)


The basic point is this: For the generic metric $(M^{2n},g)$ in dimension $d=2n$, there will not exist, even locally, $n$ independent functions $x^1,\ldots,x^n$ such that the nonvanishing $n$-form $\alpha = \mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ be co-closed. 


The reason is this: If $\mathrm{d}(\ast\alpha)=0$, then, since $\ast\alpha$ will be decomposable (and nonzero), there would have to exist, in a neighborhood of any point $p$ in the domain of the $x^i$, $n$ local functions $y^1,\ldots,y^n$ such that $\ast\alpha = \mathrm{d}y^1\wedge\cdots\wedge\mathrm{d}y^n$.


Since $\alpha\wedge(\ast\alpha)\not=0$, it follows that $x^1,\ldots,x^n,y^1,\ldots,y^n$ is a local coordinate system in a neighborhood of $p$. Moreover, the $g$-duality relation shows that the span of $\mathrm{d}x^i$ is $g$-orthogonal to the span of $\mathrm{d}y^i$. Thus, in this local coordinate system, $g$ must take the form
$$
g = g\_{ij}(x,y)\,\mathrm{d}x^i\mathrm{d}x^j 
 + h\_{ij}(x,y)\,\mathrm{d}y^i\mathrm{d}y^j.\tag1
$$
Moreover, setting $G = \sqrt{\det(g\_{ij})}$ and $H = \sqrt{\det(h\_{ij})}$, we clearly have
$$
G\,(\ast\alpha)=\ast(G\alpha) = \ast\bigl(G\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n\bigr) = H\,\mathrm{d}y^1\wedge\cdots\wedge\mathrm{d}y^n = H\,(\ast\alpha),
$$
so $G = H$. 


Thus, a metric $g$ for which a nontrivial solution to $\mathrm{d}\bigl(\ast(\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n)\bigr)=0$ exists can always be put locally in the form $(1)$ for some functions $g\_{ij}=g\_{ji}$ and $h\_{ij}=h\_{ji}$ with $\sqrt{\det(g\_{ij})}=\sqrt{\det(h\_{ij})}$. Thus, modulo diffeomorphisms (i.e., changes of coordinates), such metrics depend on at most $n(n{+}1)-1$ functions of $2n$ variables. 


However, the general metric in dimension $2n$ depends, modulo diffeomorphisms, on $n(2n{+}1)-2n = n(2n{-}1)$ functions of $2n$ variables, and, by Cartan's count, there is no normal form that can reduce this minimum number of arbitrary functions. 


Thus, when $n>1$, the generic metric $(M^{2n},g)$ does not admit any local coordinate system satisfying even the single condition
$$
\mathrm{d}\bigl(\ast(\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n)\bigr)=0.
\tag2
$$


In the OP's formulation, one is asking for the conditions that $g$ admit a coordinate system such that
$$
\mathrm{d}\bigl(\ast(\mathrm{d}x^{i\_1}\wedge\cdots\wedge\mathrm{d}x^{i\_n})\bigr)
=0.\tag3
$$
for *each* choice of $1\le i\_1<i\_2<\cdots<i\_n\le 2n$, so this is very highly overdetermined. *Update (in view of the added information above): It is now clear that this set of metrics contains the locally conformally unimodular Hessian metrics, which, of course, properly contain the conformally flat metrics.*


