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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 $|J|=k>1$.

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    $\begingroup$ Dieter Kotschick is a good person to ask. I am not sure you will get much information at MO. $\endgroup$ – Mikhail Katz May 28 '18 at 9:47
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    $\begingroup$ I'll just point out that this is a conformally invariant problem (since the Hodge star operator in the middle dimension is conformally invariant). In particular, any conformally flat manifold has such local coordinates. $\endgroup$ – Robert Bryant May 29 '18 at 11:15
  • $\begingroup$ Also, if a metric $g$ on a $d$-manifold (with $d$ even) has a (local) orthonormal coframing $\omega_i$ with the property that the wedge product of every $\tfrac12d$ of the $\omega_i$ is co-closed, then the $\omega_i$ are closed, so $g$ is flat and the $\omega_i$ are locally the differentials of coordinates. $\endgroup$ – Robert Bryant May 29 '18 at 11:41
  • $\begingroup$ @RobertBryant Thanks. Regarding your second comment: Can you please elaborate on that? (why the $\omega_i$ are closed in this case? I see why the $\frac{d}{2}$-wedge product is closed). Regarding your first comment, you are right of course; I am interested in a more general case than the conformally flat one. $\endgroup$ – Asaf Shachar May 29 '18 at 11:52
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    $\begingroup$ The proof of my second comment, though straightforward, will not fit into 600 characters. The basic idea, as you have noticed, is to work with the fact that this is equivalent to requiring that any $\tfrac12d$-fold wedge product of the $\omega_i$ be closed. Since one has $$\mathrm{d}\omega_i =-\tfrac12C_{ijk}\,\omega_j\wedge\omega_k$$ for some unique functions $C_{ijk}=-C_{ikj}$, it's a matter of showing that the above closures impose so many linear equations on the $C_{ijk}$ that they are forced to be zero. It's linear algebra (that can be simplified using a little representation theory). $\endgroup$ – Robert Bryant May 29 '18 at 13:21
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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.

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  • $\begingroup$ Thanks Robert, this is a very illuminating answer. I have learned a lot from it. Can you please clarify a few things? (1) First, the terminology "locally conformally unimodular Hessian" is your own, right? (2) You said that "by Cartan's count, there is no normal form that can reduce this minimum number of arbitrary functions". Can you give me a reference or a general direction of reading about that statement (it's a bit far from my comfort zone, so I failed in finding the right sources, I guess. Explicitly, I am not familiar with the terms "normal form" and "Cartan's count"). $\endgroup$ – Asaf Shachar Jun 7 '18 at 6:39
  • $\begingroup$ (3) Are you sure you meant $\ast_h(\mathrm{d}x^1\wedge\mathrm{d}x^2)= \alpha_3\wedge\alpha_4$? I think the right equality should be $\ast_h(\mathrm{d}x^1\wedge\mathrm{d}x^2)= \alpha_1\wedge\alpha_2$ etc (If I am not mistaken, you somehow mixed the indices in all the equations, i.e. it should be $\ast_h(\mathrm{d}x^i\wedge\mathrm{d}x^j)= \alpha_i\wedge\alpha_j$). $\endgroup$ – Asaf Shachar Jun 7 '18 at 6:39
  • $\begingroup$ @AsafShachar: (1) Maybe it's not in the literature as such, but the terminology 'Hessian metric' is, so it's not much of a stretch to take 'unimodular Hessian' to mean that the coordinate system's volume form is the metric volume form. (2) By "Cartan's count", I meant the fact that, modulo diffeomorphisms, metrics in dimension $2n$ depend on $n(2n{-}1)$ functions of $2n$ variables, and, as Cartan showed, this yields a dimension of the space of $k$-jets of metrics mod diffeomorphism that is too high to be accounted for by $k$-jets of metrics in the form I derived. It's not a deep fact. $\endgroup$ – Robert Bryant Jun 7 '18 at 7:20
  • $\begingroup$ @AsafShachar: (3) Yes, I am sure that what I wrote is what I meant. The alternative formula you propose cannot be correct. Just take $h_{ij} = \delta_{ij}$, where it's easy to compute explicitly, and you'll see that my formula is correct and yours is not. (The formula would not be correct without the assumption that $\det(h_{ij}) = 1$.) $\endgroup$ – Robert Bryant Jun 7 '18 at 7:22
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I don't have an answer to the question about obstructions. However, I can suggest some simplifications to the condition you are trying to satisfy.

Let $J_a^i = \nabla_a x^i$, where $x^i$ are your sought after coordinates. Let us restrict our attention to the case when $J_a^i$ is invertible. In the sequel, the Latin letters $a,b,c,\cdots$ may be freely raised and lowered using the metric $g_{ab}$. The equation you want to satisfy is $$ \nabla^a (\wedge^n J)_{a_1\cdots a_n}^{i_i\cdots i_n} = 0 , $$ where the dimension is $d=2n$ and the $\wedge$ operation is carried out over the $a_1,\ldots,a_n$ indices. Though as a consequence the above expression is also antisymmetric in the $i_1,\ldots,i_n$ indices. This equation can be simplified by a judicious application of the Leibniz rule: \begin{align*} \nabla^{a_1} (\wedge^n J)_{a_1\cdots a_n}^{i_i\cdots i_n} &= n g^{ba_1} (\nabla_b J \wedge (\wedge^{n-1} J))_{a_1\cdots a_n}^{i_1\cdots i_n} \\ &= n (\nabla^2 x)^{[i_1} (\wedge^{n-1} J)_{a_2\cdots a_n}^{i_2\cdots i_n]} + n(n-1) (J^{b[i_2} \nabla_b J^{i_1}) \wedge (\wedge^{n-2} J)^{i_3\cdots i_n]})_{a_2\cdots a_n} \\ &= n[((\nabla^2 x)^{[i_1} J^{i_2} + (n-1) J^{b[i_2} \nabla_b J^{i_1}) \wedge (\wedge^{n-2} J)^{i_3\cdots i_n]}]_{a_2\cdots a_n} \end{align*}

I don't have a slick argument for it, but I suspect that the $\wedge^{n-2} J$ factor may be canceled, since we are assuming the non-degeneracy of $J$. Extracting the non-trivial factor, we get the new equation \begin{align*} 0 &= J^{ka}(g_{ab} \nabla^2 x + (n-1)\nabla_b \nabla_a x)^{[i} J^{j]b} =: E^{ijk} , \end{align*} where I've added the extra $J^{ka}$ factor for symmetry. In this form, the expression $E^{ijk}$ has the symmetries $E^{[ij]k} = E^{ijk}$ and $E^{[ijk]} = 0$, which corresponds to a Young diagram of the form $$ \begin{array}{|c|c|} \hline i & k \\ \hline j \\ \hline \end{array} , $$ where, to get the right shape, you should ignore any box not filled with an index. This is an overdetermined system for the unknown coordinate functions $x^i$. Unfortunately, from this point on, I don't see an easy way to extract integrability conditions on the $x^i$, other than taking more covariant derivatives and trying to see what that gives you.

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