Can every closed differential form be expressed via constant coefficients?

Question

Let $M$ be a smooth $n$ dimensional manifold, and let $1 \le k < n$. Let $\omega \in \Omega^k(M)$ be a closed $k$-form on $M$.

Let $p \in M$. Do there exist coordinates around $p$, such that $\omega=a_{i_1i_2\dots i_k}dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_k}$, where $a_{i_1i_2\dots i_k}$ are constants?

That is, I ask whether every closed differential form be locally expressed via constant coefficients.

Edit:

I forgot to require that $\omega$ should be everywhere non-zero. Otherwise, as mentioned by Paulo Mourão, one can take $xdx$ on $M=\mathbb{R}$.

Answer 1

There is a nontrivial necessary condition -- $\omega$ must have constant rank.

The rank of differential form is defined as follows: a $k$-form is said to be decomposable if it can be written locally as a wedge product of $1$-forms. Every $k$-form can be written locally as a sum of decomposable $k$-forms, and the rank of a $k$-form at a point is defined as the minimum number of decomposable terms in any such representation. See this Wikipedia article for more.

A constant-coefficient form obviously has constant rank, so that's a necessary condition for $\omega$ to be expressed as a constant-coefficient form in some coordinates. To see that this condition is stronger than nowhere-vanishing in general, consider the following closed $2$-form on $\mathbb R^4$, with coordinates $(w,x,y,z)$: $$ \omega = w\, dw\wedge dx + dy \wedge dz. $$ This is nowhere vanishing, and it has rank $2$ wherever $w\ne 0$, but on the $w=0$ hyperplane it has rank $1$. Thus there are no coordinates in a neighborhood of the origin that will make it constant-coefficient.

For $1$-forms, constant rank just means either identically zero or nowhere-vanishing. Obviously the zero form has a constant-coefficient representation, and as @TedShifrin pointed out in a comment, the Frobenius theorem shows that a nowhere-vanishing closed $1$-form has such a representation in a neighborhood of each point. Thus the constant-rank condition is sufficient for $1$-forms.

For $2$-forms, constant rank is equivalent to the matrix of the form having constant rank in any coordinates. In this case, here's a proof that the constant-rank condition is sufficient: Suppose $\omega$ is a closed $2$-form with constant rank $p$. By the Poincaré lemma, in a neighborhood of each point there is a $1$-form $\theta$ such that $d\theta = \omega$. Then because $d\theta$ has constant rank, the the generalized Darboux theorem implies that there are local coordinates $(x^1,\dots,x^{n-p},y^1,\dots,y^p)$ in which $\theta$ is given by $$ \theta = x^1\,dy^1 + \dots + x^p\,dy^p, $$ and therefore $\omega$ has the constant-coefficient expression $$ \omega = dx^1\wedge dy^1 + \dots + dx^p \wedge dy^p. $$

I'm not sure about sufficiency for forms of degree higher than $2$, but I suspect it's true in that case as well. I would consult the book Exterior Differential Systems by Bryant et al.

Answer 2

Let me generalize somewhat Jack Lee's answer. Assume $V,W$ are finite dimensional real vector spaces and let $\mu \in \Lambda^k(V)^{*}, \nu \in \Lambda^k(W)^{*}$ be two $k$-alternating forms on $V$ and $W$ respectably. Let's say that $\mu$ and $\nu$ are equivalent if there exists an isomorphism $T \colon V \rightarrow W$ such that for all $v_1, \dots, v_k \in V$ we have $$ \nu(Tv_1, \dots, Tv_k) = \mu(v_1, \dots, v_k). $$ In order words, $\nu$ and $\mu$ are equivalent if $T^{*}(\nu) = \mu$. By functoriality of the pullback, this is an equivalence relation.

Claim: Let $\omega$ be a $k$-form on a manifold $M^n$ and let $p \in M$. A necessary condition for the existence of a coordinate system around $p$ in which $\omega$ has constant coefficients is that there exists a neighrhood $U$ of $p$ in which $\omega_{p'}$ is equivalent to $\omega_p$ for all $p' \in U$.

Proof: If $\varphi \colon V \rightarrow U \subseteq M$ is a coordinate system in which $\omega$ has constant coefficients, then $\varphi^{*}(\omega|_{V}) = \sum_{I} a_I dx^{I}$. Set $\omega_0 = \sum_{I} a_I dx^{I} \in \Lambda^k(\mathbb{R}^n)^{*}$ and let $p' = \varphi(q') \in U$. Then $d\varphi|_{q'} \colon \mathbb{R}^n \rightarrow T_{p'}M$ is a linear isomorphism which gives an equivalence between $\omega_{p'}$ and $\omega_0$. Since $p' \in U$ was arbitrary, $\omega_{p}$ is also equivalent to $\omega_0$ and so $\omega_p$ and $\omega_{p'}$ are equivalent.

When $k \in \{1, 2, n-2, n-1, n\}$, one can classify explicitly when two $k$-forms are equivalent using the rank. For the other cases and general $n$, it seems to be an open problem (you asked about it before here) but having constant rank is definitely not sufficient. For some examples showing that constant rank or local equivalence is not sufficient, see here.

Answer 3

I believe you don't have such coordinates around $0$ for $\omega=xdx$ in $\mathbb{R}$.