Let $\omega$ be a non-vanishing (for clarification: nowhere vanishing) smooth $1$-form on a smooth manifold $M$, if $\mathrm{d}\omega \wedge \omega =0$, do we already have $\mathrm{d}\omega= \sum a_i \wedge \omega$ for some $1$-forms $a_i$?

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    $\begingroup$ Hint: Extend $\omega$ to a basis for the $1$-forms. $\endgroup$ – Ted Shifrin Dec 13 '14 at 13:19
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    $\begingroup$ @Ted Could you please post a solution : I don't understand what you suggest. $\endgroup$ – Georges Elencwajg Dec 13 '14 at 13:40
  • $\begingroup$ I think I understand the hint (we would have for $d\omega= \sum a_j \omega_j \wedge \omega_k$ that $\omega \wedge d\omega= 0$ implies (by linear indepence) that all the $\omega \wedge \omega_j \wedge \omega_k$ must vanish). But, being a beginner in differential forms on smooth manifolds: can we extend to a basis globally? Or does this suffices locally? $\endgroup$ – user110071 Dec 13 '14 at 13:44
  • $\begingroup$ Dear @user110071 when you talk of basis, you must indicate of what vector space. If you consider the space of global differential $1$-forms $\Omega^1(M)$ you can choose a basis $(\omega)_i$ of it (with some $\omega_{i_0}=\omega$) but the subtle point is that the products $\omega_i\wedge \omega_j$ will definitely not be a basis of $\Omega^2(M)$ and you can't write $d\omega$ the way you do in your comment. That's why I believe that Ted's hint won't work. $\endgroup$ – Georges Elencwajg Dec 13 '14 at 13:57
  • $\begingroup$ @GeorgesElencwajg As I understand it, one wants that $\{\omega_i(p)\}$ is a basis of $T_pM^\ast$ for every $p$. Then $\omega_1 \wedge \omega_j$ is a basis of $\Omega^2(M)$ as a $C^\infty(M)$-module, isn't it? $\endgroup$ – Daniel Fischer Dec 13 '14 at 14:13

But, being a beginner in differential forms on smooth manifolds: can we extend to a basis globally?

In general, no. On parallelizable manifolds, you can extend to a global basis in the sense that $(\omega(p),\omega_2(p),\dotsc,\omega_k(p))$ are a basis of the cotangent space at every point, but not on other manifolds.

Or does this suffices locally?

If your definition of manifold includes paracompactness (or second countability, which implies paracompactness for locally Euclidean Hausdorff spaces), then there exist smooth partitions of unity subordinate to every open cover of $M$.

You can then extend $\omega$ to a local frame on every coordinate patch $U$ - meaning you find smooth $1$-forms $\omega_2,\dotsc,\omega_n$ on $U$ such that $\omega(p),\omega_2(p),\dotsc, \omega_n(p)$ is a basis of $T_p^\ast M$ for all $p\in U$ - and the condition $d\omega \wedge \omega = 0$ then implies that, with smooth functions $\beta_k$, we have

$$d\omega = \underbrace{\sum_{k = 2}^n \beta_k\cdot \omega_k}_{\alpha_U}\wedge \omega$$

on $U$. Cover $M$ with a family $\mathfrak{U}$ of such coordinate patches and find a subordinate smooth partition of unity $\{ \varphi_U : U \in \mathfrak{U}\}$. Then you have

$$d\omega = \sum_U \varphi_U\cdot d\omega = \sum_U \varphi_U (\alpha_U\wedge \omega) = \Biggl(\sum_U \varphi_U\cdot \alpha_U\Biggr)\wedge \omega.$$

  • $\begingroup$ +1 for drawing attention to the importance for global forms $\omega_i\in \Omega^1(M)$ of having as values at a point $p\in M$ linearly independent linear forms $\omega_i(p)\in T_p ^*(M)$ $\endgroup$ – Georges Elencwajg Dec 13 '14 at 14:38

No: $M=\mathbb R^2,\: \omega=xd y$

In the first version of the question "non vanishing" was in brackets and I interpreted the question as "not identically zero but maybe having zeros", as is indeed the case in my example..

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    $\begingroup$ This form is not nonvanishing. $\endgroup$ – Jack Lee Dec 13 '14 at 13:07
  • $\begingroup$ Then $d\omega = (dx/x) \wedge \omega$. $\endgroup$ – Jack Lee Dec 13 '14 at 13:14
  • $\begingroup$ Then $\mathrm{d}\omega = \mathrm{d}x \wedge \mathrm{d}y = \frac{1}{x} \mathrm{d}x \wedge (x \mathrm{d}y)$. $\endgroup$ – user110071 Dec 13 '14 at 13:15
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    $\begingroup$ I'm sorry I had the "non-vanishing" in brackets only at first, so now I see that for the statement to have a chance to hold (I still do not know if it is true), this is at least necessary. $\endgroup$ – user110071 Dec 13 '14 at 13:17

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