Let w be a 3-form on a 4-manifold, and U be the set where w is nonzero. w is closed iff for all p in U there exists a neighborhood of p where w has coordinate representation 1 dx dy dz.

To show the reverse direction I just took d and got dw=0 so it's closed. I can't figure out the forward direction. I've tried to do it by linearity but that results in a system of equations I don't know how to solve. Any ideas?

  • $\begingroup$ This can't be true. The form $dx\wedge dy\wedge dz$ always returns 0 when one of the vectors plugged in is in the fourth direction. So, for example, the form $dx\wedge dy\wedge dz+dx\wedge dy\wedge dw$ can't be represented in this way. $\endgroup$ – Amitai Yuval Jun 26 '17 at 20:13
  • 2
    $\begingroup$ @AmitaiYuval: Sure it can. It's equal to $dx\wedge dy \wedge d(z+w)$, so you just need to define new coordinates by $\big(\tilde x,\tilde y,\tilde z,\tilde w\big) = (x,y,z+w,z-w)$. $\endgroup$ – Jack Lee Jun 26 '17 at 20:33
  • $\begingroup$ @JackLee Thanks, I clearly made a mistake. But was it just a bad example, or is my previous comment completely wrong? I mean, is it true that for every 3-form $\omega$ on a 4-dimensional space $V$, there exists $v\in V$ such that $\omega(v,u,w)=0$ for every $u,w\in V$? $\endgroup$ – Amitai Yuval Jun 26 '17 at 21:31
  • $\begingroup$ @AmitaiYuval: I haven't taken the time to think it through carefully, but I suspect your basic idea is correct. You'd just need to come up with a different example. $\endgroup$ – Jack Lee Jun 26 '17 at 21:33
  1. If $dVol$ is a volume form, there exists a vector field $V$ such that $i(V) dVol=\omega$. Indeed if $\omega = a dXdYdZ+bdYdZdT+..$ and $dVol= dXdYdZdT$, take $X= a{\partial f\over \partial T}+b{\partial f\over \partial X}+..$

  2. Then, as $V\not=0$, we can choose a new system of coordinates $(x,y,z,t)$ such that $V= {\partial\over \partial t}$, and in this coordinates system, $dVol=f(x,y,z,t)dxdydzdt$, and $\omega= f(x,y,z,t)dxdydz$.

  3. Then, $d\omega=0$ means that ${\partial f\over \partial t}=0$. Therefore $\omega= f(x,y,z)dxdydz$, is basically a volume form on $\bf R^3$, and the standard "Moser lemma" produces a new system of coordinates where $\omega= dxdydz$

  • $\begingroup$ What's the Moser lemma that you're referring to? $\endgroup$ – aaron Jun 27 '17 at 5:04
  • $\begingroup$ Volume forms (eg $f(x,y,z)dxdydz$ can be made standard (eg $dudvdw$ ) by a change of coordinates . It was probably known in the 19th century, but people call this Moser lemma. See math.tecnico.ulisboa.pt/~mabreu/GD/moser.pdf $\endgroup$ – Thomas Jun 27 '17 at 5:08
  • $\begingroup$ And in fact if $\omega= f(x,y,z)dxdydz$ and $Z$ is the primitive $\int f(x,y,z)dz$, then the form $\omega $ is $dx dy dZ$. $\endgroup$ – Thomas Jun 27 '17 at 5:13
  • $\begingroup$ The question doesn't assume that the manifold is compact, connected, or oriented. Which I think, after reading some of that paper, is necessary for mosers lemma, and generally to assume dvol exists. $\endgroup$ – aaron Jun 28 '17 at 19:01
  • $\begingroup$ Fine, did you understand te argument ? $\endgroup$ – Thomas Jun 29 '17 at 4:19

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