6

Let $\omega$ be a differential form on a manifold and suppose $\omega^k$ is exact (but non-zero). Is $\omega$ necessarily exact?

Note, the converse is true. If $\omega$ is exact, then $\omega^k$ is exact. To see this, let $\omega = d\alpha$, then $$\omega^k = (d\alpha)^k = d\alpha\wedge\dots\wedge d\alpha = d(\alpha\wedge d\alpha \wedge\dots\wedge d\alpha).$$

The case I am interested in, $\omega$ is a two-form and the manifold is compact, but I'd be interested to see what can be said in general.

Michael Albanese
  • 99,526
  • 2
    For $k$ big enough, $\omega^k=0$ is always exact, isn't it? – Robert Rauch Jun 21 '13 at 01:25
  • Yes, that's true. I'll adjust my question. – Michael Albanese Jun 21 '13 at 01:42
  • That doesn't seem to fix the problem. It shouldn't be hard to find examples of manifolds $M$ with $H^n(M, \mathbb{R}) \neq 0$ but $H^{2n}(M, \mathbb{R}) = 0$, then find an $n$-form $\omega$ not trivial in $H^n$ with $\omega^2 \neq 0$. (Maybe we should take $n = 2$ and $M$ should be a non-orientable $4$-manifold, for example.) – Qiaochu Yuan Jun 21 '13 at 03:37
  • I think this is related: if $E$ is a vector bundle and $E^k$ is trivial, it does not follow that $E$ is trivial. – Adam Saltz Jun 21 '13 at 14:36

1 Answers1

2

Consider the $2$-form $$\omega = dx\wedge dy + \sin (2\pi x)\, dz\wedge dw$$ on the $4$-torus $\mathbb R^4/\mathbb Z^4$. Since $$d\omega = 2\pi \cos (2\pi x)\, dx\wedge dz\wedge dw \ne 0,$$ the form $\omega$ is not exact. On the other hand, $$ \omega\wedge\omega = 2 \sin (2\pi x)\,dx\wedge dy\wedge dz\wedge dw = d (-\pi^{-1}\cos(2\pi x)\, dy\wedge dz\wedge dw) $$

ˈjuː.zɚ79365
  • 3,004