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All:

Let $f: M \to N$ be a smooth map between manifolds, and let $w$ be a $1$-form on $M$. Under what conditions is there a $1$-form $z$ defined on $N$ so that $w=f^*z$, i.e., so that $w$ is the pullback of the form $z$ by the map $f$?

All I can think of is considering the respective cohomologies of $M$, $N$, so that, e.g., if $H^1(N)=0$, then this would not be possible, or maybe we can use the (contravariant) map induced by $f$ in cohomology, but I cannot think of more general conditions.

Thanks for any suggestions.

RickyBobby
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  • I assume you mean $w = f^z$ (instead of $z = f^w$). – Jesse Madnick Jul 26 '13 at 05:56
  • You always have pullbacks on differential forms. If $H^{1}(N) = 0$ that means there are no closed one forms which are not exact but there still are oodles of one-forms. – DBS Jul 26 '13 at 06:01
  • Thanks, Jesse ,just corrected it. – RickyBobby Jul 26 '13 at 06:02
  • DBS: true, but I don't think there is always an injection, i.e., I don't think the induced map on cohomology is necessarily a surjection (nor an injection, but that's a different issue) – RickyBobby Jul 26 '13 at 06:03
  • @DBS He's asking when one form is the pullback of another, not if forms can always be pulled back. – Potato Jul 26 '13 at 06:04
  • okay , I saw the correction afterwards. – DBS Jul 26 '13 at 06:05
  • If $M$ is a finite covering map in the sense of topology (or in more fancy language etale). – DBS Jul 26 '13 at 06:09
  • Dear @DBS, étale is not fancy language for covering map, but a much more general concept. – Georges Elencwajg Jul 26 '13 at 06:37
  • @GeorgesElencwajg In my undertanding* in the category of topological spaces or analytic spaces over $\mathbb{C}$ etale maps are (locally) covering maps. Of-course it is a huge generalization in algebraic geometry but I didn't mean that. *From Beauville's Complex Algebraic Surfaces. – DBS Jul 26 '13 at 06:42
  • Dear @GeorgesElencwajg Yes of course. I should have mentioned surjective finite '{e}tale maps. My Bad! – DBS Jul 26 '13 at 07:05
  • When the cover is regular, it is necessary and sufficient that $w$ be invariant under the deck transformation group; that is $w=\phi^*\circ w$ for every deck transformation $\phi$. – 40 votes Jul 26 '13 at 14:05
  • But I'm not sure the map I'm working with is a covering map, tho it may be so locally, i.e., on non-critical points, i.e.,if x is not a critical point of f, then there may be an open set containing x so that f restricts to a covering map. – RickyBobby Jul 27 '13 at 20:28

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