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Let $M$ be a connected manifold. One defines differential $k$-forms as sections of the $k^{\text{th}}$ exterior power of the cotangent bundle. This is a sort of sheafification of a more naive approach, which is to let $D$ be the module of differential 1-forms, considered as a $C^{\infty}(M, \mathbb{R})$-module, and then take the $k^{\text{th}}$ exterior power of $D.$

Question: Is there a connected manifold where the $k^{\text{th}}$ exterior power of $D$ does not coincide with the actual module of differential $k$-forms, for at least one $k$?

Michael Barz
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    By D you probably do not mean "the ring of differential forms" but the module thereof. – Mariano Suárez-Álvarez Jan 24 '23 at 23:54
  • @MarianoSuárez-Álvarez Yes, thanks! – Michael Barz Jan 25 '23 at 00:22
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    Notice there were two – Mariano Suárez-Álvarez Jan 25 '23 at 00:32
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    What precisely do you mean by "the module of differential forms"? – Thorgott Jan 29 '23 at 20:40
  • @Thorgott The set of all differential forms forms an abelian group, since you can add two forms and get a new one; it is also a module over the ring of smooth functions, since you can multiply a form by a smooth function. – Michael Barz Jan 29 '23 at 21:00
  • Of course, but which differential forms do you mean. Do you want $D$ to be the entire exterior algebra or only looked at $k$-forms of a fixed degree $k$? – Thorgott Jan 29 '23 at 21:52
  • @Thorgott Oh, sorry--I think the only interpretation for which my question makes sense is to interpret it as showing that kth power of 1-forms isn't the same as k-forms – Michael Barz Jan 29 '23 at 22:01
  • I do not believe your proposed counter-example works, then. Instead, these descriptions always coincide. Could you elaborate on why you think your counter-example works? – Thorgott Jan 29 '23 at 23:54
  • @Thorgott I think I'm confusing myself on the counterexample now, but I am pretty confident this claim is false-- it's essentially saying that a presheaf shouldn't always be a sheaf for free. If you claim to have a proof they coincide, I'd love to see it, but I've not seen a single geometry book make this claim and I doubt it's true. – Michael Barz Jan 30 '23 at 02:01
  • @Thorgott Ah; abstractly, the space of 1-forms on the disjoint union is free of rank 4. So it's wedge product with itself should be free of rank 6. But the space of 2-forms on the disjoint union of planes should only be rank $2,$ since a 2-form on a disjoint union is just the data of a 2-form on each plane. – Michael Barz Jan 30 '23 at 02:03
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    I disagree. The disjoint union of two planes is parallelizable. If you have an $n$-dimensional parallelizable manifold, its cotangent bundle and hence also its exterior powers are trivial. The sections of a trivial vector bundle over $M$ are a free module over $C^{\infty}(M,\mathbb{R})$ whose rank is the rank of the vector bundle. It follows that both $\bigwedge^k\Omega^1(M)$ and $\Omega^k(M)$ are free over $C^{\infty}(M,\mathbb{R})$ of rank ${n\choose k}$. – Thorgott Jan 30 '23 at 03:29
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    In general, there is a map $\bigwedge^k\Omega^1(M)\rightarrow\Omega^k(M)$ given by mapping the formal wedge product to the usual wedge product of differential forms. This should be an isomorphism. This is clear locally as locally everything is trivial and you can construct an inverse by gluing local inverses together using a partition of unity. – Thorgott Jan 30 '23 at 03:36
  • @Thorgott Ah, I'm sorry, my example does not work. But still, I would be shocked if the claim I make in my question was false; if you have a proof that these two notions coincide in general I'd love to see it. – Michael Barz Jan 30 '23 at 03:37
  • @Thorgott I am not so confident an inverse can be constructed as you claim, but mabye I am wrong -- I would assume that something goes wrong, since generally I would expect $\wedge^k \Omega^1$, constructed naively, to be a presheaf and not a sheaf – Michael Barz Jan 30 '23 at 03:39
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    Yeah, it is quite surprising. If you don't wanna get your hands dirty writing down the inverse, you can instead try and use the result that bundles over $M$ always admit complements. Try taking a look at this post, which discusses tensor products rather than exterior powers in an analogous fashion. – Thorgott Jan 30 '23 at 12:49
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    On any coordinate chart, it is clear that any compactly supported $k$-form can be written as a linear combination of wedge products of $1$-dorms. Therefore, if $M$ has a finite partition of unity subordinate to coordinate charts, the answer is yes. So a counterexample has to be non-compact. It's also clear that if there is an open $O \subset M$ with compact closure such that $M\backslash O$ can be covered by finitely many coordinate charts, then the two are the same. Given all this an obvious possible counterexample is a surface with infinite genus, – Deane Jan 30 '23 at 15:54
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    If you allow an infinite linear combination of wedge products of $1$-forms such that at each point, the sum is finite, then the two modules coincide. And, obviously, if you restrict to compactly supported forms, then the modules also coincide. – Deane Jan 30 '23 at 15:57
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    @Thorgott's answer is correct. The inverse is simply writing the $k$-form with respect to local coordinates. – Deane Jan 30 '23 at 15:59
  • Thank you both! – Michael Barz Jan 30 '23 at 23:21

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