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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 $k$?

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    $\begingroup$ By D you probably do not mean "the ring of differential forms" but the module thereof. $\endgroup$
    – Mariano Suárez-Álvarez
    Jan 24 at 23:54
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    $\begingroup$ Notice there were two $\endgroup$
    – Mariano Suárez-Álvarez
    Jan 25 at 0:32
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    $\begingroup$ What precisely do you mean by "the module of differential forms"? $\endgroup$
    – Thorgott
    22 hours ago
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    $\begingroup$ 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 ${n\choose k}$. $\endgroup$
    – Thorgott
    15 hours ago
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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    15 hours ago

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