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Let $M$ be a smooth manifold and let $A=C^\infty(M).$

We consider module of Kahler differentials $\Omega_k(A)$ and module of 1-forms $\Omega^1(M).$ Denote Kahler differential by $d_k$ and classical differential on manifold $M$ by $d$.

From universal property of $\Omega_k(A)$ we know that $\Omega_k(A)^*\cong Der(A).$ More precisely there is an isomorphism of $A-$modules $\beta:Der(A)\to\Omega_k(A)^*$ such that

$$[\beta(X)](d_kf)=X(f)$$ for all $X\in Der(A)$ and $f\in A.$

On the other hand we know that $\frak{X}$$(M)\cong Der(A)$ and $\Omega^1(M)=\frak{X}$$(M)^*.$ Hence $$\Omega_k(A)^{**}\cong Der(A)^*=\Omega^1(M).$$

Using $\beta$ we naturally get a map $\gamma:\Omega_k(A)\to\Omega^1(M)$ defined by composition: $$\Omega_k(A)\stackrel{**}{\to}\Omega_k(A)^{**}\stackrel{\beta^*}{\to}\Omega^1(M).$$ If we compute $\gamma$ we see that $$[\gamma(d_kf)](X)=[\beta^*(d_kf)^{**}](X)=(d_kf)^{**}(\beta(X))=[\beta(X)](d_kf)=X(f)=(df)(X)$$ for every $X\in Der(A)$ and $f\in A.$

Main problem. How to prove that $\gamma$ is surjective?

Georges Elencwajg in his answer to this question wrote that

There is an obvious surjective map $\Omega_k(A)\to\Omega^1(M)$ because the relations displayed above are valid in the classical interpretation of the calculus (Leibniz rule).

I do not understand this argument. In my opinion surjectivity of $\gamma$ holds for completely different reason. I would say that $\gamma$ is surjective because any 1-form $\alpha$ can be written as a finite sum $$\alpha=\sum_{i=1}^p g_idf_i$$ for some $f_1,\dots,f_p,g_1,\dots,g_p\in A.$ From this we get that $$\gamma\left(\sum_{i=1}^p g_id_kf_i\right)=\sum_{i=1}^p g_idf_i=\alpha.$$ Presenting $\alpha$ as a such finite sum I would explain using Serre-Swan theorem.

Actual question. Is there a simpler argumentation of surjectivity of $\gamma$ or do I really need to rely on specific form of every 1-form $\alpha?$

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  • $\begingroup$ What is $\mathfrak{X}(M)$? $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 25, 2016 at 19:01
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    $\begingroup$ @LaurentMoret-Bailly Module of smooth vector fields on $M.$ $\endgroup$
    – Fallen Apart
    Commented Nov 25, 2016 at 19:03
  • $\begingroup$ My impression is that "because the relations displayed above are valid in the classical interpretation of the calculus" refers to the existence of the map, not to its surjectivity (which I would prove exactly as you do). $\endgroup$
    – Denis Nardin
    Commented Nov 25, 2016 at 22:40
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    $\begingroup$ @DenisNardin I see problem, cause in order to apply $\gamma$ I need to write 1-form $\alpha$ as a finite sum. When manifold $M$ is not compact, it possibly do not admin cover with finite amout of charts. And hence gluing by partition of unity would give me infinite sum. If I remember correctly the key in Swan's proof to make vector bundles finitely generated is Whitney Embedding theorem. $\endgroup$
    – Fallen Apart
    Commented Nov 25, 2016 at 23:25
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    $\begingroup$ @DenisNardin On an $n$-manifold, one can always get a finite atlas where every "chart" is defined on a disjoint union of contractible subsets. This should be good enough for the argument. To get such an atlas, start with the first barycentric subdivision of a triangulation and consider the open stars of all vertices coming from degree $k$ simplices as one chart. Then you have $(n+1)$ types of charts, and so you can always do with a sum over $n(n+1)$ forms of the type $f\,dg$. $\endgroup$
    – Sebastian Goette
    Commented Mar 11, 2017 at 17:56

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