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Let $M$ be a manifold and let $A = \mathcal{C}^\infty(M)$ be the ring of smooth real-valued functions.

An old posting asks about the relationship of Kähler differentials and ordinary differential forms. While it's clear from the universal properties that there is a surjective $A$-module homomorphism map $\Omega^1_{A/\mathbb{R}} \to T^*(M)$, it is not necessarily an isomorphism. A counterexample for $M = \mathbb{R}$ is given as $\mathrm{d}e^x \neq e^x \mathrm{d}x$ in $\Omega^1_{A /\mathbb{R}}$.

However, it's equally clear, e.g. by this argument, that they they have the same evaluations; if $P$ is a point and we consider the homomorphism $A \to \mathbb{R} : f \mapsto f(P)$, then $$ \mathbb{R} \otimes_A \Omega^1_{A / \mathbb{R}} \cong T_P^*M $$ as vector spaces.

I interpret David Speyer's proof of the above counterexample as considering the evaluations at one of the hyperreal-valued "points at infinity", which has many nonstandard scalars, and for these, $\mathrm{d}r$ is not forced to be zero, so you can make weird things happen

If $M$ is a compact manifold, however, then there are no points at infinity; all of the points are real-valued points. In this case, do we have an isomorphism $\Omega^1_{A / \mathbb{R}} \to T^*(M)$?

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    $\begingroup$ Note that you are asking if $\Omega^1_{A/\mathbb{R}}$ is isomorphic to its double dual. $\endgroup$
    – vap
    Nov 14, 2017 at 2:23
  • $\begingroup$ There is a notion of derivation for a "Fermat theory"-algebra which yields the "right" definition of Kähler differentials, meaning the coincide with the differential $1$-forms, when applied to manifolds, see here ncatlab.org/nlab/show/Fermat+theory $\endgroup$
    – vap
    Nov 14, 2017 at 2:27
  • $\begingroup$ But, aside from identifying the differential forms, can we at least identify the homology groups they give rise to? I mean if $H^i_{\text{dR}}(M)\cong H_i((\Omega^\bullet_{A/\mathbb{R}},d))$? By Grothendieck's comparison theorem this is true for complex (analytic) manifolds, but what about smooth manifolds? $\endgroup$
    – vap
    Nov 14, 2017 at 2:31
  • $\begingroup$ Two observations: (1) The localization of $\mathcal{C}^\infty(M)$ at a point $p$ is isomorphic to the local ring $\lim_{0 \in U \subset \mathbb{R}^n } \mathcal{C}^\infty(U)$. As $\Omega$ commutes with localization, the module $\Omega^1_{A / \mathbb{R}}$ "knows about" the differentials of those local rings. (2) The rings $M=\mathcal{C}^\infty(U)$ for open $U \subset M$ are modules over $\mathcal{C}^\infty(M)$. The Kaehler differentials classifies derivations into this module $Hom(\Omega^1_{A / \mathbb{R}},M) = Der(A, M)$. $\endgroup$
    – Heinrich Hartmann
    Jul 29, 2022 at 21:01
  • $\begingroup$ The surjectivity of the map from Kaehler differentials for $C^\infty(M)$ to 1-forms on $M$ does not follow from the universal property of Kaehler differentials. You really need to show that every 1-form on $M$ is a finite linear combination of 1-forms like $f dg$. When $M$ is compact you just cover $M$ with finitely many charts and use a partition of unity argument. But for $M$ noncompact it's subtler! One approach is to use Swan's theorem for noncompact smooth manifolds, which is also subtle: mathoverflow.net/questions/255601/… $\endgroup$
    – John Baez
    Aug 29, 2023 at 23:08

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