TITLE: Kahler differentials and Ordinary Differentials
QUESTION [68 upvotes]: What's the relationship between Kahler differentials and ordinary differential forms?

REPLY [4 votes]: The key ingredient in proofs that derivations are differential forms is the ability to evaluate coefficients: e.g. to go from $\mathrm{d}f(x) = f'(x) \mathrm{d}x$ to $\mathrm{d}f(x)|_a = f'(a) \mathrm{d}x|_a$.
Evaluation maps $\theta_P : \mathcal{C}^\infty(M) \to \mathbb{R} : f \to f(P)$ are examples of ring homomorphisms; abstractly, for any ring homomorphism $\varphi : R \to S$, we can consider the corresponding "evaluation"
$$
\varphi_* : \Omega_R \cong R \otimes_R \Omega_R \to S \otimes_R \Omega_R
$$
The $S$-module $S \otimes_R \Omega_R$ is isomorphic to the quotient of $\Omega_R$ by the relations $f \mathrm{d}g = \varphi(f) \mathrm{d}g$, and $\varphi_*(f \mathrm{d}g) = \varphi(f) \mathrm{d}g$.
In the case that we use one of the $\theta_P$, $\theta_{P*}$ is evaluating the coefficients of a Kähler differential at $P$, and so the corresponding $\mathbb{R} \otimes_{\mathcal{C}^\infty(M)} \Omega_{\mathcal{C}^\infty(M) / \mathbb{R}}$ looks like the cotangent space to $M$ at $P$.
Write $\mathrm{d}f|_P$ for $\theta_{P*}(\mathrm{d}f)$.
Now, we can apply the usual argument. For $M = \mathbb{R}^n$, to any function $f$, we can take the differential of a Taylor polynomial and get
$$ \mathrm{d}f(x)|_P = f'(P)  \mathrm{d}x|_P$$
so it's clear that we really do get the cotangent space to $M$ at $P$.
If we take all Kähler differentials and identify differentials that have the same "values" at every point of the manifold, then we get the ordinary differential forms.

Conjecture If $M$ is a compact manifold, then $\Omega_{\mathcal{C}^\infty(M) / \mathbb{R}} \cong \Omega^1(M)$.

(this conjecture has been posted to Are Kähler differentials the same as one-forms for compact manifolds?)