1

Let $A$ be a $k$-algebra and let $\mathcal{M}_A$ be the set of all $A$-modules. In $\mathcal{M}_A$, there exists a universal object $\Omega_{A/k}$, called the module of Kahler differentials, and a $k$-derivation $d: A \to \Omega_{A/k}$ such that for any $k$-derivation $D: A \to M$ there exists a map $f: \Omega_{A/k} \to M$ such that $f\circ d = D$.

I'm trying to understand what this object has to be when I look at manifolds that are also smooth affine varities. I think that the answer should be the space of sections of the cotangent bundle over the ring of $C^\infty$ functions on the manifold. I'm unable to figure out how to show this.

Any help would be appreciated.

ChaPi
  • 111
  • The module of Kähler differentials algebraically is known to be different from the module of smooth Kähler differentials (for $C^\infty$-rings). This is folklore to me, but I have seen it several times, including on the nLab page on Kähler differentials. – Harry Gindi Dec 22 '18 at 02:14
  • This page should have the details you want: https://ncatlab.org/nlab/show/smooth+algebra . It uses Quillen's characterization of the module category and the Kähler differentials as a section of the module bifibration. A question I have is how does smooth Quillen cohomology relate to De Rham cohomology. – Harry Gindi Dec 22 '18 at 02:21
  • This has been answered in this question. – abx Dec 22 '18 at 05:34

0 Answers0