Kähler differentials on $C^\infty(S^1)$ I'm trying to learn a bit about "de Rham homology" for algebras. Let us take $A=C^\infty(S^1)$.
So if I understand well, the complex given by Kähler differentials and $d$ should compute de Rham cohomology of the circle.
But I don't quite see what would be the representative for the generator "$d\theta$" in degree 1. Since $\theta$ is not a function I am a little confused as to what should be the element in $\Omega^1_A$ representing that class.

While $\theta$ only exists as a "multivalued function", there is still a sense in which this analytic, and we can compute its differential. E.g. considering $S^1$ as the unit circle in the punctured plane, from
$$ \tan(\theta) = \frac{y}{x} $$
we compute
$$ \sec(\theta)^2 \mathrm{d} \theta = \frac{x \mathrm{d}y - y \mathrm{d} x}{x^2} $$
$$ \sec(\theta)^2 = 1 + \tan(\theta)^2 = \frac{x^2 + y^2}{x^2} $$
and thus
$$ \mathrm{d}\theta = \frac{x \mathrm{d}y - y \mathrm{d} x}{x^2 + y^2} $$
This last formula is a well-defined differential form, and has exactly the properties we'd want from $\mathrm{d} \theta$.

I should remark that for this ring, the module of Kähler differentials $\Omega^1_{A/\mathbb{R}}$ is uncomfortable to work with, and may be a bad idea to think deeply upon if your interests are more along the lines of algebraic geometry.
In particular, while there is a relationship with classical differential forms, they aren't the same thing. See this old (unanswered) question of mine, as well as the posts it links to.
The module $\Omega^1_{A}$ of absolute differentials would be even worse to consider, since it will contain all of the weird features of $\Omega^1_{\mathbb{R}}$.
