Let $M$-be a differentiable manifold. Then, suppose to capture the underlying geometry we apply the singular homology theory. In the singular co-chain, there is geometry in every dimension. We look at the maps from simplexes, look at the cycles and go modulo the boundaries. This has a satisfying geometric feel, though I need to internalize it a bit more(which matter I tried to address in other questions).

Now on the other hand, let $\Omega^1$ be the space of $1$-forms on the space. The rest of the de Rham complex comes out of this object, wholly through algebraic processes, ie by taking the exterior powers and also the exterior derivative. After getting this object in hand, the journey upto getting the de Rham cohomology ring is entirely algebraic.

And by the de Rham theorem, this second method is equally as good as the more geometric first method. In the second method no geometry is explicitly involved anywhere in any terms after the first term. So the module of $1$-forms somehow magically capture all the geometry of the space $M$ without need of any explicit geometry. This is amazing from an algebraic point of view since we have less geometric stuff to understand.

This makes me wonder for the conceptual reason why this is true. I know that one should not look a gift horse in the mouth. But there is the need to understand why there is such a marked difference in the two approaches to capturing the geometry in a manifold, viz, through de Rham cohomology in differential topology, and through singular homology in algebraic topology. I would be grateful for any explanations why merely looking at all the $1$-forms is so informative.

Edited in response to comments: I meant, the de Rham cohomology is as good as singular homology for differentiable manifolds. What is "geometry of rational homotopy type"? And what is "geometry of real homotopy type"?

allthe geometry of a differentiable manifold. In fact, it does not even captureallof the topology, only the rational homotopy type. $\endgroup$1more comment