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"?
Best Answer
When you say that the geometry of the whole space 'just comes out of' the space of 1-forms, there is naturally a huge amount of machinery hiding underneath the surface: $\Omega^n$ does not just drop out under successive exterior derivatives $d^n$: $d^2$ is the zero map- in fact the whole point of DeRham cohomology is that the exterior derivative is not surjective.
What's really going on is with the ring structure of $\Omega^*$- the wedge product of differential forms is designed to mimic the determinant for subspaces not neccessarily equal the whole space, so that we can calculate eg. signed 2-volume in a 4-dimensional space with ease. (This follows from the fact that the n-volume of an n-parallellopiped is [up to scalar] the unique skew symmetric n-linear form on an n-dimensional subspace).
Wedged asortments of 1 forms added together give different weightings of n-volume to different orientations of n-planes in the tangent space- giving a way of measuring n-volume locally, knitting together into a global picture across a manifold of how to measure for example an n-dimensional embedded submanifold.
Edit: okay so the real question here is why can De Rham cohomology do what homology does (ie count the k-dimensional holes in a space)? What follows will be hands-wavy, but less so than my original finale.
Suppose we have a k dimensional hole in our manifold, with no intention of providing proof or even consistent definitions I claim the following moral argument:
Let's pretend that our hole looks normalish, let's say we have a neighbourhood of the hole 'homotopic' to $\mathbb{R}^k\setminus0$
The volume form of $S^{k-1}$, '$\omega$' is a k-1-form in $\mathbb{R}^k\setminus0$ (though not yet a smooth k-1-form when considered in the ambient space)
$\alpha:=\omega(\frac{x}{||x||})$ is a smooth k-1 form $\in \Omega^{k-1}(\mathbb{R}^k\setminus0)$
We can see because of the $||x||$ bit that if we have a copy of $S^{k-1}$ which we measure with this form, expanding it or contracting it will not change a thing, homotopies that cause overlap will be sorted by the volume form bit (negative and positive orientations will cancel each other out)
Thus, morally, $d\alpha=0$
Now suppose we had a $\beta$ with $d\beta=\alpha$, well then we could use it to measure half of an ebbedded $S^{k-2}: S^{k-2}_+ $ so let $\beta(S^{k-2}_+)=b$
By stokes', as we rotated our $S^{k-2}_+$ around through the spare dimension giving say $r(S^{k-2}_+)$ then $\beta(r(S^{k-2}_+))-b$ would be the k-1 spherical volume in between them
Rotating the full way around would give a value, simultaneously of $vol_{k-1} (S^{k-1})$ and 0
so there is no $\beta$, and $\alpha \in H^{k-1}(\mathbb{R}^k \setminus 0)$
Bump it, extend it by zero, add dimensions by the pullback of the projection, pull it back through a chart and you've probably got an element of your original cohomology group
Question:So we can count the holes- do we count anything else and ruin our counting? Answer: nope- to get that kind of homotopy invariance needs a singularity- and smooth sections of the cotangent bundle don't take too kindly to singularities unless they're in a hole where the manifold isn't.