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"?

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    $\begingroup$ The de Rham complex does not capture all the geometry of a differentiable manifold. In fact, it does not even capture all of the topology, only the rational homotopy type. $\endgroup$ Commented May 20, 2010 at 16:37
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    $\begingroup$ In fact, it doesn't even necessarily capture all of the rational homotopy type but only the real homotopy type, as you can lose information about e.g. a bilinear pairing by passing from Q to R. $\endgroup$ Commented May 20, 2010 at 16:40
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    $\begingroup$ I think the key idea lies in the richness of 0-forms in contrast to 0-dimensional homology. One-forms only generate all forms because of the action of $C^\infty(M)$ on them. $\endgroup$ Commented May 20, 2010 at 16:57
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    $\begingroup$ Indeed, once one has $C^\infty(M)$ as an $\mathbb{R}$- (or $\mathbb{C}$- if you prefer) algebra, one can define $\Omega^1(M)$ etc. All this is done in detail in the book (whose title escapes me) by "Jet Nestruev". $\endgroup$ Commented May 20, 2010 at 22:33
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    $\begingroup$ @Eric: By duality, it might be fairer to talk about the richness of $n$-dimensional homology (e.g., you can make neighborhoods of all sorts of homology classes by clumping together balls) rather than the poverty of $0$-dimensional homology. Even so, functions on any space tell its story pretty effectively... $\endgroup$ Commented May 20, 2010 at 23:01

2 Answers 2


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:

  1. Let's pretend that our hole looks normalish, let's say we have a neighbourhood of the hole 'homotopic' to $\mathbb{R}^k\setminus0$

  2. 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)

  3. $\alpha:=\omega(\frac{x}{||x||})$ is a smooth k-1 form $\in \Omega^{k-1}(\mathbb{R}^k\setminus0)$

  4. 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)

  5. Thus, morally, $d\alpha=0$

  6. 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 $\beta(S^{k-2}_+)=b$

  7. 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

  8. Rotating the full way around would give a value, simultaneously of $vol_{k-1} (S^{k-1})$ and 0

  9. so there is no $\beta$, and $\alpha \in H^{k-1}(\mathbb{R}^k \setminus 0)$

  10. 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

  11. 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.


We should distinguish between thinking locally and thinking globally. Locally, any differential $k$-form is built out of wedging together $1$ forms and taking linear cmobinations. Globally, a closed $k$-form need not be a wedge of closed $1$-forms. So the statement that everything is built out of $1$-forms is only a local statement, whereas cohomology is precisely about the difference between the local and the global. (Indeed, cohomology measures the extent to which a locally exact $k$-form need not be exact.) For this reason, I don't like your statement that the de-Rham complex is all built from $1$-forms.

I want to say that there should be a similar story in singular cohomology: that any closed $k$-co-chain should be locally a wedge of closed-$1$-co-chains. The picture I want to give is that any $k$-simplex "is" the wedge of the $k$ edges coming out of a single vertex, and so singular cohomology is locally built from $1$ dimensional objects to the same extent that deRham cohomology is. But I am hazy as to how to make this precise.

  • $\begingroup$ I like this answer. I think it might also be instructive to ponder the question for 1-dimensional manifolds. $\endgroup$
    – Deane Yang
    Commented May 21, 2010 at 0:22

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