# The Unapologetic Mathematician

## De Rham Cohomology

The really important thing about the exterior derivative is that it makes the algebra of differential forms into a “differential graded algebra”. We had the structure of a graded algebra before, but now we have a degree-one derivation whose square is zero. And as long as we want it to agree with the differential on functions, there’s only one way to do it.

Why does this matter? Well, the algebra $\Omega(M)$ is the direct sum of its grades — the spaces $\Omega^k(M)$, and for each one we have a map $d:\Omega^k(M)\to\Omega^{k+1}(M)$. We can even write them out in a row:

$\displaystyle \dots\to\mathbf{0}\to\Omega^0(M)\to\dots\to\Omega^k(M)\to\dots\to\Omega^n(M)\to\mathbf{0}\to\dots$

where we have padded out the sequence with $\mathbf{0}$ — the trivial space — in either direction. This is just like a chain complex, except the arrows go backwards! Instead of the indices counting down, they count up. We can deal with this by thinking of these as negative numbers, but really it doesn’t matter.

Anyway, now we can bring all our homological machinery to bear! We say that a differential form $\omega\in\Omega^k(M)$ is “closed” if $d\omega=0$, and we write the subspace of closed forms as $Z^k(M)=\mathrm{Ker}(d)\subseteq\Omega^k(M)$. We say that $\omega$ is “exact” if there is some $\alpha\in\Omega^{k-1}(M)$ with $\omega=d\alpha$, and we write the subspace of exact forms as $B^k(M)=\mathrm{Im}(d)\subseteq\Omega^k(M)$. The fact that $d^2=0$ implies that $B^k(M)\subseteq Z^k(M)$.

And now we can define the $k$-th “de Rham cohomology space” $H^k(M)=Z^k(M)/B^k(M)$. The cohomology space $H^k(M)$ measures the extent to which it is possible to have a $k$-form on $M$ be closed without being exact. If $H^k(M)=\mathbf{0}$, then closed $k$-forms are all exact. And it’s roughly accurate to say that the rank of $H^k(M)$ counts the “number of independent ways” to set up a closed-but-not-exact $k$-form.

The really amazing thing, which we will come to understand later, is that this actually tells us a lot about the topology of $M$ itself: combinatorial information about the topology of a manifold is encoded into the algebraic structure of its sheaf of differential forms.

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July 20, 2011 - Posted by | Differential Topology, Topology

## 11 Comments »

1. Beautiful stuff. I love how these more conceptual results act as a “light at the end of the tunnel” after all the tedious computations from the past week.

Comment by Sam Alexander | July 21, 2011 | Reply

2. This material you’ve been posting on differential topology has been some of the best I’ve seen on your site in several years of devoted reading. Thanks very much.

Comment by Dan | July 21, 2011 | Reply

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4. [...] turns out that the de Rham cohomology spaces are all contravariant functors on the category of smooth manifolds. We’ve even seen how it [...]

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5. [...] on to the theorem! We know how to integrate a differential -form over a -chain . We also have a differential operator on differential forms and a boundary operator on chains. We can put these together in two [...]

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6. [...] is an exact -form. It’s not quite as nice as equality, but if we pass to De Rham cohomology it’s just as good. Advertisement Eco World Content From Across The Internet. Featured [...]

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7. [...] also has an empty boundary. Since is a top form, we know that — it’s closed in the de Rham cohomology. But we know that it cannot also be exact, for if for some -form then Stokes’ theorem would [...]

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8. [...] know that a map induces a chain map , which induces a map on the de Rham cohomology. This is what we mean when we say that de Rham cohomology is [...]

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9. [...] we’ve seen that all contractible spaces have trivial de Rham cohomology, we can’t use that tool to tell them apart. Instead, we introduce de Rham cohomology with [...]

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10. [...] We’ve seen that if a manifold is simply-connected then the first degree of cubic singular homology is trivial. I say that the same is true of the first degree of de Rham cohomology. [...]

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11. [...] but knowing that is not enough to conclude that for some . In fact, this question is exactly what de Rham cohomology is all [...]

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