# 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.

July 20, 2011