How is de Rham cohomology useful?

Question

I'm learning about smooth manifolds from the last part of Introduction to Manifolds. Despite the fact I read this part few times I feel like I can't master it because I need to understand how it is useful.

Can you give me a couple of examples explaining how de Rahm cohomology is useful? I understand the given definition, and how the computation is made easier by using the Mayer Vietoris sequence, but I really struggle to find the following:

1. Why is the de Rham cohomology useful? If we wouldn't be using it what sorts of problem would be hard? (Some people say they're useful to solve equations, but how? the few examples I've seen are quite obscure to me.)
2. Maybe a set of 5 examples with computation included that might actually be useful to understand how they're actually computed. You can even give me a web page or a Youtube video (the latter would be better) with few exercises.

Answer 1

The one-sentence explanation is that cohomology on a space $X$ answers the question of when you can promote local solutions of a problem to global solutions; that is, if you can solve a problem on each element of a cover of $X$, can you patch those individual solutions together to get a global solution?

For de Rham cohomology, the first question is one of closed versus exact differential forms. Given a closed form $\omega$ on a smooth and (let's say) compact manifold $X$, is there another form $\eta$ with $d\eta = \omega$? The answer is easy on a star-shaped patch of $X$: just integrate. (I'm glossing over reduced versus unreduced cohomology, so you have to be more careful with the dimension of $\omega$ to get a definitive answer here.) With a bit more work, the same answer holds for a coordinate patch diffeomorphic to $\mathbb{R}^n$. That means we know the answer locally on $X$, but it's not clear how to proceed to a global answer. The classical example is $\mathbb{R}^2\setminus{0}$, where the form \begin{align*} d\theta = \frac{- y\, dx + x\, dy}{x^2 + y^2} \end{align*} is closed but not exact by Stokes' theorem, since its integral over a loop around $0$ doesn't vanish. That last part is a hint that topology is involved, since we can look at $\int_{\gamma} d\theta$ as a function of $\gamma$ and consider how it varies under smooth homotopy.

Still, the idea of closed versus exact differential forms may seem a bit arbitrary. One way of resolving it is to note that de Rham cohomology group $H^0(X)$ is the space of functions that are locally constant (as there are no (nonzero, I guess) forms of negative degree). Under some reasonable assumptions on $X$, that means that $H^0(X)$ is just the $\mathbb{R}$-space of connected components of $X$. The higher groups have similar topological ramifications that would take a bit too long to describe, but the point is that despite the de Rham cohomology defined in terms of a certain set of differential equations, they're fundamentally topological objects.

To that end, the answer to the question of what the de Rham cohomology groups are good for might be that they're a particularly useful and concrete way of defining and computing the cohomology of a space. There are multiple cohomology theories (not all of which are well-defined for all spaces): singular cohomology, cellular cohomology, etc. Because of abstract nonsense, these construction generally coincide under reasonable conditions on $X$. The short explanation is that all of these start with the same $H^0$ and satisfy certain axioms that force the higher $H^*$ to agree.

Quicker and simpler explanations for the specific questions you asked:

1) Some assorted applications:

• De Rham cohomology is invariant under homotopy equivalence (in the smooth category), which allows us to easily prove that two manifolds are not homeomorphic or homotopy equivalent. Furthermore, cohomology comes with a ring structure that gives an extra condition to require for equivalence.

• Homotopy groups are hard to compute; cohomology groups are generally much easiser. The Hurewicz theorem states that the lowest homotopy and homology group of a reasonable space coincide if they're in dimension $\geq 2$; in dimension $1$, you have to abelianize $\pi_1$ first. (Homology groups can generally be obtained from cohomology via, say, Poincare duality.)

• Rational homology groups (i.e., $\pi_* \otimes \mathbb{Q})$ can be obtained from $H_*(X) \otimes \mathbb{Q}$ via minimal models, under some mild restrictions on the underlying space.

• Closed, simply-connected $4$-manifolds are determined by their intersection forms, which in the smooth case is similar to the the map $H^2(X) \otimes H^2(X) \to \mathbb{R}$ defined by $(\alpha, \beta) \to \int_X \alpha \wedge \beta$. (It's only "similar to" in part because de Rham cohomology is naturally defined over $\mathbb{R}$, while we need to work over $\mathbb{Z}$ here.) With a lot of work, this leads to proving that there exist closed, simply-connected manifolds that have no smooth structure.

• For an Eilenberg-MacLane space $K(G, A)$ with $A$ abelian, the map $f \to f^* \omega$ for some specific $\omega$ gives a bijection between the space $[X, K(G, n)]$ of maps $X \to K(G, n)$ modulo homotopy and $H^*(X, G)$. This leads to the idea of classifying spaces for vector bundles, and bundles over a given space can be classified via cohomology. This result also has deeper consequences; for example, it leads to a proof (although there are other ones available) that every closed, oriented $3$-manifold is the boundary of some compact, oriented $4$-manifold.

• For groups $G$ and $A$ with $A$ abelian, a central extension of $G$ by $A$ is a group $E$ for which $A\subset Z(E)$ and $E/A = G$. These groups are classified by the second group-homology group $H^2(G, A)$. The connection with group-cohomology and topological-cohomology is that, for reasons of abstract nonsense, $H^*(G)$ is precisely $H^*(K(G, 1))$, with $K(G, 1)$ the Eilenberg-MacLane space above. (We're now definitely dealing with, e.g., singular topology rather than de Rham cohomology, since we have no convenient smooth structure at hand, but the point is that an abstract question about group theory turns into a problem about topology.)

• In general, there are varieties of cohomology in commutative algebra, number thoery, physics, etc. At some point, it's like the concept of integral: the original impetus and most concrete application may be computing the area under a curve, but there are so many applications of it in so many different fields that it's hard to nail down one single thing that you can point and say it's designed for. Similarly, just as Riemann integrals are a first, concrete step toward a general idea of integration (leading to integrating differential forms, the Lebesgue integral, etc.), de Rham cohomology is one particular useful and concrete kind of cohomology theory.

2) Five is a lot of examples to give, and I'm not sure what exactly you're looking for here. There are a bunch of examples in Hatcher (in the topological rather than smooth case, but the principle and the axioms are effectively the same), and you can work through the known case of the genus $g$ oriented, closed surface.

Answer 2

Solutions of the equation of motion for a topological gauge field theory are given by closed form modulo exact form(the gauge symmetry). This is precisely the de Rham cohomology.

Please see 4.3 (page 20) and 4.4 (pages 20-21 ) of this reference:

https://uis.brage.unit.no/uis-xmlui/bitstream/handle/11250/2786275/no.uis%3Ainspera%3A79010903%3A35182444.pdf?sequence=1&isAllowed=y

This particular reference then goes further to show you the full story. It will enable you to give answers to physics problems without even lifting a finger to do a complicated calculation. POWER!

Extended commentary.

You might want to read part B first

Part A

There are three concepts here. Homology groups, Cohomology groups, and Equations of motion for say a Chern-Simons theory. A Chern-Simons theory is a name given to an object called an action which can be written as some combination of 3 forms at least according to Edward Witten’s formulation.

One of two central results from the Homology and Cohomology idea is that even though homology lives in the land of k-chains and linear transformations that take you from k to k-1 chains at least in Noether’s formulation, and Cohomology lives in the land of k-forms and derivatives that take you from k to k+ 1 forms, two theorems show that they measure the same holes. DeRham’s theorem says the kth homology and the kth cohomology are isomorphic. Similarly, Stokes-Cartan theorem holds.

The statement in the shared reference that “Solutions of the equation of motion are given by closed form modulo exact form (the gauge symmetry)” for the three-dimensional Abelian Chern-Simons theory is just saying that if you write down the formula for the Cohomology group and just as he does in example 14, writing down the equations of motion from the action.

Many speculations (conjectures) could probably be drawn from this. I would leave such speculations to your imagination.

PART B

I think what you are probably really first thinking about are questions about the validity of basic notions of calculus, and how to identify and measure holes.

• Is the fundamental theorem of calculus always valid?
• What does Derham cohomology look like in different dimensions?
• What happens if I walk in circles on a cylinder?
• How can I see how mappings between k chains work?
• Etc
• . . . Well, there is a short and precise video with this and most of the first applications of DeRham Cohomology.

This video:

https://youtu.be/2ptFnIj71SM?si=TrwTa4jZUxHfR0SP

Answer 3

At least for the first de Rham cohomology, it is easy to give interesting physical interpretations that students could appreciate:

The generator of the first de Rham cohomology of $\mathbb{R}^3 \setminus \operatorname{span} \{ e_z \}$ is the magnetic field of an infinitely long wire with current flowing in the $z$-direction. This tells you that path integrals of this field locally only depend on the start and finish points as long as your path does not wrap all the way around the wire.

The generator of the first de Rham cohomology of $\mathbb{R}^2 \setminus \{ (0,0)\}$ is just a cross-section of the magnetic field of a long wire with current flowing out of the page. Alternatively, we can interpret this result as telling us that there is no continuous way to assign angles to all points on the punctured plane.

Answer 4

Since all exact forms are closed (To see this, suppose $\omega$ is an exact $n-$form, then there exists some $(n-1)-$form $\eta$ such that $d\eta=\omega$ but then $d\omega=d(d\eta)=d^2\eta=0$ thus $\omega$ is closed), the de Rham cohomology measures to what degree closed forms fail To be exact (this is a quote from Terence Tao, not taking credit but this seems to be the best laymen's terms method of phrasing it). If the de Rham cohomology group is trivial all closed forms are exact. It’s a measure of when the fundamental theorem of calculus fails in higher dimensions. I.e., the $p-$th de Rham cohomology of some smooth manifold with or without boundary (oriented, for non-oriented one must add a twist but it still holds), say $M$ is

$$H_{\text{dR}}^p(M)=\{\text{closed $p-$forms on $M$ modulo exact $p-$forms on $M$}\}.$$

(To rigourously define this requires the heavy machinery of sequences which are certain "nice" maps between say $n$ forms and $n+1$ forms on a given smooth manifold with or without boundary, but in a nut shell its just closed forms modulo exact forms in the algebraic sense of modding out by a normal subgroup.)

So if this group is trivial all closed forms are exact. If it is non-trivial, then there exists closed $n-$forms on $M$ that are not exact. For instance a case where the de Rham cohomology is trivial is on a star shaped domain. This is known as the Poincaré Lemma .Recall, a space $S$ is said to be a star-shaped domain if there exists a point $s\in S$ such that the line between this point and any other points lies within $S$. The ideal picture of one in $\Bbb{R}^2$ is an actual star as the center can be connected to all other points in $S$ with a straight line such that the line still lies entirely within $S$.

Cool fact 1: for the $n-$sphere, all closed $p-$forms for $0<p<n$ are exact. And for $\Bbb{R}^n$ when $n>1$, we have that the only non trivial de Rham cohomology groups of this space minus a point are the $0$ and $n-1$ de Rham cohomology groups. I.e.,for $n>1$, if $\overline{x}\in\Bbb{R}^n$ is given by say $\overline{x}=(x^1,x^2,…,x^n)$ its coordinate representation in $n-$space, then

$$H_{\text{dR}}^p(\Bbb{R}^n\setminus \{\overline{x}\})=0$$

Except when $p=0, n-1$.

Cool fact 2: It can be used to show $\Bbb{R}^m\cong\Bbb{R}^n$ iff $m=n$. This is done so by considering the de Rham cohomology group when removing a point. That is, we suppose $\Bbb{R}^m\cong\Bbb{R}^n$, then let $U$ be an open neighborhood homeomorphic to both. Then while it is homeomorphic to $\Bbb{R}^n$,

$$H_{\text{dR}}^{n-1}(U\setminus \{\overline{x}\})\neq 0$$

but since it is also homeomorphic to $\Bbb{R}^m$,

$$H_{\text{dR}}^{n-1}(U\setminus\{\overline{x}\})=0$$

and the de Rham cohomology groups are topological invariants thus $m=n$. In other words for $U\subset\Bbb{R}^n$,

$$H_{\text{dR}}^p(U\setminus\{\overline{x}\})=0$$

when $p\neq 0,n-1$.

For an example of closed not being exact, @anomaly gave the classical example as seen in Lee, and the reason it works is our domain isn’t star shaped any longer if you remove origin from plane. Because then the line connecting $(1,1),(-1,-1)$ is not fully inside $\Bbb{R}^2\setminus\{(0,0)\}$.