Motivating the de Rham theorem - MathOverflow

Motivating the de Rham theorem

2010-10-26T16:09:27Z

In grad school I learned the isomorphism between de Rham cohomology and singular cohomology from a course that used Warner's book Foundations of Differentiable Manifolds and Lie Groups. One thing that I remember being puzzled by, and which I felt was never answered during the course even though I asked the professor about it, was what the theorem could be used for. More specifically, what I was hoping to see was an application of the de Rham theorem to proving a result that was "elementary" (meaning that it could be understood, and seen to be interesting, by someone who had not already studied the material in that course).

Is there a good motivating problem of this type for the de Rham theorem?

To give you a better idea of what exactly I'm asking for, here's what I consider to be a good motivating problem for the Lebesgue integral. It is Exercise 10 in Chapter 2 of Rudin's Real and Complex Analysis. If $\lbrace f_n\rbrace$ is a sequence of continuous functions on $[0,1]$ such that $0\le f_n \le 1$ and such that $f_n(x)\to 0$ as $n\to\infty$ for every $x\in[0,1]$, then $$\lim_{n\to\infty}\int_0^1 f_n(x)\thinspace dx = 0.$$ This problem makes perfect sense to someone who only knows about the Riemann integral, but is rather tricky to prove if you're not allowed to use any measure theory.

If it turns out that there are lots of answers then I might make this community wiki, but I'll hold off for now.

Answer by Dick Palais for Motivating the de Rham theorem

2010-10-26T16:27:34Z

Here is a really "trivial" application. Since a volume form (say from a Riemannian metric) for a compact manifold $M$ is clearly closed (it has top degree) and not exact (by Stoke's Theorem), it follows that the cohomology is non-trivial, so $M$ cannot be contractible.

Answer by Francesco Polizzi for Motivating the de Rham theorem

2010-10-26T16:39:26Z

Differential forms and cohomology are somewhat less intuitive than integration (at least for me), so maybe it is no easy to find such a neat example.

Anyway, let's try this one. Consider the 1-form

$\omega:=\frac{xdy-ydx}{x^2+y^2}$

in $X:=\mathbb{R}^2 \setminus 0$.

It provides the standard example of closed form which is not exact, and in fact it is essentially the only example on $X$, because of the following

Proposition.

Every 1-form on $X$ which is closed but not exact is of type $a\omega + \eta$, where $a \in \mathbb{R}$ and $\eta$ is an exact 1-form.

This statement makes perfect sense to everyone who understands differential forms, and at first glance it does not seem obvious at all.

On the other hand, it is an immediate consequence of De Rham theorem: in fact, since $X$ retracts on $S^1$, we have

$H^1_{DR}(X)=H^1_{sing}(X, \mathbb{R})=H^1_{sing}(S^1, \mathbb{R})= \mathbb{R}$,

with generator $[\omega]$.

Answer by Johannes Ebert for Motivating the de Rham theorem

2010-10-26T17:23:20Z

There is quite a number of surprising and deep statements that can be proven using de Rham. The examples I list are not elementary in any sense, but give a glimpse at the power of the theory. They all have in common that they employ features of the de Rham theory that are not at hand in singular theory.

Often de Rham theory is presented as simply being the quickest way to develop cohomology theory, but in my opinion this misses the point. First of all, whether the development of the theory is really simpler than singular theory is contestable, especially if you consider that you get a considerably weaker theory as long as you if you restrict your toolkit to the Eilenberg-Steenrod axioms. Secondly, the real power of de Rham theory becomes apparent when you study specific situations where you can apply different methods than that of standard homology theory. What are these specific situations? Well, I have three examples in mind, but certainly there are much more:

1.) Connections and curvature, i.e. Chern-Weil theory. This can be motivated by the Gauss-Bonnet formula, or, better, the Gauss-Bonnet-Chern theorem, equating the Euler number of a manifold with an integral of some differential form constructed from the curvature. Already the statement that this integral is an integer is pretty intriguing if you do not know about de Rham's theorem.

2.) Symmetry! If a compact group acts on the manifold, you can restrict to invariant forms. If the action is homogeneous, you are left with a finite-dimensional complex. So symmetry can be reduced to cut down the size of the de Rham complex, leading for example to the isomorphism $H^{\ast}(G) \cong (\Lambda \mathfrak{g}^{\ast})^G$ for compact $G$, which came as a real surprise to me when I saw it first. As far as I know, this is the simplest way to the real cohomology of Lie groups.

3.) Kähler metrics! The Hodge decomposition is of course even less elementary than the previous examples, but the statement that the dimension of the space of holomorphic 1-forms on a closed Riemann surface $S$ is precisely the genus (defined as the number of handles) is rather mysterious in the first place.

Answer by Orbicular for Motivating the de Rham theorem

2010-10-26T17:26:30Z

At first I always thought about the deRham theorem in terms of vector analysis and fluid dynamics. For instance, if one has a curl-free vector field, then one might want to write it as a gradient field of a function. But if your domain has holes (of a certain kind) this will not necessarily be true. The analogous statement holds true for divergence-free vector fields that you want to write as the curl of another vector field.

Answer by Deane Yang for Motivating the de Rham theorem

2010-10-26T17:56:41Z

I don't know if this counts as "elementary", but I think the whole story connecting the topology and Morse theory of a closed oriented surface with its deRham cohomology is quite pretty. I recommend the book "Differential Topology" by Guillemin and Pollack. Or Milnor's "Topology from a Differentiable Viewpoint".

Answer by Dmitri Pavlov for Motivating the de Rham theorem

2010-10-26T17:59:08Z

One can use the de Rham theorem to define the Lebesgue integral without ever using any notion of measure theory. More precisely, the integral can be defined as the composition of the following sequence of maps: C^∞_cs(Dens(M))→Hⁿ_cs,dR(M,Or(M))→Hⁿ_cs(M,Or(M))→H₀(M)→H₀(∙)=R.

Here C^∞_cs(Dens(M)) denotes the space of all smooth densities with compact support. The space C^∞_cs(Dens(M)) is mapped to Hⁿ_cs,dR(M,Or(M)) (the nth de Rham cohomology of M with compact support twisted by the orientation sheaf of M) by the obivous map given by the definition of de Rham cohomology. The space Hⁿ_cs,dR(M,Or(M)) is isomorphic to Hⁿ_cs(M,Or(M)) (the nth twisted ordinary cohomology of M with compact support) by the de Rham theorem. The space Hⁿ_cs(M,Or(M)) is isomorphic to H₀(M) by Poincaré duality. Finally, H₀(M) can be mapped to H₀(∙)=R by the usual pushforward map for homology.

More details are available in this answer: http://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38479#38479

Here is an easy application of the above definition: The easiest version of Stokes' theorem states that ∫dω=0, where ω∈Ω^n-1(M,Or(M)). Proof: ∫ factors through the map to the de Rham cohomology. The form dω is a coboundary, hence its image vanishes in the de Rham cohomology and the integral equals zero.

Answer by Georges Elencwajg for Motivating the de Rham theorem

2010-10-26T18:03:23Z

An interesting application of De Rham's theorem is to show that certain differential manifolds are not diffeomorphic. Here are two examples.

1) For $n$ even the sphere $S^n$ and real projective space $\mathbb P^n(\mathbb R)$ are not diffeomorphic since $H^n(S^n) \simeq \mathbb R$ while $H^n(\mathbb P^n(\mathbb R))=0$. Ah, you say, but I can see that with the concepts of orientation or fundamental group $\pi_1$: I don't need your Swiss's stuff! Fair enough: these are reasonable elementary alternatives.

2) Fix $N\geq 2$ and delete $k$ points from $\mathbb R ^N$: call $X_k$ the resulting manifold. Then for $k\neq l$, the manifolds $X_k$ and $X_l$ are not diffeomorphic since $dim_{\mathbb R} H^{N-1}(X_k )=k\neq l=dim_{\mathbb R} H^{N-1}(X_l )$ . However they are both orientable, and simply connected for $N\geq 3$. So the elementary tools of example 1) do not apply.

Answer by Georges Elencwajg for Motivating the de Rham theorem

2010-10-26T20:02:41Z

Dear Timothy, here is a theorem which, according to your wish, "could be understood, and seen to be interesting, by someone who had not already studied the material in that course": Brouwer's celebrated fixed point theorem!

It says that every continuous function from the n-dimensional closed ball $B^n \subset \mathbb R^n$ into itself has a fixed point. Please notice that I wrote "continuous" and didn't even mention the word "differentiable"! So how does De Rham solve the problem ?

Step 1 Reduce to showing that there is no continuous retraction to the inclusion $S^{n-1} \to B^n$ of the boundary sphere. This reduction needs only completely elementary vector (= "analytic") geometry.

Step 2 Reduce the no-retraction statement to non-contractibility of $\mathbb R^{n}\setminus O$. Again, this is easy and requires little more than the definition of contractibility

Step 3 Prove the non-contractibility of $\mathbb R^{n}\setminus O$ by showing that $H^{n-1}( \mathbb R^{n}\setminus O)\simeq\mathbb R$, whereas contractible manifolds have zero de Rham cohomolgy in positive degree. This is the step where De Rham's cohomology shines in all its splendour!

In the same vein you can also prove that the n-dimensional sphere $S^n$ has a tangent everywhere non-vanishing vector field if and only if $n$ is odd.

An excellent source for this material is Madsen and Tornehave's extremely well-written From Calculus to Cohomology (Cambridge University Press).

Answer by Johannes Ebert for Motivating the de Rham theorem

2010-10-26T21:55:56Z

More elementary than my previous post is this "de Rham for the punctured plane in a nutshell", which shows how differential forms capture essential topological information. Put $\omega:= \frac{1}{2\pi i} z^{-1} dz$, a closed $1$-form on $C^{\times}$. Given any smooth path $c$ in the punctured plane, the integral $\int_{c} \omega$ gives a lift of $c$ to $C$ (i.e., the logarithm of $c(1)/c(0)$). If $c$ is closed, you get an integer, call it $\langle \omega, c\rangle$, which is of course the usual path-lifting from covering space theory. But you can also view it as an integration of a specific form over cycles! It is not hard to show, using integration, that $c$ is nullhomotopic iff $\langle \omega,c\rangle=0$, which gives, by the way, a computation of $\pi_1 (C^{\times})$. But now you can vary $\omega$ instead. If $d\omega \neq 0$, examples show that $\langle \omega, c\rangle$ is not homotopy-invariant, so discard that case. If $\omega =df$, show that $\rangle \omega,c\rangle =0$. If $\langle \omega,c\rangle=0$ for all $c$, a likewise elementary argument shows that $\omega=df$, and you have proven de Rhams theorem for $C^{\times}$. You can do the same computation for $S^1=R /Z$ instead, but there you are tempted to integrate only over the fundamental class, which hides essential features of that yoga.

B.t.w.:Moritas book "Geometry of differential forms" contains another application cute application of the de Rham theorem: a definition of the integral Hopf invariant of a map $f:\bS^3 \to \bS^2$ in terms of differential forms (page 133). Morita also discusses Gauss' integral formula for the linking number.

Answer by Donu Arapura for Motivating the de Rham theorem

2010-10-27T15:56:43Z

I don't know if it is necessary to add yet another answer, but this theme is close to my heart. I'm not a historian, and I would be happy if someone corrects me here, but I have the impression that the idea of understanding a differential in terms of its periods, which would go back to Riemann at least, would have been a historical antecedent to de Rham's theorem. In other words, I don't think the theorem came out of a vacuum.

To explain what I mean by periods, suppose that $X$ is a compact Riemann surface of genus $g$. Then $H_1(X,\mathbb{Z})=\mathbb{Z}^{2g}$, with a basis of loops $\gamma_i$ constructed in the usual way. De Rham's theorem gives an isomorphism of the first de Rham space $H^1(X,\mathbb{C})\cong \mathbb{C}^{2g}$ by identifying a $1$-form $\alpha$ with its period vector $(\int_{\gamma_i}\alpha)$. Of course, the 19th century people would have been more interested in the case where $\alpha$ is holomorphic. In this case, the space of holomorphic forms injects into $H^1(X,\mathbb{C})$ (Proof: $\alpha=df$ implies that $f$ is holomorphic and therefore constant). This is why they could talk about this without explicitly defining cohomology first.

Answer by Steven Gubkin for Motivating the de Rham theorem

2011-03-01T22:06:59Z

$\frac{1}{4\pi} \oint_{\gamma_1}\oint_{\gamma_2} \frac{\mathbf{r}_1 - \mathbf{r}_2}{|\mathbf{r}_1 - \mathbf{r}_2|^3} \cdot (d\mathbf{r}_1 \times d\mathbf{r}_2)$

is an integer when $\gamma_1, \gamma_2: S^1 \to \mathbb{R}^3$ are non-intersecting differentiable curves.

Seriously?

This number tells you how many times $\gamma_1$ winds around $\gamma_2$ (The linking number). My wife was a math and biochem major as an undergraduate interested in applying knot theory to genomics, and she and I spent countless hours trying to make sense of this without any knowledge of cohomology. Builds character I guess.