Question

I'm soon giving an introductory talk on de Rham cohomology to a wide postgraduate audience. I'm hoping to get to arrive at the idea of de Rham cohomology for a smooth manifold, building up from vector fields and one-forms on Euclidean space. However, once I've got there I'm not too sure how to convince everyone that it was worth the journey. What down-to-earth uses could one cite to prove the worth of the construct?

Answer 1

The motivation that most appeals to me is very simple and can come up in a freshman vector calculus course.

We say that a vector field $F$ in $\mathbb{R}^3$ is conservative if $F = \nabla f$ for some scalar-valued function $f$. This has natural applications in physics (e.g. electric fields). It's easy to see this happens iff line integrals of $F$ are path independent, iff line integrals around closed loops vanish, etc. On a simply connected domain, $F$ is conservative iff $\nabla \times F = 0$ (use the freshman version of Stokes' theorem). On a non-simply connected domain, this may fail (e.g. $\mathbb{R}^3$ minus a line). The extent to which it fails is of course the de Rham cohomology of the domain. So this suggests that the de Rham cohomology is a good way to detect the "shape" of a domain, and one goes on from there.

Maybe this is too basic to be interesting, but I like it myself.

Answer 2

This is probably not "down-to-Earth" enough for your purposes, but it was one of the first uses of de Rham cohomology that I really enjoyed and I feel like I must share it. (I learned it from Bott's article "The geometry and representation theory of compact Lie groups" in the 1977 Proceedings of the SRC/LMS Research Symposium on Representations of Lie Groups.)

Theorem: If the $n$-sphere $S^n$ is a Lie group, then $n$ must be odd (or zero!).

Proof: $S^0$ is obviously a Lie group, so that's that. If $S^{2n}$ were a Lie group (and $n>0$), then we'd have a smooth multiplication map $m \colon S^{2n} \times S^{2n} \to S^{2n}$ and an identity element $e \in S^{2n}$. The map $m$ induces a ring homomorphism $m^\ast \colon H^\ast(S^{2n}) \to H^\ast(S^{2n} \times S^{2n})$. Composing this with the Kunneth isomorphism $$ H^\ast(S^{2n} \times S^{2n}) \stackrel{\sim}{\longrightarrow} H^\ast(S^{2n}) \otimes H^\ast(S^{2n}) $$ yields a map $f \colon H^\ast(S^{2n}) \to H^\ast(S^{2n}) \otimes H^\ast(S^{2n})$.

Now, $$ H^\ast(S^{2n}) = \begin{cases} \mathbb{R} & \text{if }\ast=0 \text{ or } 2n \\ 0 & \text{otherwise.}\end{cases} $$ So, writing $1$ and $\lambda$ for the generators of $H^0$ and $H^{2n}$, respectively, we have $f(1)=1\otimes1$ and $f(\lambda)=a(\lambda\otimes1)+b(1\otimes\lambda)$ for some $a,b\in\mathbb{R}$. To determine $a$ and $b$, first restrict the multiplication map to $S^{2n} \times \{e\}$, where it is the identity map. Consequently $m^*$ gives the identification $H^\ast(S^{2n} \times \{e\}) \cong H^\ast(S^{2n})$ and it follows that $a=1$. Similarly, $b=1$.

As $f$ is a ring homomorphism, we also have $$\begin{align} f(\lambda^2) = f(\lambda)^2 &= (\lambda\otimes1 + 1\otimes\lambda)^2 \\ &= \lambda^2\otimes1+ (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)+1\otimes\lambda^2.\end{align}$$ As $\lambda^2 \in H^{4n} = 0$, this reduces to $0 = (\lambda\otimes1)(1\otimes\lambda)+(1\otimes\lambda)(\lambda\otimes1)$. Recall, however, that the product structure on the tensor product $A \otimes B$ of graded rings is given by $$ (a \otimes b)(c \otimes d) = (-1)^{\deg a \deg c}(ac\otimes bd). $$ Since $\lambda \in H^{2n}$, this means that $$ (\lambda\otimes1)(1\otimes\lambda) = (1\otimes\lambda)(\lambda\otimes1) = \lambda \otimes \lambda.$$ Consequently, $\lambda \otimes \lambda = 0$, which is a contradiction! $\blacksquare$

Of course, there are several other (perhaps even better) ways of proving this theorem, but I think this proof is fairly charming.

Answer 3

The calculations of cohomology of homogeneous space $X=G/H$ is reduced to a problem in linear algebra.

[If $G$ is compact and connected then any form is cohomologous to its left shifts and therefore it is cohomologous to the avagage of all left shifts, which is a left-invariant form. Thus $H^k(X,\mathbb R)$ is isomorphic to space of $k$-forms at one point which is invariant under roation of the stabilizer.]

Answer 4

I'd split your question in 2:

(1) Motivation for (co)homonology in general

(2) Motiviation for de Rham cohomology in particular.

The anwser to (1) is the basis for algebraic topology. One reduces complicated geometry problems to the only thing mathematicians really understand aka linear algebra. A typical example would be are Brouwer fixed point theorem (or its corollary $\mathbb{R}^n \simeq \mathbb{R}^m$ $\Rightarrow$ $n=m$).

An answer to (2) is that it is usually much much easier to compute. The reason for this is that it is already more linear by definition but this efficiency comes at the cost of restricting ourselves to smooth manifolds. For example, it is obvious that for any manifold $X$ of dimension $n$, $H^i_{dR}(X) = 0$ for $i > n$ while this is highly non trivial when you look at singular cohomology or sheaf cohomology. On the other hand it is not obvious at all that de Rham cohomology is actually a topological invariant. Another example one would be that if $X$ is an affine complex algebraic variety, one can compute its de Rham cohomology using only algebraic differential forms (not an obvious fact but pretty intuitive for the audience). An audience discovering the theory might be glad to see how easy that makes the computation of the de Rham cohomology of $\mathbb{C}^\times$ (or $S^1$).

Another answer to (2) is that de Rham cohomology brings a different flavor and the interaction between Betti and de Rham cohomology leads to periods, Hodge theory etc... but this may be hard to get to in an introductory talk.

PS: I can't believe I didn't mention it (that's the what I was aiming at in the last paragraph) but I think it would make a great goal for an introductory lecture: STOKES' THEOREM!!! As an undergrad, my physics teachers kept bothering me with operators like "div" and "rot", we had a formula for a volume, one for a surface and one for a line. I had to wait several years and a differential geometry class to understand that these were all special case of the simple and elegant Stokes' formula. In modern terminology it just means that integration induces a morphism of complexes: $\int_X: \Gamma(X,\Omega_X^\bullet) \to Hom(C_\bullet(X),\mathbb{K})$.

Answer 5

Perhaps even simpler than the examples from electromagnetism in $\mathbb{R}^3$ minus some points is the following:

The angle "function" $\varphi\colon S^1 \longrightarrow \mathbb{R}$ is not really globally defined as turning aroung one time gives a discontinuity. It jumps by $2\pi$. Neverhtheless, the differential $\mathrm{d}\varphi$ is a perfectly global one-form on $S^1$. It is the usual volume form, not being exact but closed for dimensional reasons. So the non-trivial first deRham cohomology of $S^1$ is responsible for counting angles and the fact that $0 \ne 2\pi$ ;)

This can be upgraded to the more interesting statement that on a orientable compact manifold without boundary you have a non-trivial top-degree deRham cohomology: again, the reason is that we can integrate a volume form resulting in a non-zero volume. Thus (by Stokes theorem) the volume form can not be exact. It is closed without thinking about it, simply for dimensional reasons.

Answer 6

You may be able to convey the significance of de Rham cohomology to really wide audiences through electromagnetism.

I don't claim to understand all the physics (or topology for that matter), but see my friend Rob Kotiuga's book at http://library.msri.org/books/Book48/index.html. See, for instance, chapter 1D: Nineteenth-Century Problems Illustrating the First and Second Homology Groups, or pp. 30--32, "Chain complexes in electrical circuit theory."

Answer 7

Here is my attempt to explain DeRham cohomology and Hodge theory to a group of first year grad students that have no knowledge of manifolds.