# De Rham cohomology

1. Nov 14, 2007

### Cincinnatus

I have never studied topology (other than point-set topology). I am now taking a course on manifolds using Spivak volume 1.

I am having some difficulty understanding the motivation for defining de Rham cohomology, as expected probably, since I have no idea what "cohomology" is. The definition seems (to me) to be arbitrary and strange (quotient space of closed by exact forms).

It isn't at all clear to me what knowing the de Rham cohomology of a manifold is supposed to tell me about the manifold. I gather that (perhaps some of) the motivation for making this definition is that this space is isomorphic to some topological construction which I don't understand.

My professor assures me that I needn't understand cohomology to understand de Rham cohomology, so I am asking here, what is the motivation for defining de Rham cohomology?

2. Nov 14, 2007

Well, are you aware that all closed forms on the plane are exact, but that this is not true if you remove the origin from the plane? This is a standard example from multivariable analysis.

Anyway, the idea is that the failure of closed forms to be exact occurs when there are "holes" of sorts in your space (like the hole at the origin above), and so DeRham Cohomology measures these holes. Although I must admit, all this is much clearer if you have at least seen the definitions of singular cohomology.

3. Nov 15, 2007

### Hurkyl

Staff Emeritus
Have you ever wanted to "antidifferentiate" a form? e.g. you have a closed form w and wish to seek a function f such that w = df?

The cohomology group is precisely the "obstruction" to being able to do this. For any function f, the expression w - df is equivalent to w.

DeadWolfe mentioned the punctured plane; you can prove that (the equivalence class represented by) $d\theta$ generates of the cohomology group. ($\theta$ is not a function of the punctured plane) So, any closed form w can be written as
$$\omega = df + a d\theta$$

We understand exact forms very well, and we understand $d\theta$ very well. So, this decomposition allows us to understand w very well!

(a is unique. f is determined up to a closed 0-form -- i.e. up to an additive constant)

Last edited: Nov 15, 2007
4. Nov 15, 2007

### mathwonk

it tells you when you can solve df = w.

it says if there is a loop in the space that cannot be contracted to a point, then there is a closed form whose integral over that loop is one.

i.e. there is a closed form that measures how much a given loop imitates that loop.

so it meSURES how many loops in the space cannot be shrunk to a point by measuring instead how many closed forms w are not of form df.

thus it lets you use the power of differential and integral calculus to study the "holes" in your space. e.g. if the equation df = w can always be solved for all clsoed forms w, the there are no (1 dimensional) holes in your space.

or conversely if there are no holes, then dw = 0 is sufficient to solve the eqution df = w.

applications include proving the fundamental theorem of algebra, etc...

doesnt spivak include applications of de rham cohomology?

5. Nov 15, 2007

### Chris Hillman

This book is just what you want:

I. H. Madsen and J. Tornehave, From calculus to cohomology: de Rham cohomology and characteristic classes, Cambridge University Press, 1997.

6. Nov 16, 2007

### slider142

I also suggest learning a tiny bit of topology, as it will help immensely. More importantly, you'll learn what's so great about homology and where category theory came from. :D L. Christine Kinsey's "Topology of Surfaces" is one of my favorites, as it's very clear, has one chapter on higher dimensional topology, and the last chapter has applications to calculus with a short description of cohomology. Madsen's book above is great for the calculus side of things as well.

Last edited: Nov 16, 2007
7. Nov 16, 2007

### Cincinnatus

Thanks for the helpful replies everyone, My professor also recommended the Madsen book. I'll take a look when I have time.

This (and the other comments) reminds me of the winding number / residue theorem from complex analysis, is this a good thing to be reminded of here?

8. Nov 17, 2007

### Hurkyl

Staff Emeritus
A thousand times yes.

9. Nov 17, 2007

### mathwonk

yes in fact if a convex plane region has a finite number k of points a1,...ak, removed then it has k holes. hence it should have k dimensional derham cohomology. in fact a basis for the cohomology consists of the winding number forms with respect to each hole, [isn't it something like dz/(z-a) ?].

10. Nov 19, 2007

### llarsen

If you are looking for an intuitive feel for de Rham cohomology in pictoral terms, it might be useful to look at the site http://www.ee.byu.edu/forms/forms-home.html [Broken], and in particular at the document http://www.ee.byu.edu/forms/ftext.pdf [Broken].

It is designed to present differential forms as a tool for visualizing the laws of electromagnetism. The audience is undergraduate electrical engineers, so do not expect it to be too rigorous. However, it does do a nice job of presenting a visual representation of differential forms. Although de Rham cohomology is not presented explicitly, there is a discussion of stokes theorem which has close connection to the de Rham cohomology.

The paper talks about how current through a wire generates a magnetic field that can be represented as a closed but not exact one form. For purposes of the de Rham cohomology, the wire can be considered a hole in a three dimensional manifold. It also shows how the esistence of a point charge gives rise to electric flux density which is represented as a closed but not exact two form. Again, for purposes of the de Rham cohomology, the location of the charge can be taken as a hole in a three dimensional manifold, and the electric flux density characterizes that hole.

You might note that a wire carrying current can be though of as a one dimensional line (or 1-d hole) in a three dimensional space. A close but not exact one form can be used to characterize this one dimensional hole. A point charge can be thought of as a point or a zero dimensional hole in a three dimensional space. A two form can characterize this point hole in the three dimensional manifold.

The de Rham cohomology generalizes this to arbitrary dimensions, but I found the presentation of stokes theorem and the related pictures to visualize stokes theorem in three dimensions to be very helpful in understanding how the de Rham cohomology characterizes the topology of a more general manifold.

Last edited by a moderator: May 3, 2017
11. Nov 19, 2007

### mathwonk

imagine the complement of the unit disc in R^2 and draw a thin strip containing the positive x axis.

now define a function that equals zero abovew the strip and equals 1 below the strip and grows smoothly from 0 to 1 as you cross the strip.

then take d of this function. you get a one form that is locally exact, and defiend everywhere since d of zero, and d of 1, are both zero.

this one form is hence closed, and gives 1 when integrated over a loop, if and only if that loop goes once around the origin, clockwise.

on any 2 dimensional closed surface, choose a closed loop that cannot be contracted to a point and a sterip around that loop. make the same construction foa funcrtion thatg rows by one as it crosses the strip, and take d of it.

again you get a one form that is closed and integrates to 1 with anhy loop that crosses this loop once in the given direction.

this form thus integrates to 1 over loops that go around any loop transverse to the given one.

if you do this construction for all loops in a given standard homology basis you will get a basis of the derham cohomology,

and you will also see the poincare duality exhibited by the intersection pairing.

i.e. you will have constructed a one form such that integrating against it is the same as intersecting with the given loop.

this is my favorite way to "see" de rham cohomology, and i leartned it from hermann weyl's book on riemann surfaces.

you can also read an abstract sheaf theoretic proof due to andre weil, (as presented in spivaks diff geom),

if you do not want to understand anything at all, but still see a proof that cannot be faulted logically.

Last edited: Nov 19, 2007