DeRham Cohomology in Bott and Tu

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In summary, the conversation is about finding the DeRham Cohomology of a space X=R^2-P-Q, where P and Q are two different points in R^2. The exercise is confusing because it is introduced before the Mayer-Vietoris sequence. The exercise can be solved using the definition of DeRham cohomology as the quotient space of closed forms modded out by exact forms. The conversation then discusses the identification of 1-forms and vector fields, and how integrating a one form over a curve is equivalent to computing the work of the corresponding vector field. It is also mentioned that a closed 1-form is exact if and only if the corresponding vector field is the gradient of some function. The conversation
  • #1
Bacle
662
1
Hi, Everyone:

There is an exercise in the beginning of Bott and Tu's Diff. Forms in
Algebraic Topology, of finding the DeRham Cohomology of X=R^2-P-Q,
where P,Q are two different points in R^2. What is confusing is that,
at the point of the exercise, we have not yet introduced Mayer-Vietoris
sequence. And we cannot either use tricks like using the fact that X retracts
(i.e., is homotopic to) the figure-8 space, aka, wedge of two circles, and then
using algebraic topology.

Bottom line is all we seem to have available is just the definition of
DeRham cohomology as the quotient space of closed forms modded out by
exact forms.

Any Ideas.?

Thanks.
 
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  • #2
I think I figured this out.

The bottom line is that if you transform the problem into the language of vector fields according to the correspondance described in example 1.2, then this is all well known classical stuff.

I found it a little easier to do the case R² - 0 first.

So we are identifying 1-forms and vector fields in the following way: [tex]\alpha=Mdx+Ldy \leftrightarrow \mathbf{F}=M\hat{x}+L\hat{y}[/tex]. Under this identification, integrating a one form over a curve is the same as computing the work of the vector field F over that curve:

[tex]\int_{\gamma}\alpha = \int_{\gamma}\mathbf{F}\cdot d\mathbf{r}[/tex]

Also, if [tex]\alpha[/tex] is closed, then it means that

[tex]\frac{\partial L}{\partial x}-\frac{\partial M}{\partial y}=0[/tex]

and in particular, by Green's thm, if C is a contractible closed curve (i.e. here, any curve not enclosing the origin), then

[tex]\int_{C}\mathbf{F}\cdot d\mathbf{r}=0 [/tex]

And [tex]\alpha[/tex] is exact iff [tex]\mathbf{F}=\nabla\phi[/tex] for some scalar function [tex]\phi[/tex].

Note also that there is in this setting a particular case of DeRham's theorem that is well known, and that is that a vector field [tex]\mathbf{F}[/tex] is the gradient of some function iff

[tex]\int_C\mathbf{F}\cdot d\mathbf{r}=0[/tex]

for all closed curves C. (That is, iff F is conservative.)

This said, notice that there is at least one closed 1-form on R² - 0 that is not exact, and that is the 1-form [tex]\alpha[/tex] represented by the vector field of length 1/r curling around 0 in the anti-clockwise direction:

[tex]\mathbf{F}=\frac{y\hat{x}+x\hat{y}}{x^2+y^2}[/tex]

It is straightforward to check that it is closed, and it is not exact because obviously, for [itex]C_0[/itex] the unit circle around the origin oriented in the anti-clockwise direction,

[tex]\int_{C_0}\mathbf{F}\cdot d\mathbf{r}=2\pi \neq 0[/tex]

Very well, so [tex]H^1_{DR}(R^2 - 0)[/tex] is not trivial and obviously, for each real number r, [itex][r\alpha][/itex] is a different cohomology class.

Now, suppose [tex]\omega[/tex] is a closed but inexact 1-form. Since the integral of 1-form around a loop depends only on the winding number of the loop around 0, then it must be that

[tex]\int_{C_0}\omega \neq 0[/tex]

In particular, there exists a real number r with

[tex]\int_{C_0}\omega =\int_{C_0}r\alpha[/tex]

And so for C any loop of winding number n about 0, we have

[tex]\int_{C}\omega =n\int_{C_0}\omega=n\int_{C_0}r\alpha=\int_{C}r\alpha[/tex]

So

[tex]\int_C(\omega-r\alpha) = 0 [/tex]

which is to say, [itex]\omega -r\alpha[/itex] is an exact 1-form. Thus, the cohomology classes [itex]r\alpha[/itex] are the only things in the first cohomology group: [itex]H^1_{DR}(R^2 - 0) = \mathbb{R}\left<[\alpha]\right>\approx \mathbb{R}[/itex].

For the other groups, it's easy:

-H^k(R² - 0) is 0 for k>2

-H^0(R² - 0) are the functions whose derivatives vanish everywhere, i.e. the constant functions. So H^0(R² - 0)=R.

-H²(R² - 0) is the space of all 2-forms fdxdy on R² - 0 modulo the exact ones, which can be identified with the space of all smooth maps f on R² - 0 modulo those of the form

[tex]f = \frac{\partial L}{\partial x}-\frac{\partial M}{\partial y}[/tex]

for some smooth maps M and L. But all smooth maps are of this form since one can always take M(x,y)=0 and

[tex]L(x,y):=\int_{a}^x f(x',y)dx'[/tex]

(a =\=0). Thus H²(R² - 0)=0.


Now, for R² - p - q (or for R² - {any finite set of points}), it is the same thing, except in the computation of the first cohomology group, the argument is that H^1 now has at least two distinct non zero elements. Namely, those 1-forms [itex]\alpha,\beta[/itex] corresponding to the vector fields rotating anti-clockwise around p and q respectively.

And now a closed non exact 1-form [itex]\omega[/itex] and C_p, C_q loops around p and q of winding number 1, there are real numbers r and s such that

[tex]\int_{C_p}\omega =\int_{C_p}r\alpha[/tex]
[tex]\int_{C_q}\omega =\int_{C_q}s\beta[/tex]

And so for C any loop of winding number n about p and m about q, we have

[tex]\int_{C}\omega =n\int_{C_p}\omega+m\int_{C_q}\omega =n\int_{C_p}r\alpha + m\int_{C_q}s\beta = \int_{C}r\alpha+ \int_{C}s\beta[/tex],

so [itex]\omega[/itex] is cohomologous to [itex]r\alpha + s\beta[/itex] and thus H^1 is the free vector space generated by [itex][\alpha][/itex] and [itex][\beta][/itex] : [itex]H^1(\mathbb{R}^2-p-q)\approx \mathbb{R}^2[/itex].
 
  • #3
Thanks a lot, Q (Monsieur Le Q ), very complete and very helpful. If it is
O.K, I will ask a few followups ( I don't have my own computer, so I have limited
ability to reply.)
 
  • #4
no problem, B.
 
  • #5
Quasar:
I was wondering about some result I think you are using implicitly, but I am
missing some conditions on the functions f; I wonder if you know what this
theorem is: (Sorry, I will learn Tex soon.)

If two curves C,C' in a space X are homotopic, let Int_C(t) f(t)dt be the
line integral of f along C ; then, for any f (piecewise continuous, maybe.?),

Integral_C(t) fdt =Integral_C'(t) fdt

In particular, if C (therefore C') are contractible in X, then Int_C(t) f(t)dt=0

Rotman defines Curves C,C' to be homologous to each other if the above
is the case, i.e., if for every (p.wise cont.?) f , we have

Int_C-C' f(t)dt

then C,C' are said to be homologous curves in X

Do you know of this result, and, could you please refer me to a proof.?

Thanks Again for your Excellent Reply.
 
  • #6
The precise formulation of the result that I used implicitely is the following: Let [itex]\alpha[/itex] be a closed smooth 1-form (translate that in the language of vector fields if you'd like) in an open set U of R² and [itex]\gamma_0[/itex], [itex]\gamma_1[/itex] two closed curves in U homotopic by a smooth homotopy [itex]H:I\times I\rightarrow U:(s,t)\mapsto H(s,t)=H_s(t)[/itex], [itex]H_0=\gamma_0[/itex], [itex]H_1=\gamma_1[/itex], [itex]H(s,0)=H(s,1)[/itex]. This last condition just means that H is a homotopy of closed loops: for every value of the parameter s, t-->H_s(t) is a closed loop. Then,

[tex]\int_{\gamma_0}\alpha = \int_{\gamma_1}\alpha[/tex]

Probably the most direct proof uses Stokes' theorem like this:

[tex]\int_{\gamma_1}\alpha-\int_{\gamma_0}\alpha=\int_{[0,1]}\gamma_1^*\alpha-\int_{[0,1]}\gamma_0^*\alpha=\int_{[0,1]}H_1^*\alpha-\int_{[0,1]}H_0^*\alpha=\int_{\{1\}\times [0,1]}H^*\alpha-\int_{\{0\}\times [0,1]}H^*\alpha+0[/tex]

[tex]=\int_{\{1\}\times [0,1]}H^*\alpha-\int_{\{0\}\times [0,1]}H^*\alpha+\left(\int_{[0,1]\times\{1\}}H^*\alpha-\int_{[0,1]\times\{0\}}H^*\alpha\right)=\int_{\partial([0,1]\times[0,1])}H^*\alpha=\int_{[0,1]\times[0,1]}d(H^*\alpha)=\int_{[0,1]\times[0,1]}H^*(d\alpha)=0[/tex]

The thing in parenthesis is 0 because

[tex]\int_{[0,1]\times\{1\}}H^*\alpha-\int_{[0,1]\times\{0\}}H^*\alpha=\int_{[0,1]}H(\cdot,1)^*\alpha-\int_{[0,1]}H(\cdot,0)^*\alpha[/tex]

and [itex]H(\cdot,1)=H(\cdot,0)[/itex].

But there is a better proof in Fulton (Prop. 2.17) in that you understand much better how\why the cancellation takes place.
 
Last edited:
  • #7
Thanks again, Quasar, very clear and helpful.
 
  • #8
I haven't read that great looking answer, but this is one of my favoriote topics, being the one through which I first began to discover some algebraic differentiL TOLOGY FIOR MYSELF AS A YOUNG calculous instructor.

Basically, a one form is exact if and only its path integrals depend only on endpoints, iff the integral around all closed paths are zero. To prove this one direction is the FTC, and the other follows by just defining an antiderivative by path integration from any fixed starting point.

Stokes or green's theorem implies that integrating over two paths that wind around the origin the same number of times is the same.

Hence the group homomorphism from closed one forms on the punctured plane to the reals, by integrating over the unit circle, has kernel equal to the exact one forms.

then one shows it is surjective by writing down the usual angle form dtheta. the same approach computes H^1 of the n times punctured plane.
 

1. What is DeRham Cohomology?

DeRham Cohomology is a mathematical tool used in differential geometry and topology to study the topology of smooth manifolds. It is a cohomology theory based on the concept of differential forms, which are geometric objects that can be integrated over a manifold.

2. What is the significance of DeRham Cohomology in Bott and Tu?

In their book "Differential Forms in Algebraic Topology," Bott and Tu introduce DeRham Cohomology as a powerful tool for studying algebraic topology. They show that this cohomology theory can be used to prove the de Rham theorem, which states that every closed differential form on a smooth manifold can be written as the exterior derivative of another form.

3. How is DeRham Cohomology computed?

To compute DeRham Cohomology, one needs to construct the de Rham complex, which is a sequence of vector spaces and differential operators. The cohomology groups are then defined as the spaces of closed forms modulo exact forms. The cup product, which is a bilinear operation on cohomology classes, can also be used to compute the cohomology groups.

4. What are some applications of DeRham Cohomology?

DeRham Cohomology has various applications in mathematics and physics. In mathematics, it is used to study the topology of manifolds, as well as to prove important theorems such as the de Rham theorem and the Poincaré lemma. In physics, it is used to describe the topology of spacetime in theories such as general relativity and string theory.

5. How does DeRham Cohomology relate to other cohomology theories?

DeRham Cohomology is one of many cohomology theories, each with its own distinct properties and applications. However, DeRham Cohomology is unique in that it is defined purely in terms of differential forms, making it particularly useful for studying smooth manifolds. It is also closely related to other cohomology theories, such as singular cohomology and Čech cohomology, and can be used to prove results in these theories.

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