DeRham Cohomology in Bott and Tu

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Discussion Overview

The discussion revolves around the computation of DeRham Cohomology for the space X = R² - P - Q, where P and Q are distinct points in R². Participants explore the implications of this computation without the use of the Mayer-Vietoris sequence or homotopy equivalences, relying instead on the definition of DeRham cohomology as the quotient of closed forms by exact forms.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to approach the exercise without using Mayer-Vietoris or homotopy arguments.
  • Another participant suggests transforming the problem into the language of vector fields, noting that this approach is well-known and classical.
  • It is proposed that integrating a one-form over a curve corresponds to computing the work of a vector field over that curve, leading to insights about closed and exact forms.
  • A specific closed 1-form is identified that is not exact, demonstrating that H¹_{DR}(R² - 0) is not trivial.
  • Further analysis leads to the conclusion that the first cohomology group for R² - P - Q has at least two distinct non-zero elements corresponding to the vector fields around points P and Q.
  • Another participant inquires about the conditions under which certain integral results hold, referencing a theorem related to homotopy of curves and line integrals.
  • A later reply provides a precise formulation of the theorem regarding homotopic curves and their integrals, including a proof using Stokes' theorem.
  • Discussion includes a mention of the relationship between exact forms and path integrals depending only on endpoints.

Areas of Agreement / Disagreement

Participants do not reach a consensus on all points, as there are differing views on the implications of certain theorems and the conditions required for the results discussed. Some participants agree on the classical results related to vector fields and closed forms, while others raise questions about specific assumptions and theorems.

Contextual Notes

Some assumptions regarding the functions involved in the integrals and the conditions under which the results hold are not fully articulated, leading to potential gaps in understanding the applicability of the discussed theorems.

Bacle
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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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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: \alpha=Mdx+Ldy \leftrightarrow \mathbf{F}=M\hat{x}+L\hat{y}. 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:

\int_{\gamma}\alpha = \int_{\gamma}\mathbf{F}\cdot d\mathbf{r}

Also, if \alpha is closed, then it means that

\frac{\partial L}{\partial x}-\frac{\partial M}{\partial y}=0

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

\int_{C}\mathbf{F}\cdot d\mathbf{r}=0

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

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 \mathbf{F} is the gradient of some function iff

\int_C\mathbf{F}\cdot d\mathbf{r}=0

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 \alpha represented by the vector field of length 1/r curling around 0 in the anti-clockwise direction:

\mathbf{F}=\frac{y\hat{x}+x\hat{y}}{x^2+y^2}

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

\int_{C_0}\mathbf{F}\cdot d\mathbf{r}=2\pi \neq 0

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

Now, suppose \omega 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

\int_{C_0}\omega \neq 0

In particular, there exists a real number r with

\int_{C_0}\omega =\int_{C_0}r\alpha

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

\int_{C}\omega =n\int_{C_0}\omega=n\int_{C_0}r\alpha=\int_{C}r\alpha

So

\int_C(\omega-r\alpha) = 0

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

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

f = \frac{\partial L}{\partial x}-\frac{\partial M}{\partial y}

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

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

(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 \alpha,\beta corresponding to the vector fields rotating anti-clockwise around p and q respectively.

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

\int_{C_p}\omega =\int_{C_p}r\alpha
\int_{C_q}\omega =\int_{C_q}s\beta

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

\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,

so \omega is cohomologous to r\alpha + s\beta and thus H^1 is the free vector space generated by [\alpha] and [\beta] : H^1(\mathbb{R}^2-p-q)\approx \mathbb{R}^2.
 
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.)
 
no problem, B.
 
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.
 
The precise formulation of the result that I used implicitely is the following: Let \alpha 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 \gamma_0, \gamma_1 two closed curves in U homotopic by a smooth homotopy H:I\times I\rightarrow U:(s,t)\mapsto H(s,t)=H_s(t), H_0=\gamma_0, H_1=\gamma_1, H(s,0)=H(s,1). 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,

\int_{\gamma_0}\alpha = \int_{\gamma_1}\alpha

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

\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

=\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

The thing in parenthesis is 0 because

\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

and H(\cdot,1)=H(\cdot,0).

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:
Thanks again, Quasar, very clear and helpful.
 
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.
 

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