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DeRham Cohomology in Bott and Tu

  1. Jul 23, 2010 #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.
     
  2. jcsd
  3. Jul 25, 2010 #2

    quasar987

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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: [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].
     
  4. Jul 28, 2010 #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.)
     
  5. Jul 28, 2010 #4

    quasar987

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    no problem, B.
     
  6. Jul 30, 2010 #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.
     
  7. Jul 31, 2010 #6

    quasar987

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    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: Jul 31, 2010
  8. Jul 31, 2010 #7
    Thanks again, Quasar, very clear and helpful.
     
  9. Sep 20, 2010 #8

    mathwonk

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    I havent 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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