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Computing a Generating Set in Cohomology

  1. May 18, 2014 #1

    WWGD

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    Hi, this issue came up in another site:

    We want to compute ( not just ) the deRham cohomology of ## X=\mathbb R^2 - ##{p,q} , but also

    an explicit generating set for ## H^1 (X) = \mathbb Z (+) \mathbb Z## in deRham cohomology . Only explicit
    generating set I can see here is {(0, +/- 1),(+/-1,0)}. How to map this into a pair of 1-forms that generate the first deRham cohomology?

    We can get the actual cohomology using , e.g., Mayer Vietoris, but , without an explicit isomorphism, I don't see how to get a generating set. Maybe we can use the explicit maps in the
    MV sequence to get some generators (in the deRham chain complex) ?

    I was thinking of this: we compute ## H^1(X) ##, then we know all theories are equivalent, so this gives us singular cohomology , and then we use the explicit isomorphism in deRham's theorem to get some generators? We would see where the generators are sent.
     
    Last edited: May 18, 2014
  2. jcsd
  3. May 18, 2014 #2
    A standard example is the 1-form

    [tex]\omega = \frac{ydx - xdy}{x^2 + y^2}[/tex]

    This defines a closed form on ##\mathbb{R}^2\setminus \{(0,0)\}## that is not exact. So it serves as a basis for the cohomology group ##H^1(\mathbb{R}^2\setminus\{(0,0)\)##.

    Then consider the diffeomorphisms ##F,G:\mathbb{R}^2\rightarrow \mathbb{R}^2## by ##F(x,y) = (x,y) - p## and ##G(x,y) = (x,y) - q##, where ##p,q\in\mathbb{R}^2##.

    Are then ##F^*\omega## and ##G^*\omega## not generating elements of ##H^1(X)##
     
  4. May 18, 2014 #3

    WWGD

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    Thanks; if I understood you correctly, I think you need to define maps F,G on or into ## \mathbb R^2 -p ## to be able to pull back the information, because w is not defined in the entire plane.

    I was trying something similar, but the problem is that ## \mathbb R^2 -{p,q} ## retracts into a wedge of circles, but is not homeomorphic to it, and I don't see how to use a homotopy equivalence to pullback a form.

    I still think deRham's isomorphism , if we can use singular, will send generators to generators. But, again, we would need to know an explicit generating set for singular. Maybe the actual maps in Mayer-Vietoris can help get the explicit generating set. Thanks.
     
  5. May 18, 2014 #4
    Yes, you can just restrict domain and codomain to something suitable.

    Sure.

    Finding an explicit generating set for singular homology doesn't seem all that difficult. The intuition is that a circle around one of the points cannot be deformed to a point. So the idea is just to take two cycles, one surrounding ##p## and one surrounding ##q##. Then (in theory) we can find 1-forms associated with these cycles which form a basis for the deRham groups, but this seems difficult in practice.
     
  6. May 23, 2014 #5

    WWGD

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    Never mind, thanks, I found the answer. It is simpler than I thought.
     
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