Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Computing a Generating Set in Cohomology

  1. May 18, 2014 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    User Avatar
    Science Advisor
    Gold Member

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    User Avatar
    Science Advisor
    Gold Member

    Never mind, thanks, I found the answer. It is simpler than I thought.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Computing a Generating Set in Cohomology
  1. Open set (Replies: 2)

  2. Connected set (Replies: 16)

  3. Union of sets (Replies: 3)

  4. Open set (Replies: 19)

Loading...