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Is there a DeRham _Homology_?

  1. Sep 8, 2011 #1

    WWGD

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    Hi, I hope this is not too ignorant, but my Algebraic Topology is rusty:

    Is there such a thing as DeRham _homology_? I always hear and read about

    DeRham cohomology, but I have never heard of DeRham homology. Is there

    such a thing?
     
  2. jcsd
  3. Sep 9, 2011 #2

    hunt_mat

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    DeRahm cohomology is all to do with the cohomology of differential forms, you set up your Meyer-Veotoris sequence via the exterior derivative. There is alway a pairing between homology and cohomology, so I guess that there is that but there is no formal theory called DeRahm homology as far as I am aware.
     
  4. Sep 9, 2011 #3
    Why stop there with the (good) question? How about Cech homology?
     
  5. Sep 9, 2011 #4

    quasar987

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    Cech homology exists, but it is not a homology theory in the sense of the Eilenberg-Steenrod axioms. The little book by Hocking and Young talks a bit about it.
     
  6. Sep 13, 2011 #5

    lavinia

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    The connection of the De Rham cochain complex to cohomology with real coefficiants is Stokes theorem. So the duality is through the isomorphism with real cohomology.
     
  7. Sep 13, 2011 #6
    Hmm, this is an interesting question and is applicable to something I have been thinking about. Apparently there is a notion of Cech homology- but it doesn't satisfy all of the properties you'd like (mainly, it doesn't do well with giving you sequences which should be exact). However, it can be altered, to a thing called (I think) strong homology, which does give a well behaved homology theory and for which cech cohomology is the dual theory.
     
  8. Jun 26, 2012 #7
    We have the homology of simplixes (was it called simplical homology?) whose cohomology is precisely isomorphic to deRham cohomology I believe (you can show that each linear functional on the free group of simplices is precisely integration by some form of appropriate degree)
     
  9. Jun 26, 2012 #8

    lavinia

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    Smooth simplexes
     
  10. Jun 26, 2012 #9
    erm yes smooth it should be to relate it to some triangulation of the manifold. I was a bit unclear sorry, I think it should be a triangulation of the smooth manifold whose deRham cohomology you want to dualize. I never proved this so I am a bit unsure about the precise conditions.
     
  11. Jun 26, 2012 #10

    lavinia

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    It should be a smooth triangulation of the manifold.
     
  12. Jun 26, 2012 #11

    lavinia

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    I think there is a theorem which says that any real (maybe Z/2Z) homology class of a smooth manifold can be represented by an embedded submanifold. I will check it. If this is true, then embedded submanifolds might give you a de Rham homology. This is just an impression. Let's think it through.
     
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