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

Why only Closed Forms Matter in DeRham Cohomology?

  1. Sep 16, 2014 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Hi All,

    One gets homological/topological information (DeRham cohomology ) from a manifold by forming the algebraic quotients

    H^Dr (n):= (Closed n-Forms)/(Exact n- Forms)

    Why do we care only about closed forms ? I imagine we can use DeRham's theorem that gives us a specific isomorphism with Singular Homology to see why, but I cannot see an answer off-hand?
     
  2. jcsd
  3. Sep 16, 2014 #2

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    We do not care only about closed forms. But closed forms modulo exact forms give you the chronology with real coefficients. The key is Stokes Theorem.
     
  4. Sep 16, 2014 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Well yes, but that seems to beg the question ( and I think I am using that expression correctly here). Why is it that the quotient gives you information ; why is it that non-closed forms give you no information?
     
  5. Sep 17, 2014 #4
    To give an example, if you have a divergence-free vector field, then to find the line integral around some loop, you can choose any surface you want and integrate the curl vector field over that. Gives you independence of the surface. So, it's a higher-dimensional version of that.

    As to why it gives topological information, that's the universal coefficient theorem, which basically says homology and cohomology are two sides of the same coin. The example I gave is illustrating the fact that the Kronecker pairing is well defined on homology (cocycle evaluates to the same thing in a well-defined way on homology, not just on the chain-level), and that's what's involved in the universal coefficient theorem.

    So, when you mod out by exact forms, you are making the Kronecker pairing well-defined, so that you get something that's sort of dual to homology, which is the more geometrically meaningful thing, for which we understand why it should be a topological invariant. You could say the same for singular cohomology, as well as De Rham. At least, that's the way I see it.
     
  6. Sep 17, 2014 #5

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I guess I don't understand your question. De Rham's theorem proves that de Rham cohomology is isomorphic to singular cohomology with real coefficients. Are you asking how the proof works?

    A non closed form will not take on the same value ,in general, on homologous smooth cycles. Therefore such a form is not in the dual space to the singular homology. A closed form is. Two closed forms that differ by an exterior derivative will take on the same value on homologous cycles. Taking the quotient identifies them which is right because their values are the same on smooth cycles.

    But I feel that I am still begging the question.

    A good example of a form that is not closed but is information packed is a connection one form on a circle bundle. Such a form is closed only if the connection is flat.
     
    Last edited: Sep 17, 2014
  7. Sep 17, 2014 #6

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Thanks, Lavinia, Homeo; I guess it is up to me now to read the DeRham proof and your answers more carefully before asking a new question.
     
  8. Sep 17, 2014 #7
    It might help to revisit where exactly the definition for de Rham cohomology comes from. Basically the idea is given a smooth manifold M we can look at the space Kq of differential q-forms and notice that the exterior derivative gives a map d:Kq→Kq+1 that satisfies d2 = 0. This means we have a cochain complex associated to each manifold and so with a little work it makes sense to call the homology of this cochain complex the (de Rham) cohomology of M; that is, the de Rham cohomology is closed forms modulo exact forms. My hunch is that this is all pretty unsatisfying, but on a formal level at least it explains why closed forms come in.

    Now to understand the topological information this provides it is probably best to pay attention to integration and the de Rham isomorphism, as others in the thread have suggested. Try working some examples, especially in the small dimensional cases where we can actually draw the manifold, to get some intuition on exactly what topological features de Rham cohomology captures. If you want a textbook that covers this kind of stuff in detail, then Bott and Tu is a great choice.

    Lastly since you seem to be asking why pay attention to only closed q-forms (modulo exact forms) instead of all q-forms, the answer partially comes down to computability. The (co)chain complexes arising in our definitions for the (co)homology of a space often contain lots of topological information. Problem is these things are enormous and pretty much impossible to compute with. Passing to (co)homology provides a major advantage since things get smaller and more manageable and we also pick up some formal rules for computing (like excision, long-exact sequence of a pair, etc) which are tremendously helpful.
     
    Last edited: Sep 17, 2014
  9. Sep 17, 2014 #8

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I strongly suggest the beautiful proof in Singer and Thorpe.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why only Closed Forms Matter in DeRham Cohomology?
Loading...