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Cohomology with compact support

  1. Jan 23, 2012 #1
    Hi there.

    I'm looking at Poincare duality, and there's something extremely wrong with the way I'm looking at one or more of the concepts and I need to figure out which.

    When dealing with non-compact manifolds, you can fix Poincare duality by looking at something called "cohomology with compact support". It mentions it on page 7 here:
    http://www.math.upenn.edu/~ghrist/EAT/EATchapter6.pdf

    It mentions there that you can restrict your cochains (singular, simplicial, cellular) to those which evaluate only on some compact subspace.

    My confusion is that I don't understand how the boundary of a singular chain will necessarily be a singular chain with compact support (I can easily see it for the other two).

    Take, for example, the 0-cochain which evaluates to 1 on a specific point and 0 elsewhere. The boundary of this chain (I think) will be the cochain which evaluates to 1 on all lines which start at the point and to -1 on all lines which end at the point.

    So in that light I don't see how we get back something which is of compact support.
     
    Last edited: Jan 23, 2012
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  3. Jan 23, 2012 #2
    Oh, and another way to look at Poincare duality is to fix your homology theory instead - you can look at formally infinite chains for which every compact subset only intersects with finitely many chains. I can easily see here how the boundary map restricts.

    I imagine I've gone embarrassingly wrong with my interpretation of the coboundary map.
     
  4. Jan 23, 2012 #3

    morphism

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    It seems to me that you're attempting to compute the simplicial coboundary map. I think in order for compactly supported simplicial (co)homology to make sense, you need to restrict to simplices that are locally finite. This will ensure that the coboundary map is well-defined. In your computation for instance, the local finiteness ensures that there are finitely many lines coming in/out of the point of interest, so the coboundary of your "0-cochain" is indeed compactly supported.

    For actual singular cohomology, the fact that the coboundary d*a of a compactly supported cochain a is again compactly supported follows from the fact that d*a=ad so supp(d*a) \subset supp(a) as a closed subset.
     
  5. Jan 23, 2012 #4

    quasar987

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    So in singular cohomology with compact support, you consider the subcomplex of C^k(X) of those singular k-cochains f having the property

    (P) there exists a compact subset K of X such that f(c)= 0 for all k-chains c sitting outside of K.

    To see that this is indeed a subcomplex, it suffices to observe that if f is such a compactly supported k-cochain, then d*f is the cohain that evatuates on a (k+1)-chain c thus: (d*f)(c) = f (dc). If c sits outside of K, then so does its boundary dc, hence f(dc) = 0. That is, d*f has property P also for the same compact set K as f.
     
  6. Jan 23, 2012 #5
    I've found the solution to my problem. It's a very nuanced subtlety in the definition, it's mentioned here on page 5 in part of Hatcher's book:

    http://www.math.cornell.edu/~hatcher/AT/ATcapprod.pdf

    I think he anticipated the use of "compact support" that most people might naturally assume and addressed it there (indeed with exactly the same counter example that I stated).

    The solution is that instead of forcing your cochains to evaluate to non-zero values on chains inside some compact subset, what you actually want is the subset of all cochains which evaluate to zero on all chains contained outside some compact subset.

    The locally finite thing is important too though, thanks for your help.
     
  7. Jan 23, 2012 #6
    Ahha, just beat me to it quasar, thanks though!
     
  8. Jan 23, 2012 #7

    lavinia

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    I strongly recommend that you look at Bott and Tu's development of Poincare Duality using differential forms. Thom classes of normal bundles fall out of this approach easily and actually use a somewhat different idea, cohomology with compact support along the fibers.

    I find the Algebraic Topology version of this more techincal and less intuitive,
     
  9. Jan 23, 2012 #8

    morphism

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    If you want to understand X, and if you can find X in Bott and Tu, then "read Bott and Tu's presentation of X" is usually a good piece of advice. :)
     
  10. Jan 23, 2012 #9
    I'm trying to adapt something rather particular and need Borel-Moore homology, where you can have chains of non-compact support. This should be Poincare dual to (singular, simplicial, cellular) cohomology. I'll look up Bott and Tu's version anyway though, it always helps to see different perspectives on things.

    I've seen the Thom isomorphism used to prove Poincare duality before actually. Indeed, looking at the wiki page for Poincare duality now, it mentions that this is a nice way since it proves Poincare duality for any generalised theory for any such theory which has a Thom isomorphism.
     
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