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Visualising singular cohomology using cap product.

  1. Jun 1, 2012 #1
    Hello all,

    I'm trying to get my head around the cap product and singular cohomology. I've always found singular cochains rather hard to visualise (e.g. what is the fundamental cochain of a manifold? i.e. that chain which generates the Z in the top cohomology group), and I've found looking at the cap product similtaneously helpful but also confusing.

    http://en.wikipedia.org/wiki/Cap_product


    To prove Poincare Duality on a closed n-manifold, one can show that taking the cap product of the fundamental homology class of a manifold gives you an isomorphism from the i'th singular cohomology group to the (n-i)'th homology group.

    I've tried to put this into action to find a description of the fundamental cochain class of the easiest example, the circle.

    To start, I can take as my fundamental homology class a loop with winding number 1 which starts and ends at the same point. The cap product of a 1-chain and a 1-cochain is the 0-chain in the image at the start of the line segment given by the 1-chain, with coefficient given by the cochain evaluating on that 1-chain. So I'd better make sure that my fundamental cochain eats such chains and spits out the value 1.

    But I also need the right answer for all other possible fundamental classes, built from more than one segment, all joined up with winding number 1.

    Can anyone see how to continue? For example, I know that taking a 2 part segment going around the circle, my chain will need to return on one of the segments n and the other n-1 (there is already choice, and more importantly, some break of symmetry!). This means that if I have a fundamental class of 3 segments containing the chain which my cochain eats and returns n, it will need to return n-1 on the sum of the other two segments.


    Of course, I'd need to actually show that my cochain not only returns the right 0-chain, but has coboundary zero. I'd be happy for now though that it just satisfied the property I want (which might the coboundary condition anyway?) I'm starting to think that no simple description is actually available, we just know that this class exists, which is quite fascinating in itself given that the cochain simply describes the fundamental class of a circle!
     
  2. jcsd
  3. Jun 1, 2012 #2
    cf. the above with the simple description of the generator of the zeroeth cohomology group for a path connected space - the 0-cochain assigning the value 1 to each 0-chain. It is easy to see that capping with this class will just be the identity, as desired.
     
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