I'm not really sure what I want to ask here, which isn't a great start for a question, but nonetheless...

I am wondering if there are some nice results that we can get from considering (co)homology with coefficients in an arbitrary abelian group. For example Wikipedia states that

...it is common to take $A$ to be $\mathbb{Z}/2\mathbb{Z}$, so that coefficients are modulo 2.

Is this just for ease of computation, or is there some useful results that can be gathered (for any abelian group, not just modulo 2)?

Edit: I was thinking about this question because the cohomlogy of the projective spaces in the mod 2 case just become $\mathbb{Z}/2\mathbb{Z}$ - compare to the cohomlogy with coefficients in $\mathbb{Z}$,where the odd coefficients are zero

  • $\begingroup$ Well, $3$-torsion in homology is not seen in homology modulo $2$... $\endgroup$ – Mariano Suárez-Álvarez May 5 '11 at 9:18
  • $\begingroup$ @Mariano - well...yes! I guess I am after useful applications $\endgroup$ – Juan S May 5 '11 at 9:23
  • $\begingroup$ Algebraic topology is often done "one prime at a time". That is, you check out what happens mod p for each p, you check what happens rationally, and then you try to stitch the pieces back together to a coherent picture. This began with Serre's C-theory, which was the framework for his method of computing homotopy groups of spheres. Certainly the UCT played a role here! $\endgroup$ – Aaron Mazel-Gee May 5 '11 at 9:30
  • 4
    $\begingroup$ Coefficients in arbitrary abelian groups comes up naturally in obstruction theory, which is a fairly pragmatic, computable theory for solving "every day" algebraic topology problems. The first developments in obstruction theory were things like Stiefel-Whitney classes, which are UCT images of things called "primary obstruction classes". $\endgroup$ – Ryan Budney May 5 '11 at 9:53
  • $\begingroup$ possible duplicate of Why homology with coefficients? $\endgroup$ – Grigory M Dec 27 '13 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.