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Why Are Homology Groups Not MUCH Larger?

  1. May 16, 2014 #1
    So far, I think algebraic topology is turning out to be the best thing since sliced bread. However, I'm having a bit of difficulty with homology, for one particular reason.

    Consider, as an example, the first homology group of ##S^1##. The definition of the free abelian group (or, in general, the ##R##-module) of ##n##-chains is the free abelian group generated by ALL ##n##-simplices. Why do we not include ALL paths in ##S^1## in this definition? Are these not also 1-simplices?

    Thank you!
  2. jcsd
  3. May 16, 2014 #2


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    You are not talking about "paths", you are talking about equivalence classes of paths. Many different paths lie in one equivalence class.
  4. May 16, 2014 #3
    AHA! Thank you! That makes soooooo much more sense. :redface::rofl:
  5. May 16, 2014 #4


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    To add a little bit, remember that homology is a quotient be it of modules or vector spaces --cycles/boundaries -- which shows the elements to be classes, where two cycles are equivalent (geometrically *) if their difference bounds . And in the case of ##S^1## , the classes are the paths that go around n times. You can see that if you go around a non-integer "number of times" k , then the path can be deformed/homotoped to the path class |_ k _|, where |_ k _| is the greatest integer less than k .

    * You can also do this purely algebraically.
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