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Definition of the boundary map for chain complexes

  1. Jan 29, 2013 #1
    I've been poking around, learning a little about homology theory. I had a question about the boundary operator. Namely, how it's defined.

    There's two definitions I've seen floating around. The first is at:

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

    The second, at

    http://www.math.wsu.edu/faculty/bkrishna/FilesMath574/S12/LecNotes/Lec16_Math574_03062012.pdf [Broken]

    The only difference seems to be the inclusion of a factor of (-1)i inside the sums.

    My guess is that the extra factor doesn't matter, since there is some choice in how you construct chain. In other words, the fact that you're working with a FREE abelian group over the p-simplexes of your complex, flipping the signs results in an isomorphic group.

    (If that's not the case, my other guess would be that the latter only works in Z/2Z, where sign doesn't matter anyway).

    Is my reasoning sound? Or am I missing something?
     
    Last edited by a moderator: May 6, 2017
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  3. Jan 30, 2013 #2

    lavinia

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    With Z2 coefficients signs don't matter since minus 1 and one are the same. The Wikipedia definition of boundary is correct in general. You can check this with examples.
     
  4. Jan 30, 2013 #3
    Ah. Thank you.

    Now that I think about it, you can' "choose" what group you want the coefficients to be in if your generating your groups freely anyway.

    (I'm guessing that would be some quotient of the free group, determined by the type of coefficient you're interested in, but I'll worry about that later).
     
  5. Jan 31, 2013 #4

    lavinia

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    In homology I think you start with the free abelian group on simplices, define the boundary operator, then choose other coefficients than the integers by tensoring (over Z) each group with the coefficient group.You never start with a free group, always a free abelian group. It is a characterisitc of homology that the groups are always abelian, unlike the fundamental group which usually is not abelian.
     
  6. Jan 31, 2013 #5
    Yes. I meant "abelian", but omitted it to introduce some confusion :)
     
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