Definition of the boundary map for chain complexesby TacTics Tags: boundary, chain, complexes, definition 

#1
Jan2913, 06:56 PM

P: 810

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/bkri...4_03062012.pdf 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 psimplexes 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? 



#2
Jan3013, 06:41 AM

Sci Advisor
P: 1,716

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.




#3
Jan3013, 02:08 PM

P: 810

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). 



#4
Jan3113, 01:16 PM

Sci Advisor
P: 1,716

Definition of the boundary map for chain complexes 



#5
Jan3113, 01:28 PM

P: 810

Yes. I meant "abelian", but omitted it to introduce some confusion :)



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