Definition of the boundary map for chain complexes


by Tac-Tics
Tags: boundary, chain, complexes, definition
Tac-Tics
Tac-Tics is offline
#1
Jan29-13, 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 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?
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lavinia
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#2
Jan30-13, 06:41 AM
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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.
Tac-Tics
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#3
Jan30-13, 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).

lavinia
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#4
Jan31-13, 01:16 PM
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Definition of the boundary map for chain complexes


Quote Quote by Tac-Tics View Post
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).
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.
Tac-Tics
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#5
Jan31-13, 01:28 PM
P: 810
Yes. I meant "abelian", but omitted it to introduce some confusion :)


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