Change of Coefficients in Homology and Individual Classes

1. Sep 17, 2011

WWGD

Hi, If we change coefficients in Homology, we can use the universal coefficient theorem

to find the change in the homology group. Still, is there a way of knowing what happens

to _individual classes_ under this change of coefficients? I am just thinking of cases

where there is a major collapse under the coefficient change, like when we go from

Z--integers to Z/2 in, say, S^1 , the unit circle, so that H_1 goes from Z to Z/2;

I guess all classes are collapsed into 2 classes.

Is there a way of knowing what happens to individual classes under this change

?

2. Sep 17, 2011

lavinia

not sure what you mean by what happens to the classes,

3. Sep 17, 2011

Jamma

I suppose that it can only be possible to see "what collapses where" if you have a map between your groups (i.e. f:G->G'). So, for example, in your case, we can map Z to Z/2 via the obvious quotient map. This will act on the chains, so, for example, the chain which consists of going around the circle once with coefficient 2 will be mapped to 0 in Z/2. I'm guessing you can follow all of this through to the homology level and it gives a map between f*^n(X,G) -> H^n(X,G') (an isomorphism if f is onto?).

You will need to check all of this (I'm just going by intuition, not what by I know to be true). Clearly though if you don't have such a map between the groups the question is meaningless. For example, if you are changing from Z/2 coefficients to Z/5 coefficients, there is no association of elements in Z/2 with Z/5 and we can't meaningfully say what it is to collapse an element of Z/2 to Z/5.

4. Sep 17, 2011

WWGD

Thanks, both. No, I was curious, since I read a description of curves x,y such that
x~y (x is homologous to y) over Z/2, i.e., in H_1(X;Z/2) . I then wonder if x,y
would remain homologous if/when we changed coefficients, i.e., are x,y homologous
also in H_1(X,Z/5) , etc ? I never fully understood what happened geometrically
are not preserved when changing from G to G' unless, e.g., (necessary, but not sufficient)
unless G,G' have the same cardinality.

5. Sep 17, 2011

lavinia

If there is a coefficient homomorphism A -> B then this induces a map on homology so classes that are homologous with A coefficients map to classes that are homologous with B coefficients. A short exact sequence of coefficients 0 -> A -> B -> C -> 0 creates a long exact sequence of homology groups known as the Bockstein exact sequence.
One implication is that there may be homology classes with C coefficients that are not the image of homology classes with B coefficients. A simple example using the coefficient sequence

0 -> Z ->Z -> Z/2Z ->0 is the Bockstein for an unorientable compact manifold without boundary, then the sequence looks like

0 -> Z/2Z -> Hn-1(M:Z) -> Hn-1(M:Z) -> Hn-1(M;Z/2Z) -> .....

Last edited: Sep 17, 2011
6. Sep 17, 2011

Jamma

Let's take Z/3 and Z/5 instead. Where would an element such as 2.x map to? In fact, a better direction is the other way, say changing coordinates from Z/5 to Z/3. Where does 4.x map to?

This is what I mean by your question not meaning all that much unless you have a map between your groups. But then, the way to continue is clear. The way I described it seems correct from what Lavinia said (I didn't know about the Bockstein exact sequence, thanks for the info!).

7. Sep 17, 2011

lavinia

Chains that are cycles in one set of coefficients may not be cycles in another.

8. Sep 18, 2011

lavinia

In the cohomology of groups, the first and second cohomology groups have direct interpretations although these are algebraic rather than geometric. The first cohomology with coefficients in a G-module are the crossed homomorphisms from the group,G, into the coefficient module. The second cohomology is the group extensions of the coefficient module by the group,G. There is some interpretation of the third cohomology as well but I do not know what it is.

When the coefficient module is trivial, these are the usual cohomology groups of the K(G,1), the Eilenberg and Maclane space whose first fundamental group is G. For Z/2Z for instance, this is the infinite real projective space.

As an example, the second cohomology of Z/2Z, i.e. the infinite real projective space, with Z/2Z coefficients is the group extension 0 -> Z/2Z -> Z/4Z -> Z/2Z -> 0.

With Z/5Z and Z/7Z coefficients one gets the group extensions 0 -> Z/5Z -> Z/10Z -> Z/2Z -> 0 and

0 -> Z/7Z -> Z/14Z -> Z/2Z -> 0

Last edited: Sep 18, 2011
9. Sep 18, 2011

Jamma

Well, what I'm saying is that you can't even really say "what a chain is with different coefficients" unless you have a homomorphism of the coefficients. But when you have that it is clear how you associate a chain with coefficients A to a chain in coefficients in B (i.e. you let the homomorphism act on the coefficients). The boundary maps will commute with this change of coefficients yielding a well defined map on homology.

10. Sep 18, 2011

lavinia

Right

11. Sep 18, 2011

WWGD

Right, good points, both. Just for context, the whole thing came up when I saw
two chains x,y being described as being homologous in H_1(X,Z/2), and I wondered what that meant, and how it would be different from having x,y being homologous over some other coefficient . The whole issue of homology
with different coefficients seems strange in that it seems like the space has no intrinsic
property that would remain unchanged when coefficients change, and the changes
are often important, in that we may or may not have torsion; the homology may or may
not be finite, depending on the choice of coefficients.

12. Sep 18, 2011

lavinia

Not sure what you mean by "no intrinsic property that remains unchanged" under a change of coefficients. In practice, at least in the theory of manifolds, the coefficients that are almost always used are Z, Z/2Z, Q and R.

13. Sep 18, 2011

Jamma

Well, it often will lead to change. For example, using the reals doesn't detect torsion. If you choose Z/2 it won't find 3 torsion and so on.

Actually, I have always wondered about the following, maybe Lavinia knows a partial answer to this:

does knowing the homology with coefficients in Q, Z/2Z, Z/3Z,...Z/nZ,..... tell you the homology with coefficients in Z?

Coefficients in Z give you the most information in that you can effectively work out the rest from Z using the universal coefficient theorem. Would knowing the homology with coefficients in Q tell you what the "free part" is, and the rest tell you what the torsion is? And can you then combine these to figure out what the homology with coefficients in Z is?

14. Sep 20, 2011

zhentil

I think in principle the answer is yes, that if you have a finite-dimensional manifold and you know the cohomology over Q and over Z/pZ for every prime p, that you could deduce the cohomology of the manifold. But it's not as easy as you're making it - i.e. RP^3 is a Q-homology sphere, and is Z/2 in every dimension with Z/2 coefficients. This doesn't imply that the Z cohomology is Z, Z/2, Z/2, Z.

15. Sep 21, 2011

Jamma

Thanks Zhentil.

I'm not sure I quite follow though, you start by saying "yes" for finite-dimensional manifolds but then finish by saying that the homology of RP^3 can't be deduced- but RP^3 is a finite dimensional manifold.

Do you just mean that it's not as easy as just "adding up all of the coefficients"? I realise that something like a Q in Q coefficients will show up as a Z/2 in Z/2 coefficients, so we basically have to "remember to ignore the free terms with the torsion groups". Can this be made precise? (i.e. can we simply "add up" all of the coefficients, remembering to discard one of each coefficient if there is one in Q?) I suspect that even if that did work (and it'd be a long shot) it would only work for something like CW complexes- you wouldn't be able to detect groups here such as Z[1/2] (which I think can occur).

16. Sep 23, 2011

zhentil

Right, that's precisely it. Knowing the Q and Z/2 cohomology allows you to deduce what's happening at the chain level. For instance, RP^3 has a cell decomposition with one cell in each dimension. Looking at the map from C^1 to C^2, and denoting the respective generators by a and b, we see that d(a) can't be zero over Z, but must be over Z/2, hence one obtains that H^1(RP^3;Z)=0 and H^2(RP^3;Z) is torsion. Now that I think about it for a moment, though, I doubt it's enough to know it over Q and Z/pZ - for instance, this doesn't tell us that H^2(RP^3;Z) is Z/2.

17. Sep 24, 2011

Jamma

Hmm ok. My idea was more if we did know all of the homology groups with coefficients in Q, Z/2,Z/3,... then could we deduce those over Z? So for example, suppose that we can somehow (and I don't know if this can ever be done in practicality) find out that the homology of RP^3 and yet not know what it looks like over Z coefficients; its homology groups look like follows:

In Q coefficients, the homology is Q,0,0,Q
In Z/2 coefficients Z/2,Z/2,0,Z/2
And for Z/p coefficients, for any p, we get Z/p,0,0,Z/p
[we are listing the homology groups in order from dimension 0 up to 3)

Can we go on to say that in Z coefficients the homology must be Z,Z/2,0,Z? The reasoning being that the Z's correspond to the free part in the Q coefficients- these also showing up in the Z/p coefficients, so we ignore them there, and the extra remaining Z/2 shows up in dimension 1 because of our information from the Z/2 coefficient homology.

In full generality, if we have the Q coefficient homology as being a_0,a_1,a_2,a_3,... where a_n is the rank of the group, and in Z/p coefficients we have the groups p_0,p_1,p_2,.... then can we say that in Z coefficients we have:

in dimension n the homology is the direct sum of all of the groups p_n and with Z^a_n, except that we need to make sure to ignore terms which are "repeated" because they show up from the free part?

I know this is all rather wooly, and possibly useless (although I have heard to topologists trying to find the torsion homology of spaces because over Z it can sometimes be too difficult) but I think it would give me a deeper understanding of how everything fits together!