Intuition behind homology with general coefficients

Solution 1:

I am starting to think that the issue here is that I do not have a firm grasp on what homology is "counting" in general (beyond of course "cycles modulo boundaries") and why I might want to move to a different group other than comptuational ease.

(a) One way to visualize homology is to think about cellular homology. Here, we form a chain complex which is free abelian on the number of cells in each dimension, and our differential is determined by the degrees of the various attaching maps. Somehow this measures how twisted up and far away our CW-complex is from being just a wedge of spheres. In the case of $\mathbb{R}P^{n}$, for example, there's only one cell in each dimension, and the degrees of the attaching maps are either $0$ or $2$; in particular, they are always even! So if we take cellular homology with $\mathbb{Z}/2$ coefficients, the maps all become zero. In the case of $\mathbb{C}P^n$, there's too much space in between dimensions for anything to happen, so it "looks" like a wedge of spheres, according to homology. The upshot is that the algebra simplifies greatly for certain choices of coefficients because you can simplify or get rid of differentials in the cellular chain complex.

(b) Often I tend to think of $H_*(-, \mathbb{Z})$ as being singular homology, and other choices of coefficients just a convenient way to organize the algebra (this is justified by the universal coefficient theorem). Picking coefficients in a certain characteristic can either isolate phenomena in that characteristic, or make certain problems vanish. From the cellular perspective it seems to make things vanish (since you're making it closer to a wedge of spheres)... however, it seems that for other invariants you manage to isolate things. This usually has to do with torsion... It's helpful to work through some examples other than just projective space: I recommend lens spaces.

...if there is some structure in an appropriate group action on something related to some space should I expect to see corresponding structure when viewing homology groups of that space with that group as the coefficients?

(a) It's unfortunate that the two different $\mathbb{Z}/2$'s here are named the same. What's important here is that the group $C_2$ is doing the acting and its order is divisible by $2$, and so things will behave differently with coefficients in $\mathbb{Z}/2$ as opposed to $\mathbb{Z}/p$ for odd primes. This generalizes: if you have a finite group $G$ acting on a space, $X$, then things with coefficients in $\mathbb{Z}/p$ will behave differently depending on whether or not $p$ divides $\vert G\vert$. One hint at this is the following: An action of $G$ descends to an action on the singular chain complex with coefficients in whatever (say, field) $k$ you're using, and so you have a bunch of vector spaces with actions of $G$, i.e. a bunch of representations of $G$. Now, it turns out that representation theory for finite groups behaves very differently in characteristic $p$ depending on whether or not $p$ divides the order of the group.

(b) Part (a) can be made even more precise... It turns out that there is a special homology theory for $G$-spaces called (Borel) equivariant homology. I won't go into it, but it has two properties (1) It has a very close relationship with group homology and (2) sometimes the ordinary homology of $X/G$ serves as a good approximation to the equivariant homology of $X$.

And here are some miscellaneous related comments:

• There is a notion of R-orientability for any ring $R$; see Hatcher. This might help explain some things. In particular, note the sections on Poincare duality, which explains some of the regularity we see when choosing particular coefficients.
• You don't have to be too exotic in your choice of ring because of the following Exercise: Show that a map $f: X \rightarrow Y$ induces an isomorphism on all homology groups if and only if it induces an isomorphism on rational homology and an isomorphism on homology with coefficients of $\mathbb{Z}/p$ for all $p$. (In some sense this point tells you that all different choices of coefficients can do for you is pinpoint torsion at different primality.)