# Coefficients of homology

I am wondering why people use different coefficients when defining homology of simplicial complex, like homology over $R$, $Z$, $Z/2$, etc? Is one better then the other and why?

Moreover, which one(s) is(are) most useful in practical computation other than abstract theory?

Thanks a lot.

• – Watson Jun 5 '16 at 16:00
• In some sense, if you don't go to deeply in the theory, knowing with $\mathbb{Z}$ coefficients is everything. This is the content of the universal coefficient theorem. It often occurs that one can compute homology with other coefficients more easily (e.g. with Z/2 coeffients poincare duality works directly for non orientable manifolds). Sometimes the way (co) homology is defined leads one directly to different coefficients. A great example is the de Rham cohomology. – Thomas Rot Jun 5 '16 at 20:54

## 1 Answer

There is no best coefficient in general, but there is indeed a best coefficient for specific case.

Let me show you some examples.

Suppose you want to compute the Euler Characteristic of a space $X$. Suppose that for some reason, counting cell is not a viable option, let's say you don't have an explicit cell structure of your space. You are left with trying to compute the rank of the homology groups. Now you can verify that $\text{rank}H_*(X;\mathbb{Z})=\dim H_*(X,\mathbb{Q})$. The latter group is indeed a vector space, so it has no torsion and it is easier to compute (every short exact sequence splits for example). Even more is true, computing Euler Characteristic is invariant from the chosen field coefficients. In conclusion, it is natural to trying to compute $H_*(X,\mathbb{Q})$ in this case.

Suppose you have a non-orientable closed manifold $M$: When you want to study closed manifolds (from an homology viewpoint), Poincaré Duality is a powerful tool. A powerful tool with some hypothesis to verify, namely orientability. If you don't have orientability you are left with two option, trying to compute $H_*(M;\mathbb{Z}_{\rho})$ (homology with local coefficient - something pretty technical which was let's say created for dealing with this kind of situation) or more easily $H_*(M;\mathbb{Z}_2)$, why? well because every manifold is $\mathbb{Z}_2$ orientable, and therefore you have P-D with these coefficients.

Clearly, homology with $\mathbb{Z}$ coefficients contains all the homological information, since we have UCT which means that if you know homology with $\mathbb{Z}$ for a space, then you know the homology for that space with coefficient any abelian group.

What's the problem: You don't always know how to compute homology with $\mathbb{Z}$-coefficient. By its very property that it bring a lot of informations, computations in general are cumbersome: you have torsion to deal with for instance which is a pain in the ass. So we came up with some weaker version of it, easier to compute but still interesting. This is how research is done usually, you find some great invariant and then realise that it's impossible to compute (very common) and then he must find something weaker but computable and still interesting (very hard).

A last example to back up what I've said. When you will deal with characteristic classes, in particular Stiefel-Whitney ones, you will realise what I mean by sometimes computing cohomology with integer coefficients is hard. But if you turn to $\mathbb{Z}_2$ coefficients, the cohomology ring turns out to be a free graded ring and the generators are the universal classes.

• I can't really see any way of making the assertion "$\operatorname{rank} H_*(M;\mathbb{Z}) = \dim H_*(M;k)$" true if the characteristic of $k$ is positive... – Najib Idrissi Jun 6 '16 at 8:41
• maybe I rushed a little, but if I recall correctly, $\text{rank}H_*(M;\mathbb{Z})=\dim H_*(M;\mathbb{Z})\otimes \mathbb{Q}$ and then one verifies at once that euler char doesn't depend on the field you are computing it. But you are right, I need to modify it a little, since what I wrote it's formally wrong – Riccardo Jun 6 '16 at 8:44
• Yes, by the UCT $\operatorname{rank} H_k(M;\mathbb{Z}) = \dim H_k(M;\mathbb{Q})$, and yes, the Euler characteristic does not depends on the field. But for example $H_2(\mathbb{RP}^2;\mathbb{Z}) = 0$ has rank zero whereas $H_2(\mathbb{RP}^2;\mathbb{F}_2) = \mathbb{F}_2$ has dimension one. ("Formally wrong" is just another way of saying "wrong"...) – Najib Idrissi Jun 6 '16 at 8:46
• yes, and thank you for pointing out this:) – Riccardo Jun 6 '16 at 8:52