# Coefficients in homology

The singular homology of a space $$X$$ is defined to be the homology of the chain complex $${\displaystyle \ldots {\stackrel {}{\longrightarrow }}\mathbb Z[Sing_2(X)]{\stackrel {}{\longrightarrow }}\mathbb Z[Sing_1(X)]{\stackrel {}{\longrightarrow }}\mathbb Z[Sing_0(X)]}{\stackrel {}{\longrightarrow }}0.$$

More precisely, this is called the singular homology of $$X$$ with coefficients in $$\mathbb Z$$.

Question: What is a coefficient group and why is it called "coefficient"? I only know the term from elementary algebra, where the $$a_i$$ in a polynomial $$a_nx^n+\dots +a_2x^2+a_1x+a_0$$ are called coefficients. Is this concept related to the coefficients in the above sense?

How is homology in other coefficients defined and what changes? Is homology then a module instead of an abelian group?

• Homology groups, say, for a group $G$ and a $G$-module $M$ are groups $H_n(G,M)$, where the modules $M$ are called coefficients. For example, homology with trivial coefficients just means that $M$ is the trivial $G$-module (see Betti numbers). Commented Dec 22, 2021 at 12:56

If $$R$$ is a ring and $$R[x]$$ is a polynomial algebra, then you correctly state that $$R$$ are the coefficients. If now $$R\to S$$ is a ring map, $$S$$ becomes an $$R$$-module, and you can prove/note that $$S[x]$$ is equal to $$S\otimes_R R[x]$$: we have "changed coefficients" by taking a tensor product over $$R$$.

More generally, you can consider $$M\otimes_R R[x] = M[x]$$ for $$M$$ an $$R$$-module, and work with "polynomials" with coefficients in $$M$$ (caveat: you cannot multiply these.)

In the case of complexes, say $$(C,d)$$ is a complex of $$R$$-modules, we can do the same, either compute $$H(C,d)$$ (with "coefficients in $$R$$") or pick an $$R$$-module $$M$$ and compute $$H(C\otimes_R M,d\otimes 1)$$, with "coefficients in $$M$$".

In case $$R=\mathbb Z$$, the $$R$$-modules are just abelian groups $$G$$, so these are our possible coefficients, and when we choose $$C = \mathsf{Sing}_*(X)$$, we call $$H(C\otimes G)$$ the homology groups of $$X$$ with coefficients in the abelian group $$G$$.

Also note that most of your favourite results about "homology with coefficients" (like the Universal Coefficient Theorem) work in full generality for the first interpretation of coefficients (i.e. $$C\otimes G$$ for $$C$$ a complex of abelian groups and $$G$$ an abelian group, or more generally a ring of homological dimension $$\leqslant 1$$).

For a topological space $$X$$, you may define singular homology groups with coeficents in any ring $$R$$ (perhaps requiring that 0 and 1 are distinct elements both contained in the ring). These are usually denoted $$H_n(X,R)$$. The most common one to encounter is the singular homology groups with coefficents in $$\mathbb{Z}$$, however $$R$$ could in principle be any ring. Note that abelian groups are naturally $$\mathbb{Z}$$-modules. In general the homology groups $$H_n(X,R)$$ are $$R$$-modules.

When defining the singular homology groups, one first defines the singular chain groups. Recall that if $$\Delta_p$$ is a standard $$p$$-simplex, then a continious map $$\sigma: \Delta_p \to X$$ is called a singular $$p$$-simplex in $$X$$. We define the $$n$$th chain group (with coefficents in $$R$$) as a finite formal linear combination of $$p$$-simplices in $$X$$ with coefficents in some ring $$R$$, i.e. $$C_n = \{\sum_i a_i \sigma_i | a_i \in R \}$$ Then $$C_n$$ is a $$R$$-module. In the case where $$R = \mathbb{Z}$$ then $$C_n$$ is just the free abelian group generated by all singular $$p$$-simplicies. One then defines the boundary operator between chain groups and the homology groups as usual. One can then check that $$H_n(X, R)$$

Remark: It is very common that authors (or teachers) focus almost exclusively on the case of $$R = \mathbb{Z}$$. The reason is the universal coefficient theorem, which informally says that the homology groups $$H_n(X, \mathbb{Z})$$ completely determines the homology groups $$H_n(X, R)$$ , for any ring $$R$$. Other common cases one might encounter include $$R = \mathbb{R}$$ and $$R = \mathbb{Z}/2\mathbb{Z}$$. These are usually (much) easier to compute then the $$R=\mathbb{Z}$$ case.