Coefficients in homology

Question

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?

Answer 1

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 $\leqslant 1$).

Answer 2

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.