2
$\begingroup$

$\DeclareMathOperator{im}{im}\DeclareMathOperator{Sing}{Sing}$Let be a chain complex $C_\bullet$ with differentials $\partial_n$.

I know that the homology groups of $C_\bullet$ are defined via $$H_n(C_\bullet):=\ker(\partial_n)/\im(\partial_{n+1}).$$

But how are the differentials of a chain complex labelled? I saw several possibilities:

  • Wikipedia: $${\displaystyle \ldots {\stackrel {\partial _{3}}{\longrightarrow }}C_{2}{\stackrel {\partial _{2}}{\longrightarrow }}C_{1}{\stackrel {\partial _{1}}{\longrightarrow }}C_{0}{\stackrel {\partial _{0}}{\longrightarrow }}0}$$
  • Gelfand and Manin (see I.4.3 of their book on homological algebra) use $${\displaystyle \ldots {\stackrel {\partial _{n+1}}{\longrightarrow }}C_{n}{\stackrel {\partial _{n}}{\longrightarrow }}C_{n-1}{\stackrel {\partial _{n-1}}{\longrightarrow }}\dots},$$ which I guess means $${\displaystyle \ldots {\stackrel {\partial _{3}}{\longrightarrow }}C_{2}{\stackrel {\partial _{2}}{\longrightarrow }}C_{1}{\stackrel {\partial _{1}}{\longrightarrow }}C_{0}},$$ but then there is no differential $\partial_0$ -- so what is $H_0$ then?

In particular, I am interested in singular homology. If $X$ is a topological space, then its singular homology is defined to be the homology of the chain complex $${\displaystyle \ldots {\stackrel {}{\longrightarrow }}\mathbb Z[\Sing_2(X)]{\stackrel {}{\longrightarrow }}\mathbb Z[\Sing_1(X)]{\stackrel {}{\longrightarrow }}\mathbb Z[\Sing_0(X)]}$$ be a chain complex $C_\bullet$. To calculate $$H_n(C_\bullet):=\ker(\partial_n)/\im(\partial_{n+1})$$ I have to know which arrow is which $\partial_n$.

Are the differentials here labelled as $${\displaystyle \ldots {\stackrel {\partial_1}{\longrightarrow }}\mathbb Z[\Sing_2(X)]{\stackrel {\partial_1}{\longrightarrow }}\mathbb Z[\Sing_1(X)]{\stackrel {\partial_0}{\longrightarrow }}\mathbb Z[\Sing_0(X)]}$$ so that one can just apply the formula $H_n(C_\bullet):=ker(\partial_n)/im(\partial_{n+1})$ or are they labelled as $${\displaystyle \ldots {\stackrel {\partial_3}{\longrightarrow }}\mathbb Z[\Sing_2(X)]{\stackrel {\partial_2}{\longrightarrow }}\mathbb Z[\Sing_1(X)]{\stackrel {\partial_1}{\longrightarrow }}\mathbb Z[\Sing_0(X)]}$$ and $\partial_0$ is something like Wikipedia suggests: $$\partial_0\colon \mathbb Z[\Sing_0(X)]\to 0?$$

I'm really unsure about the conventions. Is this written down somewhere in a clear way?

$\endgroup$
3
  • 5
    $\begingroup$ $\partial_0=0$. So $H_0=C_0/\text{im}\partial_1$. $\endgroup$ Dec 15, 2021 at 18:33
  • 1
    $\begingroup$ A chain complex has no end. If the author stops writing the complex, it's usually safe to assume that the remaining part of the complex is $0$. $\endgroup$
    – Qi Zhu
    Dec 15, 2021 at 19:25
  • $\begingroup$ To add to Paul Frost's great answer, I just want to add that when people say "chain complex", they almost always mean the most general possible notion, i.e. of a chain complex that doesn't terminate. When they write e.g. $\cdots\to C_2\to C_1\to C_0$, really you should be thinking of a bunch of zero maps to the right of the $C_0$. This general notion subsumes all the special cases, as explained in the answer already given, $\endgroup$
    – YiFan Tey
    Dec 22, 2021 at 13:31

1 Answer 1

2
$\begingroup$

Unfortunately the notion of a chain complex is not a standardized concept in the literature. Some authors define it as Wikipedia in your above link. Other authors (for example Allen Hatcher in his book "Algebraic Topology") define it as a sequence $$\ldots C_n \stackrel{\partial_n}{\to} C_{n-1} \to \ldots \to C_1 \stackrel{\partial_1}{\to} C_0 \stackrel{\partial_0}{\to} C_{-1} = 0$$

Let us denote such objects ad hoc as chain complexes terminating at a dimension $k$. The Wikipedia-definition produces chain complexes terminating at dimension $0$, Hatcher's definition produces chain complexes terminating at dimension $-1$ with the special feature that $C_{-1} = 0$.

However, most authors define it as in this Wikipedia-article as a sequence of $C_n$ and boundary operators $\partial_n : C_n \to C_{n-1}$ with $n \in \mathbb Z$. Let us call this ad hoc an unterminating chain complex. Clearly the latter is a more general concept: We can regard each chain complex terminating at dimension $k$ as an unterminating chain complex by setting $C_n = 0$ for $n <k$ and $\partial_n = 0$ for $n \le k$.

You are right, if we work with a chain complex terminating at dimension $0$, then there is no $\partial_0$. This shows that this concept is inadequate for the purposes of homology. But you can take Hatcher's concept. In fact, you may argue that the singular chain complex of $X$ should be understood as ending at $C_0 = \mathbb Z[Sing_0(X)]$, but can be artificially transformed into a Hatcher-like complex. But formally we can also argue that $Sing_n(X) = \emptyset$ for $n < 0$ so that automatically $C_n = \mathbb Z[Sing_n(X)] = 0$ for $n < 0$. This gives you an unterminating chain complex.

I recommend to work with unterminating chain complexes because also Hatcher-like chain complexes have a serious weakness: Usually one introduces the reduced homology groups by considering the augmented chain complex $$\ldots \mathbb Z[Sing_1(X)] \stackrel{\partial_1}{\to} \mathbb Z[Sing_0(X)] \stackrel{\epsilon}{\to} \mathbb Z \to 0$$ which terminates at dimension $-2$. This is no longer a Hatcher-like chain complex, but again it can be regarded as an unterminating chain complex.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .