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If I understand correctly, chain complexes make sense in any category enriched in the world of pointed sets. In practice, there's also a notion of an augmented chain complex, where we have an extra morphism $\varepsilon : C_0 \rightarrow \mathbb{Z}$ between $C_0$ and $0$. In practice, these seem to arise in algebraic topology by taking certain generators of $C_0$ to $1_\mathbb{Z}$ and extending by linearity to get $\varepsilon$. But I don't really understand what an "augmented chain complex" actually is; I only understand how they arise. So my question is:

Question. Does "augmented chain complex" actually mean anything? If so, what? If not, what should be going through your head whenever you encounter this phrase to best understand the writer's meaning/intention?

For example, one of our algebraic topology homework questions says:

Let $\{C,\varepsilon\}$ be an augmented chain complex. Show that $H_{-1}(C) = 0$ and $$\tilde{H}_0(C) \oplus \mathbb{Z} \cong H_0(C).$$

I don't understand how to interpret this question. Am I supposed to assume that there's a simplicial complex floating around somewhere in the background? If not, what is the likely intended meaning of the question?

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As others have said, an augmented chain complex is an $(\mathbb{N} \cup \{-1\})$-graded chain complex $C_*$ with $C_{-1} = \mathbb{Z}$ (or more generally your base ring $\Bbbk$) and such that $H_{-1}(C) = 0$, equivalently the map $C_0 \to C_{-1} = \mathbb{Z}$ is surjective. The splitting result follows from the projective property of $\mathbb{Z}$ among abelian groups.

The "$C_{-1} = \mathbb{Z}$" and the "$C_0 \to C_1$ is surjective" conditions are rather is rather specific to algebraic topology, and it's probably possible that in some other contexts the conditions would be dropped. So if in general by chain complex you mean $\mathbb{N}$-graded chain complex, then an augmented chain complex would be an $(\mathbb{N} \cup \{-1\})$-graded chain complex. In some other contexts chain complexes are $\mathbb{Z}$-graded anyway, so this wouldn't even be anything special. It's all a matter of context.


To shed some light on the terminology, consider the usual notion of an augmented algebra, i.e. an algebra $A$ (say over $\mathbb{Z}$) equipped with a ring morphism $\varepsilon : A \to \mathbb{Z}$. Such a ring morphism is necessarily surjective, since $\varepsilon(1) = 1$ and there aren't many subgroups of $\mathbb{Z}$ containing $1$.

There is the usual bar construction $B_\bullet(A,A,\mathbb{Z})$, a simplicial abelian group whose realization is a cofibrant replacement of $\mathbb{Z}$ as a left $A$-module. It's defined by $$B_n(A,A,\mathbb{Z}) = A \otimes A^{\otimes n} \otimes \mathbb{Z} = A^{\otimes (n+1)}$$ and the simplicial maps are explicit. (See two-sided bar construction, the monad being $A \otimes -$. The construction $B_\bullet(R,A,L)$ is more generally well-defined for a right $A$-module $R$ and a left $A$-module $L$).

Then it turns out to actually be an augmented simplicial abelian group, i.e. a functor $\Delta_+^{\mathrm{op}} \to \mathsf{Ab}$ where $\Delta_+ = "\Delta \cup \{ \varnothing \}"$. Such an augmented simplicial object $X_{\bullet \ge -1}$ is of course the same thing as a simplicial object $X_{\bullet \ge 0}$ equipped with a morphism to the constant simplicial object $X_{\bullet \ge 0} \to \operatorname{const} X_{-1}$. In our case $\varepsilon$ induces $B_\bullet(A,A,\mathbb{Z}) \to \operatorname{const} \mathbb{Z}$, and under the realization functor this is the cofibrant replacement map.

Now there is an enhanced version of the Dold–Kan correspondence that says that an augmented simplicial abelian group is in fact the same thing as an $(\mathbb{N} \cup \{-1\})$-graded chain complex. If you apply it to $B_\bullet(A,A,\mathbb{Z})$ you obtain a chain complex that looks like $\dots \to A \to \mathbb{Z} \to 0$, with $\mathbb{Z}$ in degree $-1$. So there you have it! An augmented chain complex.


PS: The condition "$C_0 \to C_{-1}$ is surjective" is a bit superfluous. Consider that it's reasonable to say that $H_{-1}(\varnothing) = \mathbb{Z}$! This is also why the $n$Lab says a space is "$(-1)$-connected" if it is nonempty. It also explains why the empty set is not (generally considered to be) contractible, as it's too simple to be simple, and so on.

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One considers $\mathbf Z$ as a complex, the complex all of whose chain groups are $0$ except the one in degree $0$ where we have the group $\mathbf Z$. We write this complex as $\mathbf Z[0]$. An augmentation of a complex $C$ is then an epimorphism of complexes $C\rightarrow \mathbf Z[0]$.

A morphism of two augmented chain complexes $\delta\colon C\rightarrow \mathbf Z[0]$ and $\epsilon\colon D\rightarrow\mathbf Z[0]$ is a morphism of complexes $f\colon C\rightarrow D$ such that $\epsilon\circ f=\delta$. You get a category like this. I think people call this a slice category, but I'm not sure.

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    $\begingroup$ I learnt about augmented chain complexes from Munkres' book Elements of algebraic topology. There, morphisms of augmented chain complexes are assumed to induce the identity on $\mathbf Z$, if I recall correctly, but, indeed, there may very well be places in the literature where one considers more general morphisms of augmented chain complexes. $\endgroup$ – Johannes Huisman Jul 13 '16 at 10:02
  • $\begingroup$ After looking into it a bit more, I think your definition is the most typical one. $\endgroup$ – Eric Auld Jul 14 '16 at 4:14
  • $\begingroup$ That is indeed called a slice category, or the category of slices, or the category of $\mathbf{Z}[0]$-graded objects. $\endgroup$ – goblin Mar 2 '18 at 22:54
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From some googling, apparently an "augmented chain complex" is defined to be a chain complex together with a surjective homomorphism $C_0\to\mathbb{Z}$ such that the composition $C_1\to C_0\to\mathbb{Z}$ is $0$. This meaning seems compatible with your homework problem. I must confess that I had never seen this definition before now; if someone asked me what an "augmented chain complex" is I would have just said a chain complex with an extra term at the end.

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