32
$\begingroup$

The Dold-Kan correspondence gives an equivalence of categories between $SAb$, the category of simplicial abelian groups, and $Ch_{\geq 0}$, the category of non-negatively graded chain complexes of abelian groups.

Recall that the equivalence in the direction $C: SAb \rightarrow Ch_{\geq 0}$ is most naturally given $C(A)_{n}=A_n/(s_{0}A_{n-1}+\cdots s_{n-1}A_{n-1})$ (simplices modulo degenerate simplices) and taking the differential to be $d=\sum (-1)^{i}d_{i}$. The equivalence in the other direction $\Gamma: Ch_{\geq 0} \rightarrow SAb$ is most naturally given by $\Gamma(C)_{n}=Ch_{\geq 0}(C(\mathbf{Z}\Delta[n]),C)$ (the group of chain maps) where $\Gamma(C)$ gets a simplicial structure by pulling back the comsimplicial structure on the collection of complexes $C(\mathbf{Z}\Delta[n])$.

The reason that I say this is the most natural is because fits the standard pattern of constructing an adjunction between a presheaf category and another category. Namely, there is a functor $\Delta \rightarrow Ch_{\geq 0}$ sending $[n]$ to the complex of non-degenerate simplices in $\Delta[n]$ with differential given as the alternating sum of face maps. Then $C: SAb \rightarrow Ch_{\geq 0}$ is the left Kan extension of $\Delta \rightarrow Ch_{\geq 0}$ and $\Gamma$ is the obvious right adjoint to that (it can't be anything else). (Once upon a time I learned this kind of construction from Maclane-Moerdijk, but it can also be found on the nLab :http://ncatlab.org/nlab/show/nerve+and+realization.)

One would then like to show that $C$ and $\Gamma$ are inverse equivalences by showing that the (obvious) unit $Id \rightarrow \Gamma C$ and counit $C\Gamma \rightarrow Id$ are isomorphisms.

By constrast, in every textbook treatment that I have seen, one constructs instead of a quotient complex $C$ an isomorphic subcomplex $N$, complementary to degenerate simplices, but you can do this in at least two different ways. Moreover, one constructs an isomorphism $\oplus_{[n] \rightarrow [k]} NA_{k} \rightarrow A_{n}$, where the sum is over surjections $[n] \rightarrow [k]$ in $\Delta$, and defines $\Gamma$ similarly, as $\Gamma(C)=\oplus_{[n] \rightarrow [k]} C_{k}$. Ultimately, it seems that one then shows $\Gamma N(A)\simeq A$ for $A \in SAb$ and $N\Gamma(C) \simeq C$ for $C \in Ch_{\geq 0}$, but the isomorphism $\Gamma N(A)\simeq A$ is somehow in the wrong direction (before you know it's an isomorphism), since a priori you would only expect $N,\Gamma$ to form an adjunction and the isomorphism should be given by the unit $A \simeq \Gamma N(A)$.

So my question is whether there is a 'more natural' treatment somewhere in the literature, phrased in terms of the quotient complex $C$, rather than the subcomplex $N$, and in terms of $\Gamma$ given by the obvious right adjoint, not as a sum over surjections, which at first seems to be pulled out of a hat. Of course one can unwind the usual proof and see that everything matches up. For example, one should be able to identify $Ch_{\geq 0}(C(\mathbf{Z}\Delta[n],C)$ with $\oplus_{[n]\rightarrow [k]} C_{k}$, but one can see at least two ways to do this.

If there is no such treatment, can one explain how to construct one or why one shouldn't bother?

$\endgroup$
  • 2
    $\begingroup$ I think that the levelwise direct sum decomposition of simplicial abelian groups is a very nice computation based in the observation that a simplicial object contains lots of split monos and epis. Therefore, I don't think it's pulled out of a hat. Sometimes computations are really necessary. Not everything is 'categorical nonsense' (I hate this unfair expression, but I don't know a better way of putting my ideas into words right now). $\endgroup$ – Fernando Muro Feb 10 '15 at 22:54
  • $\begingroup$ @ChrisBrav Can you perhaps shed some light on one (or both) of your two ways of identifying $\operatorname{Ch}_{\ge0}(C(\mathbb{Z}\Delta[n]),C)$ and $\bigoplus_{[n]\twoheadrightarrow[k]} C_k$? I currently don’t quite see a natural approach to this. $\endgroup$ – Gaussler Jan 7 '17 at 14:40
23
$\begingroup$

The proof of the Dold-Kan theorem basically amounts to the following. Let $\mathbb{Z}\Delta$ denote the pre-additive category generated by the simplicial indexing category, so that $s\mathrm{Ab}=\mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})$, the category of additive functors.

Let $\mathcal{C}$ be the "Karoubi envelope" of $\mathbb{Z}\Delta$, the category obtained by freely adjoining splittings of idempotents. You can identify the Karoubi envelope with a full subcategory $s\mathrm{Ab}$, namely the closure of the image of the Yoneda embedding under splitting idempotents.

Then it is trivial that $\mathrm{Fun}^{\mathrm{add}}(\mathcal{C}^\mathrm{op},\mathrm{Ab}) \approx \mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})=s\mathrm{Ab}$. The Dold-Kan theorem amounts to observing that every object in $\mathcal{C}$ is a direct sum of objects $G(n)$, and that the full subcategory of $G(n)$s is the "indexing category" for chain complexes. You also have that $G(n)\approx \mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$.

The usual formulas for normalized chains arise by explicitly identifying $G(n)$ with either a (split) subobject or a (split) quotient object of $\mathbb{Z}\Delta[n]$. In fact, there is only one way (up to sign) to identify $G(n)$ as a split subobject of $\mathbb{Z}\Delta[n]$, and this translates into the "quotient by degeneracies" construction of normalized chains. There are several ways to identify $G(n)$ as a split quotient: the usual presentation of normalized chains as subobjects is induced by the projection $\Delta[n]\to\Lambda^0[n]$.

Of course, you could also be perverse: $G(n)$ is a summand of $\mathbb{Z}\Delta[k]$ for any $k\geq n$, so you could define normalized chains using your favorite idenification of $G(n)$ as a summand of $\mathbb{Z}\Delta[n+42]$, if you want!

Added. Let me try to answer Karol's question, by giving a proof which doesn't use much computation.

Set $F(n):=\mathbb{Z}\Delta[n]$ and $G(n):=\mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$. The key observation is to show that the projection $F(n)\to G(n)$ admits a section. I don't have a clever way to do that ... you just have to write down the formula. (Which is: send the "generator" $\langle0,1,\dots,n\rangle$ of $G(n)$ to $\sum \pm\langle 0,a_1,\dots,a_n\rangle$, where the sum is over sequences with $a_k\in \{k-1,k\}$, and the sign is the parity of $\lvert\{k\,|\,a_k\neq k\}\rvert$.) The nice thing is that the section turns out to be unique (this is hardly obvious to begin with).

Now filter $\Delta[n]$ by subcomplexes $\Delta^k[n]=$ the union of all $k$-simplices which contain the vertex $0$. It's easy to see that $$ \mathbb{Z}\Delta^k[n] /\mathbb{Z}\Delta^{k-1}[n] \approx G(k)^{\oplus \binom{n}{k}}. $$ But we just showed that the $G(k)$ are projective, so the filtration lifts to a direct sum decomposition $F(n)\approx \bigoplus_k G(k)^{\oplus\binom{n}{k}}$.

Now you want to compute $\hom(G(m), G(n))$. Here is a trick. It's clear that $\hom(F(m),F(n))=\mathbb{Z}^{\binom{n+m+1}{m+1}}$. Since the $G(k)$ are summands of the $F(n)$, we have $\hom(G(i),G(j))\approx \mathbb{Z}^{a_{i,j}}$ for some $a_{i,j}\geq0$. We want to show $a_{i,i}=a_{i,i+1}=1$ and all other $a_{i,j}=0$.

It's easy to exhibit that $a_{i,i}\geq1$ and $a_{i,i+1}\geq1$. Plugging the direct sum decomposition into $\hom(F(m),F(n))$ gives $$ \binom{n+m+1}{m+1} = \sum_{i,j} \binom{m}{i}\binom{n}{j} a_{i,j}. $$ But you can use standard combinatorial identities to show this is already an equality when we set $a_{i,i}=a_{i,i+1}=1$ and other $a_{i,j}=0$. So that must be the answer.

$\endgroup$
  • 3
    $\begingroup$ I went down that road when I was trying to figure out the proof and that is a nice way of thinking. What's not clear to me though is whether the fact that "every object in $\mathcal{C}$ is a direct sum of objects $G(n)$" can be proven with a significantly different or easier argument than the ones that appear in the standard approach. Is there some nice trick for that? $\endgroup$ – Karol Szumiło Feb 11 '15 at 15:12
  • $\begingroup$ I also tried working this out a little while ago and got stuck because $\mathbb{Z} \mathbf{\Delta}$ has lots of idempotents. But it is a nice perspective, because it is formally equivalent to the standard formulation of Dold–Kan and indeed even implies Dold–Kan for all additive categories (not just $\mathbf{Ab}$). $\endgroup$ – Zhen Lin Feb 11 '15 at 16:01
  • $\begingroup$ @KarolSzumiło I added the quickest proof I can think of. $\endgroup$ – Charles Rezk Feb 12 '15 at 3:59
9
$\begingroup$

I have asked myself almost exactly the same question but haven't found any answer in the literature. Then I have found a proof which ended up being very similar to the classical one. This might be the reason why this approach is not discussed anywhere, perhaps people who found it just didn't feel that it contributed any more insight than the standard proof. Besides that, I can think of only one (but important) advantage of the normalized chain complex: it makes it easy to prove that $\pi_* A \cong H_* N A$.

Here's an argument along the lines you suggest.

By the triangular identities it is enough to check that the unit $\eta_A \colon A \to \Gamma C A$ is an isomorphism and that $\Gamma$ reflects isomorphisms.

Let's do the latter first. For $m > 0$ let $D^m$ denote the chain complex freely generated by $\iota$ in degree $m$ and by the boundary of $\iota$ in degree $m - 1$ and zero elsewhere. Then chain maps out of $D^m$ classify $m$-chains. The map $D^m \to C\mathbb{Z}\Delta[m]$ classifying the identity has a retraction given by sending $\mathrm{id}_{[m]}$ to $\iota$, $\delta_0$ to the boundary of $\iota$ and everything else to zero. Complexes $D^m$ along with $\mathbb{Z}$ concentrated in degree $0$ detect isomorphisms and hence so do all $C\mathbb{Z}\Delta[m]$s.

Now consider the unit $\eta_A \colon A \to \mathrm{Ch}(C\mathbb{Z}\Delta[m], CA)$. It acts by regarding $a \in A_m$ as $a \colon \mathbb{Z}\Delta[m] \to A$ and applying $C$. I will construct the inverse map sending $f \colon C\mathbb{Z}\Delta[m] \to CA$ to $\tilde{f} \in A_m$ by induction on $m$. The main ingredient is the same as in the standard proof, i.e. a lemma saying that $A_m$ splits as a direct sum of the subgroups $NA_m = \bigcap_{i < m} \ker \delta_i$ and $DA_m = \Sigma_{i < m} \mathrm{im}\, \sigma_i$. Start with $f \colon C\mathbb{Z}\Delta[0] \to CA$ and set $\tilde{f} = f(\mathrm{id}_{[0]}) \in CA_0 = A_0$. Next, assume that we have handled all simplices below degree $m$ and let $f \colon C\mathbb{Z}\Delta[m] \to CA$. Then $f(\mathrm{id}_{[m]}) \in CA_m = A_m / DA_m$, pick $a \in A_m$ such that $f(\mathrm{id}_{[m]}) = a + DA_m$. The simplex $\tilde{f}$ that we want to construct has to live in the same coset, i.e. $\tilde{f} = a + x$ for some $x \in DA_m$. Moreover, we need $\tilde{f} \delta_i = \widetilde{f \delta_i}$ for all $i \in [m]$ where $\widetilde{f \delta_i}$s were constructed in the previous step of the induction. The lemma I mentioned before says that there is a unique $x$ such that $\tilde{f} \delta_i = \widetilde{f \delta_i}$ for all $i < m$. We need to check that we also have $\tilde{f} \delta_m = \widetilde{f \delta_m}$ and this follows by induction since both $\tilde{f} \delta_m$ and $\widetilde{f \delta_m}$ are determined by their faces and reductions modulo $DA_{m-1}$. Thus $\tilde{f}$ is a unique simplex with $\eta_A(\tilde{f}) = f$.

I should also mention that it is possible to phrase the standard argument in a similar way, except that it is easier to exhibit $N$ as a right rather than a left adjoint of $\Gamma'$. (I'm going to write $\Gamma'$ to distinguish it from $\Gamma$ above.) The unit $\eta'_X \colon X \to N \Gamma' X$ is given by $\eta'_X(x) = (\mathrm{id}, x)$ and the counit $\epsilon'_A \colon \Gamma' N A \to A$ is given by $\epsilon'_A(\sigma, a) = a \sigma$. Then the triangular identities are immediately verified and the standard proof can be arranged to say that $\eta'_A$ is an isomorphism and $N$ reflects isomorphisms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.