maybe this question does not make sense and it's just a psychological problem of mine. However I cannot understand the geometric picture of the Dold-Kan correspondence.

Let $\mathbf{Ab}$ denotes the category of abelian groups and $\mathbf{Ch}_{+} (\mathbf{Ab})$ the category of connective (non-negative) chain complexes of abelian groups. If $A \in \mathbf{Ab}^{\Delta^{op}}$, let $DA_n$ denote the degenerated simplices in degree $n$, then $DA$ is a subchain complex of $A$ with differential $d= \sum_i (-1)^i d_i$. The Dold-Kan correspondence asserts that there's an equivalence of categories $$N: \mathbf{Ab}^{\Delta^{op}} \longrightarrow \mathbf{Ch}_{+} (\mathbf{Ab})$$ $$\Gamma: \mathbf{Ch}_{+} (\mathbf{Ab}) \longrightarrow \mathbf{Ab}^{\Delta^{op}}$$, where $$N (A)_n= \bigcap_{i =0}^{n-1} Ker (d_i^n) \cong A_n/DA_n$$, where the differential is given by $(-1)^{n}d_n^n$ and $$\Gamma(C)_n = \bigoplus_{[n] \twoheadrightarrow [k]} C_k$$.

I know the proof of the Dold-Kan correspondence and I have used it many times. However I cannot get the geometrical picture of it. I thought that computing $N$ and $\Gamma$ for the case of $\mathbb{Z} [\text{Sing} (X)]$ for some nice topological space $X$ would clarify my intuition, but this was not the case (or I failed in my computations and I've got some mistakes). Anyway, it seems that $\Gamma$ and $N$ are very non-intuitive at least for me.

Now, let me clarify what I mean by a geometric picture. In the case of simplicial sets, one can draw a picture of it by gluing simplices, however in the case of simplicial abelian groups there are no sets of simplices, so the best I can think of is to think just about $\mathbf{Z} [\text{Sing} (X)]$. In this case, the Moore complex $(A, d = \sum_i (-1)^i d_i)$ makes sense. It means that one is gluing simplices according to their orientation (the $+$ and $-$ sign). However when I go to the normalized case $N(A)$, I don't know how to think about it. Now, about $\Gamma$, I don't know even how to start to think about it.

Hope my question is clear, please let me know if I made any mistake.

Thanks in advance.

  • $\begingroup$ There's this answer on MO by Ronnie Brown that suggests looking at generalizations of Dold-Kan to try to understand it. The whole thread is interesting too. $\endgroup$ – Najib Idrissi Oct 16 '15 at 11:34
  • $\begingroup$ @NajibIdrissi Thanks for the comment. However I can understand the $d^2 = 0$ part (my problem is the normalization part), maybe in the article that he cites there's something interesting, so I will look when I have more time. $\endgroup$ – user40276 Oct 16 '15 at 11:39
  • $\begingroup$ I wouldn't be so quick saying I understand "$d^2 = 0$" :) I mean the proof sure, but the intuition behind that equality... $\endgroup$ – Najib Idrissi Oct 16 '15 at 11:43
  • $\begingroup$ @NajibIdrissi As I understand this is because one defines the sum of the faces according to the orientation (by the way, this is written in my question), so the sign change means the act of gluing, therefore for a closed shape $d = 0$. My point is different, I can understand the Moore complex, however I cannot understand its normalized version. $\endgroup$ – user40276 Oct 16 '15 at 11:48

There is a distinction made in this 2015 presentation between equivalent but "broad and narrow" algebraic models. Broad models are good for conjectures and proofs, and narrow models are good for calculation and relation to classical methods. The fact, or proof, of equivalence is then a great technical tool, since it allows one to hop at will between the two models without worrying too much about the mechanics of the hop. The more difficult the proof of equivalence, the greater the value of each hop!

In the classical Dold-Kan theorem, chain complexes are "narrow" and simplicial abelian groups are "broad". Even more tricky are the nonabelian cases, which are crossed complexes and cubical $\omega$-groupoids, see the book Nonabelian Algebraic Topology, and even harder, the Ellis-Steiner "crossed $n$-cubes of groups" and "cat$^n%$-groups".

Dold's first proof of the theorem was rather combinatorial, rather like your formulation. Then Kan shed a huge light on it by defining for a chain complex $C$ the simplicial abelian group $K(C)$ as $$K(C)_n= Chn(C_*(\Delta^n),C).$$

Think of the analogy with the singular simplicial set of a topological space:

$$S^\Delta(X)_n= Top(|\Delta^n|,X).$$

For a crossed complex $C$ we gets its simplicial and cubical nerves as

$$N^\Delta(C)_n= Crs(\Pi(\Delta^n_*),C), \quad N^\square(C)_n= Crs(\Pi(I^n_*),C), $$ where the latter uses cubical sets with connections. See the above book.

I doubt I can do the Ellis-Steiner result

Ellis, G.J. and Steiner, R. "Higher-dimensional crossed modules and the homotopy groups of $(n+1)$-ads". J. Pure Appl. Algebra 46 (1987) 117--136.

in the same manner!

  • $\begingroup$ Thanks for your answer, it's important too know your opinion since you're one of the leading topologists in this field. However for me, even simplicial abelian groups are not that broad (this would mean intuitive in your definition). For instance, what's a simplicial abelian group that is not of the form $\mathbb{Z}[\text{Sing} (X)]$? One could argue that even the case of simplicial sets is not that intuitive (when considering any non Kan-fibrant object). $\endgroup$ – user40276 Oct 16 '15 at 15:58
  • $\begingroup$ More important than anything that I said before is: how would you draw a picture of a topological space representing a normalized chain complex (that is $\Gamma (N (A))$)? Or maybe other question, why Dold and Kan had these ideas of using these functors in first place? Maybe the general picture of nerve and realization (as in ncatlab.org/nlab/show/nerve+and+realization) would be a justificative. Maybe my question does not even make sense and I'm just being stubborn. $\endgroup$ – user40276 Oct 16 '15 at 16:02
  • $\begingroup$ Keep on asking! The theory arose from descriptions of $K(A,n)$. Then you try to describe the function simplicial set $K(A,n)^Y$. See papers [3,4] on my publication list. There, FD-complexes are now called simplicial abelian groups. At least the latter have a geometric realisation, a space, not so clear for a chain complex. I wanted to replace $A$ by a chain complex, but that turned out to be Dold-Kan again! $\endgroup$ – Ronnie Brown Oct 16 '15 at 20:09
  • $\begingroup$ @RonnieBrown Do you know of any source (other than Kan’s original article, which I do not find very accessible) that establishes that the definition $K(C)_n= \operatorname{Chn}(C_*(\Delta^n),C)$ is equivalent to the definition stated in the question? I do not find that very obvious. $\endgroup$ – Gaussler Jan 5 '17 at 15:24

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.