## July 12, 2010

### The Dold–Kan Theorem: Two Questions

#### Posted by Tom Leinster The Dold–Kan Theorem states that the category of simplicial abelian groups is equivalent, in a particular way, to the category of chain complexes. I was never particularly captivated by it until André Joyal explained to me that it can be viewed as a categorification of the fundamental theorem of Newton’s finite difference calculus. He spoke about this again at Category Theory 2010 in Genova last month, but there are no notes available online and that’s not what this post is about—although his talk is what reignited my interest. So that’ll just have to be a teaser.

The purpose of this post is to ask two questions. First I’ll explain why I’m asking them. Chain complexes (of abelian groups) can be regarded as strict $\infty$-categories in $\mathbf{Ab}$, the category of abelian groups. So the Dold–Kan Theorem states that in $\mathbf{Ab}$, strict $\infty$-categories are the same as simplicial objects. I’m sure other people have contemplated this: hence the following questions.

Ross Street defined a functor $J: \Delta \to \infty\mathbf{-Cat}$ assigning to $[n]$ the $n$th oriental. (Here $\infty\mathbf{-Cat}$ is the category of strict $\infty$-categories.) Since $\Delta$ is small and $\infty\mathbf{-Cat}$ cocomplete, there is an induced adjunction $- \otimes J: \mathbf{Set}^{\Delta^{op}} \stackrel{\longrightarrow}{\leftarrow} \infty\mathbf{-Cat}: Hom(J, -).$ The right adjoint $Hom(J, -): \infty\mathbf{-Cat} \to \mathbf{Set}^{\Delta^{op}}$ assigns to an $\infty$-category what is sometimes called its simplicial nerve. The left adjoint $- \otimes J$ is a kind of ‘realization’ functor. Many similar ‘realization’ functors preserve finite products.

Q1  Does this left adjoint preserve finite products?

If the answer is no, this tentative train of thought is over.

If the answer is yes, there is an induced adjunction on abelian group objects. Since abelian groups in $\infty\mathbf{-Cat}$ are chain complexes, this means that we have an adjunction $\mathbf{Ab}^{\Delta^{op}} \stackrel{\longrightarrow}{\leftarrow} \mathbf{Ch}(\mathbf{Ab})$ between simplicial abelian groups and chain complexes.

Q2  Is this adjunction the equivalence of Dold and Kan?

I don’t feel that this is a very deep or original train of thought, so I’m optimistic that someone has been through it already and knows the answers—even if they’re ‘no’.

Posted at July 12, 2010 9:28 PM UTC

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### Re: The Dold–Kan Theorem: Two Questions

I am not quite sure concerning your specific questions Q1 and Q2 , but I can offer maybe a good reason to be captivated by the Dold-Kan correspondence in the sense of: a good way to understand its relation to strict $\infty$-categories:

Strict $\infty$-groupoids are equivalently crossed complexes and this equivalence is transparent: over a basepoint $x$ the crossed complex is in degree $k \geq 2$ the group of $k$-morphisms in the $\infty$-groupoid whose source is the identity on the identity on the identity, etc. on $x$.

Chain complexes of abelian groups (in non-negative degree) are just special cases of crossed complexes. And the sequence of inclusions

$Ch_\bullet^+ \hookrightarrow Crs \simeq Str \infty Grpd \hookrightarrow \infty Grpd \simeq KanCplx$

of strict and strictly abelian into strict into general $\infty$-groupoids is the Dold-Kan correspondence.

(This is in Ronnie Brown’s collected works, of course, though I realize we are lacking a good $n$Lab summary at the moment. The best reference I have energy for digging out at this time of night is somewhat curious: take footnote 116 on page 316 of his book Nonabelian algebraic topology and work yourself backwards from there…

Posted by: Urs Schreiber on July 12, 2010 11:16 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Urs wrote:
the sequence of inclusions

……

of strict and strictly abelian into strict into general infty-groupoids is the Dold-Kan correspondence.

Is’ ?? Do you mean that this sequence of inclusions gives PRECISELY what is written in Dold-Kan?

Posted by: jim stasheff on July 13, 2010 12:35 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Thanks, Urs. But I don’t understand in what sense the overall inclusion $Ch_{\bullet}^{+} \hookrightarrow KanCplx$ can be the Dold-Kan correspondence. (This may or may not be the same as Jim’s query.)

My understanding of the Dold-Kan correspondence is that it’s an equivalence between chain complexes and simplicial abelian groups. You’ve written down an inclusion of chain complexes into Kan complexes.

The underlying simplicial set of any simplicial abelian group is a Kan complex, so we have a forgetful functor $U: \mathbf{Ab}^{\Delta^{op}} \to KanCplx$. At first I thought you might mean that $U$ is full, so that $\mathbf{Ab}^{\Delta^{op}}$ is a full subcategory of $KanCplx$, and that the inclusion $Ch \hookrightarrow KanCplx$ determines an equivalence of categories $Ch \simeq \mathbf{Ab}^{\Delta^{op}}$. But now I realize you can’t mean that, because $U$ certainly isn’t full.

Posted by: Tom Leinster on July 13, 2010 3:51 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I don’t understand in what sense the overall inclusion $Ch_\bullet^+ \hookrightarrow KanCplx$ can be the Dold-Kan correspondence.

As that remark by Ronnie says, the image of that inclusion of a chain complex is the Kan complex underlying the simplicial abelian group corresponding to the chain complex under Dold-Kan. The Dold-Kan correspondence identifies the image of that functor in $KanCplx$.

Do you mean that this sequence of inclusions gives PRECISELY what is written in Dold-Kan?

Yes, the composite functor is precisely the Dold-Kan map $Ch_\bullet \to sAb$ composed with the forgetful $sAb \to KanCplx$.

I mentioned a reference with all the details: Remark 9.10.6 of Brown’s Nonabelian Algebraic Topology.

Posted by: Urs Schreiber on July 13, 2010 11:51 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Posted by: Urs Schreiber on July 14, 2010 6:09 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I think the answer to Q1 is No, and the short reason why is the Gray tensor product. The realization of the 1-simplex $\Delta^1$ is the walking arrow $\mathbf{2}$, and $\mathbf{2}\times \mathbf{2}$ is the walking commutative square, with no nonidentity 2-morphisms, whereas $\Delta^1 \times \Delta^1$ has nondegenerate 2-simplices and so its realization will have nonidentity 2-morphisms.

Posted by: Mike Shulman on July 13, 2010 5:53 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Ah! Good. Thanks, Mike.

Posted by: Tom Leinster on July 13, 2010 6:16 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Now that that particular idea is consigned to the scrapheap, let me say something more about the general idea. Maybe someone can see a different way of making it work.

For any category $\mathcal{E}$ with finite limits, we may consider the category $\infty\mathbf{Cat}(\mathcal{E})$ of (strict) $\infty$-categories in $\mathcal{E}$ and the category $\mathcal{E}^{\Delta^{op}}$ of simplicial objects in $\mathcal{E}$.

When $\mathcal{E} = \mathbf{Ab}$, there is an equivalence between the two categories, by the Dold-Kan Theorem.

When $\mathcal{E} = \mathbf{Set}$, there is an adjunction between the two categories, arising from Street’s oriental construction.

I wondered whether the equivalence that happens for $\mathbf{Ab}$ was an analogue of the adjunction that happens for $\mathbf{Set}$. I still wonder—because although the particular idea I suggested was wrong, it might be true in some other sense.

But even if Street’s adjunction turns out to have nothing to do with the Dold-Kan equivalence, there’s still a more general question. I want to know whether it’s possible to understand the Dold-Kan Theorem by finding some relationship (e.g. an adjunction) between $\infty\mathbf{Cat}(\mathcal{E})$ and $\mathcal{E}^{\Delta^{op}}$ that works for some large class of categories $\mathcal{E}$, then specializing to the case $\mathcal{E} = \mathbf{Ab}$, then showing that in that case the relationship is an equivalence. (The approach that I suggested would have worked whenever $\mathcal{E}$ is the category of models for a finite product theory.)

Thoughts, anyone?

Posted by: Tom Leinster on July 13, 2010 3:39 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

But even if Street’s adjunction turns out to have nothing to do with the Dold-Kan equivalence,

But it does: what Brown writes “$\Pi \Delta^n$” is the (crossed module version of the) strict $\infty$-groupoid version of the $n$-th oriental. Homming the $n$th oriental into a strict $\infty$-groupoid is the same as homming this “$\Pi \Delta^n$” into the $\infty$-groupoid.

His Remark 9.10.6 in Nonabelian algebraic topology says that the Dold-Kan map $Ch_\bullet \to sAb$ is under the identification of chain complexes with certain strict $\infty$-groupoids (via crossed complexes) just forming the $\omega$-nerve of these strict $\infty$-groupoids.

Posted by: Urs Schreiber on July 13, 2010 11:58 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I was never particularly captivated by it until André Joyal explained to me that it can be viewed as a categorification of the fundamental theorem of Newton’s finite difference calculus. He spoke about this again at Category Theory 2010 in Genova last month, but there are no notes available online and that’s not what this post is about—although his talk is what reignited my interest. So that’ll just have to be a teaser.

Posted by: Eric Forgy on July 13, 2010 6:53 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Well, seeing as you ask so nicely, here’s a taste of what Joyal said…

First, here’s something about the finite difference calculus. Begin with a sequence $a_0, a_1, a_2, \ldots$ of numbers. Define $(\delta a)_n = a_{n + 1} - a_n$. Then we get a new sequence $a_0, (\delta a)_0, (\delta^2 a)_0, \ldots$ of numbers. (In fact we get a triangle, which I won’t attempt to draw, involving all the terms $(\delta^k a)_n$.) In the finite difference calculus there are formulas relating these two sequences. In one direction, the formula is $a_n = \sum_{k = 0}^n \binom{n}{k} (\delta^k a)_0.$

Next, here’s something about the Dold-Kan theorem. This is about relating simplicial abelian groups $A_0 --- A_1 --- A_2 --- \cdots$ (where $---$ indicates various back and forth arrows that I’m too lazy to typeset) to chain complexes $B_0 \leftarrow B_1 \leftarrow B_2 \leftarrow \cdots.$ One half of the equivalence is the functor $K: \mathbf{Ch} \to \mathbf{Ab}^{\Delta^{op}}$, which is (partially) described as follows: for a chain complex $B$, $K(B)_n = \bigoplus B_k$ where the sum is over all $k$ and surjections $[n] \to [k]$ in $\Delta$. How many such surjections are there? $\binom{n}{k}$.

So, in this categorification, the sequence $(a_n)$ becomes a simplicial abelian group. The sequence of differences becomes a chain complex. Newton’s formula for the original sequence in terms of the sequence of differences becomes the formula for the functor $K$. There’s more to say (e.g. about the other direction), but I won’t say it.

Posted by: Tom Leinster on July 13, 2010 7:40 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Thank you Tom. I get overly excited whenever I see something important (especially coming from you!), e.g. Dold-Kan Theorem, relating to stuff I actually understand, e.g. finite differences. I was hoping that exploring this would be a good way to learn about the Dold-Kan Theorem.

Sorry about the “terminated” comment. I meant it in a lighthearted way, but it could have come across in an unintended manner.

Posted by: Eric Forgy on July 13, 2010 8:41 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Posted by: Tom Leinster on July 13, 2010 3:13 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Does this finite difference/Dold-Kan analogy belong to a larger picture which includes the calculus/Goodwillie analogy, outline by Joyal noted here?

Posted by: David Corfield on July 13, 2010 10:45 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

That seems likely, doesn’t it? I don’t know anything more about that than you (in fact, I’m sure I know less). Thanks for pointing it out. Clearly André said only a fraction of what he might have said in that talk.

Posted by: Tom Leinster on July 13, 2010 3:17 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

The Goodwillie calculus looks like it combines a host of Café interests. Take a look at the final paragraphs of Eric Finster’s research statement:

Categorication and Combinatorics. Many aspects of the Goodwillie homotopy calculus can be stated in the language of operads. An operad is a symmetic sequence (a functor $F : \mathbb{B} \to \mathcal{S}_*$, where $\mathbb{B}$ denotes the category of finite sets and bijections) of spaces with additional structure, namely a monoid structure for the composition product. If we require the functor to take values in the subcategory of finite sets, then a symmetric sequence is what Joyal called a “species of structure” in . This notion is much used by combinatorists as sort of “categoried formal power series” for enumerating various combinatorial structures like trees, graphs, and permutations. A robust theory of these objects exists, including notions of differentiation, composition, and even combinatorial differential equations, see, for example, . In fact,  also introduces a notion of analytic functor in this context and proves that such functors can be described in terms of the preservation of certain limits in $Set$.

A long standing goal of mine has been to view Goodwillie’s homotopy calculus as a generalization of this discrete incarnation of ideas from calculus in combinatorics. In other words, to replace the topos of sets with the homotopy topos of spaces. Lurie’s definition of $\infty$-topos may be just the tool to make this idea more precise. In general, this line of thinking would aim to uncover a closer connection between the Goodwillie calculus and the program of “categorication” laid out in, for example, .

 is none other than John and Jim’s ‘Categorification’.

The nLab entry on the Goodwillie Calculus could do with some serious work if any experts are tuning in.

Posted by: David Corfield on July 14, 2010 4:23 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Greetings!

Well as a long time reader and admirer of the cafe, I am both surprised and extremely pleased to have something I wrote posted here. I don’t think these comments are probably the place for a full-fledged discussion of Goodwillie Calculus matters, so I’ve written a post on my own blog here which describs some of my thoughts on how it might be related to some of the ideas which crop up here from time to time. Have a look if you’re interested, and comments are definitely welcome.

Posted by: Eric Finster on August 2, 2010 3:29 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Posted by: David Corfield on July 13, 2010 11:25 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I see Urs was taking us two edges round the cube up here.

Posted by: David Corfield on July 13, 2010 3:59 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I see Urs was taking us two edges round the cube up here.

Yes, for the cube on page 10 the composite

$\array{ Ch_\bullet^+ &\hookrightarrow& Crs &\hookrightarrow& KanCplx \\ \downarrow^{\simeq} && \downarrow^{\simeq} && \downarrow^{\simeq} \\ StrAb Str \infty Grpd &\hookrightarrow& Str \infty Grpd &\hookrightarrow& \infty Grpd }$

are the steps

top-right-rear $\to$ top-right-front $\to$ top-left-front

in that cube.

(Maybe up to a difference between “strict stable $n$-groupoids” versus “strictly stable strict $n$-groupoids”.)

Posted by: Urs Schreiber on July 14, 2010 12:08 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Tom,

By saying that

abelian groups in ∞−Cat are chain complexes

you are implicitly using another, globular version of the Dold-Kan correspondence. It is much simpler than the simplicial one but still not completely vacuous. It works out in exactly the same way: given a reflexive globular abelian group, you can form its normalized chain complex by taking the $n$-chains to be the kernel of the source map in that dimension, and taking the differential to be the target map. The right adjoint of this functor is the “Dold-Kan” functor yielding a reflexive globular abelian group, which is automatically a strict $\infty$-groupoid in a unique way (this can be viewed as a globular version of the abelian Moore’s lemma). Simply put, any globe can be decomposed as the sum of a globe with zero source and a degenerate one. Of course you know all this since it’s in your book, but I want to put it in a way that can be compared directly with the simplicial picture.

Now, in the simplicial picture everything is very similar: just replace “source” by “0th horn” and “target” by “0th face”. What one uses here is the fact that a simplicial abelian group is not just a Kan complex but a T-complex: the thin simplices are sums of degenerate ones. So every simplex decomposes as the sum of the unique thin (i.e. degenerate) filler of its 0th horn and a simplex whose 0th horn is equal to zero (it doesn’t matter which horn one uses, since T-complexes are automatically symmetric). In fact, this gives directly a functor from simplicial to reflexive globular abelian groups (hence, to strict $\infty$-groupoids): just take the source to be the unique thin filler of the 0th horn, the target the 0th face and the identity the 0th degeneracy.

Finally, the normalized chain complex functor of a simplicial abelian group – whose right adjoint is the Dold-Kan – obviously factors through the functor I’ve just described. And this is why I doubt that DK factors through Street’s nerve. Recall that the top-dimensional cell of the $n$th oriental goes from the composite of the even faces of the $n$-simplex to the composite of the odd ones, rather than from the composite of the faces of a horn to the opposing face. However, for $n\leq 2$ the two constructions do coincide, which might lead one to jump to the wrong conclusion.

Of course, I could be missing something, in which case I’d appreciate a correction, because I also want to get to the bottom of this.

Posted by: Dmitry Roytenberg on July 14, 2010 10:49 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Sorry, of course I mean the 0th face of the unique thin filler of the 0th horn as the source.

Posted by: Dmitry Roytenberg on July 15, 2010 12:17 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

And this is why I doubt that DK factors through Street’s nerve.

Let’s see. I am thinking that the following is evident, but maybe I am making a mistake somewhere:

1. The strict $\omega$-groupoid corresponding to the fundamental crossed complex $\Pi \Delta^n$ of the $n$-simplex regarded as a filtered topological space is the free strict $n$-groupoid on the $n$-th oriental $O(n)$.

2. For a strict $\omega$-category $C$ that happens to be a strict $\omega$-groupoid we therefore have

$Hom(O(n), C) \simeq Hom(\Pi \Delta^n, C)$

3. That the Dold-Kan map $Ch_\bullet \to sAb \to KanCplx$ factors as the composite of the $\Theta$-map $\Theta : Ch_\bullet \to Crs \simeq Str \omega Grpd$ and the $\omega$-nerve $N : Str \omega Grpd \to KanCplx$ follows immediately from the above and remark 9.10.6.

What do you think?

Posted by: Urs Schreiber on July 15, 2010 8:05 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Urs,

Having slept on this, I’ve realized where my confusion lay: we’re dealing with groupoids here, not just categories. Since all cells are (even strictly) invertible, any morphism from the composite of some pasting of boundary cells to the composite of the remaining ones can be rewritten as a morphism from the composite of all but one boundary cells to the remaining one. In particular, what I was describing is in fact equivalent to Street’s orientals in the groupoid case (what you at some point called “unorientals”).

So your argument does go through, and one can even bypass crossed complexes altogether and work directly with T-complexes.

I guess all of that should be contained in the theses of Keith Dakin and Nick Ashley, which I’ve been unable to obtain, including from Ronnie Brown.

Posted by: Dmitry Roytenberg on July 15, 2010 2:57 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Since all cells are (even strictly) invertible, any morphism from the composite of some pasting of boundary cells to the composite of the remaining ones can be rewritten as a morphism from the composite of all but one boundary cells to the remaining one.

Yes, exactly. To a large extent that’s also the prescription by which one sees that a crossed complex is a condensed but equivalent encoding of a strict $\infty$-groupoid: it remembers only the $(k \geq 2)$-morphisms that start at an identity on an object. But that’s sufficient to reproduce all of them by whiskering.

In particular, what I was describing is in fact equivalent to Street’s orientals in the groupoid case (what you at some point called “unorientals”).

Oh, your remember that. Yes, that was on page 42 of some unfinished notes.

I guess all of that should be contained in the theses of Keith Dakin and Nick Ashley, which I’ve been unable to obtain, including from Ronnie Brown.

I haven’t heard of these. Would be interested in seeing them. Did you check with Tim Porter?

Posted by: Urs Schreiber on July 15, 2010 3:40 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I’ve updated the Acrobat Reader and am now having trouble viewing your unfinished notes…

I havent heard of these. Would be interested in seeing them. Did you check with Tim Porter?

Here are the references:

MR0766238
Dakin, M. K.(4-NWAL)
Kan complexes and multiple groupoid structures. Mathematical sketches, 32, Paper No. 2, xi+92 pp.,
Esquisses Math., 32, Univ. Amiens, Amiens, 1983.

and

N. Ashley, Simplicial T-Complexes: a non abelian version of a theorem of Dold-Kan, Disser- tations Math., 165, (1989), 11 – 58.

and the n-Lab entry

http://ncatlab.org/nlab/show/simplicial+T-complex

I hope Tim Porter is reading this.

Posted by: Dmitry Roytenberg on July 15, 2010 8:00 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

He is now!

I have been out of contact for a short time in places with poor or non-existent Wifi connections.

I do not know of electronic versions of Nick Ashley’s thesis nor of Keith Dakin’s. I do not think that you would find answers your questions there, but it is still early morning here and I am not yet sure of the exact questions as I have only skimmed the discussion as yet.

It may be worth looking at Pilar Carrasco’s thesis and her published paper with Antonio Cegarra. In gneral the Granada group have explored this area as much as I have.

Another very interesting paper on Dold Kan is by Dominic Bourn (for doing similar things in a more semi-abelian contexts.)

Posted by: Tim Porter on July 21, 2010 6:24 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Here is the reference for that last comment:

Moore normalisation and Dold-Kan theorem for semi-Abelian categories, D. Bourn in Categories in algebra, geometry and mathematical physics, Amer. Math. Soc., 2007, Contemp. Math.Volume 431, 105 - 124

Posted by: Tim Porter on July 21, 2010 6:28 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Thank you for the references! What I thought might be contained in one or the other thesis was a detailed argument showing the equivalence of simplicial T-complexes, crossed complexes and strict $\infty$-groupoids.

When I first saw the definition of a T-complex was when I understood conceptually why Dold-Kan holds.

Posted by: Dmitry Roytenberg on July 22, 2010 10:32 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

The other day I should have also mentioned the work of Gary Nan Tie, who was a student of Jack Duskin.

{\sc Nan Tie, G.}, A Dold-Kan theorem for crossed complexes’, {\em J. Pure
Appl. Algebra}, 56, 177-194, 1989.

{\sc Nan Tie, G.}, Iterated $W$ and $T$-groupoids’, {\em J. Pure Appl.
Algebra}, 56, 195-209, 1989

These contain some interesting results and do not seem to be well known.

The algebraic aspects of generalised Dold-Kan are neat. Look at simplicial groups and within them the subcategory of group T-complexes. These correspond to crossed complexes and are characterised by a very simple condition (for which see Pilar Carassco’s thesis, Nick Ashley’s paper, my Menagerie notes.) The group T-complexes form a variety within the category of all simplicial groups, and one can write down a set of determining conditions for it equationally.

The graded verbal subgroup corresponding to this contains all the Whitehead products using the Kan formula for Whitehead products in simplicial groups, in terms of shuffles.

Posted by: Tim Porter on July 23, 2010 7:34 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Dmitry wrote:

By saying that

abelian groups in $\infty\mathbf{-Cat}$ are chain complexes

you are implicitly using another, globular version of the Dold-Kan correspondence.

Ah, interesting. I like that point of view.

You also bring out another important point that’s recently helped me to understand this stuff a lot better, namely:

In $\mathbf{Ab}$, every reflexive globular object is a strict $\infty$-category in a unique way.

This is a souped up, infinite-dimensional version of the fact that in $\mathbf{Ab}$, every object is a monoid in a unique way.

(You can take both statements further, as you know: in $\mathbf{Ab}$, every reflexive globular object is a strictly symmetric strict monoidal strict $\infty$-groupoid in a unique way; and every object is an abelian group in a unique way.)

The equivalence between chain complexes in $\mathbf{Ab}$ and strict $\infty$-categories in $\mathbf{Ab}$ splits into two steps:

1. the equivalence between chain complexes and reflexive globular abelian groups
2. the fact that every reflexive globular abelian group is a strict $\infty$-category in $\mathbf{Ab}$ in a unique way.

The first of these is also, I think, a categorification of an ancient algebraic principle. Given a chain complex $B$, the corresponding reflexive globular abelian group $A$ is defined by $A_n = B_0 \oplus \cdots \oplus B_n.$ You describe the opposite process, which involves taking kernels. The fact that these processes are mutually inverse is analogous to the principle of telescoping sums.

Posted by: Tom Leinster on July 17, 2010 1:41 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Yes, come to think of it, the Dold-Kan theorem (and its nonabelian version involving crossed complexes) should hold for a very general class of shape categories, as long as they have enough degeneracies. This last condition is crucial, though I can’t make it precise at the moment (e.g. dearth of degeneracies is the reason DK doesn’t hold for the dendroidal category: the minimal structure is richer than just a chain complex).

I wonder if anyone has looked at this. And also at whether the “Dold-Kan” property of a given shape category has anything to do with it being a “test category” in the sense of Grothendieck.

Posted by: Dmitry Roytenberg on July 20, 2010 9:08 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

If you ever write this up, I hope very much that the phrase dearth of degeneracies will acquire a formal definition :-)

Posted by: Tom Leinster on July 21, 2010 3:49 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I’ll try to remember that if I ever get that far :)

Posted by: Dmitry Roytenberg on July 23, 2010 10:01 AM | Permalink | Reply to this

### Shapes?

I missed the birth of shapes’ in this sense, but they bring to mind the following: the adjointness of singular and realization holds for simplicial and cubical and many others. Is there a result on the shapes of maximal generality for which the adjointness holds?

Posted by: jim stasheff on July 21, 2010 12:52 PM | Permalink | Reply to this

### Re: Shapes?

It’s very general. Suppose you have a (small) category of “shapes” $S$ and a functor $\sigma: S \to Top$ which interprets those shapes as topological spaces. (For example, $S$ could be the simplex category $\Delta$, and $\sigma: \Delta \to Top$ the functor which maps each object $[n]$ of $\Delta$ to the corresponding affine simplex.)

Then there is a realization functor

$Real_{\sigma}: Set^{S^{op}} \to Top$

that is left adjoint to a singularization functor

$Sing_{\sigma}: Top \to Set^{S^{op}}$

The formula for $Sing_{\sigma}$ is just as you’d expect: if $X$ is a space, then $Sing_{\sigma}(X): S^{op} \to Set$ takes a shape $s$ to $\hom_{Top}(\sigma(s), X)$. The formula for $Real_{\sigma}$ is also pretty much as expected: given $F: S^{op} \to Set$, its realization is given by a coend

$\int^{s \in Ob(S)} F(s) \cdot \sigma(s)$

by which we mean the quotient space of

$\coprod_{s \in Ob(S)} F(s)_{discrete} \times \sigma(s)$

where $(F(f)(x), y) \sim (x, \sigma(f)(y))$ whenever $f: s \to s'$ is a morphism of $S$, $x$ is an element of $F(s')$, and $y$ is a point of $\sigma(s)$.

This adjoint-pair construction works for any small category $S$ and any functor $\sigma: S \to Top$. However, other desirable properties familiar from certain categories of shapes might not hold. For example, one very nice feature of geometric realization of simplicial sets is the fact it preserves finite products (and also equalizers), at least if $Top$ is construed as a convenient category of spaces. This result cannot be expected to hold for other general categories of shapes.

Posted by: Todd Trimble on July 21, 2010 2:33 PM | Permalink | Reply to this

### Re: Shapes?

Todd wrote:

one very nice feature of geometric realization of simplicial sets is the fact it preserves finite products

Indeed. It’s so nice, and so important, that one might almost want to reserve the phrase realization functor for those that do preserve finite products.

This is what led me to the over-optimistic hope, expressed in the original post, that the analogous realization functor into strict $\infty$-categories (rather than topological spaces) might also preserve finite products—until Mike set me straight.

Posted by: Tom Leinster on July 21, 2010 10:06 PM | Permalink | Reply to this

### Re: Shapes?

It’s very general.

And it’s on the $n$Lab: nerve and realization.

Posted by: Urs Schreiber on July 22, 2010 1:56 PM | Permalink | Reply to this

### Re: Shapes?

And it’s on the nLab: nerve and realization.

Notice, incidentally, the rather striking accomplishment of the article by Dan Kan referenced there. In that single article he

1. introduces the notion of Kan extension;

2. introduces the general notion of nerve and realization;

3. introduces the Dold-Kan correspondence as a special case of nerve and realization.

My understanding is that he got there independently of the slightly earlier published work by Dold and Puppe, who gave the Dold-Kan correspondence in a pedestrian way in components.

So it’s a rather remarkable article.

Posted by: Urs Schreiber on July 22, 2010 3:04 PM | Permalink | Reply to this

### the general notion of nerve and realization

Urs: the general notion of nerve and realization

Me:how general?

Posted by: jim stasheff on July 23, 2010 12:35 PM | Permalink | Reply to this

### Re: the general notion of nerve and realization

As general as this: given any cocomplete category $C$, and given any functor $F: S \to C$, the tremendous construction of Kan gives an adjoint pair of functors

$(R: Set^{S^{op}} \to C) \dashv (N: C \to Set^{S^{op}})$

(It’s a little more general than this; for example, one can do all this in enriched category theory instead of ordinary category theory.) This construction was described for the case $C = Top$ in my comment here.

The classical examples include:

• $F: S \to C$ is the usual functor $\Delta \to Cat$, which interprets an object of $\Delta$, i.e., a finite ordinal, as a poset and therefore a category. The corresponding $N$ is the classical nerve $N: Cat \to Set^{\Delta^{op}}$, and the $R$ assigns to a simplicial set its fundamental category.
• $F: S \to C$ is the topological simplex functor $\sigma: \Delta \to Top$. The $N$ is the singularization functor $Top \to Set^{\Delta^{op}}$ and the $R: Set^{\Delta^{op}} \to Top$ is geometric realization.

and more can be found at the nLab article Urs referred to.

The statement at the beginning of this comment makes it clear that the Kan construction is quite general: just need a functor $S \to C$, where $S$ is small and $C$ is small-cocomplete, to perform the construction.

Posted by: Todd Trimble on July 23, 2010 2:15 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

DK doesn’t hold for the dendroidal category

That’s in which sense? I mean, in which way would Dold-Kan correspondence for dendroidal abelian groups not count?

Posted by: Urs Schreiber on July 20, 2010 9:30 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Oh, I guess because it’s not plain chain complexes on one side, but “dendroidal chain complexes”? Okay, but that is reasonable: the dendroidal correspondence does not identifiy strict and strictly abelian $\infty$-categories but strict and strictly abelian “$\infty$-multicategories”, I suppose.

Posted by: Urs Schreiber on July 20, 2010 9:35 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Yes, that’s exactly what I meant (and I also meant to include a link to the paper on dendroidal abelian groups, sorry about that).

It is indeed reasonable, but do you know if anyone has proved the statement about strict $\infty$-multicategories (or defined them, for that matter)? It would shed light on the meaning of “dendroidal chain complexes” – I find those mysterious.

Anyway, the reason I brought up the dendroidal category was to contrast it with categories of “shapes” in the usual sense of the word: shapes reasonable spaces can be built up out of. If enough degeneracies are included, I expect DK to hold for such categories, in the sense of $\mathbf{Ab}$-valued presheaves on them being equivalent to chain complexes. Failure of this might be a clue to the impossibility of topologically realizing dendroidal sets.

Posted by: Dmitry Roytenberg on July 20, 2010 11:25 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

It would shed light on the meaning of “dendroidal chain complexes” – I find those mysterious.

I am expecting that the dendroidal Dold-Kan correspondence should be the monoidal Dold-Kan correspondence where instead of strictly monoidal strictly abelian strict $\infty$-groupoids – i.e. simplicial rings – we look at the corresponding $(\infty,1)$-operads (with $n$-ary operation the $n$-fold monoidal product).

(This guess of mine builds on an insight into the nature of fully Kan dendroidal sets that arose in dicussion with Thomas Nikolaus a while ago. I can tell you more in private (are you around this week?). But Thomas is apparently close to having a formal proof written out.)

Then a “dendroidal chain complex” should be another way to encode a dg-algebra, i.e. a monoid structure on the underlying chain complex, where the dendroidal chain complex on $n$-ary trees encodes the product of $n$ elements in the dg-algebra. I guess.

Posted by: Urs Schreiber on July 21, 2010 10:03 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

By saying that

abelian groups in ∞−Cat are chain complexes

you are implicitly using another, globular version of the Dold-Kan correspondence. […] The right adjoint of this functor is the “Dold-Kan” functor yielding a reflexive globular abelian group […]

I think that’s the $\Theta$-map from Nonabelian Algebraic Topology , in its restriction to chain complexes of abelian groups:

$\Theta : ChainCplx(Ab) \to CrsCplx \simeq Str \infty Grpd \,,$

the one that factors, I think, the simplicial Dold-Kan correspondence together with the $\omega$-nerve.

For notice that for $A_\bullet$ a chain complex of abelian groups (regarded as a chain complex of groupoid modules over the trivial groupoid) we have that $\Theta(A_\bullet)$ is the crossed complex whose underlying groupoid in degree 0 and 1 is

$A_0 \oplus A_1 \stackrel{\overset{p_1 + \partial\circ p_2}{\to}}{\underset{p_1}{\to}} A_0$

and whose bundle of groups $(\Theta(A_\bullet))_k$ in degree $k \geq 2$ is $A_k \oplus A_0$.

Under the equivalence $Crs \stackrel{\simeq}{\to} Str \infty Grpd$ this is mapped to the strict $\infty$-groupoid with

• $A_0$ as its group of objects

• kernel of the source map in degree $k$ being $k$-morphisms starting on the identity on $1 \in A_0$, which is the group $A_k$.

It is a pity that these statements are not made more explicitly in the book. But all the ingredients are discussed.

Posted by: Urs Schreiber on July 17, 2010 3:53 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Above I had suggested (for instance here) that Ronnie Brown’s book Nonabelian Algebraic Topology contains somewhat implicitly the proof that the simplicial Dold-Kan map

$Hom(N \Delta^\bullet, -) : ChainCplx \to KanCplx$

factors as the cubical/globular Dold-Kan correspondence identifying strict abelian cubical/globular $\infty$-groupoids and given by the map called $\Theta$ in that book, followed by the simplicial nerve.

Upon request, Ronnie Brown has now kindly made the first part of this statement explicit in a new section of the book.

You can find it in section 14.8 on page 482 of the latest version. My above claim about the $\Theta$-map is on page 483.

The statement that $\Theta$ postcomposed with the simplicial nerve is the simplicial Dold-Kan map is still in a footnote (now number 117 on p. 307) but I am lobbying for moving this out of the footnote and making it a highlighted theorem or corollary. Maybe I am lucky…

Posted by: Urs Schreiber on July 26, 2010 6:49 PM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

I have one further point to add to this discussion. We have seen that the Dold-Kan correspondence arises via the construction to which we also associate the name of Kan, applied in the context of $\mathbf{Ab}$-enriched categories, to a certain cosimplicial object $O \colon \Delta \to \mathbf{Ch}(\mathbf{Ab})$. Explicitly, $O(n)$ is the chain complex which in dimension $k$ is free on the set $\mathcal{P}_{k+1}(n)$ of $k+1$ element subsets of $\{0, \dots, n\}$, and whose differential is given by the sum of the even faces minus the odd ones. As has been remarked, this $O(n)$ can be seen as an abelian analogue of Street’s $n$th oriental.

However, more than this is true. There is a forgetful functor $\mathbf{Ab} \to \mathbf{Set}$ which preserves limits and filtered colimits, which induces a functor $\omega$-$\mathbf{Cat}(\mathbf{Ab}) \to \omega$-$\mathbf{Cat}(\mathbf{Set})$ also preserving limits and filtered colimits. Since both domain and codomain of this functor are locally presentable, we obtain a left adjoint $L$, which, under the correspondence of abelian chain complexes with abelian $\omega$-categories, we might as well view as a functor $L \colon \omega$-$\mathbf{Cat} \rightarrow \mathbf{Ch}(\mathbf{Ab})$. What I claim is that the cosimplicial object $O \colon \Delta \to \mathbf{Ch}(\mathbf{Ab})$ is the composite of this functor $L$ with Street’s oriental functor $\mathcal{O} \colon \Delta \to \omega$-$\mathbf{Cat}$.

We show this by making use of the fact the oriental $\mathcal{O}(n)$ is free on a computad. First we observe that the value of $L$ at any $\omega$-category free on a reflexive globular set $X$ is determined: it is the chain complex corresponding to the abelian reflexive globular set $\mathbb{Z} \otimes X$. In particular, the value of $L$ at an $n$-dimensional globe is the $n$-dimensional chain complex:

$\mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow \cdots \rightarrow \mathbb{Z} \oplus \mathbb{Z}$

with differential $x \mapsto (x, -x)$ in each dimension; whilst the value of $L$ on the boundary of the $n$-globe is the chain complex obtained from the above by removing the uppermost copy of $\mathbb{Z}$. Now, any $\omega$-category free on a computad may be constructed as a transfinite composite of pushouts of inclusions of boundaries into globes; and since $L$ must preserve colimits, this is enough to tell us the value of $L$ at any computad.

Explicitly, given a computad $C$, the associated abelian chain complex has for $(L C)_n$ the free abelian group on the set of generating $n$-cells of $C$; and differential given by $\partial(x) = \sum_j t_j - \sum_i s_i$, where $\{s_i\}$, respectively $\{t_j\}$, are listings of the $(n-1)$-cells appearing in the source, respectively target, of the generating $n$-cell $x$. Now direct inspection of Street’s presentation of the $n$th oriental as free on a computad shows that $L(\mathcal{O}(n)) = O(n)$ as claimed.

Posted by: Richard Garner on July 27, 2010 3:05 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

What I claim is that the cosimplicial object $O : \Delta Ch(Ab)$ is the composite of this functor $L$ with Street’s oriental functor $\mathcal{O} : \Delta \to \omega-Cat$.

Thanks, so that’s another way to Brown’s, mentioned above, to see that simplicial Dold-Kan is globular/cubical Dold-Kan followed by simplicial nerve:

rephrasing what you just wrote, we have that the simplicial Dold-Kan map assigns to a chain complex $C_\bullet$ the simplicial set

$[n] \mapsto Hom_{Ch_\bullet}(L O(n), C_\bullet) = Hom_{\omega Cat}(O(n), R C_\bullet) = N (R C_\bullet)_n \,,$

where $R C_\bullet$ is the strict $\omega$-category obtained from $C_\bullet$ by the globular Dold-Kan correspondence.

Posted by: Urs Schreiber on July 27, 2010 7:24 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Yes, though as you’ve written it, one loses the fact that Dold—Kan lands in simplicial abelian groups, rather than mere simplicial sets. However, this may be reconstructed: for $C_\bullet$ is an internal abelian group in $\mathbf{Ch}(\mathbf{Ab})$; as such, it is sent by the right adjoints $R$ and $N$ to an internal abelian group in $[\Delta^{op}, \mathbf{Sets}]$; hence to a simplicial abelian group.

Posted by: Richard Garner on July 27, 2010 9:19 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Yes, though as you’ve written it, one loses the fact that Dold—Kan lands in simplicial abelian groups, rather than mere simplicial sets.

Yes, that’s intentional: I entered the discussion way above with the statement that the Dold-Kan theorem has a nice higher-category theory interpretation as identifying among all $\infty$-groupoids those that are strict and have (or had, before forgetting it) a strictly abelian structure: for it lands not in arbitrary simplicial sets but in Kan complexes.

As David Corfield had kindly pointed out here, this is equivalently stated as saying that the simplicial Dold-Kan correspondence traverses two edges of what John Baez once called the cosmic cube of higher category theory:

the globular/cubical version of DK identifies strict $\infty$-groupoids with strict abelian structure among all strict $\infty$-groupoids (so that involves forgetting the abelian structure) and then the $\omega$-nerve embeds these strict $\infty$-groupoids in a second step in all $\infty$-groupoids.

To me it seems that this is the important conceptual message of Dold-Kan, in the context of higher category theory, and all along I had tried to point out that Ronnie Brown has a very detailed realization of this.

In fact, I still think that Ronnie’s construction is a tad stronger than what you presented: he does not just construct, I think, the free abelian chain complex on the $n$th oriental as you did above, but the free crossed complex . Which he calls $\Pi \Delta^n$, i.e. the free strict $\omega$-groupoidification of the $n$th oriental.

By the way, i put the argument you mentioned above into the $n$Lab: here.

Posted by: Urs Schreiber on July 27, 2010 10:20 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

Thanks, Richard. That’s surely an important story.

Posted by: Tom Leinster on July 27, 2010 10:18 AM | Permalink | Reply to this

### Re: The Dold–Kan Theorem: Two Questions

That’s surely an important story.

Isn’t it?

Ronnie has now moved it out of the footnote. In the latest version, installed a few minutes or hours ago, it appears in remark 9.10.6, page 306 itself.

But maybe the genuine advertisement campain for the statement has to be run elsewhere…

Posted by: Urs Schreiber on July 27, 2010 6:46 PM | Permalink | Reply to this

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