Dold--Kan Question | The n-Category Café

November 27, 2009

Dold–Kan Question

Posted by John Baez

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Lately I’ve been writing a whole bunch of issues of This Week’s Finds in parallel, trying to tell a long story and make sure it’s told in detail.

Right now I just want to make sure I understand one thing. In their book Rational Homotopy Theory, Félix, Halperin and Thomas claim:

Proposition 26.4: If $G$ is a topological monoid, then the normalized singular chain complex $C_*(G,\mathbb{Q})$ is a differential graded Hopf algebra.

Now, there’s something puzzling about this right away. Following your instincts, when $G$ is a topological group you’d naturally expect $C_*(G,\mathbb{Q})$ to be a differential graded Hopf algebra. But when $G$ is just a topological monoid, you’d only expect $C_*(G,\mathbb{Q})$ to be a differential graded bialgebra — since where would the antipode come from, if not the inverse operation in $G$ ?

This puzzle is easily resolved: Félix et al define a differential graded Hopf algebra in a way that omits any mention of the antipode! So, they’re really getting a differential graded bialgebra.

But my real question concerns the Dold–Kan theorem, which is underlying the proof of this result.

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We’ve got a functor

$X \mapsto C_*(X,\mathbb{Q})$

from spaces to rational chain complexes, sending any space to its normalized singular chain complex with rational coefficients. If this functor were strongly monoidal, it would send monoids to monoids, comonoids to comonoids, and bimonoids to bimonoids. In other words: it would send topological monoids to differential graded algebras, topological comonoids to differential graded coalgebras, and topological bimonoids to differential graded bialgebras.

Then we’d be in luck, since every topological space is a topological comonoid in a unique way, with comultiplication being the diagonal

$\Delta : X \to X \times X$

And as a spinoff, every topological monoid is a topological bimonoid in a unique way.

So, if our functor were strongly monoidal, it would send topological monoids to differential graded bialgebras.

However, in the $n$ Lab, a wonderful entry on the monoidal Dold–Kan correspondence is gradually taking shape. This concerns the extent to which the ‘normalized Moore complex’ functor from simplicial abelian groups to chain complexes is monoidal. I believe that’s the meat of the question here! And this entry seems to suggest that this functor is merely lax monoidal. Perhaps I’m reading too much into it, or the wrong things… in which case I’d like to be corrected!

If our functor

$X \mapsto C_*(X,\mathbb{Q})$

were merely lax monoidal, it would send monoids to monoids, but not necessarily comonoids to comonoids. So then — even though any space $X$ is a topological comonoid — we’d have no right to expect that $C_*(X,\mathbb{Q})$ is a differential graded coalgebra… unless some miracle intervenes, which I’d then like to understand.

In Section 4b of Rational Homotopy Theory, Félix et al describe the Alexander–Whitney map

$AW : C_*(X \times Y, \mathbb{Q}) \to C_*(X,\mathbb{Q}) \otimes C_*(Y,\mathbb{Q})$

and they say this makes $C_*(X,\mathbb{Q})$ into a differential graded coalgebra.

So what’s going on? Is our functor

$X \mapsto C_*(X,\mathbb{Q})$

strongly monoidal, or just lax monoidal? And if the latter, why is $C_*(X,\mathbb{Q})$ a differential graded coalgebra?

Posted at November 27, 2009 11:46 PM UTC

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Re: Dold–Kan Question

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I should emphasize that I’m following Félix et al — but not the $n$ Lab — in using $C_*$ to denote the normalized singular chain complex, where we mod out by degenerate simplices.

As they point out, authors often neglect to mod out by degenerate simplices when defining the singular chain complex. For the purposes of defining homology this is harmless. But if we neglected to do this, the chain complex for a point, $C_*(point, \mathbb{Q})$ , would not be just $\mathbb{Q}$ concentrated in grade zero. So then our functor

$X \mapsto C_*(X, \mathbb{Q})$

would definitely fail to be strongly monoidal.

Posted by: John Baez on November 28, 2009 1:24 AM | Permalink | Reply to this

Re: Dold–Kan Question

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Félix et al define a differential graded Hopf algebra in a way that omits any mention of the antipode!

As you may know, this is (unfortunately?) very common among homotopy theorists. I believe it’s because any $\mathbb{N}$ -graded bialgebra for which $H_0 = \mathbb{Q}$ (or whatever ground ring) admits a unique antipode. These are called “connected” bialgebras since the (co)homology of any connected space has that property. Since classical homotopy theorists (unfortunately?) think mostly about connected spaces, they tend to forget the difference between bialgebras and Hopf algebras.

Posted by: Mike Shulman on November 28, 2009 1:48 AM | Permalink | PGP Sig | Reply to this

Re: Dold–Kan Question

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Thanks for the fact about connected Hopf algebras admitting a unique antipode — I hadn’t known that! Nor did I know it was common to use the term ‘Hopf algebra’ to mean ‘bialgebra’.

I don’t mind these guys focusing on connected spaces, but it seems like when you change your focus from homology to the actual chain complex, you have to worry more about the antipode.

Maybe for a connected pointed space we could use a modified singular chain complex that’s 1-dimensional in degree zero. But Félix and company don’t do that.

Anyway, maybe you can whisper my question about the strong vs. lax monoidalness of the normalized singular chains functor into your boss’ ear. I thought you’d know the answer. He’ll probably say everyone at Chicago has known the answer for decades!

Posted by: John Baez on November 28, 2009 2:27 AM | Permalink | Reply to this

Re: Dold–Kan Question

If you want to get all historical, then you should note that in Milnor and Moore’s original paper the definition of a Hopf algebra did not include an antipode, and is what would now be called a bialgebra. Right at the end of their paper they “[introduce] and mildy [study] a canonical anti-automorphism defined for connected Hopf algebras”, using the term ‘conjugation’ rather than ‘antipode’ for this anti-automorphism.

Posted by: Simon Willerton on November 28, 2009 9:04 AM | Permalink | Reply to this

Re: Dold–Kan Question

Simon,
Indeed that’s the history; I was there at the time. In fact the published version is nothing like the original. Many years later when I mentioned it at Milnor’s birthday fest, he thanked me for explaining what was in the published version.
Remember the purpose of the article - major theorems about Hopf algebras in topology.

Posted by: jim stasheff on November 28, 2009 1:57 PM | Permalink | Reply to this

Re: Dold–Kan Question

Hopf algebras were used and studied by Hopf before Milnor and Moore, just no systematic terminology and general theory outside of topological applications did come out before that article. The notion of a comodule, on the other hand, is due to Cartier.

Posted by: Zoran Skoda on November 29, 2009 4:14 PM | Permalink | Reply to this

Re: Dold–Kan Question

Jim, I was hoping you’d take the bait and chip in (to mix my metaphors). Although I’ve heard you mention the original version of the paper before I don’t think I’ve ever seen a copy.

The main topological example that is mentioned in the published version is very close to that which John is interested in, namely it is the homology of a topological space with a unital product which is associative up to homotopy. [Such a space is a certain kind of H-space where here the ‘H’ in H-space also stands for Hopf.] The differences with John’s example are that they look at homology groups, not the chain complexes; that they work in arbitrary characteristic, not just over the rationals; and that they consider spaces which are only associative up to homotopy, not associative on the nose.

They do also mention the fact that the Steenrod algebra is a Hopf algebra with an antipode.

Posted by: Simon Willerton on November 28, 2009 4:56 PM | Permalink | Reply to this

Re: Dold–Kan Question

Admits “unique” antipode ? But antipode on a bialgebra, if it exists, is always unique.

Posted by: Zoran Skoda on November 29, 2009 4:10 PM | Permalink | Reply to this

Re: Dold–Kan Question

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But antipode on a bialgebra, if it exists, is always unique.

Sure. I just meant that the uniqueness is conceptually important here because it means that connected graded bialgebras can really be identified with connected graded Hopf algebras.

Posted by: Mike Shulman on November 29, 2009 5:07 PM | Permalink | PGP Sig | Reply to this

Re: Dold–Kan Question

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I was confused for a bit, but now I think I know what you want. When you say

This concerns the extent to which the ‘normalized Moore complex’ functor from simplicial sets to chain complexes is monoidal.

I think you mean from simplicial abelian groups to chain complexes? The point being that the normalized chain complex functor can be written as the composite $Top \overset{Hom(\Delta^*,-)}{\to} sSet \overset{\mathbb{Q}[-]}{\to} sAb \overset{N_\bullet}{\to} Ch$ where $N_\bullet$ is the normalized Moore complex. Here the first functor is representable, hence preserves cartesian products, and the second takes (levelwise) cartesian products to levelwise tensor products, so the question is whether the final one takes levelwise tensor products to the usual tensor product of chain complexes.

I’m pretty sure that the answer is no, this functor is not strong monoidal (hence the Alexander-Whitney maps etc.), but I don’t know why $C_*(X,\mathbb{Q})$ is a coalgebra. I’ll see if I can find out.

Posted by: Mike Shulman on November 28, 2009 3:06 AM | Permalink | PGP Sig | Reply to this

Re: Dold–Kan Question

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Sorry, yes: I meant simplicial abelian groups. I’ll change my original blog entry, to spare others your confusion.

As you note, the only step in the passage from a space $X$ to its normalized rational singular chain complex $C_*(X,\mathbb{Q})$ that might fail to be strongly monoidal is taking the normalized Moore complex of a simplicial abelian group. (Or simplicial vector space, just in case working over a field of characteristic zero helps somehow!)

So, this is the step I’m concentrating on.

If this functor is only lax monoidal yet still carries certain comonoids to comonoids, perhaps there’s some interesting formal reason for this.

Thanks for trying to look into this!

Posted by: John Baez on November 28, 2009 8:06 AM | Permalink | Reply to this

Re: Dold–Kan Question

The normalized chains functor from simplicial sets to chain complexes (with any coefficients) is both lax monoidal and lax comonoidal. The Eilenberg-Zilber equivalence, from the tensor product of the chains on X and on Y to the chains on the cartesian product of X and Y, provides the natural transformation that shows that the chain functor is lax monoidal. The Alexander-Whitney equivalence goes in the opposite direction and shows that the chain functor is lax comonoidal.

Since the chain functor is lax comonoidal, the normalized chains on any simplicial set is a dg coalgebra, where the comultiplication is given by the composite of the chain functor applied to the diagonal map, followed be the Alexadnder-Whitney transformation. It turns out that the Eilenberg-Zilber equivalence is actually itself a morphism of coalgebras with respect to this comultiplication. On the other hand, the Alexander-Whitney map is a morphism of coalgebras up to strong homotopy.

The A-W/E-Z equivalences for the normalized chains functor are a special case of the strong deformation retract of chain complexes that was constructed by Eilenberg and MacLane in their 1954 Annals paper “On the groups H(pi, n). II”. For any commutative ring R, they defined chain equivalences between the tensor product of the normalized chains on two simplicial R-modules and the normalized chains on their levelwise tensor product.

Steve Lack and I observed recently that the normalized chains functor is actually even Frobenius monoidal. We then discovered that Aguiar and Mahajan already had a proof of this fact in their recent monograph… :-)

As to the Hopf algebra/bialgebra question, I admit I’ve often been guilty of referring to loop space homology as a “graded Hopf algebra”, without any mention of antipode. I was happy to learn a few years ago that connected graded bialgebras admit a unique antipode, as Mike has already pointed out. On the chain level, if the simplicial group you’re considering is reduced (i.e., has a unique 0-simplex), then its normalized chain complex is a connected dg bialgebra and therefore a dg Hopf algebra.

I hope this answered your question, John. I’d be happy to provide more details, if you’d like.

Posted by: Kathryn Hess on November 28, 2009 9:08 AM | Permalink | Reply to this

Re: Dold–Kan Question

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The normalized chains functor from simplicial sets to chain complexes (with any coefficients) is both lax monoidal and lax comonoidal.

Very nice! I was starting to guess that something like that might be true. How come no one ever says it that way? I’ve been hearing about these things for years, but without a statement like that I had no chance of remembering what they all do.

The Eilenberg-Zilber equivalence, from the tensor product of the chains on X and on Y to the chains on the cartesian product of X and Y, provides the natural transformation that shows that the chain functor is lax monoidal. The Alexander-Whitney equivalence goes in the opposite direction and shows that the chain functor is lax comonoidal.

This is the opposite direction from what is called the “Alexander-Whitney morphism” on the nLab page. Is that wrong? And is the Eilenberg-Zilber equivalence the same as what is there called the “shuffle morphism”?

Steve Lack and I observed recently that the normalized chains functor is actually even Frobenius monoidal.

Also very nice! Do all Frobenius monoidal functors preserve bialgebras? I know they preserve Frobenius algebras….

Posted by: Mike Shulman on November 28, 2009 5:51 PM | Permalink | PGP Sig | Reply to this

Re: Dold–Kan Question

Hmm. It’s slightly confusing to compare my comment to the notes on the n-Lab page, since the discussion there is about the normalized *cochain* complexes, so the arrows are all reversed. It might be a good idea to rewrite the n-Lab page in terms of chain complexes rather than cochain complexes, then to dualize, if you really need to. I’ve never forgotten John Moore’s advice to me when I was a young postdoc: never dualize unless absolutely necessary…

I’m not sure whether or not all Frobenius monoidal functors preserve bialgebras, though I know that the normalized chain functor does. I’ll think about it.

Steve and I were interested in Frobenius monoidal functors, since they send objects that are modules on one side and comodules on the other to objects of the same type, which is related to the problem of preserving bialgebras, but not identical.

Posted by: Kathryn Hess on November 28, 2009 7:28 PM | Permalink | Reply to this

Re: Dold–Kan Question

Jim Stasheff sent me this reply by email, which he asked me to post for him.

“I agree totally. Similarly, I’ve never forgotten John
Moore’s same advice to me when I was a grad student.

There are two ‘evil’ influences at work here:
1. we are toilet trained with algebras not coalgebras
2. some of us are addicted to manifolds and so think of differential forms as given by God and all the rest are the works of man.”

Posted by: KH posting for JS on November 29, 2009 8:26 AM | Permalink | Reply to this

Re: Dold–Kan Question

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The nLab page on monoidal Dold-Kan correspondence has two main subsections: one on the simplicial/chain version one on the cosimplicial/cochain version.

It does actually state more detailed and powerful things about the simplicial version.

The reason that the cosimplicial version looks longer is because one author happened to like having available a place where the lax monoidalness of the Moore cochain complex functor is spelled out in detail.

One driving motivation for discussing the cochain version in more detail, eventually is still this:

for the simplicial/chain version it is known in quite some detail that the DK correspondence is oo-symmetric monoidal in both directions. In the cosimplicial/cochain case only a comparatively weak version of this statement is known.

We know that it sends commutative monoids to $E_\infty$ -monoids, and we know that its adjoint induces an equivalence of the homotopy category of dg-rings with that of cosimplicial rings.

So clearly this situation suggests that we have an $\infty$ -adjunction between $E_\infty$ -monoids on both sides. I’d like this to be spelled out in more detaiil, eventually.

Posted by: Urs Schreiber on November 29, 2009 5:07 PM | Permalink | Reply to this

Re: Dold–Kan Question

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never dualize unless absolutely necessary

Was it absolutely necessary in, say, rational homotopy theory?

Posted by: Urs Schreiber on November 29, 2009 5:16 PM | Permalink | Reply to this

Re: Dold–Kan Question

If you follow Quillen’s approach to rational homotopy theory and work with dg Lie algebra models, then you don’t need to dualize. Since dg Lie algebras are (or seem to be?) hard to calculate with, many rational homotopy theorists have prefered to work with Sullivan’s commutative cochain algebra model, at the price of working only with spaces that are rationally of finite type. The Quillen and the Sullivan models are highly complementary, so it’s useful to have both available as tools. For example, products and fibrations are easy to model via Sullivan’s approach, while the Quillen model lends itself well to wedges and cofibrations.

It was in part to avoid finite-type restrictions that John Moore advocated for not dualizing.

Posted by: Kathryn Hess on November 29, 2009 5:46 PM | Permalink | Reply to this

Re: Dold–Kan Question

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Mike wrote:

Do all Frobenius monoidal functors preserve bialgebras?

What’s a Frobenius monoidal functor? I should be able to guess, since I bet a Frobenius monoidal functor from the terminal category to a monoidal category $M$ is the same as a Frobenius monoid in $M$ . Is that right?

Posted by: John Baez on November 28, 2009 10:35 PM | Permalink | Reply to this

Re: Dold–Kan Question

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What’s a Frobenius monoidal functor? I should be able to guess, since I bet a Frobenius monoidal functor from the terminal category to a monoidal category $M$ is the same as a Frobenius monoid in $M$ . Is that right?

Yes.

But according to this paper, Frobenius monoidal functors do not, in general, preserve bimonoids. Can one isolate the properties of normalized chains which ensure that it preserves bimonoids?

Posted by: Mike Shulman on November 29, 2009 5:13 PM | Permalink | PGP Sig | Reply to this

Re: Dold–Kan Question

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Mike wrote:

Do all Frobenius monoidal functors preserve bialgebras?

Could someone check my work here?

We’ve got a functor between symmetric monoidal categories

$$N : C \to D $ It's lax monoidal, so there's a natural transformation like this: $ $F_{c,c'}: N(c) \otimes N(c') \to N(c \otimes c')$ $ and it's oplax monoidal, so there's also a natural transformation like this: $ $G_{c,c'} : N(c \otimes c') \to N(c) \otimes N(c')$ $ These are not inverses, so$ N $is not strong monoidal. But$ G $followed by$ F $is the identity: $ $F_{c,c'} \circ G_{c,c'} = 1$ $ (And suppose that$ \phi \circ \gamma = 1 $where $ $\phi : 1 \to N(1)$ $ and $ $\gamma: N(1) \to 1$ $ are also part of what it means for$ F $to be lax and oplax.) Then: Given any <a href = "http://ncatlab.org/nlab/show/PRO">PRO</a>,$ N $maps algebras of that PRO in$ C $to algebras of that PRO in$ D $. The idea is that if our PRO has some operation with 2 inputs and 2 outputs (for example), and$ c \in C $is an algebra of that PRO, then it comes with an operation $ $f : c \otimes c \to c \otimes c$ $ Applying$ N $, we get a morphism $ $N(f) : N(c \otimes c) \to N(c \otimes c)$ $ and using$ F $and$ G $, we get an operation $ $G_{c,c} \circ N(f) \circ F_{c,c} : N(c) \otimes N(c) \to N(c) \otimes N(c)$ $ which is just the sort of thing we need to make$ N(c) $into an algebra of the same PROP. But do the operations on$ N(c) $satisfy the right relations to give us an algebra of our PRO? We know the operations for$ c $satisfy these relations: a bunch of equations saying some composite of operations equals some other operation. I claim the relations for$ N(c) $follow, thanks to the fact that $ $F_{c,c} \circ G_{c,c} = 1$ $ This makes the$ F $'s and$ G $'s cancel out! The order of the$ F $and$ G$ is crucial here. Please check this, o ye mathematicians of the world!

All this is fine and dandy, if true. But alas, bialgebras — unlike Frobenius algebras — are not algebras of a PRO! They’re algebras of a PROP! Their axioms involve the braiding as well as the tensor product.

So, I’m not done understanding this stuff yet.

Posted by: John Baez on November 29, 2009 6:10 PM | Permalink | Reply to this

Re: Dold–Kan Question

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John wrote:

But $G$ followed by $F$ is the identity.

Aargh! It seems in our example it works just the other way: Eilenberg-Zilber follows by Alexander-Whitney is the identity, so in the notation I’m using above,

$F_{c,c′}:N(c) \otimes N(c′) \to N(c\otimes c′)$

followed by

$G_{c,c′}:N(c \otimes c′) \to N(c) \otimes N(c′)$

is the identity.

I’m really tired. Maybe if I have a good night’s sleep reality will have fixed itself by the time I wake up.

Posted by: John Baez on November 30, 2009 6:34 AM | Permalink | Reply to this

Re: Dold–Kan Question

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And is the Eilenberg-Zilber equivalence the same as what is there called the “shuffle morphism”?

The $n$ Lab entry currently follows for the simplicial/chain version in terminology the article by Schwede-Schipley and in the cosimplicial/cochain section it follows in terminology the article by José Burgos Gil here, as indicated.

Both these use “Alexander-Whitney” and “shuffle”

Posted by: Urs Schreiber on November 30, 2009 8:34 AM | Permalink | Reply to this

Re: Dold–Kan Question

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Hi, Kathryn! Great answers! I’ll mull them over and maybe ask some more questions later, but I think you’ve dispelled the bulk of my puzzlement.

I’m writing about rational homotopy theory. Thanks for writing an introduction to that! I got into my current mess when trying to explain how a pointed space gives a differential graded Lie algebra. At first I thought I understood it, but then it started seeming mysterious that

$X \mapsto C_*(X, \mathbb{Q})$

sends groups to Hopf algebras.

Posted by: John Baez on November 28, 2009 5:58 PM | Permalink | Reply to this

Re: Dold–Kan Question

I’m glad you find my introduction to rational homotopy theory helpful. Please just let me know if you have any questions about it that you think I might be able to answer.

Posted by: Kathryn Hess on November 28, 2009 7:37 PM | Permalink | Reply to this

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