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Stream: learning: questions

Topic: When is the bar construction nontrivial?

Oscar Cunningham (Jul 06 2025 at 12:12):

I'm trying to understand the following from the nLab's page on the bar construction:

[The bar resolution] is called a resolution of $A$ because it is equipped with a simplicial homotopy equivalence to the underlying $C$-object $U A \in C$, in a sense clarified by the "acyclic structure" of Definition 3.1. Moreover the bar resolution of $A$ has a universal property: it is initial among resolutions of $A$, as explained in Theorem 4.2.

Here we distinguish between the bar construction and the bar resolution, although the terms are often used interchangeably. It is worth noting that the bar construction $Bar_T(A)$ is not, in general, simplicially homotopy equivalent to $A$ in the category of simplicial objects in $C^T$. Only after applying the forgetful functor $U \colon C^T \to C$ do we obtain a key feature of the bar resolution: the simplicial homotopy equivalence $U Bar_T(A) \simeq U A$.

For an example, let's say that $T$ is the monad for abelian groups on $\mathbf{Set}$. Then the simplicial objects in $\mathbf{Set}^T$ can be thought of as chain complexes in nonnegative degree. My understanding is that the object '$A$ in the category of simplicial objects in $C^T$' would then just be the chain complex $\dots\to 0\to A$. So the above quote is saying that $Bar_T(A)$ might then be some other chain complex that isn't equivalent to this one. Is there a specific abelian group $A$ such that the homology groups of $Bar_T(A)$ are something other than $H_0 = A, H_1 = 0, \dots$?

John Baez (Jul 06 2025 at 12:23):

By the way, you're taking an example that's not one of the classic ones people study a lot. A 'better' example, if you want lots of people to have thought about it already, is to let $T$ be the monad for $\mathbb{Z}[G]$-modules on $\mathsf{AbGp}$, where $G$ is some group. This leads in a certain way to the subjects called 'homology of groups' and 'cohomology of groups', which were the subjects that led Eilenberg and Mac Lane to invent the bar construction in the first place.

However, your example is still a fine one to ponder!

So you are taking an abelian group $A$ and using the bar construction to build a simplicial abelian group $\text{Bar}_T(A)$. Todd is telling you this may not be homotopy equivalent to $A$ as a simplicial abelian group, though its underlying simplical set is homotopy equivalent to $A$ (or more precisely, to the free simplicial set on the underlying set of $A$).

John Baez (Jul 06 2025 at 12:37):

If I understand you right, you're hoping that the possible failure of the chain complex coming from $\text{Bar}_T(A)$ to be homotopy equivalent to $A$ may be detected by it having different homology groups from the chain complex

$A \leftarrow 0 \leftarrow 0 \leftarrow \cdots$

(I like to write chain complexes this way so the $$n$$th grade is in the $n$ th position from left to right.)

This is a fun question - I don't instantly know what happens here - but I just want to mention something. Homotopy equivalent chain complexes have the same homology, but you're sort of hoping that the converse is true: that homotopy inequivalent chain complexes have different homology.

If we work with chain complexes of vector spaces over a field this is true. For chain complexes of modules over a general ring this is not true, and the concept of quasi-isomorphism of chain complexes becomes really important: a quasi-isomorphism of chain complexes is a map that induces isomorphisms on homology groups.

I always get confused when it comes to chain complexes of modules over $\mathbb{Z}$, i.e. chain complexes of abelian groups, which is the case you're concerned with. But I seem to remember that in this case, because $\mathbb{Z}$ has homological dimension 1, any quasi-isomorphism of chain complexes is a chain homotopy equivalence.

Yes, I did not answer your question.

Oscar Cunningham (Jul 06 2025 at 12:58):

Yeah, thanks for clarifying those points. I was hoping there would be an example in chain complexes, because they're easier for me to think about than more general simplicial things. Did you have a specific example in mind for $\mathbb{Z}[G]$-modules?

Patrick Nicodemus (Jul 06 2025 at 13:00):

Oscar Cunningham said:

I'm trying to understand the following from the nLab's page on the bar construction:

[The bar resolution] is called a resolution of $A$ because it is equipped with a simplicial homotopy equivalence to the underlying $C$-object $U A \in C$, in a sense clarified by the "acyclic structure" of Definition 3.1. Moreover the bar resolution of $A$ has a universal property: it is initial among resolutions of $A$, as explained in Theorem 4.2.

Here we distinguish between the bar construction and the bar resolution, although the terms are often used interchangeably. It is worth noting that the bar construction $Bar_T(A)$ is not, in general, simplicially homotopy equivalent to $A$ in the category of simplicial objects in $C^T$. Only after applying the forgetful functor $U \colon C^T \to C$ do we obtain a key feature of the bar resolution: the simplicial homotopy equivalence $U Bar_T(A) \simeq U A$.

For an example, let's say that $T$ is the monad for abelian groups on $\mathbf{Set}$. Then the simplicial objects in $\mathbf{Set}^T$ can be thought of as chain complexes in nonnegative degree. My understanding is that the object '$A$ in the category of simplicial objects in $C^T$' would then just be the chain complex $\dots\to 0\to A$. So the above quote is saying that $Bar_T(A)$ might then be some other chain complex that isn't equivalent to this one. Is there a specific abelian group $A$ such that the homology groups of $Bar_T(A)$ are something other than $H_0 = A, H_1 = 0, \dots$?

I think the answer is no, but I'm pretty tired. John has already explained why: because having the same homology groups isn't the same as being homotopy equivalent.

If $A$ is an Abelian group and $\operatorname{Bar}_T(A)$ is the bar construction of $A$ as a simplicial object, then $\operatorname{Bar}_T(A)$ is homotopy equivalent to the constant simplicial object at $A$ in the category of sets. A homotopy equivalence induces an isomorphism of homotopy groups, and the homotopy groups of a simplicial Abelian group are isomorphic to the homology groups of the associated chain complex. In other words, thinking of these things as chain complexes, the map $\operatorname{Bar}_T(A)\to A$ is a quasi-isomorphism.

Patrick Nicodemus (Jul 06 2025 at 13:01):

This is good news, if we want $\operatorname{Bar}_T(A)$ to be a resolution of $A$, so we can use it to compute homology or cohomology with coefficients in some other group $B$.

Patrick Nicodemus (Jul 06 2025 at 13:04):

because $\mathbb{Z}$ has homological dimension 1, any quasi-isomorphism of chain complexes is a chain homotopy equivalence

@John Baez Really? You may very well be right but that's surprising to me. Even if the groups in the complexes themselves have all sorts of ugly torsion? Wow. I can believe it by appeal to "wow, aren't PID's great?"

John Baez (Jul 06 2025 at 13:10):

I'm not sure I'm right: James Dolan told me something like this once and I understood it at the time, but that was about 20 years ago. I would love to settle this issue...

... as well as Oscar's original question: let's not forget that!

Patrick Nicodemus (Jul 06 2025 at 13:13):

I posted a message, thought about it for a second, deleted it because I was unsure, but now I am again confident in the original claim I made.

The resolution map from $C_\bullet= 0\to\mathbb{Z}\to\mathbb{Z}\to 0$ to $D_\bullet =0\to\mathbb{Z}/2\mathbb{Z}\to 0$ is a quasi-isomorphism. However, it does not have a chain homotopy inverse. There are no nontrivial maps $\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}$, so in particular there cannot be any map from $D_\bullet$ to $C_\bullet$ inducing an isomorphism of homology groups.

Patrick Nicodemus (Jul 06 2025 at 13:15):

So the answer to Oscar's final question is no. There is no such Abelian group. The canonical map of chain complexes $\operatorname{Bar}_T(A)\to A$ is a quasi-isomorphism. It will not be a chain homotopy equivalence in general, unless $A$ is projective.

John Baez (Jul 06 2025 at 13:18):

Patrick Nicodemus said:

The resolution map $0\to\mathbb{Z}\to\mathbb{Z}\to 0$ of $0\to\mathbb{Z}/2\mathbb{Z}\to 0$ is a quasi-isomorphism. However, it does not have a chain homotopy inverse. There are no nontrivial maps $\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}$, so in particular there cannot be any map from the second complex to the first inducing an isomorphism of homology groups.

Thanks - that's simple enough! I wonder what I was thinking. There's something surprisingly nice that happens with chain complexes over $\mathbb{Z}$ because $\mathbb{Z}$ has homological dimension $1$ - surprising if you're just learning this stuff, I mean. But I forget what it is!

John Baez (Jul 06 2025 at 13:18):

Patrick Nicodemus said:

So the answer to Oscar's final question is no. There is no such Abelian group. The canonical map of chain complexes $\operatorname{Bar}_T(A)\to A$ is a quasi-isomorphism. It will not be a chain homotopy equivalence in general, unless $A$ is projective.

Very nice! Thanks again. I hadn't noticed at first that you settled his question.

Patrick Nicodemus (Jul 06 2025 at 13:19):

John Baez said:

Patrick Nicodemus said:

The resolution map $0\to\mathbb{Z}\to\mathbb{Z}\to 0$ of $0\to\mathbb{Z}/2\mathbb{Z}\to 0$ is a quasi-isomorphism. However, it does not have a chain homotopy inverse. There are no nontrivial maps $\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}$, so in particular there cannot be any map from the second complex to the first inducing an isomorphism of homology groups.

Thanks - that's simple enough! I wonder what I was thinking. There's something surprisingly nice that happens with chain complexes over $\mathbb{Z}$ because $\mathbb{Z}$ has homological dimension $1$ - surprising if you're just learning this stuff, I mean. But I forget what it is!

Yeah, I don't know what you have in mind. It could be that torsion-free is equivalent to projective, that's a convenient hom alg fact about modules over a PID.

Patrick Nicodemus (Jul 06 2025 at 13:22):

As a side note, I think that the name "bar construction associated to a monad" in this n-lab page is really unfortunate. It just adds too much complexity to naming conventions and makes you think you're looking at one thing when you're really looking at the other thing.

I think I'd prefer to call the "bar construction associated to a monad" the following construction: one has a monad $T$ on a category $C$ and one iteratively applies it to an object $A$ of $C$, which yields a cosimplicial object augmenting $A$. Then the "cobar construction associated to a comonad" would be the evident dual, where one has a comonad $G$ and iteratively applies it to an object $B$ to get a simplicial object augmenting $C$. But this is the opposite of what is described in this article, as one is really working with the comonad on $T$-algebras arising from the adjunction $C \iff C^T$, to get a simplicial object augmenting $A$ in $C^T$.

Patrick Nicodemus (Jul 06 2025 at 13:25):

Take the Godement construction which associates to every sheaf of Abelian groups a cosimplicial resolution of flasque sheaves. I would like to call this the "bar construction of the flasque sheaf monad". But this is a cosimplicial object, not a simplicial one. Should it be a cobar construction?

John Baez (Jul 06 2025 at 13:30):

Patrick Nicodemus said:

I think I'd prefer to call the "bar construction associated to a monad" the following construction: one has a monad $T$ on a category $C$ and one iteratively applies it to an object $A$ of $C$, which yields a cosimplicial object augmenting $A$. Then the "cobar construction associated to a comonad" would be the evident dual...

That seems reasonable, but I don't think it's what is used in most of the literature, is it?

I've indeed been repeatedly confused by the fact that the bar construction mainly makes use of the comonad, while various people act as if it's mainly connected to the monad.

Patrick Nicodemus (Jul 06 2025 at 13:32):

You're probably right, it's just a bit unfortunate. My mind goes "okay you iteratively apply a monad, and the simplex category is the walking monoid, so there's a canonical functor $\Delta \to C$ given by blah blah blah" and this is the opposite of what you want to do

John Baez (Jul 06 2025 at 13:33):

It's exactly what you want to do... in the opposite category.

Oscar Cunningham (Jul 06 2025 at 17:00):

Oh, I like the sound of a cocobar construction :chocolate:

Kevin Carlson (Jul 07 2025 at 05:21):

John Baez said:

Thanks - that's simple enough! I wonder what I was thinking. There's something surprisingly nice that happens with chain complexes over $\mathbb{Z}$ because $\mathbb{Z}$ has homological dimension $1$ - surprising if you're just learning this stuff, I mean. But I forget what it is!

I seem to remember a result that every chain complex over a PID is quasi-isomorphic to the sum of its homologies, and perhaps even the argument: put each map in the complex in Smith normal form to see that the complex is actually isomorphic to the sum of a bunch of complexes with only two nontrivial entries, and then check in that case. But there are two details here that don't look clean yet, namely, can I do Smith normal form simultaneously on both sides of each chain group, and do I need the length-2 complexes to be injective to get the desired result after reducing?

Oscar Cunningham (Jul 07 2025 at 05:48):

Does Smith Normal Form work for modules that aren't finitely generated?

John Baez (Jul 07 2025 at 06:01):

Kevin Carlson said:

I seem to remember a result that every chain complex over a PID is quasi-isomorphic to the sum of its homologies...

Yes, that's what I was dimly remembering! Thanks! I don't know the argument but that's the thing that surprised me: there's no "juice" to a chain complex of abelian groups except its homology groups, if by "juice" you mean information that's invariant under quasi-isomorphism.

(Of course a quasi-isomorphism is defined to be a map that preserves homology so one might naively find this result unsurprising.)