## January 24, 2007

### Cocycle Category

#### Posted by Urs Schreiber

Here is another guest post by Bruce Bartlett.

Luckily, Bruce is still at Fields in Toronto, attending the Thematic Program on Geometric Applications of Homotopy Theory.

Here he reports on something very interesting that is intimately related to our discussion of anafunctors.

Hi guys,

I’m no genius in homotopy theory but I think I stumbled across a really cool fact today which I’d like to bounce off you, perhaps you’d like to put it on the $n$-café. I’m so excited about it, I almost want to keep it secret… except that if true, it’s probably too small an observation to warrant a paper.

As you know, Rick Jardine is giving lectures on Simplicial Presheaves.

For $n$-café purposes, some of these notes can be found in the preprints section of his homepage. I have also taken notes and perhaps I can scan them in. But the important point is that the entire lecture today was based on his Cocycle categories preprint which can be found at that link.

What he does is he considers a model category $M$. Given two objects $X$ and $Y$ in $M$, you define a cocycle from $X$ to $Y$ as a pair of maps

$\array{ Z &\to& Y \\ \downarrow \\ X } \,,$ where the map to $X$ is a weak equivalence. A morphism between cocycles is just an obvious map making the obvious diagram commute. Then $H(X,Y)$ is called the cocycle category.

Of course, us $n$-category theorists recognize this immediately - he is simply building the 2-category $\mathrm{NiceSpan}(M)$ of ‘nice’ spans in $M$, whose objects are the same as $M$, whose morphisms are cocycles (spans where the projection map is a weak equivalence) and whose 2-morphisms are 2-morphisms between spans! (We know this because Urs taught us this from one of his recent posts at the n-cafe). Rick Jardine never mentions the word 2-category, though.

[The definition of the ordinary bicategory of spans is recalled here, aspects of the 2-category of anafunctors are discussed here. For more details check out M. Makkai’s paper and Toby Bartels’ thesis. -urs]

Here’s the great fact. He proves that for decent model categories $M$ (when $M$ is a right proper closed model structure in which weak equivalences are preserved by finite products), the canonical map $\phi : \pi_0 H(X,Y) \to \mathrm{Ho}(M) (X,Y)$ is a bijection! Here $\pi_0 H(X,Y)$ is just the iso classes of morphisms, and $\mathrm{Ho}(M)$ is the homotopy category of $M$. The map $\phi$ sends $(f,g)$ to $f^{-1} g$ (you know what I mean).

As $n$-café regulars, we get excited because we immediately see two things:

(1) He has shown that the homotopy category $\mathrm{Ho}(M)$ is nothing but $\pi_1$ of the 2-category $\mathrm{NiceSpan}(M)$! (I know that there is already a ‘$\pi_1$ of a 2-category’ interpretation of $\mathrm{Ho}(M)$, but doesn’t that only work for cofibrant fibrant objects?)

(2) It is obvious that in the ordinary model structure on $\mathrm{Cat}$, the cocycles $X \to Y$ are precisely the anafunctors! In other words, morphisms in the homotopy category $\mathrm{Ho}(\mathrm{Cat})$ are nothing but isomorphism classes of anafunctors!!!

Finally, if you read his preprint above on cocycle categories, you’ll see that the entire motivating example is nothing but the cocycle associated to a principal $G$-bundle… in other words it’s nothing but the stuff you guys needed to do for transport. This is why ‘anafunctors’ etc comes out so naturally.

I’m really excited about this… but I’m going to sleep now. Hopefully you guys will bring me down gently if I’ve gotten confused here, or if its all old hat.

Yours, Bruce

Posted at January 24, 2007 10:30 AM UTC

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### Re: Cocycle Category

[…] he is simply building the 2-category NiceSpan(M) of ‘nice’ spans in M, whose objects are the same as M, whose morphisms are cocycles (spans where the projection map is a weak equivalence) and whose 2-morphisms are 2-morphisms between spans [morphisms of cocycles]!

He hasn’t quite finished building this until he defines composition of cocycles. This is straightforward, of course, using pullbacks just as in the ordinary composition of spans. While you need all pullbacks to define composition of arbitrary spans, here you only need pullbacks involving weak equivalences, but you also need that the pullback of any weak equivalence be a weak equivalence. (I forget if this condition is part of the definition of model category, but I won’t be surprised if it is.)

I don’t find any of this surprising, although it’s certainly good that people are thinking about it. My concern with model categories (which I’d heard of before going to Toronto this month but which always seemed too complicated to learn, to the point that I’d never even looked carefully at the definition) as a tool for understanding higher categories (not that they are only that) is that they are more complicated than necessary for that purpose. Much can be done with only a good notion of weak equivalence. I don’t know if there any axiomatisations of the concept of category-with-weak-equivalences.

Posted by: Toby Bartels on January 24, 2007 3:37 PM | Permalink | Reply to this

### Re: Cocycle Category

I don’t know if there any axiomatisations of the concept of category-with-weak-equivalences.

There is at least one.

There’s a recent (2004) book with the rather intimidating title Homotopy Limit Functors on Model Categories and Homotopical Categories, by Dwyer, Hirschorn, Kan and Smith. It’s divided into two parts: the first part is about model categories, and the second part covers ‘homotopical categories’.

A homotopical category is defined to be a category equipped with a distinguished class of maps called weak equivalences, such that:

1. Every identity map is a weak equivalence,
2. given composable maps r, s and t, if ts and sr are weak equivalences then so are r, s, t and tsr.

The authors develop a fair bit of the theory of homotopical categories, including defining well-behaved notions of homotopy limit and homotopy colimit. It feels like a promising approach to me, though I haven’t done any serious work on it.

Posted by: Robin on March 6, 2007 6:53 PM | Permalink | Reply to this

### Re: Cocycle Category

Robin: That looks great, thanks!

Posted by: Toby Bartels on March 7, 2007 10:38 PM | Permalink | Reply to this

### Re: Cocycle Category

Dwyer and Kan — the masterminds of simplicial homotopy theory — have never been fully wedded to model categories, although they recognize their usefulness.

When they constructed a simplicial category from a model category back in the 1980s, they first did it much more generally for any category with a class of weak equivalences:

• William G. Dwyer and Daniel M. Kan, Simplicial localizations of categories, J. Pure Appl. Algebra 17 (1980), 267-284.

Then they showed how the construction could be done in an equivalent, more tractable way (the ‘hammock localization’) when you actually had a model category:

• William G. Dwyer and Daniel M. Kan, Calculating simplicial localizations, J. Pure Appl. Algebra 18 (1980), 17-35.

The book by Dwyer, Hirschorn, Kan and Smith is the result of decades of work. There have been dozens of radically different drafts. You can see signs of this here.

Unfortunately the authors are the old-fashioned sort of people who take their work off their websites when it finally gets published. With such people, you have to grab stuff while it’s available! Someone should tell them about the arXiv.

Posted by: John Baez on March 8, 2007 5:36 PM | Permalink | Reply to this

### Re: Cocycle Category

Interesting comment - yeps you’re right about the composition. I thought hard about the stuff I said above, discussing it with Tom Fiore, Rick Jardine and others, and taking various books out the library… including the book by Gabriel and Zisman (try to find *that* in your local library!). I think I found out that the observation’ I made above wasn’t so cool after all :-) My main mistake was in thinking that $\pi_0 (C)$ of a category $C$ was the set of isomorphism classes of $C$. It’s not - its just the set of path components ($a \sim b$ if there exists a path $a \rightarrow \leftarrow \cdots \leftarrow b$). Phrased in this way, the observation that for nice model categories $M$, $\pi_0 H(X,Y)$ is in bijective correspondence with $Ho(M) (X,Y)$ is very similar to the calculus of fractions’. Its essentially the same trick which people who do derived categories have been doing for a long time.

In other words, this changes the statement I made about anafunctors from Morphisms in the homotopy category of Cat are isomorphism classes of anafunctors’ to Morphisms in the homotopy category of Cat are path components of the space of anafunctors’, which, to me at least, is a lot less cool.

Posted by: Bruce Bartlett on January 24, 2007 11:35 PM | Permalink | Reply to this

### Re: Cocycle Category

In other words, this changes the statement I made about anafunctors from ‘Morphisms in the homotopy category of Cat are isomorphism classes of anafunctors’ to ‘Morphisms in the homotopy category of Cat are path components of the space of anafunctors’, which, to me at least, is a lot less cool.

I see what you mean. Of course, the real lesson there is that we should be using a 2category.

Posted by: Toby Bartels on January 25, 2007 1:01 AM | Permalink | Reply to this

### Re: Cocycle Category

I’m a bit jealous of you up there in Toronto, Bruce! You wrote:

Hopefully you guys will bring me down gently if I’ve gotten confused here, or if it’s all old hat. [Necessary apostrophe added free of charge by JB]

I don’t think you’re confused. And I don’t think your observations are old hat — but I don’t think they’re brand new hat, either.

Rick Jardine told me about his paper on cocycle categories after I gave a talk at Chicago last spring which concluded with a discussion of Schreier theory. (For details see the section on ‘Grothendieck’s dream’ in my paper on cohomology and n-categories, based on these talks.)

Schreier theory classifies extensions of a group $B$ by a group $F$: that is, short exact sequences

$1 \to F \to E \to B \to 1$

with fixed $F$ and $B$.

In fact, there’s a nice category $EXT(B,F)$ of such extensions. An object in here is an extension of $B$ by $F$, and a morphism is a homomorphism $E \to E'$ making the obvious diagram commute. (It’s difficult to draw here, but see my paper.)

What’s $EXT(B,F)$ actually like?

The beautiful answer is that this category is equivalent to

$hom(B, AUT(F))$

where $AUT(F)$ is the automorphism 2-group of $F$.

This fact is an example of the general clasifying space philosophy I was pushing in these talks, which you’ve recently come to share. If we think of a short exact sequence

$1 \to F \to E \to B \to 1$

as a kind of ‘bundle over $B$ with fiber $F$’, we’re saying these bundles can be classified by maps from $B$ to $AUT(F)$.

Unfortunately, the usual proof of this result requires that we pick a splitting of the sequence

$1 \to F \to E \to B \to 1$

not by a group homomorphism

$s : B \to E$

(impossible in general) but by a mere function (always possible if we use the axiom of choice).

Jardine’s ‘cocycle category’ construction allows us to avoid this arbitrary splitting, which is especially good in contexts where the axiom of choice doesn’t apply — e.g. when we’re dealing with topological groups.

And, when I thought about it a while, I realized it was just another instance of Toby’s philosophy of working with anafunctors!

I forget the details now, but I think you’ve seen the same thing and are excited for the exact same reason! Or at least a very similar reason. Namely: even though he wasn’t thinking of it quite this way, Jardine was secretly tackling a certain aspect of higher gauge theory — and he was independently forced to use the same `anafunctor’ idea that Toby was forced to think about in his work on this subject.

By the way, there should really be a 2-category $EXT(B,F)$ of extensions of $B$ by $F$, since $hom(B,AUT(F))$ is a hom of 2-categories: the 2-group $AUT(F)$ is a 2-category with one object, and the group $B$ is a 2-category with one object and only identity morphisms. However, it’s common to treat $EXT(B,F)$ and $hom(B,AUT(F))$ as mere 1-categories, and these are already equivalent. In fact, I haven’t even checked that there’s an equivalence of 2-categories here.

I should add that what I’m now calling $EXT(B,F)$ is usually called the nonabelian 2nd cohomology of $B$ with coefficients in $F$.

Posted by: John Baez on January 25, 2007 3:14 AM | Permalink | Reply to this

### Re: Cocycle Category

In fact, there’s a nice category $EXT(B,F)$ of such extensions. An object in here is an extension of $B$ by $F$, and a morphism is a homomorphism $E\to E'$ making the obvious diagram commute. (It’s difficult to draw here, but see my paper.)

Y’mean:

(1)$\begin{matrix} 1\to& F &\to& E &\to& B & \to 1 \\ &\Vert& &\darr& &\Vert& \\ 1\to& F &\to& E' &\to& B &\to 1 \end{matrix}$

Since I don’t understand the general extension problem, let me specialize to $F$ abelian. In that case, we know that extensions (1) are classified by $\Gamma=\mathrm{Ext}^1(B,F)$. This is an abelian group. Is the category structure on $EXT(B,F)$ the same as the obvious category structure on $\Gamma$?

Posted by: Jacques Distler on January 25, 2007 4:56 AM | Permalink | PGP Sig | Reply to this

### Re: Cocycle Category

Is the category structure on $\mathrm{EXT}(B,F)$ the same as the obvious category structure on $\Gamma$?

I’d need to remind myself, admittedly, of some of this $\mathrm{Ext}$-business, but I would think that this are not the same category structures.

The category $\mathrm{EXT}(B,F)$ has one object per extension of $B$ by $F$, whereas $\Gamma$, regarded as just an abelian group, has a single object only.

Posted by: urs on January 25, 2007 1:15 PM | Permalink | Reply to this

### Re: Cocycle Category

The category $EXT(B,F)$ has one object per extension of $B$ by $F$, whereas $\Gamma$, regarded as just an abelian group, has a single object only.

Umh, … no.

$\Gamma$ has $|\Gamma|$ objects, one for each $g\in\Gamma$. And, for any pair of objects, $g_1,g_2\in\Gamma$, there is precisely one morphism $g_1\to g_2$ (determined by the group law).

Posted by: Jacques Distler on January 25, 2007 2:52 PM | Permalink | PGP Sig | Reply to this

### Re: Cocycle Category

$\Gamma$ has $|\Gamma|$ objects, one for each $g\in \Gamma$. And, for any pair of objects, $g_1,g_2 \in \Gamma$, there is precisely one morphism $g_1 \to g_2$ (determined by the group law).

Okay, good. While not the obvious category structure on an abelian group $\Gamma$, this is the obvious category structure on $\Gamma$ regarded as a strict 2-group coming from the crossed module $(\Gamma \stackrel{\mathrm{Id}}{\to} \Gamma)$.

So, is this so? You are saying that there is precisely one diagram of the sort you have drawn for any ordered pair ($E$, $E'$) of extensions of $B$ by the abelian group $F$?

Because if so, then I guess the answer to your original question would be “yes”.

(But maybe I am missing something.)

Posted by: urs on January 25, 2007 3:16 PM | Permalink | Reply to this

### It was a question

So, is this so? You are saying that there is precisely one diagram of the sort you have drawn for any ordered pair $(E,E')$ of extensions of $B$ by the abelian group $F$?

No, that’s not what I was saying. I was asking a question.

When I try to follow John’s definition, I seem to get something quite different (and considerably more boring) in the case of $F$ abelian.

(As I said, when $F$ is nonabelian, I know absolutely nothing about the extension problem. So, for the purpose of understanding John’s proposal, I’d like to stick to the terra firma of the abelian case.)

Posted by: Jacques Distler on January 25, 2007 3:29 PM | Permalink | PGP Sig | Reply to this

### Re: It was a question

Sorry. Yes, I formulated that badly. I was just trying to indicate what would have to hold if the identification of category structures we talked about were in fact right.

It is a familiar fact that on abelian cohomology there is a group structure, which is not present on general nonabelian cohomology.

The reason is this:

when $F$ is abelian, then its automorphism 2-group $\mathrm{AUT}(F)$ is actually equipped with an additional, invertible product operation (meaning you can suspend it to a 3-group).

Using this, we get a group structure on $\mathrm{hom}(B,\mathrm{AUT}(F))/\sim$

(I write $\sim$ for taking equivalence classes, since $\mathrm{Ext}^1(B,F)$ is the group of equivalence classes of extensions of $B$ by $F$)

so it should be true that $\mathrm{Ext}^1(B,F) \simeq \mathrm{hom}(B,\mathrm{AUT}(F))/\sim$ and the group structure is that additional one.

Posted by: urs on January 25, 2007 5:37 PM | Permalink | Reply to this

### Clarification

The following comment of mine is a corollary to my recent my more detailed comment.

Urs essentially wrote:

I write $/\sim$ for taking isomorphism classes, since $\mathrm{Ext}^1(B,F)$ is the group of isomorphism classes of extensions of $B$ by $F$.

Actually it’s the group of isomorphism classes of abelian extensions of $B$ by $F$.

When people are talking about $Ext$ groups this restriction is so obvious that they often don’t bother to mention it. But, since I began this conversation by discussing the category of all extensions of $B$ by $F$, we need to be very clear about this.

so it should be true that

$\mathrm{Ext}^1(B,F) \simeq \mathrm{hom}(B,\mathrm{AUT}(F))/\sim$

No: when $B$ and $F$ are abelian we have

$Ext^1(B,F) \subseteq \mathbb{H}_{nonab}^2(B,F) \simeq \mathrm{hom}(B,\mathrm{AUT}(F))/\sim$

where $\mathbb{H}_{nonab}^2(B,F)$ is the 2nd nonabelian cohomology of $B$ with coefficients in $F$. This is defined to be the set of isomorphism class in the category $\mathrm{hom}(B,\mathrm{AUT}(F))$, but it’s in 1-1 correspondence with the set of isomorphism classes of extensions of $B$ by $F$. Not just abelian extensions — all extensions!

when $F$ is abelian, then its automorphism 2-group $\mathrm{AUT}(F)$ is actually equipped with an additional, invertible product operation (meaning you can suspend it to a 3-group).

Is that really true? You’re saying that $AUT(F)$ is not just a 2-group, but a braided 2-group. I have trouble believing this in the case where $F$ is, for example, $\mathbb{Z}^3$. Then the usual group $Aut(F)$ is $GL(3,\mathbb{Z})$, which is nonabelian. So, it’s hard for me to believe that the 2-group $AUT(F)$ is braided.

Posted by: John Baez on January 26, 2007 6:46 AM | Permalink | Reply to this

### Re: Clarification

Is that really true?

No, it’s not! Sorry. I was thinking about embedding $\mathrm{AUT}(U(1))$ into $\mathrm{AUT}(\Sigma(U(1)))$. But only part of it fits in there!

But let’s see. Something along these lines should be right:

Where does the group structure on $\mathrm{Ext}^1(B,F) \subset \mathrm{hom}(B,\mathrm{AUT}(F))$ and that on $H^2(B,F)$ come from? It must come from an extra group structure on $\mathrm{AUT}(F)$ that appears when we suitably cut down $\mathrm{AUT}(F)$ to the appropriate special cases.

That is, at least, the mechanism that governs the phenomenon that abelian cohomology is a group, while nonabelian cohomology is a mere set:

If a triangle $\array{ & \bullet \\ {}^{g_{ij}}\nearrow && \searrow^{g_{jk}} \\ \bullet &\stackrel{g_{ik}}{\to}& \bullet }$ is labeled by elements in $U(1)$, then I can multiply two such triangles and get another one (still commuting).

This does not work if the $g_{ij}$ take nonabelian values.

And this really is a consequence of the fact that $\Sigma(G)$ is a 2-group if and only if $G$ is abelian.

I was thinking that something similar must hold for the cases of group extensions that we are talking about.

In fact, I am now inclined to expect that the answer is my other mistake which you pointed out, namely assuming that $\mathrm{Ext}^1(B,F) = \mathrm{hom}(B,\mathrm{AUT}(F))/\sim$, while we really just have $\mathrm{Ext}^1(B,F) \subset \mathrm{hom}(B,\mathrm{AUT}(F))/\sim$.

Could it be possible to find $Q \subset \mathrm{AUT}(F)$ such that $\mathrm{Ext}^1(B,F) = \mathrm{hom}(B,Q)/\sim$ ?

And that this $Q$ is braided? And that this is where the group structure on $\mathrm{Ext}^1(B,F)$ comes from?

Something like this should be right. And it should be the answer to Jacques’ question: which structure in $\mathrm{hom}(B,\mathrm{AUT}(F))$ corresponds to the group structure on $\mathrm{Ext}^1(B,F)$?

Posted by: urs on January 26, 2007 9:30 AM | Permalink | Reply to this

### Re: Clarification

I wrote:

Could it be possible to find $Q \subset \mathrm{AUT}(F)$ such that $\mathrm{Ext}^1(B,F) = \mathrm{hom}(B,Q)/\sim$ ?

Inside the 2-group $\mathrm{AUT}(F)$, for $F$ abelian, we have $\Sigma(F) \subset \mathrm{AUT}(F) \,,$ which is the category with a single object (the trivial automorphism of $F$) and one morphism per element of $F$ (the inner automorphisms (all acting trivially for abelian $F$) relating the trivial one to itself).

Now, $\Sigma(F)$ is in fact a braided 2-group, meaning that we may think of it as a 3-group $\Sigma(\Sigma(F)) \,.$

This means that $\mathrm{hom}(B,\Sigma(F))/\sim$ is equipped with a group structure, which comes from multiplying any two pseudofunctors $\omega : \Sigma(B) \to \Sigma(\Sigma(F))$ “pointwise” (i.e. morphism-wise).

Now, such a pseudofunctor is nothing but a group 2-cocycle on $B$. A morphism between two such pseudofunctors is nothing but a group coboundary interpolating between two group cocycles.

So $\mathrm{hom}(B,\Sigma(F))/\sim = H^2(B,F) \,.$

And the group structure on that is that coming from the 3-group-structure on $\Sigma(F)$, if you like.

Same for $\mathrm{Ext}^1(B,F) \subset H^2(B,F)$.

Same holds when we allow $B$ to be a groupoid.

For instance if $B$ is the transition groupoid of some good cover of some space $X$, then $H^2(B,\Sigma(F))$ are the classes of line bundle gerbes on $X$.

Posted by: urs on January 27, 2007 2:55 PM | Permalink | Reply to this

### Re: Clarification

More generally, we can think of $\mathrm{AUT}(F)$, for any $F$, as sitting inside $F$-bitorsors $\mathrm{AUT}(F) \subset F\mathrm{BiTor} \,,$ with each automorphism defining a twisted bitorsor action of $F$ on itself.

(In fact, this inclusion is an equivalence, at least at the level of categories internal to sets.)

So we may think of a pseudofunctor $\Sigma(B) \to \Sigma(\mathrm{AUT}(F) )$ as an enrichment of $B$ by $F$-bitorsors, in the sense that an enriched category is precisely a lax functor on its bare morphisms.

That peudofunctor may be read as assigning to each element $b \in B$ the $F$-extension living over it. The compositor of the psuedofunctor gives the composition law on these.

And if $Y^{[2]}$ is a transition groupoid of a cover of a space $X$, then we have, in analogy to the line bundle gerbe example above, the statement that $F$-extensions $Y^{[2]} \to \Sigma(\mathrm{AUT}(F))$ are precisely (Aschieri-Jurčo-) bibundle gerbes on $X$.

Posted by: urs on January 28, 2007 12:07 PM | Permalink | Reply to this

### Re: Cocycle Category

For those readers that are homological algebra quasi-ignorants like me (and of course for my own benefit), I just note that the proof of the end of the Wikipedia entry on the extension problem, here, provides some helpful details about the relation between the $\mathrm{Ext}$-group and the extensions it classifies.

Posted by: urs on January 25, 2007 5:44 PM | Permalink | Reply to this

### Re: Cocycle Category

Urs wrote:

For those readers that are homological algebra quasi-ignorants like me (and of course for my own benefit), I just note that the proof of the end of the Wikipedia entry on the extension problem, here, provides some helpful details about the relation between the $\mathrm{Ext}$-group and the extensions it classifies.

Alas, Wikipedia is a great source of information about mathematics iff you know enough to spot when it’s wrong. The article on the extension problem is full of nice information about classifying central extensions — but it uses the term $Ext^1(H,K)$ in a completely inaccurate way!

Contrary to what this article says, $Ext^1(H,K)$ does not classify central extensions of $H$ by $K$. It classifies abelian extensions of $H$ by $K$. Central extensions are classified by the ‘group cohomology’ $H^2(H,K)$.

I will try to clean up this article in a while. In the meantime, I urge folks to learn homological algebra from reputable sources, especially:

• Mac Lane, Homology
• Cartan and Eilenberg, Homological Algebra
• Rotman, An Introduction to Homological Algebra
• Weibel, An Introduction to Homological Algebra

These are all great! Weibel’s is the most modern, and broad in scope, while Rotman’s is the easiest way to learn (say) the difference between $Ext^n$ and the group cohomology $H^n$, and what they’re good for.

In case anyone is wondering, I did check with someone to see if I’d gone off the deep end before asserting that the Wikipedia is wrong here. They wrote (modulo some changes of notation to make it easy to compare this with the Wikipedia article):

To be generous, the author was probably remembering something about $Ext^1$ classifying short exact sequences of abelian groups and applying it to this situation. But no, you’re right. To be less generous, if we’re considering central extensions of the form $1 \to K \to G \to H \to 1$ then we may think of $H$ as acting trivially on the abelian group $K$, and in this situation central extensions are classified by second group cohomology $H^2(H, K)$. If $H$ happens to be abelian, we do have a monomorphism $Ext^1(H, K) \to H^2(H, K)$, where the $Ext^1$ corresponds to extensions where $G$ is abelian.

Speaking of mistakes, these remarks also point out an embarrassing mistake in my previous long comment on this subject, which I will now fix.

Posted by: John Baez on January 26, 2007 7:30 PM | Permalink | Reply to this

### Re: Cocycle Category

Wouldn’t the statement
“There is precisely one diagram (as above) for every ordered pair of extensions”
actually be “just” the 5-lemma for sufficiently nice categories?

And since groups are being fixed to being abelian in the discussion, won’t we be at least in the vicinity of a category where the 5-lemma holds?

Posted by: Mikael Johansson on January 28, 2007 10:36 PM | Permalink | Reply to this

### Re: Cocycle Category

And I don’t think your observations are old hat — but I don’t think they’re brand new hat, either.

Whatever age this hat is, it is the kind of hat I enjoy seeing in our $n$-Café!

Posted by: urs on January 25, 2007 12:58 PM | Permalink | Reply to this

### Re: Cocycle Category

John wrote :

I forget the details now, but I think you’ve seen the same thing and are excited for the exact same reason! Or at least a very similar reason. Namely: even though he wasn’t thinking of it quite this way, Jardine was secretly tackling a certain aspect of higher gauge theory — and he was independently forced to use the same ‘anafunctor’ idea that Toby was forced to think about in his work on this subject.

Indeed… thanks for this explanation. This was precisely what I was most excited about: that the ideas of pages 1-3 of Rick Jardine’s paper on ‘Cocycle categories’ seem to be essentially the same ideas which leads you guys to talk about anafunctors when describing higher gauge theory.

Posted by: Bruce Bartlett on January 25, 2007 8:13 PM | Permalink | Reply to this

### Re: Cocycle Category

John wrote:

In fact, there’s a nice category $EXT(B,F)$ of such extensions. An object in here is an extension of $B$ by $F$, and a morphism is a homomorphism $E\to E'$ making the obvious diagram commute. (It’s difficult to draw here, but see my paper.)

Jacques wrote:

Y’mean:

(1)$\begin{matrix} 1\to& F &\to& E &\to& B & \to 1 \\ & \Vert & &\darr& & \Vert & \\ 1\to& F &\to& E' &\to& B &\to 1 \end{matrix}$

?

Yeah!

(In my paper I drew it a different way, with some diagonal-slanting arrows. But the diagram you just drew says the same thing. Nice.)

Since I don’t understand the general extension problem, let me specialize to $F$ abelian. In that case, we know that extensions (1) are classified by $\Gamma=\mathrm{Ext}^1(B,F)$.

Actually, only extensions with $E$ abelian are classified by $Ext^1(B,F)$. The thing I’m talking about classifies all extensions, even those where $E$ is not abelian.

It took me about 45 years to figure this stuff out, even though it’s not that complicated, so it’s probably worth saying a bit about it before I kick the bucket. There are three popular gadgets that classify three kinds of extensions of groups. These gadgets are special cases of three different general constructions, named in italics below:

• Ext groups. If $B$ and $F$ are abelian groups, $Ext^1(B,F)$ classifies abelian extensions of $B$ by $F$. In other words, elements of $Ext^1(B,F)$ are in 1-1 correspondence with isomorphism classes of extensions

$1 \to F \hookrightarrow E \stackrel{p}{\to} B \to 1$

where $E$ is abelian.

• Group cohomology. If $B$ is a group and $F$ is an abelian group, $H^2(B,F)$ classifies central extensions of $B$ by $F$. In other words, elements of $H^2(B,F)$ are in 1-1 correspondence with isomorphism classes of extensions

$1 \to F \hookrightarrow E \stackrel{p}{\to} B \to 1$

where $F$ is in the center of $B$.

• Nonabelian cohomology. If $B$ and $F$ are groups, $H^2_{nonab}(B,F)$ classifies extensions of $B$ by $F$. In other words, elements of $H^2_{nonab}(B,F)$ are in 1-1 correspondence with isomorphism classes of extensions

$1 \to F \hookrightarrow E \stackrel{p}{\to} B \to 1$

These three gadgets are in general different even when $B$ and $F$ are abelian. The first two are groups, while the last one is a mere set.

The last one is the set of isomorphism classes in the category of extensions of $B$ by $F$ which I was calling $EXT(B,F)$ in my previous comment. That notation was somewhat nonstandard, and may have helped throw you off. A more standard notation is something like $\mathbb{H}^2_{nonab}(B,F)$. This is supposed to remind us of $H^2_{nonab}(B,F)$ but look a bit scarier, since it’s a category instead of a set.

Posted by: John Baez on January 26, 2007 6:22 AM | Permalink | Reply to this

### Re: Cocycle Category

(In my paper I drew it a different way, with some diagonal-slanting arrows. But the diagram you just drew says the same thing. Nice.)

I can do the slanty thing, too. \begin{aligned} 1 \to F & \to & E & \to & B \to 1 \\ &\searr&\darr&\nearr& \\ & & E' & & \end{aligned} But I think my original diagram looks better. And it conveys the idea better (each row is an object and the vertical arrows — subject to the commutativity condition — are a morphism).

Anyway, from what you say, it doesn’t sound like I am going to understand the category $\mathbb{H}^2_{\text{nonab}}(B,F)$ by looking at particular subcases, like abelian extensions, or central extensions (or whatever).

Posted by: Jacques Distler on January 26, 2007 8:36 AM | Permalink | PGP Sig | Reply to this

### Re: Cocycle Category

Jacques Distler wrote:

I can do the slanty thing too.

Good! My slanty thing had reflection symmetry around the $x$ axis. But anyway, I’m not knocking your original rectangular thing. In fact, I’ll steal it in a minute…

Anyway, from what you say, it doesn’t sound like I am going to understand the category $\mathbb{H}^2_{\text{nonab}}(B,F)$ by looking at particular subcases, like abelian extensions, or central extensions (or whatever).

Actually one can understand a bunch about it from these cases. Take your favorite abelian group $A$ (say the real numbers) and consider the trivial extension

$0 \to A \to A \oplus A \to A \to 0$

This gives an object in the category of extensions of $A$ by $A$ — the category I was calling $EXT(A,A)$, which we are now calling $\mathbb{H}^2_{\text{nonab}}(A,A)$ in a desperate attempt to make it seem as intimidating as possible.

What are the morphisms from this object to itself? They are commuting diagrams

(1)$\begin{matrix} 0\to&A&\to&A\oplus A&\to&\A&\to 0\\ &\Vert& &\darr f& &\Vert&\\ 0 \to&A&\to&A\oplus A&\to&A&\to 0 \end{matrix}$

What can this $f$ be like? If you think about it, it’s just any triangular $2 \times 2$ matrix with entries in $A$! (Either upper or lower triangular, depending on your conventions.)

So, even boring extensions can have interesting endomorphisms. This information is recorded in the category of extensions, but lost in the set of isomorphism classes of extensions.

The beauty of nonabelian cohomology — or at least the portion we’re looking at here — is that the category of extensions of $B$ by $F$ (which can now be nonabelian groups) is none other than

$hom(B,AUT(F))$

where $AUT(F)$ is the 2-group of symmetries of $F$.

Posted by: John Baez on January 26, 2007 7:18 PM | Permalink | Reply to this

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