The n-Category Café

Urs wrote:

I am glad this article is finally seeing the day of light. I remember that Danny in particular had started thinking about this long ago.

Yes, Danny had this 3-lemma proof almost worked out at least a year ago. Then I got involved and wanted a ‘cleaner, more conceptual’ proof. We spent the spring of 2007 working on that proof… but it became clear that while certain experts on model categories could carry it out, we weren’t them.

So, when Danny came back to Riverside over Christmas, we decided it was time to switch back to his original strategy. It turned out a few details needed to be straightened out, involving nasty subtleties like the difference between locally trivial fiber bundles and fibrations (and how the former isn’t always the latter), and the difference between locally contractible topological groups, groups where the inclusion $1\hookrightarrow G$ is a closed cofibration (aka ‘NDR pair’) and groups where the inclusion $1\hookrightarrow G$ is a strong NDR pair. Technical things I never really wanted to think about! Luckily, in the end Peter May saved us — and in the end, everything worked out perfectly.

All this may sound a bit scary and unpleasant, but it’s actually not, in the end. Milgram, Segal and Steenrod noticed you could build a classfying space $BG$ for a topological group $G$ by taking the geometric realization of the nerve of $G$ (viewed as a topological category with one object). They also got a space $EG$ and a map $EG\to BG$. The question then arises:

When is $EG\to BG$ a fibration?

and the nicest known answer seems to be

When $1\hookrightarrow G$ is a closed cofibration!

This should be in some textbook somewhere, but we found it in some classic papers by Peter May, in somewhat concealed form.

As a general comment, I would like to better understand the relation between

- the “standard” transformations between ana 2-functors $X\stackrel{\simeq }{<\leftarrow }Y\stackrel{g}{\to }BG$ which you use, and which follows the principle summarized in Morphisms of anafunctors

- and the “concordance” point of view on transformations between anafunctors, which Nils Baas, Tore Kro and M. Böksted use in studying classifying spaces of 2-categories.

Notice that concordance is designed exactly such as to deal elegantly with the somewhat subtle issue in transformations of anafunctors: the case where you have a transformation between two of them which are not defined on the same cover $Y$.

Yes, this subtle issue is also what my ‘cleaner, more conceptual’ proof was supposed to tackle head-on… but in a superficially different way than you suggest, using model categories rather than higher categories:

In the entry Concordance (pdf) I observe that there should be a nice relation between the two points of view, both being related by the Hom-adjunction in $\omega$-categories (if we restrict attention to strict $\mathrm{\infty }$-groups for simplicity and for the moment).

In the model category approach, I believe this adjunction should manifest itself as a ‘Quillen equivalence’: a specially nice adjunction between model categories. I’ve gotten an email from a young homotopy theorist who thinks he can work out the technical issues involved, so we’ll see what happens.

I haven’t got any feedback on this contribution so far, which probably just means that it’s not relevant.

It’s probably very relevant! Alas, every time you’ve told me about this, my internal reaction was aargh, Danny and I should just finish writing this paper… I didn’t want to think about yet another ‘cleaner, more conceptual’ strategy for solving a problem, when we had one solution almost ready to write up.

One more thing:

Having criticized it a little, I should add a few words in defense of Danny’s 3-lemma strategy for proving this theorem:

Theorem. Suppose that $G$ is a well-pointed topological 2-group and $M$ is a paracompact Hausdorff space admitting good covers. Then there is a bijection $${\check{H}}^1(M,G)\cong [M,B\mid G\mid ]$$ where the topological group $\mid G\mid$ is the geometric realization of the nerve of the topological groupoid $G$.

The main thing I like is Lemma 1, which gives a very practical concrete description of $\mid G\mid$ in terms of the crossed module $(G,H,t,\alpha )$ corresponding to the 2-group $G$. Namely:

$$\mid G\mid \cong (G⋉EH)/H$$

Segal showed that $EH$ is a topological group. $G$ acts on $H$ via $\alpha$, so it acts on this group $EH$. This lets us define the semidirect product $G⋉EH$. There’s a way to see $H$ as a normal subgroup of $G⋉EH$, via $h\mapsto (t(h),h^{-1})$. So, we can take the quotient $(G⋉EH)/H$, and that turns out to be $\mid G\mid$.

If you remember, Danny was using some very similar tricks already in our paper ‘From Loop Groups to 2-Groups’.

Note: This business of e.g. ‘without the Pontryagin class’ follows from a classic Gysin sequence argument, using crucially that the extension is by S¹.

For other groups, life will be much more interesting.

the Lie $\mathrm{\infty }$-algebraic Cartan-Ehresmann kind of way to say that String bundles have the same characteristic classes as the underlying $G$-bundles but without the Pontrjagin class corresponds to theorem 3, p. 15.

There is now a more pronounced statement of this, figure 8 on p. 48.

“just as a group is a set with multiplication and inverses, a 2-group is a category with multiplication and inverses.”

A “set” is a 0-category, and a “category” is a 1-category, so wouldn’t it be more logical to call a group, a “0-group”, and a 2-group a “1-group”?

Also, if you say that a category is a set with morphisms, a 2-category is a category with 2-morphisms between morphisms, a 3-category is a 2-category with 3-morphisms between 2-morphisms, etc, wouldn’t it be more logical to define “2-group” as follows? A category is a set with morphisms between elements. A group is a set with binary operations between elements (with the identity and inverses). Just as a 2-category is a category with 2-morphisms between morphisms, a 2-group would be a group with 2-binary operations between binary operations. In other words, let’s say you had a group with several binary operations, and it was a group under each of those binary operations. Then let’s say you could perform a 2-binary operation on two of those binary operations, and get one of the other binary operations. In fact these these binary operations would themselves form a group, where the various binary operations would be elements of the group, which would then be a group under the 2-binary operation.

Jeffery winkler

Non-abelian cohomology can also be defined via a generalization of abelian gerbes (in terms of 2-categories). This is explained in this paper about a tower of n-gerbes by A. Tsemo.

I haven’t absorbed that article yet. But I know the following way to characterize nonabelian cohomology by collections of higher abelian cohomologies:

Given a classifying space $\mid G\mid$ for some $n$-group $G$, we can consider for each ordinary degree $k$ cohomology class of $\mid G\mid$ the $(k-2)$-gerbe = $(k-1)$-line bundle classified by it.

Hence, given any classifying map $$X\to \mid G\mid$$ we can not only pull back the universal (in general non-abelian) $G$-bundle to $X$ and get a class in nonabelian cohomology, but we can also pull back all these higher abelian line bundles. Their classes know all about the characteristic classes of the nonabelian $G$-bundle. That characterizes the nonabelian cocycle to a great extent (though in general not entirely).

This is proposition 35 on p. 64 in the paragraph Line $n$-bundles on classifying spaces here.

I can’t tell yet if that is at all related to what the paper you pointed to is addressing, but the term “tower of $n$-gerbes” vaguely reminded me of this.

Sure
Never seen it said that way but
t:H –> G
determines the k-invariant K(H,n+1)–>K(G,n+1)

David wrote:

Is there a notion of Eilenberg-MacLane spaces for 2-groups? So $K(G,n)$ has homotopy only in dimensions $n$ and $n+1$ given by the 2-group $G$, and is unique in some suitable sense.

This works, but you have to be a bit careful. You’ll like this!

First let’s warm up a bit:

$K(G\mathrm{,1})$ makes sense for any group $G$.

$K(G\mathrm{,2})$, $K(G\mathrm{,3})$ etcetera make sense when $G$ is abelian.

Backing up, $K(G\mathrm{,0})$ makes sense whenever $G$ is a mere set!

So: set, group, abelian group, abelian group… what we’re seeing here is the first column of the periodic table!

A $k$-tuply groupal $n$-groupoid is an $(n+k)$-groupoid with only one $j$-morphism for $j$ less than $k$. The periodic table shows the pattern:

...............................................
               k-tuply groupal n-groupoids 



             n = 0       n = 1          n = 2



k = 0        sets        groupoids      2-groupoids
     



k = 1        groups      2-groups       3-groups



k = 2        abelian     braided        braided
             groups      2-groups       3-groups



k = 3        "     "     symmetric      sylleptic
                         2-groups       3-groups



k = 4        "     "     "      "       symmetric
                                        3-groups



k = 5        "     "     "      "       "      "    
....................................................

According to the homotopy hypothesis, $k$-tuply groupal $n$-groupoids are secretly the same as spaces with only ${\pi }_k,\dots ,{\pi }_{n+k}$ nontrivial.

Now you’re asking about the second column. Let $G$ be a 2-group. According to the homotopy hypothesis, a 2-group is secretly a space with only ${\pi }_1$ and ${\pi }_2$ nontrivial. You’re calling this space $K(G\mathrm{,2})$.

What about $K(G\mathrm{,3})$? According to the periodic table, this should make sense when $G$ is a braided 2-group. And, it should be a space with only ${\pi }_2$ and ${\pi }_3$ nontrivial.

Indeed, that’s right: braided 2-groups classify spaces with only ${\pi }_2$ and ${\pi }_3$ nontrivial. A proof can be found lurking rather deeply here.

But here’s the interesting subtlety, which you will surely enjoy. For a group to be abelian is a mere property. But, for a 2-group to be braided is an extra structure.

So, you can’t get $K(G\mathrm{,3})$ unless you equip your 2-group $G$ with a braiding.

I leave the case of $K(G\mathrm{,4})$ as an exercise.

It takes work to prove some of these things, but the pattern is simple and beautiful! It could be taught in high school.