## September 12, 2007

### Obstructions for n-Bundle Lifts

#### Posted by Urs Schreiber

I am in the process of preparing some slides which are supposed to eventually provide an overview of the current state of the art of the second edge, with an emphasis on understanding String and Chern-Simons Lie $n$-algebras, the Lie $n$-algebra cohomology and the corresponding theory of bundles with Lie $n$-algebra connection in terms of the differential Lie analog of parallel transport $n$-functors, like String 2-bundles and Chern-Simons 3-bundles.

Here a Chern-Simons 3-bundle associated to a $G$-bundle should be the obstruction to lifting the $G$-bundle to a $\mathrm{String}(G)$ 2-bundle.

I know how to say this in a pedestrian way. But now I want to say it in the most elegant possible way. I want to understand how it really™ works.

I am too tired to do the topic justice. But all the more grateful I’d be for comments on the following considerations.

2-bundles measuring the lift of a 1-bundle through a central extension

Before formulating the general setup, consider the simple example of a lifting line 2-bundle.

Let $U(1) \to \hat H \to H$ be an exact sequence of groups, hence a central extension $\hat H$ of the group $H$.

Given an $H$-bundle, we may try to lift it to a $\hat H$-bundle.

The way to do this is to first of all notice that an $H$ 1-bundle is the same as a $(1 \to H)$-2-bundle.

Here I write $(G_1 \to G_0)$ for the strict 2-group coming from a crossed module of two groups $G_1$ and $G_0$.

And then to notice that the crossed module $(U(1) \to \hat H)$ is equivalent to $(1 \to H)$: $(U(1) \to \hat H) \simeq (1 \to H) \,.$ The $U(1)$ worth of automorphisms of every object in $(U(1) \to \hat H)$ exactly cancel the central part of the group of objects $\hat H$.

Therefore, every ordinary $H$-bundle is automatically a $(U(1) \to \hat H)$ 2-bundle: $H \stackrel{=}{\to} (1 \to H) \stackrel{\simeq}{\to} (U(1) \to \hat H) \,.$

More concretely, when we hit the 1-cocylce

$\array{ && \bullet \\ & {}^{\pi^*_{12} g}\nearrow && \searrow^{\pi^*_{23}g} \\ \bullet &&\stackrel{\pi^*_{13} g}{\to}&& \bullet }$

of the $H$-bundle – which is a diagram in $\Sigma H$, hence a diagram in $\Sigma (1 \to H)$ – with any of the equivalences

$(1 \to H) \stackrel{\simeq}{\to} (U(1) \to \hat H)$

(and for the cognoscenti: I am simply glossing over all smoothness or structure-ness issues for the moment)

we obtain the 2-cocycle (this “2” is my counting: some people count differently here – doesn’t matter)

$\array{ && \bullet \\ & {}^{\pi^*_{12} \hat g}\nearrow &\Downarrow^{\pi^* h}& \searrow^{\pi^*_{23}\hat g} \\ \bullet &&\stackrel{\pi^*_{13} \hat g}{\to}&& \bullet }$

(imagine this making the obvious tetrahedron 2-commute, which I won’t draw right now).

So up to this point we have simply puffed up the available 0-categorical (or 1-categorical, depending on how you count) information of an ordinary $H$-bundle to an equivalent 1-categorical (or, 2-categorical, respectively) piece of data.

The point being that the addition wiggle room we obtain this way allows some nifty moves.

Namely, what we are really interested in is whether the original $H$-cocycle can be lifted to a $\hat H$-cocycle. Now this is equivalent to asking if our $(U(1) \to \hat H)$-cocycle can be reduced to a $(1 \to \hat H)$ cocycle: does it factor through

$(1 \to \hat H) \hookrightarrow (U(1) \to \hat H)$ ?

That’s what we would like to know. To check if it does, we should find a measure of how this might fail.

In the simple case we are dealing with here, it is easy to construct this measure “by hand”: clearly, we should simply check if the image of our 2-cocycle under the projection

$(U(1) \to \hat H) \to (U(1) \to 1)$

is a trivializable $(U(1) \to 1)$ 2-cocycle or not. When this is trivializable, we may gauge it away and thereby indeed reduce our $(U(1) \to \hat H)$ 2-coycle to the desired $(1 \to \hat H)$ 2-cocycle. If not, the class of this $(U(1) \to 1)$ 2-cocycle is a good measure for the failure of the original $H$-bundle to lift to a $\hat H$-bundle.

And this class is of course nothing but the class of any line 2-bundle (which you may perhaps better know in its incarnation as a line bundle gerbe) having that 2-cocycle.

So, this way a line 2-bundle is the obstruction to lifting a principal bundle through a central extension of its structure group.

Fine. But how does this generalize? What is the power tool that achieves in general what we could achieve using a hand drill and simply guessing that we should look at the projection $(U(1) \to \hat H) \to (U(1) \to 1)$ ?

Clearly, since in general $(G_1 \to 1)$ doesn’t even exist (unless $G_1$ is abelian), the naive guess here is wrong.

general setup

I guess it’s clear that what is really going on is that we form a quotient:

$\array{ (1 \to \hat H) \\ \downarrow \\ (U(1) \to \hat H) &\simeq & (1 \to H) \\ \downarrow \\ (U(1) \to 1) &=& (U(1) \to \hat H)/(1 \to \hat H) }$

in a suitable sense.

And this “suitable” is where, I think, it gets tricky. Once we want to formulate this very generally.

Eventually, I would like to have a statement applicable to general $n$-groups. Like this:

For $K \to G \to B$ an exact sequence of $n$-groups, there is an $(n+1)$-group $(K \to G)$ which is equivalent to $B$ $(K \to G) \simeq (1 \to B) \,.$

Then we can form the quotient $(n+1)$-group (a cokernel of sorts) $\array{ (1 \to G) \\ \downarrow \\ (K \to G) &\simeq & (1 \to B) \\ \downarrow \\ \hat B_K &:=& (K \to G)/(1 \to G) }$

such that the obstruction to lifting a $B$-$n$-bundle through $G \to B$ is a $\hat B_k$-$(n+1)$-bundle.

(Here I haven’t said which construction the notation $(K \to G)$ means in detail. At the moment this is just notation for some $(n+1)$-group. But I expect that this needs to be read as the mapping cone of $K \to G$. (Compare definition 4 for the mapping cone of a morphism of two strict 2-groups.))

special case: inner automorphism $(n+1)$-groups

As a fun special case, whose understanding probably helps understanding the general situation, consider sequences of $n$-groups whose left morphism is the identity:

$G_{(n)} \stackrel{\mathrm{Id}}{\to} G_{(n)} \to 1 \,,$

where 1 is the trivial $n$-group.

Here clearly $(G_{(n)} \stackrel{\mathrm{Id}}{\to} G_{(n)})$ is precisely the mapping cone, hence, according to the considerations in the inner automorphism 3-group of a strict 2-group, we take $(G_{(n)} \to G_{(n)}) := \mathrm{INN}_0(G_{(n)}) \,.$

Indeed, this is equivalent to the trivial $n$-group $\mathrm{INN}_0(G_{(n)}) \simeq 1$

by general properties of tangent categories.

If $G_{(n)}$ is “sufficiently abelian”, we do know what the cokernel of $\array{ G_{(n)} \\ \downarrow \\ \mathrm{INN}_0(G_{(n)}) }$ should be: $\Sigma G_{(n)}$.

(If $G_{(n)}$ is not “sufficiently abelian”, I suspect we find the quotient has to be an $n$-group for potentially very high $n$: related to the highest degree of $\mathrm{Lie}(G_{(n)})$-invariant polynomials.To see what I am thinking of here, you might go to the part titled The algebra of $g_{(n)}$-invariant polynomials in those slides. This is a subtle aspect which I would dearly appreciate comments on.)

Given that, as I kept emphasizing, the $(n+1)$-curvature of a $G_{(n)}$ $n$-transport takes values in $\mathrm{INN}(G_{(n)})$, we seem to be able to interpret this, in the present context, in terms of the following somewhat entertaining slogan:

The $(n+1)$-curvature of an $n$-transport is the lifting $(n+1)$-bundle with connection of the lift of an $n$-bundle with trivial structure group to $G_{(n)}$.

I need to think more about how to make this precise. But this looks like it could actually be a very useful way to look at the construction illustrated by that “diagram movie with subtitles” in the part “$n$-categorical background” of those slides.

finally: Lie $n$-algebra

I am thinking that the really™ good way to handle Chern-Simons 3-bundles along the lines of the second edge as described in those slides is to pretty much follow the above construction, with Lie $n$-algebras instead of Lie $n$-groups. Then by expressing these Lie $n$-algebras dually in terms of their dual quasi-free differential graded commutative algebras (qDGCAs), the $(K \stackrel{t}{\to} G)$-construction should be precisely the mapping cone construction of the corresponding $t^*$ on the level of qDGCAs.

I had energy to make some quick consistency checks that this indeed gives the right answer. For instance when applied to the exact sequence

$\mathrm{Lie}(\Sigma U(1)) \to g_{\mu_k} \to g$

for $g$ an ordinary Lie algebra and $g_{\mu_k}$ the corresponding Baez-Crans type String Lie 2-algebra, it seems that the mapping cone construction indeed correctly eliminates the vanishing of the first Pontryagin class in the collection of characteristic classes of the corresponding $g_{(n)}$-bundles. That’s exactly what should happen for the Chern-Simons 3-bundle which measure the obstruction of the lift from a $g$-bundle to a $g_\mu$-bundle to exist, which is nothing but this Pontryagin class.

But I need to spell this out in more detail after I have recharged my batteries.

Meanwhile, all comments are very welcome.

Posted at September 12, 2007 10:01 PM UTC

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### Re: Obstructions for n-bundle lifts

I believe I made some progress with understanding this issue, at the level of Lie $n$-algebras.

In the file

Obstructions and cokernels of Lie n-algebra morphisms

I describe:

- how the (strict) cokernel of a morphism of semistrict Lie $n$-algebras $f_{(n)} \to g_{(n)}$ looks like; how it is in general a Lie $n'$-algebra for $n' \gt n$

- how the cokernel of the canonical embedding $g_{(n)} \to \mathrm{inn}(g_{(n)})$ is, as a special case of a general statement, indeed the Lie $n'$-algebra which I have started calling $b g_{(n)}$, characterized by the fact that its dual is generated from all the classes of invariant $g_{(n)}$ polynomials

- how in special cases the cokernel of morphisms $f_n \to g_n$ of Lie $n$-algebras is itself a Lie $n$-algebra and then equivalent to the mapping cone Lie $(n+1)$-algebra $(f_n \to g_n)$

- how all this can be used to determine the codomain of obstructions to extensions of structure Lie $n$-algebras of $g_{(n)}$-bundles .

I list a couple of examples illustrating this, containing also the Lie-version of the lifting gerbe construction described in the entry above.

The discussion of the Chern-Simons 3-bundles as obstruction to lifts through the string extension $\mathrm{Lie}(\Sigma U(1)) \to g_\mu \to g$ is then a straightforward generalization of this discsussion, left as an exercise (to both, reader and author ;-).

Posted by: Urs Schreiber on September 13, 2007 5:32 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

This stuff sounds nice. I think homotopy theorists all agree that the mapping cone of a map between chain complexes

$F: X \to Y$

is the nice homotopy-invariant version of the concept of ‘cokernel’. In $n$-category jargon, you’d call it the ‘weak cokernel’.

In the ordinary cokernel, we take guys in $Y$ and set them equal to zero if they’re in the image of $F$. In the mapping cone, we make them equivalent to zero — that is, they become $d$ of something.

You can see this easily if you keep in mind the ‘witch’s hat’ picture of a mapping cone of a map between pointed spaces

$F : X \to Y$

Here, for each guy in the image of $F$, we sew in a path to the basepoint, which plays the role of zero.

So, if you’re comparing the ordinary cokernel of $F$ to its mapping cone, and asking when they’re equivalent, you’re asking a very popular sort of question in homotopy theory: when is a ‘weak’ or ‘homotopy-invariant’ construction equivalent to its simpler ‘strict’ version. And, I think the popular answer would be: when $F$ is a cofibration.

I’ve thought more about the dual question, where you have a map $F: X \to Y$ and you ask when the ‘strict’ fibers $F^{-1}(y)$ are equivalent to the ‘homotopy fibers’. The popular answer here is: when $F$ is a fibration.

So, it sounds like you’re saying that “in special cases” a morphism of Lie $n$-algebras is a cofibration.

There’s probably a model category structure on Lie $\infty$-algebras that would make this precise….

By the way: Urs will remember what I mean by ‘witch’s hat’, but I should explain for everyone else, so you don’t all think I’m insane. Often mathematicians draw spaces $X$ and $Y$ as circular blobs, and a map $F: X \to Y$ as a bunch of parallel arrows going down from $X$ to $Y$. Then the mapping cylinder of $F$ looks like the stove-pipe hat Abe Lincoln wore. Even better, the retraction from the mapping cylinder down to $X$ is just the process of popping down the hat! (Some top hats were collapsible.)

On the other hand, the mapping cone of $F$ looks like the hat a witch wears! Every point in the image of $X$ is connected by a path to the tip of the hat.

I don’t know if the map from the mapping cylinder to the mapping cone can be told as a fun story in which Abe Lincoln suddenly transforms into a witch.

Trivia question: who was the last president to wear a top hat to his inauguration?

Posted by: John Baez on September 14, 2007 10:15 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Thanks for these comments! Great. Just what I needed.

In $n$-category jargon, you’d call it the ‘weak cokernel’.

All right, let me see. In my notes, def 3 and prop 4, I consider the strict cokernel (actually I consider the kernel of the dual, which amounts to the same thing) in the sense that it is the 1-categorical cokernel, i.e. the corresponding universality diagram with no 2-morphisms and higher around.

So, there is a 1-category of Lie $\infty$-algebras and part of the fun (and, as it turns out, important in applications) insight here is that the strict cokernel – in this sense – of a morphism of Lie $n$-algebras may be a Lie $n'$-algebra for $n'$ much larger than $n$. Possibly infinite, even.

This, for instance, translates into the following statement about bundles:

a line bundle has a single characteristic class, with real image being a 2-form. Hence a line bundle trivializes a line 2-bundle.

but a general principal bundle has an entire tower of characteristic classes. Each of them a $d$-form which may be thought of as the $d$-form connection of a trival $d$-bundle.

The reason is that, the cokernel of the injection $G \hookrightarrow EG$ to be read as $G \hookrightarrow \mathrm{INN}(G)$ in the world of Lie $n$-algebras $g \hookrightarrow \mathrm{inn}(g)$ namely what I call

$g \stackrel{i}{\hookrightarrow} \mathrm{inn}(g) \to \mathrm{coker}(i) := b g$

is a Lie 2-algebra (like $\mathrm{inn}(g)$) only if $g$ is abelian. Otherwise it is a Lie $n$-algebra with $n$ being the degree of the highest degree class of generators of the algebra of invariant polynomials of $g$.

(A few weeks ago I would have been startled by a statement like this. The fun thing is how simple and hands-on this stuff becomes when using the $\mathrm{inn}(\cdot)$-technology paired with the passage to dual differential algebras.)

Anyway. At this point I did certainly wonder what would have happened had I tried to determine the weakened universality diagram for the cokernel of morphisms of Lie $n$-algebras.

Things become subtle here, because of the elusive nature of the details of higher morphisms of Lie $n$-algebras. But, if what I wrote in ATDT: Higher Morphisms is on the right track, then I actually do have a handle on this.

So, I could in principle try to think about weak colimits (though, if I recall correctly, there are a bunch of different flavours of weak colimits(?)), where the usual conditions hold only up to higher coherent equivalence (actually, in the $(\infty,1)$-category of Lie $n$-algebras all $n \geq 2$-morphisms should be strictly invertible).

So my question (boldface here, since I am once again writing too much and becoming afraid that my question gets lost among all my rambling): when you say weak cokernel above, would it be this kind of weak universal construction?

Urs will remember what I mean by “witch’s hat”

Yes, I do, with gratification. A very good picture to keep in mind.

(The other nice thing you taught me that day in the Café was how cohomology is (hope I am getting this right) about strict free $\omega$-categories, while homotopy groups are about weakly free $\omega$-categories (or whatever the right term is). You should explain that great way to understand why homotopy is hard while cohomology is easy one day in a TWF (unless you already have…?).)

Posted by: Urs Schreiber on September 14, 2007 10:52 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

This seems to call attention to the fact that n-Lie algebras are special L_\infty-algebras and it’s the latter that form a nice category.

Compare: H-spaces and H-maps do NOT form a category or at least not a nice one.

Posted by: jim stasheff on September 14, 2007 11:58 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

This seems to call attention to the fact that $n$-Lie algebras are special $L_\infty$-algebras and it’s the latter that form a nice category.

Yes.

By the way: I keep saying Lie $n$-algebra instead of $n$-Lie algebra.

One reason: John and Alissa called them this way!

The other reason: Lie $n$-algebras are monoidal $n$-categories with extra structure and properties. But monoidal $n$-categories are $n$-algebras. And the extra structure and property is the Lie structure and property. So from this point of view we should speak of Lie $n$-algebras.

But I guess one could also find support for a point of view which would justify the term “$n$-Lie algebra”.

Posted by: Urs Schreiber on September 15, 2007 11:48 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Let’s keep n-Lie for the case of an n-ary bracket and no others. There exists significant literature.

Posted by: jim stasheff on September 16, 2007 3:34 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Jim accepted the term ‘Lie $n$-algebra’ a long time ago, precisely because ‘$n$-Lie algebra’ means something else. He just slipped this time.

I like ‘Lie $n$-algebra’ because there are also ‘associative $n$-algebras’ and so on. Unfortunately I went along with Kapranov and Voevodsky on their ‘2-vector space’ terminology instead of talking about a ‘vector 2-space’, which really makes more sense, at least for ‘Baez–Crans vector 2-spaces’, which are a special case of smooth 2-spaces, just as vector spaces are a special case of smooth spaces.

Posted by: John Baez on September 21, 2007 4:27 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Urs wrote:

So my question (boldface here, since I am once again writing too much and becoming afraid that my question gets lost among all my rambling): when you say weak cokernel above, would it be this kind of weak universal construction?

Well, let’s see. Let me talk about chain complexes at first, since I understand them better than $L_\infty$-algebras, and the basic issues should be similar.

For me, the ‘weak cokernel’ of a map between chain complexes

$f : X \to Y$

is nothing but the mapping cone of $f$. For some reason the mapping cone is clearly the right concept — a thousand algebraic topologists cannot be wrong! If we accept this, the question of what universal property the weak cokernel has becomes something we can investigate as follows.

First, we recall what the mapping cone is.

As a graded vector space, the mapping cone $C f$ is the direct sum of $Y$ with a shifted-up version of $X$:

$C f = X[1] \oplus Y$ But, we make it into a chain complex by equipping it with a differential like this:

$\partial (x,y) = (\partial(x), f(x) - \partial y )$

If I could draw pictures more easily here, I’d show everyone how I remember this formula simply by pondering a picture of a witch’s hat! But alas…

Anyway, the mapping cone should have some sort of universal property like this:

1. there are maps of chain complexes:

$X \stackrel{f}{\to} Y \stackrel{i}{\to} C f$
and while it’s not true that $i \circ f = 0$, there’s a chain homotopy from $i \circ f$ to zero.
2. $C f$ is universal with respect to property 1. That is, if we have any other setup

$X \stackrel{f}{\to} Y \stackrel{j}{\to} Z$
with property 1, we get a map from $C f$ to $Z$ making a bunch of diagrams commute.

I said ‘commute’, but perhaps I meant ‘commute up to chain homotopies’? I’m too lazy to check it, but the point is: one can simply work it out and see! If the necessary diagrams really commute on the nose, we have a strict universal property; if they commute up to chain homotopy we have a weak one. In the latter case, you may wish to determine whether these chain homotopies are themselves unique (at least up to further chain homotopies).

If the universal property of the mapping cone is ‘stricter than you might expect’, with some equations where you might only expect chain homotopies, don’t be too upset. There’s a strange tendency in algebraic topology for certain constructions to work ‘more strictly than you might expect’. For example, the product of spaces is a product ‘on the nose’, not just up to homotopy, and similarly for the coproduct. I’m sure there’s a general theory of when this unexpected strictness happens, but I’m not sure anyone knows this theory yet — I sure don’t. This is why I’m willing to accept the apparent fact that $C f$ is the ‘right’ concept of weak cokernel, and use this to deduce what properties the weak cokernel should have.

I bet most of what I said applies to $L_\infty$-algebras too, but Jim Stasheff might know better.

Posted by: John Baez on September 21, 2007 4:28 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

As John well knows, in homotopy theory proper the Eckmann-Hilton dual concept is the homotopy fiber so in your language the ?weak kernel? I’m in favor of less ambiguity in terminology os I much prefer homotopy cofiber instead of weka cokernel.
It does satisfy a universal property up to homotopy.

As for $L_\infty$-algebras as for $A_\infty$ algebras, the issue is in what category the discussion takes place. The easiest way to punt is to define these structures in terms of the universal constructions and state the homotopy uniqueness at that level.

Posted by: jim stasheff on September 22, 2007 1:38 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

You meant from the mapping cylinder to the witches hat.

Coincidence: In work with Herman Gluck and Fred Cohen, the Barrtatt-Puppe sequence of homotopy sets has come up. the witches hat plays a key role.

Posted by: jim stasheff on September 14, 2007 11:55 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

I wrote:

I don’t know if the map from the mapping cone to the mapping cylinder can be told as a fun story in which Abe Lincoln suddenly transforms into a witch.

Jim wrote:

You meant from the mapping cylinder to the witch’s hat.

You’re right, sorry. To avoid confusing people, I’m going to doff my wizard’s hat, turn on my super-powers, and rewrite history: I’ll fix that sentence in my comment.

Posted by: John Baez on September 21, 2007 4:32 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Trivia question: who was the last president to wear a top hat to his inauguration?

For some reason I want to say Franklin D. Roosevelt.

I think I remember correctly that Kennedy was the first American president not to wear a hat, top or otherwise, to his inauguration (which at the time was considered the death knell for the hat industry), and I believe no president has since. Somehow I just don’t picture Truman in a top hat – he just seems too earthy for that. I’m guessing the same for Eisenhower for vaguely similar reasons.

(Huh – I sound just like a contestant on Who Wants to be a Millionaire? “Final?” “No, wait Regis, I’d like to call my friend Jim Dolan in California.” Except that Jim would probably use up the minute without giving a definite answer.)

Posted by: Todd Trimble on September 15, 2007 3:55 AM | Permalink | Reply to this

### Last President to Wear a Top Hat at His Inauguration

According to Wikipedia:

The last American president to wear a top hat to an inauguration was Richard Nixon. Gerald Ford was not inaugurated at the Capitol and Jimmy Carter abolished the use of morning dress for inaugurations. It was reinstated, minus a top hat, by Ronald Reagan but not worn by any later presidents to date.

Later, Nixon loosened up:

Posted by: John Baez on September 21, 2007 3:39 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

In case anyone actually looks at the details of Obstructions and cokernels of Lie n-algebra morphisms: there are a couple of typos and things in need of fixing, towards the end.

For instance the description of $\mathrm{ker}(t^*)$ is not really good. I should just say:

The graded commutative algebra underlying the strict qDGCA kernel of a morphism $f_n^* \stackrel{t^*}{\leftarrow} g_n^*$ of qDGCAs is $d_{g_n}^{-1} (\wedge^\bullet ( \mathrm{ker}_{\mathrm{Vect}}(t^*) ) )|_{\mathrm{ker}_{\mathrm{Vect}}(t^*)} \,.$

Then, what deserves more discussion is how the very point where all the interesting information concerning obstructions to extensions enters is in the morphism

$\mathrm{ker}(t^*) \leftarrow (f_n^* \stackrel{t^*}{\leftarrow} g_n^*)$

This picks up, in degree 2 (which are the elements dual to the compositor of the dual pseudofunctor), precisely the failures of the lift.

When I find the time I’ll provide a more detailed discussion.

Posted by: Urs Schreiber on September 15, 2007 11:43 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Just a simplicial thought. I have not yet checked back through all the preceding parts of this but if you look at things simplicially, then the lifting problem is fairly well represented by the Puppe sequence of a map (especially if it is a fibration). Breen’s old (1994) results on Puppe sequences of classifying spaces of simplicial groups give a nice set of useful tricks in the bottom dimensions, i.e. for 2-groups.

I had an old project/research proposal in which I looked at some of the tools of rational homotopy theory (dga, dg Lie algbras and their dash/ homotopy coherent analogues, as trial coefficients for TQFT type constructions, with similar motivation to yours here (but with less clear diff. geom linkage (this was nearly 10 years ago)(. (Unfortunately the proposal was not only not funded but was vicously sneered at by a referee!! Heigh ho! c’est la vie!)

There was a clear link with even older material on obstructions to traingulations and smoothing and the Haupvermutung work (Kirby and Siebenmann etc, where the key to the results is the fact that the fibre of a map of classifying spaces has non trivial \pi_2. I wonder if resurrecting some of those old results might give a set of tools for comparing discrete and differential models for space-time.

But that is getting off subject!

Posted by: Tim Porter on September 15, 2007 12:50 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

The Barratt-Puppe sequence is for a map and it mapping cone. The Hilton-Eckmann dual’ for the homotopy fibre is due to Nomura:

MR0132545 (24 #A2385)
Nomura, Yasutoshi
On mapping sequences.
Nagoya Math. J. 17 1960 111–145.
55.40

Eckmann-Hilton dual’ to Barrett-Puppe
i.e. for [W, -]
rather than [-, W]
and the homotopy fibre rather than the mapping cone

Posted by: jim stasheff on September 16, 2007 3:39 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Tim writes:

Breen’s old (1994) results on Puppe sequences of classifying spaces of simplicial groups give a nice set of useful tricks in the bottom dimensions, i.e. for 2-groups.

Jim: reference or precise coordinates please!

I had an old project/research proposal in which I looked at some of the tools of rational homotopy theory (dga, dg Lie algbras and their dash/ homotopy coherent analogues, as trial coefficients for TQFT type constructions, with similar motivation to yours here (but with less clear diff. geom linkage (this was nearly 10 years ago)

Do you have a copy tyou could share?

(Unfortunately the proposal was not only not funded but was vicously sneered at by a referee!! Heigh ho! c’est la vie!)

Just ahead of your time - cf. my first attempt to publish my thesis.

Posted by: jim stasheff on September 16, 2007 3:44 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Final shot for the moment: weak cokernel is not necessarily the best or right’ term. Homotopy theorists from way back used weak’ to mean existence but not necessarily uniqueness of the induced map in such contexts. Homotopy cokernel is now the standard term within homotopy theory so when looking/Googling at the interaction of these problems with the constructs of homotopical and homological algebra a search on homotopy cokernel is likely to be more fruitful than one on weak cokernel. Of course all the witches hats, etc are examples of homotopy colimits but do not forget homotopy limits either as they are what Breen used for his Puppe sequences, that is resolving the right hand part of the $[X,BG]$ rather than the left hand.

Posted by: Tim Porter on September 15, 2007 1:17 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Tim Porter wrote:

Of course all the witches hats, etc are examples of homotopy colimits…

Hmm! You’re reminding me of things. Now suddenly my impromptu reply to Urs’ question about the universal property of the mapping cone is looking very naive.

I suddenly remember the time when, as a raw youth with a head full of dream about $n$-categories, I made a pilgrimage to Wales. I met Ross Street and John Power for the first time on the train there — I overheard them talking about ‘cosmoi’. And I met you and Ronnie Brown when I arrived in Bangor. At one point you sat me down in your office and told me a thing or two about the wisdom of simplicial methods. I’ve been struggling to learn more about those ever since.

In theory, I now understand the Dwyer–Kan theory of homotopy colimits. It all clicked when Todd Trimble explained how it was a special case of the abstract bar construction. In Vienna I explained this construction to Urs, and being a smart cookie he may be able to figure out how it works with just a teeny bit of nudging. I hope the following is right — I’m pretty tired:

When we take the homotopy colimit of a diagram of spaces, we get a simplicial space where an $n$-simplex is the same thing as a length-$n$ string of composable morphisms in our diagram:

$X_0 \stackrel{f_1}{\to} X_1 \stackrel{f_2}{\to} \cdots \stackrel{f_n}{\to} X_n$

together with a point in $X_0$.

A homotopy cokernel is a homotopy colimit of a diagram like this:

$X \stackrel{\to}{\to} Y$

where one arrow here is any map $f: X \to Y$, and the other is the constant map.

So, the only nondegenerate simplices in the homotopy cokernel are the 0-simplices and the 1-simplices. The 0-simplices are the points on brim of the witch’s hat, together with its tip. The 1-simplices are the line segments going from the tip to the brim:

(slightly bent in this picture, but we’re doing topology).

Anyway, regardless of whether I explained them correctly just now, homotopy colimits were worked out by Dwyer and Kan eons ago. Knowing the formidable thoroughness of those gentlemen, they must have worked out the universal property of any homotopy colimit. It also follows from the universal property of the general bar construction as explained by Todd.

So, I was wrong when I said:

There’s a strange tendency in algebraic topology for certain constructions to work ‘more strictly than you might expect’. For example, the product of spaces is a product ‘on the nose’, not just up to homotopy, and similarly for the coproduct. I’m sure there’s a general theory of when this unexpected strictness happens, but I’m not sure anyone knows this theory yet — I sure don’t.

In fact, this general theory should follow from the theory of homotopy colimits, and in principle I should be able to work it out from stuff I know!

Posted by: John Baez on September 21, 2007 7:10 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

there is always Lurie!

Posted by: jim stasheff on September 22, 2007 1:41 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

For a Welsh version of Abe’s hat see

http://en.wikipedia.org/wiki/Welsh_hat

Posted by: Tim Porter on September 15, 2007 1:47 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

“… A derived meaning of Welsh hat is an ancillary stack…”

That sounds categorical.

Posted by: Jonathan Vos Post on September 16, 2007 12:37 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Greetings from Yale. Hisham Sati is so kind to let me use his computer for a moment, while he is lecturing.

This gives me a chance to get back to this discussion here. Thanks to all those who dropped a comment. I do heartily appreciate it. And I will come back to some of the things that have been mentioned when time permits.

While I still haven’t found the time to go over the notes I linked to, let me here give a more concise and refined version of one of the main statements.

Obstructions to extensions to Baez-Crans/String-type Lie $n$-algebras

Let $g$ be a Lie algebra and $\mu$ an $(n+1)$ cocycle on it, which is transgressively related to the invariant polynomial $k$. Recall that this gives rise to the (weakly exact) sequence

$g_{\mu} \to \mathrm{cs}_k(g) \to \mathrm{ch}_k(g)$

of Lie $(n+1)$-algebras.

Given a $g$-bundle $P \to X$ we want to characterize the obstruction for $\array{ \Omega^\bullet(X) &\stackrel{(A,F_A)}{\leftarrow}& \mathrm{inn}(g)^* \\ \uparrow^{p^*} &\Downarrow& \uparrow \\ \Omega^\bullet(X) &\leftarrow& b g^* }$ to factor through the dual $\mathrm{Lie}(\Sigma^{n-1}U(1))^* \stackrel{t^*}{\leftarrow} g_\mu^* \leftarrow g$ of the “String extension” sequence $\mathrm{Lie}(\Sigma^{n-1}U(1)) \stackrel{t}{\to} g_\mu \to g \,.$

For $(\mathrm{Lie}(\Sigma^{n-1}U(1)) \stackrel{t}{\to} g_\mu )$ denoting the Lie $n$-algebra obtained from the mapping cone of $t$, this is measured by the pullback along $(\mathrm{Lie}(\Sigma^{n-1}U(1)) \stackrel{t}{\to} g_\mu )^* \leftarrow ker(i^*)$

dual to $g_\mu \stackrel{i}{\to} (\mathrm{Lie}(\Sigma^{n-1}U(1)) \stackrel{t}{\to} g_\mu ) \to \mathrm{coker}(i) \,.$

Here one finds $\mathrm{coker}(i) = \mathrm{Lie}(\Sigma^n U(1)) \,.$

Then one checks the following

Proposition

We have a canonical 2-isomorphism $\array{ \mathrm{inn}(g)^* &\stackrel{\simeq}{\leftarrow}& \mathrm{inn}(\mathrm{Lie}(\Sigma^{n-1}U(1)) \to g_{(n)})^* &\leftarrow& \mathrm{inn}(\Sigma^{n}U(1))^* \\ \uparrow &&&& \uparrow \\ \mathrm{ch}_k(g) &&\Downarrow&& \uparrow \\ \uparrow &&&& \uparrow \\ b g^* &\leftarrow&\leftarrow&\leftarrow& b \mathrm{Lie}(\Sigma^n U(1)) }$ which “vanishes on the fibers”, meaning that $\array{ g^* \\ \uparrow \\ \mathrm{inn}(g)^* &\stackrel{\simeq}{\leftarrow}& \mathrm{inn}(\mathrm{Lie}(\Sigma^{n-1}U(1)) \to g_{(n)})^* &\leftarrow& \mathrm{inn}(\Sigma^{n}U(1))^* \\ \uparrow &&&& \uparrow \\ \mathrm{ch}_k(g) &&\Downarrow&& \uparrow \\ \uparrow &&&& \uparrow \\ b g^* &\leftarrow&\leftarrow&\leftarrow& b \mathrm{Lie}(\Sigma^n U(1)) }$ vanishes.

The obstruction we are looking for is our $g$-bundle $\array{ \Omega^\bullet(X) &\stackrel{(A,F_A)}{\leftarrow}& \mathrm{inn}(g)^* \\ \uparrow^{p^*} &\Downarrow& \uparrow \\ \Omega^\bullet(X) &\leftarrow& b g^* }$ pasted on the right with this 2-morphism

$\array{ \Omega^\bullet(P) &\stackrel{(A,F_A)}{\leftarrow}& \mathrm{inn}(g)^* &\stackrel{\simeq}{\leftarrow}& \mathrm{inn}(\mathrm{Lie}(\Sigma^{n-1}U(1)) \stackrel{t}{\to} g_{(n)})^* &\leftarrow& \mathrm{inn}(\Sigma^{n}U(1))^* \\ \uparrow^{p^*} && \uparrow &&&& \uparrow \\ \uparrow && \mathrm{ch}_k(g) &&\Downarrow&& \uparrow \\ \uparrow && \uparrow &&&& \uparrow \\ \Omega^\bullet(X) &\stackrel{(\{K_i = k_i(F_A)\})}{\leftarrow}& b g^* &\leftarrow&\leftarrow&\leftarrow& b \mathrm{Lie}(\Sigma^n U(1)) } \,.$

Here the horizontal composite on the bottom is the classifying morphism of a $\Sigma^n U(1)$ (n+1)-bundle (an abelian $n$-gerbe, if you wish). The diagram says that this is given by the $(n+2)$-from $P := k(F_A) \,.$

For $\mu$ the canonical 3-cocycle on a semisimple Lie algebra $g$, $g_\mu$ is the Baez-Crans type String Lie 2-algebra and $P = k(F_A) = \langle F_A \wedge F_A\rangle$ the first Pontryagin class of the given $g$-bundle. It classifies the Chern-Simons 3-bundle obstructing the lift of $P$ to a String bundle.

Posted by: Urs Schreiber on September 19, 2007 6:36 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Greetings from Yale.

Whoa, you’re visiting Yale now? I live in the area, and would love to get together if you’re around for a while!

Posted by: Todd Trimble on September 19, 2007 9:47 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Hi Todd,

would love to get together if you’re around for a while

Oh, didn’t know you are here in the area! I’ll be here at the Yale math department for tomorrow, Friday, 21 and maybe Saturday morning, talking to Hisham Sati. Would be great to see you. Drop me a note and let us know where we can meet you!

Posted by: Urs Schreiber on September 21, 2007 12:58 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Greetings from Yale.

You’re at Yale? Darn, I just missed you by an academic year. Say hi to Hisham for me, and Dr. Zuckerman as well (who’s finally responding to my calls).

And Todd lives near New Haven? And he didn’t tell me this while I was there? >:(

Posted by: John Armstrong on September 20, 2007 1:34 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Todd,
Sorry for communicating this way. I’m lacking your e-address. So you are near Yale!! come on down and give a talk here at Penn. Like wise anyone else within a similar radius OR can cover their own transportation beyond that.

jim

Posted by: jim stasheff on September 20, 2007 5:21 PM | Permalink | Reply to this

### Email contact

Jim –

Thanks for the kind invitation; for anyone who might want to contact me, my email is trimble one at optonline dot net.

For the record, I live in Redding CT where I have a full-time (or almost full-time) job as a stay-at-home parent; my children are still very young (ages 6 and 3). I do math here at home as time allows, and communicate almost entirely via the internet; my opportunities for things like talks are decidedly limited these days. (Some time back I took the opportunity to meet Nick Gurski, who is now at Yale – something I’d very much like to do again sometime soon. I’m sorry I didn’t get to meet John Armstrong while he was there.)

If any of the cafe regulars pass through Yale, I’d enjoy getting together for lunch or a chat or something. Feel free to drop me a line.

Posted by: Todd Trimble on September 20, 2007 6:14 PM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

Well, over the winter break I’ll be heading back up to Maryland. Check out my “Shameless Self-Promotion” post and see if there’s anything you think the folks at Penn would like :D

Posted by: John Armstrong on September 21, 2007 1:46 AM | Permalink | Reply to this

### Re: Obstructions for n-Bundle Lifts

I just want to quickly say thanks for all the further comments. I’ll reply to them later. Currently I am awefully jet-lagged and awefully off-line, generally, being on just a few hours stop at home in between a trip from Yale to Split/Croatia.

The theory of obstructions of $n$-bundles has considerably developed in the meantime. There is one big nice diagram which controls it all. Todd Trimble has seen a hand-drawing in Hisham Sati’s office already. When time permits, I will need to LaTeXify this and explain it all in detail.

Gotta run now.

Posted by: Urs Schreiber on September 23, 2007 2:40 PM | Permalink | Reply to this
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