Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Bundles with fibre F are classified by BDiff(F)?

Chris Grossack (they/them) (Oct 03 2024 at 20:07):

I'm watching Lurie's lectures on TQFTs and around the 14m30s mark, he says that $\text{Hom}_{\text{Bund}_1}(\emptyset, \emptyset)$ is the "classifying space for 1-manifolds" (with empty boundary). In particular it has connected components corresponding to some number of circles.

When we look at the component for a single circle, to me it seems like we should get $B\text{Diff}^+(S^1)$, but Lurie says that we get the classifying space for circle bundles, $\mathbb{CP}^\infty$. He does also say a few seconds later that he means $B\text{Diff}^+(S^1)$, but it's not obvious to me that these are the same space.

Should it be obvious? Maybe they're merely homotopy equivalent? Is it true more generally that $B\text{Diff}(F)$ should classify bundles with fibre $F$?

Thanks for the help! ^_^

David Corfield (Oct 03 2024 at 20:25):

Something like: $Diff(S^1) = O(2)$, so $BDiff(S^1) = BO(2)$. Then $BO(2) = BU(1) =  \mathbb{C}\mathbb{P}^{\infty}$.

John Baez (Oct 03 2024 at 23:34):

Hi David! $\mathrm{Diff}(S^1) \simeq \mathrm{O}(2)$, but $\mathrm{O}(2) \ncong \mathrm{U}(1)$ so this argument needs to patched a little. In fact $\mathrm{O}(2) \cong \mathrm{U}(1) \times \mathbb{Z}/2$ where the two components correspond to orientation-preserving and orientation-reversing rotations.

But notice Lurie mentioned $\mathrm{Diff}^+(S^1)$, the topological group of orientation-preserving diffeomorphisms of the circle. We have $\mathrm{Diff}^+(S^1) \simeq \mathrm{SO}(2) \cong \mathrm{U(1)}$ and then yes, $\mathrm{BU}(1) \simeq \mathbb{CP}^\infty$.

John Baez (Oct 03 2024 at 23:52):

It may seem trivial that $\mathrm{Diff}^+(S^1) \simeq \text{SO}(2)$, but this is a bit stronger than just saying every orientation-preserving diffeomorphism of the circle is smoothly homotopic to a rotation: it's saying we can do that homotopy in a kind of 'uniform' way.

It's a lot harder to show $\mathrm{Diff}^+(S^2) \simeq \text{SO}(3)$. For some reason I was interested in these things once, and a proof is section 7.3 of Kuipers' Lectures on Diffeomorphism Groups of Manifolds

But then things get really hard. There are 'weird' diffeomorphisms of the $n$-sphere, and gluing together two $(n+1)$-balls along their boundary using such a weird diffeomorphism you can get an exotic $(n+1)$-sphere: one that's homeomorphic but not diffeomorphic to the usual $(n+1)$-sphere. For example, $\mathsf{Diff}^+(S^6)$ has 28 connected components, corresponding to the 28 exotic 7-spheres. So we can't have $\mathsf{Diff}^+(S^6) \simeq \mathrm{SO}(7)$, since $\mathrm{SO}(n)$ always has just one connected component!

John Baez (Oct 03 2024 at 23:55):

Smale and Cerf showed that for $n \ge 6$, $\pi_0(\mathrm{Diff}^+(S^n))$ is isomorphic to the group of (orientation-preserving diffeomorphism classes of) exotic $(n+1)$-spheres. The first interesting case is

$\pi_0(\mathrm{Diff}^+(S^6)) \cong \mathbb{Z}/28$

But this is really hard math, and I'm just reporting results that I don't understand.

John Baez (Oct 03 2024 at 23:56):

So be glad Lurie was working with the lowly 1-sphere, @Chris Grossack (they/them)!

Chris Grossack (they/them) (Oct 04 2024 at 00:02):

1. This is all extremely interesting! And the fact that nobody has quickly told me why BDiff(F) should classify F-bundles also makes me feel better that maybe Lurie misspoke quickly (or more likely, that maybe I misunderstood what he said) and in fact this is all something special about $S^1$.

Chris Grossack (they/them) (Oct 04 2024 at 00:05):

Though there is some evidence for the more general claim. For instance, see Andy Putman's comments on this MO post, or the Theorem 6.2 in Christensen and Wu's Smooth Classifying Spaces

Kevin Carlson (Oct 04 2024 at 00:15):

John Baez said:

It may seem trivial that $\mathrm{Diff}^+(S^1) \simeq \text{SO}(2)$, but this is a bit stronger than just saying every orientation-preserving diffeomorphism of the circle is smoothly homotopic to a rotation: it's saying we can do that homotopy in a kind of 'uniform' way.

In particular, every oriented diffeomorphism of the circle is smoothly homotopy to a point, so there's something interesting to show!

Is it just me, or is the evaluation-at-1 map $\mathrm{Diff}^+(S^1)\to S^1$ actually a trivial fiber bundle, whose fiber is the group of orientation-preserving diffeos of the circle which fix $1$? This latter group has a pretty explicit contraction if you view it as the space of increasing (smooth) bijections $[0,1]\to [0,1]$, which gives the result. In fact, I guess it doesn't matter whether the bundle is trivial--any bundle with contractible fibers is homotopy equivalent to the base!

Kevin Carlson (Oct 04 2024 at 00:32):

Chris Grossack (they/them) said:

1. This is all extremely interesting! And the fact that nobody has quickly told me why BDiff(F) should classify F-bundles also makes me feel better that maybe Lurie misspoke quickly (or more likely, that maybe I misunderstood what he said) and in fact this is all something special about $S^1$.

I do think it's generally true that $B\mathrm{Diff}(F)$ is the classifying space for bundles with fiber $F$ for $F$ a smooth manifold. One way of constructing $B\mathrm{Diff}(F)$ is as the quotient of the space of embeddings of $F$ in $\mathbb R^\infty$ by its action of $\mathrm{Diff}(F),$ since the space of embeddings is contractible (the space of maps into $\mathbb R^\infty$ is contractible since $\mathbb R^\infty$ is, and there's lots of room to wiggle any non-embedding of $F$ or of $F\times I$ a tiny bit to get an embedding.) Thus a map $M\to B\mathrm{Diff}(F)$ is a smooth choice of subspace of $\mathbb R^\infty$ diffeomorphic to $F$ for every point of $m.$ At least if $M$ is itself a manifold, then any $F$-bundle over $M$ will be too, and so you can embed the whole bundle into $\mathbb R^\infty$ (again, uniquely up to contractible choice), which gives a map from $F$-bundles over $M$ to $[M,B\mathrm{Diff}(F)].$ This isn't a proof but I think it's the vibe Lurie was hoping you'd have.

Kevin Carlson (Oct 04 2024 at 00:35):

An even more core way of stating the vibe is that $B\mathrm{Diff}(F)$ is "the space of all possible copies of $F$, up to diffeomorphisms." It's just that there's no actual universal space in which all copies of $F$ live, so we pick $\mathbb R^\infty$ as a sufficiently-universal space if $F$ is a smooth manifold insofar exactly as we get the contractible embedding space; any other universal space with that property also works.

John Baez (Oct 04 2024 at 00:37):

Chris Grossack (they/them) said:

1. This is all extremely interesting! And the fact that nobody has quickly told me why BDiff(F) should classify F-bundles also makes me feel better that maybe Lurie misspoke quickly (or more likely, that maybe I misunderstood what he said) and in fact this is all something special about $S^1$.

What do you mean by an F-bundle? If you mean a bundle where all the fibers are diffeomorphic to F and we glue together the bundle from trivial F-bundles using continuous Diff(F)-valued cocycles on the overlaps $U_\alpha \cap U_\beta$ of some good cover of the base space, it's more or less an automatic consequence of general results about classifying spaces that F-bundles are classified by BDiff(F)... well, at least F-bundles over a paracompact base space! (There is some annoying technical dirt like that in the general theory of classifying spaces, at least the old stuff I know.)

If we want the diffeomorphisms to be smooth cocycles (as is natural in differential topology) there may be extra things to check.

But even so - if that's all we wanted, we'd be studying circle bundles where the fibers are mere manifolds, not $\mathrm{U}(1)$-torsors. If we're trying to classify principal $\mathrm{U}(1)$ bundles, which are much more rigid, we should be using $B\mathrm{U}(1)$. So then we run into the question of how $B\mathrm{Diff}(S^1)$ is related to $B\mathrm{U}(1) \cong B\mathrm{SO}(2)$. And then we wind up wanting to know

$\mathrm{Diff}^+(S^1) \simeq \mathrm{SO}(2)$

which is a very special fact (since the analogous fact doesn't hold for spheres of every dimension).

Kevin Carlson (Oct 04 2024 at 00:39):

Once you believe in this vibe, then no matter what kind of a thing $F$ is you might believe that maps $X\to B\mathrm{Auto}(F)$ are $F$-bundles in whatever category $\mathrm{Auto}(F)$ comes from are going to be maps from $X$ into the "space of all possible copies of $F$ up to automorphisms", and then if you define such maps to be bundles with fiber $F$ then you can just try to apply a representability theorem to define that space of all possible $F$ s as $B\mathrm{Auto}(F)$, without futzing with any universal embedding space for $F.$ You can actually do this in an appropriate context with Brown representability, if you didn't know.

John Baez (Oct 04 2024 at 00:51):

Kevin Carlson said:

Is it just me, or is the evaluation-at-1 map $\mathrm{Diff}^+(S^1)\to S^1$ actually a trivial fiber bundle, whose fiber is the group of orientation-preserving diffeos of the circle which fix $1$?

How are you getting the trivialization? Let's see. I'll treat $S^1$ as a subgroup of $\mathrm{Diff}^+(S^1)$ in the obvious way: it consists of rotations of the circle.

Then I'll try to write any diffeomorphism $f: S^1 \to S^1$ as $g: S^1 \to S^1$ times $h: S^1 \to S^1$ where $g$ is the rotation that maps $1 \in S^1$ to $f(1)$, and $h$ is an orientation-preserving diffeomorphism that fixes $1$.

Is that the idea?

(By "times" I might mean composition in the diffeomorphism group, or I might mean "pointwise times" using the fact that $S^1$ is a group. I also refuse to say which in which order I'm multiplying $g$ and $h$, on the grounds that it might incriminate me.)

This latter group has a pretty explicit contraction if you view it as the space of increasing (smooth) bijections $[0,1]\to [0,1]$, which gives the result.

Okay. Yes, I vaguely recall that the result is roughly this hard: nothing mind-boggling. As we start studying higher-dimensional spheres the study of $\mathrm{Diff}^+(S^n)$ quickly becomes much harder, reaching a maximum around the scary dimensions $n = 3, 4$ and then turning into a bunch of homotopy-theoretic calculations connected to algebraic K-theory.

In fact, I guess it doesn't matter whether the bundle is trivial--any bundle with contractible fibers is homotopy equivalent to the base!

Okay, all that work I did for nothing.

Kevin Carlson (Oct 04 2024 at 00:57):

John Baez said:

Is that the idea?

Yeah, something like that. What I was specifically thinking is that the fiber over $e^{i\theta}$ is naturally identified with the smooth strictly increasing self-bijections of $[e^{i\theta},e^{i\theta}+2\pi],$ so the trivialization just adds $e^{i\theta}$ to the input and output of a self-map of $[0,2\pi].$ So, multiplying by that rotation, yeah.

John Baez (Oct 04 2024 at 01:02):

Okay, good. By the way, there's something funky about your use of quotes here. (It's always a pain dealing with quotes here.)

Kevin Carlson (Oct 04 2024 at 01:19):

Thanks, that should do it.

Chris Grossack (they/them) (Oct 04 2024 at 03:13):

Iiiiii see. The point is that if you believe there's a contractible space of Fs, then you end up with $B\text{Aut}(F) = \star \big / \text{Aut}(F)$ being the space of Fs up to automorphism.

Chris Grossack (they/them) (Oct 04 2024 at 03:13):

Thanks!

John Baez (Oct 04 2024 at 06:05):

Yeah, that's the easy beautiful conceptual part. The fun particular part is showing the topological group $\mathrm{Diff}^+(S^1)$ is equivalent as an $\infty$-group to the circle! So that's what Kevin showed.

Kevin Carlson (Oct 04 2024 at 07:25):

Chris Grossack (they/them) said:

Iiiiii see. The point is that if you believe there's a contractible space of Fs, then you end up with $B\text{Aut}(F) = \star \big / \text{Aut}(F)$ being the space of Fs up to automorphism.

Specifically a contractible space of $F$ s on which the automorphism group acts freely, eg you can’t just take a point.

Mike Shulman (Oct 04 2024 at 15:23):

You can just take a point if by "$/$" you mean a homotopy quotient.

Mike Shulman (Oct 04 2024 at 15:24):

Terminologically, I would say that $B \mathrm{Aut}(F)$ is "the space of $F$ s".

John Baez (Oct 04 2024 at 20:25):

Have people developed enough 'cohesive homotopy type theory' (or whatever they call it) so we can easily do things like study the de Rham cohomology of $B \mathrm{Aut}(F)$ when $F$ is a Riemannian manifold - so if $F$ is the round $(n+1)$-sphere, $\mathrm{Aut}(F) = \mathrm{O}(n)$ and $B \mathrm{Aut}(F)$ is a smooth model of $B\mathrm{O}(n)$? I'd like to make it not hard to show $B\mathrm{O}(n)$ is "equal" to the space of $n$-dimensional linear subspaces of an infinite-dimensional Hilbert space, or something like that.

John Baez (Oct 04 2024 at 20:26):

I can do all this stuff using ordinary math, but it's a bit onerous at certain points, and I can imagine it becoming easier.