My fundamental picture: $E$ an $F$-bundle over $X$, and we have section $s : X^n \to E$. Obstruction cochain $\psi \in C^{n+1}(X; \pi_n F)$ defined by: take the attaching map $\varphi : S^n = \partial e^{n+1} \to X$, then compose $s \circ \varphi : S^n \to E$, which homotopes to a map $\tilde{\varphi} : S^n \to F$ because $\varphi$ was nullhomotopic downstairs, so $s \circ \varphi$ homotopes inside the fiber upstair.

This is the element of $\pi_n F$ that $e^{n+1}$ gets sent to; i.e, this element is $\psi(e^{n+1})$

I find it more helpful that the pullback and piecemeal section picture, though they are equivalent

I was confused too about why $\psi$ is not a coboundary, because we pass to boundary in defining $\varphi$, so it feels like it should be

But we lift $\varphi$ to $E$ using a partially defined section on $X$, is the point, I think

If we had a fully defined section, it'd sure as hell be zero (stronger than a coboundary)

So the fact that $\varphi$ was a boundary of something (the characteristic map $e^{n+1} \to X$ bounded it) is completely forgotten upstairs in $E$

because the section is not defined over it

Is this better?

room mode changed to Gallery: anyone may enter, but only approved users can talk

Is there a geometric framing of the notion of coboundary, then?

For elements of $C^{n+1}(X; \pi_n F)$? Good question!

Hmmmmm

Because it would help me see why it doesn't apply above I guess

Yeah.

I guess I'm stumbling because if I start with an abstract cochain $\phi$ of $C^n(X; \pi_n F)$ and take $\delta \phi$ it might not have any geometric meaning because $\phi$ doesn't have any geometric meaning because it has nothing to do with obstruction

The context of the section and stuff is not relevant anymore

Maybe we could work out an example of a bundle where it's not a coboundary

Uhh, what's a simple example of a fiber bundle over a 2-complex with no sections, but which trivializes over the 1-skeleton?

$T_1 S^2$ over $S^2$?

So, give $S^2$ the cell structure of a circle with two disks attached (northern/southern hemispheres). Call this cell complex $\mathscr{S}$.

$\mathscr{S}^{(1)}$ is merely $S^1 \subset S^2$ (:P @ notation), and the bundle $E = T_1 S^2$ over $\mathscr{S}$ trivializes over $\mathscr{S}^{(1)}$.

Take a section $s : \mathscr{S}^{(1)} \to E$ which is the constant unit vector along the equator.

If I am not wrong, the unit tangent vector along the equator of the sphere, as free loop in $T_1 S^2 = SO(3)$, homotopes to the generating element of $\pi_1 F$

Think of shrinking the equator to the north pole. Then it's unit tangent vector field shrinks to the fiber $F$ over it

So $\psi(e_1)$ is the generator $1$ of $\pi_1 F = \Bbb Z$. Similarly $\psi(e_2)$ is $-1$. Then that cochain $\psi \in C^2(\mathscr{S}, \pi_1 F) = C^2(S^2, \Bbb Z)$ is precisely the top dimensional cellular cochain representing the generator of $H^2(S^2)$, so not the coboundary

No, $\psi(e_2)$ is $1$ again. No difference happens in that computation, 'cuz the attaching map of the bottom cell is still via identity.

I was really scared because $\psi(e_1) = 1$ and $\psi(e_2) = -1$ would have actually meant it was $\delta \phi$ where $\phi \in C^1(S^2; \Bbb Z)$ takes $\phi(S^1) = 1$ on the equator $S^1 \subset S^2$.

(I hope I am right on the sign issues)

$T_1 \Sigma_2$ could also be good

I think the obstruction cocycle should be 2

I guess the point is that $e_1 - e_2$ is the fundamental class, and $\psi(e_1 - e_2) = 2$

Oh yeah

wait no

What is $\partial e_2$?

That's what I'm unsure about. Feels like it should be $-[S^1]$

Right?

Which would mean that $e_1 + e_2$ is the fundamental class, so $\psi$ is $0$ on it, and hence... is a coboundary

I'm confus

I don't understand signs

I guess we can present the sphere as the double of the disc along its boundary

So $e_1$ and $e_2$ are both attached to $S^1$ by the identity map on the boundary. I think that means $\psi(e_1) = \psi(e_2) = 1$, because the above computation has the same parity for $e_1$ and $e_2$. Now, what should be the fundamental class?

So we think of both $e_1$ and $e_2$ as having the boundary as identity mpa

Yeah

But then... the fundamental class is $e_1 - e_2$

Fucking fuck

I guess that confuses me.

I am not sure the above computation has the same parity for $e_1$ and $e_2$

I bet the difference between them has to do with clutching functions or whatever

Hmmm

it's vector fields, this should be easy :(

This sucks. I don't get it

I would say "trivialize $T^1S^2$ over $D^2$" but that means picking a unit vector field over $D^2$. Then you compare the unit vector field you chose to the unit vector field we obtain here

So for the top hemisphere flatten it out and pick the vector field pointing "left" one unit

Ah, you want to pullback the bundle to the disk like you were doing, and choose a section. Ahh this might explain it.

Then compared to that, our vector field (which is the tangent field to the disc) rotates one time counter-clockwise

Agreed.

And I guess we parameterize $S^1$ with exponential, so that means +1

Nod.

Why does this change over the southern hemisphere?

I think this is a canonicality issue. The element of $\pi_n F$ you get depends on the choice of the section on the disk, which is what initially had me stuck on my fiber exact sequence computation.

What section over the southern hemisphere will you choose?

How would you compare the two sections?

Well we need some way

Oh, I want your picture. I think you want to take the unit vector field, and then null-homotop that circle through $D^2$ to a point in the boundary

Yeah.

(I'm imagining a 1-parameter family like the Hawaiian earrings, collapsing the boundary to a point)

Yup yup yup

I am convinced that when you do that in the top hemisphere you get +1 and in the bottom hemisphere -1

Huh...

The point we contract to is say $(1,0,0)$

So you'd want the homotopy to stay in the disk you're choosing to compute $\psi$ on? It wasn't clear to me that you did. I guess that's the choice.

and in the top hemisphere a loop just before the end will be the obvious one going right, up, left, down, right, ...

Yeah, I just mean I drew a picture in my head that convinced me, and now want to communicate it

Please do. I think I see it too.

I am imagining on the sphere, a way to take the map $S^1 \to T^1 S^2$ living above the equator, and take it to a map $S^1 \to T^1_{(1,0,0)} S^2$

To do that, I want to "pull" this vector field along the equator to that point - one way is above the north half, the other way is along the bottom half

Yeah. And the results will be different.

Near the end of this pulling process, on the north hemisphere, we have the vector field that starts at $(1,0,0)$, and starts going right, and then goes into the north hemisphere

so is CCW

On the southern hemisphere, we have the vector field that starts at $(1,0,0)$, goes right, and into the southern hemisphere, so is clockwise

So maybe kinda like this

Yeah, one is the top and one is the bottom of the four loops in that dipole

if that parses

Right

oh I parsed the visual incorrectly but I think what I said can be fixed

we kept trivializing the bundle and trying to compute in "the trivial bundle" without paying attention to our trivialization

anyway this only sort of clarifies things but I'm okay with it

(w/r/t why it's not a coboundary, I mean)

Yeah I get it, think of the great circle passing through the singular point of the picture above that cuts the ball in half from your perspective, the left and right hemispheres are the top/bottom. Clearly the "limit vector fields" (which lie in the fibers over the singular point) given by sliding the equatorial vector field are on the opposite direction

I was freely homotoping everything to a single point, which is perhaps wrong and breaks canonicality in some way

I think there's some distinction between $TS^2 \big|_{N} \cong TD^2$ and $TS^2\big|_{S} \cong TD^2$

perhaps orientations

An observation: Kirk-Davis assumes that their range space while doing obstruction theory is simple.

Is this relevant?

$SO(3)$ is simple, right? No problem.

$\pi_1$ acts on $S^3$ by an orientation-preserving homeomorphism, isotopic to the identity

so should act trivially on all homotopy groups

Right...

Hmm, why is the nontrivial deck transformation of $S^3$ (as a 2-fold cover of SO(3)) isotopic to identity?

homotopic is because deg = 1

Oh, duh.

I have forgotten everything

isotopic by thinking of them as sitting inside $SO(4)$

I guess here is what I really wanted to say though the whole time. I'm not sure we ever needed to know that the obstruction class was a cocycle to discuss this.

Suppose the obstruction class is a coboundary. Do you see how to define a section over $X^{(n+1)}$, which perhaps only agrees with the existing section on $X^{(n-1)}$?

Hmmm

@MikeMiller Tangentially, I sifted through K-D to where it says obstruction theory of fibrations. They want the fiber space to be $n$-simple (when the problem is to extend section from $n$-skeleton to $(n+1)$-skeleton)

In our case, $S^1$ is not $1$-simple :D

I bet this isn't a problem

$\pi_1$ does act trivially here, the action is by commutator

Oh, so just demanding the monodromy action is trivial should be enough? I see

But I don't see how they propose to fix that

I think they're claiming this assumptions removes the possibility of different results for elts of $\pi_n F$

Ionno

I don't get the obstruction class

This is a confusing mess.

I brought it up to you because the story should be quite clean

but... I am not remembering a clean story

The idea is quite beautiful actually. But this technical bit about how to set up the map $S^n \to F$, to do which one seemingly has to compare different sections of the pullback of the bundle over the cells, is really frustrating...

$\psi(e^{n+1})$ should just depend on the attatching map of $e^{n+1}$, (the lift of) which one should be able to homotope however they want to

The n-simple hypothesis is there to say that a map $S^n \to F$ uniquely determines an element of $\pi_n F$

Oh ok

Because the action of $\pi_1 F$ on $\pi_n F$ preserves free homotopy classes but not pointed homotopy classes, this is the problem

The assumption that $\pi_1 B$ acts trivially on $\pi_n F$ is there, I think, to say that you can't end up with different free homotopy classes $S^n \to F$ depending on the fuckery you did in the process of homotoping them into $F$

Yeah I suppose

Look at Mosher-Tangora page 6

Downloading it

Also I just realized that the previous issue with $T_1 S^2$ was an issue of the bundle, not an issue of what we were doing.

If we think $S^2$ as the double of the disk, as a complex, the unit tangent bundle of $S^2$ as an abstract bundle on this 2-complex pulls back to different things on the two disks

Right, that's what I was trying to get at (but didn't quite understand) about the statement that the restriction to the disc - and then trivialization - $TS^2\big|_N \cong TD^2$ depends on a choice of trivialization

and the "obvious" one is different for the two hemispheres

Note that $\psi(e_1) - \psi(e_2)$ is precisely the clutching function's value

yeah I just understood that comment. Good point about the clutching function

So the consistent choice is to fix a trivialization of $TS^2$ on $e_1$ and use the same trivialization on $e_2$, which wasn't what we were doing when confusing ourselves, presumably

I think I just said the opposite of what I meant

lol

Yah I got you though

But yeah I realize it was just the bundle being unfortunate and not simply an arbitrary bundle over a 2-complex. It's a bundle that depends on the 2-complex being an oriented manifold

I'm reading page 6