Here's what I was proposing earlier, btw. Suppose $X$ is a cell complex. Say $c \in H_n(X^{(n)}, X^{(n-1)})$ is a cellular $n$-chain in $X$.

That's a formal linear combination $c = \sum_i c_i e_i^n$ where $e_i^n$ are the $n$-cells in $X$ and $c_i$ are the integral coefficients. If $X$ is a connected cell complex, consider the map $\varphi : \bigsqcup_i D^n_i \times \{0, 1, \cdots, c_i\} \to X$ where $\varphi$ restricts to $D^n_i \times \{0, 1, \cdots, c_i\} \stackrel{\pi}{\to} D^n_i \stackrel{char}{\to} X$ on each of the disjoint pieces ($\pi$ = first projection, $char$ = characteristic map).

So basically let $\varphi$ map a $n$-disk to $e_i^n$ with multiplicity $c_i$, corresponding to what happens in $c$

If $X$ is connected, $X^{(0)}$ is just a point, and the boundary of all these n-disks map to $X^{(n-1})$, so contains that 0-cell in the image.

Realizing $\sqcup_i D_i^n \times \{0, 1, \cdots, c_i\}$ as a disjoint union of disks, this means $\varphi$ induces a map from a wedge of $n$-disks: $\varphi : \bigvee_i (\bigvee_{j = 1}^{c_i} D_{j_i}^n) \to X$

I'm writing the full wedge as a wedge of wedges because I don't want to loose track of the multiplicity information in $c$

Oh, and the wedge point is on the boundary of all these disks

If all this is right, this seems to be an analogous description of singular chains as maps from simplicial complexes: cellular chains are maps from wedge of disks

Ah your objection was what if $c_i$ is negative for some $i$. I think the point is, if $c_i = -1$, $\varphi$ would restrict to $D_i^n \to D_i^n \stackrel{char}{\to} X$ where the first map is the one which reverses the orientation of the disk. So on the boundary $\partial D_i^n$, it's the attaching map with the "inverse word"

For example, for $S^2$ with the equator and two 2-cells attached to it by the identity map, the map corresponding to $e_1 - e_2$ (fundamental class) is $\varphi : D^2_1 \vee D^2_2 \to S^2$ where $\varphi|D^2_1$ is attaching map of $e_1$ and $\varphi|D^2_2$ attaching map of $e_2$ composed with the reflection of the disk

Seems natural, right?

And since you can get a wedge of disks by crushing diameters in a disk, you can realize any cellular $n$-chain as a map from a single disk $D^n \to X$. You'd need to keep track of the diameters to crush to recover the formal linear combo back from it though

Sure, fair enough, I agree with that

So I think the obstruction cocycle $\psi \in C^{n+1}(X; \pi_n F)$ of an $F$-bundle $E$ over $X$ with a section $s : X^{(n)} \to E$ over the $n$-skeleton is defined by, for some cellular $(n+1)$-chain $\xi$, let $D^{n+1} \to X$ be the map from the disk that represents it, take the boundary $S^n \to X$, compose with section to $S^n \to E$, and homotope it to the fiber $F$. That's $\psi(\xi)$.

Now, this should somehow make $\delta \psi = 0$ an apparent consequence of $\partial^2_{cellular} = 0$

I am just always paranoid about an argument that says $\delta \psi = 0$ because $\partial^2 = 0$, because it seems like you must be showing it's a coboundaryu

or needing that locality argument Steenrod used

my picture is that $\psi$ is something obtained from a boundary after forgetting that it bounded something (because $s$ doesn't extend to $X^{(n+1)})$, so $\delta \psi = \psi \circ \partial$ should be something obtained from a boundary of a boundary after forgetting that it bounded something. But boundary of boundary = 0 so it's going to be 0 nonetheless

if that makes any sense

So it's unclear to me why I'd end up arguing $\psi$ is a coboundary

Yeahhhhh I get that

I am just skeptical for no good reason

I understand your skepticism. It's valid until I write an actual proof :)

I am currently curious how many cells the Poincare sphere has in its cell decomposition as a dodecahedron with face identifications

But I am too lazy to do the combinatorics

so one 3-cell, six 2-cells, one 0-cell.

no idea how many 1-cells

ah I see the one 0-cell is easy to follow because the quotient group acts transitively on the 0-skeleton

yup

i want to know the number of 1-cells but fuck it

Maybe dualize to an icosahedron, identify antipodal vertices instead of faces, and then count the edges

maybe

no

lol

rating of MT?

It looks very concise and clean. I'm going to read the proof the theorem that if $[\psi] =0$ then the section extends from codimension 2

Theorem 3

I think the presentation there is more or less what I remember of the subject

Including the lack of proof of cocycle

hah

observation: for each pair of antipodal faces, the swap-and-fifth-twist is an order ten operations acting simply transitively on the edges

So you get at most one per pair of antipodal faces in the quotient, and then there's some extra work in looking at I guess adjacent faces and seeing what survives?

I don't want to counttt

every edge on the backside is equivalent to an edge on the frontside

every edge on a given front face is equivalent to any other edge on that front face

I forget what's the angle you twist by

pi+2pi/5?

pi/10 I thought

note that the opposite side is not actually aligned with the first side

oops yes

so just the little twist you need to align it with the opposite

i was thinking pi + pi/10 for some reason

that would be ok too I'm sure

The main property is simple transitivity which follows because the action swaps the two sides and it squares to a nontrivial thing on the one side

I keep getting that I'm left with one edge but that ain't right

en.wikipedia.org/wiki/Seifert%E2%80%93Weber_space interesting

I'm painting edges in MS paint and I'm confused

Do you agree there's only one edge left in the end

Let me complete the drawing, I certainly get a lot of identifications on a single face

i am also getting only one face so i must be doing something wrong

I think I'm doing more identifications than I'm allowed to do

You can't identify the face to itself by a pi/5 rotation...

No, I was doing something wrong

I think that's the orbit of a single edge

3??

Seems so

Am I incorrect?

So the rule is, looking through the dodecahedron through the edge's vantage point, you rotate the face it's on by pi/10

And that's order 2 through any given face

Through the top face it gives you the edge on the other side

Have to be... dodecahedron has 20 edges. 3 for each of the 5 gives 3*5 = 15... wut

Ah here's the issue

surely I am missing one other

You didn't do the rotation from the perspective of the top orange guy

wait no

That gives the blue edge on the bottom-left

yeah

I guess edges should be thought of better as hinges between two faces

If we place the given hinges, there are 6 untouched face

s

each in "triads"

The hinges we place form a chain of 6 pentagons around the guy

which we are just rotating I guess

and the triads outside get rotated too

this sucks too much

I get your picture of 6 hinged pentagons

I am not sure there's a group action going on here as opposed to just various identifications

I guess what I'm a little confused about is, if you rotate the front face pi/10 clockwise, the blues edge matches with the blue edge of the back face. But if you rotate the back face pi/10 clockwise....

Maybe that's a perspective issue

yeah it is

It is, I think this must be why this is something possible in 3d but not 2d

Doing a pi/10 rotation from one side is -pi/10 from the other

ahhh

I only see 3 damnit

Sorry I'm a douchebag. There are 30 edges. Lol.

there are 30

yes.

I also just got this

5 edges from two non-adjacent, non-antipodal faces

yeah...

1 vertex, 10 edges, 6 faces.

yup

whew

at least the relations are quite elementary

ams.org/publicoutreach/feature-column/fc-2013-10

Wow what a great hands-on article

Tony Phillips is the master of concrete computations