@BalarkaSen you might like this

Challenge, which I won't think about since I should be working on other stuff: find a proof that does this much more quickly

I suspect you an very quickly reduce to the $\Bbb Z/2^\ell$ case using the fact that the group action contains the action of $-1$, which should force the cohomology to be 2-torsion

I would also like to see the LHS argument removed but this really is an extension and not a product so I cannot use Kunneth

It seems for the case of $GL_n \Bbb Z$, set $\Gamma_n(2) \subset GL_n \Bbb Z$ to be the subgroup of matrices which reduces to the identity mod 2. (This is not what is usually called the "congruence subgroup" since I ignore the determinant.) I think I need to know $H^*(\Gamma_n(2); \Bbb F_2)$.

I also don't know anything about GL_n(Z/2^k)

I suspect those corners are totally untouchable

$M$ is a retract of $TM$ is it not

I'm referring to the zero vector field

will it work to just contract every vector to zero?

Do you mean a deformation retract?

Retract should be easy since you just project

Deformation retract... I do feel like scaling every vector to 0 should do it but I want to be sure that there's no problem with orientability

why do I need an orientation?

I think he's worried that the image of determinant under O(n) is disconnected, but that's not relevant here

or smth

anyway the "cleanest" definition of vector bundle makes this clear rather quickly, and you should be able to prove that your def'n is equivalent to this: a vector bundle is a pair of smooth manifolds $E,M$ with a surjective submersion $\pi: E \to M$, and smooth maps $+: E \times_M E \to E$ and $\lambda: \Bbb R \times E \to E$ which satisfy the vector space axioms fiberwise

(for topological spaces you need to figure out what 'submersion' should mean topologically)

So I want $TM \times I \to TM : (D_{p,v},t) \mapsto (1-t)D_{p,v}$

looks fine to me

with this notation, $f_t(v) = \lambda(1-t, v)$ is a (smooth) homotopy between the projection to the zero section and the identity

which is what you wrote

indeed nothing to do with orientations

thanks

sure, it gave me an excuse to mention that nice(r) definition

Basically, I'm wondering if trying to scale everybody to 0 can be done on each tangent space simultaneously if you're not orientable. But I guess you could use the LES of homotopy groups and the projection induces a WHE so I buy that it works

:P

And lemme read what you wrote

@Daminark I'd like to understand why you think that so I can help clarify

WHE?

weak homotopy equivalence, ignore him

he's using big tools because he was never taught to just draw the deformation retraction :)

<3

Lmao

Okay so I think I see what was messing me up a bit, I had thought that when you tried to scale every vector down, then if you chose different bases that'd give a matrix of possibly different determinant signs, so you needed to do something consistently

can I find SO3 and S3 inside TS^2?

That's what I was mumbling about O(n)

I think Balarka gave me SO3 inside TS^2

@LeakyNun S^3 in a dumb way (it's a 4-manifold so choose a coordinate chart), SO(3) because you can prove that's the unit tangent bundle

oh it's called the unit tangent bundle, lol

if you want the map S^3 -> S^2 to be the Hopf map then no but i dont wanna prove it

ok thanks

yeah by "inside TS^2" I mean a homomorphism of bundles

they are all bundles on S^2

that is not quite what I would call a homomorphism of bundles but it's clear what you mean

a map in the arrow category

ok thanks

am i killing the fun by responding to all these questions?

if so i can stop

not at all!

the fun is paramount

ok! :)

so you always have a lot of smooth functions

but you can have no top form at all

careful, you can have no nonvanishing top form

you always have lots of sections just like you have lots of smooth functions

oh...

aha

so what happens on a non-oriented manifold is like, you can find some nice section of $\Lambda^n T^*M$, and you can make it nonzero most everywhere, but (in the nicest case) there will be some codimension 1 submanifold that your section is zero on

the non-orientability says that when you try to build a nonvanishing section you get "caught" on that submanifold (more or less), just like when you try to find an antiderivative of $1/z$ on $\Bbb C^\times$ - as you walk around the loop there is an obstruction

ok so someone told me that an orientation is a non-vanishing top form

and I'm trying to think about how this is related to normal vectors

they are right!

aha

and I think I came up with an alternative definition but I'm not sure if I'm right

I want to define an orientation as a section of the "$\mu_2$-bundle"

Ah yeah I had just gotten to that theorem in Bott-Tu but trailed off since classes started

where $\mu_2 = \{-1,1\}$

@Daminark which theorem?

Orientation = non-vanishing top form

then what is an orientation?

is there also monodromy for orientation lol

Well, Bott-Tu had defined being orientable as having an atlas with transition maps being orientation-preserving. Neves was like, smooth choice of basis on each tangent space, and he had this orientation double cover going on

oh

I've heard that smooth choice of basis is equivalent to tangent bundle being trivial instead

(which is not equivalent to orientability)

@LeakyNun heck yeah

@LeakyNun That's correct, how do you define that bundle?

@MikeMiller locally, I guess

I can give you a better definition if you like, or I can help you massage the one you have in mind

I don't really know how to define bundles

Ahhh

(can you do both? =P)

In fact I decided after your last comment I should, morally

thanks a million

Clean approach: given a vector bundle $E \to M$ of rank $n$, consider the action of $GL_n$ on $E \setminus 0$ (which is a $\Bbb R^n \setminus 0$ bundle), and on $\mu_2$ by the action of the determinant. Then the "orientation double cover of $E$" is the double cover $\Lambda(E) \to M$, where $$\Lambda(E) = (E \setminus 0) \times_{GL_n} \mu_2.$$

That is, all you're saying is "Forget the bases, just remember the orientation", and there is a way to make this happen just by taking products and quotients - insert a factor for arbitrary orientations, and then quotient by the rule saying a basis is equivalent to the corresponding orientation

Approach that is better for your conceptual understanding:

A vector bundle is specified by the following data: an atlas $\mathcal U$ on $M$, and maps $\rho_{ij}: U_i \cap U_j \to GL_n$, with $\rho_{ii} = 1$ and $\rho_{ij} \rho_{jk} = \rho_{ik}$, possibly up to me having put these in the wrong order.

We imagine that over each $U \in \mathcal U$ we have fixed isomorphisms $E\big|_U \cong U \times \Bbb R^n$, and the functions $\rho_{ij}$ are mediating the difference between these fixed isomorphisms over each $U_i \cap U_j$

but $\mu_2$ isn't a vector space

are you responded to "a vector bundle is..." ?

I'm not done

ok sorry

no worries

This is usually called the cocycle description of vector bundles; this data is precisely what is called a Cech cocycle in $Z^1(M; \mathcal{GL}_n)$, where the latter is the sheaf of groups given by $\mathcal{GL}_n(U) = \text{Map}(U, GL_n)$; one can perfectly make sense of first cohomology with coefficients in sheaves of nonabelian groups, and what the above description says is $$H^1(M; \mathcal{GL}_n) \cong \text{Vect}_n(M).$$

that mathcal looks ugly but ok.

There's no real reason here to restrict to $GL_n$. Here is the general definition: a fiber bundle $E \to M$ with fiber $F$ and structure group $G \subset \text{Aut}(F)$ (automorphisms continuous or smooth or whatever is appropriate here) is equipped with local trivializations $E \big|_U \cong U \times F$ as above, so that the transition maps $U_i \cap U_j \to \text{Aut}(F)$ always factor through $G$

Then a $G$-valued cocycle as above gives you via the same recipe a bundle $F \to E \to M$, determined up to isomorphism by the corresponding cohomology class $H^1(M; \mathcal G)$ - aka, essentially uniquely determined by the cocycle

For you, $F = G = \Bbb Z/2$

How does $GL_n$ act on $\mu_2$? I'm a bit slow

$A \cdot x = |\det(A)| x$

oh...

it seems I was being sloppy above, I apologize

it is indeed not $(E \setminus 0) \times_{GL_n} \mu_2$ as I guess you were confused about

I should have taken that to be $\text{Fr}(E) \times_{GL_n} \mu_2$, where the first is the fiber bundle with $$\text{Fr}_x(E) = \text{Space of bases for } E_x.$$

That's a pretty big mistake, sorry

In the second description, here is what's going on

We have a group homomorphism $GL_n \to \mu_2$, $A \mapsto |\det A|$

Therefore, if we have a bundle described by an atlas with transition functions $\rho_{ij}: U_i \cap U_j \to GL_n$ above, we may replace these by $|\det \rho_{ij}|: U_i \cap U_j \to \mu_2$

Because $|\det A|$ is a group homomorphism, these new transition functions still satisfy the cocycle law $\rho_{ij} \rho_{jk} = \rho_{ik}$

you mean $\operatorname{sgn}(\det A)$?

sorry

I'm trying to follow

So Fr(E) is a bundle of M?

I guess it's just the GL(n)-bundle of M, if that even makes sense?

@LeakyNun oh... yes, another mistake :(

it's alright...

@LeakyNun Exactly, these are called "Principal G-bundles", and we took a vector bundle and spat out a principal GL(n)-bundle

It's easier to pass between different principal bundles by the formal operations above, so it is nice to pass to that first

In general, "Fiber bundle with structure group $G$" and "principal $G$-bundle" carry the exact same data; you can go back and forth from one to the other (there's an eq. of categories)

The transition function stuff is mainly how I think about this in practice though... maybe in the last year I have converged towards the cleaner language but that's mostly because I haven't had to compute much lately or think in serious detail

and what does $\times_{GL_n}$ mean?

oh, sorry, that one I didn't realize I was being sloppy about

if $X$ has a right $G$ action and $Y$ has a left $G$-action, the "balanced product" $X \times_G Y := (X \times Y)/G$, with the equivalence relation precisely being $(xg, y) \sim (x, gy)$

the function for us above: replace the $GL_n$ fiber with a $\mu_2$ fiber, as $GL_n \times_{GL_n} X = X$

what is $H^1(S^2;\mathcal{GL}_2)$?

heh, so here the fancy language fails me - when I say that, I just think of it as "cocycle descriptions of vector bundles, up to equivalence". To answer that I would need to just classify vector bundles, instead of really calculating any cohomologies by hand like you might want

I would instead calculate $\text{Vect}_2(S^2)$, which I know how to do, but uses homotopy-theoretic technology

(Early theorem: there is a space, called $BGL(2)$, so that $[S^2, BGL(2)] = \text{Vect}_2(BGL(2))$, where the first set means "homotopy classes of maps"

It is easier to compute the pointed homotopy classes and then determine the action of $\pi_1(BGL(2)) = \pi_0 GL(2) = \mu_2$ on this set

lol

v sweet

Then the point is that $\pi_2 BGL(2) = \pi_1 GL(2) = \Bbb Z$, while the action of $\Bbb Z/2$ on this is induced by the conjugation-action of some element in the other component, and for us that reflects the identity component

So $\text{Vect}_2(S^2) = \Bbb Z/(x \sim -x) = \Bbb N$ (as sets, not groups, of course)

this bijection comes geometrically through intersection numbers: given a bundle $E$, choose a section $s$ which intersects the $0$ section transversely, and the map sends $E \mapsto \big| \#(s(x) = 0)\big|$, where the # is signed intersection

we take absolute values because the actual final sign depends on an orientation of $E$, but every bundle is isomorphic to its orientation-reversal

I'll be back later feel free to keep talking/thinking