instantly stars the declaration of instantaneous starring, while also instantaneously starring the declaration of instantaneous starring of the declaration of instantaneous starring so nobody else can instantaneously star this message
If $H$ is closed, $H \to G \to G/H$ is a fiber bundle, right? Pick a small nbhd $U$ of $G/H$, then $p^{-1}(G/H)$ is foliated by the $H$-cosets in $G$ varying over $U$.
If $G$ is a simply connected topological group and $H$ is a discrete closed normal subgroup, then $\pi_1(G/H,1)\cong H$. This generalizes $\pi_1(S^1) = \mathbb{Z}$
So I've seen Peter do the bar construction once but this was before I knew any algebra or topology (I knew Sylow theorems and group actions but not enough to verify in my mind what he was doing)
So let's say for all intents and purposes that I don't
(Also in the general list of things to do, if you'd be down for going through stuff after that, would be to do the LES of a pair, of a fibration, and the business about fiber bundles => fibrations)
So let's say $X$ is a topological space and $p : E \to X$ is a fiber bundle over $X$ with fiber space $F$. This means, remember that there are open sets $U_\alpha$ covering $X$ such that there is a commutative diagram of maps $p : p^{-1}(U_\alpha) \to U_\alpha$, $h_\alpha : p^{-1}(U_\alpha) \to U_\alpha \times F$ (the trivialization) and $\pi_1 : U_\alpha \times F \to U_\alpha$ (first projection).
So for any two overlaping $U_\alpha, U_\beta$, there is this map $h_\alpha \circ h_\beta^{-1} : U_{\alpha \beta} \times F \to U_{\alpha \beta} \times F$ where $U_{\alpha \beta} := U_\alpha \cap U_\beta$. This map can be written as $(x, f) \mapsto (x, \varphi_{\alpha \beta}(x)f)$ where $\varphi_{\alpha \beta} : F \to F$ is what the map does in the second component. This is the transition function.
So for every pair of overlaping $U_\alpha, U_\beta$, there is a map $\varphi_{\alpha \beta} : U_{\alpha \beta} \to \text{Homeo}(F)$ which is the transition function over that overlap. These determine how the trivializations over $U_\alpha$'s are glued togather to make $E$, so these functions in fact determine the fiber bundle.
Just to recall basic stuff
Suppose $G \subset \text{Homeo}(F)$ is a subgroup of homeomorphisms of the fiber space $F$. An $F$-bundle $E$ over $X$ is said to have "structure group $G$" if all the transition functions $\varphi_{\alpha \beta}$ takes values in $G \subset \text{Homeo}(F)$.
Yeah, so I didn't see the transition functions but that's basically gonna be the statement that your local trivialization and a bunch of open sets are compatible with each other
To give an example of where the transition functions are relevant, consider a rank $n$ vector bundle over $X$. That's a fiber bundle with fiber space $\Bbb R^n$, but more than that: the transition functions are linear - they take values in $\text{GL}_n(\Bbb R) \subset \text{Homeo}(\Bbb R^n)$
Because that's what it takes to incorporate the vector space structure of $\Bbb R^n$ in the story - demand that the compatibility is linear wrt that vector space structure
Okay so the business and the structure group, maybe this'll be answered later but would the transition functions not form a group themselves?
Right now the wording of the statement about structure group strikes me as allowing any group which contains all the transition functions to count here, while if they did form a group themselves I would guess that we'd just call that the structure group
Transition functions are maps from the overlaps $U_{ij}$ to the group $\text{Homeo}(F)$, for $i, j$ varying over the cover of $X$ so it's unclear to me how you'd make that into a group.
The structure group is just the subgroup of Homeo(F) in which the transition functions takes their value in
Because in principle, let's say the transition functions' images live in some group G, well they also live in any larger group. So are we just taking the group they generate? Do they form a group themselves? Etc
Yeah I see what you mean. I don't think it's correct to think that there is a smallest possible structure group, no. You can actually enlarge and shrink them all the time.
Consider a rank $n$ vector bundle $E$ over $X$. This has structure group in $GL_n(\Bbb R) \subset \text{Homeo}(\Bbb R^n)$. But if you give this vector bundle a fiberwise Riemannian metric, so that each $\Bbb R^n$ fiber is an inner product space now (and the inner product varies fiber to fiber continuously), then this is the same thing as to shrink the structure group to $O(n) \subset \text{GL}_n(\Bbb R)$
So imposing more structure on the vector bundle => shrinking the structure group
More or less
Eg, you if your rank $n$ vector bundle is orientable, the structure group shrinks to $\text{GL}_n^+(\Bbb R) \subset \text{GL}_n(\Bbb R)$, to give another example (that subgroup is the subgroup of $n\times n$ invertible matrices with $\det > 0$)
A-priori the structure group is in GL_n(R), but once you claim that it's orientable and actually choose an orientation fiberwise, it shrinks to GL_n^+(R)
Ah, so it's basically just what acting group we restrict attention to. So if we're not adding structure we're just gonna say Homeo(F) is the structure group, etc
For any closed group $G \subset \text{Homeo}(F)$, $BG$ is the unique space upto homotopy equivalence such that $F$-bundles over $X$ with structure group $G$ are in 1-1 correspondence with homotopy classes of maps $X \to BG$.
The phrasing here suggests otherwise but if you were to give me some random group, it doesn't seem like there's a God-given $F$. Or can we just choose however we please and the answer won't depend on the choice?
@Daminark Yup, the choice of the fiber space doesn't actually matter!
Here's a more clarifying parsing. Consider the functor $\mathscr{F} : \mathsf{Top} \to \mathsf{Set}$ where $\mathscr{F}(X)$ is the set of $F$-bundles on $X$ with structure group $G$.
Then there is a space, which we call $BG$, that represents $\mathscr{F}$. Namely, there is the functor $[-, BG] : \mathsf{Top} \to \mathsf{Set}$ which sends a space $X$ to $[X, BG]$, and these two functors are naturally isomorphic
@Mike Is this right?
Also yeah Mike's point about the god-given F being G is really important. Please note that.
@Daminark The point is, if you have an $F$-bundle on $X$ with structure group $G$ and if you have an $F'$-bundle on $X$ with structure group $G$ and they have the same transition functions, then they both correspond to the "principal $G$-bundle" where the fiber space is $G$, and the same transition functions are used.
The crucial slogan is, a fiber bundle is nothing but (1) + (2) where (1) is choice of a fiber space (2) is choice of transition functions
So for classification of fiber bundles with structure group $G$, leaving the fiber space unspoken shouldn't matter
Cool. I can tell more stuff on $BG$ but I think doing this while I am sick is not such a great idea.
So instead we can idly chat about what we have thought so far, and I can ask my question:
If $H \subset G$ is a closed subgroup of a topological group $G$, we had the fiber bundle $H \to G \to G/H$. Now, this is a principal $H$-bundle over $G/H$... so it's classified by a map $G/H \to BH$, right?
@MikeMiller Is it correct that this is the first arrow of the fibration $G/H \to BH \to BG$?
@Daminark The cool thing is that if $0 \to H \to G \to K \to 0$ is a short exact sequence, then there is a fibration $BH \to BG \to BK$. So for example, our SES $0 \to H \to G \to G/H \to 0$ gives a fibration $BH \to BG \to B(G/H)$.
I was wondering if we can splice these togather to make $H \to G \to G/H \stackrel{?}{\to} BH \to BG \to B(G/H)$
where the question marked arrow is the map I suggested
Also $G\to G/H$ doesn't strike me as much of a fiber inclusion. Of course Mike's gonna know this better than I do but it's at least tough to swallow that this is a fibration
It's just quotient of the fiber inclusion G --> EG by H, yeah.
So it classifies pullback of EG --> EG/H = BH by G/H --> BH
Which should be nothing but G --> G/H
(side-quest: Now I am wondering if I can continue this (and this might sound like black magic @Daminark, but I can explain what $B^n$ means later) to $\cdots B^{n-1}(G/H) \to B^n H \to B^n G \to B^n (G/H) \to \cdots$ where $B^{-n} = \Omega^n$ in the negative direction too, and what does feeding that to the functor $[X, -]$ mean? It should give a gigantic LES for homology of $\Sigma^n X$ and $\Omega^n X$ with various coefficients)
Hmm, would it be a good idea to try to take a hint from algebra and define a map $S^1\times S^1 \to S^2$? Then show that distinct points have the same image when they're in the wedge?
Actually it might be helpful to think of $S^1$ as $I/\partial I$
I know. I'm just saying, I don't get Ronnie Brown's explanation of why the fundamental groupoid approach is useful by demonstrating how it can compute $\pi_1 S^1$. That's not much of an accomplishment. What else can you do with it?
I am sure you can do more, I just haven't seen a single inspiring example
Can you compute some higher homotopy groups that I can't? That'd convince me.
@BalarkaSen I guess this doesn't prove new stuff (at least, as far as I know?), but I like that with groupoids, we can model a covering map between spaces with a covering map of groupoids
it feels like the algebra is closer to the topology
$[\Sigma^2 X, Y]= [\Sigma X, \Omega Y]$. $\Omega Y$ is a $h$-group and $\Sigma Y$ is a $h$-cogroup, so that will be an abelian group by some general result
Okay so, the argument I saw for higher homotopy groups being abelian used this thing about how if you have two operations $*,*'$ on a set which distribute over each other, then they agree and are abelian
Generally if $X$ is a $h$-cogroup, then $[X,Y]$ is a group. If $Y$ is a $h$-group, then $[X,Y]$ is also a group. If we have both, then both induced group operations on $[X,Y]$ coincide and they are abelian
Okay so, lemme be sure about this $h$-cogroup. So you said pinch the meridian? Meaning $\Sigma X = X\times I /(\{x_0\} \times I \cup X\times \partial I)$, and we're squashing $X\times \{\frac{1}{2}\}$
lmao this actually seems like a good exercise to familiarise oneself with some of the stuff in etale cohomology
i'm slightly confused to the part where you pass to integer coefficients - that seems to suggest to me that you can do this in general but i don't think it's true in general that $H^*_{et} (X,\mathbb{Z}) = H^*_{sing}(X(\mathbb{C}),\mathbb{Z})$?
And in general if you have a map $G\to G\times G$, we want to define multiplication of homotopy classes of maps outta $G$ to be, so given $f:G\to H$ and $g:G\to H$, we define the map $G\to_{\text{belt}} G\vee G \to_{f\vee g} H$
so etale coh lacks the thing (universal coefficients probably) that allows you to transfer your result from $p$-adic coefficients to $\mathbb{Z}$-coefficients as in the singular case here
etale with $\Bbb Z$-coefficients exists in theory, but it's never used
because it lacks any good properties
you prove all result for torsion coefficients and then you take an inverse limit (outside the cohomology, not inside the coefficients, this is not the same!) to lift it to $\Bbb Z_p$
Oh I guess that "$\times$" was supposed to be $\vee$. Then if you have $G\times G\to G$, then we want the maps into it to be sending $X\to_{f\times g} G\times G \to_{\text{groups tho}} G$. Of course all this is up to homotopy. So now what's left is to show that $\Omega Y$ is an h-group and that the two operations on $[\Sigma X, \Omega Y]$ satisfy the requirement for Eckmann-Hilton
hmm ok that's pretty cool. I always thought that the fact that etale coh is only an isom with singular coh with finite coefficients is somewhat 'lacking' - but you're saying that in fact this recovers everything about singular coh
note that this is not special to the homotopy category. If $C$ is any category, then $Hom_C(X,Y)$ is a group of $Y$ is a group object and also if $X$ is a cogroup object and if we have both, then both group operations coincide and are abelian
@loch my argument for that worked only using the fact that the cohomology is finitely generated
@Daminark Stuff: Suppose you have a principal $G$-bundle on CW complex $\Sigma X$. This means a fiber bundle on $X$ with fibers homeomorphic to $G$ such that the transition functions take values in $G$ (i.e., structure group is $G$). There are two inclusion maps $CX \to \Sigma X$ - taking the upper cone, or the lower cone. Pull the bundle back by these inclusion maps (or rather, restrict to the cones). $CX$ is contractible, so the $G$-bundle trivializes over $CX$
So you have two independent copies of $CX \times G$
So we want to create a map $\Omega Y \times \Omega Y \to \Omega Y$. So given $f,g:I\to Y$ such that $f(0) = f(1) = g(0) = g(1)$ we just concatenate paths, basically it's the idea of $\pi_1(X)$ being a group?
You want to glue them back to the original $G$-bundle on $\Sigma X$. To you glue them cones $CX \times \{0\}$ and $CX \times \{1\}$ along the bottom $X$ of each. But who's to say how the fibers over $X \times \{0\}$ and $X \times \{1\}$ are glued? Well, it's a principal $G$-bundle, so they better be glued by multiplication by some $g$: $G \to G$, $x \mapsto gx$.
So for each $x \in X$, the fibers are glued above by multiplication by $g_x \in G$.
Aka, you have a map $X \to G$, $x \mapsto g_x$.
This is called the clutching function construction. It can be shown that homotopy clutching functions give isomorphic bundles. So principal $G$-bundles on $\Sigma X$ are in 1-1 correspondence with homotopy classes of maps $X \to G$. More adeptly, $\text{Princ}_G(\Sigma X) \cong [X, G]$.
From previous discussion we also know that $\text{Princ}_G(\Sigma X) \cong [\Sigma X, BG]$ - that's the classifying map construction! We defined $BG$ that way.
So $[\Sigma X, BG] \cong [X, G]$. Apply loopspace-suspension adjunction on the first thing: $[X, \Omega BG] \cong [\Sigma X, BG] \cong [X, G]$.
This is a natural isomorphism of functors. Yoneda lemma kicks in.
$\Omega BG \cong G$.
That's why $B$ is oftentimes called the "delooping" operator. Loopspace of $BG$ is homotopy equivalent to $G$
Okay, so let me completely absorb what's going on here. So you pulled back your bundle $G\to E \to \Sigma X$ twice to get two copies of $G\to CX\times G \to CX$. And the process of gluing bundles together works how?
