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Mike Miller
Mike Miller
9:19 PM
@BalarkaSen Spin structures are obvious in dimension >2
Let $G$ be a (well-behaved) topological group. Then it has an $n$-connective cover $G\langle n\rangle$, a topological group such that $\pi_k(G\langle n\rangle) = 0$ for $k \geq n$, equipped with a group homomorphism $p: G\langle n\rangle \to G$ which is an isomorphism on homotopy groups of degree $k > n$.
$G\langle n\rangle$ is universal in some sense, whatever.
A $G$-bundle is classified by a map $X \to BG$. A lift of the structure group to $G\langle n\rangle$ is a lift of this map to the top of the fibration $B(G\langle n\rangle) \to BG$.
Let's do some quick obstruction theory. Suppose I have a map $f: X \to BG$ which has a lift to $B(G\langle n\rangle)$. Because $\pi_k B(G\langle n\rangle)$ is zero for $k \leq n+1$, we see that $\tilde f: X \to B(G \langle n\rangle)$ is null-homotopic when restricted to $X_{n+1}$.
And in particular, the same is true of $f$.
I made some mistakes in the above, let me re-write it.

$\pi_k(G\langle n\rangle) = 0$ for $k < n$, equipped with $p: G\langle n\rangle \to G$ such that $p_*: \pi_k G\langle n\rangle \to \pi_k G$ is an isomorphism for $k \geq n$.
A map $f: X \to BG$ which has a lift to $B(G\langle n\rangle)$ necessarily has $f: X_n \to BG$ null-homotopic, because $B(G\langle n\rangle)$ is $n$-connected.
Conersely, if $f: X \to BG$ has $f_n: X_n \to BG$ null-homotopic, we may construct a lift $\widetilde f_n: X_n \to B(G\langle n\rangle)$. Consider the problem of extending this to a lift $f: X \to B(G\langle n\rangle)$.
Write $F$ for the fiber of $B(G\langle n\rangle) \to BG$. Then $\pi_k F = 0$ for $k \geq n$ and $\pi_k F = \pi_k G$ for $k < n$.
So long as $f_n$ is null-homotopic, our next obstruction to a lift lies in the group $H^{n+1}(X;\pi_n F) = 0$, because $\pi_n F = 0$; and similarly all higher obstructions vanish.
Thus: a $G$-bundle lifts to a $G\langle n\rangle$ bundle if and only if the $G$-bundle trivializes over the $n$-skeleton.
For $G = O(k)$ and $k \geq 1$, we have $G\langle 1\rangle = SO(k)$, so that an orientability is induced by a trivialization over the 1-skeleton.
For $k \geq 3$ we have $G\langle 2\rangle = \text{Spin}(k)$, so that spinnability is equivalent to trivializability over the 2-skeleton.
Now, using obstruction theory, one should further be able to prove that if $f: X \to B(G\langle n-1\rangle)$ is a lift of some $G$-bundle, then there is only one obstruction to $f$ lifting to $B(G\langle n\rangle)$, and it lies in $H^n(X;\pi_{n-1} G)$.
Further, so long as you choose one, lifts of $f$ to $B(G\langle n\rangle)$ are in bijection with classes in $H^{n-1}(X;\pi_{n-1} G)$.
This matches the story for orientations and spin structures.
For an only slightly more exotic example: take $G = U(n)$. Then $G\langle 2\rangle = SU(n)$. $G\langle 1\rangle = G$, so there is a single obstruction to lifting a $U(n)$-bundle to an $SU(n)$-bundle, and it lies in $H^2(X; \pi_1 U(n)) = H^2(X;\Bbb Z)$. This is $c_1$.
The "trivialization over the 2-skeleton" is what we need to trivialize the determinant line bundle.
Let's do this for the stable group $U$. We know that $\pi_k U$ is 2-periodic, with $\pi_0 = 1$ and $\pi_1 = \Bbb Z$.
I will observe that therefore, $\pi_k U\langle 2n\rangle$ is first nonzero in degree $2n+1$, where it is equal to $\Bbb Z$.
Therefore $H^{2n+2}(B(U\langle 2n\rangle);\Bbb Z) = \Bbb Z$ by Hurewicz. Consider the map $H^{2n+2}(B(U\langle 2n\rangle);\Bbb Z) \leftarrow H^{2n+2}(BU)$. What is the image of this map?
I would bet anything that the image is $(n+1)!\Bbb Z$, with $(n+1)!$ being hit by $c_{n+1}$.
I will maybe try to work this out on paper.
(This smells related to that one fact about Bott periodicity we talked about a while back.)
 
Balarka Sen
Balarka Sen
10:17 PM
Oh shoot I have to catch up
 
Mike Miller
Mike Miller
Jan 27 at 20:49, by Mike Miller
Now the differential $x -> y$ is instead encoded as four summands like $(x,o) -> (y,o)$, counting the intersection points with orientation, such that if you swap orientations on either side you negate this count by -1
I want to phrase this from a while back in a better way. Let $\Lambda$ be a 2-element set, with the obvious free $\Bbb Z/2$ action. There is a group $\Bbb Z_\Lambda = \Bbb Z \otimes_{\Bbb Z[\Bbb Z/2]} \Bbb Z^\Lambda,$ where $\Bbb Z/2$ acts on $\Bbb Z$ by negation. If we write $\Lambda = \{e_1,e_2\}$, then $\Bbb Z_\Lambda$ consists of pairs $(m,n)$ modulo setting $(m,n) = -(n,m)$.
Then a map $\Bbb Z_\Lambda \to \Bbb Z_\Lambda$ is best encoded as follows: it's a quadruple of maps, $f_{xx}: \Bbb Z_x \to \Bbb Z_x$, similarly $f_{xy}, f_{yx}, f_{yy}$. And these maps satisfy that $f_{xx} = -f_{xy} = -f_{yx} = f_{yy}.$
The moral of this is that whenever you have to make some choice, but the maps defined via that choice obey some sign rule, then instead of making a choice you can instead encode $\Bbb Z$ as $\Bbb Z_{\Lambda}$ --- and define a map between these by looking at all possible choices and showing they're compatible.
 

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