I do, I am actually thinking about it (along the lines of how I was thinking of it before). But I'll be honest that I don't really have an answer for your structure group question.
@MikeMiller: I have a guess. I think it also explains why you could give me counterexamples for dimension 2 for the thing I conjectured (about the bundle being trivial on 2-skeleton).
You probably want to resolve the 1-homotopy type of the structure group (from O(n) to SO(n) we resolved the 0-homotopy type).
That is, take it's universal cover for n > 2. For n = 2, it's a circle, that's why you can't resolve and this should also be the reason for having counterexamples in dimension 2.
Modular forms are sections of a certain vector bundle equivariant under the action of the "modular group". The modular group is a group of matrices whose entries are 1 mod N.
@BalarkaSen A spin structure on an oriented vector bundle is a choice of lift $X \to B\text{Spin}(n)$ of the classifying map to $BSO(n)$. There is a fiber sequence $B(\Bbb Z/2) \to B\text{Spin}(n) \to BSO(n)$. Let's try to construct the lift cell-by-cell. Of course these classifying spaces are simply connected, so you can lift the one-skeleton.
Also modular group is SL(2, Z). The thing Mike is referring to are their subgroups. IIRC modular forms are certain 1-forms on the quotient of the upper half plane by those.
@BalarkaSen Consider a 2-cell as a map $c: S^1 \times [0,1] \to X \to BSO(n)$. Lift this upstairs (fitting with the lift you already chose for $S^1 \times \{0\}$ inside the 1-skeleton). Then the lift of $c(S^1 \times 1)$ lies inside the fiber, where it's either null or not null. Assign $o(c)$ to be $1$ if that lift is non-null, otherwise zero.
@TedShifrin So basically, the rule is either the number needs to satisfy the clock (in this case 16, since that is -1 from 17) or be -1 from a multiple of 17?
Wenn der reelle Theil von s negativ ist, kann das Integral, statt positiv um das angegebene Gro¨ssengebiet auch negativ um das Gro¨ssengebiet, welches s¨ammtliche u¨brigen complexen Gro¨ssen entha¨lt, erstreckt werden, da das Integral durch Werthe mit unendlich grossem Modul dann unendlich klein ist. Im Innern dieses Gro¨ssengebiets aber wird die Function unter dem Integralzeichen nur unstetig....
This defines a cellular cochain $o \in C^2(X)$. It's actually a cellular cocycle, defining a cohomology class $[o] \in H^2(X)$. If $[o]$ vanishes, it's a reasonably easy fact to see that the whole 2-skeleton has a lift upstairs (though you might need to fiddle with the 1-skeleton's lift).
@BalarkaSen Now, basically as a corollary of the fact that the fiber is aspherical, you can then lift all the higher skeleta. So $w_2$ vanishes iff the bundle has a spin structure (given that $w_1$ already vanished).
When $n>2$, and $X$ is a 2-complex, then a Spin(n)-bundle over X is automatically trivial. So if $w_1 = w_2 = 0$ and $n>2$ the bundle trivializes over the 2-skeleton.
A spin structure is a homotopy class of non-vanishing vector fields on the one skeleton which extend to the two skeleton. This is the only way I think about these things anymore.
There is a story about going up the Whitehead tower of O(n), but it stops being about Stiefel-Whitney classes, and you're no longer working with Lie groups, so it's hard to apply to differential geometry. It shows up in algebraic topology but I'm mostly ignorant of it.
I think I ever end up teaching beginner algebraic topology, I would like to spend an above average amount of time explaining how to actually construct homotopies, instead of expending nearly all of the time manipulating various obstructions.
It's instanton homology, and you don't pour it in your water, you snort it.
@TedShifrin The word 'handwavy' should only be used if something is not rigorous. Topologists are rigorous. They just approach the mathematics they do in a different way.
So, why does one care about Spin structures? $w_2 = 0$ doesn't sound like a satisfying enough reason, but it's still interesting that such (oriented) bundles are classified by maps to BSpin
@BalarkaSen Differential geometry perspective: Spin structures give us Clifford algebras and Dirac operators. A corollary is Rohlin's theorem: A spin 4-manifold has signature divisible by 16. You can also prove that a simply connected 4-manifold has a spin structure iff its intersection form is even. As an immediate corollary, Freedman's E8 manifold is not smoothable.
@PVAL-inactive It's a paper but everything else is right.
There's a whole nice book called Spin geometry by Lawson and Michelson that goes through the topological, geometric, and analytic aspects of spin structures.
Rohlin's invariant of 3-manifolds has you pick a spin 4-manifold that bounds the 3-manifold, divide its signature by 8, and take that mod 2.
Pick another one. Glue them together along the boundary. Signature is additive under gluing along a homology sphere. We see that the signature of the two things must differ by a multiple of 16 by Rohlin's theorem.
It's an open question whether the homomorphism $\beta: \Theta^3_{\Bbb Z} \to \Bbb Z/2$ lifts to a homomorphism to $\Bbb Z$. It's known that there is a map $\beta$ with the property that $\beta(-Y) = - \beta(Y)$ that lifts Rohlin, but it's not a homomorphism. (My advisor proved this, and the triangulation conjecture is false as a corollary.)
@DanielFischer The integrals $\int_{0}^{\pi/4} \frac{\sin(xt^{2})}{2x} \left(\frac{2 \sec^{2}(t) \tan(t)}{t} - \frac{\tan^{2}(t)}{t^{2}} \right) \,dt$ and $\int_{0}^{\pi/4} \frac{\sin(xt^{2})}{2x} \, dt$ should behave somewhat similarly as $x \to \infty$. But how would you argue that both integrals have the same leading-order behavior?
The reason I ask is because the leading-order behavior of $-\int_{0}^{\pi/4} \frac{\sin(xt^{2})}{2x} \, dt$ is exactly the second term in the asymptotic expansion of $\int_{0}^{\pi/4} \cos(xt^{2}) \tan^{2}(t) \, dt$ as $x \to \infty$.
@RandomVariable Consider $$F(t) = \frac{2\tan t}{t\cos^2 t} - \frac{\tan^2 t}{t^2} - 1.$$ Note that at $0$ we have $F(t) \in O(t^2)$. Thus you can integrate $\int_0^{\pi/4} F(t)\sin (xt^2)\,dt$ by parts. Show that that has lower order than $\int_0^{\pi/4}\sin (xt^2)\,dt$.
@BalarkaSen No, that's not what he said. A spin structure is a choice of nonzero vector field on the 1-skeleton, chosen up to isotopy, that extends to a nonzero vector field on the 2-skeleton.
A spin structure is a homotopy class of non-vanishing vector fields on the one skeleton which extend to the two skeleton. This is the only way I think about these things anymore.
A spin structure is a homotopy class of non-vanishing vector fields on the one skeleton which extend to the two skeleton. This is the only way I think about these things anymore.
Anyway, I think of it as exactly what it is. A lift of the structure group to Spin(n). Some people think about it in terms of a Clifford module structure on the tangent bundle.
I think they're interested in knots which do surger to lens spaces. But starting to ask questions about 3-manifolds that carry such knots would be interesting.
That seems extremely difficult though. Take eg $S^3$. If you could prove that a 3-manifold $Y$ didn't have a knot that surgered to $S^3$ wouldn't you prove it's not surgery on a knot, by going backwards?
if you could do that people would be very interested, since the only existing proofs of that are either heegaard floer theoretic or come from Taubes periodic ends theorem
@MikeMiller You gotta talk about 2-frames to show $TY^3$ is trivial. I'm not sure how you do that with spin structures as the orthogonal complement of a nonzero v.f is not necessairly trivial.
@MikeMiller Maybe its obvious that every orientable 2-bundle over a 2-skeleton is trivial after summing with the trivial bundle. I don't remember how to do that though.
I want to evaluate $$\int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx$$ using residues
How can I let the starting point for the corresponding contour integral be $z=-\infty$?
I.e turn it into an improper integral
Maybe $$\int_{-\infty}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx = 2 \int_{0}^{\infty} \frac{x^2}{(x^2+1)(x^2+4)} dx$$ since the integrand looks pretty even
"For a different approach, recall that Milnor [5] observed that a spin structure on an oriented vector bundle over a CW–complex is equivalent (after stabilizing if necessary) to a trivialization over the 1–skeleton that can be extended over the 2–skeleton"
I guess its really a trivialization of the entire vector bundle after stabilizing and not just a cross-section as Milnor seems to state.
Maybe this is old terminology playing with me and a "crossection of an SO(3)-bundle" is really a trivialization of the corresponding vector bundle.
If I have a singularity in the upper plane and a singularity in the lower plane and I want to enclose these singularities in two connected semicircular contours, do these contours necessarily have different directions?
@TedShifrin I'm supposed to compute the integral by employing residue theory on $$p.v \int_{-\infty}^{\infty} \frac{e^{i3x}}{2(x^2+4)} dx + p.v \int_{-\infty}^{\infty} \frac{e^{-i3x}}{2(x^2+4)} dx$$
First of all, @Lozansky, just do the real part of the first (without the 1/2). No need to do two integrals. Now you must pick a contour that guarantees $e^{i3z}$ decays ... If you do that, $e^{-i3z}$ will blow up, so you can't win trying to do both simultaneously.
