Vtyjkyuk

• The first Stiefel–Whitney class $w_1$ is zero if and only if the bundle is orientable. In particular, a manifold $M$ is orientable if and only if $w_1(TM) = 0$.




• The first and second Stiefel–Whitney classes are zero, $w_1(TM) =w_2(TM) = 0$, if and only if the bundle admits a spin structure.




• The third integral Stiefel–Whitney class is zero if and only if the bundle admits a spin$^c$ structure.

Questions: Are there obvious ways to relate and encode the above obstructions in terms of group extension languages? For example, the failure of the pullback from a group $G$ to a group $G'$. And how do $w_1(TM)$, $w_1(TM)^2$, $w_2(TM)$ and the third integral Stiefel–Whitney class enter into the homomorphism map in the exact sequences?

My attempts:

1. It looks to me that for $w_1(TM)^2$, is has something to do with the classifying the extensions:
   $$
   1 to mathbbZ_2 to SO(3) rtimes mathbbZ_4 to O(3) to 1
   $$
   or for the odd $n$ (how about the even $n$)
   $$
   1 to mathbbZ_2 to SO(n) rtimes mathbbZ_4 to O(n) to 1
   $$
   What is the extension and obstruction for classifying $w_1(TM)$?




2. It looks to me that for $w_2(TM)$, is has something to do with the classifying the extensions:
   $$
   1 to mathbbZ_2 to Spin(n) to SO(n) to 1
   $$



3. It looks to me that for the integral $tilde w_3(TM)$, is has something to do with the classifying the extensions:
   $$
   1 to mathbbZ_2 to Spin^c(n) to SO(n)times U(1) to 1
   $$
   which we may view (?) the
   $$
   Spin^c(n) =fracSpin(n)times U(1)mathbbZ_2.
   $$

You might be thinking about the Whitehead tower of the orthogonal group. Given a manifold $M$, its tangent bundle classifies a map $tau_M: M to BO$. If we are given a group homomorphism $G to O$, we can ask whether the structure group of the tangent bundle reduces to $G$. For example, asking for a $SO$-structure is the same as requiring $M$ to be orientable. In terms of classifying maps, this means a lift of the tangent classifier along $BG to BO$:
$$beginarrayccc
& & BG \
& nearrow & downarrow \
M & xrightarrow[tau_M] & BO
endarray$$

In the case $G = SO$, we have that $BSO$ is the fiber of a map $w_1: BO to K(mathbbZ/2,1)$. So the existence of a lift $M to BSO$ is equivalent to asking that the composite $w_1(M): M to BO to K(mathbbZ/2,1)$ is nullhomotopic. This means that $M$ is orientable iff $w_1(M) = 0$.

We can continue up the Whitehead tower for $BO$. Suppose $M$ is orientable. To lift the tangent classifier to $BmathrmSpin$ and endow $M$ with a spin structure is to ask that $w_2(M): M to BSO to K(mathbbZ/2,2)$ is zero.

$$beginarrayccccc
& & downarrow \
& & BmathrmSpin & xrightarrowfracp_12 & K(mathbbZ,4) \
& & downarrow \
& & BSO & xrightarroww_2 & K(mathbbZ/2,2) \
& & downarrow \
M & xrightarrow[tau_M] & BO & xrightarroww_1 & K(mathbbZ/2,1)
endarray$$

In this diagram, each "L"-shaped part (e.g., $BmathrmSpin to BSO to K(mathbbZ/2,2)$) is a fiber sequence that we use to rephrase the problem of reducing to a structure group in terms of cohomology classes. These fiber sequences are incarnations of the group extensions you've described, e.g.:
$$1 to SO to O xrightarrowdet mathbbZ/2 to 1$$
$$1 to mathbbZ/2 to mathrmSpin to SO to 1.$$