# Stiefel-Whitney classes and lifts of structure groups

Let $M$ be a compact, smooth Riemannian manifold with tangent bundle $TM$. I will not distinct between $TM$ and the associated $O(n)$-frame bundle. I believe the following statements are true, but if they are not, please correct me.

1) An orientation of $M$ is an equivariant lift of $TM$ to a $SO(n)$-bundle under the natural inclusion $SO(n) \hookrightarrow O(n)$. The only obstruction to the existence of orientation of $M$ is the first Stiefel-Whitney class $w_{1}(TM) \in H^{1}(M, \mathbb{Z} _{2})$. Moreover, if one exists, then the set of orientations is a $H^{0}(M, \mathbb{Z} _{2})$-torsor.

2) A spin structure on an oriented manifold is an equivariant lift of the $SO(n)$-bundle of oriented frames in $TM$ under the natural projection $Spin(n) \rightarrow SO(n)$. The only obstruction to the existence of a spin structure on a manifold is the class $w_{2}(TM) \in H^{2}(M, \mathbb{Z}_{2})$. Moreover, if one exists, then the set of spin structures is a $H^{1}(M, \mathbb{Z}_{2})$-torsor.

Are these phenomena somehow naturally related? How can one see this? Do there exist some "higher structures" on the tangent bundle such that higher Stiefel-Whitney classes are obstructions to their existence?

I have been trying to push this analogy even further and realized, that $SO(n) \rightarrow O(n)$ is the universal recipient of group homomorphisms from connected topological groups $G$ into $O(n)$. Moreover, by covering theory, $Spin(n) \rightarrow SO(n)$ is the universal recipient of group homomorphisms from 1-connected topological groups G into $SO(n)$. I have no idea how to extend this viewpoint further, since $\pi_{2}$ always vanishes for Lie groups, in particular it vanishes for $Spin(n)$.

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This is discussed on p. 82 of Michelsohn and Lawson's spin geometry. As you point out, $\pi_2(G)$ always vanishes for Lie groups and if $\pi_3(G) = 0$ then $G$ is contractible (so any $G$-bundle is trivial). – Eric O. Korman Feb 7 '13 at 17:12

You're pushing the analogy in the correct way. The relevant construction here is the Whitehead tower of the orthogonal group. Going up a step in the Whitehead tower kills the lowest homotopy group. Now we can't continue getting finite-dimensional covering groups once we get to $\mathrm{Spin}(n)$ in the Whitehead tower for $\mathrm{O}(n)$, but we can get an infinite-dimensional $3$-connected Lie group $\mathrm{String}(n)$. Then a string structure on a spin vector bundle is a lift of the structure group through $B\mathrm{String}(n) \longrightarrow B\mathrm{Spin}(n)$. The obstruction to the existence of a string structure is not a Stiefel-Whitney class as the analogy may suggest, but it is instead the fractional Pontrjagin class $\tfrac{1}{2}p_1$.
You can go a step further and form the $7$-connected cover $\mathrm{Fivebrane}(n)$ of $\mathrm{String}(n)$. Then a fivebrane structure on a string vector bundle is a lift of the structure group through $B\mathrm{Fivebrane}(n) \longrightarrow B\mathrm{String}(n)$. The obstruction to the existence of a fivebrane structure on a string vector bundle is the fractional Pontrjagin class $\tfrac{1}{6}p_2$.