An isomorphism between obstruction classes and Stiefel-Whitney/Euler/Chern/Pontryagin classes

Question

In Milnor & Stasheff it is shown that some of the obstruction classes for constructing a section of vector bundle are equivalent to Stiefel-Whitney classes or the Euler class. Now according to this answer, there are obstruction classes that coincide with fractional Pontryagin classes. Are all obstruction classes for a vector bundle actually isomorphic to one of the characteristic classes mentioned in the title?

Answer 1

No, the Stiefel-Whitney, Euler, Chern, and Pontryagin classes together do not cover all possible obstruction classes for the existence of sections of a vector bundle.

In fact, there is a non-trivial vector bundle for which all of these classes vanish: Since $\pi_8 SO(10) \cong \mathbb{Z} / 2$, the unique non-trivial homotopy class $S^8 \to SO(10)$ gives rise to a 10-dimensional non-trivial real vector bundle over $\xi$ over $S^9$ via clutching. Since $H^k(S^9)$ vanishes in all positive degrees $\neq 9$, there can be no non-trivial Chern (if $\xi$ happens to admit the structure of a complex vector bundle in the first place) or Pontryagin classes here, nor can the Euler class be nontrivial. As for the Stiefel-Whitney classes, the only candidate is $w_9(\xi) \in H^9(S^9; \mathbb{Z} / 2)$ which vanishes since it satisfies $$ w_9(\xi) = w_1(\xi) w_8(\xi) + \mathrm{Sq}^1 w_8(\xi) = 0 $$ by the Wu formula¹.

We can use this to produce a bundle for which all of these characteristic classes vanish but which doesn't admit a nowhere non-zero section: Indeed, let $\zeta$ be such that $\xi \cong \zeta \oplus \mathbb{R}^k$ with $k$ maximal. Then all the Stiefel-Whitney, Chern, and Pontryagin classes of $\zeta$ vanish since they equal those of $\xi$ by stability (or, alternatively, by the exact same arguments as above), and the Euler class is necessarily zero for degree reasons. But $\xi$ is nontrivial, hence so must $\zeta$ be, and the assumption that $k$ is maximal implies that $\zeta$ does not admit a nowhere vanishing section.

As a final observation, $\pi_8 SO(10)$ is in the stable range for $SO(n)$, i.e. $\pi_{8 + k} SO(10 + k) \cong \pi_8 SO(10)$ for all $k \geq 0$ via the inclusions. This implies that $\xi$ is in fact stably non-trivial, i.e. $\xi \oplus \mathbb{R}^k$ is never a trivial bundle, so the collection of classes you named cannot even detect stable triviality!

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¹which says that $$ \mathrm{Sq}^a w_b(\xi) = \sum_{t = 0}^a \binom{b - a + t - 1}{t} w_{a - t}(\xi) w_{b + t}(\xi) $$ for any vector bundle $\xi$.