I hope this is not too trivial.

Let $M$ be a compact oriented manifold and $x\in H^2(M, \mathbb{Z}/2\mathbb{Z})$. My question is that if there exists a real oriented vector bundle $V$ over $M$ such that the second Stiefel-Whitney class of $V$ is $x$.

  • $\begingroup$ If the class is integral, the answer is yes (in fact, there is a complex line bundle: just use the exponential exact sequence). $\endgroup$ – Alex Degtyarev Nov 12 '14 at 12:50
  • $\begingroup$ I know that fact. Here I am considering the $\mathbb{Z}/2\mathbb{Z}$ coefficient case, which seems more complicated. Thank you all the same. $\endgroup$ – Bei Liu Nov 12 '14 at 13:02
  • 1
    $\begingroup$ By "integral" I mean $\beta x=0$. If $\beta x\ne0$, an oriented bundle cannot be of dimension $2$: in fact, it must have $w_3=\beta x$. But I cannot be of any further help :) $\endgroup$ – Alex Degtyarev Nov 12 '14 at 13:14
  • $\begingroup$ I see. Thank you for your explanation:) $\endgroup$ – Bei Liu Nov 12 '14 at 13:27

Not necessarily. A necessary condition is that $x^2$ must be the reduction of an integral class. This is because if $V$ is an oriented bundle then $\rho(p_1(V))=w_2(V)^2$, where $\rho$ denotes reduction mod 2.

There are closed oriented manifolds and classes $x$ for which this condition is not satisfied. Since this is also a necessary condition for the Poincaré dual of $x$ to be realized by an immersion, the examples in http://arxiv.org/abs/1111.0249 should work.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.