# Realization of second Stiefel-Whitney class

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 $x$.

• If the class is integral, the answer is yes (in fact, there is a complex line bundle: just use the exponential exact sequence). – Alex Degtyarev Nov 12 '14 at 12:50
• 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. – Bei Liu Nov 12 '14 at 13:02
• 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 :) – Alex Degtyarev Nov 12 '14 at 13:14
• I see. Thank you for your explanation:) – 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.