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I hope this is not trivial.

Let $B$ be a nice topological space (paracompact, CW-complex or whatever you think is nice)

For $i=1,\ldots,n$ let $x_i \in H^i(B,\mathbb{Z}_2)$ be certain cohomology classes.

Does there exists a vectorbundle $E$ of rank $n$ with $w_i=x_i$?

Of course $w_i$ is the $i$-th Stiefel Whitney class of $E$.

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    $\begingroup$ You are asking whether $Z/2$-cohomology as a functor from the homotopy category of spaces to $Z/2$-algebras is full on certain morphism sets. This is wrong in general. First there is more structure on the algebraic invariants (they are algebras over the Steenrod algebra). Even with this modification, this is wrong in general, without assumptions on the spaces, as additional invariants can show. $BO(n)$ is not so special here. $\endgroup$
    – nsrt
    Commented Apr 22, 2014 at 16:48

2 Answers 2

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No, because the Wu formulae express $\mathrm{Sq}^j(w_i)$ in terms of $w_k$'s, so if the $x_i$ you choose don't satisfy this formula, they cannot possibly arise as Stiefel--Whitney classes.

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    $\begingroup$ What is if they satisfy Wu's formula? $\endgroup$
    – Oliver Straser
    Commented Apr 22, 2014 at 8:42
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    $\begingroup$ This gives a potential obstruction. To make it into an answer one needs to produce an example of $B$ and $x_i$'s for which the Wu formula is not satisfied. $\endgroup$
    – Igor Belegradek
    Commented Apr 22, 2014 at 12:57
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    $\begingroup$ @IgorBelegradek: There's no two-dimensional bundle over $\Sigma RP^2$ with $w_1=0,w_2\ne 0$. $\endgroup$
    – nsrt
    Commented Apr 22, 2014 at 13:05
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    $\begingroup$ @nsrt: nice example, and it does not require Wu formula: since $w_1=0$ the bundle is orientable, so $w_2\in H^2(B;\mathbb Z_2)\cong H^1(RP^2;\mathbb Z_2)\cong\mathbb Z_2$ must be a reduction of the Euler class which lives in $H^2(B)=H^1(PR^2)=0$. $\endgroup$
    – Igor Belegradek
    Commented Apr 22, 2014 at 13:32
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    $\begingroup$ @IgorBelegradek: The simplest example is $B=S^3$ with $x_3$ the nontrivial element of $H^3(S^3;\mathbb{Z}/2)$. As $w_3 = \mathrm{Sq}^1(w_2)$ this cannot be realised. $\endgroup$
    – Oscar Randal-Williams
    Commented Apr 22, 2014 at 14:51
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In addition to Wu's formula mentioned in Oscar's answer, there are further obstructions coming from the fact that the mod 2 reduction of $p_i$ (the $i$th Pontryagin class) is $w_{2i}^2$. So your classes $x_{2i}$ would have to satisfy $\beta(x_{2i}^2)=0$, where $$\beta: H^{\ast}(B;\mathbb{Z}/2)\to H^{\ast+1}(B;\mathbb{Z})$$ is the Bockstein associated to the coefficient sequence $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\to 0$.

(Incidentally the class $\beta(x_{2i}^2)$ is also an obstruction to $x_{2i}$ being realized by an immersion, see here.)

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