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 $E$.

  • 1
    $\begingroup$ 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 Apr 22 '14 at 16:48

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.

  • 6
    $\begingroup$ What is if they satisfy Wu's formula? $\endgroup$ – Oliver Straser Apr 22 '14 at 8:42
  • 1
    $\begingroup$ 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 Apr 22 '14 at 12:57
  • 1
    $\begingroup$ @IgorBelegradek: There's no two-dimensional bundle over $\Sigma RP^2$ with $w_1=0,w_2\ne 0$. $\endgroup$ – nsrt Apr 22 '14 at 13:05
  • 2
    $\begingroup$ @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 Apr 22 '14 at 13:32
  • 7
    $\begingroup$ @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 Apr 22 '14 at 14:51

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.)


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.