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Consider $SO(n)$ bundles over smooth manifolds. Then using the fact that the Stiefel-Whitney classes are the modulo 2 reductions of the Chern classes, one can prove $w_{2i}^2(E) = p_i(E) \bmod 2$. Now consider an $SO(3)$ bundle over a 4-manifold. Since the particular case I am studying concerns K3 surfaces, let us assume that $H^2(M,\mathbb{Z})$ contains no torsion. Then it is stated in various places that more is true:

$w_2^2(E) = p_1(E) \bmod 4$.

I may be missing something elementary here, but where does the mod 4 comes from?

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    $\begingroup$ I think the square is the Pontryagin square. See eom.springer.de/p/p073810.htm $\endgroup$
    – BS.
    Commented Apr 8, 2011 at 10:40
  • $\begingroup$ See also this MO question and answsers mathoverflow.net/questions/59593/… $\endgroup$
    – BS.
    Commented Apr 8, 2011 at 10:41
  • $\begingroup$ A reference is Dold and Whitney's 1959 Annals paper on bundles over 4-dimensional complexes (perhaps someone can consult this paper and post the explanation?). $\endgroup$
    – Tim Perutz
    Commented Apr 8, 2011 at 19:59
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    $\begingroup$ @BS, Yes, the assertion is that $P(w_2)=p_1$ mod 4. $P(w_2)=w_2^2$ mod 2 is a property of the Pontriyagin square, and when $H^2(M;Z)$ has no 2-torsion, $w_2$ lifts to a $Z$ class, answering Manuel's question. $\endgroup$
    – Paul
    Commented Apr 9, 2011 at 19:09

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