I am looking for an example of an oriented rank 5 (or lower) real vector bundle $V$ over an oriented manifold such that the cup product $w_2(V) w_3(V)$ of Stiefel-Whitney classes does not vanish. It would be best if the manifold had dimension 7 or lower.


2 Answers 2


As far as I know the Wu manifold $X=SU(3)/SO(3)$ is orientable and has mod 2 cohomology ring $H^*(X;\mathbb{Z}_2)=\Lambda(\omega_2(X),\omega_3(X))$. Thus $\omega_2(X)\cdot\omega_3(X)\neq 0$, and in fact generates $H^5(X;\mathbb{Z}_2)$.

  • $\begingroup$ Great thanks! Do you have a handy reference where this is proven? $\endgroup$ Sep 11, 2016 at 17:17
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    $\begingroup$ For a closed oriented $5$-manifold $M$, the Stiefel-Whitney number $\langle w_2 w_3, [M] \rangle$ can be easily deduced from the isomorphism type of the torsion subgroup of $H_2(M;\mathbb{Z})$. See the second page of the paper [G. Lusztig, J. Milnor & F. Peterson, Semi-characteristics and cobordism. Topology 8 (1969) 357--359], where the example of Wu's manifold is mentioned. $\endgroup$ Sep 11, 2016 at 19:50
  • $\begingroup$ Thanks, this helped. I also found the answer to this MO question useful: mathoverflow.net/questions/164714/… $\endgroup$ Sep 13, 2016 at 11:28

I think you can take the manifold to be $M = \mathbb{RP}^5$ and the vector bundle $V$ to be the direct sum of $3$ copies of the tautological line bundle $\gamma$ on $M$. Let $x = w_1(\gamma)\in H^1(M,\mathbb{Z}/2\mathbb{Z})$. Then by the Whitney product formula you get $w_2(V) = x^2$ and $w_3(V) = x^3$, so $w_2(V)w_3(V) = x^5$. But this is non-zero in the mod 2 cohomology of $\mathbb{RP}^5$, which is, as an algebra, isomorphic to $\mathbb{Z}/2\mathbb{Z}[x]/(x^6)$.

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    $\begingroup$ Here the vector bundle is not orientable. $\endgroup$
    – Mark Grant
    Sep 11, 2016 at 15:50
  • $\begingroup$ Thanks! But I was asking for an orientable bundle, indeed. $\endgroup$ Sep 11, 2016 at 17:18
  • $\begingroup$ Ah, missed that part. $\endgroup$
    – Tony
    Sep 11, 2016 at 18:20

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