It is extremely well-known that $H^*(BQ_8;\mathbb{Z})=\mathbb{Z}[\alpha,\beta,\gamma]$ with relations $2\alpha=2\beta=8\gamma=\alpha^2=\beta^2=\alpha\beta-4\gamma=0$, $|\alpha|=|\beta|=2$ and $|\gamma|=4$. I was wondering if anyone can given an explicit reference showing that $c_2(\sigma)$ generates $H^4(BQ_8)\cong \mathbb{Z}_8$, where $\sigma: Q_8 \hookrightarrow SU(2)$ is the standard representation (i.e., $c_2(\sigma)=c_2(E_\sigma)$, $E_\sigma= S^\infty\times_{Q_8}\mathbb{C}^2)$.

Any help would be greatly appreciated, as I would like to avoid the following reasoning if possible:

$E_\sigma$ is the pullback of the canonical quaternionic line bundle $E=S^\infty\times_{SU(2)}\mathbb{C}^2\rightarrow \mathbb{H}P^\infty$ by the projection $S^3/Q_8\hookrightarrow BQ_8 \stackrel{\pi}{\rightarrow}\mathbb{H}P^\infty$, and Leray-Hirsch gives $H^4(\mathbb{H}P^\infty;\mathbb{Z}_2)\cong H^4(BQ_8;\mathbb{Z}_2)\cong\mathbb{Z}_2$. Looking at $H^4(\mathbb{H}P^\infty;\mathbb{Z})\cong \mathbb{Z}\rightarrow H^4(BQ_8;\mathbb{Z})\cong \mathbb{Z}_8$ then shows that $c_2(\rho)=\pi^*(c_2(E))$ must therefore represent an odd number in $\mathbb{Z}_8$.


It is enough to show that the reduction mod 2 of that class is not zero. However, reducing $c_2$ mod 2 gives the Stiefel-Whitney class $w_4$ of the underlying real representation. That this class is non-zero is classical, and explicitly stated and proved for example in Quillen's "The Mod 2 Cohomology Rings of Extra-special 2-groups and the Spinor Groups", Math. Ann. 194, 197--212 (1971). (A somewhat overkill reference, proving much more).

  • $\begingroup$ PS you can prove that $w_4$ is non-zero based on the fact that the mod 2 cohomology ring is finitely generated over the subring of all Stiefel-Whitney classes. See the introduction to my own paper, P. Guillot, The computation of Stiefel-Whitney classes, Ann. Inst. Fourier vol 59, 2009. $\endgroup$
    – Pierre
    Jun 20 '14 at 7:02

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.