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Recall that 2-dimensional complex vector bundles over $S^4$ are classified by $\pi_4(BU(2))=\pi_4(BU)=\mathbb Z$. For any integer $\lambda$ one can consider projectivisation of the corresponding bundle, $M_\lambda^{\mathbb C}$ — which fibers over $S^4$ with fiber $\mathbb CP^1\cong S^2$. Serre spectral sequence computing $H^*(M_\lambda^{\mathbb C})$ degenerates by dimention argument, so additively $H^*(M_\lambda^{\mathbb C})=H^*(\mathbb CP^3)$.

The construction has obvious quaternionic analogue (we still have $\pi_8(BSp(2))=\pi_8(BSp)=\mathbb Z$), which gives fibration $S^4\to M_\lambda^{\mathbb H}\to S^8$ and additively $H^*(M_\lambda^{\mathbb H})=H^*(\mathbb HP^3)$.

Question. What is multiplicative structure in cohomology of $M_\lambda^{\mathbb C}$ and $M_\lambda^{\mathbb H}$?

(For example, for $\lambda=0$ corresponding spaces are just products, and $H^*(\mathbb CP^3)\neq H(S^4\times\mathbb CP^1)$ as rings, so the answer indeed depends on $\lambda$.)

(Some motivation/background. One example of the situation from the first paragraph is Hopf fibration $S^2\cong\mathbb CP^1\to\mathbb CP^3\to\mathbb HP^1\cong S^4$. On the other hand, there seems to be no Hopf fibration $S^4\cong\mathbb HP^1\to\mathbb HP^3\to\mathbb OP^1\cong S^8$…)

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    The Leray-Serre spectral sequence is a spectral sequence of algebras. Since the filtration on the limit $H^\bullet(M_\lambda^\mathbb C)$ is trivial, this means that $E_\infty$ is $H^\bullet(M_\lambda^\mathbb C)$ as a *ring*. You can then read the algebra structure from $E_2$, which is the obvious one (see Proposition 5.6 in McCleary's book)2011-05-29
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    @Mariano I don't quite understand your argument. It's, indeed, easy to compute $E_\infty=E_2=H(S^4\times\mathbb CP^1)$, but since $H(S^4\times\mathbb CP^1)\neq H(CP^3)$ (as rings) story doesn't end here...2011-05-29
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    Why are bundles over $S^4$ classified by $\pi_4$? I thought it should be $\pi_3$, by the clutching construction.2011-06-01
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    @Aaron G-bundles on $S^4$ are classified by $\pi_3(G)=\pi_4(BG)$.2011-06-01
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    Oh right of course. Thanks. I knew I was getting something wrong, but I'm not sure what I was thinking.2011-06-01

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