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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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    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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Generally, for a complex vector bundle $E$ the ring structure of $H^\ast(P(E))$ is determined by the characteristic classes of $E$: one has

$ H^\ast(P(E),{\mathbb{Z}}) = H^\ast(B)[x]/(x^n+c_1x^{n-1}+\cdots+c_{n-1}x + c_n) $

where $c_k\in H_{2k}(B;\mathbb{Z})$ is the $k$th Chern class and $x$ restricts to the generator of $H^2(\mathbb{C}P^n)$.

In your case one has $H^\ast(B) = \mathbb{Z}[t]/(t^2)$, so

$ H^\ast(P(M_\lambda)) = \mathbb{Z}[x,t]/(t^2,x^2+c_2(M_\lambda)) $

as rings. Here $c_2(M_\lambda) = \rho\cdot t$ for some $\rho\in\mathbb{Z}$ and it remains to identify $\rho:\pi_4(BU(2)) \rightarrow \mathbb{Z}$. This might be addressed in your source for $\pi_4(BU(2)) \cong \mathbb{Z}$.

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    Just to rephrase the a$n$swer: $\operator$n$ame H(M^{\mathbb C}_\lambda)=\operator$n$ame H(\mathbb CP^1)[\sqrt{\lambda t}]$. Nice.2011-06-02