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 $S^4\cong\mathbb HP^1\to\mathbb HP^3\to\mathbb OP^1\cong S^8$…)

  • $\begingroup$ 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) $\endgroup$ – Mariano Suárez-Álvarez May 29 '11 at 16:13
  • $\begingroup$ @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... $\endgroup$ – Grigory M May 29 '11 at 16:23
  • $\begingroup$ Why are bundles over $S^4$ classified by $\pi_4$? I thought it should be $\pi_3$, by the clutching construction. $\endgroup$ – Aaron Mazel-Gee Jun 1 '11 at 15:36
  • $\begingroup$ @Aaron G-bundles on $S^4$ are classified by $\pi_3(G)=\pi_4(BG)$. $\endgroup$ – Grigory M Jun 1 '11 at 18:41
  • $\begingroup$ Oh right of course. Thanks. I knew I was getting something wrong, but I'm not sure what I was thinking. $\endgroup$ – Aaron Mazel-Gee Jun 1 '11 at 23:21

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 $\pi_4(BU(2)) \cong \mathbb{Z}$.

  • $\begingroup$ Oh, right (can't believe, I forgot about this) — thank you. And $\rho=1$ (in general, $c_n\colon\mathbb Z=\pi_{2n}(BU)\to H^{2n}(S^{2n})=\mathbb Z$ is multiplication by $(n-1)!$ — see cor. 4.4 in Hatcher's book). But generalization to quaternionic case doesn't seem quite obvious... $\endgroup$ – Grigory M Jun 1 '11 at 11:15
  • 1
    $\begingroup$ Just to rephrase the answer: $\operatorname H(M^{\mathbb C}_\lambda)=\operatorname H(\mathbb CP^1)[\sqrt{\lambda t}]$. Nice. $\endgroup$ – Grigory M Jun 2 '11 at 7:32

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.