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

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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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... –  Grigory M Jun 1 '11 at 11:15
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Just to rephrase the answer: $\operatorname H(M^{\mathbb C}_\lambda)=\operatorname H(\mathbb CP^1)[\sqrt{\lambda t}]$. Nice. –  Grigory M Jun 2 '11 at 7:32
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