Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I cannot find any references with how to handle this tautological line bundle $V \rightarrow P(\mathbb{H}^2)$ where $\mathbb{H}$ are the Hamilton quaternions.

I know it is a complex rank 2 vector bundle, so the total chern class is $1 + e(V_\mathbb{R})$ and the top chern class is $c_2(V) = e(V_{\mathbb{R}})$, so $c_0(V) = 1, c_1(V) = 0$ but I want to know what $c_2(V)$ is.

Does anyone have an idea for approaching this?

I think it I should use the splitting principle. From Bott and Tu, we can use the complex projectivization of V as a splitting manifold. And we know this is diffeomorphic to $P(\mathbb{C}^4)$

share|improve this question
    
Dear mathdragon, Just to check, $\mathbb H$ denotes the Hamilton quaternions? Regards, –  Matt E May 17 '13 at 2:30
    
Yeah, sorry. I'll edit to add that. –  mathdragon May 17 '13 at 2:32
    
Dear mathdragon, One more thing: if the fibres are $\mathcal H$-lines in $\mathcal H$ (which I would guess they are, since this is the usual definition of the tautological bundle), then aren't they $2$-dimensional over $\mathbb C$, making $V$ a rank $2$ complex vector bundle? (You write line bundle in your post, which is why I am asking.) Regards, –  Matt E May 17 '13 at 2:36
    
Matt, you're right. Again, sorry. I should have read that over more before posting. –  mathdragon May 17 '13 at 2:38
    
I'll add this to the top, but I think it I should use the splitting principle. From Bott and Tu, we can use the complex projectivization of V as a splitting manifold. And we know this is diffeomorphic to $P(\mathbb{C}^4)$ –  mathdragon May 17 '13 at 2:40

1 Answer 1

I'm not really $100\%$ sure my solution is correct, also, there remains an indeterminacy in the end result: I don't quite identify the class you are looking for in $P^1\Bbb H\simeq \Bbb S^4$, only up to sign.


Let $Q$ be the projectivisation of $p:V\rightarrow P^1\Bbb H$. As you note, $Q$ is isomorphic to $P^3\Bbb C$. To be precise, we consider $\Bbb C\subset \Bbb H$ through $1,i\mapsto 1,i$ respectively. Thanks to the standard left $\Bbb H$-vector space structure on $\Bbb H^2$, $\Bbb H^2$ can be seen as a complex vector space isomorphic $\Bbb C^4$. Now by definition of $Q$ we get a homeomorphism $$\begin{array}{rcl}Q=\coprod_{S\subset\Bbb H^2}P(S)&\longleftrightarrow &P^3\Bbb C\\ l&\mapsto&l\\l\in P(\Bbb H\cdot l)&\gets&l\end{array}$$ The direct sum is taken over all quaternionic (left) lines $S$, and for every such line $S$, $P(S)$ is the complex projective space on $S$.

The pullback bundle of $V$ over $Q$ splits into two complex line bundles $L\oplus L'$ ver $Q$, with the fiber of $L$ over $l$ being $l$ itself, so that (modulo the above isomorphism), $$L\text{ is isomorphic to the tautological line bundle over }P^3\Bbb C$$

@Matt E gave a convincing argument for the vanishing of the first Chern class of $V$. Computing the total Chern class of the pullback bundle gives $$\pi^*(1+c_2(V))=c(L\oplus L')=c(L)c(L')=1+c_1(L)+c_1(L')+c_1(L)c_1(L')$$ The class $c_1(L)+c_1(L')$ vanishes, and it follows (I believe) that $L'\simeq L^*$, since both are line bundles, and thus completely caracterized by their first Chern class. If $c$ is the standard degree $2$ generator of $H^3(Q)=H^2(P^3(\Bbb C))$ (i.e. the first Chern class of the tautological line bundle over $P^3(\Bbb C)$), then $$\pi^*(c_2(V))=-c^2.$$


It remains to understand $\pi$. The canonical map $Q\to P^1\Bbb H$ is a fibration. Actually, it is obtained from the Hopf fibration $\Bbb S^3\hookrightarrow\Bbb S^7\hookrightarrow\Bbb S^4$ by quotienting out the action of $\Bbb S^1$). $$\begin{array}{rc}\Bbb S^2\simeq P^1\Bbb C\hookrightarrow & P^3(\Bbb C)\\&\downarrow\\ &\Bbb S^4\end{array}$$ The associated spectral sequence collapses at rank $2$ because of how the nonzero nodes are placed, and this tells us that the cohomology of $\Bbb S^4$ in degree $4$ is isomorphic to that of $P^3\Bbb C$ in degree $4$ through $\pi^*$. Since $\pi^*(c_2(V))=-c^2$ is a generator in degree $4$, we necessarily have $c_2(V)=$ one of the two generators of $H^4(\Bbb S^4)$ . I don't know which one this is.

share|improve this answer
    
Dear Olivier, This looks good to me; I was also a bit unsure about the sign. Part of the issue perhaps is that one has to figure out how $P(\mathbb H^2)$ is oriented in the first place. I guess the quaternionic situation is something like the complex situation, where one gets a canonical orientation (?) . Regards, –  Matt E May 17 '13 at 4:55
    
@MattE I would put my money on the opposite of the canonical generator of $H^4(\Bbb S^4)$, but I'm going to bed right now ^^ –  Olivier Bégassat May 17 '13 at 4:58
    
Dear Olivier, I agree, based on the analogy with complex projective case. Cheers, –  Matt E May 17 '13 at 5:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.