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 $P(\mathbb{C}^4)$

  • $\begingroup$ Dear mathdragon, Just to check, $\mathbb H$ denotes the Hamilton quaternions? Regards, $\endgroup$ – Matt E May 17 '13 at 2:30
  • $\begingroup$ Yeah, sorry. I'll edit to add that. $\endgroup$ – mathdragon May 17 '13 at 2:32
  • $\begingroup$ 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, $\endgroup$ – Matt E May 17 '13 at 2:36
  • $\begingroup$ Matt, you're right. Again, sorry. I should have read that over more before posting. $\endgroup$ – mathdragon May 17 '13 at 2:38
  • $\begingroup$ 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 $P(\mathbb{C}^4)$ $\endgroup$ – mathdragon May 17 '13 at 2:40

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.

  • $\begingroup$ 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, $\endgroup$ – Matt E May 17 '13 at 4:55
  • $\begingroup$ @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 ^^ $\endgroup$ – Olivier Bégassat May 17 '13 at 4:58
  • $\begingroup$ Dear Olivier, I agree, based on the analogy with complex projective case. Cheers, $\endgroup$ – Matt E May 17 '13 at 5:03

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