I am just starting out learning about characteristic classes (Euler, Chern, etc.) from Bott and Tu's book, and I had the following question.

Let $E$ be an oriented $S^2$ bundle over $M$ with structure group $SO(3)$. In general, I understand that there may not be a global cohomology class in $H^2(E)$ which restricts to the generator on each fiber: this will happen if(f) the global angular form is closed (that is, the Euler class is zero).

On the other hand, now suppose $E$ is a rank-two complex vector bundle over $M$. Then the projectivized bundle $\pi: P(E) \rightarrow M$ is a fiber bundle over $M$, all of whose fibers are spheres (being $CP^1$'s). Now in this case there always exists an element of $H^2(P(E))$ which restricts to the generator on each fiber: in the construction of the Chern class, we take the Euler class of the universal subbundle of $\pi^{-1}E$, and this is such an element.

So now I am a bit confused: it seemed naively to me that $S^2$ bundles and $CP^1$ bundles should be the same, but this can't be the case. Do the structure groups differ? Can someone explain if/why $CP^1$ bundles are "special" cases of $S^2$ bundles? Is it possible to see more directly why the second kind of $S^2$ bundle always admits a global cohomology class restricting to the generator on each fiber?



1 Answer 1


The structure groups differ: in the second case, because you're taking fiberwise projectivizations starting from a rank $2$ complex vector bundle, the structure group has been lifted to $\text{U}(2)$. This implies, for example, that $w_2$ vanishes for such bundles.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .