General question: We know that the top Chern class $c_n(\xi)$ of an $n$-dimensional complex vector bundle $\xi$ is its Euler class, while the first Chern class, $c_1(\xi)$, is the Euler class of its determinant bundle. Can the intermediate classes also be expressed as Euler classes of bundles related to $\xi$?

Let me also ask a much more specific question: We know that $H^\ast(BU(3);{\mathbb Z}) = {\mathbb Z}[c_1, c_2, c_3]$. Is there a complex 2-plane bundle over the classifying space $BU(3)$ (or $BU(n)$ for $n\geq 3$) whose Euler class is $c_2$?

Motivation: Long story short, I'm trying to construct an analogue of $c_2$ in a context where I know that the usual constructions I'm familiar with won't work. But I do have Euler classes in this context, so if I could express $c_2$ as an Euler class as above, I might be able to use that in my context. On the other hand, if anyone has a proof that there is no such bundle over $BU(3)$, then that will save me time chasing down that particular rabbit hole.

  • 3
    $\begingroup$ That's one of the constructions I'm familiar with, but note that only the top class is defined directly as an Euler class of a bundle on the given space. The construction carries over to my context, but gives the "wrong" classes for reasons too complicated to explain in this comment. $\endgroup$ – Steve Costenoble Jan 5 at 4:02
  • 2
    $\begingroup$ If I am not mistaken, reduced mod 2 we have $Sq^2 c_2 = c_1 c_2 + c_3$ so the answer to the specific question is no. $\endgroup$ – Gustavo Granja Jan 5 at 10:48
  • 1
    $\begingroup$ Maybe I misunderstood the question. Such a bundle would be classified by a map to $BU(2)$ which would pullback $c_2 \in H^2(BU(2))$ to $c_2 \in H^2(BU(3))$. That is precluded by the formula for the action of $Sq^2$ on $H^*BSU(3)$. $\endgroup$ – Gustavo Granja Jan 5 at 16:30
  • 2
    $\begingroup$ @GustavoGranja: Yes, of course, now I see. If you want to submit your comment as an answer I'll accept it. Many thanks. $\endgroup$ – Steve Costenoble Jan 5 at 16:47
  • 2
    $\begingroup$ @SteveCostenoble I thought there might a virtual bundle operation (a la Adams operations, etc) generalizing the "top exterior power" and "identity" operations that gives you the first and last chern class, but i tried a few things and nothing worked...it seems there's something special about the first and last symmetric polynomials! $\endgroup$ – John Greenwood Jan 7 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.