It is well-known that the Chern class of a line bundle in $H^2(M,\mathbb Z)$ fully determines the bundle up to isomorphism. However, in this wikipedia entry on Calabi-Yau manifolds it is stated that there are manifolds that have a non-trivial canonical bundle, but a vanishing integral Chern class.

What is happening here? The only thing I can think of is that a differentiably trivial line bundle doesn't have to be holomorphically trivial. Is that the case? How should one think of such bundles that are topologically but not analytically trivial?

  • $\begingroup$ The Chern classes of a bundle of dimension $\ge2$ don't necessarily determine it up to isomorphism. $\endgroup$ – Lord Shark the Unknown Aug 31 '18 at 22:22
  • $\begingroup$ @LordSharktheUnknown I know, but in dimension 1 apparently not either $\endgroup$ – doetoe Aug 31 '18 at 22:24
  • 6
    $\begingroup$ Possible duplicate of If a holomorphic bundle is smoothly trivial, is it holomorphically trivial? $\endgroup$ – user99914 Aug 31 '18 at 22:25
  • 1
    $\begingroup$ @Mohan Does that mean that divisors of the same degree on a curve give diffeomorphic line bundles, even if the divisors are not linearly equivalent? $\endgroup$ – doetoe Sep 1 '18 at 17:43
  • 1
    $\begingroup$ Yes, linear equivalence is a holomorphic condition in the usual situation. $\endgroup$ – Mohan Sep 1 '18 at 17:57

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.