# How can it be that the Chern class fully determines a line bundle, but having Chern class zero doesn't imply a line bundle is trivial?

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?

• The Chern classes of a bundle of dimension $\ge2$ don't necessarily determine it up to isomorphism. – Lord Shark the Unknown Aug 31 '18 at 22:22
• @LordSharktheUnknown I know, but in dimension 1 apparently not either – doetoe Aug 31 '18 at 22:24
• Possible duplicate of If a holomorphic bundle is smoothly trivial, is it holomorphically trivial? – user99914 Aug 31 '18 at 22:25
• @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? – doetoe Sep 1 '18 at 17:43
• Yes, linear equivalence is a holomorphic condition in the usual situation. – Mohan Sep 1 '18 at 17:57