4
$\begingroup$

By the GAGA principle we know that a holomorphic vector bundle E->X is analitically isomorphic to an algebraic one, say F->X, and by definition F is locally trivial in the Zariski topology. But since the isomorphism between E and F is analytic, I fail to see if this implies that E is Zariski locally trivial too.

I hope the answer is not "trivially yes" for some stupid reason, but I cannot guarantee that.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
7
$\begingroup$

You get (analytic) trivializations of E over Zariski-open sets just by composing a trivialization of E with the isomorphism between E and F. Of course, you do not get algebraic trivializations, but for this you would need an algebraic structure on E in the first place.

Improve this answer
$\endgroup$
1
  • $\begingroup$ It seems to work, and it seems that my question was trivial, as I suspected. Thank you anyway, for your clarification :-) $\endgroup$
    – Andrea Ferretti
    Commented Nov 5, 2009 at 0:02
5
$\begingroup$

It's probably worth noting that etale-locally trivial principal GL(n)-bundles are automatically Zariski-locally trivial. This isn't necessarily true for general G.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I don't see how the situation depends on the group G; can you explain a bit? $\endgroup$
    – Kevin H. Lin
    Commented Nov 5, 2009 at 2:24
0
$\begingroup$

My memory is fuzzy, but when you compute algebraic bundle cohomology, and it happens to be 0 here because of GAGA, I think, on all affine subschemes, wouldn't it automatically mean the bundle is locally trivial?

Improve this answer
$\endgroup$

You must log in to answer this question.

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