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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.

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3 Answers 3

up vote 7 down vote accepted

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.

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It seems to work, and it seems that my question was trivial, as I suspected. Thank you anyway, for your clarification :-) –  Andrea Ferretti Nov 5 '09 at 0:02

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.

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I don't see how the situation depends on the group G; can you explain a bit? –  Kevin H. Lin Nov 5 '09 at 2:24

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?

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