5
$\begingroup$

Serre’s GAGA theorem gives an equivalence of categories between algebraic and analytic coherent sheaves over a complex projective variety. The proof relies on the finiteness of the cohomologies of coherent analytic sheaves over compact complex manifolds, which is a non-trivial analytic result.

I was wondering that maybe GAGA theorem could be proven quite more easily in the case of vector bundles over a Riemann surface $X$ by using the existence of meromorphic functions as the underlying analytic result and “Grothendieck” theorem about the classification of vector bundles on the Riemann sphere.

The idea is to consider a meromorphic function $f:X \rightarrow \mathbb{P}^1$ and, for any vector bundle $E\rightarrow X$, consider the pushforward $f_*E$, which is a vector bundle over $\mathbb{P}^1$, which by the Grothendieck theorem is algebraic.

My question is if this idea can be used to show that, since $f_*E$ is algebraic, then $E$ is also algebraic.

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ Well, the existence of non-zero meromorphic functions is itself a non-trivial analytic result. $\endgroup$
    – Francesco Polizzi
    Jul 14, 2020 at 8:15
  • $\begingroup$ @FrancescoPolizzi Of course, but imho is quite more simple than the other one I mention. $\endgroup$
    – G. Gallego
    Jul 14, 2020 at 8:23
  • $\begingroup$ $f_*E$ is not locally free in general $\endgroup$
    – Chris
    Jul 14, 2020 at 8:59
  • 1
    $\begingroup$ Yes it is, since $f$ is finite and flat and so $R^if_*E=0$ for all $i >0$. $\endgroup$
    – Francesco Polizzi
    Jul 14, 2020 at 9:25
  • 2
    $\begingroup$ I think this can work. There is a natural transformation $E \to f^{\ast} f_{\ast} E$, and I think that in the curve case I can show that it is injective and $E$ even splits off as a summand. Your argument easily shows that $f^{\ast} f_{\ast} E$ is algebraic, so we'd want to know that a summand of an algebraic bundle is algebraic. $\endgroup$
    – David E Speyer
    Jul 14, 2020 at 11:45

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.