1
$\begingroup$

Let $f:X\rightarrow Y$ be a birational map of smooth projective varieties over complex numbers. Let $E$ be a vector bundle on $X$. Will $f_*E$ be a reflexive sheaf. Is it possible to impose some additional conditions to ensure that the direct image is a reflexive sheaf?

$\endgroup$
  • $\begingroup$ you can take double dual $(f_∗E)^{**} $. $\endgroup$ – user21574 Jun 13 '16 at 9:47
  • $\begingroup$ @Hassan Jolany Sorry I don't understand your comment. Can you elaborate? Even $(f_*E)^*$ will be reflexive right?. But I am interested to know when $f_*E$ itself is reflexive $\endgroup$ – gradstudent Jun 13 '16 at 9:49
  • 1
    $\begingroup$ read this paper of Hartshorne gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002320002 $\endgroup$ – user21574 Jun 13 '16 at 9:57
  • 6
    $\begingroup$ Take for $f$ the simplest example of birational morphism, the blowing up of a point $p$ in a surface; let $E$ be the exceptional divisor. Then $f_*\mathcal{O}(-E)$ is the ideal sheaf of $p$, which is not reflexive. $\endgroup$ – abx Jun 13 '16 at 10:30
  • 1
    $\begingroup$ @ abx - I think your comment is so elegant you really ought to consider making it an answer so that people searching this will see it. $\endgroup$ – meh Jun 13 '16 at 14:36
3
$\begingroup$

As requested I put my comment into an answer: take for $f$ the simplest example of birational morphism, the blowing up of a point $p$ in a surface; let $E$ be the exceptional divisor. Then $f_*\mathcal{O}(-E)$ is the ideal sheaf of $p$, hence is not reflexive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.