2
$\begingroup$

i would like to ask you a question i can not answer myself, i hope this is not too trivial and i'm not missing something too basic.

Let's suppose we have $X$ and $Y$ Kahler manifolds and $f:X\rightarrow Y$ a bimeromorphic map such that $f^*:H^2(Y)\rightarrow H^2(X)$ is an isomorphism and also an Hodge isometry. Then, given $\omega$ the class of the Kahler form on $X$ and $\omega'$ the class of the Kahler form on $Y$, is it possible $f^*(\omega')=-\omega$? Following the article i'm reading the answer seems to be no, but i can't figure out why..

$\endgroup$
5
  • 1
    $\begingroup$ You probably want $X$ and $Y$ compact. And could you explain what you call a Hodge isometry? $\endgroup$
    – abx
    May 13, 2014 at 14:59
  • $\begingroup$ yes, i'm sorry, so let's say $X$ and $Y$ are compact Kahler surfaces and $f$ is an isometry with respect to the intersection form which also respect the Hodge decomposition (this is what i mean by Hodge isometry) $\endgroup$ May 13, 2014 at 15:17
  • 1
    $\begingroup$ Just look at what happens to the pullback in local coordinates. You'll find that the pullback of a positive $(1,1)$-form by a holomorphic map is always semipositive. Your morphism is only bimeromorphic, but this is enough to see that we can't get $f^*(\omega') = -\omega$ on an open dense set. $\endgroup$ May 13, 2014 at 16:23
  • 1
    $\begingroup$ @Gunnar: the question is whether $f^*(\omega')=-\omega$ in $H^2(Y)$. How does your argument apply? $\endgroup$
    – abx
    May 13, 2014 at 16:37
  • 1
    $\begingroup$ The cohomology class of $f^*\omega'$ should be both pseudoeffective (even big) and anti-Kähler, so it is impossible. $\endgroup$
    – Henri
    May 14, 2014 at 7:45

1 Answer 1

4
$\begingroup$

It is impossible, because the birational (movable) nef cone is mapped to birational nef cone, where birational nef cone is a cone of all classes which are non-negative on all curves which move in families covering the whole manifold. Clearly, $\omega$ belongs to the movable nef cone, and $-\omega$ does not.

This is for projective (or Moishezon) manifolds; if you need to see the same in non-projective situation, you need to use the pseudoeffective cone, which is the cone of all classes which can be represented by positive, closed (1,1)-currents. Demailly-Paun has shown that this cone (which coincides with the movable nef in projective situation) is a bimeromorphic invariant.

$\endgroup$

Your Answer

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

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