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..



Hodge isometry sending the Kahler class to its opposite
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.


