bio  website  math.jussieu.fr/~diverio 

location  Paris  
age  35  
visits  member for  4 years, 10 months 
seen  1 hour ago  
stats  profile views  3,101 
I am a researcher in complex analytic and algebraic geometry at the Institut de MathÃ©matiques de Jussieu.
2d

comment 
Isotrivial families with nonzero Kodaira spencer map
If all fibers are isomorphic, then by the FischerGrauert theorem the family is locally trivial. Then, the KodairaSpencer map being "local" should be zero, isn't it? 
Jul 14 
awarded  Nice Question 
Jul 2 
awarded  Popular Question 
Jun 12 
comment 
Moduli space of linebundle holomorphic structures and sections over a KahlerHodge manifold
Chern connection with respect to which hermitian metric? 
Jun 3 
revised 
Noncohomological proof that the pullback of an ample bundle by a finite morphism is ample
added the étale hypothesis in order to be more precise. 
Jun 3 
comment 
Noncohomological proof that the pullback of an ample bundle by a finite morphism is ample
@Misaka01034, you can look at any of the books by Demailly about transcendental methods in algebraic geometry (you'll find them freely on his web page)! Anyway for semiampleness you don't have a metric characterization (otherwise it would be a numerical property, and it is not), but just that semiampleness implies the existence of a smooth metric with positive semidefinite Chern curvature. While bigness is equivalent to posses a (possibily) singular hermitian metric whose curvature current is bounded below by a strictly positive $(1,1)$form. 
Apr 22 
comment 
Is dimension invariant under blowups?
This is certainly true if $X$ is a locally Noetherian integral scheme and you blowup a nonzero quasicoherent sheaf of ideals. In this case, the blowup morphism is proper and birational, and $X'$ integral. Finally, if $f\colon Z\to Y$ is any proper birational morphism, where $Y$ is a locally Noetherian integral scheme, then $\dim Z=\dim Y$. 
Mar 30 
revised 
Examples of Brody hyperbolic affine varieties which are not Kobayashi hyperbolic
added a quasiprojective example, a reference to a book of S. Lang, a hint where to look at in order to answer the question 
Mar 3 
revised 
Applications of Liouville's theorem
added the "teaching" tag 
Mar 3 
comment 
Bertini theorem for big divisors and klt pairs
Wilson's theorem is Theorem 2.3.9 in Lazarsfeld's book "Positivity in algebraic geometry I". The corollary Hacon is talking about is Proposition 11.2.18 in "Positivity in algebraic geometry II". 
Mar 3 
revised 
Why can we not always take a Kähler class to be in rational cohomology?
modified the last part following the comments of ACL 
Mar 3 
comment 
Why can we not always take a Kähler class to be in rational cohomology?
@ArtiePrendergastSmith: it was my pleasure! I didn't want to "steal" your answer, but at some point I saw that you weren't answering and so... Anyway, maybe I shall modify my $\mathbb Q^2$ in $\mathbb Q^3$, to avoid the problem pointed out by ACL. 
Mar 3 
awarded  Nice Answer 
Mar 2 
comment 
Why can we not always take a Kähler class to be in rational cohomology?
Ahhhhh ok! Now I see the point!! Yes, indeed! My exemple was somehow artificial just to explain what was going on... You are right! 
Mar 2 
comment 
Why can we not always take a Kähler class to be in rational cohomology?
Right, there is no pure Hodge structure of weight 2 with $\dim V^{1,1}_\mathbb R=1$. But I am still missing something in order to understand your comment above... 
Mar 2 
comment 
Why can we not always take a Kähler class to be in rational cohomology?
@ACL. Sorry but I am not sure I understand well your comment... The solution to what? And what do you mean about the $V^{01}$? Cheers, Simone. 
Mar 2 
revised 
Why can we not always take a Kähler class to be in rational cohomology?
slight change following YangMills comment. 
Mar 2 
revised 
Why can we not always take a Kähler class to be in rational cohomology?
slight change following YangMills comment. 
Mar 2 
comment 
Why can we not always take a Kähler class to be in rational cohomology?
Dear YangMills, I think this is a misunderstanding: I was speaking about a general dense subset, saying in parenthesis to keep in mind the case of $V_\mathbb Q$. But I admit that the way I wrote it can be misleading. I try to fix it. 
Mar 1 
answered  Why can we not always take a Kähler class to be in rational cohomology? 