bio | website | math.jussieu.fr/~diverio |
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location | Paris | |
age | 35 | |
visits | member for | 4 years, 10 months |
seen | 12 hours ago | |
stats | profile views | 3,118 |
I am a researcher in complex analytic and algebraic geometry at the Institut de MathÃ©matiques de Jussieu.
Aug
2 |
reviewed | Approve Product of Positive Matrices |
Aug
2 |
reviewed | Approve What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem? |
Jul
30 |
comment |
Isotrivial families with non-zero Kodaira spencer map
If all fibers are isomorphic, then by the Fischer-Grauert theorem the family is locally trivial. Then, the Kodaira-Spencer 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 line-bundle holomorphic structures and sections over a Kahler-Hodge manifold
Chern connection with respect to which hermitian metric? |
Jun
3 |
revised |
Non-cohomological 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 |
Non-cohomological 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 semi-ampleness you don't have a metric characterization (otherwise it would be a numerical property, and it is not), but just that semi-ampleness implies the existence of a smooth metric with positive semi-definite 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 blow-ups?
This is certainly true if $X$ is a locally Noetherian integral scheme and you blow-up a non-zero quasi-coherent sheaf of ideals. In this case, the blow-up 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 quasi-projective 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?
@ArtiePrendergast-Smith: 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. |