bio | website | dm.unipi.it/~pardini |
---|---|---|
location | Italy | |
age | ||
visits | member for | 4 years |
seen | Nov 16 at 11:48 | |
stats | profile views | 2,069 |
My name is Rita Pardini. I work in classical algebraic geometry, mainly surfaces.
Nov 8 |
answered | Kummer Coverings |
Nov 6 |
awarded | Yearling |
Sep 24 |
awarded | Autobiographer |
Jul 8 |
comment |
smooth quotient out of a singular variety?
I forgot irreducible, sorry! |
Jul 8 |
comment |
smooth quotient out of a singular variety?
A variation of this answer in order to satisfy Libli's request that the example be quasi-projective: take $X$ the union of two planes of $\mathbb C^4$ intersecting at the origin and let $\mathbb Z_2$ act by identifying the two planes via any linear isomorphism. $X$ is affine and behaves as Dan's example. |
Jul 2 |
awarded | Curious |
Jun 28 |
awarded | Popular Question |
Mar 4 |
comment |
Does $h^1(D)=0$ imply numerical connectedness on K3 surfaces?
Ciao Pigna! welcome to MO! |
Mar 4 |
comment |
Does $h^1(D)=0$ imply numerical connectedness on K3 surfaces?
No. For instance if $A$ is a a rational curve with $A^2<0$, then $h^0(\mathcal O_A(-A))>0$. |
Dec 29 |
comment |
(Non-)Existence of curves of low degree on affine and projective varieties
The Noether Lefschetz theorem says that the Picard group of a very general surface of degree $d\ge 4$ of $\mathbb P^3$ is cyclic generated by the hyperplane section, so in this case all curves have degree divisible by $d$. ("Very general" means "in the complement of a countable union of Zariski closed subsets"). |
Nov 6 |
comment |
Etale covers of products of curves
For completeness, one needs to show that the unramified cover is not itself a product of curves. If, for instance, $C_1'$ and $C'_2$ have the same genus and $|G|=2$, this can be seen by computing $h^1(\mathcal O)$ and $h^2(\mathcal O)$. |
Nov 6 |
awarded | Yearling |
Oct 14 |
comment |
Complete Linear system on Del Pezzo surfaces
My feeling is that the statement might be correct (I thought about it a bit couldn' find a counterexample), but I wouldn't know how to prove it. |
Oct 12 |
revised |
Complete Linear system on Del Pezzo surfaces
added 11 characters in body |
Oct 12 |
revised |
Complete Linear system on Del Pezzo surfaces
added 110 characters in body |
Oct 12 |
comment |
Complete Linear system on Del Pezzo surfaces
If you are interested in a precise example of linear system, it may be possible to answer the question by using numerical considerations, but asking about the intersection of the components in general does not make much sense. |
Oct 12 |
answered | Complete Linear system on Del Pezzo surfaces |
Oct 11 |
awarded | Constituent |
Oct 1 |
awarded | Caucus |
Sep 26 |
comment |
Quotient of an abelian surface by an antisymplectic involution
@Misha: the map that switches factors is antisymplectic, since $dx\wedge dy$ goes to $dy\wedge dx$. |