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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
4
votes
Accepted
equations for a bidouble cover
As abx note in the comments, you miswrote the equations. Still the singular point remains.
The point is that the "reduced" data works well algebraically, as indeed you can deduce $L_3$ from the othe …
7
votes
Accepted
Genus in weighted projective space
If p, q and r are pairwise coprime the generalized Fermat curve is embedded in the weighted projective space ${\mathbb P}(qr, pr, pq) $ which is isomorphic to a straight ${\mathbb P}^2$ via $(x:y:z)\m …
3
votes
Does every finite map of smooth varieties birationally embed as a smooth hypersurface of 1d ...
I may have misunderstood something. However my impression is that, since you are asking $Y'$ smooth, the answer is negative.
The simplest counterexamples are probably compact Riemann surfaces, since b …
3
votes
Accepted
On surfaces with $p_g=0$, $q=1$, and $K^2=-3$
Xiao Gang is taking a configuration of six lines in the plane with 4 triple points $x,z_1,z_2,z_3$ and three double points $y_1,y_2,y_3$. He considers a general quartic through the seven points which …
2
votes
Accepted
Pullback of line bundles and divisors from $Kum(C)$ to $Jac(C)$
1) is correct, 2) is not.
Indeed, if $i(C)=C$, then the map $C \rightarrow C'$ is a double cover, and $f^*{\mathcal O}_Y(C')={\mathcal O}_X(C)$ since in a neighbourhood of a general point of $C$ the …
0
votes
Accepted
Uniform position property and general hyperplanes
The answer to both questions is negative, as pointed out by aginesky in the comments. To complete his answer, let me give a simple concrete counterexample.
Consider a canonical curve $C$ of genus 4 w …
2
votes
Finding an algebraic equation given divisors
abx has already pointed out that your divisors are not the canonical divisors. Still, your (not-so-clear) question was maybe slighlty different, on "how to find the equation"?
If I understand your qu …
1
vote
Accepted
The geometric genus under a generically finite to one rational map
Mohan' answer in the comments is correct: this is more a comment to his answer.
If you are only interested in the inequality you do not need Riemann-Hurwitz formula as I will explain in the following …
3
votes
Torsion of the Picard group for surfaces isogenous to a product
I think the answer to the question as it is stated is negative: if I do understand correctly the question I can prove that, if $G$ is abelian and both curves $C_i/G$ are rational, this is never true …
3
votes
a net of quadrics and the corresponding intersection
(1) Yes.
Consider a net of reducible quadrics (here the dimension of the projective space is not relevant): since all the associate quadratic forms have rank at most 2, and since they vary linearly, …
5
votes
Accepted
Will any two linearly equivalent ample divisors on an abelian variety intersect?
Of course not.
If $L$ is very ample, $D_1$ and $D_2$ are two hyperplane sections for some embedding in a projective space. Therefore their intersection is at most codimension 2 in $X$, intersection o …
1
vote
Accepted
Relation between curves in a complete linear system contained in another
In the following I suppose that by "curve" in a linear system you mean "effective divisor" vithout any claim about being irreducible and/or reduced.
1) Curves in |L'| are exactly the pull-back of cur …
2
votes
Accepted
Global sections of coherent sheaves on determinantal hypersurfaces in $\mathbb{P}^n$
This is meant to be an integration to Yusuf answer.
Consider a section
$$
\stackrel{\rightarrow}{\alpha}=
\begin{pmatrix}
\alpha_1\\
\vdots\\
\alpha_r
\end{pmatrix} \in {\mathbb C}^r\cong H^0({\math …
4
votes
Isotrivial fibrations over $\mathbb P^1$
If the genus of the fibre is not 0, by Theorem 2.1 in Serrano's paper "Fibrations on algebraic surfaces" any isotrivial fibration is birational to $(A \times B)/G \rightarrow B/G$ where $G$ is a finit …
2
votes
Accepted
Is each rationally chain connected surface rational?
By the comments of abx you only need an answer to question $3$. You get easily an answer if you assume the results in the classical book of Beauville on algebraic surfaces.
Namely, if $S$ is uniruled …