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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.

15 votes
Accepted

Enriques surfaces over $\mathbb Z$

A preprint by Stefan Schröer came out today with the answer to this question: arXiv:2004.07025. No such Enriques surface exists. In fact, there is no classical Enriques surface over $\mathbb F_2$ wit …
Davide Cesare Veniani's user avatar
9 votes
1 answer
714 views

Surjective morphism from $X$ to itself is finite

Let $X$ be a projective variety. Why is any surjective morphism from $X$ to itself finite?
Davide Cesare Veniani's user avatar
7 votes
1 answer
589 views

Discriminant locus of elliptic K3 surfaces

Given a complex elliptic K3 surface $\pi\colon X\rightarrow \mathbb P^1$, its discriminant locus is the divisor $$D = \sum_{i = 1}^s n_i P_i$$ on $\mathbb P^1$ such that $n_i$ is equal to the Euler-Po …
Davide Cesare Veniani's user avatar
6 votes
2 answers
400 views

adjacency matrix of a graph and lines on quartic surfaces

Suppose you are given a smooth quartic surface $X$ in $\mathbb P^3$. I would like to find an upper bound for the number of lines on $X$ in the case that there is no plane intersecting the curves in fo …
Davide Cesare Veniani's user avatar
5 votes
0 answers
167 views

Explicit Enriques involutions on the Fermat quartic surface

Let $X$ be the complex Fermat quartic surface defined by the polynomial $x^4+y^4+z^4+w^4$. By results of Sertöz, we know that the surface $X$ admits at least one Enriques involution, i.e. an involuti …
Davide Cesare Veniani's user avatar
5 votes
2 answers
275 views

Singular abelian surfaces that can be defined over $\mathbb Q$

An abelian surface $A$ is called singular if it has maximal Picard number $\rho(A) = 4$. By work of Shioda-Mitani, any singular abelian surface $A$ is the product $A = E_1 \times E_2$ of two isogeno …
Davide Cesare Veniani's user avatar
5 votes
1 answer
281 views

Euler number for base change of a K3 surface

Suppose you have a K3 surface $S$ containing a smooth rational curve $C$ and suppose you have an elliptic fibration $S \rightarrow \mathbb P^1$ that restricts to a morphism $C \rightarrow \mathbb P^1$ …
Davide Cesare Veniani's user avatar
5 votes
0 answers
993 views

Base change of integral scheme of finite type over a field

Let $X$ be an integral scheme of finite type over a field $k$. If $k'\supseteq k$ is a field extension, then $X' = X\otimes_k k'$ is not necessarily integral. Why does each irreducible component o …
Davide Cesare Veniani's user avatar
3 votes
0 answers
215 views

Cohomology of the Hilbert square of a degree 14 K3 surface [Beauville-Donagi]

I have a question about the article by Beauville-Donagi called La variété des droites d'une hypersurface cubique de dimension 4 (C. R. Acad. Sc. Paris, t. 301, Série I, n° 14, 1985). Their construct …
Davide Cesare Veniani's user avatar
2 votes
Accepted

A question regarding lines on a cubic surface

To each automorphism of the binary form there correspond actually $d$ lines, not only one. In case $d = 3$, since there are always 6 automorphisms, we get $6\cdot 3 = 18$ lines, which summed to the 9 …
Davide Cesare Veniani's user avatar
2 votes
0 answers
97 views

Monodromy operators on hyperkähler varieties

Let $X$ be a hyperkähler variety. In an article (Conjecture 2.1) from some years ago, Markman conjectured that any monodromy operator acting trivially on $H^2(X,\mathbb Z)$ is the identity operator, a …
Davide Cesare Veniani's user avatar
1 vote

Linear system corresponding to rational curves on a K3 surface

A base-point free linear system has non-negative self-intersection, so at least one of the two components of $C+D$ must lie in the base locus of $|C+D|$. But if one component were moving it would have …
Davide Cesare Veniani's user avatar
1 vote
0 answers
104 views

Action on cohomology by automorphisms of ihs manifolds

For all known deformation types of irreducible holomorphic symplectic manifolds (which I call K3, K3$^n$, Kum$^n$, OG$^3$, OG$^5$, the exponent being half the complex dimension), it is known that the …
Davide Cesare Veniani's user avatar