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

7 votes

Varieties where every non-zero effective divisor is ample

It is a general fact that on any simple abelian variety, an effective divisor is ample. The following result underlies the usual algebraic proof of the projectivity of abelian varieties (defined initi …
Vesselin Dimitrov's user avatar
8 votes
3 answers
2k views

When are isotrivial families split by a finite base-change?

A well-known theorem of Grauert and Fischer states that a smooth proper family of complex manifolds is a locally trivial fibration as soon as all the fibers are isomorphic. It is also easy to obtain a …
Vesselin Dimitrov's user avatar
26 votes

irreducibility of discriminant

The discriminant locus has the following geometric interpretation, given in the introductory chapter of [Gelfand, Kapranov, Zelevinsky: Discriminants, Resultants and Multidimensional Determinants]. L …
Vesselin Dimitrov's user avatar
5 votes
Accepted

Reducibility of resultants

All such resultants and/or discriminants are geometrically irreducible in characteristic zero, and a power of an irreducible in general. This is actually covered by the geometric argument I quoted in …
Vesselin Dimitrov's user avatar
11 votes
Accepted

Nakai-Moishezon theorem for abelian varieties

On an abelian variety (regardless of the characteristic), an effective divisor with positive self-intersection is ample. To be more precise, it suffices here to recall that on any simple abelian varie …
Vesselin Dimitrov's user avatar
2 votes

How do I find the set of all lines lying on a general quadric in $\mathbb{CP}^3$?

You can show that a smooth quadric in $\mathbb{P}^3$ is GL-equivalent to the product $\mathbb{P}^1 \times \mathbb{P}^1$ embedded by the Segre map. Once you know this, you're done. (A curve on $\mathb …
Vesselin Dimitrov's user avatar
1 vote

The height of an orbit under rational self-maps

Thank you very much for this answer! This is not going to fit in a comment, so I'm writing another answer. [EDIT: I had hastily written some incorrect statements, now corrected in the text below. M …
Vesselin Dimitrov's user avatar
9 votes
Accepted

Geometric Lang conjecture - reference

abx's comment was made while I was writing this, but I am posting it as an answer anyway. There has not been a proof of this conjecture of Lang, which remains a wide open problem. Lu and Miyaoka's pa …
Vesselin Dimitrov's user avatar
10 votes
2 answers
717 views

The height of an orbit under rational self-maps

I have this basic question on which, strangely enough, the algebraic dynamics literature appears to be silent. But the question does not appear to be totally trivial or uninteresting to me - am I wron …
35 votes
Accepted

Are rational varieties simply connected?

Yes! (I assume it was implicit in your question that the variety be projective?) More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book …
Vesselin Dimitrov's user avatar
4 votes

Variety acquiring rational point over any quadratic extension

Will Sawin and Michael Stoll have noted that, as a consequence of Faltings's "Big Theorem," a hyperelliptic equation $y^2 = f(x)$ with $\deg{f} > 6$ (genus $> 2$) and not admitting a degree $2$ non-co …
Vesselin Dimitrov's user avatar
1 vote

Colmez conjecture and endomorphism rings

In Colmez's formulation, it is necessary that the endomorphism ring be the maximal order $\mathcal{O}_k$. It is then proved only in special cases ($k/\mathbb{Q}$ abelian), or on average over the CM ty …
Vesselin Dimitrov's user avatar
6 votes
2 answers
360 views

The kernel of a nef line bundle

Let $V$ be a complex projective variety and $L$ a nef line bundle on $V$ (i.e., $L$ is non-negative on every curve in $V$). Denote, as usual, $\deg_LX = c_1(L)^{\dim{X}}.[X]$ for $X$ a subvariety of $ …
Vesselin Dimitrov's user avatar
13 votes

Smoothness of the "Archimedean special fiber" in Arakelov geometry

In Arakelov geometry, the conventional wisdom is that the ``closed fibre at $\infty$'' should be viewed as totally degenerate. This is the extreme opposite of smoothness. A visualization in the case o …
Vesselin Dimitrov's user avatar
6 votes
1 answer
536 views

Generalizations of de Franchis and function field Mordell

The classical de Franchis theorem, as generalized by S. Kobayashi and T. Ochiai ("Meromorphic mappings onto compact complex spaces of general type," Inventiones, 1975), states that if $X$ is a complex …
Vesselin Dimitrov's user avatar

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