Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Good example teaches sometimes more than couple of theorems. I wonder what are your favourite examples in complex algebraic geometry, the ones that were astonishing for you, the simpler (at least available for people without PhD in AG) the better.

share|improve this question
7  
The second sentence is a question. The rest is a blog post. I am not sure that this is the way to go with MO questions. –  Yemon Choi Apr 17 '10 at 2:36
    
I'm not going to vote to close, only because I've been known to ask very long questions myself. (I hope, though, that my questions are both much more focused and also formatted so that it's clear what is the question and why I provide the background I do.) But I agree with Yemon Choi: there's no question here. Then again, I don't care for "what is your favorite example" questions in general, although there have been some successful ones here on MO. –  Theo Johnson-Freyd Apr 17 '10 at 5:15
2  
@Karol: Here is what I think what you should've done to make this a better question. First, post a question (make it nice and clear what kind of examples you are looking for, ideally make it clear why they'd be interesting). Look what answers other people give. At some point, post your own example as an answer. Ideally, try to format it so that it's clearer what exactly it shows, and what the logical steps are (at the moment, it's even hard to see how many examples there are in you post, let alone what the logic is). As it is, the potentially interesting math gets lost in the confusion. –  Ilya Grigoriev Apr 17 '10 at 8:44
    
@Palka: is the goal of your example to construct a singular compact algebraic non-projective complex surface? Also: what's the "intersection matrix of a divisor"? –  Qfwfq Apr 17 '10 at 9:44
1  
Note: the comments above no longer make sense, as the question was modified. Karol's answer below was originally part of the question. –  Scott Morrison Apr 21 '10 at 0:29

1 Answer 1

up vote 3 down vote accepted

I will give an example which I was looking for a long time by myself. The question concerns CONTRACTIBILITY OF CURVES ON SMOOTH SURFACES: Having a smooth (hence projective) surface $X$ and a divisor $E$ when can $E$ be contracted into an algebraic singularity (i.e. such that the quotient $X/E$ is an algebraic surface)?

Background: It is known that a divisor on a smooth projective surface can be contracted in the analytic category if and only if its intersection matrix is negative definite (Grauert). However, the question in the algebraic category is much more subtle. There are criteria (numerical) by Artin for contractibility of bunch of rational curves to a rational singularity but no general ones. The following two examples show that it is not possible to give a general numerical criterion for contractibility (people refer to Zariski, The theorem of Riemann-Roch for high multiples of an effective divisor on an algebraic surface. Ann. of Math. (2) 76, 1962, 560--615).

Background for the example: Let $C$ be an elliptic curve and let $P_0$ be a point in $C$. The linear system $|3P_0|$ gives an embedding of $C$ into $\mathbb{P}^2$, in particular for some hyperplane section $H$ we have $H\cap C=3P_0$. The bijection $P\to \mathcal{O}_C(P-P_0)$ gives $C$ the group structure from $Pic^0(C)$.

Example 1: Take 12 points $P_1,\ldots,P_{12}\in C$. Let $p:X\to \mathbb{P}^2$ be the blowup in these points, denote the proper transform of $C$ by $C'$. Since $C'^2=9-12=-3$, it can be contracted analytically (to a singular point), let $\pi:X\to Y$ be the contraction. We show that for a general choice of $P_i$'s each divisor on $Y$ meets the singular point, so $Y$ cannot be algebraic.

Suppose $D$ is a divisor on $Y$ which does not meet the singular point. The divisor $B=p_*\pi^*D$ meets the cubic only in the chosen points, so scheme-theoretically for some integers $k_i$ we have $\sum k_i P_i=B\cap C\sim deg B\cdot H\cap C=3deg B\cdot P_0$. we see that $\sum k_iP_i$ is the zero of the group, which does not happen for a general choice of $P_i$'s. (Note that since $C$ is uncountable, one can easily find $12$ $\mathbb{Z}$-independent points on it).

Example 2: In the example above take for $P_i$'s the points of intersection of some quartic $Q$ with our cubic. The linear system $\mathfrak{s}$ spanned by $Q$ and quartics of type $C+line$ is transformed birationally by $p$ to a free linear system $p^{-1}\mathfrak{s}$. We see easily that for an irreducible $U\subseteq X$ we have $U\cdot (C'+p^{-1}line)=0$ if and only if $U=C'$, so the new system realizes the contraction $X\to Y$ algebraically.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.