10
$\begingroup$

One often finds statements of the sort "and one can contract this subvariety $E\subset X$ to a point in the projective variety $W$," without any explanation of the reasons such a contraction exists, let alone why the variety $W$ is projective. I was wondering if there are any known criteria (obviously necessary and sufficient criteria would be the best) for when one can do this. Staying algebraic/projective is my main issue here since I'm aware that Grauert has many theorems allowing this in the analytic category.

I'm aware of Ishii's paper, "Some Projective Contraction Theorems" which pretty much sums up the story very nicely in the case of divisors on projective varieties, and everything there remains in the projective category.

I was wondering about contractions of higher codimension subvarieties. Here are two examples from the literature of contractions which are claimed to exist, but no reason is given. Is there a well-known theorem (just unknown to me) lurking in the background here?

1) In Namikawa's paper "Mukai flops and derived categories", he takes a subvariety $Y\cong \mathbb P^n$ on a smooth $2n$-dimensional projective variety $X$ with normal bundle $N_{Y/X}\cong \Omega^1_{\mathbb P^n}$. Then he blows $X$ up along $Y$ and blows down the exceptional divisor in the other direction to obtain another subvariety $Y^+$ in $X^+$ satisfying the same conditions as the "non-plus" versions. Then he claims that $Y$ and $Y^+$ can be contracted down to the same point on a projective variety $\overline{X}$.

2) In Kawamata's paper "D-equivalence and K-equivalence", he has something probably easier to explain: the smooth $2m+1$-dimensional projective variety $X$ contains a subvariety $E\cong \mathbb P^m$ with normal bundle $N_{E/X}\cong \mathcal O_{\mathbb P^m}(-1)^{m+1}$. Again he blows up $X$ along $E$ and then blows down the exceptional divisor in the other direction to obtain $F\subset Y$ which again satisfy the same conditions as $E\subset X$. Again there is a claim that both $E$ and $F$ can be contracted in $X$ and $Y$, respectively, to points in the same projective variety $W$.

These are the kind of contractions I'm wondering about. Why do these two contractions exist (and specifically why in the projective/algebraic category), and why do both subvarieties in each case get contracted to the same point in the same projective variety?

$\endgroup$
4
$\begingroup$

My understanding is that there are essentially no results of the form you are looking for, i.e. there are no local criteria for contractibility in the category of projective varieties.

The problem is: assume such a contraction exists; it would be induced by a nef divisor L with L.C = 0 for any curve C in Y. But then the map induced by L would also contract any curve numerically equivalent to C. This is an inherently global phenomenon.

In fact, regarding your examples 1): Namikawa assumes to start with that the variety $X^+$ is also projective. There are counter-examples to such a Mukai flop being projective, see e.g. section 4.4 in Yoshioka, arXiv:math/0009001: there is a projective flop of $X$, given by simultaneously flopping a number of projective subspaces in $X$; the lines in the different projective spaces all have the same numerical class $l$ on $X$. Thus, in the projective category, no single $\mathbb P^r$ can be flopped individually (then $l$ and $-l$ would be effective curve classes, contradicting the existence of ample divisors), and no single $\mathbb P^r$ can be contracted individually (by the argument in the previous paragraphs). Other examples could be constructed using section 14 of my own paper with E. Macrì, arXiv:1301.6968, where we give examples of flopping contraction whose exceptional locus has arbitrary many connected components; I would be surprised if the formal local structure of these components is distinguishable from cases where the flopping contraction has just one irreducible component in the exceptional locus.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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