47
$\begingroup$

I've been asked this question by a colleague who's not an algebraic geometer; we both feel that the answer should be "no", but I don't have a clue how to prove it. Here's the question: let $X$ be a smooth rational variety (over the complex numbers, say). Is it true that every point of $X$ has a Zariski open neighbourhood that is isomorphic to an open subset of ${\mathbb P}^n$?

$\endgroup$
7
  • 8
    $\begingroup$ Qiaochu: every point. (I almost asked the same question, by the way.) $\endgroup$
    – user5117
    Commented Jun 8, 2012 at 20:40
  • 24
    $\begingroup$ I have discussed this over the years with several people. It is expected to be false, but it is open. $\endgroup$ Commented Jun 8, 2012 at 21:16
  • 1
    $\begingroup$ Trivial in dimension 1. True in dimension 2 because the minimal models have this property and blow-ups preserve it. Is it true in dimension 3? $\endgroup$
    – Will Sawin
    Commented Jun 8, 2012 at 21:52
  • 6
    $\begingroup$ I remember discussing this with Joe Harris many years ago. The problem seems to have been around for a while. $\endgroup$
    – Angelo
    Commented Jun 9, 2012 at 6:59
  • 4
    $\begingroup$ In dimension 2 the result is true and is in fact stronger: we can assume that the open neighbourhood is isomorphic to $\mathbb{A}^2$. Do you have a counterexample in dimension $3$ of this? By the way, I would have thought that the answer to your question (not the stronger one) should be "yes", but as many people think the converse, I am now confused. If a point admits no such neighbourhood, it implies that every birational maps $X\to \mathbb{P}^n$ is either not defined at $x$ or contracts something through $x$. Do you have some candidate for $x$ and $X$? $\endgroup$ Commented Oct 10, 2012 at 21:08

2 Answers 2

14
$\begingroup$

Some partial results related to this open problem can be found in the very recent preprint by F. Bogomolov and C. Böhning On uniformly rational varieties, see arXiv:1307.0102.

According to the authors (see the Introduction) this question was first raised by M. Gromov in his paper

Oka's principle for holomorphic sections of elliptic bundles, Journal of the American Mathematical Society 2, Vol. 4 (1989).

$\endgroup$
1
  • $\begingroup$ Grazie Francesco! $\endgroup$
    – rita
    Commented Jul 3, 2013 at 19:45
8
$\begingroup$

Here is a preprint of Ilya Karzhemanov constructing counterexamples in dimensions $n\geq 4$:

http://research.ipmu.jp/ipmu/sysimg/ipmu/1588.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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