Skip to main content
11 events
when toggle format what by license comment
Feb 22, 2016 at 21:18 answer added John Pardon timeline score: 8
Jul 3, 2013 at 15:14 history edited Charles Staats
edited tags
Jul 2, 2013 at 8:28 answer added Francesco Polizzi timeline score: 14
Oct 10, 2012 at 21:08 comment added Jérémy Blanc 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 $X$?
Jun 9, 2012 at 6:59 comment added Angelo I remember discussing this with Joe Harris many years ago. The problem seems to have been around for a while.
Jun 9, 2012 at 5:55 comment added rita Jason, thank you, this puts my question in perspective.
Jun 8, 2012 at 21:52 comment added Will Sawin 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?
Jun 8, 2012 at 21:16 comment added Jason Starr I have discussed this over the years with several people. It is expected to be false, but it is open.
Jun 8, 2012 at 20:40 comment added user5117 Qiaochu: every point. (I almost asked the same question, by the way.)
Jun 8, 2012 at 20:32 comment added Qiaochu Yuan I don't understand. Isn't this true by definition?
Jun 8, 2012 at 20:20 history asked rita CC BY-SA 3.0