Timeline for Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 3
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Oct 30, 2011 at 14:40 | history | edited | Peter Scholze | CC BY-SA 3.0 |
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| Oct 30, 2011 at 14:39 | answer | added | Peter Scholze | timeline score: 11 | |
| Oct 12, 2011 at 18:43 | comment | added | Peter Scholze | Of course. For this reason, the question is formulated in the weakest possible sense -- instead of $X$ you might deform any morphism $Y\rightarrow \mathbb{P}^n$ with set-theoretical image equal to $X$, for example. It might also be that one can answer this question via some kind of approximation method, i.e. without using deformation theory... Or, the answer might be no, even for this weak question. | |
| Oct 12, 2011 at 17:40 | comment | added | M P | Some singularities are non-smoothable, so at least if you want the variety $\tilde{X}$ to be in a family containing $X$, I think that the answer is that what you ask does not happen in general. | |
| Oct 12, 2011 at 16:14 | history | edited | Peter Scholze | CC BY-SA 3.0 |
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| Oct 11, 2011 at 22:01 | history | asked | Peter Scholze | CC BY-SA 3.0 |