Timeline for Smoothings of isolated non-irreducible surface singularities
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2018 at 15:04 | vote | accept | user131261 | ||
| Nov 16, 2018 at 12:40 | answer | added | inkspot | timeline score: 1 | |
| Nov 15, 2018 at 21:57 | comment | added | user131261 | Thank you. Yes, that solves my issue. I guess this question was obvious for an expert. But still if you want write your comment as an answer, I will check it as valid. | |
| Nov 15, 2018 at 21:20 | comment | added | inkspot | In your case $X$ is a connected component of the normalisation $Z$ of $Y$. On p. 255 of dam.brown.edu/people/mumford/alg_geom/papers/… Mumford attributes to Rim the result that $Z$ is local (so that $Z=X$) if $Y$ is smoothable. This seems to resolve your issue. | |
| Nov 15, 2018 at 18:11 | comment | added | Jason Starr | Certainly non-S2 singularities can admit smoothings, e.g., when two disjoint 2-planes in affine 4-space coalesce. In that case, the central fiber is not S2. Also, the central fiber is not S1. | |
| Nov 15, 2018 at 18:07 | comment | added | user131261 | There are non-normal isolated singularities (therefore not $S2$) that admit smoothings. The space $(Y,0)$ has a versal deformation. So again, I do not understand the sentence "exclude deformations of $(Y,0)$". | |
| Nov 15, 2018 at 17:51 | comment | added | Jason Starr | I mean exclude deformations of $(Y,0)$. The scheme you have described is S1 but not S2. | |
| Nov 15, 2018 at 16:43 | comment | added | user131261 | @JasonStarr I am sorry. Maybe it is the english that I don't understand. You said that I should be able to rule out (exclude) deformation of singularities of $(Y,0)$. Do you mean exclude some (a priori possible) deformations of $(Y,0)$? Or exclude deformations of some of the irreducible components of $(Y,0)$? | |
| Nov 15, 2018 at 16:32 | comment | added | Jason Starr | No, that is not what I am saying. | |
| Nov 15, 2018 at 16:07 | comment | added | user131261 | I am not sure I understand your comment. Do you mean that $(Y,0)$ does not admit a smoothing because one of its components does not? | |
| Nov 15, 2018 at 15:00 | comment | added | Jason Starr | If $Y$ is reduced near $0$, then you should be able to rule out deformations of singularities of $(Y,0)$. | |
| Nov 15, 2018 at 14:53 | comment | added | user131261 | Yes, I assume that. I edited the equestion. Although, I would be happy with $Y$ being reduced along a component isomorphic to $(X,0)$. | |
| Nov 15, 2018 at 14:51 | history | edited | user131261 | CC BY-SA 4.0 |
added 15 characters in body
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| Nov 15, 2018 at 13:38 | comment | added | Jason Starr | Are you assuming that $Y$ is reduced near $0$? | |
| Nov 15, 2018 at 6:08 | history | asked | user131261 | CC BY-SA 4.0 |