Timeline for Uniqueness of smooth compactification upto a smooth morphism
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 15, 2016 at 13:04 | comment | added | Mikhail Bondarko | I have several papers on this subject (arxiv.org/find/math/1/au:+Bondarko_M/0/1/0/all/0/1) including the survey arxiv.org/abs/0903.0091 that is unfortunately far from being perfect. If you don't like my style and/or Voevodsky motives then you can also read the (somewhat "classical") paper Gillet H., Soulé C. Descent, motives and K-theory// J. reine und angew. Math. 478, 1996, 127–176. | |
| Apr 15, 2016 at 12:22 | comment | added | Anandam Banerjee | @MikhailBondarko Yes, I seem to be have went in the wrong direction in asking this question. If I want to understand how the non-uniqueness is captured in the weight filtration of motives (as you mentioned in your comment), can you point me to some reference? | |
| Apr 15, 2016 at 7:29 | vote | accept | Anandam Banerjee | ||
| Apr 14, 2016 at 15:41 | comment | added | Mikhail Bondarko | So, it seems that your question is not a "right" one. Are you interested in other results in this direction (say, in motivic or cohomological ones)? | |
| Apr 14, 2016 at 15:34 | answer | added | Sándor Kovács | timeline score: 9 | |
| Apr 14, 2016 at 9:02 | comment | added | Anandam Banerjee | Thanks for the example! Isn't the non-uniqueness coming from the choice of a resolution of singularities, so that if $\bar{X}$ is already smooth (following the notation of the question), then the resolution of singularities $\pi$ will be an isomorphism? | |
| Apr 14, 2016 at 6:57 | comment | added | Ariyan Javanpeykar | The answer is no, I think. Take $X$ to be a smooth projective minimal surface (of positive Kodaira dimension to be safe). Let $D$ be a smooth irreducible curve in $X$, and let $U =X\setminus D$. Then $X$ is a smooth compactification of $U$. Any other smooth compactification of $U$, $X'$ say, will map uniquely to $X$, and this morphism is smooth if and only if it is an isomorphism. Is that convincing? | |
| Apr 14, 2016 at 4:42 | comment | added | Anandam Banerjee | As you point out, a morphism between smooth compactifications is not in general smooth. My question can also be framed as : given 2 smooth compactifications, does there exist a third with smooth morphisms to the first two? | |
| Apr 13, 2016 at 19:48 | comment | added | Ariyan Javanpeykar | Let $\bar{X}$ be a smooth compactification of $X$ with (non-empty) boundary $D$. Now, blow-up a point in $D$. Write $\bar{X}'\to \bar{X}$ for this blow-up. Then $\bar{X}'$ is another compactification of $X$. This is your $g$, right? This morphism $g$ is not smooth. So what am I missing in your question? | |
| Apr 13, 2016 at 9:50 | history | asked | Anandam Banerjee | CC BY-SA 3.0 |