Timeline for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 24, 2011 at 18:28 | comment | added | Ravi Vakil | As of the time I write this, there are a number of different answers (in two families) that have given me a great deal of enlightenment. Thank you Georges for this one! | |
| Jun 22, 2011 at 11:39 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
pure typography : suppressed "\" and added ":"
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| Jun 22, 2011 at 11:26 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
Added Second edit. displayed all calculations of Pic
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| Jun 22, 2011 at 7:43 | comment | added | Georges Elencwajg | Dear Dmitry, the (analytic !) isomorphism is proved by pulling back the projective $\mathbb P^1$ bundle $L\to E$ to the analytic covering space $\mathbb C \to E$ where it becomes trivial.Then you see that the pull back of $U$ is just $\mathbb C \times \mathbb C$ and $U$ is the quotient of that $\mathbb C \times \mathbb C$ by a suitable action of $\mathbb Z \times \mathbb Z$. The quotient is $\mathbb G_m \times \mathbb G_m$ and this is why $U$ is isomorphic to $\mathbb G_m \times \mathbb G_m$ | |
| Jun 21, 2011 at 22:05 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
Added reference To Hartshorne's Lecture Notes. Added constancy of regular functions on U
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| Jun 21, 2011 at 21:12 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added three lines to my "edit"
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| Jun 21, 2011 at 20:20 | comment | added | Georges Elencwajg | Dear Torsten, yes the indefinite article is ambiguous : here it doesn't mean "any" but "a certain bundle that I'm not describing in detail". As I wrote, details are to be found by following the link given, but ultimately I am responsible for any lack of clarity in my answer: so I both apologize and thank you for your constructive comment. | |
| Jun 21, 2011 at 20:10 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added "edit"
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| Jun 21, 2011 at 20:03 | comment | added | Torsten Ekedahl | The indefinite article in "a $\mathbb P^1$-bundle" is a little bit ambiguous as you can not just take any $\mathbb P^1$-bundle with any section, you have to choose them so that the complement is a non-trivial $\mathbb A^1$-bundle as in ulrich's answer (see my comment to it to get an explicit isomorphism). | |
| Jun 21, 2011 at 19:58 | comment | added | Dima Sustretov | Sorry, what's the isomorphism between $U$ and $\mathbb{G}_m \times \mathbb{G}_m$? | |
| Jun 21, 2011 at 19:36 | history | answered | Georges Elencwajg | CC BY-SA 3.0 |