Timeline for When is the analytification of a variety homeomorphic to $\mathbb C$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 9, 2014 at 19:34 | comment | added | Donu Arapura | Yes to the follow up question, because $\mathbb{A}^1$ can be characterized as the unique nonsingular simply connected curve which isn't compact. | |
| May 9, 2014 at 17:29 | comment | added | Ste3an | @JasonStarr Thank you. If we ask $V$ to be smooth, then $X$ is isomorphic to $\mathbb A^1$ I think. Is that true? | |
| May 9, 2014 at 17:15 | comment | added | Jason Starr | Yes, of course: the analytification of $\{(z,w)\in \mathbb{C}^2 : z^2 = w^3\}$ is homeomorphic to $\mathbb{R}^2$. | |
| May 9, 2014 at 17:06 | history | edited | Ste3an | CC BY-SA 3.0 |
my other question was not precise enough
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| May 9, 2014 at 8:43 | history | edited | Ste3an | CC BY-SA 3.0 |
deleted 31 characters in body
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| May 9, 2014 at 8:28 | history | edited | Ste3an | CC BY-SA 3.0 |
added 301 characters in body; edited title
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| May 8, 2014 at 18:36 | review | First posts | |||
| May 8, 2014 at 18:41 | |||||
| May 8, 2014 at 18:16 | history | asked | Ste3an | CC BY-SA 3.0 |