Timeline for Analytic function avoiding elements of the modular group
Current License: CC BY-SA 3.0
17 events
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| Apr 28, 2018 at 3:00 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 8, 2013 at 1:34 | comment | added | Venkataramana | Wow! Thanks, Alexandre! I did not expect the bounty! | |
| Feb 8, 2013 at 0:43 | vote | accept | Alexandre Eremenko | ||
| Feb 8, 2013 at 0:43 | history | bounty ended | Alexandre Eremenko | ||
| Feb 3, 2013 at 15:31 | comment | added | Alexandre Eremenko | I think I can prove that there are no fractional-linear $f$ satisfying the conditions. This implies that Aakumadula's example is not fractional-linear. | |
| Feb 3, 2013 at 8:42 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 3, 2013 at 5:59 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 3, 2013 at 5:42 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 3, 2013 at 5:22 | comment | added | Alexandre Eremenko | I deleted my comment because I understood why it was true:-) It is indeed unlikely that your function $f$ is fractional-linear, but I would like to see a proof. | |
| Feb 3, 2013 at 5:13 | comment | added | Venkataramana | Alexandre, Thanks for the edit: i meant $z\mapsto \lambda ^{-1}(x(az+b))$. | |
| Feb 3, 2013 at 4:40 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 3, 2013 at 4:35 | comment | added | Venkataramana | $z\mapsto x(z/3):H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty\}$ is a holomorphic map from a simply connected space, and hence, by the lifting criterion lifts to a map $f:H\rightarrow H$ to the universal cover such that $ x(z/3)=\lambda (f(z))$. | |
| Feb 3, 2013 at 4:33 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Feb 3, 2013 at 4:27 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Feb 3, 2013 at 4:12 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 3, 2013 at 4:00 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 3, 2013 at 3:06 | history | answered | Venkataramana | CC BY-SA 3.0 |