Timeline for Is any continuous group homomorphism from R to C* an exponential map? [closed]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30, 2014 at 19:30 | comment | added | Qiaochu Yuan | And every smooth homomorphism between Lie groups is real analytic! | |
| Jan 29, 2014 at 20:27 | comment | added | archipelago | Even better. Every measurable homomorphism between locally compact groups is continuous, so every measureable homomoprhism between lie groups is smooth. | |
| Jan 29, 2014 at 3:46 | vote | accept | Hiro | ||
| Jan 29, 2014 at 0:12 | history | closed |
Qiaochu Yuan YCor Ricardo Andrade Stefan Kohl♦ Anthony Quas |
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| Jan 28, 2014 at 23:43 | answer | added | Keenan Kidwell | timeline score: 8 | |
| Jan 28, 2014 at 23:26 | answer | added | Marty Isaacs | timeline score: 5 | |
| Jan 28, 2014 at 23:20 | review | Close votes | |||
| Jan 29, 2014 at 0:15 | |||||
| Jan 28, 2014 at 23:05 | comment | added | Qiaochu Yuan | Yes. In fact much more is true: any continuous homomorphism between Lie groups is automatically smooth, and hence must be induced by the corresponding map of Lie algebras. | |
| Jan 28, 2014 at 23:04 | comment | added | Peter Crooks | I would suggest lifting $\varphi:\mathbb{R}\rightarrow\mathbb{C}^*$ to the universal cover $\exp:\mathbb{C}\rightarrow\mathbb{C}^*$. | |
| Jan 28, 2014 at 22:56 | history | asked | Hiro | CC BY-SA 3.0 |