Timeline for Is the condition ``adjoint action does not have eigenvalue $-1$" dense in a Lie group?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 20, 2014 at 11:20 | vote | accept | Xin Nie | ||
| Jun 20, 2014 at 11:16 | vote | accept | Xin Nie | ||
| Jun 20, 2014 at 11:20 | |||||
| Aug 6, 2012 at 2:45 | answer | added | Will Sawin | timeline score: 4 | |
| Aug 6, 2012 at 0:33 | answer | added | Peter McNamara | timeline score: 7 | |
| Jul 13, 2012 at 8:54 | comment | added | Xin Nie | This actually solves the problem... You should put it as an answer. | |
| Jul 13, 2012 at 7:31 | history | edited | Xin Nie | CC BY-SA 3.0 |
added 260 characters in body
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| Jul 13, 2012 at 5:57 | comment | added | Peter McNamara | I feel the proof should proceed by showing that the function f(g)=char poly of Ad(g) evaluated at -1 is an analytic function on G. Then on each connected component, it either is identically zero, or nonzero on a dense set. Since f doesn't vanish at identity, QED for connected G. | |
| Jul 13, 2012 at 2:13 | comment | added | Peter McNamara | At least an assumption that G is connected should be included to rule out the orthogonal group O(2). | |
| Jul 12, 2012 at 21:58 | history | edited | Evan Jenkins | CC BY-SA 3.0 |
Fixed braces (using backticks)
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| Jul 12, 2012 at 21:45 | history | asked | Xin Nie | CC BY-SA 3.0 |