Timeline for Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?
Current License: CC BY-SA 2.5
12 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Aug 4, 2010 at 16:25 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
added a comma
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| Aug 4, 2010 at 13:05 | comment | added | Keivan Karai | Nice; to me it seems correct. | |
| Aug 4, 2010 at 7:32 | comment | added | Pierre-Yves Gaillard | The above two comments refer to a previous version of the post. | |
| Aug 4, 2010 at 7:25 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Streamlined the proof.
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| Aug 3, 2010 at 19:09 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
corrected spelling
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| Aug 3, 2010 at 18:50 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Tried to complete the proof.
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| Aug 3, 2010 at 17:42 | comment | added | Pierre-Yves Gaillard | Thanks a lot! I edited the post. I hope it is correct now; but even if it is correct, it doesn't answer the question. --- I agree that $Ad(G)$ may not have semisimple elements. But if $x\in M_n(\mathbb R)$ is not semisimple, then $exp(\mathbb Z x)$ is infinite and discrete. (Look at Jordan's blocks.) Don't hesitate to tell me if this is false! | |
| Aug 3, 2010 at 17:31 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Corrected a mistake.
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| Aug 3, 2010 at 16:26 | comment | added | Keivan Karai | You need to be more careful. What if $Ad(G)$ is compact? Even if $Ad(G)$ is not compact, it may not have any semi-simple elements. Consider just the case that $G$ is a linear nilpotent group, then all of the eigenvalues in question will be zero. | |
| Aug 3, 2010 at 13:21 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added the general case.
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| Aug 3, 2010 at 6:04 | history | answered | Pierre-Yves Gaillard | CC BY-SA 2.5 |