Timeline for Existence of at least one compact orbit on the sphere
Current License: CC BY-SA 3.0
18 events
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| Aug 18, 2018 at 17:01 | comment | added | guido giuliani | Thank you very much for your answer! From what I understand one considers the Iwasawa decomposition $G=KAN$ of a semisimple group $G$, therefore if we take $M$ the centralizer of $A$ in $K$, then $MAN$ is a minimal parabolic subgroup of $G$. The factor $AN$ now is cocompact solvable and connected, as desired. Am I right? -Guido- | |
| Aug 14, 2018 at 10:38 | comment | added | Uri Bader | OK, I understand that what you miss is the fact that every semisimple Lie group has a solvable cocompact subgroup. A quick way to see that is by taking the minimal parabolic subgroup in it, which is cocompact, and to note that the minimal parabolic itself has a normal cocompact solvale subgroup. These two facts should appear in any good book regarding semisimple Lie groups. | |
| Aug 13, 2018 at 14:26 | comment | added | guido giuliani | The last part using Lie Kolchin is clear. Many thanks in advance -Guido- | |
| Aug 13, 2018 at 14:25 | comment | added | guido giuliani | I know that any lie Group contains a cocompact subgroup, but in general we cannot guarantee that it is connected, so this is where I am a bit confused. In the other hand, at the beginning of your last paragraph, it seems implicitly stated that with the previous arguments we should find a cocompact solvable connected real group. Am I wrong? -Guido- | |
| Aug 13, 2018 at 14:24 | comment | added | guido giuliani | My main issue is the connectedness of the cocompact solvable subgroup that we find. The theorem I found is the following (on the complex field): let H be a closed subgroup of G: then G/H cocompact iff H contains a maximal solvable connected subgroup (Borel-Tits IHS, prop 9.3). The equivalence follows using the fixed point theorem. I understand well that we find what we are looking for on C. How can we guarantee the existence of such a real group? | |
| Aug 13, 2018 at 14:17 | comment | added | Uri Bader | Are you concerned with the proof of the fact that every connected group has a cocompact solvable subgroup, or with the stuff comes after that? | |
| Aug 13, 2018 at 13:44 | comment | added | guido giuliani | Dear professor Bader, sorry to bother you again, but I realized that there is still a small point in the proof which puzzles me. Namely, I've understood that we can reduce everything to the case of real algebraic linear groups. Nonetheless to apply correctly the Borel Fixed point theorem we need to complexify to work on an algebraically closed field. The cocompact solvable connected subgroup that we find is thus a complex subgroup. But we would like to have it real. Is it automatic the passage to a real subgroup having all the required properties, or am I missing something? Bests, -Guido- | |
| Apr 4, 2018 at 11:34 | comment | added | guido giuliani | I cannot express enough my gratitude. It was a pleasure to have this discussion with you! My bests -Guido- | |
| Apr 4, 2018 at 10:29 | comment | added | Uri Bader | The 1-2 dimensiinality is an artifact of the reduction steps in the proof. (varying the acting group) you can have compact orbits of unbounded dimension, as the example of the standard rep of SL_n shows. | |
| Apr 4, 2018 at 9:58 | comment | added | guido giuliani | Thank you very much. Your explanation clarifies exactly all the points where I was a bit confused. I've got a last (small question). The fact that all these representation are 1 or 2 dimensional does not imply anything on the dimension of the compact orbits. Am I right? I mean, we can have, in general compact orbits of any dimension (I am thinking of irreducible groups where it seems strange to have circles). BTW, I am really grateful for your explanation. Again thank you very much! | |
| Apr 4, 2018 at 9:54 | vote | accept | guido giuliani | ||
| Apr 3, 2018 at 23:04 | comment | added | Uri Bader | @guido, I added details. Please let me know if anything remains unclear. | |
| Apr 3, 2018 at 23:02 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 2, 2018 at 13:00 | comment | added | guido giuliani | From what I understand, the cocompact solvable subgroup should be the solvable radical, am I right? Then I have some difficulties concerning the statement: the unipotent radical of the Zariski closure should be trivial (here is my inexpertise, unfortunately). The rest is somewhat clear, since we are complexifying, the representation is either 1 or 2 dimensional (linked to the fact that the eigenvalues are complex or real, and the Lie Kolchin theorem), and then one studies the orbits (I believe we identify the projective line with the circle). This is what I understand, sorry for the mess. | |
| Apr 2, 2018 at 7:07 | comment | added | Uri Bader | @guido, With pleasure. My answer was indeed brief, as I am currently traveling and do not have much time, but I will provide more details when I am back. It will help me if you could let me know the less clear parts on your side. | |
| Apr 1, 2018 at 12:49 | comment | added | guido giuliani | Thanks for your reply. I've awarded the bounty to your answer already, but I would like to take some time before accepting it, because I would like to check myself your explanation. This will take me some time, since I'm not an expert. If you feel like adding some more details, please go ahead, that would be greatly appreciated. Thank you again! -Guido- | |
| Apr 1, 2018 at 12:47 | history | bounty ended | guido giuliani | ||
| Mar 31, 2018 at 15:42 | history | answered | Uri Bader | CC BY-SA 3.0 |