Timeline for How bad can $\pi_1$ of a linear group orbit be?
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13 events
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| May 16, 2015 at 19:45 | comment | added | YCor | I think your modification ensures that $j(G)$ is closed in $L$, but $j(G)L_v$ is not necessarily closed. | |
| May 16, 2015 at 15:06 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| May 16, 2015 at 15:05 | comment | added | Will Sawin | @YCor Then I think my amended answer, together with your answer, shows that it is actually finite-by-abelian. | |
| May 16, 2015 at 14:46 | comment | added | YCor | @WillSawin: "finite-by-abelian" was defined line 3 of my answer. In my answer I claim to prove finite-by-(2-step-nilpotent), which is weaker than finite-by-abelian, but rules out many virtually abelian (= abelian-by-finite) groups such as the infinite dihedral group. Note that the intersection of the classes (virtually abelian) and (finite-by-nilpotent) is equal to the class (finite-by-abelian). | |
| May 16, 2015 at 14:26 | comment | added | Will Sawin | @FrancoisZiegler Indeed. | |
| May 16, 2015 at 14:25 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| May 16, 2015 at 14:11 | comment | added | Francois Ziegler | @WillSawin: Thanks. Where you write "its $\pi_1$ vanishes" you really mean $\pi_2$, right? | |
| May 16, 2015 at 14:10 | comment | added | Will Sawin | @YCor I believe I have fixed the problems in the first part of my answer. And, for the second part, obviously my example is not connected. | |
| May 16, 2015 at 14:06 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| May 16, 2015 at 13:58 | comment | added | Will Sawin | @YCor For the first part, yes you are right. For the second part, what's the difference between virtually abelian and finite-by-abelian? I'm not familiar with the terminology. It seems, looking it up, that virtually abelian is the same as abelian-by-finite. Do you really prove finite-by-abelian and not abelian-by-finite? | |
| May 16, 2015 at 13:34 | comment | added | YCor | For the converse at the end of your answer, it is for the moment in contradiction with my answer. For instance, according to my answer, if $\Gamma$ appears and it's center-free, then it's finite-by-abelian. In particular we cannot get the infinite dihedral group. Also this discards center-free torsion-free virtually abelian groups, such as the fundamental group of the Klein bottle. I have to think twice to figure out where the error is located. | |
| May 16, 2015 at 13:31 | comment | added | YCor | For the first part of your answer: I need to take $L$ the unit component in the Lie topology to ensure that $[L,L]$ is contained in $j(G)$. It's not a problem in the sequel. On the other hand $j(G)$ need not be closed in the Lie topology, and probably $j(G)L_v$ can be non-closed, so we cannot say that $L/j(G)L_v$ is an abelian Lie group. | |
| May 16, 2015 at 13:11 | history | answered | Will Sawin | CC BY-SA 3.0 |