Timeline for Discrete central subgroup of a connected Lie group is finitely generated
Current License: CC BY-SA 4.0
11 events
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| Aug 7, 2019 at 22:03 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, moved main statement to main text, formatting
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| Jan 2, 2018 at 21:28 | comment | added | nfdc23 | @HJRW: The reason is that when I learned algebraic topology by self-study from books, I learned the finiteness theorem for homology of compact manifolds before anything about CW complexes. So I have no idea how most people think about the logical development of the subject. | |
| Jan 2, 2018 at 14:23 | comment | added | HJRW | @nfdc23, a tiny comment on your nice argument. Why bother going via the homology? The homology is finitely generated because K has the structure of a finite CW complex, and for exactly the same reason, $\pi_1K$ is finitely generated. | |
| Jan 2, 2018 at 1:00 | comment | added | nfdc23 | @QiaochuYuan: Fair point. Those results on maximal compact subgroups (at least for connected Lie groups) are so central to how the structure in general is demystified (at least in my own experience with self-education in the subject) that my instinct is to use them whenever convenient. | |
| Jan 1, 2018 at 23:58 | comment | added | YCor | @QiaochuYuan actually when $G$ is a closed linear group there's a somewhat immediate argument using Zariski closure. | |
| Jan 1, 2018 at 22:55 | comment | added | Qiaochu Yuan | @nfdc23: this was the first argument I thought of but it felt like overkill to me. The special case $G = \mathbb{R}^n$ can be done in a much more elementary way, for example. Is there no hope of an elementary argument in general? If $G$ is a real algebraic group perhaps one could take the Zariski closure of $\Gamma$ or something like that, at least, and argue in that closure. | |
| Jan 1, 2018 at 20:55 | vote | accept | Calamardo | ||
| Jan 1, 2018 at 20:09 | answer | added | YCor | timeline score: 4 | |
| Jan 1, 2018 at 20:05 | comment | added | YCor | Yes basically I had to rely on simple connectedness of the quotient by a maximal compact subgroup. It's even enough to know the existence of maximal subgroup and their conjugacy. I'll finish writing a post based on this. | |
| Jan 1, 2018 at 19:52 | comment | added | nfdc23 | Let $G$ be a connected Lie group, $\Gamma$ a discrete central subgroup. Then $G\to H:=G/\Gamma$ is a connected Galois cover with $\Gamma$ its group of deck transformations, so $\Gamma$ is a quotient of $\pi_1(H)$. Thus, it suffices to show $\pi_1(H)$ is finitely generated for any connected Lie group $H$. But a maximal compact subgroup $K$ of $H$ is a deformation retract (structure of connected Lie groups), so $\pi_1(H)=\pi_1(K)$. Since $\pi_1(K) = {\rm{H}}_1(K,\mathbf{Z})$, and the integral homology of any compact manifold is finitely generated, we are done. | |
| Jan 1, 2018 at 18:55 | history | asked | Calamardo | CC BY-SA 3.0 |