Timeline for map of endomorphism operad
Current License: CC BY-SA 4.0
15 events
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| Jan 14, 2021 at 20:41 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 9, 2021 at 10:11 | comment | added | Qiaochu Yuan | @LSpice: hmm, the argument I had in mind has a gap. A compact connected abelian group has Pontryagin dual a torsion-free discrete abelian group, which is a filtered colimit of $\mathbb{Z}^n$'s; so a compact connected abelian group is a cofiltered limit of tori $T^n$. What I would like to conclude at this point is that its fundamental group is the corresponding cofiltered limit of fundamental groups, but I don't actually see how to prove that. I don't think it's generally true that $\pi_1$ preserves cofiltered limits, unfortunately (is it?). | |
| Jan 8, 2021 at 15:47 | comment | added | LSpice | @QiaochuYuan, sorry for the stupid question, but why is a contractible, compact, Abelian group trivial? | |
| Jan 8, 2021 at 3:58 | comment | added | Qiaochu Yuan | @Yemon: Oh, yes. Is it something like every LCA group is a direct product of $\mathbb{R}^n$ and an extension of a compact group by a discrete group? I guess it follows that a contractible LCA group is $\mathbb{R}^n$ for some $n$ (also a fun fact) and then we are done. | |
| Jan 8, 2021 at 3:53 | comment | added | Yemon Choi | @LSpice There's a structure theorem for LCA groups, whose details I never learned properly, but I think I've seen a more or less compete proof in the book of Deitmar and Echterhoff which doesn't need to invoke GMZ; I think somehow one can split off the non-Lie weird compact bits by hand using facts special to the Abelian group setting. If one used that structure theorem as a black box then I think one would end up showing that the original topological group had to be iso to ${\mathbb R}$ as a topological group, but I admit I haven't sat and thought through the details | |
| Jan 8, 2021 at 3:45 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 6, 2021 at 0:24 | comment | added | Qiaochu Yuan | @LSpice: good question! I haven't thought about it. I don't know much about the proof so I don't know if it substantially simplifies for $\mathbb{R}$. I guess the first thing I'd try is to move the identity to the origin and see if the group operation admits anything like a "Taylor expansion" near the origin? | |
| Jan 6, 2021 at 0:21 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 6, 2021 at 0:18 | comment | added | LSpice | Also, this is a lovely proof, and that fact about topological rings is indeed fun, but it's amazing to hit a fact about $\mathbb R$ with the full GMZ hammer! Do you know if one can streamline the proof in that special case? | |
| Jan 6, 2021 at 0:17 | comment | added | Qiaochu Yuan | @LSpice: yes, thank you. | |
| Jan 6, 2021 at 0:17 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 6, 2021 at 0:11 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Jan 6, 2021 at 0:07 | history | undeleted | Qiaochu Yuan | ||
| Jan 6, 2021 at 0:06 | history | deleted | Qiaochu Yuan | via Vote | |
| Jan 6, 2021 at 0:04 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |