Timeline for Finite dimensional compact abelian group that is not a product of connected and a totally disconnected
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2023 at 16:35 | comment | added | Yanko | @YCor Do you know if any short exact sequence $0\rightarrow G_0\rightarrow G\rightarrow G/G_0\rightarrow 0$ split if we know that $G/G_0$ has some bounded torsion (say torsion $p$ so $G/G_0 = \prod \mathbb{Z}/p\mathbb{Z}$)? (Here $G$ compact and abelian) | |
| Nov 9, 2018 at 10:42 | comment | added | TopGroups | Thanks again for your last edit, my question is finally solved. I appreciate your effort a lot. | |
| Nov 4, 2018 at 18:43 | history | edited | YCor | CC BY-SA 4.0 |
clarified statements
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| Nov 4, 2018 at 16:56 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo (a previous edit replaced a typo by another typo)+ added strengthened example
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| Nov 4, 2018 at 16:43 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo (a previous edit replaced a typo by another typo)
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| Nov 4, 2018 at 14:12 | history | edited | Qfwfq | CC BY-SA 4.0 |
added 6 characters in body
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| Nov 4, 2018 at 14:11 | history | edited | YCor | CC BY-SA 4.0 |
addendum
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| Nov 4, 2018 at 13:47 | comment | added | YCor | I'll edit my post if I can answer your other question. | |
| Nov 4, 2018 at 13:46 | vote | accept | TopGroups | ||
| Nov 4, 2018 at 13:46 | comment | added | TopGroups | Oh I see, in this case I will accept this answer. But will leave the comments just to ensure nobody gets confused. I will try to use this example to construct what I need. Thanks | |
| Nov 4, 2018 at 13:45 | comment | added | YCor | Your question was somewhat unclear: I answered the question suggested by the title, for which you provided a wrong example in your post. | |
| Nov 4, 2018 at 13:40 | comment | added | TopGroups | Right I see now. You find a compact abelian group $K$ such that $K/K_0$ is a direct product, $\prod_{i\in I}\mathbb{Z}/p_i\mathbb{Z}$ and $K \not =K_0\times \prod_{i\in I}\mathbb{Z}/p_i\mathbb{Z}$. But this doesn't answer my question, is it? To answer my question I think you need to find a group $K$ and a direct product of finite groups $\Delta$ such that $K/\Delta$ is a torus but $K\not = K_0 \times K/K_0$. | |
| Nov 4, 2018 at 13:27 | comment | added | YCor | I fixed a typo ($K$ is dual of $H$, not of $G$). Except in the first and last paragraph, i.e. when I construct $H$, I consider no topology on the groups. | |
| Nov 4, 2018 at 13:26 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Nov 4, 2018 at 13:20 | comment | added | TopGroups | When you say "isomorphic to $\mathbb{Q}^k$ or some vector space over $\mathbb{Q}$" you mean isomorphic as abstract groups am I right? Because the direct sum is not a closed subgroup. I'm also confused about the last paragraph, the dual group of $G$ is the direct sum, which is discrete, then how come it admits a connected group as a closed subgroup? | |
| Nov 4, 2018 at 12:23 | history | answered | YCor | CC BY-SA 4.0 |