Timeline for When does a LCA group not contain a (closed) infinite cyclic subgroup?
Current License: CC BY-SA 3.0
12 events
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Jan 15, 2014 at 22:10 | comment | added | YCor | This holds iff $G$ is the union of its compact open subgroups, iff $G$ is the filtering union of its compact open subgroups. The latter condition (for $G$ not necessarily abelian) has various names in the literature, including "topologically locally finite" (yuk!), "locally elliptic", "elliptic". | |
Jul 23, 2011 at 17:44 | vote | accept | Iian Smythe | ||
Jul 23, 2011 at 17:44 | vote | accept | Iian Smythe | ||
Jul 23, 2011 at 17:44 | |||||
Jul 23, 2011 at 3:06 | vote | accept | Iian Smythe | ||
Jul 23, 2011 at 17:44 | |||||
Jul 21, 2011 at 20:33 | comment | added | Marc Palm | The correct statement that $\mathbb{Z}$ maps into $A$, iff $\widehat{A}$ maps onto $\mathbb{T}$. Pontryagin duality switches the arrow;) Sorry for the confusion -.- | |
Jul 21, 2011 at 20:16 | answer | added | Iian Smythe | timeline score: 2 | |
Jul 21, 2011 at 20:10 | history | edited | Iian Smythe | CC BY-SA 3.0 |
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Jul 21, 2011 at 19:34 | comment | added | Marc Palm | You'r completely right, I was overlooking that not everything splits. I deleted my previous comment, but please have a look at my final answer. | |
Jul 21, 2011 at 19:06 | comment | added | Iian Smythe | I don't see this since $\mathbb{R}$ contains $\mathbb{Z}$ as a closed subgroup, but is self-dual. If however, we are considering groups without copies of $\mathbb{R}$ inside, then this, via duality, would say that the only way $\mathbb{Z}$ can occur (closed) in such a group is as a direct factor. Is this the case? | |
Jul 21, 2011 at 18:18 | comment | added | Mark | I guess that this is true iff the dual group doesn't have $\mathbb{S}^1$ as a factor or something like that. | |
Jul 21, 2011 at 18:01 | answer | added | Marc Palm | timeline score: 3 | |
Jul 21, 2011 at 17:29 | history | asked | Iian Smythe | CC BY-SA 3.0 |