Timeline for Action on a normal subgroup where each coset acts freely
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2017 at 3:10 | comment | added | Michael Cotton | The coset representatives were for the fixed N because the assumption for them is that the subgroup being modded out is closed topologically. You have them for a specified subgroup but don't have them for all subgroups. So that's a difficulty. But I think that difficulty can be dealt with in many cases. At least w.r.t locally compact abelian groups, the duality properties seem to help make it work. I haven't got it all going yet, but I bet it'll chooch at least some of the time and that there's going to be something to be gained here. :) | |
| Oct 8, 2017 at 22:08 | comment | added | Michael Cotton | Thank you very much for all the help! I'll have to spend some time on seeing if it can be generalized and if it meshes with the topological side of things, but you've got me in what seems like the right ballpark for an approach. So maybe there's some hope! :) | |
| Oct 8, 2017 at 22:02 | vote | accept | Michael Cotton | ||
| Oct 8, 2017 at 1:03 | comment | added | Aaron Meyerowitz | I'm not sure in your descriptive set theory work what you accept as "constructive." SInce you grant that a subgroup always has a usable set of coset representatives, it might be pretty general. At any rate, I have given a last example which seems to me to work, modulo being allowed a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space (i.e. a set of coset representatives). | |
| Oct 8, 2017 at 0:59 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Oct 7, 2017 at 17:20 | comment | added | Michael Cotton | Can something similar be done on something more like $\mathbb{R}$ and the subgroup $\mathbb{Z}$, or are we leaning very hard on the structure of the permutation group here? | |
| Oct 7, 2017 at 7:26 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Oct 7, 2017 at 6:40 | comment | added | Aaron Meyerowitz | Now that I know you want infinite groups I will rewrite the answer. All I meant by arbitrarily identify was "pick a constructive bijection, any one you want" | |
| Oct 7, 2017 at 0:59 | comment | added | Michael Cotton | Also, what's a representation? I'm not an algebraist. | |
| Oct 7, 2017 at 0:43 | comment | added | Michael Cotton | The problem is "arbitrarily identify". I need to be able to describe or define what's going on. | |
| Oct 6, 2017 at 20:36 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Oct 6, 2017 at 17:13 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Oct 6, 2017 at 15:18 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Oct 6, 2017 at 15:09 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |