Timeline for Upper density of subsets of an amenable group
Current License: CC BY-SA 4.0
14 events
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Jul 11, 2022 at 20:32 | comment | added | Gabe Conant | @ARG Yes, that's correct. | |
Jul 10, 2022 at 11:33 | comment | added | ARG | @GabeConant thanks! so if $A$ is such that $\overline{d}(A^{\mathsf{c}}) =0$, then $A$ is thickly syndetic (but the converse fails)? | |
Jul 9, 2022 at 18:38 | comment | added | Gabe Conant | @ARG Yes this notion fits with the others. A set is thickly syndetic if and only if it's complement is not piecewise syndetic (somewhat analogous to how a set is thick if and only if its complement is not syndetic). | |
Jul 7, 2022 at 10:34 | comment | added | ARG | So Thickly syndetic sets fit in this picture? $A$ is thick if for any finite $F \subset G$, $\cap_{f \in F} fA \neq \emptyset$. The notion "thickly syndetic" is defined by: $A$ is thickly syndetic if for any finite $F \subset G$, $\cap_{f \in F} fA$ is a syndetic set. | |
May 6, 2020 at 14:32 | comment | added | Diego Martinez | Just to add something to the already well written answer, there's an account of the relation between Szemeredi's theorem and Furstenberg-Zimmer's theorem in the excellent book of Kerr and Li about ergodic theory, where they not only treat $\mathbb{Z}$, but general discrete groups acting on locally compact Hausdorff spaces. | |
May 6, 2020 at 13:47 | comment | added | Otto | yeah that definition completely ignores the Haar measure, which does not seem to do justice to many things. | |
May 6, 2020 at 13:26 | comment | added | Gabe Conant | I guess one thing is that, if your precise definition of $\bar{d}(A)$ is well-defined in the sense that for any finite $H$ and $\epsilon>0$ there is such a $K$, then $G$ is amenable as a discrete group. So $\bar{d}(A)$ is the upper Banach density with respect to the counting measure, and can be applied to all subsets of $G$. | |
May 6, 2020 at 13:16 | comment | added | Gabe Conant | That's a good question. There may be some discussion of this in Paterson's book "Amenability". Unfortunately my copy is in my office, which I can't access at the moment. I see now that I didn't fully appreciate this part of your question. So perhaps someone will post a better answer. | |
May 6, 2020 at 13:09 | comment | added | Otto | thanks for the info. I was thinking maybe we can do outer measure in the definition of $\bar{d}_\mathcal{F}$, then it will make sense for all subsets. But maybe this will yield too many paradoxical examples. | |
May 6, 2020 at 13:08 | history | edited | Gabe Conant | CC BY-SA 4.0 |
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May 6, 2020 at 13:04 | comment | added | Gabe Conant | Yes, probably. I have to admit that my familiarity with this largely focused in the discrete case (I'll add this for safety). | |
May 6, 2020 at 13:03 | history | edited | Gabe Conant | CC BY-SA 4.0 |
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May 6, 2020 at 13:01 | comment | added | Otto | so $\bar{d}(A)$ is only defined for $\eta$-measurable $A$'s? | |
May 6, 2020 at 12:34 | history | answered | Gabe Conant | CC BY-SA 4.0 |