Timeline for Cohomology for extension problems in symbolic/topological dynamics?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2022 at 7:39 | answer | added | Ville Salo | timeline score: 3 | |
| Jun 5, 2022 at 12:37 | comment | added | Ilkka Törmä | @VilleSalo Yes, it is. And the second note holds more generally. | |
| Jun 5, 2022 at 9:34 | comment | added | Ville Salo | @IlkkaTörmä: To clarify, isn't this is an answer to the technical question for AFT? I.e. it's decidable for them. | |
| Jun 4, 2022 at 19:42 | comment | added | Ilkka Törmä | A couple of possibly helpful notes. First, if $Y$ is of almost finite type, then such a "nice" factor map exists iff the minimal SFT cover is an example, since every factor map to $Y$ factors through the minimal cover (Boyle & Kitchens & Marcus 1985: A note on minimal covers for sofic systems.) Second, it's decidable whether a given factor map $\pi$ is nice, and equivalent to the existence of a section on the periodic points of $Z$ that behaves well in terms of eventually periodic tails (Salo & Törmä 2015: Category theory of symbolic dynamics. Theorem 1) | |
| Jun 4, 2022 at 10:04 | comment | added | Dan Rust | For minimal subshifts, Cech cohomology of the suspension (equivalently, the dimension group of the associated Bratteli-Vershik system) is often a useful invariant that can tell you something about extensions/factors in this way. Unfortunately, Sofic shifts are rarely minimal, which means the Cech cohomology will be infinitely generated, so difficult to handle. The Bowen Franks group is a kind of cohomology group associated with SFTs which I feel might be closer to an invariant relevant for this setting. I'm not sure if there's an analogue that is defined for Sofic shifts. | |
| Jun 4, 2022 at 5:43 | history | edited | Sophie M | CC BY-SA 4.0 |
correcting silly mistake in response to comment
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| Jun 4, 2022 at 5:33 | comment | added | Ville Salo | Slight nitpick, there are no obstructions if $Z=Y$, because then $Y$ is SFT. Take $X=Y$ and $\pi$ identity. | |
| Jun 3, 2022 at 23:39 | history | asked | Sophie M | CC BY-SA 4.0 |