Timeline for Given a quasi-convex subgroup $H$ of hyperbolic $G$, can we decide if two elements $x,y \in G$ lie in the same double coset of $H$?
Current License: CC BY-SA 4.0
11 events
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| Mar 11, 2022 at 18:48 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| Mar 11, 2022 at 18:41 | comment | added | Benjamin Steinberg | You can do it the same way as for a free group. See my answer based on Derek's comment to mathoverflow.net/questions/391055/… | |
| Mar 11, 2022 at 11:11 | comment | added | YCor | A remark: define $f_H(n)$ as the the sup over all $x,y$ in the $n$-ball and in the same $H$-doublecoset of the minimal $k$ such that there exist $h,h'$ in the $k$-ball with $y=hxh'$. Then deciding whether 2 elements are in the same $H$-doublecoset is the same as asking whether $f_H$ is a computable map, or also as whether it is bounded above by a computable map. Then for $H$ quasi-convex in $G$ hyperbolic, one certainly expects $f_H$ not only to be bounded above by a computable function, but to actually grow linearly (= be bounded above by a polynomial of degree 1). | |
| Mar 11, 2022 at 10:09 | comment | added | Derek Holt | Yes that's OK but I won't be able to respond until next week. | |
| Mar 11, 2022 at 9:36 | comment | added | jpmacmanus | Thanks @DerekHolt - would it be okay if I contacted you via email to discuss this? | |
| Mar 11, 2022 at 7:58 | comment | added | Derek Holt | I have more or less convinced myself that the argument outlined in my comments works, and If necessary I can expand it to a more detailed answer, but that will have to wait until Monday - it really needs a picture! | |
| Mar 10, 2022 at 22:32 | comment | added | HJRW | In free groups, this problem is solved by computing the fibre product of core graphs. Kharlampovich, Myasnikov and Weil have a paper generalising core graphs to the context of quasi convex subgroups of hyperbolic groups, so it’s conceivable that their techniques can be used to solve this problem. | |
| Mar 10, 2022 at 20:00 | comment | added | Derek Holt | In my previous comment, for the cutting out part to work, we need to be using words that are labelled by generators of $H$. While these words are not generally geodesic words in the Cayley graph, the quasiconvexity of $H$ means that they can be chosen to lie uniformly close to geodesic words, so the argument should still work. | |
| Mar 10, 2022 at 19:33 | comment | added | Derek Holt | If you look at a geodesic quadrilateral with paths labelled by geodesic words for $x$, $y$, $h$ and $h'$ then, if the words for $h$ and $h'$ are much longer than those for $x$ and $y$, won't you get a thin bit in the middle of the $h$ and $h'$ words, and hence a repeating short connecting path, which would enable you to cut of out bits of these words and shorten them? | |
| Mar 10, 2022 at 17:30 | history | edited | YCor |
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| Mar 10, 2022 at 16:58 | history | asked | jpmacmanus | CC BY-SA 4.0 |