Timeline for Membership to double cosets in free groups
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2021 at 18:49 | comment | added | Benjamin Steinberg | @HJWR I also added Stallings users the coset version of the core in his proof of M Hall's Theorem | |
| Apr 24, 2021 at 18:48 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Apr 24, 2021 at 18:36 | comment | added | Benjamin Steinberg | @HJRW, I add a comment that the product of inverse automata is the fiber product of the corresponding immersions and emphasized inverse automata are immersions over bouquets. | |
| Apr 24, 2021 at 18:35 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Apr 24, 2021 at 18:32 | comment | added | HJRW | Ah yes, I found it deep in the comment thread! I think it’s a shame that it isn’t mentioned prominently though, since the topological point of view in terms of graphs and fibre products is the primary modern way of thinking about subgroups of free groups. Perhaps I’ll add another answer to that effect... | |
| Apr 24, 2021 at 17:22 | vote | accept | Ashot Minasyan | ||
| Apr 24, 2021 at 17:03 | comment | added | Benjamin Steinberg | @HJWR, I also mentioned the fiber product viewpoint in my comments on Derek's answer | |
| Apr 24, 2021 at 16:46 | comment | added | HJRW | Note that, topologically, this is precisely the fibre product that Stallings describes in his famous paper. So the algorithm is easily stated in terms of graphs: fold to construct core graphs for $H$ and $K$ (big enough to contain the elements of interest); construct the fibre product explicitly; now check whether or not the points corresponding to the elements you’re interested in are in the same component of the graph. | |
| Apr 24, 2021 at 16:06 | comment | added | Ashot Minasyan | I think that both of your answers are useful: this answer is good if you need to check whether just one element belongs to the double coset, so you build an automaton specifically for that element. Your other answer allows to build a single automaton, which works for all inputs! | |
| Apr 24, 2021 at 15:52 | comment | added | Benjamin Steinberg | you are welcome. But I think it is what Derek was saying | |
| Apr 24, 2021 at 15:40 | comment | added | Ashot Minasyan | Thank you, this is great! | |
| Apr 24, 2021 at 15:39 | comment | added | Benjamin Steinberg | My first answer is overkill because it solves a much more general problem and I didn’t bother to think if your problem was inherently easier. | |
| Apr 24, 2021 at 15:36 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Apr 24, 2021 at 15:34 | comment | added | Benjamin Steinberg | Yes, I got it backwards | |
| Apr 24, 2021 at 15:22 | history | answered | Benjamin Steinberg | CC BY-SA 4.0 |