Timeline for Finitely presented group containing every $\mathrm{GL}_n(\mathbb{Z})$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2021 at 23:24 | comment | added | Matt Zaremsky | @YCor Indeed, and actually in my original question I almost included, "I'm also curious about the analogous question using $Aut(F_n)$," but decided that was too many sub-questions! | |
| Oct 11, 2021 at 21:24 | comment | added | YCor | Indeed the point seem to be essentially that relators of the different $\mathrm{SL}_n(\mathbf{Z})$ have "the same form", with all generators playing "the same role". We might expect something similar with presentation of (outer) automorphism groups of free groups? Mapping class groups? assuming there is a uniformly bounded number of "types" of generators and relators... | |
| Oct 11, 2021 at 19:28 | comment | added | Matt Zaremsky | Aha! Very nice. I would think if the action has finitely many orbits in $S^n$ for every $n$, and the stabilizer of any finite subset of $S$ is type $F_\infty$ (so like, $F$ acting on dyadics) then the resulting group is type $F_\infty$, but I would also think that this would take quite a lot more work to prove. | |
| Oct 11, 2021 at 19:26 | vote | accept | Matt Zaremsky | ||
| Oct 11, 2021 at 19:23 | comment | added | AGenevois | I was taking a look at Milnor's book and was going to write an answer. But you are faster! It seems reasonable to think that examples of type $F_n$ can be constructed for arbitrarily large $n$, but what about examples of type $F_\infty$? (EDIT: After a thought, finally it should also be doable without too much trouble if it already works for arbitrarily large $n$.) | |
| Oct 11, 2021 at 19:14 | history | answered | YCor | CC BY-SA 4.0 |