Timeline for Growth of the number of generators in hyperbolic groups
Current License: CC BY-SA 3.0
12 events
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| Sep 24, 2021 at 1:17 | comment | added | David Roberts♦ | @IanAgol here's an updated link to the paper Rank gradient, cost of groups and the rank versus Heegaard genus problem by Abert and Nikolov: arxiv.org/abs/math/0701361 (published as doi.org/10.4171/JEMS/344) (please upvote this so it appears above the fold) | |
| Jan 6, 2015 at 18:29 | comment | added | HJRW | @Pablo, all I meant was that, for a fi subgroup H of a 2-dimensional group G, $b_1(H) \geq 1-|G:H|\chi(G)$, by the multiplicativity of Euler characteristic. Thanks for your e-mail - I'll reply soon! | |
| Jan 6, 2015 at 11:53 | comment | added | Pablo | @HJRW regarding your first comment, if my group is a $2$-dimensional group with negative Euler characteristic, then I will have some formula for the rank of finite index subgroups, like Schreier's index formula in the free group, or at least the limit (and not only liminf) of $(d(H) - 1)/[F : H]$ will exist as $[F : H]$ tends to infinity, right? | |
| Dec 24, 2014 at 9:37 | comment | added | Pablo | @HJRW thanks for that. Indeed, I am interested in the positivity of the rank gradient (or even a sublinear asymptotic lower bound) for finitely generated groups in general and not only for hyperbolic groups. | |
| Dec 24, 2014 at 8:22 | comment | added | HJRW | It's a fact that the rank gradient of an infinite group is bounded below by the first $l_2$ Betti number. See, for instance, arxiv.org/abs/0905.1322 . | |
| Dec 24, 2014 at 6:39 | comment | added | Pablo | @IanAgol do you have an idea how to show that the number of generators grows asymptotically linearly with the index in the case that the first $l_2$ betti number is not zero? | |
| Dec 24, 2014 at 4:42 | comment | added | Ian Agol | I suspect that locally quasi-convex residually finite hyperbolic groups have the property that the rank grows with the index. However, I also think that the only known examples of this type might have non-zero 1st $l_2$-betti number. | |
| Dec 24, 2014 at 4:29 | comment | added | Ian Agol | More generally, if the 1st $l_2$-betti number is non-zero. A related concept is cost/rank gradient. front.math.ucdavis.edu/0701.5361 | |
| Dec 23, 2014 at 22:51 | comment | added | HJRW | On the other hand, some of the arguments of this paper may be applicable: arxiv.org/abs/1212.4192. | |
| Dec 23, 2014 at 22:45 | comment | added | HJRW | The obvious comment is that the free group argument generalises to any 2-dimensional group with negative Euler characteristic. On the other hand, as Sam points out, with zero Euler characteristic this is usually false. | |
| Dec 23, 2014 at 21:16 | comment | added | Sam Nead | The existence of hyperbolic HNN extensions (eg some free-by-cyclic groups) makes this tricky. Does it help to assume that the hyperbolic group has trivial abelianization? | |
| Dec 23, 2014 at 18:34 | history | asked | Pablo | CC BY-SA 3.0 |