Timeline for Groups with trivial rational homology and their finite index subgroups
Current License: CC BY-SA 3.0
10 events
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| Oct 23, 2015 at 9:41 | comment | added | Todd Leason | @Qiaochu Yuan: Yes, having no finite-index subgroup is essential for making the example with Thompson's T work because it forces the group to be a normal subgroup as observed in your 1st comment ahead. In general, if $(H:G)<\infty$, then we still know that the image of the restriction $res^H_G$ in rational cohomology is the subring of invariants of $H^\ast(G;\mathbb{Q})$. But, if $G$ isn't normal, these depend on the cohomology of the subgroups $G \cap hGh^{-1}$ which is usually hard to get at. | |
| Oct 22, 2015 at 18:22 | comment | added | Qiaochu Yuan | Completeness is unnecessary anyway. As Todd Leason's answer suggests, if I'm reading it correctly, you can replace this condition with the condition that the automorphism group of the rational cohomology has a nontrivial fixed point. (Still need no finite-index subgroups for this approach though.) | |
| Oct 22, 2015 at 16:10 | comment | added | Qiaochu Yuan | It was just the first example of an infinite complete group I could find. | |
| Oct 22, 2015 at 9:17 | comment | added | HJRW | Yes, it's rather odd to single out $\mathrm{Aut}(F_2)$ just because it's complete. I don't know if there's a proof in the literature of this exact statement, but completeness is certainly a 'generic' property of groups (in Gromov's density model, say); it's very common indeed. | |
| Oct 21, 2015 at 17:10 | comment | added | Jim Conant | Just a remark, $\mathrm{Aut}(F_2)$ has no rational homology. The first time $\mathrm{Aut}(F_n)$ has rational homology is $n=4$, when $H_4(\mathrm{Aut}(F_4);\mathbb Q)=\mathbb Q$. | |
| Oct 20, 2015 at 11:03 | comment | added | HJRW | There seems to be a 1996 paper of Brin showing that Out(T) has order 2. So it's not quite what's needed, but very close; I wonder if your argument can be adapted slightly to handle this, using Brin's description of Out(T)? | |
| Oct 20, 2015 at 10:58 | comment | added | HJRW | It seems that V is acyclic, but that T (not U, apologies) has non-trivial rational cohomology. The cohomology of T is a paper of Ghys--Sergiescu and for V it's a recent arXiv preprint of Szymik. | |
| Oct 20, 2015 at 10:37 | comment | added | HJRW | This is a good strategy. There's every reason to think that such groups should exist. (Groups with no finite-index subgroups are plentiful, and the other properties are sufficiently generic that they should be possible to ensure.) Finitely presented candidates include Wise, Bhattacharjee and Burger--Mozes' examples of groups with no proper finite quotients, and Thompson's infinite simple groups U and V. I think all of these have trivial centre, but the questions of computing the cohomology and outer automorphism groups might require some work. | |
| Oct 20, 2015 at 7:54 | comment | added | Qiaochu Yuan | Oh, hmm. If you additionally require that $G$ itself has no finite index subgroups, then it must be normal in any $H$ in which it's finite index. So if anyone can find $G$ with this additional property then that's a counterexample. Maybe the automorphism group of an infinite set? | |
| Oct 20, 2015 at 7:05 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |