Timeline for Are finitely generated subgroups of $\text{GL}_n(\mathbb{Q})$ virtually special?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18 at 12:23 | comment | added | Mark Hagen | Slightly more detail though edit window expired: in any CAT(0) cube complex, up to cubically subdividing, non-elliptic isometries have combinatorial geodesic axes. It follows easily that in an f.g. group acting on a CAT(0) cube complex (no restriction on dimension), distorted cyclic subgroups act elliptically. (One subtlety is that when the dimension is infinite, one can't make this argument using CAT(0) translation lengths, since the CAT(0) and combinatorial metrics are no longer QI.) | |
| Jul 18 at 12:12 | comment | added | Mark Hagen | @HJRW, the proof by Haglund is Theorem 1.3 here: arxiv.org/pdf/0705.3386 | |
| Jul 14 at 7:13 | comment | added | YCor | @HJRW This is originally due to Haglund with a geometric argument, and I proved it by elementary means (in terms of commensurating actions, so eventually it's a fact about abstract actions of $\mathbf{Z}$ on sets) in my FW paper (Corollary 6.A.3). | |
| Jul 14 at 7:09 | comment | added | HJRW | @YCor: OK, so what's the argument for that!? (This is what I was asking for originally...) | |
| Jul 14 at 7:09 | comment | added | YCor | @HJRW I don't think it's a problem, because these groups just don't act properly at all on any CAT(0) cube complex. A group with a distorted $\mathbf{Z}$ cannot (no cocompactness or even finite-dimensionality assumption). | |
| Jul 14 at 7:05 | comment | added | HJRW | @YCor: thanks, that clarifies things. However, it calls into question the validity of the accepted answer. Virtually special is a little weaker than virtually compact special. Our discussion shows that translation-length considerations can only be used in the cocompact case. (Actually looking again at the accepted answer, perhaps this is what is intended...) | |
| Jul 14 at 7:02 | comment | added | YCor | @HJRW yes, sorry it's not cocompact (with a finite volume fundamental domain). | |
| Jul 14 at 6:59 | comment | added | HJRW | @YCor: I am puzzled, because a proper cocompact action on any CAT(0) space (not just a cube complex) should rule out having a solvable Baumslag--Solitar subgroup: cf. Corollary II.7.6 of Bridson--Haefliger. What am I missing? | |
| Jul 13 at 22:26 | comment | added | YCor | @HJRW no, it acts properly cocompactly on the product of a tree (Bruhat-Tits of $\mathrm{SL}_2(\mathbf{Q}_p)$ and the hyperbolic plane (symmetric space of $\mathrm{SL}_2(\mathbf{R})$). It doesn't have Property PW because it has a distorted infinite cyclic subgroup (inside a copy of a solvable Baumslag-Solitar group). | |
| Jul 13 at 17:04 | comment | added | HJRW | @YCor: what’s the argument that $SL_2(\mathbb{Z}[1/p])$ can’t act properly on a cube complex? I thought I recalled that it acts on a product of trees with finite stabilisers, but I don’t have my copy of Serre to hand to check. | |
| Jul 12 at 6:19 | comment | added | YCor | Yes, there are, e.g., finite virtual cohomological dimension. But eventually this is not what you asked. | |
| Jul 12 at 2:48 | comment | added | user2357 | @YCor Ah, I see. This isn't an area I'm terribly familiar with, but I was basically wondering if there are much stronger properties satisfied by f.g. subgroups of $\text{GL}_n(\mathbb{Q})$ vs those of $\text{GL}_n(\mathbb{C})$. | |
| Jul 12 at 2:21 | vote | accept | user2357 | ||
| Jul 11 at 23:40 | comment | added | YCor | Virtually special implies the much weaker PW property (acts properly on a CAT(0) cube complex). Plenty of f.g. Q-linear groups fail to have Property PW, e.g., those with Kazhdan's Property T (e.g. $\mathrm{SL}_3(\mathbf{Z})$), those with a distorted copy of $\mathbf{Z}$ such as $\mathrm{SL}_2(\mathbf{Z}[1/p])$ or (as already mentioned) solvable Baumslag-Solitar groups, non-abelien nilpotent torsion-free f.g. groups, etc. | |
| Jul 11 at 22:41 | answer | added | Matt Zaremsky | timeline score: 8 | |
| Jul 11 at 22:11 | comment | added | Matt Zaremsky | That's right for things like BS(2,3), but the BS(1,n) actually are Q-linear. I can write something more helpful soon, when I'm not just on my phone. | |
| Jul 11 at 22:03 | comment | added | user2357 | @MattZaremsky aren't these not $\mathbb{Q}$-linear? They're not generally residually finite are they? | |
| Jul 11 at 21:53 | comment | added | Matt Zaremsky | baumslag-solitar groups, right? | |
| Jul 11 at 21:05 | history | asked | user2357 | CC BY-SA 4.0 |