Timeline for Gluing hexagons to get a locally CAT(0) space
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Mar 25, 2019 at 21:46 | comment | added | HJRW | @DylanThurston -- the "converse" I was thinking of is "non-hyperbolic implies 3-manifold" for 2-generator, 1-relator groups. This is false because the $(2,3)$--Baumslag--Solitar group is non-hyperbolic but also not a 3-manifold group. But there's no non-positive curvature there, so perhaps that observation isn't helpful. | |
| Mar 25, 2019 at 14:14 | comment | added | Dylan Thurston | @HJRW, sorry, can you spell out what you mean by "converse" (and $BS(2,3)$)? | |
| Mar 24, 2019 at 12:10 | comment | added | HJRW | @DylanThurston -- it's not a complete coincidence: any 2-generator, 1-relator 3-manifold group $G$ can't be hyperbolic, since $\chi(G)=0$, so the boundary of the 3-manifold would give a $\mathbb{Z}^2$ subgroup. The converse doesn't hold, of course (eg $BS(2,3)$). | |
| Nov 4, 2018 at 16:26 | comment | added | Ian Agol | @DylanThurston: Maybe what I was outlining earlier would show that if all of the hexagon flat planes are complete, then attaching horoballs gives $\mathbb{H}^3$, and hence it's commensurable with the Gieseking example. | |
| Nov 4, 2018 at 14:53 | comment | added | Dylan Thurston | Is it a coincidence that the obstruction to development is the same as the obstruction to thickening to a 3-manifold, or is this an instance of some more general fact? | |
| Nov 3, 2018 at 5:56 | comment | added | Ian Agol | Yeah, I should delete the previous comment, since I hadn't thought about the fact that development might be branched. Your interpretation make it more clear. Anyway, I wouldn't be surprised if this group is in the literature (surely it is known to Dani Wise). | |
| Nov 2, 2018 at 20:22 | comment | added | Dylan Thurston | OK, I thought about it a little more: the obstruction to extending a flat is actually the same obstruction to thickening this complex to a 3-manifold. If you take a hexagon in $\widetilde{C}$ and look at a neighborhood of its boundary (outside of the hexagon), you get a Möbius strip rather than an annulus. On the one hand this is the obstruction to thickening a simple complex to a 3-manifold, and on the other hand it means that when you try to develop a flat you get some inconsistency. | |
| Nov 2, 2018 at 19:23 | comment | added | Dylan Thurston | @IanAgol, how does the obstruction work? Your earlier argument that the flats exist was pretty convincing. I guess I should try it myself. | |
| Nov 1, 2018 at 23:02 | comment | added | Ian Agol | Okay, now I actually tried carrying out the development of flats, and they are obstructed (there are no complete flats). By Bridson's flat plane theorem, this implies that $\pi_1(C)$ is $\delta$-hyperbolic. | |
| Nov 1, 2018 at 22:37 | comment | added | Ian Agol | It's easy to see that any pair of adjacent hexagons in $C$ "develop" to a unique flat (there is a unique hexagon that fits in the corner adjacent to two, since the link of the vertex is $K_4$). This pattern of flats in $C$ must be different, and presumably are stabilized by virtually $\mathbb{Z}^2$ subgroups. So I think that $C$ has "isolated flats", and is hyperbolic relative to these flats. In fact, one ought to be able to attach horoballs to the flats to obtain a $\delta$-hyperbolic space (maybe $CAT(-1)$). But there must be some branching so that it is not isometric to $\mathbb{H}^3$. | |
| Nov 1, 2018 at 22:34 | comment | added | Ian Agol | @DylanThurston: A few more observations about the 4th complex (let's call it $C$). $C$ cannot be commensurable with the Gieseking complex (basically by Mostow rigidity), so the universal cover of $C$ must be different from that of the Gieseking. In the Gieseking case (as you observe, the normalizer of $PGL_2(\mathbb{Z}[\omega])$), there are flats corresponding to horocusps. So the flats in $C$ must have a different pattern, even though the complexes are locally equivalent. | |
| Nov 1, 2018 at 22:17 | history | edited | Ian Agol | CC BY-SA 4.0 |
added 109 characters in body
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| Nov 1, 2018 at 13:34 | comment | added | Dylan Thurston | That free group automorphism is pretty simple, it looks like it should be a standard example. | |
| Nov 1, 2018 at 13:33 | comment | added | Dylan Thurston | Thanks, that's great. We might get another description of the third example from the theory of arithmetic groups: the group will be commensurable with $\mathit{PSL}(2; \mathbb{Z}[\omega])$ where $\omega^3 = 1$. | |
| Nov 1, 2018 at 13:29 | vote | accept | Dylan Thurston | ||
| Nov 1, 2018 at 8:59 | comment | added | HJRW | Oh sorry, right, the relation is a one-sided curve on the boundary of a non-orientable handlebody. | |
| Nov 1, 2018 at 7:28 | comment | added | Ian Agol | @HJRW I don’t think that it thickens to be a 3-manifold, since it would then have to admit an ideal triangulation, and hence is the Gieseking. | |
| Nov 1, 2018 at 7:13 | comment | added | HJRW | In the fourth example, since the link of the vertex is planar, it should also thicken to a 3-manifold. Do you know which one? | |
| Nov 1, 2018 at 5:46 | history | answered | Ian Agol | CC BY-SA 4.0 |