Timeline for Virtually large groups of small rank (related to 3-manifolds)
Current License: CC BY-SA 4.0
16 events
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| May 27, 2021 at 15:16 | comment | added | lemon314 | to clarify - I meant the Euler number of the Seifert manifold; to have $\mathbb{H}^2\times\mathbb{R}$ geometry, the orbifold Euler characteristic must be negative, and the Euler number must be zero - and I meant that $x/2+y/3+z/7$ is not an integer for the admissible $x,y,z$, and so that second condition would fail; | |
| May 26, 2021 at 8:08 | comment | added | HJRW | @lemon314 -- I'm so glad to been of some assistance! I'm slightly confused about your comment about the (2,3,7) orbifold; this is just the smallest hyperbolic orbifold with underlying surface the sphere, and indeed any hyperbolic orbifold will have negative (orbifold) Euler characteristic. But anyway, it doesn't matter. | |
| May 25, 2021 at 15:16 | comment | added | lemon314 | @HJRW, I finally convinced myself that I understand where I went wrong, and edited the question accordingly. Though I think that (2,3,7) is not a good example, as it won't have the Euler characteristic zero - if this denotes the sphere with three singular points of the appropriate degree. Nevertheless, thank you so very much for pointing me in the right direction. | |
| May 25, 2021 at 15:09 | history | edited | lemon314 | CC BY-SA 4.0 |
edited after counterexamples to the premise of the question were given
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| Jan 15, 2021 at 18:06 | answer | added | Moishe Kohan | timeline score: 5 | |
| Jan 15, 2021 at 9:15 | comment | added | HJRW | Could you explain how it follows from Boileau and Zieschang that the rank is at least 3 when the geometry is H^2 x R? If I remember correctly, there are Seifert fibred manifolds with Euler number zero that fibre over a hyperbolic triangle orbifold (2,3,7), say. These would have H^2 x R geometry, and Boileau and Zieschang seem to imply that their rank is 2. | |
| Jan 14, 2021 at 23:55 | comment | added | lemon314 | @HJRW, corrected, thanks. Copied and pasted the condition from a book without thinking. | |
| Jan 14, 2021 at 23:53 | history | edited | lemon314 | CC BY-SA 4.0 |
added 110 characters in body
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| Jan 14, 2021 at 23:26 | comment | added | HJRW | Oh, I see from the comment thread that you do want the manifold to be closed. Perhaps you should update the question? | |
| Jan 14, 2021 at 23:20 | comment | added | HJRW | In the second paragraph you talk about “closed 3-manifolds”, but in the first paragraph you allow $F$ to be free, which corresponds to the case with toroidal boundary. Could you clarify which case you care about? In the closed case, a finite extension is a 3-manifold group iff it’s torsion-tree, so it is just an algebraic condition. I suspect the same is true in the case with boundary, but am less certain. | |
| Jan 14, 2021 at 20:47 | comment | added | lemon314 | ...but by @MoisheKohan's hint (that should follow from the trefoil group being the extension of $SL(2,\mathbb{Z})$?) it seems that being a fundamental group of a closed 3-manifold is indeed a restriction in this case. | |
| Jan 14, 2021 at 20:43 | comment | added | lemon314 | @YCor, Boileau-Zieschang show that Seifert fibered space of prescribed symbol (as per the classification by genus + Euler number + twisting of exceptional fibers) must have fundamental group of a specific rank (2 times genus + number of fibers - 1). Then, using that (by geometrization) a 3-manifold has a fundamental group virtually $\mathbb{Z}\times F$ iff it fits in one of the cases in their paper, we get that any finite index overgroup of $\mathbb{Z}\times F$ that is a 3-manifold group has generating rank $\geq 3$. I wasn't sure if the 3-manifoldness was necessary or if it was algebraic... | |
| Jan 14, 2021 at 19:54 | comment | added | Moishe Kohan | Hint: Consider the exterior of the trefoil knot. Do you know a presentation of its fundamental group? Can you prove that this group is virtually $Z\times F$? | |
| Jan 14, 2021 at 19:54 | comment | added | YCor | Just to be sure: you mean that in the Boileau-Zieschang paper there's a claim that any finite index overgroup of such a $\mathbf{Z}\times F$ group has generating rank $\ge 3$, and you want an explanation of this claim? | |
| Jan 14, 2021 at 19:53 | history | edited | YCor |
edited tags
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| Jan 14, 2021 at 19:14 | history | asked | lemon314 | CC BY-SA 4.0 |