Timeline for Do once-punctured torus bundles have integral traces?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2021 at 1:03 | vote | accept | wandersam | ||
| Oct 1, 2021 at 8:44 | comment | added | wandersam | Thank you! Yes, on the Bruhat-Tits tree of a finite extension of a $\mathbb{Q}_p$. Do you know where to find the second statement? (It seems to be rather delicate because in the construction of the surface: nontrivial action on tree -> nontrivial splitting over the quotient graph -> nonempty incompressible surface, the quotient graph already is no longer a tree.) | |
| Sep 30, 2021 at 16:32 | comment | added | Neil Hoffman | I think the key point is that a non-integral trace corresponds to an action on a R-tree. This action is associated to an orientable surface. | |
| Sep 30, 2021 at 16:30 | comment | added | Neil Hoffman | Non-integral traces would produce a closed, incompressible, orientable surface, so non-orientable surfaces might be possible, but they are a slightly different beast. In fact, any Z/2Z homology would allow for a one-sided incompressible surface where the complement of the surface is a compression body. (see Rubinstein's work projecteuclid.org/journals/pacific-journal-of-mathematics/…). | |
| Sep 28, 2021 at 15:10 | comment | added | wandersam | I have a new objection: the article you cite classifies only orientable closed incompressible surfaces, however according to J. H. Przytycki: "Nonorientable, incompressible surfaces in punctured-torus bundles over S1" (arxiv.org/pdf/1812.11228.pdf), Thm. 2.13 (a)(ii) and example 3.1, nonorientable closed incompressible surfaces can exist and are not ruled out as long as the Stallings-Epstein-Waldhausen construction (i. e. [1] Thm. 5.2.2) does not make claims as to the orientability of the incompressible surface. Do you have an idea how we can fix this? | |
| Sep 16, 2021 at 15:46 | vote | accept | wandersam | ||
| Sep 28, 2021 at 15:03 | |||||
| Sep 16, 2021 at 15:46 | comment | added | wandersam | Upon rereading the statement of Bass's Theorem in [3], the fiber is ruled out due to the condition for unipotent elements that the HNN or amalgamated free product decomposition needs to satisfy. I now trust this proof, thank you! | |
| Sep 13, 2021 at 8:57 | comment | added | wandersam | [1] Thm. 5.2.2 derives the existence of a embedded incompressible surface from the fact that the fundamental group of the manifold is a nontrivial HNN extension. This holds for once-punctured torus bundles by definition, but will only give back the fiber i.e. a non-closed incompressible embedded surface. To my mind, it is just by considering the manifold as a compact manifold with boundary that closedness of the embedded incompressible surface can be reached (which would be a one-holed torus with circle boundary in this case). | |
| Sep 9, 2021 at 13:59 | comment | added | Neil Hoffman | In a once-punctured torus bundle, the fiber is a once-punctured torus (think removing just a point), which is not a closed surface. The convention here is that a once-punctured torus bundle is the complement of a knot in torus bundle, it has a torus cusp. This is consistent with considering a once-punctured torus bundle as a quotient of H^3 as in [1, Thm. 5.2.2]. | |
| Sep 8, 2021 at 23:34 | comment | added | wandersam | If we follow @lee-mosher there, then the fiber would have to be counted as a closed embedded incompressible surface. It seems to me like switching between the hyperbolic manifold and the manifold with its cusp chopped off makes the fiber surface disappear from the list, although it is precisely this fiber bundle structure that makes its fundamental group a nontrivial HNN-extension and hence the assumption of [1] Theorem 1.5.2 true (which in turn crucially underlies [1] Thm. 5.2.2). | |
| Sep 8, 2021 at 23:34 | comment | added | wandersam | Thank you. Yes, this is the argumentation I had tried in the question linked above. But I am still unsure about the "closed surface" part of the argument. See my question here: math.stackexchange.com/q/4243973/478924. | |
| Sep 8, 2021 at 17:03 | history | answered | Neil Hoffman | CC BY-SA 4.0 |