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Once-punctured torus bundles are a well-studied class of hyperbolic 3-manifolds, but unfortunately I have been unable to find out whether they always have integral traces (in the sense of [1], Def. 5.2.1). Is this already known?

In several dozens of examples that I have calculated with the help of Snappy or the approach in [2], this seems to be the case. Moreover, for bundles with monodromy $LR^n$ this seems to follow from the calculations in [2].

Bass's Theorem ([3]; [1] Thm. 5.2.2) sometimes helps in this situation, but I am not enough of a topologist to tell (and am inclined to say it does not). See my question (and attempt of answer) on MSE: https://math.stackexchange.com/q/4237616/478924

When calculating the trace field with the approach of [2], there is a finite set of algebraic solutions one of which belongs to the geometric structure of the manifold. It might be worthwhile to mention that in all my examples, also the other, non-geometric solutions define a (non-discrete) representation with integral traces.

Related question discussing closedness in Bass's Theorem: https://math.stackexchange.com/q/4243973/478924

[1] MacLachlan, Reid: The Arithmetic of Hyperbolic 3-Manifolds.

[2] Rehmann, Vinberg: On a Phenomenon Discovered By Heinz Helling. Transformation Groups 22, 259–265 (2017).

[3] Bass, Hyman: Chapter VI Finitely Generated Subgroups of Gl2. Pure and Applied Mathematics Vol. 112, 1984, pp 127-136.

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Floyd and Hatcher give a classification of essential surfaces in once-punctured torus bundles. In particular, there are no closed incompressible surfaces (other than the boundary torus), if there were they would persist in high parameter Dehn fillings, contradicting Corollary 1.2:

Floyd, W.; Hatcher, Allen E., Incompressible surfaces in punctured-torus bundles, Topology Appl. 13, 263-282 (1982). ZBL0493.57004.

Then the version of Bass's theorem you cite (see Theorem 5.2.2 of the reference below aka [1] above) which shows that for hyperbolic manifolds with non-integral traces contain a closed incompressible surface. Together, these results confirm your computations more broadly. Hyperbolic once-punctured torus bundles have integral traces.

Maclachlan, Colin; Reid, Alan W., The arithmetic of hyperbolic 3-manifolds, Graduate Texts in Mathematics. 219. New York, NY: Springer. xiii, 463 p. (2003). ZBL1025.57001.

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  • $\begingroup$ 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. $\endgroup$
    – wandersam
    Commented Sep 8, 2021 at 23:34
  • $\begingroup$ 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). $\endgroup$
    – wandersam
    Commented Sep 8, 2021 at 23:34
  • $\begingroup$ 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]. $\endgroup$
    – Neil Hoffman
    Commented Sep 9, 2021 at 13:59
  • $\begingroup$ [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). $\endgroup$
    – wandersam
    Commented Sep 13, 2021 at 8:57
  • $\begingroup$ 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! $\endgroup$
    – wandersam
    Commented Sep 16, 2021 at 15:46

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