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I am interested in when the trace field of a knot complement has the form $F(\sqrt{-d})$ for $F\subset\mathbb{R}$ and $d\in F^+$ (squarefree). Does this occur for infinitely many choices of pairs $(F,d)$?

For knot complements, the trace field and invariant trace field are the same, and the trace field is equal to the field generated by the shape parameters of an ideal triangulation. So phrased more topologically, this asks if there are infinitely many knot complements, no two of which share a finite-sheeted cover, where for every shape parameter $s\in\mathbb{C}\setminus\mathbb{R}$, we have $s^2\in\mathbb{R}$.

Update: As @NeilHoffman has pointed out below, the answer to this is not known. So let me ask a weaker question. Are there infinitely many different hyperbolic knot complements with trace fields satisfying the conditions, where we don't mind if only finitely many of the $F$ are realized? This may or may not be any easier, for instance I'm not sure if it's known even if there can be an infinite family sharing the same field.

Update 2: If we relax the original condition so that we allow the manifolds to be link complements rather than just knot complements, the answer is yes. In the following paper by Chesebro and Deblois, that is done with $F=\mathbb{Q}(\sqrt{2})$ and $d=1$: http://www.math.umt.edu/chesebro/AIMCLC.pdf

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  • $\begingroup$ I am better versed in general hyperbolic $3$-manifolds than in knot complements. So, if this question is not interesting from a knot theory perspective, I would be interested to know why. $\endgroup$
    – j0equ1nn
    Commented Mar 13, 2016 at 23:01

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I think this question is open and even in this narrowly framed context the question still has a number of interesting seemingly weaker questions.

First, as a reference Long and Reid addressed a similar question in section 6 of this paper:

D. D. Long and A. W. Reid, Fields of definition of canonical curves, Interactions between hyperbolic geometry, quantum topology and number theory, Contemporary Math. 541, 247–257, A.M.S. Publications, (2011). (pdf)

As part of there construction of finding ($S^3$) complements with invariant trace fields having class number bigger than 1, they find knot complements that having invariant trace fields which are a quadratic extension of $\mathbb{Q}(\cos(\pi/p))$ ($=\ell_p$ defined above Theorem 1.1 in their paper). Having said that, I don't think it gives a clean, complete answer to your question even for $F=\ell_p$.

Finally, the original question restricted to the case that $F=\mathbb{Q}$ is open. Although A. W. Reid showed the figure eight knot complement is the only knot complement with integral traces and invariant trace field of the form $\mathbb{Q}\sqrt{-d}$ for $d$ a square-free positive integer, the question is wide open for knot complements (in $S^3$) when the integral trace condition is relaxed.

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  • $\begingroup$ I guess that's a satisfying answer because I now feel much less stupid for not knowing how to resolve this. $\endgroup$
    – j0equ1nn
    Commented Mar 14, 2016 at 1:18
  • $\begingroup$ What is the convention on if a posted question is open? Should I accept an answer advising me that the problem is open, or should I interpret that to mean that it is impossible for there to be an acceptable answer? $\endgroup$
    – j0equ1nn
    Commented Mar 14, 2016 at 5:41
  • $\begingroup$ Well it's been a few days, I think this is as satisfying an answer as one could hope for. And secretly I have what I really needed for my current interests. $\endgroup$
    – j0equ1nn
    Commented Mar 19, 2016 at 23:53

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