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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.

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

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 $s^2\in\mathbb{R}$.

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.

This question would provide half of an answer to this question (and @IanAgol has given me a good boost toward the other half): How many non-commensurable non-arithmetic manifolds have a quaternion algebra like this?

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). I want to know if this occurs infinitely often. It may be sufficient to just consider when $F=\mathbb{Q}$, but in this problem we are free to choose any real number field for $F$.

For knot complements, the trace field and invariant trace field are the same, so phrased more topologically, this asks if there are infinitely many knot complements satisfying my trace condition, no two of which share a finite-sheeted cover. Since all but one of the knot complements are non-arithmetic, it would give an answer to the non-compact case of the linked question.

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 $s^2\in\mathbb{R}$.