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I am interested in realizing commensurability classes of hyperbolic $3$-manifolds whose quaternion algebra (note: not invariant quaternion algebra) is isomorphic to one of the form $\Big(\frac{a,b}{F(\sqrt{-d})}\Big)$, where $F\subset\mathbb{R}$ and $a,b,d\in F^+$. (The reason why is rather lengthy but if you really want to know you can check out the slides from my last talk: http://joequinn.ws.gc.cuny.edu/files/2016/03/2016-03-Iowa-University.pdf)

There are infinitely many non-commensurable examples of such manifolds. Moreover this can be achieved using only non-compact arithmetic manifolds, or using only compact arithmetic manifolds. To see this for non-compact arithmetic manifolds, just consider torsion-free finite index subgroups of Bianchi groups. To see this for compact arithmetic manifolds, let $B=\Big(\frac{a,b}{\mathbb{Q}(\sqrt{-d})}\Big)$ with $a,b,d\in\mathbb{Q}^+$ chosen so that $B$ is ramified, then take an order $\mathcal{O}\subset B$, and take a finite-index torsion free subgroup of $\mathrm{P}\mathcal{O}^1$.

I am interested in non-arithmetic examples. Are there infinitely many non-commensurable examples satisfying my condition among non-arithmetic non-compact manifolds? How about among non-arithmetic compact manifolds?

My feeling is that the answer is yes, because $F$ can be literally be any real number field (it's okay if it has complex embeddings, or if the algebra is split at some of its places), leaving many possibilities. In the non-compact case, it would be sufficient to find an infinite family of knot groups satisfying the trace field condition (for knots and links, the quaternion algebra and the invariant quaternion algebra are the same). The compact case may be more difficult because even if the quaternion algebras are different, the invariant quaternion algebras might be the same. On the other hand, in the compact case one has infinitely many different algebras over any one field.

Many people expect that for every quaternion algebra, there is a manifold having that as its invariant, but that is not known and of course is a harder problem than this.

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  • $\begingroup$ Note that since the field includes $\sqrt{-d}$, the condition is unaffected if we allow $a,b$ to be negative as well. $\endgroup$ – j0equ1nn Mar 12 '16 at 22:53
  • $\begingroup$ Some alternative motivation for this question, from Walter Neumann: arxiv.org/pdf/1108.0062.pdf $\endgroup$ – j0equ1nn Mar 13 '16 at 1:59
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    $\begingroup$ I think one could use the inbreeding technique to prove this for the arithmetic ones which contain totally geodesic surfaces. arxiv.org/abs/math/0612290 That is, starting with an arithmetic group containing a totally geodesic surface, one should be able to manufacture infinitely many incommensurable examples with the same quaternion algebra. $\endgroup$ – Ian Agol Mar 13 '16 at 14:20
  • $\begingroup$ @IanAgol A compact arithmetic one with trace field $\mathbb{Q}(\sqrt{-d})$ will satisfy this Hilbert symbol condition. It will also have infinitely many noncommensurable Fuchsian subgroups, giving diverse geodesic subsurfaces. So I think you're saying to look at the result of gluing pairs of these subsurfaces. Would it be reasonable to conclude that choosing those pairs appropriately gives non-commensurable and non-arithmetic examples? $\endgroup$ – j0equ1nn Mar 13 '16 at 19:51
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    $\begingroup$ yes, that's right. The doubling procedure ("inbreeding") described in the above paper ought to preserve the quaternion algebra, since the reflection involution lies in the commensurator. One may make a sequence of non-congruence manifolds this way with arbitrarily small injectivity radius, and hence in infinitely many commensurability classes (Margulis' theorem implies that a non-arithmetic group has discrete commensurator, hence a lower bound on the systole). $\endgroup$ – Ian Agol Mar 13 '16 at 21:12
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The comments from @IanAgol above lead to an affirmative answer in the compact case. This paper gives an affirmative answer in the non-compact case: http://www.math.umt.edu/chesebro/AIMCLC.pdf

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