Number of Fuchsian groups with same trace field Let $\Gamma,\Sigma\subset \mathrm{SL}_2({\mathbb R})$ be cocompact arithmetic subgroups. They are called commensurable in the wider sense, if there exists
$g\in \mathrm{SL}_2({\mathbb R})$, such that the intersection of $\Gamma$ and  $g\Sigma g^{-1}$ has finite index in both.
The trace field of $\Gamma$, denoted ${\mathbb Q}(\mathrm{tr}\,\Gamma)$ is the field extension of $\mathbb Q$ generated by all traces of elements of $\Gamma$.
Next let $\Gamma^{(2)}$ be the subgroup of $\Gamma$ generated by all squares $\gamma^2$ with $\gamma\in\Gamma$. The invariant trace field is defined as  $I(\Gamma)={\mathbb Q}(\mathrm{tr}\,\Gamma^{(2)})$.
The (invariant) trace field is a number field and commensurable groups have the same invariant trace field.
My question is this:
For a given number field $K$, is it true that there is only a finite number of commensurability classes $[\Gamma]$ with $K=I(\Gamma)$?
 A: No, this follows from a result of Bogwang Jeon. He showed that given a number field $K$ and quaternion algebra $A$ over $K$ with $A\otimes_K \mathbb{R} \cong M_2(\mathbb{R})$, one can find a fuchsian surface of genus $g$ having $K$ as its invariant trace field and $A$ as the invariant quaternion algebra. Now one observes that there are infinitely many isomorphism classes of quaternion algebras over $K$ which split over $\mathbb{R}$. See e.g. Theorem 7.3.6 of MacLachlan-Reid,  which states that two quaternion algebras over $K$ are isomorphic iff they have the same ramification set of places, and that any admissible ramification set is realized by a quaternion algebra. Hence there are infinitely many quaternion algebras from the infinitude of prime ideals in the ring of integers of a number field (corresponding to the non-Archimedean places).
Since the invariant quaternion algebra and trace field are commensurability invariant (Theorem 3.3.4 and Corollary 3.3.5 of MacLachlan-Reid) , this implies the infinitude of incommensurable fuchsian groups with a given invariant trace field.
