# 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 $$K=I(\Gamma)$$?

• No, in fact, one may construct infinitely many with a given trace field. This is certainly true in the arithmetic case when K is totally real - see Maclachlan & Reid for the classification in terms of quaternion algebras. Otherwise, one can construct examples by hand with the same trace field and of fixed genus. Then by Margulis’ arithmeticity theorem the commensurator is discrete, so these give infinitely many commensurability classes. – Ian Agol Jul 28 at 13:06

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).