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 $K=I(\Gamma)$?