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Is there a natural example of a discrete subgroup $\Gamma\leq PSL_2(\mathbf{R})$ such that

(1) $\Gamma$ has finite covolume

(2) $\mathfrak{h}/\Gamma$ is not compact ($\mathfrak{h}$ being the upper half-plane)

(3) $\Gamma$ is not commensurable to a conjugate of $PSL_2(\mathbf{Z})$.

I cannot think of any such example but I don't see any reason why they should not exist. Note that if we further require that the trace of all matrices of $\Gamma$ lie in a some number field $K\subseteq\mathbf{R}$, then I can show that no such example exists.

Is there a natural example of a discrete subgroup $\Gamma\leq PSL_2(\mathbf{R})$ such that

(1) $\Gamma$ has finite covolume

(2) $\mathfrak{h}/\Gamma$ is not compact ($\mathfrak{h}$ being the upper half-plane)

(3) $\Gamma$ is not commensurable to a conjugate of $PSL_2(\mathbf{Z})$.

I cannot think of any such example but I don't see any reason why they should not exist.

Is there a natural example of a discrete subgroup $\Gamma\leq PSL_2(\mathbf{R})$ such that

(1) $\Gamma$ has finite covolume

(2) $\mathfrak{h}/\Gamma$ is not compact ($\mathfrak{h}$ being the upper half-plane)

(3) $\Gamma$ is not conjugate to $PSL_2(\mathbf{Z})$.

I cannot think of any such example but I don't see any reason why they should not exist. Note that if we further require that the trace of all matrices of $\Gamma$ lie in a some number field $K\subseteq\mathbf{R}$, then I can show that no such example exists.

Is there a natural example of a discrete subgroup $\Gamma\leq PSL_2(\mathbf{R})$ such that

(1) $\Gamma$ has finite covolume

(2) $\mathfrak{h}/\Gamma$ is not compact ($\mathfrak{h}$ being the upper half-plane)

(3) $\Gamma$ is not commensurable to a conjugate of $PSL_2(\mathbf{Z})$.

I cannot think of any such example but I don't see any reason why they should not exist. Note that if we further require that the trace of all matrices of $\Gamma$ lie in a some number field $K\subseteq\mathbf{R}$, then I can show that no such example exists.