I have the following questions: Are all Fuchsian groups of signature $(0;2,2,2,\infty)$ arithmetic? What is known about the trace fields of these groups?

Best, K.


2 Answers 2


The answer is NO. Any four times punctured sphere admits two involutions, the quotient by which is an orbifold of the signature you describe. Similarly, you can take a punctured torus, and the quotient by the elliptic involution is one of your surfaces. Conversely, you can cover one of them by a torus or a four-times-punctured sphere. Since very few tori are arithmetic, same is true of your class of groups.


Here is another way to see the answer the first question is no and the answer to the second question is tricky because the deformation space of Fuchsian groups of signature $(0;2,2,2,\infty)$ is one two dimensional.

The (decorated) Teichmüller space of an orientable 2-orbifold $O$ with $\chi(O)<0$ is dimension $-3\chi(|O|)+2k$ where $|O|$ is the underlying space (in this case the disk) and $k$ is the number of cone points (see Thurston's notes Chapter 13, Corollary 13.3.7).

As mentioned above a careful reading of the argument before this corollary (motivated by the sanity check in the cases of a cusped 2-manifold like the thrice punctured sphere and compact 2-orbifold like a sphere with three cone points) shows, that is gives the dimension of decorated Teichmüller space is case of cusped orbifolds. Here decorated Teichmüller space is the deformation space of hyperbolic structures together with parameterized horoball packings. Here there is one cusp so the deformation space of horoball packings is 1 dimensional, meaning the the (undecorated) Teichmüller space should have dimension -3(1)+6-1=2 (more generally $-3\chi(|O|)+2k-c$ where $c$ is the number of cusps).

Since the group is simple enough we can also do this computation explicitly in this case to get generators for the trace field. We can think of this group and being generated by three rotations $\{R1,R2,R3\}$ of order 2 such each one fixing a point on the edge of ideal triangle. If we conjugate the triangle to have vertices at $\{0,1,\infty\}$ and enforce the condition that $R1.R2.R3$ is parabolic (this condition is why we subtract out $c$ above), we then have

$$R1 = \pmatrix{0& -r1\\ \frac{1}{r1} & 0}$$ which fixes $r1\cdot i$.

$$R2 = \pmatrix{\frac{1}{r2} & -\frac{1}{r2} - r2\\ \frac{1}{r2} & -\frac{1}{r2}}$$ which fixes $1+r2\cdot i$.

$$R3 = \pmatrix{\frac{r1}{r2} & -\frac{r1}{r2}\\ \frac{r1}{r2} + \frac{r2}{r1} & -\frac{r1}{r2}}$$ which fixes $\frac{r1}{(r1 + r2\cdot i)}$.

Now we get $$P = R1.R2.R3 =\pmatrix{1 & \frac{1 + r1^2 + r2^2}{r1^2}\\ 0 & 1}.$$

Also, we have that $\frac{tr(P.R3)}{tr(P.R1)}=\frac{1}{r2}$ and $\frac{tr(P.R3)}{tr(P.R1)}=\frac{1}{r1}\cdot\frac{1+r2^2}{r2}$

Thus, we have that the trace field generators are $r1$ and $r2$.


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