Where are the vertices of the universal cover of the genus 2 torus octagon?

The universal cover of the genus 2 torus is hyperbolic plane and the fundamental domain is a octagon. Here is a picture, which I took from here. Is there a closed form for the points of set of the vertices of these octagons? I am thinking of the disk sitting the in euclidean plane, centered at the origin with radius 1. So the point $(0,0)$ we have, where are all the others? If this is not known, maybe there is a way that won't be too arduous that someone knows.

• What do you mean by "closed form"? $\pi_1(M_2)$ acts on $\Bbb H^2$ so that the quotient is $M_2$. The collection of all the vertices of the octagonal triangulation above is then merely the orbit of the origin under the action of $\pi_1(M_2)$. Does this answer your question? – Balarka Sen Dec 6 '15 at 22:07
• @BalarkaSen I will make an analog with a much simpler example. If we were doing this is a genus 1 torus, we would have $\mathbb{Z}^2$ acting on the origin, and the orbit is just $\{(x,y)\vert x,y\in \mathbb{Z}^2\}$. That is much simpler, but is there a nice presentation of the set for my question? – N. Owad Dec 7 '15 at 18:33
• Write down the complex matrices $M_a, M_b, M_c, M_d$ representing the four Mobius transformations that generate the action of $\pi_1(M_2)$ on $\mathbb{H}^2$ (using the Poincare disc model as your picture indicates, each matrix representing a fractional linear transformation of the complex plane that preserves the Poincare disc).
• Enumerate a set of words $w$ in the generators $a,b,c,d$ representing each element of $\pi_1(M_2)$ (using an automatic structure, for example).
• For each word $w = w_1 w_2 … w_L$ in the enumeration (where each of the $w_i$ equals one of $a,b,c,d$ or their inverses), let $M_w$ be the corresponding product of the matrices $M_a,M_b,M_c,M_d$ or their inverses.
• Compute $M_w \cdot \pmatrix{0 \\ 1} = \pmatrix{x_W \\ y_W}$ and take $z_w$ to be the quotient $x_w / y_w$.