I am finding the explicit description of genus $2$ surface as the upper half plane modulo group of Deck Transformation. I did't find it anywhere. Also, I found a similar question here. But, there is no answer. My question is same as the question given in the link. Please help me to find the answer explicitly.
Thanking in advanced.
Looking at the link you give, it seems that you want an explicit representation of $\pi_1 = \pi_1(S)$, the fundamental group of the genus two surface, into $\mathrm{PSL}(2, \mathbb{C})$ so that the upper half plane, modulo the induced action of $\pi_1$, is a "hyperbolic structure" on the topological surface $S$.
That is a long question (so perhaps I've guessed incorrectly). The answer is somewhat involved, and requires some hyperbolic trigonometry, which is why nobody is about to just write down the answer for you. I have done this at least once - but it is buried deep in a program that I wrote ages ago.
So here is a sketch -- you will have to work out the details for yourself.
Step one: Draw a (cartoon of a) regular octagon $R$ in the upper half plane model of the hyperbolic plane. We want the one that has angle $2\pi / 8$ at each of its vertices. Do this so that the centre is at $i$, and so that two opposite sides of $R$ are perpendicular to the imaginary axis. Let $a$ and $b$ be the midpoints of the two sides crossing the imaginary axis.
Step two: Chop $R$ into sixteen triangles. Each triangle meets the centre (the point $i$), a vertex, and an edge midpoint. All of these triangles have angles $(\pi/2, \pi/8, \pi/8)$ at their vertices; thus the triangles are all congruent (and isosceles!)
Step three: Do the hyperbolic trig to find the side-lengths of this triangle. What you really want is $\ell_{m}$, the hyperbolic distance from the edge midpoint (of the octagon) to the centre $i$.
Step four: Produce the Mobius transformation $\rho$ that preserves the imaginary axis and takes $a$ to $b$ (the edge midpoints). This is easy since you have its hyperbolic translation distance ($2 \ell_m$) in your hands. Excellent - you have your first generator.
Step five: Find the Moebius transformation $\sigma$ that rotates the hyperbolic plane about $i$ through an angle of $2\pi/8$.
Step six: Conjugate. That is, build $\rho_i = \sigma^{-i} \rho \sigma^{i}$. This gives you eight Moebius transformations. These are the images of the four generators of $\pi_1$ and their inverses.
Step seven: Check your work by checking that $\rho_i^{-1} = \rho_{i + 4}$. Also, show that the elements $\rho_i$ satisfy the defining relation of $\pi_1$.
There are many details in this github repository: https://neilstrickland.github.io/genus2/. In particular, there is a monograph at https://neilstrickland.github.io/genus2/genus2.pdf and you can look in Section 4 for discussion of the relevant formulae. The most important are as follows. We fix $b\in(0,1)$ and put $b_+=\sqrt{1+b^2}$ and $b_-=\sqrt{1-b^2}$. We then define Mobius transformations $\beta_k$ for $k\in\mathbb{Z}/8$ by \begin{align*} \beta_0(z) &= \frac{b_+z+1}{z+b_+} \\ \beta_1(z) &= \frac{b_+^3z+(2+i)b^2-i}{((i-2)b^2+i)z+b_+^3} \\ \beta_{2n}(z) &= i^n \beta_0(z/i^n) \\ \beta_{2n+1}(z) &= i^n \beta_1(z/i^n). \end{align*} These generate a group $\Pi$, subject only to the relations $\beta_k\beta_{k+4}=1$ and $\beta_0\beta_1\beta_2\beta_3\beta_4\beta_5\beta_6\beta_7=1$. Moreover $\Pi$ acts freely on the open unit disc, and the quotient has genus two.
As well as the PDF monograph, the repository contains extensive Maple code for working with these surfaces.
