How using the elliptic integrals, explicitly, one can transform Riemann surface to a complex plane. I have the following polynomial \begin{equation} P(z)=(z-a)(z-1)(z+1)(z+a) \end{equation} where $a(k_{y})>1$, $z$ is a complex number and $k_{y}\in S^{1}$. The paper says that using elliptic integral \begin{equation} w=\int_{z_{0}}^{z}\frac{dt}{\sqrt{P(t:k_{y})}} \end{equation} one can transform the Riemann surface of the roots of $\sqrt{P(z)}$ in to a plane. I don't really get this argument. Thanks in advance for comments.
@ Jeremy Upsal:Thanks a lot for the useful comments. I have edited my question. Actually you are completely right. The paper, I am following, outline a sequence of mapping that finally map the Riemann surface to a torus. The first map, which I have defined above, transform the Riemann surface to a rectangle formed by $0,w_1,w_1+w_2$ and $w_2$ in the complex plane. Then using a GL(2,$\mathbb{R}$), they map $w$ to $\tilde w$ to map the rectangle to the square $S$ bounded by $0,1,1+i$, and $i$. Finally, for $\tilde w=\theta+i \phi$, they embed the torus in three-dimensional Euclidean space as $x_{1}=(R+\sin \phi)\cos(\theta), x_{2}=(R+\sin \phi)\sin(\theta), x_{3}=\cos \phi$. I want to do it explicitly and I think I am missing some gap in the argument. Also, I don't understand how does this mapping depends on specific form of $a$. Thanks. The Riemann surface is constructed from the roots of $P(z)$ not from the roots of $\sqrt{P(z)}$. This is to study the analytical structure of the $\sqrt{P(z)}$.
Sorry if my question seems a bit vague, I lack understanding on the topic.
First note that $$ \int_{z_0}^z \frac{dt}{\sqrt{P(t)}} $$ is an elliptic integral since $P(t)$ is a quartic polynomial. Byrd and Friedman or the dlmf have good resources for working with these integrals. Here we will use the Weirstrass-p function ($\wp$). The formula for integrating this expression explicitly in terms of the $\wp$ function involve cubics. You can introduce a linear fractional transform to rewrite your elliptic curve $Q^2 - (z-a)(z-1)(z+a)(z+1)=0$ to something of the form $v^2 - 4(u-u_1)(u-u_2)(u-u_3)=0$ (see this discussion for instance). Therefore for the rest of this discussion I will assume you have transformed your quartic to this form.
Now $v^2 - 4(u-u_1)(u-u_2)(u-u_3)=0$ defines a genus 1 Riemann surface, $X$. For any point $(u,v) \in \mathbb{C}^2$ on $X$, the map $$ z = z(P) = \int_{-\infty}^P \frac{du}{v} $$ (with integration carried out along the curve by paramaterizing) takes a point on the curve to a point $z$ on the torus. The above integral defines $z$ on the torus $\mathbb{T}^2$ since it is unique up to a linear combination of integrals around certain paths called the cycles, $a$ and $b$ on the Riemann Surface. i.e. for integers $n,m$ $$ \int_{-\infty}^P \frac{du}{v} + n\oint_a \frac{du}{v} + m\oint_b\frac{du}{v} = z(P) + 2n\omega_1 + 2n\omega_2 $$ defines the same $z$ as above mod $2\omega_1 + 2\omega_2$. So we now have a map from the Riemann Surface defined by the curve to the torus which is topologically equivalent to the lattice defined by the lines from $0$ to $\omega_1$ to $\omega_1 + \omega_2$ to $\omega_2$ and back to $0$ upon enforcing periodicity. But what are $\omega_1$ and $\omega_2$? Note that the $\wp$ function is doubly-periodic. So if we construct a $\wp$ function defined on this lattice we have $\wp(z+2n\omega_1 + 2m\omega_2) = \wp(z)$. Conveniently we have the nice differential equation for $\wp$ which says $$ (\wp')^2 = 4\wp^3 - g_2 \wp - g_3 $$ where you can find $g_2$ and $g_3$ by comparing this with your cubic equation. Associating $(v,u)$ with $(\wp'(z),\wp(z))$ we see that the point $(\wp'(z),\wp(z))$ lies on our Riemann surface. Now we can get $\omega_1$ and $\omega_2$ from $g_2$ and $g_3$ (see the DLMF). This gives us a map back the other direction, from the torus to our elliptic curve. i.e. we now have an isomorphism from the torus to the Riemann surface X, this is how we define equivalency. From the torus $\mathbb{T} = \mathbb{C}^2 / \{2n\omega_1 + 2n\omega_2\}$ you can use a conformal map/linear fractional transformation to map the points $(0, \omega_1, \omega_1 + \omega_2, \omega_2) \to (0,1,1+i,i)$ and get the unit torus. You can do this by noting that the linear fractional transform $$ q = f(z) = \frac{\alpha z + \beta}{\gamma z + \delta} $$ defines new coordinates $q \in \mathbb{C}$. You want $$ 0 = f(0)\\ 1 = f(\omega_1)\\ 1+i = f(\omega_1 + \omega_2)\\ i = f(\omega_2) $$ which you should be able to compute by hand (this is the action by $GL(2,\mathbb{C})$.
I realize now that I did not give an answer using the method that they apparently use in the paper, but they are equivalent. Specifically I did not finally embed the torus in 3-space, but I think you can do that using the map you described above.
Interestingly this is one of the few cases that you can actually do by hand. Specifically genus 1 Riemann Surfaces can be uniformized by elliptic functions. For higher genus surfaces you need theta-functions which are harder to compute. You may be interested in looking at the software packages algcurves for maple and abelfunctions for sage. These packages actually allow for some computation on Riemann Surfaces, though it is costly. Can I ask why you are interested in the explicit map?