0)$ with a metric \beq ds^2 = \frac{dX^2+dY^2+dZ^2}{Z^2}= \frac{dZ^2+d\bar{\X}d\X}{Z^2}, \ \ \ \ \X=X+iY \label{m1} \eeq Given a contour $(X(s),Y(s),Z=0)\equiv(\X(s)=X(s)+i Y(s),Z=0)$ at the boundary $Z=0$ we seek a surface of minimal area ending on it and parameterized as \beq Z(z,\bz), \ \ X+iY=\X(z,\bz) \label{m2} \eeq where the coordinates $(z,\bz)$ parameterize the complex plane. In that plane we find a closed curve $z(s)$ such that $Z(z(s),\bz(s))=0$. Such curve maps to the contour in the boundary of (EAdS$_3$) space, and the interior maps to a minimal area surface. At this time there is no generic solution to such problem of finding the minimal area surface for an arbitrary boundary. However we can give an infinite parameter family of minimal area surfaces and the corresponding contours where they end. It is an open problem if such solutions approximate all contours (and if so, in which sense). The construction starts by choosing an auxiliary hyperelliptic Riemann surface $\cM$, known as the spectral curve and given, as a subspace of $\mathbb{C}^2$ by \beq \mu^2 = \lambda \prod_{j=1}^{2g}(\lambda-\lambda_j) \label{m3} \eeq where $(\mu,\lambda)$ parameterize $\mathbb{C}^2$ and $\lambda_i\neq \lambda_j$ if $i\neq j$. There is a reality condition that requires the set of branch points $\{0,\infty,\lambda_{j=1\ldots 2g}\}$ to be symmetric under the involution $T:\lambda\leftrightarrow -1/\bar{\lambda}$. A basis of one-cycles $a_i,b_i$ is chosen such that the non-trivial intersections are $a_i\circ b_j=\delta_{ij}$ and that under the involution behave as \beq T a_i = - T_{ij} a_j , \ \ \ T b_i = T_{ij} b_j \label{m4} \eeq where $T_{ij}$ is a $g\times g$ symmetric matrix such that $T^2=1$. As a matter of notation, a generic point in the Riemann surface $\cM$ is denoted as $p_i=(\mu_{p_i},\lambda_{p_i})\in\cM$. The coordinate $\lambda_{p_i}$ is called the projection of $p_i$ on the complex plane. The origin and infinity play an important role and are denoted as $p_1=0$, $p_3=\infty$. Except for the branch points, given a point $p_4$ in one sheet there is another point denoted as $p_{\bar{4}}$ on the other sheet of the Riemann surface such that $\lambda_{p_4}=\lambda_{p_{\bar{4}}}$ and $\mu_{p_4}=-\mu_{p_{\bar{4}}}$. The relation between $p_4$ and $p_{\bar{4}}$ is the hyperelliptic involution associated with $\cM$. Now we consider the basis of holomorphic differentials \beq \nu_j = \frac{\lambda^{j-1}}{\mu(\lambda)} d\lambda, \ \ \ \ \ i,j=1\ldots g \label{m5c} \eeq and define the matrices \beq C_{ij} = \oint_{a_i} \nu_j, \ \ \ \ \tilde{C}_{ij} = \oint_{b_i} \nu_j \label{m6c} \eeq and the basis of normalized holomorphic differentials $\omega_{i=1\cdots g}$ \beq \omega_i = \nu_j (C^{-1})_{ji} = \sum_{j=1}^g C^{-1}_{ji} \lambda^{j-1} \frac{d\lambda}{\mu(\lambda)} \label{m7c} \eeq such that \beq \oint_{a_i} \omega_j = \delta_{ij}, \label{m5} \eeq which defines the periodicity matrix \beq \oint_{b_i} \omega_j = \Pi_{ij}, \label{m6} \eeq and the theta function \beq \theta(\zeta) = \sum_{n\in \mathbb{Z}^g} e^{i\pi\, n^t\Pi n +2\pi i\, n^t \zeta}, \ \ \ \ \zeta\in \mathbb{C}^g \label{m7} \eeq It also defines the Jacobi map $\phi: \cM \rightarrow \mathbb{C}^g$ for which the standard notation is \beq \phi(p_4) = \inte{p_1}{p_4} \omega = \inte{1}{4} \label{m8} \eeq \looseness=-1 where the point $p_1=0$ is chosen as a distinguished point, $p_4$ is an arbitrary point and $\inte{1}{4}$ is just notation for the same function. There is an ambiguity in choosing the path of integration. Since two different paths from $p_1$ to $p_4$ can only differ by a closed cycle, $\phi(p_4)$ changes by \beq \phi(p_4) = \inte{1}{4} \ \ \rightarrow \ \ \ \inte{1}{4} + \epsilon_2 + \Pi \epsilon_1 \label{m9} \eeq where $\epsilon_1$ and $\epsilon_2$ are vectors with integer components. Further, if $p_4$ is another branch point, the path $1\rightarrow 4$ can be traced back on the lower sheet defining a generically non-trivial cycle. By choosing a certain path from $p_1=0$ to $p_3=\infty$ we define the vectors $\Delta_{1,2}\in \mathbb{Z}^g$ through \beq \inte{1}{3} = \half \Delta_2 + \half \Pi \Delta_1 \label{m10} \eeq The path should be chosen such that $\Delta^t_1.\Delta_2$ is an odd integer. With this vector we define another function \beq \that(\zeta) = \exp\left\{2\pi i\left[\frac{1}{8}\Delta_1^t\Pi\Delta_1+\half\Delta_1^t\zeta+\frac{1}{4}\Delta_1^t\Delta_2\right] \right\} \theta\bigg(\zeta+\inte{1}{3}\bigg) \label{m11} \eeq Because $\Delta^t_1.\Delta_2$ is odd it follows that $\that(-\zeta)=-\that(\zeta)$ and in particular $\that(0)=0$. \looseness=-1 Further, consider the vector $\omega=(\omega_1,\cdots,\omega_g)$ and expand it around $p_1=0$. Since $p_1=0$ is a branch point an appropriate coordinate can be chosen as $y =-2i\sqrt{\lambda}$, namely $\lambda=-y^2/4$. It follows that \beq \omega = (\omega_1 + y^2 \omega_{12} + y^4 \omega_{14} +\cdots ) dy \label{m12} \eeq \looseness=-1 for some constant vectors $\omega_1$, $\omega_{12}$, etc.\ that can be computed from \eqn{m7c} by expanding $\mu(\lambda)$. Near infinity, on the other hand, an appropriate coordinate is $\tilde{y}=\frac{2}{\sqrt{\lambda}}$ and the expansion is \beq \omega = (\omega_3 + \tilde{y}^2 \omega_{32} + \tilde{y}^4 \omega_{34} +\cdots) d\tilde{y} \label{m13} \eeq The vectors $\omega_1, \omega_3, \omega_{12}, \omega_{32},\ldots \in\mathbb{C}^g$ play an important role and it is convenient to introduce a notation for the gradient of a function along those directions \beqa D_1 F(\zeta) &=& (\omega_1 . \nabla) F(\zeta) \\* D''_1 F(\zeta) &=& (\omega_{12} . \nabla) F(\zeta) \\ D_3 F(\zeta) &=& (\omega_3 . \nabla) F(\zeta) \\ D''_3 F(\zeta) &=& (\omega_{32} . \nabla) F(\zeta) \label{m14} \eeqa Notice that $D''_1$ is a first derivative. Second derivatives are denoted \eg\ as $D_1^2 F(\zeta) = (\omega_1 . \nabla) (\omega_1 . \nabla) F(\zeta)$. With all these ingredients in place, the solutions are given by \beqa Z&=& \left|\frac{\that(2\inte{1}{4})}{\that(\inte{1}{4})\theta(\inte{1}{4})}\right| \frac{|\theta(0)\theta(\zeta)\that(\zeta)||e^{\mu z+\nu \bz}|^2}{|\that(\zeta-\inte{1}{4})|^2+|\theta(\zeta-\inte{1}{4})|^2} \label{m15a} \\ X+iY &=& \X= e^{2\mu z+2\nu \bz} \frac{\theta(\zeta+\inte{1}{4})\overline{\theta(\zeta-\inte{1}{4})}-\that(\zeta+\inte{1}{4})\overline{\that(\zeta-\inte{1}{4})}}{|\that(\zeta-\inte{1}{4})|^2+|\theta(\zeta-\inte{1}{4})|^2} \ \ \ \ \ \label{m15} \eeqa where the vector $\zeta(z,\bar{z}) \in \mathbb{C}^g$ is given by \beq \zeta = 2\omega_1 \bz + 2\omega_3 z \label{m16} \eeq and the constants $\mu$, $\nu$ are given by \beq \mu = -2 D_3\ln\theta\bigg(\inte{1}{4}\bigg), \ \ \ \ \nu=-2D_1\ln\that\bigg(\inte{1}{4}\bigg) \label{munu} \eeq which uses the directional derivatives $D_{1,3}$ defined in \eqn{m14}. Notice that \eqn{m16} implies $\partial_z F(\zeta(z,\bz)) = 2 D_1 F(\zeta(z,\bz))$ and $\partial_{\bz} F(\zeta(z,\bz)) = 2 D_3 F(\zeta(z,\bz))$. The solution contains an arbitrary parameter $p_4=(\mu_{p_4},\lambda_{p_4})\in\cM$. For \eqn{m15} to define a solution it is necessary that $|\lambda_{p_4}|=1$. Therefore, each spectral curve $\cM$ actually defines a one real parameter family of surfaces. It was shown in~\cite{IKZ} that all those surfaces have the same area. \looseness=-1 Since the boundary is at $Z=0$, to find the curve where the surface ends we need to find the zeros of $Z$ or, equivalently, the zeros of $\that$. Typically the zeros are given by isolated curves $z(s)$ on the world-sheet from which one can be chosen. The solution \eqn{m15} maps such curve $z(s)$ to a curve $\X(s)$ at the boundary $Z=0$ defining the shape of the Wilson loop. The region of the world-sheet inside the curve $z(s)$ is mapped, by \eqn{m15} to the minimal area surface ending at the curve $\X(s)$. ]]>

1$. Unlike the $g=1$ case, even if we take $\lambda=1$, the solutions are not automatically periodic; we must pick a suitable value of $a>1$ as we move $b$ to get periodic concentric curves. \looseness=-1 The ratio of Riemann theta functions in \eqn{m21} is periodic if the argument shifts by an integer vector, also, the exponential function is periodic under a $2\pi$ shift in the imaginary direction. Both periods have to match. This can be achieved in practice by moving the branch points around. We found values for $a$ and $b$ such that we can match 2,3,4 and 5 periods of the theta function into a single period of the exponential. In each case we obtained concentric Wilson loops. The curves deviate from a circle in a way such that the periodicity is manifest as can be seen in figures~\ref{mc1} and~\ref{mc2}. The corresponding positions of the branch points is shown in table~\ref{tab:tab1} below. \begin{table}[position specifier] \centering \begin{tabular}{ | l | l | l | l | l |} \hline & n=2 & n=3 & n=4 & n=5 \\ \hline & $a=1.28088$ & $a=1.102149$ & $a=1.035312$ & $a=1.0304752$ \\ \hline & $b=0.5+0.01 i$ & $b=0.5+0.1 i$ & $b=0.7+0.2 i $ & $b=0.7+0.3 i$ \\ \hline \end{tabular} ]]>