Minimize the area of a maximal surface in R^{2,1} with boundary on the unit sphere Let $\mathrm{H}$ be the unit sphere in the Minkowski space $\mathbb{R}^{2,1}$ (i.e., a one-sheeted hyperboloid $x_1^2+x_2^2=x_3^2+1$). Assume that $\gamma\subset \mathrm{H}$ is a closed space-like curve that "makes one full turn on $\mathrm{H}$" (the unit circle in the plane $x_3=0$ is the simplest example).
Let a maximal (i.e., the one maximizing the area) surface $S_\gamma$ be the solution to the Plateau problem with boundary $\gamma$. Could it be true that the minimum of the area of $S_\gamma$, taken over all such $\gamma$, is attained on planar curves, more precisely on the unit circle mentioned above and its images under $\mathrm{O}(2,1)$? (Note that a priori it is not even clear that the infimum is strictly positive.)
PS I am not familiar with the subject, so even a reference to, say, existence and uniqueness of solutions to this Plateau's problem would be helpful (though this is certainly not what the question itself is about).
Updated Nov 7th: From an explicit computation for the symmetric polyline contour with vertices at $(\cos\frac\pi{2n})^{-1}\cdot(\cos\frac{k\pi}{2n};\sin\frac{k\pi}{2n};(-1)^k\sin\frac\pi{2n})$, $k=0,...,2n-1$, it seems that the area of $S_\gamma$ actually increases when $n$ grows, not decreases. Thus, the original post seems to be terribly wrong: provided I did not mess this computation up, the minimizer could be provided by the symmetric $4$-segments contour (though I have no idea about how to prove that) but not by the unit circle, in a sense the picture is just the opposite. Sorry for a misleading question.
 A: That's a very interesting question but I don't think that the estimate as it is proposed can hold (as suggested also in the update at the end of the question). (We just discussed this with Nathaniel Sagman before seeing the update, so I'll just add this argument here in case someone is interested.)
Assume that the maximal surface $S$ contains $0$ -- this can be achieved for instance using a symmetric curve $\gamma$, for instance by taking $\gamma$ invariant under the isometry of $R^{2,1}$ which is the composition of the rotation of angle $\pi/2$ around the axis (0z) and the symmetry across the plane of equation $z=0$.
Let $x\in S$, consider the vertical plane $P_x$ (parallel to $Oz$) through $0$ and $x$. The length of the segment of $P_x\cap S$ between $0$ and $x$ has length at most equal to the distance from $0$ to $x$ in $R^{2,1}$, because $P_x$ is isometric to $R^{1,1}$. So the intrinsic distance between $0$ and $x$ on $S$ is also at most $\| x\|$ (the Minkowski distance between $0$ and $x$).
So in $S$, with its intrinsic distance, every point is at distance at most $1$ from $0$.
However the Gauss formula indicates that the induced metric on $S$ has curvature $K\geq 0$.
Standard estimates then imply that the area of $S$ (for the induced metric) is at most $\pi$, with equality only when $S$ is totally geodesic.
