I would like to study what the quotient $$T^2 / \Omega $$ of a closed compact Riemann surface with $g=1$ handles, once a complex structure is chosen, over an antiholomorphic involution $\Omega,$ can be, for various choices of $\Omega.$
In particular, which involution does give the annulus $$ A=\{ z\in \mathbb{C} | \ a < |z| < c \},$$ with $0<a<c$ real numbers ?
Here are the three examples of quotients of the flat torus. In each case, the torus $T$ will be obtained by identifying ends of the product cylinder $A=S^1\times [-1,1]$ via an isometry $f=(h,id): S^1\times (-1)\to S^1 \times (1)$. In the first two examples, $h$ will be the identity map, in the third example, $h(z)=-z$, where I identify $S^1$ with the unit circle in the complex plane.
The (isometric) involution $\tau: T\to T$ is the projection of the reflection $\sigma_1: A\to A$, $\sigma_1(z,t)=(z, -t)$. Note that the reflection $\tau$ has two fixed circles, projections of the circles $S^1\times (1)$ and $S^1\times 0$. The quotient annulus $T/\tau$ is identified with the sub-annulus $S^1\times [0,1]$.
The involution $\tau: T\to T$ is the projection of the orientation-reversing involution $\sigma_2: A\to A$, $\sigma_2(z,t)= (-z, -t)$. Then $\tau$ has no fixed points and, hence, $T/\tau$ is the Klein bottle (since torus can cover only one nonorientable surface, the Klein bottle).
This is the most interesting example. In this case, $\tau$ is again the projection of the involution $\sigma_2$ as in (2), but now $\tau$ has a fixed circle, projection of the circle $S^1\times (1)$ to the torus $T$. Hence, $T/\tau$ has exactly one boundary component, hence, it has to be the Moebius band.
By looking at the Euler characteristic, one can see that every quotient of $T$ by an orientation-reversing involution has to have zero Euler characteristic, hence, it is either the annulus, or the Klein bottle or the Moebius band.