Is there a conformal map that preserves the Poincaré disc and maps two points $z_1, z_2$ to different points $z_1', z_2'$?

Question

Or does such a map not even exist? I think the wanted map has to merely translate/rotate/maybe reflect the underlying lattice. I would like to tessellate the hyperbolic plane. My idea is to draw the fundamental polygon in the center of the disk, and then map two vertices at z1 and z2 that are connected by an edge to $z_1'$=-1/2, $z_2'$=1/2, then mirror the fundamental polygon and finally apply the inverse transformation to the lattice. I know I am looking for some möbius transformation, but I fail to find its parameters such that it satisfies my needs.

Answer 1

No, such a map need not exist.

Also, your problem is worded somewhat ambiguously. At first it seems that $z_1,z_2,z'_1,z'_2$ are points in the Poincare disc $\mathbb D$ itself which is the open disc of radius $1$ centered on $z=0+0i$ and does not contain its boundary circle. But later you refer to $0$, $1$, $\infty$ which are often regarded as points on the boundary "line" of the upper half plane model of the hyperbolic plane (which is metrically isomorphic to the Poincaré disc model, although it is a different point set).

To give a consistent answer, I am going to stick with the Poincare disc $\mathbb D$, and assume that $z_1,z_2,z'_1,z'_2 \in \mathbb D$.

Conformal maps from the Poincaré disc onto itself preserve the hyperbolic metric, i.e. the Poincaré metric, which is given by the line element $$ds = \frac{2 \sqrt{dx^2 + dy^2}}{1 - (x^2+y^2)} $$ Any two points $z_1,z_2 \in \mathbb D$ are endpoints of a unique hyperbolic line segment $\overline{z_1 z_2}$, i.e. the unique shortest length path with endpoints $z_1,z_2$, where (hyperbolic) length is determined by integrating $ds$ along that path. Assuming that $z_1 \ne z_2$, that line segment can be visualized in one of two ways: if $z_1,z_2$ lie on a diameter of $\mathbb D$ then $\overline{z_1 z_2}$ is a subsegment of that diameter; otherwise $\overline{z_1 z_2}$ is a circular subarc of the unique circle passing through $z_1,z_2$ that intersects the boundary circle $S^1 = \partial \mathbb D$ in right angles.

Those, basically, are the elements of hyperbolic geometry: points, (hyperbolic) line segments, (hyperbolic) length. Furthermore, the (orientation preserving) rigid motions of this geometry are precisely the conformal transformations of the Poincare disc.

Now one proves that this geometry satisfies all the axioms and theorems of Euclidean geometry, except of course the parallel postulate and theorems which depend on it. In particular, the following theorem is just as true in hyperbolic geometry as it is in Euclidean geometry:

Two line segments are congruent if and only if they have the same length.

where for purposes of the present discussion, we can take "congruence" to mean existence of a rigid motion taking one line segment to the other.

So, to answer your question, given $z_1,z_2,z'_1,z'_2 \in \mathbb D$, there exists a conformal map taking $z_1 \mapsto z'_1$ and $z_2 \mapsto z'_2$ if and only if the line segments $[z_1,z_2]$ and $[z'_1,z'_2]$ are congruent, if and only if they have the same hyperbolic length.

For example, consider the hyperbolic line segment $[z'_1,z'_2]$ with endpoints $z'_1 = -\frac{1}{2}$ and $z'_2 = \frac{1}{2}$, and the line segment $[z_1,z_2]$ with endpoints $z_1 = - \frac{3}{4}$ and $z_2 = \frac{3}{4}$. Both of those line segments lie on the $x$-axis diameter of $\mathbb D$. The line segment $[z'_1,z'_2]$ is a proper subsegment of $[z_1,z_2]$, so $[z'_1,z'_2]$ is therefore shorter than $[z_1,z_2]$, and those two line segments are therefore not congruent. So there is no conformal transformation taking $z_1,z_2$ to $z'_1,z'_2$.