Poincare disk and Poincare half plane

Question

My book claims that the Möbius transforms are isometries of the Poincaré half plane model. Thus, the metric is preserved under these maps. But I know that the Poincaré disk can be derived from applying such a transform to the half plane model. Now, I think it is well-known that the metric tensors of the half plane model and the disk are very different. This means that there is something wrong with my understanding of the statement given in my book.

Answer 1

If a Möbius transformation $$\phi: z \mapsto \frac{a z + b}{c z + d}$$ preserves the upper half-plane, then by continuity (as a self-map of the Riemann sphere) it must map $\mathbb{R} \cup \{\infty\}$ to itself, and some algebra forces that (possibly after canceling a common nonreal factor) $a, b, c, d$ are all real. Any such map maps the u.h.p. either to itself or to the l.h.p., and it does the former iff $a d - b c > 0$. Like you say, direct checking shows that any such transformation preserves the Poincare metric, and also the orientation of the upper half-plane.

On the other hand, a typical Möbius transformation that maps the u.h.p. to the unit disk is $$z \mapsto \frac{z - i}{z + i},$$ and no such transformation has all real coefficients.

Remark For any invertible complex matrix $A := \begin{pmatrix}a & b\\c & d\end{pmatrix}$ and $\lambda \in \mathbb{C}^*$, the matrix $\lambda A = \begin{pmatrix}\lambda a & \lambda b\\ \lambda c & \lambda d\end{pmatrix}$ determines the Möbius transformation $$ \phi: z \mapsto \frac{\lambda a z + \lambda b}{\lambda c z + \lambda d} = \frac{a z + b}{c z + d} , $$ namely the same transformation determined by $A$, but some algebra shows that this is the only redundancy, so we can think of the group of Möbius transformations as $GL(2, \mathbb{C}) / \mathbb{C}^* \cong PSL(2, \mathbb{C})$. The above shows that the group of (oriented) isometries of the Poincare half-plane is the proper subgroup $PSL(2, \mathbb{R})$.

Answer 2

The isometry group of the upper half plane model are Möbius transforms with real coefficients. You can check that, if you have complex coefficients, then the upper half plane is not mapped to itself. The Möbius transform that takes the upper half plane to the unit disk necessarily will have complex coefficients. For example $ z \rightarrow \frac{z-i}{z+i} $.