Poincaré disk model

The Poincare disc carries the metric $4|dz{|}^2/(1-|z{|}^2{)}^2$ on $|z|<1$.

For $|a|<1$, the Möbius transformation $f_a(z)=(z-a)/(1-\overline{a}z)$ is an isometry of the Poincare disc model sending $a$ to zero. Rotations combined with these transformations describe its orientation-preserving isometries.

In curvature minus one, a hyperbolic disc of radius $R$ has areaIn the Poincare disc this follows by integrating $4r(1-r^2{)}^{-2}drd\theta$ up to Euclidean radius $\tanh (R/2)$.

If a domain of finite hyperbolic area $A$ contains a hyperbolic disc of radius $r$ in curvature $-1$, then $2\pi (\cosh r-1)\le A$, hence $r\le \mathrm{arcosh}(1+A/(2\pi ))$. For an interior point of a hyperbolic triangle, the open disc with radius equal to its distance from the boundary lies inside the triangle. Its area is at most $\pi$ by Gauss-Bonnet theorem, giving the elementary bound $r\le \mathrm{arcosh}(3/2)$; this need not be the sharp triangle inradius bound.

Hyperbolic discs of area $\pi /2$ have radius $\log 2$. Starting at the origin and placing their centers successively along one radial geodesic gives center distance $n\log 4$ and Euclidean coordinate

An isometry sends the centers of a collinear family of equal hyperbolic discs to points on one hyperbolic geodesic. In the upper-half-plane model no geodesic contains three distinct points on a horizontal Euclidean line, since geodesics are vertical lines or semicircles orthogonal to the boundary.

For a piecewise smooth curve $z(t)$ in the Poincare disc, its hyperbolic length is

The geodesics are diameters and arcs of circles orthogonal to the unit circle. A radial segment is length minimizing becausewith equality for constant angle and monotone radius. Hence the distance from the origin to $z$ is $2\mathrm{artanh}|z|$.

For $z_1,z_2$ in the Poincare disc model, putTheir hyperbolic distance is

For a proper arc $I$ of the boundary of the unit disc, $Q(I)$ is the side of the hyperbolic geodesic between its endpoints adjacent to $I$. For short arcs this is a curved boundary cap. For arcs longer than a semicircle it contains the center. With endpoints $e^{\pm i\alpha }$ and $0<\alpha <\pi /2$, the geodesic is the circle with center $\sec \alpha$ and radius $\tan \alpha$, orthogonal to the unit circle. It is a diameter at $\alpha =\pi /2$.

For a short arc $I$ of normalized length $\ell$, every point of its hyperbolic geodesic cap lies within Euclidean distance $C\ell$ of its center boundary point. Conversely, for any fixed $b>0$, the region with angular projection in $I$ and depth $1-|z|\le b\ell$ lies in $Q(\kappa I)$ for a fixed enlargement $\kappa$ depending only on $b$, when $I$ is sufficiently short. To verify this, rotate the midpoint to one and write the endpoints as $e^{\pm i\alpha }$. The cap inequality is $2\mathrm{Re}z>(1+|z{|}^2)\cos \alpha$. Its radial depth at the midpoint is $1-\sec \alpha +\tan \alpha$, comparable to $\alpha$ for small $\alpha$; its angular projection lies in $I$. Enlarging $\alpha$ by a sufficiently large fixed factor makes the inequality hold throughout the indicated rectangle. Large arcs use a finite cover of the disc. Thus Carleson boxes and hyperbolic geodesic caps define equivalent Carleson measure conditions.

The rectangle containment can also be checked directly. Write $z=r{e}^{i\theta }$ with $|\theta |\le \pi \ell$ and $1-r\le b\ell$. Its membership in the enlarged cap is equivalent toFor small enough $\ell$, $r\ge 1/2$. The right side is at most $b^2{\ell }^2$, while the left side is at least $({\kappa }^2-1){\pi }^2{\ell }^2/8$, using the sine product formula and $\sin x\ge x/2$ for small positive $x$. Thus $\kappa =4+4b/\pi$ suffices.

If $P$ is at hyperbolic distance $\rho$ from the origin, the Poincare-disc geodesic through $P$ perpendicular to $OP$ is represented by a Euclidean circle of radius $\mathrm{csch}\rho$ whose centre is at Euclidean distance $\coth \rho$ from the origin.