Give an example of a homeomorphism \[ f:\Delta\to\Delta \] that is distance-decreasing for the hyperbolic metric but does not extend continuously to \(S^1\).
Hint: in polar coordinates, try \[ f(r,\theta)=(R(r),T(r)+\theta), \] with \(T(r)\to\infty\) as \(r\to1\).
Solution
It is convenient to use hyperbolic polar radius rather than Euclidean radius. Put \[ t=2\operatorname{artanh}r =\log\frac{1+r}{1-r}, \qquad r=\tanh(t/2). \] Define \[ S(t)=\operatorname{arsinh}t \] and \[ T(t)= \begin{cases} 0,&0\le t\le2,\\[1mm] \frac12\log(t/2),&t\ge2. \end{cases} \] In hyperbolic polar coordinates define \[ F(t,\theta)=(S(t),\theta+T(t)). \] Equivalently, in Euclidean polar coordinates, \[ f(re^{i\theta}) =\tanh\!\left(\frac12\operatorname{arsinh} \left(2\operatorname{artanh}r\right)\right) e^{\,i(\theta+T(2\operatorname{artanh}r))}. \]
1. In the coordinate \(t=2\operatorname{artanh}r\), the hyperbolic metricon \(\Delta\) is\[ ds^2=dt^2+\sinh^2t\,d\theta^2. \]
The hyperbolic metric on \(\Delta\), of curvature \(-1\), is \[ ds=\frac{2|dz|}{1-|z|^2}. \] In Euclidean polar coordinates, \[ |dz|^2=dr^2+r^2d\theta^2. \] Since \[ r=\tanh(t/2), \qquad \frac{dr}{dt}=\frac12(1-r^2), \] the radial part becomes \[ \frac{4dr^2}{(1-r^2)^2}=dt^2. \] Also \[ \frac{2r}{1-r^2}=\sinh t, \] so the angular part is \(\sinh^2t\,d\theta^2\).
2. The map \(F\) defines a homeomorphism \(\Delta\to\Delta\).
The function \[ S(t)=\operatorname{arsinh}t \] is a strictly increasing homeomorphism \([0,\infty)\to[0,\infty)\). The function \(T\) is continuous and vanishes near \(0\).
Thus away from the origin the map \[ (t,\theta)\longmapsto(S(t),\theta+T(t)) \] has the continuous inverse \[ (s,\phi)\longmapsto \left(\sinh s,\phi-T(\sinh s)\right). \] At \(t=0\), both source and target angular coordinates collapse to the origin, and \(S(t)\to0\), so the map and its inverse extend continuously there. Hence \(F\) is a homeomorphism of the disk.
3. For \(0<t<2\), the differential of \(F\) has hyperbolic operator normat most \(1\).
Use the orthonormal frames \[ e_t=\partial_t, \qquad e_\theta=\frac1{\sinh t}\partial_\theta \] in the source and the analogous frame at radius \(S(t)\) in the target.
For \(t<2\), \(T'(t)=0\). Therefore the differential is diagonal in these frames, with factors \[ S'(t)=\frac1{\sqrt{1+t^2}}\le1 \] and \[ \frac{\sinh S(t)}{\sinh t} =\frac{t}{\sinh t}\le1, \] because \(\sinh S(t)=t\) and \(\sinh t\ge t\). Hence the operator norm is at most \(1\).
4. For \(t>2\), the differential of \(F\) has hyperbolic operator normstrictly less than \(1\).
For \(t>2\), \[ T'(t)=\frac1{2t}. \] In the same orthonormal frames as in 3, the differential matrix is \[ A(t)= \begin{pmatrix} \dfrac1{\sqrt{1+t^2}}&0\\[3mm] \sinh(S(t))T'(t)&\dfrac{\sinh S(t)}{\sinh t} \end{pmatrix} = \begin{pmatrix} \dfrac1{\sqrt{1+t^2}}&0\\[3mm] \dfrac12&\dfrac{t}{\sinh t} \end{pmatrix}. \]
For \(t\ge2\), \[ \frac1{1+t^2}\le\frac15. \] Also \[ \sinh t \ge t+\frac{t^3}{6}, \] so \[ \frac{t}{\sinh t} \le\frac1{1+t^2/6} \le\frac35. \] Thus the squared Frobenius norm satisfies \[ \|A(t)\|_{\mathrm F}^2 \le\frac15+\frac14+\frac9{25} =\frac{81}{100}<1. \] The operator norm is at most the Frobenius norm, hence \[ \|A(t)\|_{\mathrm{op}}<1. \]
5. The homeomorphism \(F\) is distance-decreasing for the hyperbolicmetric.
The map is locally Lipschitz and piecewise \(C^1\); the only seam is \(t=2\). By 3 and 4, its differential has hyperbolic operator norm at most \(1\) almost everywhere.
Therefore the hyperbolic length of the image of any rectifiable path is at most the length of the original path. Taking infima over paths joining two points gives \[ d_{\Delta}(F(x),F(y))\le d_{\Delta}(x,y). \] Thus \(F\) is distance-decreasing.
6. The map does not extend continuously to \(S^1\).
Along the radial ray \(\theta=0\), let \(t\to\infty\). Since \[ S(t)=\operatorname{arsinh}t\longrightarrow\infty, \] the Euclidean radius of \(F(t,0)\) tends to \(1\). But \[ T(t)=\frac12\log(t/2)\longrightarrow\infty, \] so its angle keeps winding around the circle.
More explicitly, take \[ t_n=2e^{4\pi n} \qquad\text{and}\qquad s_n=2e^{4\pi n+2\pi}. \] Then \[ T(t_n)=2\pi n, \qquad T(s_n)=2\pi n+\pi. \] Both source sequences approach the same boundary point \(1\in S^1\) along the positive real radius, but their images satisfy \[ F(t_n,0)\longrightarrow1, \qquad F(s_n,0)\longrightarrow-1. \] Hence there is no limit at the boundary point \(1\), so no continuous extension to \(S^1\) exists.