Hyperbolic distances and geodesics in disc and half-plane

For every $w\in \mathbb{C}$ one has the identity $\mathrm{\mid }w+i{\mathrm{\mid }}^2-\mathrm{\mid }w-i{\mathrm{\mid }}^2=4\mathrm{Im}w$; hence for $w\in \mathbb{H}$, $\mathrm{\mid }C(w){\mathrm{\mid }}^2=\mathrm{\mid }w-i{\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }w+i{\mathrm{\mid }}^2<1$, and for $\mathrm{\mid }\zeta \mathrm{\mid }<1$ the point $w:=C^{-1}(\zeta )$ satisfies $\mathrm{Im}w=(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }1-\zeta {\mathrm{\mid }}^2>0$. Direct substitution gives $C(C^{-1}(\zeta ))=\zeta$ and $C^{-1}(C(w))=w$, so $C$ is a bijection of $\mathbb{H}$ onto $\mathbb{D}$.

The radial segment $\gamma (t)=t$, $0\le t\le r$, from $0$ to $r$ has, by [F6] and [F7], ${\mathrm{\ell }}_{\mathbb{D}}(\gamma )={\int }_0^r\frac{2dt}{1-t^2}=2\mathrm{artanh}r$, since $\frac{d}{dt}2\mathrm{artanh}t=\frac{2}{1-t^2}$ and $2\mathrm{artanh}0=0$. On the other hand [F3] gives ${\phi }_0(r)=(0-r)\mathrm{/}(1-0)=-r$, so [F2] gives $d_{\mathbb{D}}(0,r)=2\mathrm{artanh}\mathrm{\mid }-r\mathrm{\mid }=2\mathrm{artanh}r$. Hence the radial segment attains the distance.

Let $u\in \mathbb{D}\setminus \{0\}$ and let $\gamma :[a,b]\to \mathbb{D}$ be piecewise $C^1$ with $\gamma (a)=0$ and $\gamma (b)=u$; put $\rho :=\mathrm{\mid }\gamma \mathrm{\mid }$ and $f:=2\mathrm{artanh}\rho$. At every point where both derivatives exist one has $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }$, and $f$ is absolutely continuous with $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }=2\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}(1-{\rho }^2)$, so ${\mathrm{\ell }}_{\mathbb{D}}(\gamma )={\int }_a^{b}2\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}(1-{\rho }^2)dt\ge {\int }_a^{b}2\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}(1-{\rho }^2)dt\ge \mid {\int }_a^b{f}^{\mathrm{\prime }}\mid =\mathrm{\mid }f(b)-f(a)\mathrm{\mid }=2\mathrm{artanh}\mathrm{\mid }u\mathrm{\mid }$. By [F2] the last quantity is $d_{\mathbb{D}}(0,u)$, so every curve from $0$ to $u$ has length at least $2\mathrm{artanh}\mathrm{\mid }u\mathrm{\mid }$.

The Blaschke factor satisfies ${\phi }_a^{\mathrm{\prime }}(v)=-(1-\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{/}(1-\bar{a}v{)}^2$ and $1-\mathrm{\mid }{\phi }_a(v){\mathrm{\mid }}^2=(1-\mathrm{\mid }a{\mathrm{\mid }}^2)(1-\mathrm{\mid }v{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }1-\bar{a}v{\mathrm{\mid }}^2$ for $v\in \mathbb{D}$; hence $2\mathrm{\mid }{\phi }_a^{\mathrm{\prime }}(v)\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }{\phi }_a(v){\mathrm{\mid }}^2)=2\mathrm{/}(1-\mathrm{\mid }v{\mathrm{\mid }}^2)$, so ${\mathrm{\ell }}_{\mathbb{D}}({\phi }_a\circ \gamma )={\mathrm{\ell }}_{\mathbb{D}}(\gamma )$ for every piecewise $C^1$ curve $\gamma$ in $\mathbb{D}$.

Fix $a\in \mathbb{D}$. If $a=0$, put $S_0:=\mathbb{R}\cup \{\mathrm{\infty }\}$; since ${\phi }_0(t)=-t$, this is the extended image of the real line. If $a\mathrm{\ne }0$, put $S_a:=\{{\phi }_a(t):t\in \mathbb{R},1-\bar{a}t\mathrm{\ne }0\}\cup \{1\mathrm{/}\bar{a}\}$. Then $S_a$ is a Euclidean circle or line meeting the unit circle at right angles. Indeed, since ${\phi }_a$ is an involution [F4], for $u\mathrm{\ne }1\mathrm{/}\bar{a}$ one has $u\in S_a$ if and only if ${\phi }_a(u)\in \mathbb{R}$, and expanding ${\phi }_a(u)=(a-u)\mathrm{/}(1-\bar{a}u)$ gives $2i\mathrm{Im}{\phi }_a(u)=[(a-u)(1-a\bar{u})-(\bar{a}-\bar{u})(1-\bar{a}u)]\mathrm{/}\mathrm{\mid }1-\bar{a}u{\mathrm{\mid }}^2=[(a-\bar{a})(1+\mathrm{\mid }u{\mathrm{\mid }}^2)+(1-a^2)\bar{u}-(1-{\bar{a}}^2)u]\mathrm{/}\mathrm{\mid }1-\bar{a}u{\mathrm{\mid }}^2$, so $u\in S_a$ if and only if $(a-\bar{a})(1+\mathrm{\mid }u{\mathrm{\mid }}^2)+(1-a^2)\bar{u}-(1-{\bar{a}}^2)u=0$. If $a\in \mathbb{R}$, this reads $(1-a^2)(\bar{u}-u)=0$, so $S_a=\widehat{\mathbb{R}}$ is the real line, which meets the unit circle at $\pm 1$ at right angles. If $a\mathrm{\notin }\mathbb{R}$, dividing by $a-\bar{a}=2i\mathrm{Im}a$ and writing $A:=(1-a^2)\mathrm{/}(a-\bar{a})$ turns the equation into $1+\mathrm{\mid }u{\mathrm{\mid }}^2+A\bar{u}+\bar{A}u=0$, that is $\mathrm{\mid }u+A{\mathrm{\mid }}^2=\mathrm{\mid }A{\mathrm{\mid }}^2-1$. Since $\mathrm{\mid }A{\mathrm{\mid }}^2-1=(\mathrm{\mid }1-a^2{\mathrm{\mid }}^2-\mathrm{\mid }a-\bar{a}{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }a-\bar{a}{\mathrm{\mid }}^2=(1-\mathrm{\mid }a{\mathrm{\mid }}^2{)}^2\mathrm{/}\mathrm{\mid }a-\bar{a}{\mathrm{\mid }}^2>0$, this is a circle with centre $-A$ and radius $\rho >0$; a point $u\mathrm{\ne }1\mathrm{/}\bar{a}$ lies on that circle exactly when $u\in S_a$, and $1\mathrm{/}\bar{a}\in S_a$ is the limit ${\lim }_{\mathrm{\mid }t\mathrm{\mid }\to \mathrm{\infty }}{\phi }_a(t)$ of points of the circle, so $S_a$ is exactly this circle. Finally $\mathrm{\mid }-A{\mathrm{\mid }}^2-{\rho }^2=1$, and a circle with centre $c$ and radius $\rho$ that meets the unit circle meets it orthogonally precisely when $\mathrm{\mid }c{\mathrm{\mid }}^2-{\rho }^2=1$: at an intersection point $w$ the tangents are perpendicular exactly when the radius vectors $w$ and $w-c$ are perpendicular, that is $\mathrm{Re}(\bar{w}(w-c))=0$, which with $\mathrm{\mid }w\mathrm{\mid }=1$ says $\mathrm{Re}(\bar{w}c)=1$, and substituting this into $\mathrm{\mid }w-c{\mathrm{\mid }}^2={\rho }^2$, namely $1-2\mathrm{Re}(\bar{w}c)+\mathrm{\mid }c{\mathrm{\mid }}^2={\rho }^2$, gives $\mathrm{\mid }c{\mathrm{\mid }}^2={\rho }^2+1$. So $S_a$ meets the unit circle at right angles.

Let $M(z)=(az+b)\mathrm{/}(cz+d)$ have real $a,b,c,d$ with $ad-bc>0$. For $s\in \mathbb{R}$ one computes $M(is)=(b+ais)\mathrm{/}(d+cis)=(bd+ac{s}^2+is(ad-bc))\mathrm{/}(c^2{s}^2+d^2)$. If $c=0$, then $\mathrm{Re}M(is)=b\mathrm{/}d$ is constant and the image of the imaginary line is the vertical line $\mathrm{Re}w=b\mathrm{/}d$. If $c\mathrm{\ne }0$ and $d=0$, then $\mathrm{Re}M(is)=a\mathrm{/}c$ is constant and the image is the vertical line $\mathrm{Re}w=a\mathrm{/}c$. If $c\mathrm{\ne }0$ and $d\mathrm{\ne }0$, then multiplying out shows that every $M(is)$ satisfies $(X-p{)}^2+Y^2={\sigma }^2$ with $X=\mathrm{Re}M(is)$, $Y=\mathrm{Im}M(is)$, centre $p:=\frac{1}{2}(\frac{a}{c}+\frac{b}{d})\in \mathbb{R}$ and radius $\sigma =\mathrm{\mid }ad-bc\mathrm{\mid }\mathrm{/}(2\mathrm{\mid }c\mathrm{\mid }\mathrm{\mid }d\mathrm{\mid })>0$; the real points $a\mathrm{/}c$ and $b\mathrm{/}d$ of this circle are $p\pm \sigma$, so it is a Euclidean circle with centre on $\mathbb{R}$. In every case the image of the imaginary line meets $\mathbb{R}$ at right angles: vertical lines do so plainly, and a circle with centre on $\mathbb{R}$ has vertical tangents at its two real points.

For $w\in \mathbb{H}$ the quotient rule [F7] gives $C^{\mathrm{\prime }}(w)=2i\mathrm{/}(w+i{)}^2\mathrm{\ne }0$, and step 1.1 gives $1-\mathrm{\mid }C(w){\mathrm{\mid }}^2=(\mathrm{\mid }w+i{\mathrm{\mid }}^2-\mathrm{\mid }w-i{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }w+i{\mathrm{\mid }}^2=4\mathrm{Im}w\mathrm{/}\mathrm{\mid }w+i{\mathrm{\mid }}^2$; therefore $2\mathrm{\mid }{C}^{\mathrm{\prime }}(w)\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }C(w){\mathrm{\mid }}^2)=(4\mathrm{/}\mathrm{\mid }w+i{\mathrm{\mid }}^2)\cdot (\mathrm{\mid }w+i{\mathrm{\mid }}^2\mathrm{/}(4\mathrm{Im}w))=1\mathrm{/}\mathrm{Im}w$.

In step 1.3 equality holds if and only if $\gamma$ is a monotone reparametrisation of the radial segment from $0$ to $u$. Indeed, equality in the second inequality forces $f$, hence $\rho$, to be nondecreasing; equality in the first forces $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }$ almost everywhere, which combined with ${\rho }^{\mathrm{\prime }}\ge 0$ gives ${\gamma }^{\mathrm{\prime }}=\lambda \gamma$ with $\lambda \ge 0$ real wherever $\gamma \mathrm{\ne }0$, so the unit vector $\gamma \mathrm{/}\rho$ is constant on each interval on which $\rho >0$. Since $\rho$ is nondecreasing from $\rho (a)=0$ to $\rho (b)=\mathrm{\mid }u\mathrm{\mid }>0$, the set $\{\rho >0\}$ is an interval $(t_0,b]$, and continuity of $\gamma \mathrm{/}\rho$ there gives $\gamma (t)\mathrm{/}\rho (t)=u\mathrm{/}\mathrm{\mid }u\mathrm{\mid }$ for $t>t_0$. Thus $\gamma$ traces the segment $[0,u]$ with nondecreasing modulus, and conversely every such parametrisation realises equality.

Let $z\in \mathbb{D}$ and let $e\in \mathbb{C}$ with $\mathrm{\mid }e\mathrm{\mid }=1$. For all $v\in \mathbb{C}$ with $1-\bar{z}ev\mathrm{\ne }0$ one has, multiplying numerator and denominator by $\bar{e}$, ${\phi }_z(ev)=(z-ev)\mathrm{/}(1-\bar{z}ev)=e(z\bar{e}-v)\mathrm{/}(1-\overline{z\stackrel{‾}{e}}v)=e{\phi }_{z\bar{e}}(v)$. Hence the image of the line $\{te:t\in \mathbb{R}\}\cup \{\mathrm{\infty }\}$ under ${\phi }_z$ is $e{S}_{z\bar{e}}$, a rotation of the circle or line of step 1.5. Multiplication by $e$ preserves Euclidean circles and lines, fixes the unit circle, and carries a circle of centre $c$ and radius $\rho$ to the circle of centre $ec$ and radius $\rho$, so $e{S}_{z\bar{e}}$ is again a Euclidean circle or line meeting the unit circle at right angles.

Define ${\mathrm{\ell }}_{\mathbb{H}}(\gamma ):={\int }_a^b\mathrm{\mid }{\gamma }^{\mathrm{\prime }}(t)\mathrm{\mid }\mathrm{/}\mathrm{Im}\gamma (t)dt$ for piecewise $C^1$ curves $\gamma :[a,b]\to \mathbb{H}$, and $d_{\mathbb{H}}(p,q):=\inf {\mathrm{\ell }}_{\mathbb{H}}(\gamma )$ over such curves from $p$ to $q$. By the chain rule and step 2.1, ${\mathrm{\ell }}_{\mathbb{D}}(C\circ \gamma )={\mathrm{\ell }}_{\mathbb{H}}(\gamma )$ for every piecewise $C^1$ curve $\gamma$ in $\mathbb{H}$, and likewise ${\mathrm{\ell }}_{\mathbb{H}}(C^{-1}\circ \sigma )={\mathrm{\ell }}_{\mathbb{D}}(\sigma )$ for every piecewise $C^1$ curve $\sigma$ in $\mathbb{D}$; since $C$ and $C^{-1}$ transport curves in both directions, taking infima gives $d_{\mathbb{H}}(p,q)=d_{\mathbb{D}}(C(p),C(q))$ for all $p,q\in \mathbb{H}$, and $C$ is a biholomorphism of $\mathbb{H}$ onto $\mathbb{D}$.

Let $z,w\in \mathbb{D}$ with $z\mathrm{\ne }w$, and put $\phi :={\phi }_z$ and $u:=\phi (w)\mathrm{\ne }0$. By [F4], $\phi$ is an automorphism of $\mathbb{D}$, $\phi (z)=0$, and ${\phi }^{-1}=\phi$. A piecewise $C^1$ curve $\sigma$ from $z$ to $w$ has ${\mathrm{\ell }}_{\mathbb{D}}(\sigma )=d_{\mathbb{D}}(z,w)$ if and only if $\phi \circ \sigma$ has length $d_{\mathbb{D}}(0,u)$, because step 1.4 gives ${\mathrm{\ell }}_{\mathbb{D}}(\phi \circ \sigma )={\mathrm{\ell }}_{\mathbb{D}}(\sigma )$ and [F2] gives $d_{\mathbb{D}}(0,u)=d_{\mathbb{D}}(\phi (z),\phi (w))=d_{\mathbb{D}}(z,w)$. By steps 1.3 and 2.2 the length-minimising curves from $0$ to $u$ are exactly the monotone reparametrisations of the radial segment $[0,u]$. Consequently the length-minimising curves from $z$ to $w$ are exactly the images under $\phi$ of those curves, that is, the images of the radial segment from $0$ to $u={\phi }_z(w)$.

For $y>0$ one has $C(i)=0$ and $C(iy)=(iy-i)\mathrm{/}(iy+i)=(y-1)\mathrm{/}(y+1)$, a real number of modulus $\mathrm{\mid }y-1\mathrm{\mid }\mathrm{/}(y+1)<1$. Steps 3.1 and 1.2 and [F6] therefore give $d_{\mathbb{H}}(i,iy)=d_{\mathbb{D}}(0,\mathrm{\mid }y-1\mathrm{\mid }\mathrm{/}(y+1))=2\mathrm{artanh}(\mathrm{\mid }y-1\mathrm{\mid }\mathrm{/}(y+1))=\mathrm{\mid }\log y\mathrm{\mid }$: for $y\ge 1$ the logarithm formula gives $2\mathrm{artanh}\frac{y-1}{y+1}=\log \frac{(y+1)+(y-1)}{(y+1)-(y-1)}=\log y$, and for $0<y<1$ replacing $y$ by $1\mathrm{/}y$ gives the same identity with $\log (1\mathrm{/}y)=\mathrm{\mid }\log y\mathrm{\mid }$. The vertical segment $\gamma (t)=it$, $t$ running from $1$ to $y$, has ${\mathrm{\ell }}_{\mathbb{H}}(\gamma )=\mid {\int }_1^{y}ds\mathrm{/}s\mid =\mathrm{\mid }\log y\mathrm{\mid }$ by [F9], so it attains the distance.

Now let $z,w\in \mathbb{D}$ with $z\mathrm{\ne }w$, put $u:={\phi }_z(w)\mathrm{\ne }0$ and $e:=u\mathrm{/}\mathrm{\mid }u\mathrm{\mid }$. The radial segment $[0,u]$ is contained in the line $\{te:t\in \mathbb{R}\}$, so by step 3.2 the length-minimising curves from $z$ to $w$ are the images under ${\phi }_z$ of that segment, and these lie on ${\phi }_z(\{te\}\cup \{\mathrm{\infty }\})$, which by step 2.3 is a Euclidean circle or line meeting the unit circle at right angles.

Let $y>0$ and let $\gamma :[a,b]\to \mathbb{H}$ be piecewise $C^1$ from $i$ to $iy$; put $\rho :=\mathrm{Im}\gamma >0$ and $g:=\log \rho$. Then $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }$ wherever both derivatives exist, so ${\mathrm{\ell }}_{\mathbb{H}}(\gamma )={\int }_a^b\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}\rho dt\ge {\int }_a^b\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}\rho dt\ge \mid {\int }_a^b{g}^{\mathrm{\prime }}\mid =\mathrm{\mid }\log y-\log 1\mathrm{\mid }=\mathrm{\mid }\log y\mathrm{\mid }$, with $g$ absolutely continuous by [F9]. Equality in the second inequality forces $\rho$ to be monotone, and equality in the first (where $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}{\mathrm{\mid }}^2=(\mathrm{Re}{\gamma }^{\mathrm{\prime }}{)}^2+{\rho }^{\mathrm{\prime }2}$) forces $x:=\mathrm{Re}\gamma$ to satisfy $x^{\mathrm{\prime }}=0$ almost everywhere, hence to be constant; so the curves attaining $\mathrm{\mid }\log y\mathrm{\mid }$ are exactly the monotone parametrisations of the vertical segment from $i$ to $iy$. In particular the vertical segment is the unique minimiser, and step 4.1 shows its length is $d_{\mathbb{H}}(i,iy)=\mathrm{\mid }\log y\mathrm{\mid }$.

Let $p,q\in \mathbb{H}$ with $p\mathrm{\ne }q$, put $z^{\mathrm{\prime }}:=C(p)$, $w^{\mathrm{\prime }}:=C(q)$, $u^{\mathrm{\prime }}:={\phi }_{z^{\mathrm{\prime }}}(w^{\mathrm{\prime }})$ and $r:=\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\in (0,1)$. Let $R(\zeta ):=\zeta \overline{u^{\mathrm{\prime }}}\mathrm{/}r$ be the rotation of $\mathbb{D}$ carrying $u^{\mathrm{\prime }}$ to $r$, and put $g:=R\circ {\phi }_{z^{\mathrm{\prime }}}$ and $m:=C^{-1}\circ g\circ C$. Then $g$ is an automorphism of $\mathbb{D}$ with $g(z^{\mathrm{\prime }})=0$ and $g(w^{\mathrm{\prime }})=r$, so $m$ is a biholomorphic self-map of $\mathbb{H}$ with $m(p)=C^{-1}(0)=i$ and $m(q)=C^{-1}(r)=i(1+r)\mathrm{/}(1-r)=:iy$ with $y>0$. By [F5] one has $m(z)=(az+b)\mathrm{/}(cz+d)$ with $a,b,c,d$ real and $ad-bc>0$, and likewise $m^{-1}(z)=(dz-b)\mathrm{/}(-cz+a)$ has real coefficients and determinant $ad-bc>0$. For such a map the quotient rule [F7] gives $m^{\mathrm{\prime }}(z)=(ad-bc)\mathrm{/}(cz+d{)}^2$ and $\mathrm{Im}m(z)=((ad-bc)\mathrm{Im}z)\mathrm{/}\mathrm{\mid }cz+d{\mathrm{\mid }}^2$, hence $\mathrm{\mid }{m}^{\mathrm{\prime }}(z)\mathrm{\mid }\mathrm{/}\mathrm{Im}m(z)=1\mathrm{/}\mathrm{Im}z$ for $z\in \mathbb{H}$; so $m$ preserves ${\mathrm{\ell }}_{\mathbb{H}}$ and transports minimisers to minimisers.

By steps 5.1 and 6.1 the minimisers from $p$ to $q$ are the images under $m^{-1}$ of the monotone parametrisations of the vertical segment from $i$ to $iy$, and these images lie on $m^{-1}(\{is:s\in \mathbb{R}\}\cup \{\mathrm{\infty }\})$. Since $m^{-1}$ is again given by real Mobius coefficients with positive determinant, step 1.6 shows that this set is a vertical line or a Euclidean circle with centre on the real axis, meeting $\mathbb{R}$ at right angles. So the geodesic segment from $p$ to $q$ is an arc of such a circle or line.

Steps 3.1, 1.2, 4.1, 4.2 and 7.1 establish the four clauses of the Example. All curves and maps in the argument are given by explicit formulae; in particular the disc and half-plane geodesics are obtained by inverting explicit biholomorphisms, and the infima in [F1] and step 3.1 are taken over explicitly parametrised families, so no choice principle is used. When $z=w$ in $\mathbb{D}$ or $p=q$ in $\mathbb{H}$ the constant curve has length $0=d_{\mathbb{D}}(z,z)=d_{\mathbb{H}}(p,p)$, and steps 2.2 and 5.1 identify the strict minimisers only in the nondegenerate case.